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Organ0hemist

Patt ISections I-IV

Section IStructure, Bonding, and Reactivity

Section IIStructure Elucidation

Section IIIStereochemistry

$ection IVHydrocarbon Reactions

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Speci alizing in II{CAT Preparation

Section IStructure,tsonding,

and

Reactivityby Todd Bennett

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Nomenclaturea) IUPAC Nomenclatureb) General Nomenclature

Bonding and Molecular Orbitalsa) Lewis Dot Structuresb) Bonding Modelc) Covalent tsondsd) Molecular Orbitals and Bonds

i. Single Bondsii. Double Bondsiii. Triple Bonds

e) Molecular Structuresf) Octet Rule (HONC Shortcut)g) Charged Structures

Hybridizationa) Hybridization of Atomic Orbitals

i. sp-Hybridzationii' sP2-HYbridzationiii' sP5-nybridzation

b) Common Shapes

Bond Energya) Bond Dissociation trnergyb) Ionic Bonds

Intramolecular Featuresa) Resonanceb) Inductive Effectc) Steric Hindranced) Aromaticity

Fundamental Reactivitya) Proton Transfer Reactionsb) Lewis Acid-Base Keactionsc) Acid and Base SLrength

i. Primary Effectsii. Secondary Effectsiii. Values and Terminology

d) Electrophiles and Nucleophiles

Physical Propertiesa) l-lydrogen-Bondingb) Polarityc) Van der Waals Forcesd) Solubility and Miscibility

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Organic Chemistry Molecular Structure Introduction

The perfect place to start any review of organic chemistry is the basics ofmolecular structure, which traditionally include bonding, hybridization, andelectronic distribution. We shall consider a chemical bond to be the result ofatomic orbitals overlapping to form molecular orbitais. We shall consider allbonds involving carbon to be covalent in nature. A covalent bond is thought toinvolve the sharing of electrons between two adjacent nuclei. According to therules of electrostatics, the region between two nuclei offers a highly favorableenvironment for electrons, where they can be shared by neighboring atoms.

However, there are several other factors to consider in bonding. if bonding werepurely a matter of electrostatics, then all of the eiectrons would be found betweentwo neighboring nuclei, not just the bonding electrons. The sharing of electronsmay be either symmetric (when the two atoms of the bond are of equalelectronegativity) or asymmetric (when the two atoms of the bond are of unequalelectronegativity). Sharing of electrons occurs when the atoms of a bond lack acompiete valence electron shell. By sharing electrons, each atom moves closer tocompleting its shell This is the driving force behind the formation of stablecovalent bonds.

Having looked briefiy at electron distribution, we can introduce the idea ofelectronic orbitals, which are three-dimensionai probability maps of the locationof an eiectron. They represent the region in space where an electron is found95% of the time. We shall consider the orbitals and the overlap of orbitals todescribe the electronic distribution within a molecule. Once one has establisheda foundation in bonding, the classification of molecules can be made based onsinrilarities in their bonding of particular atoms, known as functional groups. Eachfunctional group shall be considered in terms of its unique electron distribution,hvbridization, and nomenclature. Nomenclature, both that of the InternationalUnion of Pure and Applied Chemists (IUPAC) and more general methodsdescribing the substitution of carbon within a functional group, shall be used todescribe a particuiar organic molecule. The review of nomenclature iscontinuous throtrghout all sections of this book.

Then, we shall consider the factors that affect the distribution of eiectron densitywithin a molecule, including resonance, the inductive effect, steric hindrance,aromaticity, and hybridization. The distribution of electron density can be usedto explain and predict chemical behavior. The simplest rule of reactivity inorganic chemistry is that regions of high electron density act as nucleophiles bysharing their electron cloud with regions of low electron density, which act as

electrophiles. If you can correctly label a molecule in terms of the region thatcaries a partially negative charge (the electron-rich environment) and the regionthat carries a partially positive charge (the electron-poor environment), you canunderstand chemical reactions better.

And so begins your review of organic chemistry. Fortunately, much of organicchemistry is taught from the perspective of logic, which makes preparing fororganic chemistry on the MCAT easier. In organic chemistry courses you arerequired to process information and reach conclusions based on observations,which is also required on the MCAT. Reviewing and relearning this materialwill help you develop criticai thinking skills, which will carry over into yourrerriew for other portions of the exam. Despite what you may have perceivedwas a girth of information when you initially studied organi.c chemistry, youdon't need to review that much material to prepare successfully for the MCAT.

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Organic Chemistry Molecular Structure Nomenclature

IUPAC Nomenclature (Systematic Proper Naming)IUPAC Nomenclature is an internationally used system for naming molecules.Molecular names reflect the structural features (functional groups) and thenumber of carbons in a molecule. In IUPAC nomenclature, the name is based onthe carbon chain length and the functional groups. The suffix indicates whichprimary functional group is attached to the carbon chain. Table 1-1 lists prefixesfor carbon chains between one and twelve carbons in length. Table 1-2 lists thesuffices for various functional groups. Be aware that "R" stands for any genericalkyl group. When R is used, it indicates that the carbon chain size is irrelevantto the reaction. Table 1-3 summarizes the nomenclature process by listing severalfour-carbon compounds.

Carbons Prefix

1 meth-

2 eth-

J prop-

4 but-

Carbons Prefix

5 pent-

6 hex-

7 hept-

8 oct-

Table 1-1

Carbons Prefix

9 non-

10 dec-

11 undec-

12 dodec-

Functionality Compound Name Bonding

R-CH3 Alkane C-C & C-H

R-O-R Ether C-O-C

R-CO-H Aldehydeoil

C-C- H

R-CH2-OH Alcohol C-O-H

R-CO-R Ketoneoil

C_C-C

R-CO-OH Carboxylic acidoil

C_C_ OH

Table 1-2

Formula IUPAC Name Structural Class

H3CCH2CH2CH3 Butane Alkane

H3CCH=CHCH3 Butene Alkene

H3CCH2CH2CHO Butanal Aldehyde

H3CCH2COCH3 Butanone Ketone

H3CCH2CH2CH2OH Butanol Alcohol

H3CCH2CH(oH)CH3 2-butanol Alcohol

H3CCH2CH2CH2NH2 Butanamine Amine

H3CCH2CH2CO2H Butanoic acid Carboxylic acid

Table 1-3

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Organic Chemistry Molecular Structure Nomenclature

Figure 1-1 shows examples of IUPAC nomenclature for four organic compoundswith variable functional groups:

o

3-methylpentanoic acid

Longest chain: 5 carbons

Carboxylic acid groupMethyl substituent at C-3

3-ethylcyclopentanoneRing of 5 carbons

Ketone groupEthyl substituent at C-3

C1

4-chloro-S-methyl-3-heptanol

Longest chain: 7 carbons

Alcohol groupChloro substituent at C-4

Methyl substituent at C-5

3,3-dibromobutanalLongest chain: 4 carbons

Aldehyde group2 Bromo substituents at C-3

n-Propyl chloride(1-Chlropropane)

o

Figure 1-1

General Nomenclature (Common Naming Based on Substitution)In addition to the IUPAC naming system, there is a less rigorous method ofnaming compounds by functional group and carbon type (based on carbonsubstitution). Carbon type refers to the number of carbon atoms attached to thecentral carbon atom (carbon atom of interest). A carbon with one other carbonattached is referred to as a primary (1') carbon. A carbon with two other carbonsattached is secondary (2"). A carbon with three other carbons attached is tertiary(3"). Figure 1-2 shows some sample structures.

H CH.\ .s r"'r,iu.y carbon

r_zt{./-\HgC CHg H3CH2C

HCH"HH

l}}i""oary carbon \r)yn"t carbon

./'\ -/-\oH H3CH2C ClIsobutane

(2-Methylpropane)Sec-butanol(2-Butanol)

Figure 1-2

Nomenclature is an area of organic chemistry best learned through practice andexperience. We will deal with nomenclature throughout the course, as weintroduce each new functional group. Understanding nomenclature is especiallyimportant in MCAT passages where names rather than structures are given. Be

sure to know the Greek prefixes for carbon chain lengths up to twelve carbons.


Organic Chemistry Molecular Structure Bonding and Orbitals

Lewis Dot Structures (Two-Dimensional Depiction of Molecules)Lewis dot structures represent the electrons in the valence shell of an atom orbonding orbitals of a molecule. Typically, we consider the Lewis dot structuresof elements in the s-block and p-block of the periodic table. For every valenceelectron, a dot is placed around the atom. Single bonds are represented by a pairof dots in a line between the atoms, or by a line itself. A double bond isrepresented by a double line (implying that four electrons are being shared.)Likewise, a triple bond is represented by a triple line (implying that six electronsare being shared.) Lewis dot structures are familiar to most chemistry students,so recognize the exceptions to the rules, as they make good test questions.

Example 1.1

V\4rat is the Lewis dot structure for HcBF?

B.A.

D.C.

: H- S-'F :t".ry

:H-B-F:t"u

B-F :t"H

H-ri-r:t"H

H-

SolutionBoron has only three valence electrons, hence it can make only three bonds.There is no lone pair on the boron atom, eliminating choices A and C. Hydrogenhas only one electron, which is in the bond to boron, so there is never a lone pairon a bonded hydrogen. This eliminates choices A (already eliminated) and B.Fluorine has a completed octet, so it makes one bond and has three lone pairs, asdepicted in choice D, the best answer.

Bonding ModelBonding is defined as the sharing of electron pairs between two atoms in eitheran equal or unequal manner. As a general rule, a bond is the sharing of twoelectrons between two adjacent nuclei. The region between two nuclei is themost probable location for an electron. In most cases, with the exception ol ligandbonds (kaown also as Leruis ttcid-base bonds), one electron from each atom goes intoforming the bond. When electrons are shared evenly between two atoms, thebond is said to be a coualent bond. Ylh,en electrons are transferred from one atomto another, the bond is said to be an ionic bond. The difference between a covalentand ionic bond is measured in the degree of sharing of the electrons, which canbe determined from the dipole. The more evenly that the electrons are shared,the less the polarity of the bond. The relative electronegativity of two atoms canbe determined by measuring the dipole of tl-re bond they form. When thedifference in electronegativity between two atoms is less than 1.5, then the bondis said to be covalent. When the difference in electronegativity between twoatoms is greater than 2.0, then the bond is said to be ionic. When the differencein electronegativity between two atoms is greater than 1.5 but less than 2.0, thenthe bond is said to be polar-covalent (or partially ionic).

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Example L.2Which of the following bonds is MOST likely to be ionic?

A. C-OB. N-FC. Li-HD. Li-F

SolutionA bond is ionic when the difference in electronegativity between the two atomsexceeds 2.0. This means that the bond that is most likely to be ionic is the onebetween the two atoms with the greatest difference in electronegativity. Lithiumis a metal and fluorine is a halide, so they exhibit the greatest electronegativitydifference of the choices listed. The best answer is therefore choice D.

Covalent BondsBonds can be classified in one of three ways: ionic, polar-covalent, and covalent.A covalent bond occurs when electrons are shared between two atoms havingIittle to no difference in electronegativity. As the difference in electronegativitydecreases, the covalent nature of the bond increases. There are two types of:ovalent bonds: sigma bonds (o), defined as having electron density sharedbetvr'een the nuclei of the two atoms; and pi bonds (n), defined as having noelectron density shared between the nuclei of the two atoms, but instead onlyabove and below the internuclear region. Sigma bonds are made from manytypes of atomic orbitals (including hybrids), while pi bonds are made exclusivelyof parallel p-orbitals. In almost all cases, the sigma bond is stronger than the pibond, with molecular fluorine (F2) being a notable exception. Figure 1-3 shows a

generic sigma bond. You may notice that within a sigma bond, only about eightyto ninety percent of the electron density lies between the nuclei, not all of it.

.Nuclei.

o@oElectron density

Figure L-3

Example 1.3Which drawing depicts the electron density of a carbon-carbon sigma bond?

SolutionA sigma bond has its electron density between the two nuclei, which eliminateschoice D. The two atoms in the bond are identical, so the electron density shouldbe symmetrically displaced between the two nuclei. This eliminates choice B.

Most of electron density is between the nuclei, so choice A is a better answer thanchoice C. These drawings are ugly, so focus on the concept, not the pictures.



Nuclei

Electron DensitvJ

Figure 1-4 shows a generic n-bond. Within a n-bond, there is no electron densitybetween the two nuclei. The electron density in a ru-bond results from electronsbeing shared between the adjacent lobes of parallel p-orbitals.

Figure 1-4

In organic chemistry, covalent bonds are viewed. in great detail, taking intoaccount hybridization and overlap. In alkanes, carbons have sp3-hybridiiationand all of the bonds are sigma bonds. In alkanes, there are two typbs of bonds:c

- H lorps-r bonds) and c- c (o'sp3-sp3 bonds). In alkenes, t-h-"." uru sigma

and pi bonds plesent. The n-bond Conslsts of p-orbitals side by side, und it,carbons have sp2-hybridization. The C=Cbond is made up of aoroz-rozbondand a n2r-2p bond. Bond length varies with the size of the orbitals ih ttre bond.For instance, a sigma bond composed of an sp2-hybridized carbon and an sp3-hybridized carbon is shorter than a sigma bond comprised of two sp3-hyb ridized,carbons. Hydrogens use s-orbitals to form bonds. Figure 1-5 shows thiee sigmabonds with their relative bond lengths. The longer bond is associated wittrthelarger orbitals (bond radii: dz > dy > dil.

Figure 1-5

Figure 1-5 confirms that most of the electron density lies between the two nucleiin sigma bonds, no matter what the orbitals are from which the sigma bondoriginates. In pi bonds, electron density does not 1ie between the two nlclei. Thelength of a bond is defined as the distance between the nuclei of the two atomsmaking the bond. Figure 1-6 shows an example of a n-bond between two Zpr-orbitals, which is typical for nearly all n-bonds encountered in organic chemistry,because carbon, nitrogen, and oxygen have 2p-orbitals in their valence shells.

n2p,-2p,ffi

eo I{

{

iI

or-rp3 orp2-rp2 osp3_sp3

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Figure 1-6

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Pi bonds are found as the second bond present in double bonds and the secondand third bonds present in triple bonds. The first type of bond to form betweenatoms is usually the sigma bond. Once a sigma bond exists between two carbonatoms, then pi bonds can form between the atoms. Fluorine gas is an exceptionto the "sigma bond first" rule. Molecular fluorine (F2) has only one rc-bond, withno o-bond present. This is attributed to the small size of fluorine and the inter-nuclear repulsion associated with a typical single bond. This is why the bonddissociation energy of F2 is less than the bond dissociation energy of Cl2, eventhough chlorine is below fluorine in the periodic table.

Molecular OrbitalsMolecular orbital is a fancy way of describing a bond or anti-bond that existsbetween two atoms. An anti-bond is a molecular orbital that results in bond-breaking when coupled with a bonding orbital. It is important to recognize theshape and location of electron density in molecular orbitals. Figure 1-7 shows thecommon bonding and anti-bonding orbitals associated with organic chemistry.

sp3

Sigma bonding molecular orbital

fB n r ffic5+6 tr= d6P P lt*rr-r,

Pi anti-bonding molecular orbital

Jsp

D*.c@ + D@c@

Ip

Ip

.@.@o + cffi'corp3-rp3

, ,3 - -3 ox .-3 .-isp" sp" ' sp--sp-

Sigma anti-bonding molecular orbital

+Pi bonding molecular orbital

Figure 1-7

The shading of the lobes in each orbital represents the direction of spin for theelectron. h:r order for electron density to overlap, the electrons must have thesame spin. This is analogous to driving on the freeway. If you join a freeway inthe same direction as traffic is flowing, you can easily blend into traffic. This is afavorable interaction. If you join a freeway in the opposite direction as traffic isflowing, you cannot easily blend into traffic. This is an unfavorable interaction.

frzp-zp

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l\lolecular Bondsof greater interest than the sigma-bonds and pi-bonds are the single, double, andtriple bonds present between atoms. single, double, and triple bond nature isdiscussed more so than the sigma and pi nature of bonds. In trganic molecules,there are only single, double, and triple bonds. Between like atoms, thedescending order of relative strengths of boncls is triple bond > double bond >single bond. Another rule to consider is that for bonds between like atoms, theionger the bond, the less the electron density overlaps between nuclei, and thusthe weaker the bond. This is summarized as: longer ionds are rueaker bonds.

single Bonds: single bonds are composed of only one sigma bond between thetwo atoms. single bonds are longer than double and triple bonds between twoatoms, even though fewe_r electrons are present. Ethane has sigma bonds onlyand is shown in Figure 1-8 in both stick figures and with the rele')ant orbitals.

HH

C-C./v"HH

Illtt,,, ..rr\HH7: \n %o="#

@rffi%

H

/ffi"porp3-5p3 \"

H

Figure 1-8

H

OC

Double Bonds: Double bonds are composed of one sigma bond and one pi bondbetween two adjacent atoms. Ethene (czH+) has f6ur sigma bonds betweencarbon and hydrogen, a sigma bond between the two carf,ons, and a pi bondpresent between the two carbons to compiete the carbon-carbon double bond.Ethene is shown in Figure 1-9 in both stick figures and with the relevant orbitals.

ftp^-p^

Hosp2-roz

Figure 1-9

Triple Bonds: Tripie bonds are composed of a sigma bond. and two pi bondsbetween two adjacent atoms. Triple bonds are shorter than either iingte ordouble bonds. Ethyne (CzHz) has two sigma bonds between carbon andhydrogen, a sigma bond between the two carbons, and two pi bonds between thetwo carbons to complete the carbon-carbon triple bond. Ethyne is shown inFigure 1-10 in both stick figures and with the relevant orbitals.

fipv-pv

@.- .€PH

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Figure 1-10

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Example 1.4\A/hat is the relative bond strength of carbon-carbon bonds in the molecule shownbelow?

,rr.. ooloo

,r,bondaS\ l, /C*C

bond c _-_*/ \

(H3.)2HC

""Y Tbond

A. Bond a > Bondb > Bond c > Bond dB. Bondb > Bond a > Bond c > Bond dC. Bond d > Bond a > Bond c > BondbD. Bondb > Bond c > Bond a > Bond d

H-\ ^CH^CH".v z r

C

CHe

d

SolutionThere strongest C-C bond is a double bond, bond b, so choices A and C areeliminated.

-Bond d is the weakest, because it is between two sp3 carbons. Bond

c is stronger than bond a, despite both sharing an sp2 and an sp3 carbon, becausebond c contains the more highly substituted carbon. Choice D is the best answer.

Molecular StructuresWe shall continue from the fundamental concept that a valence electron can be

shared between two nuclei rather than being isolated to just one nucleus, becausethe attractive force of two positive sites is greater than the attractive force of one.This is the basic, perhaps oversimplified, perspective of a chemical bond. Thesharing of electrons is what characterizes a covalent bond. One of the first rulesof organic chemistry that you must understand is the octet rule. It is valid forcarbon, nitrogen, and oxygen atoms. To understand organic chemistry, it isimportant that you recall VSEPR theory, which applies to bonding (in particular,to the subgroups of covalent bonding like single, double, and triple bonds andtheir component o-bonds and n-bonds). Table 1-4 shows the skeletal structuresof molecules that contain carbon, nitrogen, oxygen, and hydrogen.

AtomValenceElectrons

To CompleteShell

Number of Bonds in NeutralCompounds

Carbon (C) .c.4. 4 e- needed

In -tY' -L_

Nitrogen (N) .N.5. 3 e- needed 3 '";

-N' i5:- -''Y', \

Oxygen (O) - .ct.b .. 2 e- needed , _-6-_ :O.

Hydrogen (H) 1H. 1 e- needed 1H-

Table 1-4

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Organic Chemistry Molecular Structure Bonding and Orbitats

Octet Rule and the HONC Shortcut:Every molecular structure should have atoms that obey the octet rule (eightvalence electrons for C, N, and o). The numbers of electrons needed to completethe shell in the Table 7-4 are derived from the electrons needed to obey the bctetrule. All neutral structures have atomic arrangements as described inTable 1.4.To complete the octet valance shell, carbon requires four electron pairs in theform of bonds, nitrogen requires one lone pair in addition to the thiee bonds itmakes, and oxygen requires two lone pairs in addition to the two bonds it makes.You must be able to recognize valid structures by applying the bonding rules(HONC-1234). In a neutral compound, hydrogen makes onebond., oxygen makestruo bonds, nitrogen makes threebonds, and carbon makes four bonds. Neutralstructures always obey this rule. Figure 1-11 shows examples of valid andinvalid structures and a brief description of the bonding to the component atoms.

(

C:

-.{j

:,:::::.-ar-'-')-l

T

L-L/\

HHAll carbons have 4 bonds.

All hydrogens have 1 bond.

Good Structure

HsC- C- C- CH2CH2CH3

All carbons have 4 bonds.A1l hydrogens have 1 bond.

Good Structure

H CH"'\ /-"3

1\-L\

CHs

Ali carbons have 4 bonds.A1l hydrogens have 1 bond.

Nitrogen has 3 bonds.Good Structure

H ,CHZ\//^-AL-L

/\HH

Most carbons have 4 bonds,but one carbon has 5 bonds.All hydrogens have 1 bond.

Bad Structure

/CH3H"C-C:

cH2cH3

AII carbons have 4 bonds.Ali hydrogens have 1 bond,

but oxygen has 3 bonds.Bad Strttctttre

tt /tnt

N-C

Dn

:-..!i

tl__ _

:_ _

:_rtra,

H

CHs

All carbons have 4 bonds.A11 hydrogens have 1 bond,

but nitrogen has 4 bonds.Bad Structure

H

Figure 1-11

You can validate molecular structures by seeing whether they satisfy bondingrules (HoNC-7234) and conventions with regard to the number of bonds andlone pairs. If a structure does not satisfy the rules, then there must be a chargepresent. Generally, having too many bonds in a molecule results in a cation andtoo few bonds results in an anion, except with carbon. For instance, if oxygenmakes three bonds and has one lone pair, it carries a positive charge. wnu.tnitrogen makes two bonds and has two lone pairs, it carries a negative charge.\Alhen carbon makes three bonds, the charge depends on the presence or absenceof a lone pair (presence yields an anion, r,t'hile absence yields a cation).

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Charged StructuresFormal charges (charged sites on a molecule) occur when there is an excess, orshortage of electrons on an atom. For instance, an oxygen atom typically has sixvalence electrons and wishes to have eight. This means that oxygen makes twobonds to complete its valence shell (and thus satisfy the octet rule). However, ifan oxygen atom had only five valence electrons, it would be short one electron

from its original six and would consequently carry a positive charge. Havingonly five valence electrons, the positively charged oxygen would need to makethree bonds (one more than its standard two) to complete its octet. We can

conclude that oxygen with three bonds carries a positive charge. Table 1-5 shows

somecommon organic ions to commit to memory:

Drawing Molecular StructuresDrawing molecular structures from a given formula requires following the octetrule for aII atoms except hydrogen. On occasion, thete will be charged atomswithin the compound, but the number of charged atoms within the structureshould be minimized. Figure 1-12 shows some samPle structures for a fewcommon molecules.

"'*\\.g*-H

Number of Bondsto Neutral Atom

Number of Bonds toCationic Atom

Number of Bonds toAnionic Atom

Carbon (C)

4 .c.3l

-+ -

r+--/t\

? ../--

-

1---'Y',-\Nitrogen (N)

t'

3 'N'4

z\':N\-5+=t -t-

Oxygen (O)

2 .o.'-ij\,:t :o*= t

-p]'

Table 1-5

C_H

.(

)YtH

H

H

H_ C-N_ C_ H

Molecular andStructural Formula

c2H60cH3cH2oH

c2H7NcH3NHCH3

c2H50*

cH3cHo+H

Lewis Structure

HHll

H_C_C- O- HtlHH

Figure 1-12

.C_C\t\H.; .

^- rrH .v II

.C_H

"irH

3D Structure

.C-C

UH liu\it

H:

N'HHHrtlt..lHH

HHll

H-C-C: O*- Ht..H

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Organic Chemistry Molecular Structure Ilybridization

HybridizationHybridization of Atomic OrbitalsHybridization entaiis relocating electron density in atomic orbitals prior tobonding, in order to minimize the repulsion between electron pairs and therebyallow for bonding between atoms. There are three main types bi nyuria orbitalsto consider: sp,sp2, and sp3 hybrids. Hybrid orbitals are atomic orbitals that areinvolved in making bonds between atoms. Listed in Table 1-6 are some pertinentfacts and structural features for each of the three types of hybridization. lable 1-6represents general trends that are observed in neariy all molecules withhybridized orbitals involved in their molecular orbitals.

Hybrid sp sp2 sp3

Atomic Orbitals s+p s+p+p s+P+p+pAngle 180' 720" 109.5'

Shape linear trigonal pianar tetrahedralo-bonds and e- pairs 2 3 4

r-bonds 2 1 0

Table 1-6

The number of n-bonds in Table 1-6 is for typicai molecules that obey the octetrule' A class of molecules that is an exception to the features in Table 1-6 is theboranes, such as BH3, BF3, and BR3. In boranes, the boron atom has only threevalence electrons, so a neutral boron cannot satisfy the octet rule. The result isthat boron has sp2-hybridization for its three sigma bonds, but no pi bond.

Structures with OrbitalsLewis structures are used as shorthand representations of molecules. Flowever,in organic chemistry, molecules should be visualized in three dimensions, whichhybridization helps to faciiitate. Determining the three-dimensional shape of amolecule requires first assuming a shape based on hybridization of the centralatoms, then applying valence shell electron pair repulsion (vsEpR) theory.Figure 1-13 shows molecular structures witl-r orbitals and three-dimensionalorientation. Structures should be drawn with and correct bond lengths and bondangles should be based on hybridization, steric hindrance, and vsEpR rules.

:r

i!

s,4

j-

sp2 -hybrid,ized S...'n,

BH: H-p-10-n

,Hsp'-hybridized

IcH+ H-- \,,,*-,H

E

H

IL

5

:*i

sp3-hybridized,H

sp'-hybridized IqlNHe r-Ut'un

H

Hzo

"t€

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Figure 1-13

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Organic Chemistry Molecular Structure Hybridization

The orbital shown for BH3 in Figure 1-13 is actually an empty p-orbital, while the

other orbitals depicted in Figure 1-13 are hybrid orbitals. The p2-orbital of BH3 is

devoid of electrons, so the hybridization is sp2. \Mhile an empty p-orbital does

not actually exist, we consider the region where an electron pair could be

accepted. The three hybrid orbitals are detailed in Figures I-1.4,1.-75, and 1-16.

sp-Hybridization: sp-hybtidization is the result of the mixing of the s-orbital and

the pporbital.

.&.o,Two atomic orbitals

@&r.@Two sphybrid orbitals

+Hybridizesto become

Figure 1-14

st'ayarialzation: sp2-hybrtdtzation is the result of the mixing of the s-orbital,the pa-orbital, and the py orbital.

Hybridizesto become

three sp2-irybrid orbitals

,o3 ffi'I'^^, [email protected] ,or.m

Figure L-16

Example 1.5What is the hybridization of each carbon in propene (H2C=CH-CH3)?

A. sp, sp, and sp3^B. sp/,sp, and sprC. tp?,tp\, and sp3D. sp', spr, and spJ

SolutionThere are three carbons in propene. The first two carbons are involved in a rc-

bond, so they are each sp2-ltybridized. This makes the best answer choice C. The

last carbon is not involved in any n-bonds, so it has sp3-hybridization.

Py o.. s^[email protected] r)

Figure 1-15

tt'-nyAnalzation: sp3-hybtidi"ation is the result of the mixing of the s-orbital,the pa-orbital, the py-orbital, and the p2-orbital.

Three atomic orbitals

i'rery.Four atomic orbitals

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Organic Chemistry Molecular Structure Hybridization

Common Three Dimensional ShapesHybridization theory supports the notion that there are recurring molecularshapes (tetrahedral, trigonal planar, and linear) that can be seen within differentmolecules. This means that there is a electronic expianation for the structuresthat are observed within various molecules. Hybridization is a theoreticalexplanation to rationalize why electron pairs in the valence shells of bondingatoms assume orientations as far from one another as possible. Hybridization isused to explain bond lengths and bond angles. Figures r-r7,1.-rg, and 1-19 showstructures with their corresponding geometry and structural features.

Tetrahedral and sp3 -nybridizationA central atom with four electron pairs (any combination of bonds and lonepairs) has tetrahedral orientation of the electron pairs about the central atom.This does not mean that the shape is tetrahedral, but that the orientation ofelectron pairs about the central atom (geometry) is tetrahedral. Figure 1-17shows different structures with tetrahedral geometry about the central itom, butdifferent molecular shapes.

Tetrahedral Structure Trigonal pyramidal StructureH

l-/- / ------^..

t-t-'Y.tt I H/NY'H-Hlu4 atoms/0lone pairs I 3 atoms/1 lone pair

Because of -symmetry, all bond I uecause of rone pair repulsion, bondlengths and bond angles are equal./ angles decrease. N is r*ullu. than c, so

I Oond length N-H is less than C-H.

C-H: 1.10A & < HCH: 109.5" / N-H: 1.00A & < HNH: 102.3"

IlI Bent Structure\\ rr\r'n'

------* ..

H/NY,HH

1o-ru2 atoms/21one pairs

Because of lone-pair repulsion, bondangles decrease. O is smaller than N, sobond length O-H is less than N-H.

O-FI: 0.96A & < HoH: 104.5'

Figure 1-17

Trigonal Planar an d sp2 -HybridizationA central atom with three other atoms, two olher atoms and one lone pair, or oneother atom and two lone pairs attached has trigonal p?ur,ut geometry

-of th" th.""

substituents (or electron pairs) about the central atom. This does not mean thatthe shape is trigonal planar, but that the orientation of electron pairs about thecentral atom is trigonal planar. Figure 1-18 shows the planar structure andspatial representation of the bonds in ethene. The stick and ball representationshows the three-dimensional perspective for ethene.

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Organic Chemistry Molecular Structure Ilybridization

Planar Structure Spatial Representation

sn2-hvbridizationH'/\' H\i \ /

t,- L/\

HH3 atoms/0 lone pairs

Each carbon in ethene

has sp2-hybridization.

Figure L-18

Line ar and sp-Hybridiz ationA central atom with two other atoms or one other atom and one lone pairattached has linear geometry of the two substituents (or electron pairs) about thecentral atom, This does not mean that the shape is linear (although in most cases

it is), but that the orientation about the central atom is linear. Figure 1-19 showsthe linear structure and spatial representation of the bonds in ethyne. The stickand ball representation shows the three-dimensional perspective for ethyne'

Linear Structure

sr-hvbridization' tt'/\t\C::

2 atoms/0lone pairs

Each carbon in ethynehas sp-hybridization.

Spatial Representation

Figure L-19

Example 1.6The hydrogen-carbon-hydrogen bond angle in formaldehyde (H2CO) is BEST

approximated by which of the following values?

A. 108.3'B. 771.7"c. 118.5'D. 121.5'

SolutionThe first feature to look at is the hybridization of carbon. Carbon is involved inone n-bond, so the hybridization is sp2. The bond angle about an sp2-hybridized.carbon is predicted to be 120". The question here is whether the angle is slightlygreater or slightly less than 120". Because there are two pairs of electrons on theoxygen/ the electron density repels the electrons in the two carbon-hydrogenbonds. This forces the two bonds closer together, which compresses thehydrogen-carbon-hydrogen bond angle. According to valence shell electron pairrepulsion (vsEPR) theory, the angle shouid be slightly less than 120". The bestanswer is thus choice C._:

180.0"

k\,

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Organic Chemistry llolecular Structure Bond Energies

Bond EnergyBond Dissociation EnergyL:r organic chemistry, the energy required to cleave a bond in a homolytic fashionis commonly used to compare reiative bond strengths. Homolytic cleivage refersto the breaking of a chemicai bond into two fiee radical fiagments. This istypically viewed in the gas phase or an aprotic, nonpolar solvenl, where ions aretoo unstable to exist. It is important that you recall that energy is released whena bond is formed and that energy must be absorbed by the ilolecule to break abond. By subtracting the energy released upon forming new bonds from theenergy required to break bonds, the enthaipy of a reaction can be determined.This is shown in Equation 1.1.

AHReaction = lEnergy&onds broken) - Energy66nds formed) (1.1)

If the enthalpy of a reaction is known, then the bond dissociation energies forbonds that are formed and broken during the course of a reaction can bedetermined. It is this method that allows for the comparison of bonds betweenidentical atoms within different morecules. For instance, the theory ofaromaticity is supported by the excess energy that is reieased. upon the formationof a n-bond that completes the aromatic iing. The release of excess energyimplies that the molecule is more stable than expected from the standard bonddissociation energies, so some other factor must be involved. Table I-T lists somebond dissociation energies for typicat bonds in some common organic molecules.

Bond Dissociation Energies for A-B Bonds (Kcal/mole)

A B=H Me Et i-Pt t-Bu Ph OH NHzMethyl 105 90 89 86 84 102 93 85

Ethyl 101 89 B8 6/ 85 101 95 85

n-Propyl 101 89 88 86 85 101 95 85

Isopropyl 98 89 87 B5 82 99 96 B5

f-Butyl 96 87 95 82 77 99 96 B5

Phenyl 111 702 100 99 96 115 111 702

Benzyl 88 76 75 /+ 73 90 81 77

Allyl 86 74 70 70 67 N/A 7B 69

Acetyl 86 81 79 77 75 93,5 107 96

Ethoxy L04 83 85 N/A N/A 101 44 N/AVinyl 1I2 102 101 100 95 105 N/A N/AH 704.2 105 i01 98 96 111 IT9 1.07

Table 1-7

A greater value in Table 1-7 implies that the bond is stronger. you may note thatthe weakest bond listed in Tabie 1-7 is an o-o single bond within a peroxidemolecule (Eto-oH). Because this bond is so r.r'eak, peroxides are highly reactivespecies, often used to oxidize other compounds. The data in Table 7-7 also revealthat the substitution of the carbon and the position of the bond within themolecule affect the bond energy. The effect of hybridization can also be extractedwhen comparing bond energies betl-een r-inr.l and methyl substituents.

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Organic Chemistry Molecular Structure Bond Energies

Example 1.7According to the data in TabIe 1.-7, which of the following carbon-carbon singlebonds is the MOST stable?

A. An sp2-carbon to a primary sf -carbonB. An spt-carbon to a second.ary spr-carbonC. A secondary sp3-carbon to a primary sf -carbonD. A secon dary sp3-carbon to a iecondary sp3-carbon

SolutionThe most stable bond is the strongest bond. The strongest bond has the greatestbond dissociation energy, so to solve this question, the bond energies from Table1-7 must be referenced. An spt-hybridized carbon is found in the double bond ofan alkene. This is described as a vinylic carbon, so the vinyl entry in Table 1-7 isnecessary for choices A and B. Considering that we are looking at carbon-carbonbonds, a primary carbon (with only one bond to a carbon) would have to comefrom a methyl group. This value is necessary for choices A and C. Likewise, a

secondary carbon would come from a group such as ethyl or n-propylConsidering only Et is listed as a substituent in Table 7-7, the value for Et isnecessary in choices B, C, and D.

Choice A is found by looking at the entry for Vinyl-Me, which is 102 kcal/mole.Choice B is found by looking at the entry for Vinyl-Et, which is 101 kcal/mole.Choice C is found by looking at the entry for Et-Me, which is 89 kcal/mole.Choice D is found by looking at the entry for Et-Et, which is 88 kcal/mole. Themost stable bond is the one that requires the greatest energy to break. Thegreatest bond dissociation energy among these choices is 102 kcal/mole, sochoice A is the best answer.

Ionic BondsIonic bonds are bonds formed between two oppositely charged ions. They arecommon between metals and nonmetals. The strength of an ionic bond can bedetermined using Coulomb's law, Equation 1.2. Coulomb's law states that theforce between two charged species is equal to a constant k times the charge oneach ion, divided by the square of distance between the two charges, which are

treated as point charges:

p - v t 1z - 7 q1q2 0.2\t-^rt -4"% c

where F = force, q = charge, r = distance, andto= 8.85 x rcjP -C*." N.m2

The greater the charge on the ion, the stronger the bond; and the closer the ionsare to one another, the stronger the bond. Ionic bonds are typically stronger thancovalent bonds. However, because ions can be solvated in a polar, protic solvent,ionic bonds are often cleaved more readily than covalent bonds in a proticenvironment. In other words, despite the strength of ionic bonds, they are easilybroken by adding water to the ionic lattice. This implies that the Coulombicattraction of the ions to water is comparable to the attraction of the ions to oneanother.


Organic Chemistry Molecular Structure Intramolecular Features

Intramolecular FeaturesIntramolecular features encompass anything that affects the stability of amolecule and the sharing of electron density beyond the localized regionbetween two neighboring, bonding atoms. There are various factors that dictatethe chemical reactivity of a compound and explain the distribution of electrondensity within a molecule. I like to call them the "five excuses" to explain organicchemistry. They are reslnence, the inductiae effect, steric interactions, aromnticity,and hybridization. I/y'e have already examined hybridization and seen the effect ithas on the structure of a molecule in terms of bond angles. Besides consideringthe three-dimensional position of the atoms within a molecule, we will considerelectron density and thus establish reactive sites within a molecule. We shallstart by considering the ever-so-loved resonance theory.

ResonanceResonance is an intramolecular phenomenon whereby electron density is shiftedthrough regions of the molecule via ru-bonds. Resonance is defined as thedelocalization of electrons through a continuous array of overlapping p-orbitals(n-bonds and adjacent lone pairs). Resonance theory can be used to determinethe stability of a structure. There are three rules to follow to determine thestability of a resonance form prioritized according to importance from most toleast:

1. The resonance structure should contain atoms with filled octets(excluding hydrogen).

The best structure minimizes the number of formal charges throughoutthe molecule.

3a. if the molecule contains a negative charge, it is best placed on the mostelectronegative atom.

3b. If the molecule contains a positive charge, it is best placed on the leastelectronegative atom.

Figure 1-20 shows two resonance forms for an amide compound that obey theoctet rule, and a resonance hybrid that shows the composite effect. Theresonance hybrid is an average of all the major resonance contributors.

g|rtaI{':

fi

q

,!5#'s#

fr

.&-

e*.$r-

St-orfl*f,*mqfiI

fti

2.

oo-

,rr.A*rr.- ,r.4.**r,More stable form Less stable form

o6-t:

,rr.A'r'?",Resonance hybrid

Figure 1-20

The resonance structure farthest to the left in Figure 1-20 is more stable than themiddle structure, because there is no separation of charge. You must be able torank the stability of resonance structures and decide whether it is a majorcontributor. Typical questions based on resonance include determining wherecertain molecules are most reactive. You should be able to apply resonancetheory to other features of chemical structure and reactivity. For instance, whenviewing an amide, the electron-rich oxygen is the most nucleophilic site on themolecule. When protonating an amide, it is the oxygen that gets protonated.When amides form hydrogen bonds, the oxygen is the electron-donating site.This has a significant impact on molecular structure in protein folding.

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Figure 1-21 shows four examples of resonance structures and the arrow pushingnecessary to convert between resonance forms. To draw resonance structuresthat are stable, it is often helpful to start with a lone pair and push those electronsinto a n-bond. The electrons from the adjacent n-bond turn into a new lone pair.

roi H (ot H :ij:o Hll l^<-+ \ll I -* I I

H,C/t- g,Qfl*],. - tq.z ts nil,. - to r- \ r"'lv'L'l

HHH

Resonance Hvbrid: o 6-

T/C1.6;2 Ct s-HsC

f NH

H

CH"t'",P'q8

-*CH"t'

Hrc?5l,', {-*

CH"

t"3r.-t\.,,,

CH"

l"*rr3-c\c*r,

Figure 1-21

.. oioiI

6zt-'dl

o6-ti

aoy'c:.oo-

Resonance Hybrid

CH"t",.liz't'l3;,

Resonance Hybrid

CH"t""lll"'13.,

Resonance Hybrid

In Examples 1",2, and 3, shown in Figure 7-2I, the negative charge moves everyother atom between resonance forms. The lone pair becomes a n-bond, and then-bond becomes a lone pair two atoms away. This is true when the number oftotal charged sites remains constant. In Example 1, there is only one chargedatom in each of the resonance forms. You must look for the all-octet resonancestructures. A11 of the resonance forms except the carbocations in Example 4 areall-octet resonance forms. This satisfies Rule 1 on the list of resonance rules.Neither structure in Example 4 satisfies the octet rule. The resonance hybrid is acomposite of the individual resonance contributors. The most stable resonancestructures (major resonance contributors as they are called) have the greatest effecton reactivity and structure for a compound exhibiting resonance.

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Example L.8The C-O bond length is LONGEST in the compound on the:

5-" *A. left, because nitrogen donates electrons through resonance.B. right, because nitrogen donates electrons through resonance,C. left, because nitrogen withdraws electrons through resonance.D. right, because nitrogen withdraws electrons through resonance.

SolutionFor this question, the resonance forms of the lactam should be drawn first:

Partiallysingle bond

€ +The all-octet resonance form on the right has the carbonyl bond in single bondform. The single-bond resonance is caused by the donation of a lone pair ofelectrons by nitrogen. This means that the C-O bond is longer in the compoundon the left, because nitrogen donates electrons through resonance. The correctanswer is thus choice A. The effect of resonance outweighs the inductive effect.The inductive effect predicts that the nitrogen would be electron-withdrawing,because it is more electronegative than carbon.

Inductive EffectThe inductive effect, as the name implies, induces charge separation in amolecule, just as induction in physics refers to the creation of charged sitesthrough induction. From a chemist's perspective, the inductive effect is thedelocalization of electrons induced by electronegative atoms. The inductiveeffect involves the transfer of electron density through the sigma bonds. Ahighly electronegative atom pulls electron density from its neighbor, which inturn pulls electron density from its neighbor. The effect dissipates over distance,but it can affect bonds between atoms up to three or four atoms away. We mostoften consider the inductive effect when a molecule has a halogen.

The inductive effect increases with the electronegativity of the atom. Fluorine isthe most electronegative atom found in organic molecules, so it pulls electronsfrom the carbon to which it is attached in a strong manner. This makes thatcarbon electron-poor, so it in turn pulls electrons from its neighbor. Ultimately,as we see with polarity, the electron density in the molecule is pulled towards themost electronegative atom in the compound. An electronegative atom thereforewithdraws electron density and thus can increase a compound's acidity, increaseits electrophilicity, decrease its basicity, or decrease its nucleophilicity.

oo

6-

$-"o



For the relative electronegativity of common atoms in organic chemistry, thefollowingrelationshipholds: F > O> N > Cl > Br > I > S > C > H. Justrecall"Fonclbrisch" and you'll be in good shape. It may seem strange, but alkyl groupsare electron-donating by the inductive effect, because hydrogen is lesselectronegative than carbon. Figure 1-22 shows an example of the inductiveeffect as it applies to the nucleophilicity of amines reacting with an alkyl halide.

H

t.t:a"*-rt

I Z-roftate=L44HH

Methylamine

H

t.

I Z- -log rate = 4.31

FF

t-.a*l*,

Trifluoromethylamine

Rate for H3CNH2 > rate for F3CNH2Less nucleophilic due to the electronegative fluorine atoms

Figurel-22

The withdrawal of electron density by the fluorine atoms decreases thenucleophilicity of the amine compound by pulling electrons away from thenitrogen atom. As electron density is removed, the compound becomes electron-poor and thus a worse electron donor. This can be verified by the reaction rate ina substitution reaction. As the negative log of the rate increases, the rate of thereaction decreases.

Example L.9\\hich of the following compounds undergoes a nucleophilic substitutionreaction with ethyl chloride at the GREATEST rate?

-{. H3CCHFNH2B. FH2CCH2NH2C. H3CCHCINHzD. CIHzCCH2NH2

SolutionThe greatest reaction rate (the fastest reaction) is observed with the bestnucleophile. Each answer choice has one halogen, so all the choices have slowerrates than ethyl amine. The question asks which experiences the least inductive"-cithdrawal. Chlorine is less electronegative than fluorine (Fonclbrisch), sochoices A and B are eliminated. The inductive effect diminishes with distance, sotFre least electron withdrawal is observed with choice D. You must consider bothproximity and electronegativity when looking at the inductive effect.

-\lthough not applicable in Example !.9, you must also consider whether theinductive effect involves electron donation or electron withdrawal. For instance,methyl amine is more nucleophilic than ammonia (NHg), because the methylgroup is electron-donating. Varying the R-group changes the inductive effect. Italso changes the size of the molecule, so steric hindrance can affect the reaction.For instance, trimethyl amine ((H3C)3N) is less nucleophilic than dimethyl amine(H3C)2NH), because the electron donation by the additional methyl group does

not compensate for the increase in molecular size.

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Steric HindranceSteric hindrance occurs any time two atoms attempt to be in the same place at thesarne time. It is repulsive in nature and increases as the atoms draw closer. Noone is certain about the nature of the force, but it is believed to be electron cloudrepulsion. The effects are similar to what is observed in general chemistry withVSEPR (valence shell electron pair repulsion) theory, except that it is consideredonly when two separate atoms or functional groups interact. Electrons move tobe as far apart as possible, so lone pairs and bonds spread out to accommodatethe geometry that spaces the greatest distance between electrons. Figure 1-23demonstrates the effects of steric hindrance on a couple of organic molecules.Because the alkene is planar, the substituents on the alkene carbons have atendency to collide with one another.

H.C

/\I L- LJ

'""/HsC

Larger C-C-C bond angle

H.C CHs//\ /I

^-^t L-L_\,"/ \HsC CHe

Smaller C-C-C bond angle

HHH/"t\

H"C CH.

{ \.-./ \'\. \/H"C CHN-\__/

H_H H

Reduced bond angle in dimethylbutene,because the methyl group hydrogens repel.

Figure 1-23

Example 1.10Which of the foilowing functional groups is MOST likely to be found in theequatorial position on cyclohexane?

A. -OCH3B. -OCH2CH2CH2CH1C. -OCH(CH3)CH2CH3D. -oC(CH3)3

SolutionOn cyclohexane, substituents with axial orientation experience greater sterichindrance than substituents with equatorial orientation. Because of sterichindrance, the substituent most likely to assume the equatorial orientation is thebulkiest. The tert-butoxide substituent, choice D, has the most crowded alkylgroups/ resulting in the greatest steric hindrance. This makes the best answerchoice D.



AromaticityAromaticity is stability generated from having 4n + 2 ru electrons in a continuous,overlapping ring of p-orbitals, where n is any integer including 0. This is knownas the Hilckel rule. The stability is rooted in the molecular orbital model, wherean energy level is completely filled when there are 4n + 2 n-electrons in the cyclicn-network. Figure 1-24 lists experimental values for the enthalpy ofhydrogenation of a series of alkenes. The large deviation associated withbenzene is attributed to its aromatic stability.

o#o AH = -2g ktul/.ol.

o#o AH = -56 ktul/*ol"

o#o AH = -54 ktul/-n,"

o#* o LH = -49 ktul/r',or"

Despite the presence of three n-bonds, the 1,3,5-cyclohexatriene(benzene) yields far less heat from hydrogenation than expecteddue to its aromatic stability.

Figure 1-24

The first two entries show that the enthalpy of hydrogenation of an alkene is -28kcal/mole per n-bond. The third entry shows that conjugation results instability, reducing the amount of heat released upon hydrogenation, but only byabout 2 kcals/mole. Based on -28 kcal/mole for each n-bond, benzene should beexpected to have a AH of approximately -84 kcal/mole. The difference of 35kcal/mole (84 - 49) cannot be attributed to conjugation alone, hence it is said tobe due to aromatic stability.

)'lot all cyclic, conjugated polyenes show such a large deviation from theexpected value for the enthalpy of reaction. 1,3,5,7-Cyclooctatetraene (CgHg)shows a deviation of only 8 kcal/mole from its expected value of -112 kcal/mole.This implies that conjugation is useful for only a small fraction of the 35kcal/mole difference observed with benzene between its expected and actualva-lues. Because benzene has 6 n-electrons in a continuous n-cycle, it obeys theHuckel rule (it has 4n + 2 n-electrons where n = 1), whiIe1",3,5,7-cyclooctatetraene(CsHs) has 8 n-electrons in a continuous n-cycle and does not obey the Hrickelrule. This lack of aromaticity results in a less stable reactant, so more heat isgenerated in the hydrogenation reaction.

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Example 1.1LThe hydrogen-carbon-hydrogen bond angle about the terminal carbon in thefollowing alkene is BEST approximated by which of the following values?

H"CH"C H" '\ /af-- a- Iv-v

/\/HsC H

A. 108.3"B. 711..7"

c. 118.5"D. 121.5"

SolutionFirst we must consider the hybridization of carbon. It is involved in one n-bond,so the hybridization is sp2. The bond angle about an sp2-hybtidized carbon ispredicted to be 120". The question heri: is whether the angle is slightly greater orslightly less than 120". Because there are two alkyl groups on the other carbon ofthe alkene, the electron density repels the electrons in the two carbon-hydrogenbonds. This forces the two C-H bonds closer together, which compresses thehydrogen-carbon-hydrogen bond angle. According to steric repulsion theory,the angle should be slightly less than 120'. The best answer is thus choice C.

Example 1.L2Which of the following explanations accounts for the pKu of L,3-cyclopentadienebeing only 15, while the pKu for hydrogen on other spr-carbons is around 49?

A. The strain of the five-membered ring forces the proton off.B. The proton is involved in resonance.C. The conjugate base is aromatic.D. The steric hindrance of the sp3-carbon weakens the C-H bond on that carbon.

SolutionThe acidity of a compound capable of losing a proton ({+) can be determined bythe stability of its conjugate base. When an ordinary spr-hybridized carbon (one

that is not stabilized by resonance or the inductive effect) is deprotonated, the

carbanion that is formed is unstable. Carbon is not electropositive, so it does notreadily lose a proton. With 1,3-cyclopentadiene, however, the carbanion that is

formed upon deprotonation has both resonance and aromatic stabilization once itloses the proton. The cyclopentadienyl anion that is formed is aromatic. Thismakes choice C the best choice. The reaction is drawn below:

H

Six conjugated n-electrons in a continuousplanar arrangement of p-orbitals is aromatic.

Because it includes the word tesonance, choice B may at first seem apPealing. Buta proton, having no electron pair, cannot be involved in resonance. Be careful ofwording like this, because it is easy to pick resonance without thinking about it.

H'

HH

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Organic Chemistry Molecular Structure Fundamental Reactivitys

I,is

)r

rfn.e

€

F dffifitffi:xiRd#Ctffit$Fundamental Reactions in Organic ChemistryIn organic chemistry, perhaps the most common class of reaction is nucleophilicattack. In the simplest sense, a nucleophilic compound (one with an electron-richsite) attacks an electrophilic compound (one with an electron-poor site) to form anew bond. In some instances a leaving group is discarded, while in others a n-bond is broken. No matter what the result, the reaction has the sametundamental drive and mechanics. The reactions can be viewed as Lewis acid-base reactions, so organic chemistry starts with a thorough look at Lewis acidsand bases. Prior to that, we shall review Bransted-Lowry acid-base chemistry.

Proton Transfer Reactions (Brsnsted-Lowry Acid-Base Reactions)Brsnsted-Lowry acid-base reactions involve the transfer of a proton (H+) fromthe acid (defined as the proton-donor) to the base (defined as the proton-acceptor). This means that to be a Brsnsted-Lowry acid, the compound musthave a hydrogen that can be lost as H+. A hydrogen like this is often referred toas a protic hydrogen or proton. Throughout this section, we will be using the termgrotonation to describe the gain of an H+. A hydrogen atom has one proton in itsnucleus and one orbiting electron (in a 1s-orbital). when hydrogen loses anelectron to become H+, all that remains is a proton. This is to say that H+ is a

Flroton, and thus the gain of H+ can be referred to as protonation. Deprotonation isthe loss of H+.

To be a Brsnsted-Lowry base, the compound must have electrons available thatcan form a bond to H+. Because a lone pair of electrons is necessary to form abond to the proton, all Bronsted-Lowry bases are also Lewis bases. Figure 1-25shows a proton-transfer reaction, a one-step reaction.

trt, Bond forming, a-\N:, +..' N:' + Hf-.Cl ' =F- ... N- HHrc\\Y son;/;iins Hrc'{,:*:,

HBase

(Proton-acceptor)

Bond breaking

Acid(Proton-donor)

H"C"\o

ConjugateAcid

-I formedH

..O+ :cl:Bond "

broken

ConjugateBasev

te

re

)tis

itis

Figure 1-25

ln the reaction in Figure 1-25, you should note that an arrow going from a lonepair to an atom becomes a bond in the product, and an arrow going from a bondto an atom becomes a lone pair on that atom in the product. This is a standardconvention in drawing mechanisms. The reaction shown in Figure 1-25 is veryfavorable, as indicated by the asymmetric equilibrium arrow. The favorability isattributed to the fact that HCI is a strong acid. Proton-transfer reactions proceedfavorably (AG < 0) from the side with the stronger acid and stronger base to theside with the weaker acid and weaker base. This is to say that a favorablechemical reaction proceeds from the less stable species to the more stable species.There are five strong acids used in organic chemistry that you should recognize:H2SO4, HNO3, HCl, HBr, and HL An important fact to know is that as aBronsted-Lowry acid gets stronger, it loses a proton more readily, so itsconjugate base is less willing to gain a proton. The result is:

\Alhen comparing two conjugate pairs, the pair withthe stronger acid has the weaker conjugate base.

lrtof

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Organic Chemistry Molecular Structure Fundamental Reactivity

Lewis Acid-Base ReactionsLewis acid-base reactions involve the transfer of an electron pair from the base(defined as the electron-pair donor) to the acid (defined as the electron-pairacceptor). This means that for a compound to be a Lewis base, it must haveelectrons available that can form a bond to an electron deficient atom (such as,

but not exclusively, H*). A Lewis acid can have a protic hydrogen, but a Lewisacid may have an empty valence shell capable of accepting electrons. TypicalLewis acids include BF3, AlCl3, FeBr3, and SOCI2. Figure 1-26 shows a Lewisacid-base reaction, where ammonia is the Lewis base and BF3 is the Lewis acid.

Bond forming

Base Acid(Electron Pair Donor) (Electron Pair Acceptor)

Figure 1,-26

The role of a base is essentially the same in both the Lewis and Bronsted-Lowrydefinitions. A base donates a lone pair of electrons to form a bond to an acid,whether the acid is a Brsnsted-Lowry acid or a Lewis acid. In organic chemistry,the terminology varies, and Lewis bases are frequently referred to as nucleophiles.

Nucleophile means "nucleus loving", which implies that nucleophiles seek outpositively charged sites (referred to as electrophiles). The simple guide toorganic chemistry is that negative charges seek out and bond to positive charges.

AcidityAcidity is defined by three definitions: the Arrhenius definition, the Brsnsted-Lowry definition, and the Lewis definition. The Arrhenius definition is that an

acid yields HgO* when added to water. The Bronsted-Lowry definition is thatan acid is a proton (H+) donor. The Lewis definition is that an acid is an electronpair acceptor. The strength of an acid depends on the effects of intramolecularforces on the bond to the acidic proton. These are the electronic forces within a

molecule. They are responsible for the distribution of valance electrons, whichaccounts for the chemical behavior (such as acidity) of the molecule. An acid is

stronger when an electron-withdrawing group is attached to the backbone of theacid, because the molecule is electron-poor, and thus a better electron-pairacceptor. An acid is weaker when an electron-donating group is attached to itsbackbone, because the molecule is electron-rich, and thus a wolse electron-pairacceptor. The primary task associated with evaluating organic acid strength is todecide which groups are electron-donating and which are electron-withdrawing.Figure 1-27 shows some common organic acids and their pKu values.

OHPKu

TInflilW,tdqilrilml[tr!

H^C E:"'\o eti:'

"r."'.?'13,"#;\..H .F:

H.C

.N:H,CV

H

nrrrt

4.&.

4&

&t.

o

.A';il"Carboxylic acid

=10 H

I o*'=e-10

o-*VtH

Alkyl ammonium cation

o

|x:=17'20

Carbonyl cr-protonPhenol

PKa = 15

R- OH

Alkoxide

.ffii;'...:

Figure L-27

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3lSE

airrve

?S,

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BasicityBasicity is most easily thought of as the opposite of acidity. Basicity is alsodefined by three definitions: the Arrhenius definition, th-e Bronsted-Lowrydefinition, and the Lewis definition. The Arrhenius definition is that a bas-evields OH- when added to water. The Brsnsted-Lowry definition is that a base isa proton (H+) acceptor. The Lewis definition is that a base is an electron-pairdonor. The rules that you use for acidity can be applied to basicity, but with theopposite effect. Electron-donating groups increase basicity (while they decreaseacidity) and electron-withdrawing groups decrease the basicity. As a result, thestrength of a base or acid can be determined from the stability of its conjugate.The more stable the conjugate, the weaker the conjugate and the strongei therespective compound (either acid or base). Figure 1-28 shows some commonorganic bases.

dPhenoxide

=4oJl prn =e-71

-/\RO-

Carboxylate

PKa= 4

o/*Y,'H

Alkyl amine

? PKu=-6--a

RACH,-Anionic g-carbon

PKu = -1

R-O-:

Alkoxide

Figure 1-28

AJthough you are not required to memorize exact pK6 values, it is a good idea toknow the "-5-10-75-20 general rule" for organic acids. The pKn for i carboxylicacid is about 5, for a phenol it's about 10, for an alcohol it's about 15, and for aproton alpha to a carbonyl it's about 20. These are close enough for goodguessing.

Example 1.13tr\hat is the pK6 for p-nitrophenoi and the pK6 for its conjugate base?A. O2NC6H4OH has pKa = 7.2; O2NC6H4O- has pKU = t2.8B. O2NC6HaOH has pKa = 11.6; O2NC6H4O- has pKU = 8.4C. O2NC6HaOH has pKa = 7.2; O2NC6H4O- has pKb = 6.8D. O2NC6H4OH has pKa = 11.6; O2NC6H4O- has pKb=2.4

SolutionThe nitro group is electron-withdrawing, which makes nitrophenol a strongeracid than phenol. As an acid becomes stronger, its pKu value decreur"r. Thitmeans that the pKu for nitrophenol is likely tobe7.2 rather than 11.6. Thiseliminates choices B and D. The pKu and pK6 sum to L4, so choice C is the bestanswer. This question could also be answered from a base perspective. The nitrogroup is electron-withdrawing, which makes nitrophenoxide a weaker base thanphenoxide. As a base becomes weaker, its pK5 value increases. This means thatT" prr, for nitrophenoxide is greater than 4,0. This eliminates choice D. Again,the pKu and pK5 sum to 14, so choice C is the best answer. pick C andlainrncredible satisfaction doing what you should do.

Pick the perspective (acid or base perspective) that works best for you, and use itrvith these questions.

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Choosing Where to Protonate and DeprotonateA common question in organic acid-base chemistry is, "Which is more acidic?"This question can refer to two separate compounds, or two sites within the samecompound. There is also the complementary question, "Which is more basic?" InFigure I-29,we are dealing with two sites on the same molecule, so the questionis, "Which lone pair on the molecule is most basic?" It shows the two possibleproducts when acetic acid (CH3CO2H) is treated with a strong acid. Acetic acid(a carboxylic acid) has two sites with lone pairs (two oxygen atoms). Decidingwhich oxygen gets protonated (which forms a more stable protonated species)requires thai you consider many different factors, including the resonancestability associated with each protonated product. Resonance can help tostabilize the excess positive charge on the compound, and thus make the speciesmore stable.

protonation at oxygen ^r,o} tt oxygen u :p- Ho*ysJl';"' !/) fuuc-h <+ H3c-c

\oxygena [. \i6-11 .g^-H

// All-octet resonance form, #akingH.C- C"\

Oxygenb i9-11 o://)

H1C-u <-'>" \^Oxygen t !g-- g

/H

this the more stable product.

:'6:o/

H"C- CO" \^Oxygen bi6;'- 11

/H

Non-octet resonance form with separation ofcharge, making this the less stable product.

Figure 1-29

Because there is resonance stabilization when the compound is protonated atOxygen a, and there is no resonance stabilization when the compound isprotonated at Oxygen b, it can be concluded that carboxylic acids are protonatedat the carboxyl oxygen (C=O), not the hydroxyl oxygen (OH). This means thatthe carbonyl oxygen is more basic than the alcohol oxygen. In reality, we rarelysee carboxylic acids acting as bases, but amides are similar in structure, and it iscritical in biochemistry that we know where they are protonated and thus wherethey exhibit hydrogen-bonding.

Example 1.14Which of the following statements BEST explains why amides are protonated atoxygen rather than nitrogen?

A. The oxygen is less electronegative than nitrogen, so it donates electrons morereadily"

B. The oxygen is larger than nitrogen, so its electron cloud attracts protons morereadily.

C. Oxygen carries a partial positive charge due to resonance withdrawal fromthe nitrogen.

D. Oxygen caries a partial negative charge due to resonance donation from thenitrogen.

Protonation atOxygen b



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SolutionAs shown below, the nitrogen of the amide donates electron density to thecarbonyl oxygen through resonance. This places a partial negative charge onoxygen (increasing its basicity), and a partial positive charge on nitrogen

'decreasing its basicity). Choice A is eliminated, because oxygen is moreelectronegative than nitrogen. Choice B is eliminated, because oxygen is smallerJran nitrogen, not iarger. Because of resonance, oxygen carries a partial negative:harge, while nitrogen carries a partial positive charge. This means that choice C-: false and choice D is true. This explains the basicity of amides.

: O : 9 O ' *ot" basic due to negative charge'

ll-l^Ar,- Aa lessbasicduetoabsenceofelec{rons.R NHz R- -NHz

{cid Strength Factors- ,'-ctors affecting the strength of an acid or base can be broken down into primary=:iects and secondary effects. Primary effects depend on the bond directly to the

',-cic proton. The weaker the bond to the acidic proton, the more readily it:::aks, and consequently more acidic the acid. With acids, the bond breaks in a--:rerolytic fashion, forming ions. Primary effects include size, electronegativity,.:.i hybridization of the atom directly attached to the acidic proton. Secondary,.-.;ls depencl on the effect of the molecule on the atom bonded to the acidic:.,rtor-r. The more electron-rich that atom, the less acidic the proton. The more= .:tron-poor that atom, the more acidic the proton. secondary effects include-:Surflorc€ (the delocalization of electrons within a molecule through ru-bonds),-: - nductive effect (the delocalization of electrons within a molecule through o-: -::.1s), and electron cloud repulsion (deformations of a molecule and elongation:: londs within the molecule). There is also aromaticity to consider, but that will:= :-Cdressed as a special case. Secondary effects involve intramolecular forces,

-.-:h dictate where the electron density within a molecule lies, and thus they,:,::atethereactivityof acompound. weshalllookathowthesefeaturesaffect--- electron distribution within a molecule, starting with the primary effects.l,:r-e 1-B shows common acids and bases in organic chemistry, listed according- :elative strength (in both the acids and the bases).

Strong Acids H2SOa > HI > HBr > HCI > HNO3

i\-eak Acids HF > HCO2H > H3CCO2H > H2CO3 > H3CSH > H3CNH3CIH5C6OH > H3COH

: trong Bases CH3(CH2)3Li > NaNH2 > KH > NaOCH2CH3 > NaOH = KOH

l\'eak Bases H3CNH2 > NaHCO3 > H3CCO2Na > HCO2Na > H3COH

Table 1-8

Fimary Acid Strength Factors: ,::-.arv factors directly affect the strength of the bond to the protic hydrogen.--. ,'. i.'ond weakens (homolytic and heterolytic bond dissociation energy lessens),, -. j strength increases and conjugate base strength decreases. The three primary:::ors to consider are atomic size (when comparing acids involving atoms-:rin the same column of the periodic table), electronegativity (when

- --.raring acids involving atoms within the same row of the periodic table), and' ::r;lization (when comparing acids where hydrogen is on the same atom).

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Strcngth Dependence on Atomic SizeWhen comparing acids within a column of the periodic table, the strength of theacid is dictated by the size of the atom to which hydrogen is bonded. Smalleratoms form shorter bonds to hydrogen than their larger counterparts. Given thatlonger bonds are weaker bonds (in both a homolytic and heterolytic fashion), theacidity of a compound directly depends on the length and strength of the bond'A significant difference in atomic size ultimately determines the relative strengthof acids, because larger atoms make stronger acids.

A prime example of this involves proton exchange (acid-base) chemistry betweenthiols and alcohols, where thiols are stronger acids than alcohols. This is because

sulfur (found in a thiol) is a larger atom than oxygen (found in an alcohol). inacid-base chemistry, you may also consider the reactivity from the base

perspective. To apply this atomic size theory to bases, compare the distributionof electron density about a small atom velsus a larger atom. A conjugate base is

more stable if electron density is spread out over more space. Because largeratoms stabilize negative charge more readily, they are not as reactive, and thusnot as basic. This is often referred as polarizability. Il you keep in mind that themore the negative charge is spread out,.the harder it is for a proton to find thenegative charge, then you see that the compound is not as basic. The thiol-alcohol example is shown in Figure 1-30.

HSPKa = 10'4

HH\r,A-

CH:

Stronger AcidPKu = 15.7

HO CHg

Weaker Acid

Figure 1-30

You should understand that atomic size (polarizability of the conjugate base) can

be applied only when the protic hydrogen is directly bonded to the larger atom.The atomic size argument is used to explain relative acidity for haloacids.Relative haloacid strength for haloacids is: HI > HBr > HCI > HF, confirming thatsize is more important than electronegativity for elements within the same

column of the periodic table. This contradicts what you would expect if youwere to apply the rules of electronegativity. The relative strength in the case ofhaloacids is dictated by atomic size of the halogen. As mentioned before, it isbecause the bond is longest between hydrogen and the largest halogen, thusmaking it the weakest and most readily broken hydrogen-to-halogen bond. Itcan also be considered that the conjugate base (halide) is more stable as it gets

larger, because of the greater space over which the negative charge is distributed.

In the halogen case, the less concentrated (more diffuse) negative charge on

iodide (i-) is not as readily shared as the negative charge on bromide (Br-),

chloride (Cl-), or fluoride (F-). The weaker the conjugate base, the stronger theconjugate acid, This is to say that because I- is the most stable anion of thehalides, it is the weakest base of the halides. Therefore, HI (the conjugate acid ofI-) is the strongest of the haloacids. This theory is also applied when looking at

the relative reactivity of halogen containing organic compounds in reactions

where the halide is a leaving gfouP. The relative acidity of haloacids should be

familiar to you. The mathematics of acidity and basicity is important in general

chemistry, but we shall consider only values such as pK6 to compare the relativestrengths of acids. In organic chemistry, by quantifying acidity with pKa values,

we can support or disprove relationships.



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Str ength D ep en dence on El ectr oneg atia ityWithin a row of the periodic table, the strength of the acid is dictated by theelectronegativity of the atom to which hydrogen is bonded. Given that atomswithin the same row of the periodic table are approximately equal in size, thebond strength depends more on the distribution of electrons within the bondthan it does on the length of the bonds. Because the electronegativity of the atombonded to hydrogen affects the distribution of electron density within a bond, theelectronegativity of the atom directly bonded to the acidic proton dictates theacidity of the compound.

The relationship is reasonable in that as an atom bonded to hydrogen has agreater electronegativity, the electrons are pul1ed more towards that atom, ratherthan toward hydrogen. This allows the bond to be cleaved in a heterolyticfashion rather easily. The result is that the more electronegative the atom, thestronger the acid.

Comparing electronegativity applies only for hydrogensbonded to atoms in the same row of the periodic table,not the same column of the periodic table.

The greater acidity associated with alcohols than amines of equal alkylsubstitution can be explained by invoking the fact that the electronegativity ofoxygen is greater than that of nitrogen and that oxygen and nitrogen haveroughly equivalent atomic sizes. Figure L-31 shows the relative acidity whencomparing a hydrogen bonded to fluorine, oxygen/ nitrogen, and carbon.Because fluorine is more electronegative than oxygen, which is in turn moreelectronegative than nitrogen, the following relationship holds true:

H-F

Strongest Acid

PKu = 3'2

HH\rt .4.HO CHs HzNn

L

i.

rt

e

tlfsS

tt

st.

n,),

e

erf

Itseiles/

Amphoteric Very Weak AcidpKa = t57 pKa = 33

Figure 1-31

PKu = 10'8

HH\l HH\rCHg H CH:

Weakest Acid

PKo= 49

The same rationale used to explain relative acidity in Figure 1-31 can be used toerplain relative basicity in the conjugate bases amines, alcohols, and hydrofluoricadd. Because nitlogen is less electronegative than oxygen and fluorine, it morercadiiy donates an electron pair to a proton, so the trend in basicity shown inFigure 1-32 holds true:

ui.io cH.,

ir<o = -rl iro = -r.s

Figure L-32

Amides (RZN-) are some of the strong bases that are used in organic chemistry.They are used to remove alpha protons (the protons bond to the carbon that isalpha to a carbonyl) and the terminal hydrogen of an alkyne, both with pKuvalues above 17.

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Strength D ependence on HybridizationThe hybridization of an atom affects the bond length and the distribution ofelectron density within a bond, so the hybridization of an atom directly bondedto the acidic proton affects the acidity of the compound. The relationship is notobvious in that it is not true that longer bonds lead to stronger acids, as is the casewith most other acids. In fact, the relationship between length and acid strengthis exactly the opposite. As the hybrid orbital gets smaller, the electrons are heldcloser to the nucleus of the atom bonded to hydrogen, so the bond can be cleavedin a heterolytic fashion more easily.

The result is that the more s-character in the hybrid orbitalof the atom bonded to hydrogen, the stronger the acid,

This results in the relative acidity being sp > sp2 > sp3. It is most commonlyobserved with carbon acidity, but can also be observed with nitrogen andoxygen. Figure 1-33 shows the comparison of acids where hybridization explainsthe difference in acid strength.

o>e:ir":

::r: :r:-

:a:*ii lnt-:,:

r"ai:r:

*C*,

: -*,--.re5

:,*rl"

tiril-ril:

-HCH"sPL ^ \ / 'pKa=26 H- C= C- CH3 spr CrC: Ca

PKo=36 H CHsStronger Acid

oK"=26 HI

^"\r'cH3SP.N I

HStronger Acid

Weaker Acid

PKo=33 H

SP"I

N ;N--.cu,rr lz-

HHWeaker Acid

L*

: ::--{ -

Figure 1-33

Example 1.15A compound with which of the following atoms would be the STRONGESTbase?

A. A weakly electronegative atom carrying a negative chargeB. A highly electronegative atom carrying a negative chargeC. A weakly electronegative atom carrying no chargeD. A highly electronegative atom carrying no charge

A base is a compound that donates electrons. The best electron-donating groupwould be an atom that readily shares its electron density, so the atom should notbe very electronegative. The stronger base carries a negative charge, rather thanno charge, so it can more easily donate electrons to an H+. The best answer ischoice A,

l:.i:.:.:-?;

* i -!..-

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Organic Chemistry Molecular Structure Flrndamental Keactivity

Secondary Acid Strength FactorsSecondary factors indirectly affect the strength of the bond to the protichydrogen. Secondary effects involve the electronic environment of the atombonded to the protic hydrogen. Simply put: If the atom has electron densitydonated to it from the rest of the molecule, it will not pull away electrons,fromthe bond to hydrogen as readily, so the compound is less acidic. if the atom haselectron density withdrawn from it by the rest of the molecule, it will pull awayelectrons from the bond to hydrogen more readily, so the compound is moreacidic. Secondary factors to consider for now are resonance and the inductiveeffect. Resonance is a more significant factor than the inductive effect.

Strength Dependence on ResonanceResonance, being ihe delocalization of electrons through an overlapping string of;-bonds, can be either an electron donating effect or an electron-withdrawingeffect. It is important to distinguish between electron-donating groups andelectron-withdrawing groups, because they affect acidity in opposite ways.

When looking at resonance, an adjacent atom makinga n-bond is electron-withdrawing, while an adjacentatom with a lone pair is electron-donating.

Trpical examples of organic acids that are affected by resonance include phenolsand carboxylic acids. Figure 1-34 shows the resonance effect of an electron-donating group (OCH3) on a carboxylic acid. It decreases the acidity, raising thepKu only when it can delocalize n-electrons.

o!\ ith no n-bond, therecan be no donation*uough resonance

Methoxy lone pair candonate to acid groupthrough the n-bond

Pf"=56

Figure 1-34

The methoxy substituent (OCH3) has a lone pair of electrons on the first atomadjacent to the carboxylic acid. The oxygen atom of the methoxy group donateseleckon density to the system, and thereby decreases the acidity of the carboxylicacid. An electron withdrawing substituent has the opposite effect, because itgrakes hydrogen electron deficient, and thus more protic. Electron-withdrawingsubstituents increase the acidity of a compound, as shown in Figure 1-35.

".$.{'"o_Qo*Jo o

pKa=10.0

Figure 1-35

It is important to know that electron-withdrawing groups make the compound a5etter electron-pair acceptor (Lewis acid), so the acidity increases with electron--.'.ithdrawing substituents. Electron-donating groups make the compound alt orse electron-pair acceptor, so the acidity decreases with electron-donatingsubstituents. This holds true with all effects (not just resonance). It is easy to getcaught up in the idea that any resonance makes a compound more stable, but thisl-c not always true. Use the electron-donating or electron-withdrawing;haracteristics to determine the effect on the acidity or basicity.

oH, uuco: FoH



Noz t

Strength Dependence on the lnductioe EffectThe inductive effect is the delocalization of electrons through the sigma bonds of

a molecule. Fluorine is the most electronegative atom found in organicmolecules. It withdraws electron density from the carbon to which it is attached,

which makes that carbon electron-poor. As a result, the electron-poor carbon

pulls electron density from its neighbor. Ultimately the electron density in the

molecule (as with polarity) is pulled towards the most electronegative atom and

away from the other side of the molecule. Hydrogen is often at the other end ofthe molecule, so it becomes electron-deficient and becomes a better electron-pairacceptor (more acidic). An electronegative atom therefore withdraws electron

denslty and thus increases an acidic compound's acidity or decreases a basic

compound's basicity. For common atoms in organic chemistry, the order ofelectronegativityyou should recallis: F > O > N > Cl > Br > I > S > C > H' This

helps you predict the relative acidity due to the inductive effect. The carboxylicacids in Figure 1-36 show the effect of electronegativity.

cl>oo

,"\"^o\,HH HH

PKa = 3.17 PKa= 4.74

HO

Figure 1"-36

The inductive effect depends not only on the electronegativity of the

withdrawing atoms, but also on their proximity. In Figure 1.-37, the proximity ofthe chlorine atom to the acidic group affects the acidity of the carboxylic acid.

The effect on the compound's acidity drops with distance and is negligible when

they are over four atoms apart. The closer the electron-withdrawing substituentto the acid functionality, the greater the increase in the acidity.

CHg HO CHe HO cH2cl Ho

HHPKa = 1'58

HH=2.59PKo

HH HHpKo = 2.88 PKa=2.92

Cl H

PKa = 2'88

HHPKa= 4'07

H]PKa = 3'78 PKa = 434

o

'no\'"nuHH

PKa = 4'82

HHPKa = 4'51

HHPKa = 4'82

Figure 1-37

It may seem strange, but alkyl groups are electron-donating. This is attributed to

the fact that hydrogen is less electronegative than carbon. When the substituent

is electron-donating, the inductive effect decreases the acidity. As a result, alkylgroups reduce thJ acidity of a compound. In Figure L-38, the compounds

become less acidic due to the electron-donating methyl group'

o

HoA.r ' ,to


Figure 1-38

The Berkeley Review

itv Organic Chemistry Molecular Structure Fundamental Reactivity

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Values, Terminology, and ApplicationsConceptuai questions in acid-base chemistry can often be reworded into: "\zVhichacid is stronger?" The answer can tell you the sign of values and the nature of aleaction. The following terms are ways in which an increase in a compound'sacidity can be observed. As a compound becomes more acidic:

Ku of the acid increases

:Ku of the acid decreases

acid dissociation increases

pH of a 1.0 M solution decreases

-\G for reaction with a base decreases conjugate base strength decreases

fKA of the conjugate base increases the stability of the conjugate base increases

i ou must also be able to use the qualitative concepts. A proton transfer reaction

;"rcs favorably from the side with the stronger acid to the side with the weaker:cid; a favorable reaction proceeds from stronger acid to weaker acid. Tojetermine the K"O for a proton-transfer reaction, first decide which of the acids:l conjugate bases) is stronger. From there, it is a matter of determining the

;rection of the equilibrium. If the equilibrium lies in favor of the reactants, theK*" < 1. If the equilibrium lies in favor of the products, the K"q t 1.

Example 1.16r\irich of the following compounds is the STRONGEST acid?

-4- H3CCH2OHm. H3CCO2HC F3CCH2OHD. F3CCO2H

S'olution

-::boxylic acids are stronger acids than alcohols, due to the withdrawing of.!e;ton density by the carbonyl oxygen through resonance. This eliminates::--..ices A and C. Fluorine is highly electronegative, so it withdraws electrons::n the acidic hydrogen via the inductive effect, thus increasing the acidity.::,e skongest acid is the carboxylic acid with fluorines attached, choice D.

Etample 1.17i: --r,' can the difference in aciditv between trifluoroacetic acid and trichloroaceticr-: be explained?

".4. Fluorine is larger than chlorine, so trifluoroacetic acid is a stronger acid.$" Chlorine is larger than fluorine, so trichloroacetic acid is a stronger acid.C lirichloroacetic acid is a stronger acid, because chlorine is more

eleckone gative than fluorine.D. lirifluoroacetic acid is a stronger acid, because fluorine is more

e.iectronegative than chlorine.

iolutionF:-rcrine is smaller than chlorine, so choice A is eliminated. The acidic proton is:rar :,onded to the halogen, so atomic size does not dictate the acid strength. This*:::nates choice B. Fluorine is more electronegative than chlorine, so F,r-'.:-i.lrarvs electron density more than Cl, making trifluoroacetic acid more*!"-:Ton poor and a stronger acid (better electron pair acceptor). The pKu of:'iuoroacetic acid (F3CCO2H) is 0.18, while the pK3 of trichloroacetic acidi-irCCQH) is 0.64. A lower pKu value confirms that trifluoroacetic acidF:CCO2H) is the stronger of the two acids. Choice D is the best answer.

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Organic Chemistry Molecular Structure Flrndamental Reactivity

Example 1.18Which of the following statements is true as it pertains to pKu values?

A. A functional group that is electron-withdrawing by resonance lowers thepKu value, while a functional group that is electron-withdrawing by theinductive effect raises the pKu value.

B. A functional group that is electron-withdrawing by resonance lowers thepKu value, while a functional group that is electron-donating by theinductive effect raises the pKu value.

C. A functional group that is electron-donating by resonance lowers the pKuvalue, while a functional group that is electron-withdrawing by the inductiveeffect raises the pKu value.

D. A functional group that is electron-donating by resonance lowers the pKuvalue, while a functional group that is electron-donating by the inductiveeffect raises the pKu value.

SolutionRegardless of the effect (whether it is resonance or the inductive effect), electron-withdrawing groups increase acidity and lower the pKu while electron-donatinggroups decrease acidity and raise the pKu. This eliminates choices A, C, and Dand leaves choice B as the best answer. The effect of an electron-withdrawinggroup can be seen in the following trend: H3CCO2H (pKa = 4.74) is less acid thanH2CICCO2H (pKa = 2.85), which is less acidic than HCI2CCO2H (pKa = 1.26),which in turn is less acidic than CI3CCO2H (pKu = 0.64). As the amount ofelectron-withdrawal increases, the acidity increases.

Example 1.19Which nitrogen atom in the following molecule is the MOST basic?

dNHz

A. Nitrogen aB. Nitrogen bC. Nitrogen cD. Nitrogen d

SolutionThe most basic nitrogen atom is the one most capable of sharing its lone pair witha proton or electrophile, which means that the nitrogen where the lone pair isleast shared within the molecule is the most basic. Nitrogens a and b havereduced basicity, because the electron pair on nitrogen is being donated to thearomatic ring through resonance. Nitrogen d has reduced basicity, because theelectron pair is being donated to the carbonyl group through resonance. OnlyNitrogen c is free to share its electrons (which are not being delocalized

""*^"r" -t,^t^ -" *

b

NHz O

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Organic Chemistry Molecular Structure F-undamental Reactivity

Example 1.20l\hat is observed when histidine is protonated on its side chain?

A. The imine nitrogen gets protonated, because it is more electronegative thanthe amine nitrogen.

B. The imine nitrogen gets protonated, because it is less electronegative than theamine nitrogen.

C. The amine nitrogen gets protonated, because the imine nitrogen shares itslone pair of electrons with the n-bonds in the ring through resonance/ thus itcannot be protonated.

D. The imine nitrogen gets protonated, because the amine nitrogen shares itslone pair of electrons with the n-bonds in the ring through resonance, so itcannot be protonated.

Solutiontsoth nitrogen atoms are equally electronegative, because neither carries a charge.This eliminates choices A and B. One could argue that having different:"'bridization makes the electronegativity different, but the goal here is to find-ie best answer, and if it is necessary to stretch the definition of terms, you're:etter off finding a better answer choice. The histidine is protonated on the imine:rtuogen, because the lone pair on the amine nitrogen is being shared with the:;clic rc-system through resonance. This makes the ring aromatic, so the lone:air on the amine nitrogen is not available to be donated as a base. The bestins\{,er is choice D, as shown below.

H CH"

H*

-> r\-j@ R"ronur,ce is still possibie, so

the n-system remains aromatic.

oHeN

@H.N

oor

H CH,,t-4;,\:'

lmine nitrogen

Ax)^,J

H

oOilt'*#,r'

Ir---H CH?

l-H

7;;"\ 7 Amine nitrogenNT-

.I\-

@

HsN

No resonance is possible, so then-system is no longer aromatic.

H CH?

1x,^\Jz

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Summary of Acids and Bases EquationIn acid-base chemistry, it is impossible to avoid equations, aithough generalchemistry typically involves more math than organic chemistry. Some equationsdefine concepts, while others help in caiculations. Equations 1.3 through 1.10 aredefining equations, and each should be tattooed in your memory bank:

hfrhu6-{llrttu;miln$mfr

iMMsMl q

,m$Mfl,rm&dMd!

rymhq$h'

ffiilbttrrdffiffi@dQury'Fffiflmrmmmffi

pH = -log[H3O+]

pOH = -log [OH-]

PKa = -1og Ku

PKb = -1og K6

[HaO+] = 10-PH (1.4)

pH + pOH = 14 (at 25"C) (1.6)

Ku = 16-PKu (1.8)

pKa + pKb = 74 (at25"C) (1.10)

(1.11)

(1.12\

(1.1,4)

(1.3)

(1.5)

(1".7')

(1.e)

Given pKu values, the AG" and K"O for a proton transfer-reaction can becalculated. Listed below are four equations that you should be able to workwith. Equation 1.11 is the Henderson-Hasselbalch equation. It is used todetermine the pH of buffered solutions and is derived from the fundamental aciddissociation reaction. Equations 1.11 and 1.13 are common, and should becommitted to memory:

pH=pKu+logl{l- tHA]

1o(Pu-pKu;- [A-]

IHA]

ForHA + H2O -:i+ H3O+ + A- ro=[H:-O*]-[A] (1.13)tHAI

Keq = 10Pkn(product acid) - PKa(reactant acid)

Equation 1.14 is derived from the fundamental reaction:

HA + B- + A- + HB, wirere K". = [HB][A-]lHAttB-t

'm@wffi@r4ryq0

By definition:

Ka(HA) =

=#F

= 10-PKa(HA) and Ku1Hs,= t"i#it '

= 1o-pKa(HB)

Using equilibrium:

,, tHBltA-l tA l

tHAltB-l tHAllHBl =[H3o+][A] xtB-l tHAl

1 - Ka(HA)

Ka(HB) Ka(HB)

irlllH3o+llB-l

= Ka(HA) x

Thus:

Ko. =Ka(HA) = lO-pKa(HA) = 1g-pKa(HA) * 16+praluB) = 16(pKoen)-pKa111a))Ka(HB) 1g-PKa(HB)

HB is the product acid and HA is the reactant acid, so by strbstitution we deriveEquation 1.13:

Keq = 1gpKa(product acid) - pKa(reactant acid)

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Organic Chemistry Molecular Structure Fundamental Reactivity1

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Reaction TypesLr organic chemistry, there are a few fundamental reactions that describe thenajority of reactivity in organic chemistry. In addition to acid-base chemistry,there are substitution and addition reactions. All of these reactions involve theaCdition of an electron pair to an electron-poor site. If you are able to identify the:lectron pair that can be shared and the electron-poor site within the reactants,'"-ou will be successful at predicting chemical reactivity. Addition and:ubstitution reactions are often further classified according to their nucleophile,rl eiectrophile, but for now, we shall consider only a generic addition reaction to::,iroduce reaction pathways, mechanistic logic, transition states, intermediates,

'nd energy diagrams.

Electrophilic Addition Reactions;-ectrophilic addition reactions are common irr alkene chemistry. The alkene::cleophile donates its most available electron pair (the electrons of the n-bond):r an electrophile. The reaction involves a strong acid, often a haloacid, as the

=-ectrophiie. If the alkene is asymmetric, then you need to consider orientation:.:tors in the reaction in terms of minimizing steric hindrance in the transition

':ate and stability of the intermediate. When looking at an alkene, the term.;:'-.:,tnrctric refers to a state where the two carbons have an unequal types of:-'-:bons bonded to each. When the reactant alkene is asymmetric, the reaction-:,-ion's Markovnikov addition rules. The electrophilic substituent adds to the.=ss hindered carbon of the n-bond, according to the rules associated withl, f:rkovnikov addition.

\lechanisms (Reaction Pathways)l,le,chanisms are the step-by-step account of all species thought to exist as

:=--tants proceed to form products. Mechanisms are commonly referred to as':-;--;irrr pathzoays. Mechanisms focus on intermediates, propose transition states,::rl break reactions into steps. Few reactions in organic chemistry occur in one: -:. so mechanisms often involve many steps before reaching a product. You:-.=.,' recall from general chemistry that the rate of a reaction depends on the rate-:=:ermining step. Whenever there is more than one step in a reaction:re;hanism, there exists a rate-determining step.

-l:: mechanism for a reaction is proposed based on the products that are formed,: :;-,' fast the reaction proceeds, and which changes in conditions alter the rate. It:::. be supported by monitoring isotopically labeled atoms on a molecule. A:.=:hanism can never be proven, only disproved. Because reaction data are::=ed on the rate-determining steps, rate data are critical in supporting a

:ri:-hanism. Mechanisms are the most probable pathway that accounts for the-::a presented. Most reactions in organic chemistry involve a nucleophiler-:-;krng an electrophile, which is often referred to as nucleophilic substitution.ie:-.re any mechanistic studies are encountered, be sure that you have a strong;35p of the definitions of the terms.

lL= reason for thoroughly addressing a reaction using arrow-pushing techniquer. :o learn themes in reactions that are repeated in other reactions. In other,,'.':ls, understanding mechanisms allows you to predict products of unknown:=::tlons. For instance, if a carbocation is formed during one reaction:,:iranism, an analogous reaction is likely to form a carbocation as well. From--:*. perspective, almost all organic chemistry reactions can be classified as Lewisl::i-base reactions. These include all reactions that involve a nucleophile:::"cking an electrophiie.

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Example L.21A carbocation can be classified as:

A. an intermediate.B. a transition state.C. a product.D. a catalyst.

SolutionA carbocation is an intermediate, because it is not stable enough to be present atthe completion of a reaction, eliminating choice C. It is not present at the start ofa reaction, and it does not form an activation energy-lowering complex duringthe reaction, so it is not a catalyst. This eliminates choice D. It is present for afinite period of time, so it is not a transition state, meaning choice B is eliminated.It has a finite lifetime and is semi-stable, so the best answer is choice A.

Arrow-Pushing in MechanismsWhen drawing mechanisms, the atrow starts with the electron pair destined toform a bond. If a bond is broken, then the arrow starts at the bond being brokenand finishes as either a new bond being formed, or as a lone pair on anotheratom. Figure !-39 explains the arrow-pushing associated with the electrophilicaddition of hydrobromic acid, HBr, to a symmetric alkene, E-2-butene.

Step 1: There is no lone pair in the alkene, so the arrow starts from themost available electron pair (the rc-electrons) and goes to the partiallypositive hydrogen of HBr. The bond between H and Br is broken.

a^

*a^"").l.iYg.fil --

,/n-Bond\ Bondbrbakingf{ breaking CH3

H"C Bond H"Jj:"3>" ,

/\H CHs

..o. r,t.

Nucleophile(e- Pair Donor)

Electrophile(e- Pair Acceptor)

Planar carbocation Bromide anion(Electron poor) (Electron rich)

Step 2: A lone pair on bromine goes to the positive carbon of thecarbocation intermediate to form a sigma bond between C andBr. The product is an alkane with bromine (alkyl bromide).

H.C H" \^ ;f n

":',J-'t.",Alkyl bromide product

.i_

-H CHaPlanar carbocation Bromide anion

(Carbon is electron-poor) (Bromine is electron-rich)


Figure 1-39

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Organic Chemistry Molecular Structure Fundamental Keactivity

Competing MechanismslVhen an asymmetric alkene reacts with hydrobromic acid, there are multiplepotential products. The reaction can produce either a secondary or tertiary alkylbromide product. The pathway leading to the tertiary alkyl bromide has a loweractivation energy, forms a more stable intermediate, and results in the morestable product. The pathway leading to the more stable product yields thelhermodynamic product. The pathway with the lower activation energy yieldsfie kinetic product. Some reactions may involve a competition between thepathway leading to the thermodynamic product versus the pathway leading tothe kinetic product. Figure 1-40 shows the treatment of an asymmetric alkenervith hydrobromic acid.

Observed Reaction(Leads to the More Stable Product by way of the More Stable Intermediate)

H"C

HsCElectrophile

(Strong haloacid)

A

-

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Bromide anion(Electron-rich)

Figure 1"-40

CH:Nucleophile

(Alkene ru-bond)

H.CA'J4.}H +

/\HeC CHe

H:C CHs

Nucleophile Electrophile(Alkene n-bond) (Strong haloacid)

Bromide anion(Electron-rich)

Minor Product Reaction(Leads to the Less Stable Product by way of the Less Stable Intermediate)

H:C CHg3'Carbocation Bromideanion

(3" > 2" in stability) (Leaving group)

HeC CHs

2'Carbocation Bromideanion(2' < 3" in stability) (Leaving group)

Hsc. \H ..lP'':-.. \_ \,

H\t"'/ \HsC CHs

2" Alkyl bromide product(Less stable than 3" product )

H.C H'\^ ;f u... L- L

H.c\7 \: Bi CH:

S' efiyf bromide product(More stable than 2'product )

/,\

CHs

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Organic Chemistry Molecular Sllrrclrrre Fundamental Reactivity

Energy DiagramsAn energy diagram accounts for the energy of the system as a reaction proceeds.As energy is added to the system (to break bonds or reorient the molecule to aless stable structure), the energy goes up. As energy is released from the system(upon making bonds or reorienting to a more stable structure), the energy goes

down. The overall difference from start to finish represents the energy changefor the reaction (either AG or AH.) The diagram reaches an apex with a transitionstate, and a localized nadir in the middle with an intermediate. The startrepresents the energy of the reactants, and the end represents the energy of theproducts. Each apex represents a transition state in the reaction, which results ina step in the reaction, Figure 1-41 shows the two energy diagrams associatedwith the two possible reactions when adding HBr to an asymmetric alkene.

Products2": (H3C)2CHCHBTCH33': (H3C)2CBICH2CH3

AG"12.; = -7.63kcar / mole

aGr(3") = -8.77kcar / mole

T2(3')

>'borro

AJO)li

trr

Reactants(H3C)2C=CHCHa + HBr

Reaction coordinate

Figure 1-41

Energetics and KineticsWill a reaction proceed or not? This(overall energy of the process) andReactivity can be analyzed as follows:

1. Identify and label the reactants

question deals with thermodynamicskinetics (overall rate of the process).

(in most cases as nucleophile andelectrophile).

2. Predict the products of the reaction. If the reaction occurs in multiple steps,predict products for each step.

3. Show the flow of electrons in each of the steps. This is known as drawing themechanism.

4. Evaluate whether the reaction is overall thermodynamically favorable. AG"is the standard free energy. When negative (AG" < 0), the reaction isfavorable in the forward direction. When AG' is negative, Ksq (the

equilibrium constant measuring the ratio of products to reactants) is greaterthan 1 (K"q > 1). Because AG" = AH'- TAS", and most organic reactions takeplace in sblution where entropy is relatively unchanging, AH" can beapproximated from AG". In conclusion, to determine whether a reaction isthermodynamically feasible, look at O6', KeO, and AH".


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i. Evaluate whether the reaction can proceed at the given temperature. Thereare favorable reactions that never take place, because the activation energy(E6s1) is too great. When looking at reactions, there is AGX to consider, theactivation free energy for each step. There must be enough free energypresent that a reasonable number of reactants can overcome the activationbarrier. This factor is hidden in the rate constant (k.*), which takes intoaccount the frequency of collision, the temperature, and the orientation ofmolecules during collision. In conclusion, look at: T (high temperatureequals fast reaction), Eact (low Eagl equals fast reaction), and intermediates(stable intermediates equal fast reaction).

Effect of Temperature on Reaction RateThe rate of a reaction increases as temperatute increases, because there is more

=ee energy available in solution for reaction. Temperature is part the rate:onstant, kr*, mathematically expressed Equation 1.15.

k.* = A "-Eut'/RT

(1.1s)

-,q-here A is the Arrhenius constant and E661 is the activation energy.

The energy diagrams in Figure 1-42 show the change in energy level as the:eaction proceeds and Figure 1-43 shows the molecular energy distributioniroughout the entire system at two different temperatures (T1 and T2).

Reaction coordiante

Figure 1,-42

Kinetic energy

Figure 1-43

At T2, the average kinetic energy of the molecules is greater than it is at T1. Thus,at T2, more molecules have enough energy to overcome the activation energy'rarrier

of the reaction.

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rl]

state (f)

TztTr Minium Eo.1 for reaction

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Organic Chemistry Molecular Structure Physical Properties jPhysical,, Propefties,Physical Properties and Intermolecular ForcesIt is important to have an understanding of how molecules interact with oneanother. By understanding these interactions, it is easier to predict what willtake place in a chemical reaction or physical change, CommonMCAT questionsinvolving intermolecular forces include determining the boiling points of tworelated compounds, as well as a compound's solubility propeitils or meltingpoint. The rule is simple: the greater the forces, the higher the boiling point. wewill use relative boiling points to verify the relative intermoleCular forcesbetween compounds. The following are intermolecular forces that affect theboiling points,listed in descending order of strength.

Hydrogen bondingHydrogen bonding is a weak bond (approximately 4 to g kcals/mole) that existsbetween a lone pair of electrons and a hydrogen that carries a substantial partialpositive charge. A hydrogen has a substantial partial positive charge when it isbonded to a small electronegative atom such ai nitrogen, o*yg".r,-o, fluorine.There are no hydrogen bonds involving hydrogens thai are covalently bonded tocarbon! You should be aware that not all hydrogen-bonds have the samestrength. For instance, an amine lone pair binds the protic hydrogen of analcohol more tightly than an alcohol lone pair binds the protic hydro-gen of anamine. The strength of a hydrogen bond can be estimated from thu bur"properties of the lone pair donor and the acid properties of the hydrogen donor.PoIar lnteractionsPolar interactions are the Coulombic interaction between partially chargedparticles (approximately 1 to 3 kcals/mole). Negatively chirged sites attiactpositively charged sites. The greater the partial charge on the site of themolecule, the stronger the force between opposite chargei. The strength of theforce also increases as the distance between oppositely charged sites d"creases.A typical example of a polar interaction is the dissolving of ions and polarspecies into water.

Van der Waalsvan der waals forces exist between all compounds. van der waals forces areconsidered only when no other forces exist to any extent. They are the result ofthe attraction between temporary dipoles (a very weak force). They are theweakest of the three intermolecular forces between molecules that we shallconsider. They are considered to be less than 1 kcal/mole.

The intermoiecular forces are the primary consideration when approximatingphysical properties, when forces are not enough to determine ihe physicalproperties such as boiling and melting point, then structural features such asmolecular mass and molecular rigidity become the determining factors. Theheavier the compound, the harder it is to remove it from a lower energy phaseand place it into a higher energy phase (i.e., liquid to gas). what is meJnt by"molecular flexibility" is the ability to twist and conform to allow for moresurface area, and thus more intermolecular interactions. Van der Waals forcesare rarely used to explain anything except why there is not a complete absence ofintermolecular force. Figure L-44 shows the different forces.

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Organic Chemistry Molecular Structure Physical Properties

Hydrogen-bonding:

Dip ole - Dip ole I nte ractio n :

-rr.H

HeC

F_ kry

gu-?o\

CHs

H3CV'06-6+l

HaC

Van der Waals interactions:

:-.$e?.H

T.)a%'"H

Tr--$T,{,H

Figure L-44

F-xample 1.22'ir{-hv is methanol (CH3OH) a liquid at room temperature, while ethaneCII3CH3) is a gas?

3" Ethane is more polar than methanol.E- Methanol is significantly heavier than ethane.C Methanol has stronger van der Waals interactions than ethane.D. Methanol has hydrogen-bonding, while ethane does not.

Solution!fiethanol (CH3OH) is a liquid at room temperature, while ethane (CH3CH3) is a

Sas, so methanol has the greater boiling point. Ethane is a nonpolar molecule, soCpice A is eliminated. Methanol has a molecular mass of 32 grams/mole, whilee$lane has a molecular mass of 30 grams f mole, so methanol is only slightly (notsignificantly) heavier than ethane, so choice B is eliminated. The van der WaalsrnEractions are roughly equal for all molecules, so choice C is eliminated.hfiethanol has hydrogen-bonding, while ethane has no protic hydrogen andSr,erefore no hydrogen-bonding. The higher boiling point is due to thehr-drogen-bonding of CH3OH, so choice D is all yours!

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Hydrogen-BondingHydrogen-bonding is the strongest of the common intermolecular forces. It canbe thought of as a weak covalent bond between a hydrogen that carries a partialpositive charge and the lone pair on a nearby atom. The strength of a hydrogenbond varies between 4 and 8 kcals per mole. Hydrogen bonds are similar to theinteraction of a base with an acidic proton in the transition state of a protontransfer reaction. This is where the term protic comes from, as a hydrogencapable of hydrogen-bonding is also slightly acidic. The partial positive onhydrogen is strong enough to form hydrogen bonds when the hydrogen isbonded to either fluorine, oxygerL or nitrogen (highly electronegative atoms).

The strength of the hydrogen bond is best approximated by the acidity of theproton and the basicity of the lone pair donor. The strongest hydrogen bondexists between hydrogen on fluorine and a lone pair from nitrogen. Hydrogenbonding in alcohols is stronger than in amines, as supported by the greaterboiling points of alcohols relative to amines with comparable mass. The bestexplanation for this is the greater acidity of the protic hydrogen of an alcoholthan an amine. compounds that form hydrogen bonds are polar, so when acompound has hydrogen-bonding, it also has dipole-dipole interactions. Thismeans that when you are asked to compare boiling points of compounds, youshould first look for hydrogen-bonding.

Figure 1-45 shows the structures and boiling points of butanol and butanal.Butanol is capable of hydrogen-bonding, while butanal is not. For this reason,the forces between butanol molecules are stronger than the forces betweenbutanal molecules. The result is that butanol molecules bind one another moretightly, resulting in a higher boiling point. This comparison of the two isreasonable, because the two molecules are of roughly equal mass.

,,^-'AtButanal (b.p. = 76.1,'C)

Figure 1-45

Example 1.23Which of the following compounds has the HIGHEST boiling point?A. (H3C)3NB. (H3C)2NHC. (H3C)3CHD. H3COCH3

SolutionThis question centers on intermolecular forces, particularly hydrogen-bonding.As a rule, the compound with the greatest intermolecular forces has the highestboiling point. Hydrogen-bonding is the strongest of the intermolecular forces,and if a compound has hydrogen-bonding, it also has dipole-dipole interactions,so the compound with hydrogen-bonding has the greatest intermolecular forces.To form a hydrogen bond, both a lone pair of electrons and a hydrogen on ahighly electronegative atom (N, O, or F) are required. Choice C does not containa lone pair, so it is eliminated. Choices A and D have all of their hydrogens oncarbon, so they are both eliminated. This leaves choice B as the only moleculethat forms hydrogen bonds.

OH

I

/ \//Butanol (b.p. = 717.4' C)

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Folarity?olarity is defined as an asymmetric distribution of electron density within a

:roiecule. Perhaps an easier way to think of this is as a tug-o-war for the

=-ectrons between atoms. The more electronegative atoms pull the electrons::ore tightly. If the molecule has more of one atom on one side of the molecule-:ra-n another--that is, if it is asymmetric about a ceniral point-- then it is polar.:-:ure 1-46 shows a series of chlorinated methane derivatives that demonstrate::-arity (or lack of polarity) based on structure.

f'.ztyrn

H

I

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,r'"{'Jn H

'',,: l:O1af

CI

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ct o./Y.,Nonpolar

H

I

CI

I

Polar

H

I

H

I

'/Y,.'ct/V.l'--.npoiar Polar Polar Polar Nonpolar

Figure 1-46

-. : --iar compound has a dipole, which for a1l intents and purposes is a line:::,1 r'r form the positive side of the molecule to the negative side of the moleculeri- : i".-av that sums up the polarity vectors of each bond in the molecule. The linerL'-:ates the direction in which the electron density shifts. Figure 7-47 shows them : 'e;ules from Figure 7-46, wlth dipoles now drawn in.

T

CI/IV:,Polar

Figure 1,-47

-r,e :,:elaction of the dipoles between two nearby molecules accounts for a weakr'ur,:: l-- shown in Figure 1-44. The alignment of dipoles is best when the partialJ,"$itr-- e of one molecule aligns near the partial negative of another molecule.

{x.mrnie 1.24rL,r,'tu:: :i the following molecules has a dipole moment of zero?

4.lr* -::ion monoxideil" -l::*oromethane: -:-DchloropropaneI l:':r.-1,1-dichlorocyclohexane

$uluruc.ou-r

1r :a ^: a dipole moment of zero, the molecule must be symmetric. Carbon:ru n r,:, :ie i C=O), dichloromethane (see Fi gure 1,- 47 ), and 2,2- dichlor oprop ane are.sil$r r:r-;-=h'x; and thus polar. The trvo carbon-chlorine bonds of trans-1,4-lLr:-:r!:::.:i'clohexane are on opposite sides of the molecule, symmetricallylrsrJ:,i-n,i. about a central point in the molecule, so their dipoles cancel out.

-]ilu "L':t D rs the best answer.

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Van der Waals InteractionsVan der Waals forces are weak intermolecular forces that exist between allmolecules' These weak attractive forces account for the minimal attractionbetween hydrocarbons. In biochemical discussions of hydrophobic interactions,it is in fact van der Waals forces that are being considered.

Figure 1-48 shows why lard (a saturated fat) has a higher melting point thanvegetable oil (an unsaturated fat). Because the molecules in laid are moreflexible, they are better able to interact with another molecule (tie knots aroundanother molecule, if you will) than is vegetable oil. Perhaps it is easier to picturelard as a pile of strings that can tie knots around themselves, while vegetible oilis like a pile of straws. If you were to build two separate piles, one a pile ofstrings and the other a pile of straws, then intertwine each within itself, fromwhich would it be easier to remove a piece? It would be far easier to remove astraw from the straw pile, because the straws are not tangled up. Their rigidityprevents interactions between the straws. This is why saturited fats 1*hosemolecules are flexible like strings) are solid at room temperature, whileunsaturated fats (whose molecules are rigid like straws) are llquid at roomtemperature. The greater the number of rc-bonds in a compound, the lower itsmelting point, and the greater the odds it is a liquid at room temperature. The n-bonds in the fatty acids in a phospholipid bilayer affect the fiuidity of a cellmembrane.

t"Vegetable OiI (More rigid structure) Lard (More flexible structure)

Figure 1-48

Example 1.25Cell membranes are composed of many molecules, including phospholipids. Aphospholipid is a molecule with a glycerol backbone (HocH2-cH(oH)CH2oH),plus two fatty acids, and a phosphate attached to the oxygen atoms of glyceroi.A cell membrane would be most rigid if both of its fatty acids were:A. completely saturated and short molecules.B. completely saturated and long molecules.C. uqsaturated and short molecules.D. unsaturated and long molecules.

SolutionFor the membrane to be rigid, there must be many interactions between the lipidchains. The maximum degree of interaction occurs with long, saturated fittyacids. Pick B and feel good yet again.

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Copyright O by The Berkeley Review 50 The Berkeley Review


Solubility and MiscibilitySolubility is defined as the ability of a solute (solid) to dissolve into solution.Miscibility describes the ability of a liquid to mix (dissolve) into another liquid.We shall look at both of these abilities in terms of physical properties andintermolecular forces. The intermolecular forces of greatest concern here arehydrogen-bonding and polarity. The basic rule of polarity governing miscibilityand solubility is that like dissolaes like. This means that a polar molecule dissolvesmost readily in a polar solvent, and a nonpolar molecule dissolves most readilyin a nonpolar solvent. Miscibility and solubility can be used as diagnostic tests tohelp determine the identity of an unknown substance. There are threecombinations of properties that a solvent may have. It may be polar and protic(capable of hydrogen-bonding), polar and aprotic (no hydrogen-bonding, but hasdipole-dipole interactions), and nonpolar and aprotic (weak intermolecularforces). Types of solvents and their properties are described in the Table 1-9.

Type Intermolecular Forces Examples

Polar, ProticH-bonding, dipole-dipole,and van der Waals

Water. Alcohols

Polar, Aprotic Dipole-dipole and vander Waals

Ketones, Ethers, Alkyl halides

Nonpolar, Aprotic van der Waals Oils, Petroleum

Table 1-9

The followrng three solubility observations can be explained by the solubilityrules in Table 1-9.

Salts dissociate into water, because ions are stabilized by the proticnature of water.

Sugar dissolves into alcohol, because of the large amount of hydrogenbonding.

Wax dissolves into oil, because it is entropically favorable to randomize.Individual van der Waals interactions are weak; but over a longmolecule, they quickly become significant.

It is important that you have a good understanding of which common solventshave which properties. The solvent properties come into play when dealing withchromatography and extraction. In chromato graphy, the greater the solubility ofa solute in the solvent, the greater the affinity of the particle for the mobile phase,meaning it can travel farther and faster. In extraction, solutes are separated fromone another by their relative solubility (or miscibility) in two solvents. Thedifference in solubility is attributed to functional groups on both the solvent andsoiute. Micelles can be employed to increase the apparent solubility of a solute ina solvent in which it is said to be insoluble.

How can a nonpolar particle dissolve into water? Soaps help to make anonpolar, aprotic species dissolve into water. For a soap (surfactant) to do this, itmust be both hydrophilic (water-soluble) and hydrophobic (water-insoluble)simultaneously. Such molecules contain a polar (or charged) end (referred to as

the head) and a alkyl chain (referred to as the tnil). Originally, soaps were madeby treating animal fats with strong base to convert the ester to a carboxylic acid,and then further convert the carboxyiic acid to carboxylate (its conjugate base), a

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Organic Chemistry Molecular Structure Physical Properties ospecies with an organic tail and a charged head. soap molecules form micelleswhen placed into water. Micelles are little pockets (spherical in shape) with anorganic core and polar heads sticking out from the core to interact with thewater. They are shuttle pods for nonpolar, aprotic species through water.

How do candles burn? This question requires an understanding of how phasechanges with temperature. Is it the solid form or the liquid form of the wax thatburns? what is the melting point of the wax in a candle? Is heat distributedevenly in a candle (i.e., is it a thermal conductor or a thermal insulator?) Thesequestions all ask about the physical properties of wax, a conglomerate of carbonchains usually containing between 31 and 50 carbons. As heat is applied, waxmelts to form an oil. The melting point for wax i.s greater than roomtemperature, but less than the temperature of the flame. The oil and vapor thatform burn when exposed to heat from the flame and to oxygen. only the oil nearthe flame burns, so heat is not evenly distributed in the system. The workings ofa candle can be explained using fundamental principles in science. euite often,the simplest things in life are organic chemistry in action,

Example 1..26

\Alhich of the foiiowing compounds would make the BEST micelle?A. H3C(CHz)gCOzHB. H3C(CH2)3CO2-C. H3C(CH2)1aCO2HD. H3C(CHz)tqCOz-

SolutionThe best micelle has an ionic (charged) head and a long carbon chain for theorganic tail. Choices A and c are eliminated, because they have unchargedheads. Although a carboxylic acid group is polar and protic, a charged site isbetter, because it is more hydrophilic when charged. Choice D is better thanchoice B, because it has a longer organic tail. Pick D to score big on this question.

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Copyright O by The Berkeley Review 52 The Berkeley Review C{

Organic Chemistry Molecular Structure Section Summary

Key Points for Molecular Structure (Section 1)

-\omenclature1. IUPAC Nomenclature (Names are assigned systematically based on functional

groups and carbon chain length)a) Name Root (Assigned according to the longest chain)b) Suffices (Assigned according to functional group--generally, the most

oxidized functional group gets top priority)c) Prefixes (Assigned to note stereochemistry; i.e., R or S, E or Z, and a or B)

d) Common Nomenclature (Prefixes based on substitution and relativepositions of functional groups; i.e., geminal diol and secondary alcohol)

Bonding and Orbitals1. Bonding (An attractive interaction between neighboring atoms)

a) Covalent Bond (The sharing of electrons between atoms; carbon makescovalent bonds)

i. Single bonds are made of a sigma-bond; they are weaker than bothdouble and triple bonds.

ii. Double bonds are made of one sigma-bond and one n-bond; they areweaker than triple bonds, but stringer than single bonds.

iii. Triple bonds are made of one sigma-bond and two n-bonds; they arestronger than both single and double bonds.

iv. Sigma-bonds share electron density between nuclei while n-bondsshare electron density in the plane above and below the nuclei.

b) Molecular Orbitalsi. Like atoms, bonds have electrons in regions of high probability, so

there are sigma and pi orbitals to describe molecular bonds.ii. Sigma bonding orbitals are more stable than pi bonding orbitals, while

sigma anti-bonding orbitals are less stable than pi anti-bondingorbitals. They fill 62n4v*46*2

iii. Anti-bonding orbitals have no overlap between atoms

c) Structural Rules (Atoms obey predictable behavior when making bonds)i, Octet Rule (Atoms in the second row of the periodic table seek to

complete their valence shell by obtaining eight electrons.)ii. HONC Rule (In neutral molecules, H makes one bond, O makes two

bonds, N makes three bonds, and C makes four bonds. If an atomdeviates from these values, it carries a charge.)

Hybridization1. The mixing of atomic orbitals (s and p in organic chemistry) to form hybrid

orbitals capable of combining to make molecular orbitals.a) sp-hybridrzation results in linear compounds, often with two ru-bonds, a

180" bond angle, and the shortest of all hybrid orbitals.b) sp2-hybridi"ation results in trigonal planar compounds, often with one n-

bond, a 120" bond angle, and a medium sized hybrid orbital.c) sp3-hybridlzation results in tetrahedral compounds, with no n-bond.s, a

109.5'bond angle, and the longest of all hybrid orbitals.

Bond Dissociation Energy

1. The energy required to break a bond in a homolytic fashion (into free radicals)is the bond dissociation energy. Higher BE refers to a stronger bond.a) BDE depends on the atoms, the substitution, and electron distributionb) Ionic bonds, rare in organic chemistry, break in a heterolytic fashion.

5tCopyright O by The Berkeley Review Exclusive MCAT Preparation

Organic Chemistry Molecular Structure Section Summary

Intramolecular Features

1. Forces Affecting Electron Distribution within a Moleculea) Resonance (Sharing of n-electrons through an array of p-orbitals)

i, Most stable resonance structure has an octet for all atoms buthydrogen, minimal charged sites, and if there must be a charge, it sitson an atom of appropriate electronegativity.

ii. Also know as conjugation when dealing with just n-bondsiii. Atoms with lone pairs are generally electron-donating while atoms

with a n-bond and no lone pair are generally electron-withdrawing.b) Inductive Effect (The pull of electron density through sigma bonds)

i. Depends on electronegativity of atomsii. Diminishes over distance, becoming negligible after four carbons.

c) Steric Hindrance (Repulsion of two atoms at the same location)d) Aromaticity (Stability for cyclic systems with a set number of n-electrons)

i. Must contain a continuous cycle of overlapping p-orbitalsii. Must have 4n + 2 n-electrons, where n is 0, 1, 2, etc... (Hrickel's Rule)

Fundamental Reactivity1. Organic Chemistry at it foundation is Lewis Acid-Base Chemistry

a) Acid-Base Chemistryi. Bronsted-Lowry deals with proton transfer while Lewis deals with the

accepting and donating of electron pairs.ii. Electron pair donors are Lewis bases and nucleophiles, while electron

pair acceptors are Lewis acids and electrophiles.

b) Determining Acid Strength (stronger acids have lower pKu values)i. Primary factors affecting acid strength are the size, hybridization and

electronegativity of the atom to which H is bonded.ii. Secondary factors affecting acid strength are resonance and induction.iii. Base strength goes in the opposite fashion as an acid.

c) Acid and Base Terminology and Facts

Acids: pH = -log[H3O+], [H3O+] = 10-PH, pKa = -log Ka, K6 = lQ-pK2

As acid strength increases: 1) acid dissociation increases, 2) Kuincreases, 3) pKu decreases, 4) pH in an aqueous solution decreases,and 5) conjugate base strength decreases and stability increases

Bases: pOH = -log [OH-], [OH-] = 10-POH, pKb = -Iog K6, K6 = 1g-pK6As base strength increases: 1) base hydrolysis increases, 2) KUincreases, 3) pKU decreases, ) pH in aqueous solution increases, and 5)

conjugate acid strength decreases and stability increases

d) Mechanisms involve tracking the pathway of electron transfer

Physical Properties

1. The physical properties of a compound are affected by intermolecular forces

a) Hydrogen Bonding (Occurs between H on N, O, or F and a lone pair)i. Increases boiling point and melting point by increasing attraction.

b) Polarity (Interaction between dipoles of adjacent compoundsi. Polar interactions are weaker than a hydrogen bonds

c) Van der Waals Forces (Weak interaction between temporary dipoles)i. Small impact on physical properties

d) Solubility and Miscibility (Ability to dissolve into a solvent)i. Based on the idea that "Like dissolves like."ii. Solid into solvent is solubility while liquid into solvent is miscibility

t.

t1,

111.

iv.


,1fu

REKI{ELEYL)R E.v.r.n.w'

Stmcture, Bonding,and Keactivity

Passages15 Passages

I OO Questions

Suggested Structure, Bonding, and Keactivity Passage Schedule:I: After reading this section and attending lecture: Passages I - III & VI - VIII

Grade passages immediately after completion and log your mistakes.

II: Following Task I: Passages IV V & IX, (20 questions in 26 minutes)Time yourself accurately, grade your answers, and review mistakes.

III: Review: Passages X - XIII & Questions 92 - IOOFocus on reviewing the concepts. Do not worry about timing.

Speciahzrng in MCAT Preparation

I. Bond Dissociation Energies

II. Structure of Caffeine

III. Solubility of Dyes and Soap

IV. Amide Protonation

V. Physical Properties and Intermolecular Forces

VI. Molecular Structure and Polarity

VII. Micelles

VIII. IR Determination of 1", 2', and 5"Alcohols

IX. PetroleumDistillation

X. Acidity and Hybridization

XI. Ksq and Acidity

XII. Alkoxides and Alkyl Sulfides

XIII. Nucleophilicity and Basicity

Questions not Based on a Descriptive Passage

Structure, Bonding, and Reactivity Scoring Scale

Kaw Score MCAT Score

84 - 100 l5 l566-43 10 l247 -6s 7 -934 46 4-6t-53 t-3

(l -8)

(e - 14)

(rs - 2L)

(22 - 28)

(2e - 34)

(55 - 42)

(43 - 4e)

(5o - 56)

(57 - 63)

(64 - 70)

(7r - 77)

(78 - 84)

(85 - el)(92 - lOO)

betnengtis toandcarbofbinaobood

IIt'cl"iElnrcfutlEtrl'"r.EI,nrct**.l'"r.@

TguuryThcrt@ThErcwr€qETanilE0fbil

rm a&

lb'@ilm ilE

&s,

I

=I

Fassage I (Questions 1 - 8)

,\s an approximation when determining the enthalpy of**n;tion fiom bond energies, it is assumed that a bondrri'aeen two atoms has a fixed value for its bond dissociationsme'rsv, regardless of the substituents on the molecule. This$ :,J say that one assumes a covalent bond between carbonnr,i iodine always has the same bond energy, whether the;*rron is a teftiary or primary carbon. A more critical viewrr :ond energies, however, shows this assumption to bel:rlru;urate. Table I lists a series of energies for commonrmml: in a wide range of organic molecules.

Bond BElleel;mole

H3C-CH3 88

H5C2-CH3 85

(H3C)2CH-CH3 84

(H3C)3C-CH3 81

H2C=CH-CH3 91

HrC-H t04H5C2-H 98

(H:C)zCH-H 95

(H:C)rC-H 91

H2C=CH-H 108

H3CO-H t02H5C2O-H 103

Table 1

T-l: r'alues in Table i demonstrate the effect of alkylg{lrum$ -rn neighboring atoms and the bonds that they fbrm.^i'brmry s a correlation between carbon substitution and bond1mruilu:- ls rvell as between atomic size and bond strength.lllhmm ,s also a correlation between hybridization and bonditimmrsfir- L'ut it is not substantiated by the limited data inTliutulre :" nhich presents too few examples of varying degreesmi m.uqrrir;lzation to reach a solid conclusion about the effect offllilril$mflnliilriiion on bond strength. The effect of substitution on,1111p rtiil[,,rrrc on bond strength can also be evaluated using bondrflrilllfiHrHmr-i. Table 2 lists the enthalpy of reaction for theiltltlltrNflrurgfl:r.tion reactions of various alkenes. The dilferencesirilttil t'r,fri r!611-ts5 tbr the variOus hydrogenation reactions are duerui' lm* {re.Jt of alkyl groups on the strength of a n-bond.

Alkene AH lteal;

H2C=CH2 -32.6

RHC=CHz -30.2

cis-RHC=CHR -28.s

R2C=CH2 -28.3

trans-RHC=CHR -27.4

R7C=CHR -26.1

R2C=CR2 -26.4

Table 2

J;nrlght @ by The Berkeley Review@

BE 1-keal-;mole

ffilJ:C-I 3.

The presence of an alkyl group on an alkene strengthensits rc-bond. Alkyl groups on vinylic carbons are consideredto be electron donating, so fi-bonds must be electronacceptors.

I . What bond dissociation energy would you expect for thebond between carbon-l and hydrogen and the onlycarbon-carbon single bond in H-C:C-CH:?A . C1-H 92 kcal/mole; CZ-CZ 86 kcal/moleB. C1-H 1 16 kcal/mole; C2-C3 86 kcal/moleC . C1-H 92 kcallmole; C2-C3 110 kcal/moleD . C1-H I 16 kcal/mole; C2-C3 1 10 kcal/mole

2. Bromine would make the STRONGEST bond withwhich type of carbon?

A. MethylB. PrimaryC. Secondary

D. Tertiary

What can be concluded about the relationship betweenatomic size and bonding?

C.

Smaller atoms form longer, stronger bonds thanlarger atoms.

Smaller atoms form longer, weaker bonds thanlarger atoms.

Smaller atoms form shorter, stronger bonds thanlarger atoms.Smaller atoms form shorter, weaker bonds thanlarger atoms.

4. The GREATEST amount of energy is released by theoxidative cleavage of an alkene that is:

A. unsubstituted.B. monosubstituted.C. disubstituted.D. trisubstituted.

5. The difference in enthalpy ofhydrogenation between the

cis and trans alkenes can be attributed to a difference in:

A. resonance.

B. hybridization.C. the electronegativity of carbon.D. steric hindrance.

A.

B.

D.

GO ON TO THE NEXT PAGE

6. The hybridization of the carbons in H2C=CH-CH:can best be described as:

A . sp, sp, and sp2.B. sp, sp,andsp3.C . tp3, sp3, and sp2.D, tp2, sp2, and sp3.

7 . Which of the following single bonds is the strongest?

A. H3C-IB. H3C-CIC. (H3C)2CH-ID. (H3C)2CH-C1

8 . The GREATEST amount of energy is required to breakwhich of the following carbon-carbon bonds?

A. H3C-CH3B. (H3C)3C-C(CH3)3C. H2C=CH2D. (HgC)zC=C(CHl)z

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Passage ll (Questions I - 14)

Caffeine, a drug extracted from tea leaves, coffee beans,and cacao plants, exists in two forms, depending on the pH atwhich it is processed: the neutral (freebase) form or theprotonated (acid salt) form. Caffeine has medicinal uses as astimulant and as a coagulant. For years, beveragescontaining this drug have been popular for their stimulanteffects and in some cases flavor. Caffeine can be extractedfrom tea leaves by employing acid-base extraction usingstandard extraction techniques on a pulverized conglomerate ofthe leaves. Caffeine isolated from the leaves in this way isrelatively pure.

Once extracted, either the acidic form or basic form ofcaffeine can be isolated. Both forms are air-stable in theircrystalline solid state and in solution, but they exhibitdifferent physical properties due to the ionic nature ofthe acidsalt form. The acid salt form is more water-soluble and has ahigher melting point than the free-base form. The acid saltform of caffeine can be converted into the free-base form bytreating the acid salt with a strong base. Equally, the free-base form can be converted into the acid salt form by treatingthe free base with a strong acid. Drawn in Figure 1 are thefree-base and hydrochloric acid salt forms of caffeine.

Freebase

oAcid salt

o cl-HrC-\N

(+NI

H

HrC.\N

(N

'\*--CHr

*Ao\*z cH:

*AoI

cHrFigure L Freebase and acid salt forms of caffeine

Because the reactivity of caffeine varies with its form,predictions about its reactivity must be based on the pH of itsenvironment. The conjugate acid form of caffeine has a pK6

value that is slightly higher than 6, making caffeine a weakacid. It exists primarily in its deprotonated form at a pH of 7in an aqueous solution. The extraction of caffeine from tea

leaves can be carried out using vinegar to lower the pH ofaqueous layer and using an ether solvent (organic layer) as thesecond layer of the biphasic system. The caffeine cationdissolves into the aqueous layer, while the other orgicomponents of the leaves dissolve preferentially intoorganic layer.

9 , The acid salt form of caffeine can be converted tofree-base form most readily by adding which offollowing reagents?

A. HCIB. NaClC. CaCO3

D. NaOH

I

CH:

beans,

rpHator the9SASA

)ragesnulantractedusing

rrate ofway is

rrm ofn their:xhibitre acidJhasarid saltrrm bye free-reating

ue the

cl-

- CH3

'o

e

; form,{ of itsa PKa

r weakrHofTom tea

tof the

) as thecation

rrganicrto the

l to the

of the

I 0. What is the hybridization of the imine nitrogen of thepurine ring that is protonated in the acid salt form ofcaffeine?

A. rpB. tp2

c. tp3D. Nitrogen atoms do not exhibit hybridization.

i 1. Which of the following strucrural descriptions BESTdescribes the relationship between the four nitrogens incaffeine?

A . Perpendicular planarB. Coplanar

C. Tetrahedral

D . Inverted planar

I L The length of the carbonyl bonds (C=O) in the caffeinemolecule are BEST described by which of the followingsLatements?

A . Both C=O bonds in caffeine are longer than theC=O bond in cyclohexanone.

B. Both C=O bonds in caffeine are shorter than theC=O bond in formaldehyde.

C . Both C=O bonds in caffeine are shorter than theC=O bond in carbon dioxide.

D . All C=O bonds are of equal length, regardless of thecompound.

Compared to the acid salt form of caffeine, the meltingpoint of the free-base form is:

A . higher, because it is the more polar form.B . lower, because it is the more polar form.C . higher, because it is the less ionic form.D . lower, because it is the less ionic form.

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14. The compound formed by replacing the N-CH3 groupbetween the two carbonyl carbons in caffeine with an Oatom is classified as an:

A. acidanhydride.

B. acid ester.

C. ester.

D. lactone.

I

ll:'

i

Passage lll (Questions 15 - 21)

Many common household items are the products of basicorganic chemistry. Dissolving one or more colored dyes intoa volatile organic solvent, such as isopropanol, for instance,makes ink. Paint, like ink, is the combination of a dye and a

solvent. When ink is applied to a porous surface such as

paper, the pores of the material absorb the solution. Then as

the volatile organic solvent evaporates away, the solid dye isleft bound to the pores of the material. This is why ink can

smear when initially applied, but once it has dried (once thesolvent has evaporated away), the ink does not smear.

It is possible to remove dried ink from paper by treatingit with organic solvent. A problem with this method is thatthe solvent diffuses radially across the paper, taking thedissolved dye with it as it travels. This is commonly referredto as running and is the basis of paper chromatography. Inksthat run when water is spilled onto the paper to which theyare bound are made out of water-soluble dyes. The erasercapable of removing erasable ink has a surface to which thedye in the ink adheres more tightly than it adheres to thepaper.

Another common household product derived from organiccompounds is soap. Each soap molecule has a hydrophilic(waterJoving) end and a hydrophobic (water-fearing) end. Inwater, the hydrophobic portions of several soap moleculesform an aggregate pore in which nonpolar, hydrophobicspecies (dirt and oil) can gather. This pore or nticelle (thespherical cell formed by several aligned and coagulated soap

molecules) is water-soluble, because the hydrophilic end ofeach molecule composing it solvates in the water. A micelleis lemoved by continuous exposure to running water, intowhich it dissolves and migrates.

One of the most common industrial soaps is sodiumdodecyl sulfate (SDS), found in many commercial shampoosand hand soaps. Soaps can be made by treating animal lard(fatty-acid triglycerides) with a strong base (such as NaOH).This forms glycerol (HOCH2CH(OH)CHZOH) and

carboxylate anions (fatty-acid carboxylates) by a reactionreferred to as s(tponification. Carboxylic acids oncedeprotonated form carboxylates (the conjugate base of theacid). The organic chain of a soap molecule is most usefulwhen it contains at least eight carbons. Longer carbon chainsare common in soaps that are used to remove oils havinglonger carbon chains.

15. A11 of the following would be ideal properties for a

solvent used to dissolve a dye within an ink EXCEPT:

A . exerting a high vapor pressure at room temperature.

B. containing functional groups similar to the dye.

C. being highly reactive with cellulose.

D. having a boiling point slightly above roomfemperature.

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1 6. Which of the following would be the BEST solvent toremove an ink dye that has hydroxyl (OH) groupsattached to a carbon backbone?

A. Propanone

B. Propanol

C. Propanal

D. Propanoic acid

17. What is the IUPAC name for the following compound?

C H 3 C H 2 C H 2CH 2CH 2CH2CH2CH2CH 2C O 2H

A. Nonionic acidB. Decanoic acidC. Undecanoic acidD. Dodecanoic acid

I 8. Some kinds if ink run when water is spilled onpaper to which they adhere. This can best be explaiby which of the following reasons?

A . The organic solvent of the ink is miscible in water.

B. The organic solvent of the ink is immiscible iwater.

C . The dye of the ink is soluble in water.

D . The dye of the ink is insoluble in water.

19. Which of the following would be the BEST soap?

A. CH3CH2CO2H

B. CH3CH2CO2Na

C . CH3 CH2CH2CH2CH 2CH2CH2CO 2HD. CH3CH2CH2CH2CH2CH2CH2CO2Na

20. Which of the following compounds is MOST solin water?

A. CHqCHzCO2H

B. CH3CH2CO2K

C. CH3CH2CH2CH2CH2CO2H

D . CH3CH 2CH2CH2CH2CO 2K

21. Which of the following reactions forms CH3CO2Na?

A. CH3CO2H + CH3MgCl

B. HCO2H + CH3MgCl

C. Ethanoic acid + NaOHD . Propanoic acid + NaOH

ound?

,2H

on therlained

water.

ible in

soluble

2Na?

Passage lV (Questions 22 - 28)

For years, chemists pondered whether amides were;rotonated on the nitrogen or oxygen atom. The amide is an.nalog to peptide linkages, so the root of this question is- runded in the chemistry of proteins. By determining the site-: protonation, conclusions about hydrogen-bonding in the

';;ondary structure of proteins can be made. Before the.jvent and advancement of x-ray crystallography, protein!-iucture could only be hypothesized. Due to the importance. - nydrogen-bonding in protein structure, detemination of the::rtonation site was critical. Figure 1 shows the structural:-:3cts of protonation on the oxygen atom of the amide.

o ao.n o.HLl H-- ii <+ i

--c\N-R R,,\N-R *-c\p.,R'ttlHHH

Figure 1 Protonation of amide on oxygen

. =rre 2 shows the structural effects of protonation on ther.:osen atom of the amide.

oil

-c\ /R'"NI

H

olt

*- \g,,R'/\

HH

No resonance

H+_>

Figure 2 Protonation of amide on nitrogen

Frotonation at the oxygen site is favored because of the'rri.-:rlnce stabilization of the protonated product, similar to,;":;- is observed when protonating esters. Despite the greater'r ':.:-t)' of nitrogen relative to oxygen (oxygen is less basic,'f ir;:-se it is rnore electronegative), the resonance stability is-r::i:: -nough to favor O-protonation. This manifests itself ini':.:.n structure through the formation of hydrogen-bonds!1,- :hs carbonyl oxygen (lone-pair donor) to the nitrogenr-::-r (partially positive proton). Support for this

'r: :sion is found in the planar B-pleated sheets and helicesrr,::. ed in the secondary structure of proteins.

l:;ause of the resonanca structures with O-protonation,:ir :.{ atoms of the amide are all coplanar. This is due to:: r :- :-h1'bridization of carbon, oxygen, and nitrogen.

L l . Alich of the following statements CANNOT be true?

I The C=O bond of an amide is shorter than the C=Obond ol a ketone.

I The C-N bond of an amide is shorter than the C-Nbond ol a plimary amine.

fi. Amides are more basic that aldehvdes.

\. I onlyB. III onlyC . I and II onlyD. I and III only

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23. We know that amides are protonated at oxygen ratherthan nitrogen, because oxygen:

A. is less electronegative than nitrogen, so it donateselectrons more readily.

B. is larger than nitrogen, so it's electron cloud attractsprotons more readily.

C. carries a partial positive charge due to resonancewithdrawal of n-electrons by the nitrogen.

D. carries a partial negative charge due to resonancedonation of n-electrons fr:om the nitrogen.

2 4. Which of the following statements correctly describesthe geometry of the molecule shown below?

olt

-C-. /HHTH

A. The nitrogen has trigonal pyramidal geometry, sothe two hydrogens are outside of the plane createdby the other four atoms.

B. The nitrogen has tetrahedral geometry, so the twohydrogens are outside of the plane created by theother four atoms.

C. The carbon has tetrahedral geometry, so the carbonhydrogen is outside ofthe plane created by the otherfive atoms.

D. The six atoms are coplanar.

25. What is the MOST basic site on the followingmolecule?

nbUr$sn'A. Site a

B. Site b

C. Site c

D. Site d

26. The MOST stable hydrogen-bond between amidesextends from the:

A. carbonyl oxygen to the H on nitrogen.B. the carbonyl oxygen to the H on carbon.

C . amide nitrogen to the H on another nitrogen.D. amide nitrogen to the H on carbon.

co

bo

2 7. Which of the following is NOT a resonance structure ofan amide?

A. oil

*,.c-*.,R'I

H

c. o-I

R,,clN,,RI

H

D. o-I

**.. C-*,,R'

I

H

B. o-I

*, C--**. R'

I

H

28. Which arrangement accurately relates the boiling pointsof acetamide (H3CCONH2), acetone (H3CCOCH3), and

propane to each other in descending order?

A. BPnsslsmi6s ) BPgsslene ) BPpropane

B. BPsssl6ng ) BPssslam;6. > BPpropane

C. BPpropans ) BPas6l6mide ) BPu."ion"D. BPpropans ) BP6ss16ng ) BPsssl2mi6s

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Passage V (Questions 29 - 34)

The boiling point of a compound is defined as thetemperature at which the vapor pressure of the compoundequals the atmospheric pressure. It is also thought of as thehighest temperature at which a compound may still beobserved in a liquid state. Boiling points vary withatmospheric pressure, so when comparing the boiling pointsof different compounds, a standard pressure is referenced.Under standard pressure, a compound's boiling pointcorelates with the heat energy required to vaporize a moleculeof it from solution. As the heat energy of vaporization(AHuaporirulion) increases, the boiling point for the

compound increases.

Two chemists speculate about the reasons for thedifferences they observe in the boiling points of variousorganic compounds.

Chemist lChemist I proposes that the difference in boiling poin

for two similar organic compounds is related to thedifferences in their molecular masses. The heaviermolecule, the more energy that must be required to liberatethe compound into the gas phase from the liquid phase.

liberate the molecules into the gas phase, heat energy mbe added to the solution, which increases the temperaturethe solution. Chemist 1 concludes that heavier moleculhave higher boiling points than lighter molecules.

Chemist 2

Chemist 2 proposes that the boiling point ofcompound depends primarily on the strength of the attractiintermolecular forces between molecules in solution.stronger the attractive intermolecular forces betmolecules, the harder it must be to remove a molecule fthe solution to the gas phase. As it becomes more difficto liberate a molecule from its liquid phase into its gas

more heat energy is required to carry the process out.result is that the boiling point of a compound increases as

molecules bind to each other more tightly in solution. Wthe intermolecular forces are greater, fewer molecuvaporize, so the boiling point of the compound increases,the vapor pressure of the compound decreases.

The hierarchy in attractive intermolecular forceshydrogen-bonding first, polarity is second, and van der Wforces rank third in strengths. Hydrogen bonds containgreatest amount of energy of these three forces, but nothydrogen-bonds are equal in strength. For instance, alcohave stronger hydrogen bonds than amines, because thydrogen of the alcohol is more acidic than the hydrogenthe amine. The greater acidity allows the hydrogen toelectron density more readily. Any compound capableforming hydrogen bonds is also polar. Polar attractionsstronger than van der Waals forces, the weakest ofintermolecul ar forces.

therunds the

lbewithointsrced.

rointxuleltion

the

- therious

'orntsr ther theterate

r. Tomustrre ofcules

ofarctive

Theweenfrom

Ticultrhase,

Theas theWhen:cules

s, and

:es isWaals.in therot all:oholsse thegen ofaccept

ble ofIns arg

of the

19. All of the following observations support Chemist l'stheory EXCEPT:

A. (H3C)2CHOH has a higher boiling poinr rhanH3CCH2OH.

B. H3CCH2OH has a higher vapor pressure thanH3CCO2H.

C . H3CCH2CH2OH has a higher boiling point thanH3CCH2OCH2CH3.

D . H 3 COH has a higher vapor pressure thanH3C(CH2)6CH3.

3tt. 11sw would Chemist 2 rank the following compoundsaccording to their boiling points?

L H3CCH2OH

tr. H3COCH3

Itr. H3CCH2NH2

A. H3CCH2OH> H3CCH2NH2 > H3COCH3B. H3CCH2OH> H3COCH3 > H3CCH2NH2C. H3CCH2NHz > H3CCH2OH> H3COCH3D. H3CCHzNHz > H3COCH3 > H3CCH2OH

: ll. The hydrogenation of an eight-carbon diene has whichof the following effects on the physical properties of thecompound?

A. Both the molecularincrease.

B. The molecular masspoint decreases.

C. The molecular masspoint increases.

D. Both the moleculardecrease.

According to Chemist 2, as intermolecular hydrogen-bonding increases, which of the following trends shouldbe observed?

A. Both the boiling point and the vapor pressureincrease.

B. The boiling point increases, while the vaporpressure decreases.

C . The boiling point decreases, while the vaporpressure increases.

D. Both the boiling point and the vapor pressuredecrease.

mass and the melting point

increases, while the melting

decreases, while the melting

mass and the melting point

,AGE 63 GO ON TO THE NEXT PAGE

3 3. As you climb higher in the mountains, the amount ofgases in the atmosphere decreases. This affects theboiling point of propanol such that it:

A. decreases, because the amount ofhydrogen-bondingdecreases.

B. decreases, because the amount ofhydrogen-bondingrncreases.

C. decreases, becauseincreases.

D. decreases, becausedecreases.

34. How do the boiling points of the following threechlorohydrocarbons compare with each other?

HsQ cH3

ct

ct-t 3

CI GI

A. Compound I > Compound fII > Compound IIB. Compound I > Compound II > Compound IIIC . Compound II > Compound I > Compound IIID. Compound III > Compound I > Compound II

the atmospheric pressure

the atmospheric pressure

cl

ctII.

m.

Passage Vl (Questions 35 - 42)

The dipole of a bond can be fbund by considering thedistribution of charge within the bond and the length of thebond. The larger the difference in electronegativity betweenthe two atoms forming the bond, the greater the magnitude ofthe partial charges on each atom, resulting in a larger overalldipole. When considering the dipole associated withmolecules, the electron density of the entire structure is

determined by the symmetry of the structure. Each bond is

treated individually, and the sum of the component vectors is

the approximate dipole. The estimated dipole is good enoughto predict chemical behavior. Figure 1 shows examples of apolar and a nonpolar cyclohexane derivative.

Nonpolar Polar

Figure 1 Polar and nonpolar disubstituted cyclohexanes

The magnitude of a dipole is measured by placing the

compound between the two charged plates and observing thedrop in voltage. A large voltage drop in the capacitor impliesthat the dielectric constant for the compound is large, so themolecule has a large dipoie. This technique works because ofthe ability of a neutral polar compound when added to an

electric field to align with the fleld. If there is a net chargeon the molecule, it migrates toward the capacitor plate withthe opposite charge.

35. Which of the ibllowing compounds, when added to thegap between the two plates of a capacitor, produces theGREATEST reduction in voltage?

A. ArHFB. C6H6c. NzD. CO,I

3 6. In which of the following reactions is it possible toform a nonpolar organic product?

A. Hydrolysis ofan alkeneB . Halogenation of an alkaneC . Hydrogenation of an alkeneD . Reduction of an amide

3 7. Which change does NOT result in an increased dipolemoment?

A. Replacing iodine with brornine on an alkyl halideB. Oxidizing a primary alcohol into a carboxylic acid

C . Replacing iluorine with chlorine on an alkyl halideD. Adding HBr anti-Markovnikov to an alkene

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3 8. Infrared spectroscopy involves radiating a compoundwith electromagnetic radiation of a known wavelengthand observing any changes in the lengths of bondswithin that molecule as they stretch. When the bondsare stretched, the dipole moment changes and thus canbe detected. Which of the following would show theLEAST change in dipole moment?

A. Stretching a carbonyl bond in an asymmetricmolecule

B. Bending a carbonyl bond in anmolecule

C. Stretching a carbonyl bond in a

moleculeD . Bending a carbonyl bond in a symmeric molecule

3 9. Which is the BEST description of the nonpolar structuof Fe(NH3)aC12?

A. Octahedral shape with the two Cl ligands cisB. Octahedral shape with the two Cl ligands trans

C. Tetrahedral shape with the two Cl ligands cis

D. Tetrahedral shape with the two Cl ligands trans

4 0. Which of the following compounds shows a dielectconstant of zero when placed in a capacitor?

A. 1,1-dichloroethaneB. cis-1,2-dichloroetheneC. trans-1,2-dichloroetheneD. E-1-chloro-2-fluoroethene

41. Which compound has the LARGEST dipole moment?

A . I,1,2,z-tetrafluoroethaneB . 1,1,2,2-tetrafluoropropaneC . 1,1-difluoro-2,2-dichloroethaneD . I , I -difluoro-2,2-dichloropropane

42. Which of the following statements CANNOT be true?

I. For a tetrahedral structure, if any of the four liare not equivalent to the others, the moleculepolar.

II. A11 1,4-disubstituted cyclohexane moleculespolar.

il. A11 optically active molecules are polar.

A. I onlyB. II onlyC . I and II onlyD . II and III only

asymmetilc

symmetric

poundrlengthbondsbonds

ius canrw the

metric

metric

iecule

mlcture

;

NS

5

lns

rrl,lecule i

Passage Vll (Questions 43 - 49)

A common problem facing pharmacists is developing

-rugs in a form that can be easily ingested by human beings,:.rticularly the problem of getting organic compounds to:-ssolve in water. As a general rule, organic compounds arert water-soluble, so it is difficult for them to migrate

,.rough the bloodstream. Because most organic compounds:..aibit little to no hydrogen-bonding, they are referred to as

'.^Trophobic (Greek for "water fearing"). To overcome the- . Jrophobic nature of organic compounds, one of two.;hniques can be employed.

.,lnique ITechnique 1 involves the use of micelles, three-

: rensional bulbs composed of compounds that are partly.:rc and partly hydrophobic (organic). A prime example of

- r.mpound that forms a micelle in water is the conjugate::.: of a fatty acid (H3C(CH2)nCO2- Na+). The micelle is a

. *erical membrane that forms when the organic tails,-ire-qate as shown in Figure l.

Fizure 1 Aqueous arrangement of molecules in a micelle

.\n organic compound (such as an antibiotic) prefers the, : of the micelle over the aqueous solution. Overall, the-:;l1e is water-soluble due to the polar heads of ther- "idual fatty-acid carboxylate anions. After migrating- : an aqueous environment to a hydrophobic environment

..1), a micelle turns itself inside out and releases the::'iric compound in its core. This mechanism is what

:-::les water-insoluble drugs to be transported through the- , jstream (an aqueous environment) to hydrophobic targetj - rxs of the body (such as lipid membranes).-. ::'.r:ique 2

Technique 2 involves converting the compound into a,'..:r-soluble derivative that decomposes into its active,

"..:r-insoluble form once inside the body. This often" ,lves converting neutral organic compounds into ions by: .:.lr protolotion or deprotonation or altering a functional""-. -:. such as converting an alcohol into an ester by reacting

;.:h an acyl group. The drug returns to its original active, -:r it is exposed to physiological conditions.

- -:r right @ by The Berkeley Review@

Pol Hydrophobic tails

4 3. Which of the following compounds should be used inorder to make a dication more soluble in an organicsolvent?

A. H3C(CH)nCO2HB. H3C(CH)nCOz'C. H3C(CHz)nNHz

D. H3C(CHz)nNH:+

4 4. How would a micelle appear in an organic solvent?

4 5. Which of the following compounds could MOST likelybe taken into the body through respiration?

A. (H3C)2CHOCH3B. (H3C)2CHCH2OHC. (H:C)zCHNHCH3D. (HrC)zCHCH2NH2

4 6. Which of the following compounds would be MOSTsoluble in water?

lil:l:l:

B. H

"\y::D.

b5 GO ON TO THE NEXT PAGE

4 7. Which of the following compounds would make theBEST micelle?

A. H3C(CH2\CO2HB. H3C(CH)zCoz'C. H3C(CH)13CO2HD. H3C(CH)nCOz-

4 8. Which of the following compounds would require a

micelle to make it water-soluble?

A. An alcohol (RCH2OH)

B. A carboxylic acid (RCOZH)

C. An amine (RCHZNHZ)

D. An alkene (R2C=CR2)

4 9. What force holds the organic tails of a micelle together?

A . 'Van der Waals forces

B. Polar attractions

C . Hydrogen-bonding

D. Covalent bonding

66Copyright O by The Berkeley Review@ GO ON TO THE NEXT PAGE

3'Alcohol

Passage Vlll (Questions 50 - 56)

A hydrogen bond is formed between an atom able todonate a lone pair of electrons and an electropositive hydrogen(an H covalently bonded to nitrogen, oxygen, or fluorine). Ahydrogen capable of forming a hydrogen bond is said to beprotic. A protic hydrogen can form one covalent bond andone hydrogen bond. As the hydrogen bond becomes stronger,the covalent bond becomes weaker. This is to say that as alone pair is donated to a protic hydrogen, the originalcovalent bond to hydrogen weakens.

Covalent bonds can be studied using infraredspectroscopy. Different bonds have different characteristicabsorbances based on their bond strength and atomic masses.Because the degree of hydrogen-bonding affects the strength ofthe covalent bond, a hydrogen bond can be seen indirectly inthe IR stretch of the hydroxyl peak. Figure I shows the IRabsorbances associated with four different hydroxyl groups.

,01",,

Figure 1 IR signals for hydroxyl functional groups

Because the covalent bond is weakened by hydrogen-bonding, the IR signal of a covalent bond between atomsinvolved in hydrogen-bonding broadens as the degree ofhydrogen-bonding increases. Not all hydrogen bonds are

equivalent, so the signal becomes a range of absorbances thatappear as one broad band. The wave number of theabsorbance lowers, because the energy decreases. Theabsorbances in Figure 1 show that as hydrogen-bondingincreases, the IR signal broadens and the maximumabsorbance occurs at a lower wave number.

5 0 . According to the IR absorbances in Figure 1, which ofthe following compounds exhibits the GREATESTamount of hydrogen-bondin g ?

A . The tertiary alcohol

B. The secondary alcohol

C . The primary alcoholD . The carboxylic acid

1" AlcoholVt

3396crnl

2'AlcoholV,oL.*,

53.

Carboxylic acid

C.

D.

Which of the following amine compounds should showthe BROADEST signal above 3000 cm-l?

-{. AmmoniaB. Propylamine

C. Dipropylamine

D. Tripropylamine

How is the absorbance value in the IR for a covalentbond between oxygen and hydrogen affected by the bondtrength and hydrogen-bonding to other atoms?

.\ . As the bond length increases, the wave number(cm-l) ofthe absorbance decreases; so as the degreeof hydrogen-bonding increases, the bond lengthincreases and the wave number (cm-l) of theabsorbance decreases.

B. As the bond length increases, the wave number(cm-l; of the absorbance increases, so as the degreeof hydrogen-bonding increases, the bond lengthincreases and the wave number (cm-1) of theabsorbance increases.

As the bond length increases, the wave number(cm-l) ofthe absorbance decreases; so as the degreeof hydrogen-bonding increases, the bond lengthdecreases and the wave number (cm-l) of theabsorbance increases.

As the bond length increases, the wave number(cm-l; of the absorbance increases; so as the degree

of hydrogen-bonding increases, the bond lengthdecreases and the wave number (cm-1) of theabsorbance decreases.

Hydrogen-bonding occurs within which of the followingcompounds?

.{. Aldehydes

B. Esters

C. Ketones

D. Primary amines

The STRONGEST hydrogen bond is formed between:

A . the lone pair of O and a hydrogen bonded to O.

B . the lone pair of N and a hydrogen bonded to O.

C . the lone pair of O and a hydrogen bonded to N.

D. the lone pair ofN and a hydrogen bonded to N.

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5 5. As dimethyl sulfide is mixed into a pure sample of analcohol, the O-H absorbance:

A. broadens and shifts to a lower value on the wavenumber scale.

B. broadens and shifts to a higher value on the wavenumber scale.

C. sharpens and shifts to a lower value on the wavenumber scale.

D. sharpens and shifts to a higher value on the wavenumber scale.

5 6. Which of the following statements CANNOT be true?

I. The IR absorbance of a covalent bond involving anatom engaged in hydrogen-bonding is not affectedby the hydrogen-bonding.

II. The bond length of the covalent bond to the protichydrogen increases with hydrogen-bonding.

m. The acidity of a proton is increased by hydrogen-bonding.

A. I onlyB. II onlyC. I and II onlyD. II and III only

Passage lX (Questions 57 - 63)

The petrolcurn industry provides roughly forty percent ofthe annual cnergy needs of the United States. Crude oil is amixture of hydrocarbons that is refined to produce tuels,including heating oil and petroleum. Many lightweight,alkene by-products fron.r the refinement of crude oil are used

as raw materials in making polylners. The industrial process

fbr refining crude oil into useful components is relerred to as

cracking and is sirnilar to fractional distillation. Figure 1

shows a schematic lepresentation of the cracking column used

to refine crudc oil and the fi'agrnents collected at differentlevels of refinement.

- Vapor

Petroleum25"C-t]5'C----+-

___>

_+

Kerosene175'C-280'C

Heating oil250'C-350"C

Lubricating oil300'c-375"C

-------->Tar

Figure 1 Cracking column used to refine crude oil

Petroleum distillaie is sold as gasoline, the luel mostcommonly used in internal combustion engines. 'Ihe best

air-petroleum mixture for such engines is the one thatproduces the most unilbrm distribution o1' heat over the

pcriod of timc that the piston is doing work. This allows foran even expansion of the gas in the piston, which results in

more useful wolk. The result is a smooth lifting of the

piston, rather than an cxplosive jerk. Engine efficiencydepends on tl-rc unilbrmity ol heat distribution within it, so

the choice of fLrel influenccs engine etTiciency.

Gasoline is given an octane rating that is based on itscombr-rstion rate. An octiine rating is e measule of a fuel'stendcncy to caLtse knocking (non-unifbnrr combustion.) The

scale is set using 2,2,4-tlimethylpentane, wl-rich is assigned

an octane rating of 100, and n-heptanc, which is assigned an

octane rating of zcro. A higher octane rating implies a better

fuel. Table 1 lists the octane ratings and boiling points lbrsome compollents of pctroleurn distillate.

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Hydrocarbon Octane Rating Boilins Point

2-Methvlbutane 93 28'C

Benzene r06 80'cn-Hexane 25 69"C

Toluene 120 104'c

n-Heptane 0 98'C

2-Methvlhexane 12 88'C

2.2.3-Trimethvlbutane t25 82'C

2,2, 4 -T r imethy I pen tane 100 104"C

Table 1

5 7. Which of the following eight-carbon hydrocarbons has

the GREATEST octane rating?

A. 2-Methylheptane

B. n-Octane

C. 2,2-DimethylhexaneD . 2,2,4-Trimethylpentane

5 8. Which of the fbllowing components is MOST likely a

component of kerosene?

A. n-Octane

B. n-Decane

C. 2,2-Dimethyloctane

D, 2,2,4,4-Tetramethyldecane

5 9. Which is NOT an effect of branching in a hydrocarborchai n ?

A . An increase in octane rating

B. A decrease in boiling pointC. A increase in density

D. An increase in hydrogen-bonding

6 0. The cracking (refining) column operates according to the

principle that:

A. more dense hydrocarbons rise higher than less dense

hydrocarbons.

B. hydrocarbons with lower boiling points rise highe:than hydrocarbons with higher boiling points.

C . hydrocarbons with higher boiling points rise higherthan hydrocarbons with lower boiling points.

D . aromatic hydrocarbons rise higher than non-aromatic hydrocarbons.

6'1"

fr}..

6 1. The efficiency (octane rating) ofa fuel depends on the:

A. enthalpy of combustion.B. entropy of combustion.

C . ratio of CO2 to water in the exhaust.

D . r'ate o1'combustion.

6 2. Which of the following statements must be true?

I. Aromaticity incleases octane rating.

II. Ethylbenzene has an octane rating of less than 100.

il. 2,2,3-Trintethylbutane is a good fuel additive toincrcase fuel eff iciency.

A. I onlyB. III only

C. I and II only

D. I and III only

6 3 . The hybridization of carbon in the aerobic cornbustionof 2,2,4-trimethylpentane changes from :

A. rp3 to sp2.

B. rp2 to sp3.

C. sp to sp3.

D. sp3 to sp.

oly a

to the

dense

righer

righer

non-

AGE Copyright @ by The Berkeley Review@ GO ON TO THE NEXT PAGE

Passage X (Questions 64 - 70)

The acidity, bond strengths, and bond lengths ofhydrocarbons depend on the hybridization of the carbonswithin the compound. Hybridization is defined as the mixingof atomic orbitals to fbrm new hybrid orbitals that are

correctly aligned to make up the covalent bonds. Hybridorbitals are oriented to align the atoms within the moleculeinto the least sterically hindered position fbr bonding. Theorientation of electrons allows the molecule to formstructures with a central carbon that has either tetrahedral

1sp3-hybridized), tri gonal planar 1sp2 -hybridized), or linear(sp-hybridized) geometry. Although the geometry dictates the

hybridization, the hybridization oI a carbon within a moleculecan be used to predict the structure of the molecule.

The more p-character thele is in the hybrid, the longerthe hybrid orbital is, and thus the further away the electronsale f}om the nucleus. This variation in length can be used to

explain differences in chemical reactivity and physicalproperties. When estimating properties of a bond, one must

consider that acidity results frorn heterolytic cleavage, whilebond energies are determined from homolytic cleavage,Figure 1 shows both heterolytic and homoiytic cleavagc Ibrthe C-H bond of a telminal alkyne.

Hetreolytic Cleavage (into ions)

R-C:n' __ .o ..@R-C:C: + H

Homolytic Cleavage (into free radicals)

-_^E^:Lr =: o-^E^. a u.

Figure I Heterolytic and homolytic cleavage of a C-H bond

Acidity can be explained in tenns of heterolytic cleavage.

The closer the electrons of a carbon-hydrogen bond are to the

carbon nucleus, the more acidic the hydrogen on that carbon.This is to say that as the electrons in the bond get closer to

the carbon nucleus, the bond is easier to break heterolytically,and thus the acidity of the hydrogen increases. Electrons get

closer to the nucleus of carbon when the bond is shorter-

However, as the bond gets shorter, it becomes more difficultto break the bond in a homolytic fashion. It is more difficultfbr hydrogen to remove one bonding electron from the bondto carbon. This means that as the hydrogen becomes moreacidic, the homolytic bond energy increases.

The less s-character in the carbon hybrid, the longer the

length of the bond between carbon and the atom to which itis bonded. As the bond becomes longer, it becomes weaker

in a homolytic sense.

64. The MOST acidic hydrogen on 3-methyl-1-penryne ison which carbon?

A. Carbon-lB. Carbon-2

C. Carbon-3

D. Carbon-4

6 5. The LARGEST Ku is associated with which of thefollowing compounds?

"Go. ur\o,

.\*.\

66. Which of the following organic compounds is theSTRONGEST base?

A. CH3CH2CH2CH2Na

B. CH3CH2CH=CHNa

C. CH3CH2CNa=CH2

C. CH3CH2C=CNa

6 7. NaNH2 is a base strong enough to deprotonate the firsthydrogen on a terminal alkyne. Which of the followinghydrogens could it also deprotonate?

A. H on carbon-l of 2-methyl-1-buteneB . H on carbon-2 of 2-methyl-1-buteneC . H on carbon-l of 2-methyl-1-butanolD . H on oxygen of 2-methyl-l-butanol

6 8. The LONGEST carbon-carbon bond can be found inwhich of the following compounds?

A. H- CE C- Ct-ls B. H-C= c- H

"'*..-r n-n./u3 D' H:^-

^//H,/" '- o, ,/u:

t\ n

Copyright O by The Berkeiey Review@ GO ON TO THE NEXT PAGE

6 9. Which of the following compounds has the WEAKESTcarbon-carbon single bond?

A' H-c:c-H B. Hsc-cEc-cr-rs

g . H:.. ^_ ^.- Hs

H/u t-

'

D' \^-^.zH

K^- ^/ t- t-

H('v- t- ,

fucil

otrErtqiffiEr

ThEBfli:t$ fiisaugnhflGf,

from ,

wmsmmufutr lwt

Trrerirj iuquilil&][!DeEl|

*sastiilfum

Eqlu0msh

Kq"

,l\ rcl'ick$

Thi.r dEEWurftlEmesrmtfuErudlmd rcediiffieruryid cmcqldliihimfirm* fffiuc rtpr

70. The LOWEST pK6 is associated with which of thefollowing nitrogen containing compounds?oG*,, B.

-NH(,\*,

D.

G,"C.

Passage Xl (Questions 71 - 77)

Most reactions using organic reagents require a solvent:.her than water, so acid-base chemistry must be viewed frour::irer the Brpnsted-Lowry definition or the Lewis definition.lie Brgnsted-Lowry definition of an acid is a compound that.!;:s as a proton donor, while the Lewis definition of an acidi a compound that accepts electron pairs. In an organic:'::t'ent, acid-base reactions involve the transfer of a proton::rn one reactant to another. There is an equilibrium;,nstanl associated with this process that is predictable based- r the pKz values of the two acids in the reaction (one acid is

i :eactant, and the other acid is a product).

To determine the Ku value for organic acids, an organic

:;r,l is added quantitatively to an organic base. Thel:rilibrium constant (K"q) is determined from the

--ncentration of each species, once equilibrium is reached.

-r-;;ction 1 is a generic reaction between an organic acid and

.:,e conjugate base of a second organic acid

HA + B- ==::\

A- + HB

Reaction 1

Equation 1 can be used to determine the equilibrium: , rstant for Reaction I .

H* = tA-llFIBl -

Ka (acid HA) = tg(pKa (HB) - pKa (HA))

' tHAltBl Ka (acid HB)

Equation I

A series of six different organic acids are treated with 1,3-

-r;lopentadienyl anion, as shown in Reaction 2.

HA+CsHs-5+A-+C5H6

Reaction 2

The structur"e of 1,3,-cyclopentadienyl anion is shown inL:ure 1,

o

Figure 1 1,3-Cyclopentadienyl anion

The concentration of each organic species at equilibrium. determined using UV-visible spectroscopy whenever:,:ssible. In cases where no n-bond is present in both the::actant acid and the product acid, the concentrations are

::termined using gas chromatography. The conjugate base

trd reactant base concentlations are determined by the:rference between initial acid concentration and equilibrium::id concentration. The concentrations are used to determinerquilibrium constants. The calculated values are compared to,:lues found using pKu numbers in Equation 1. It is found

iat the error is greatest when Kst is greater than 104.

Jopyright O by The Berkeley Review@ 7l

71.

Table 1 lists the theoretical equilibrium constants for thesix acid-base reactions patterned after Reaction 2.

Organic Acid (HA) Equilibrium Constant (Ksq)

H3CCOCH3 2.Ox1O4H3CCOCH2COCH3 1.2 x 106

H3COH 3.9 x 10-i

H3CCH25H 5.2 x 104

CI3CH 8.0 x 10-9

H3CNO2 8.2 x I04

Table 1

Which of the following reactions has an equilibriumconstant greater than 1?

A . CI3CH + H3CCH2S- -+ C13C'+ H3CCH2SH

B. H3COH + H2CNO2- + H3CO- + H3CNO2

C . H3CCOCH3 + H3CO- *H3CCOCH2-+H3COH

D. H3CCOCH2COCH3 + C13C- +H3CCOCHCOCH3- + CI3CH

Which of the following compounds can deprotonatec5H6?

A. H3CCOCH3

B. H3CCH2SH

C. H3CO-

D. HzCNOz-

7 3. For the following reaction:

H3CCH2SH + H3Co- -- H3CCH2S- + H3COH

what is true about the relative concentrations of each

species at equilibrium, if the reactants are mixed inequal molar portions?

A. [H3CO-] > [H3CCHzS-]; [H:Co-] > [H:CoH];[H3CCH2SH] > [H:CO-]

B. [H3CCH2S-] > [H:CO-]; [H:COH] > [H:CO-];[H3CCH2SH] > [H:CO-]

C. [H3Co-] > [H:CCHzS-]; [H:CO-] > [H:COH];[H3CCH2SH] = [H3CO-1

D. [H3CCH2S-] > [H3CO-]; [H:COH] > [H:CO-];[H3CCH2SH] = [H3CO-J

72.


74, The acidity of the C5H6 is abnormally high forhydrocarbons, because:

A. it is aromatic.B. it has an aromatic conjugate base.

C. its hydrogens withdraw electron density through theinductive effect.

D. in the conjugate base, the hydrogens withdrawelectron density through the inductive effect.

7 5. Which of the following relationships accurately showsthe relative pKa values for the given acids?

A. pKalct3cH) > pKa(cH3oH) > pKa(HqcNoz)

B. pKagr3cNoz) > pKalgu3oH) > pKalgr3cH)

C. pKalcn3oH) >pKalgl3CH) > pKalu3cNo2;

D. pKalcu3oH) > pKalg3cNo2) >pKalcr3cH)

7 6. Which of the following compounds is NOT an exampleof a Lewis acid?

A. CI3CHB. H3CNO2

C. NaCH3

D. BFr

77. Which of the following acids has a pKa value close10.0, given that the pKa for C5H6 is 15.0?

A. H3CCOCH3

B. CI3CH

C. H3COH

D. H3CNO2


D.

Passage Xll (Questions 78 - 84)

The effect of atomic size on reactivity is perhaps mostpronounced when comparing the reactivity of thiols andalcohols. In protic solvents, such as water, alkoxides (RO-)are less nucleophilic than alkyl sulfides (RS-), becausealkoxides are capable of forming hydrogen bonds. In aproticsolvents, alkyl sulfides are less nucleophilic than alkoxides,because they are less basic and thus less able to donate theirlone pair of electrons to an electrophile. To compare thenucleophilicity of alkoxides and alkyl sulfides, Reaction 1 iscarried out with a range of combinations of one solvent andone nucleophile.

79. !

I

^l

I(

I

H:C

tlrf

Nuc- ITffi n.c4ff. +r

-9" Td&

.lB

C

D

it fr. tf:F

ni{

AaCD

s 1. -\,rfol

A.B.C.D.

Reaction I

Table I lists the negative logs of the reaction rates forReaction 1 observed in a series of solvents reacting with a

series of nucleophiles. In each case, the reaction is carriedout at 30'C, and with an initial concentrarion of 0.10 M forthe nucleophile and of 0.1 1 M for the electrophile.

Nucleophilic Solvent -Log rate

H3CO- Ether t.44H3CS- Ether 1 9',7

HO' Ether 1.03

HS- Ether 1.16

H3CO- Ethanol 3.19

H3CS- Ethanol 2.12

HO- Ethanol 3.35

HS- Ethanol r.96

H3CO Water 4.22

H:CS Water.t

^'7

HO- 'Water 5.62

HS Water 2.14

Table IBecause the value in Table 1 is the negative log of the

rate, the magnitude of the effect of nucleophile and solvent on

the reaction rate is not immediately apparent. The smallerthe negative 1og of the rate, the greater the rate. Thedifference between the rate of SH- and the rate of RS- isattributable to differences in their molecular size. Thedifference in reaction rates b€tween the various solvents isattributable to a change in the mechanism from Sy2-like toSyl-like (as the solvent changes from ether to water). The

more the solvent binds the nucleophile, the less rapidly thenucleophile can attack the electrophile and thus the slower the

rate of the nucieophilic substitution reaction. This affects the

reaction rate of nucleophilic substitution in a protic solvent.

ft'cSe

-{.

B.

C.

)st

nd

I)ise

tic3S,

eirhe

isnd

- 8. Which of the following sets of conditions results in theFASTEST reaction rate?

A. An alkoxide in an aprotic solventB . An alkoxide in a protic solventC . An alkyi sulfide in an aprotic solventD. An alkyl sulfide in a protic solvent

- 9. What value should be expected for the negative log ofthe reaction rate, if ethyl sulfide (CHjCH2S-) were

added to 2-iodopropane in ether solvent?

A. 1.05

B. 2.04c. 3.09

D. 4.52

fornaiedfor

thetonrllerThe)lS

Thers isetoThe

ther the

; the

:nt.

GE

\ill If iodine was replacedshould expect that lhewould:

A . increase, and the reaction rate would increase.

B . decrease, while the reaction rate would increase.

C . increase, while the reaction rate would decrease.

D. decrease, and thc reaction rate would decrease.

r 1. According to the data in Table 1, which of thelbllowing bonds is the WEAKEST?

.{. c-HB. C-IC. C-OD. C-S

From the data in Table 1, what can be concluded aboutthe effect of hydrogen-bonding on reaction rate?

A. Hydrogen bonding hindels nucleophilic attack and

thus lowers the reaction rate.

B. Hydrogen bonding enhances nucleophilic attack and

thus lowers the reaction rate.

C. Hydrogen bonding hinders nucleophilic attack and

thus increases the reaction rate.

D. Hydrogen bonding enhances nucleophilic attack and

thus increases the reaction rate.

with bromine in Reaction 1, onenegative log of the reaction rate

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8 3. Which of the following changes results in an increase inthe reaction rate?

A. Changing the nucleophile from HS- to H3CS-

B . Changing the solvent from ether to water

C . Changing the reaction temperature from 30'C to25"C

D. Changing the solvent from water to ethanol

8 4. In comparing the properties of alkoxides and alkylsulfides dissolved in ether, one notes that alkoxides have

A . higher pK6 and react faster than alkyl sulfides.

B. higher pK6 and react slower than alkyl sulfides.

C . lower pK6 and react faster than alkyl sulfides.

D . lower pK6 and react slower than alkyl sulfides.

Passage Xlll (Questions 85 - 91)

Esters are semi-reactive carbonyl compounds thatundergo substitution chenristry at the carbonyl carbon.Several biological reactions (including transesterification andtransamination) proceed through standard carbonyl chemistry.The reactivity of the carbonyl depends on both thenucleophile and the leaving group. A researcher set out todetermine the reactivity of three dil'f'erent nucleophiles in a

standard substitution reaction using an ester electrophilc. Forthe reaction in Figure 1, three diff'erent compounds(Compound A, Compound B, and Compound C) were used.

ol,curo

.-------->

1

N-I

CH

Figure I Deacylation of an ester

Contpound A has the forniula C13H24S, Compound Bhas the lormula C9H1gO, ancl Cornpound C has the folrnulaC7H9N. Figure 2 shows three graphs ciepicting the changein reaction rate of the deacyltrtion reaction as a function of thesolution pH fbr each of the thlee separate compounds.

6.0 1.0 8.0 6 0 7.0 8.0 6 0 7.0 8.0

pH pH pH

Figure 2 Reaction rate as a lunction ol'solution pH

Each reaction obeys standard mechanistic behavior fbrcarbonyl substitution. They arc believed to proceed throLrgh a

mechanism where the nucleopl-rile attacks the calbonylcarbon, breaking the C=O n-boncl and forming a tetrahedralintermediate. A lone pair ol clectrons on oxygen thenreforms the n-bond, ejecting lhe leaving group. Althoughthe nucleophilicity of the dil'fcrent compounds is not equal,the similar mechanisms make the reactions comparable. Atlow pH, the carbonyl compound can be pl'otonated, making ita better electrophile. This ncgates the effbct of decreasingnucleophilicity of alcohols and thiols, beceruse they remainuncharged at low pH values.

8 5 . Which of the tbllowing types ol conipounds is tl.re

\lOST basic'l

.{, Primary alcohols

B . Esters

C . Secondary amines

D. Teltiary thiols

Compound Cornpound Cornpound C

Coprright O by The Berkeley Reviewo GO ON TO THE NEXT PAGE

86. The nucleophilicity ofeach reagent in aqueous solution:

A . decreases as the pH is increased.

B . is best when the species is a cation.

C , is best when the species is neutral.

D . depends only on the size of the nucleophile.

Which of the following compounds would be theMOST reactive nucleophile at pH = 9.0?

A. H3CCH2CH2OCH3

B. H3CCH2CCO2CH3

C. H3CCH2CH2CONH2

D. H3CCH2CH2CH2NH2

If the pKu fbr H3CNH3+ is 10.3, which of thefollowing is the BEST approximation for the pK6 forCl3CNH3+?

A. 17.3

B. 12.3

c. 8.3

D. 1.0

If the pKu for NH4+ is 9.25, then pK6 for NH3 must

be equal to which of the fbllowing?

A. 9.25

B. 7.00c. 5.75

D. 4.15

How does the hybridization of the carbonyl carbonchange during the reaction'J

A, It changes fiom tp2 to tp3 and back to sp2.

B. It changes from tp3 to t1t2 and back to spj.

C . It rernains sp2 throughout the reaction.

D. It remains spJ throughout the reaction.

9 1. Aniline and benzylamine, drawn below, are both:

Aniline

A. primary amines.

B. aromatic amines.

C . conjugated amines.

D . nonalkyl amines.

illi

87.

88.

89.

90.

NHz

m

:

I

I

@irqplifi

Benzylamine

Questions 92 throughdescriptive passage.

4 2. Which of the following compounds has the HIGHESTboiling point?

A. 2-pentene

B. Diethyl etherC. HeptanalD. Cyclohexanol

I ,1 . The correct IUPAC name for the following molecule is:

A . 2,4-diethyl-3,5-dimethylheptaneB . 4-ethyl-3,5,6-trimethyloctaneC . 3-ethyl-5,6-dimethylnonaneD. 3,5,6-trimethyldecane

r 4. Which of the following compounds exhibitsconjugation?

I. 1,4-cyclohexadiene

II. 3-ethylcyclohexene

m. 2-methyl-1,3-cyclopentadiene

A. I onlyB. II onlyC . III onlyD . I and III only

'r 5. Which of the following compounds is MOST stable?

A . 2-methyl-I,4-pentadieneB. 3-methyl-1,4-pentadieneC . 2-methyl-l,3-pentadieneD. 1.S-hexadiene

i 6. Which of the following molecules would have a dipolemoment NOT equal to zero?

L Z-1,4-dichloro-2-butene

II. E-1,4-dichloro-2-butene

m. cis-1,2-dichlorocyclopentane

A. Compound I onlyB. Compound II onlyC. Compound III onlyD . Compounds I and III only

Copyright @ by The Berkeley Review@

97. Which of the following isomers has the HIGHESTboiling point?

A. H3CCH2OCH2CH3B. H3CCH2CH2OCH3C. (H3C)2CHoCH3D. H3CCH2CH2CH2OH

9 8. Which of the following molecules is NOT polar?

A . cis-1,3-dichlorocyclopentaneB . trans- 1 ,3-dichlorocyclopentaneC. E-1,4-dichloro-2-buteneD . 1,2,2,3-tetrabromopropane

9 9. Which of the following compounds releases the

GREATEST amount of heat upon combustion?

A.\ -\

D.C. I

o100. Which of the following functional

C2H5 CH(oCH: )C (o) CH( CH:) z?

A. AldehydeB. EsterC. KetoneD. Oxirane

groups is found in

l.D 2.A 3.C 4.A 5.D 6.D'7.8 8D 9.D 10.8 11.8 12.A

13. D 14. A 15. C 16. B r1. B 18. C19. D 20. B 2t. C 22. A 23. D 24. D25. C 26. A 21. D 28. A 29. C 30. A31. A 32. B 33. D 34. A 35. A 36. C31. C 38. C 39. B 40. C 41. D 42. B43. B 44. A 45. A 46. B 47. D 48. D49. A 50. D 51. A 52. A 53. D 54. B

55, D 56. A 51. D 58. D s9. D 60. B61. D 62. D 63. D 64. A 65. D 66. A61. D 68. C 69. C 70. A 11. D ',72. C13. D 14. B 15. A 16. C 17. D 78. A79. B 80. C 81. B 82. A 83. D 84. C85. C 86. C 87. D 88. C 89. D 90. A91. A 92. D 93" B 94. C 95. C 96. D91. D 98. C 99. B 100. C

/5 iCHEMICA ES FIN!

Structure, Bonding, and Reactivity Passage Answers

)

J.

4.

5.

6.

7.

Choice D is correct. Table 1 does not list any aikynes, so the bond energies must be estimated from trends in thedata. According to data in Table 1, a bond between two spJ-hybidized carbons has a bond dissociation energy,BDE, between 81 and 88 kcals/mole. A bond between an sp2-hybridized carbon and an sp3-hybtidi"ed carbonhas a BDE ol 97 kcals/mole. This means that the bond between C2 and C3, a bond between an sp-hybridizedcarbon and an sp3-hybridired carbon, should have a BDE greater than97 kcals/mole. This eliminates choicesA and B. A bond between a hydrogen and an sp3-hybtidi"ed carbon has a BDE between 91 and 104 kcals/mole.A bond between a hydrogen and an sp2-hybtidited carbon has a BDE of 108 kcals,imole. This means that thebond between hydrogeir and an sp-hybridized carbon should have a BDE greater than 108 kcals/moie. Thiseliminates choice C and makes choice D the best answer.

Choice A is correct. According to the bond dissociation energies listed in Table 1, iodine and chlorine both maketheir strongest bonds to methyi carbons. It thus can be assumed that bromine would exhibit the same behavioras these other halides, and that the strongest bond to bromine is formed by a methyl carbon. Choice A is best.

Choice C is correct. Table 1 shows an increase in bond strength for bonds formed between equivalent carbons andatoms of decreasing size (I, C1, ar-rd O). From this, it can be concluded that shorter bonds are generally strongerthan longer bonds, and that as atomic size decreases, the bond length to a neighboring atom decreases. No bond-length data are provided in the table, but this can be inferred from the passage. Choice C is best.

Choice A is correct. The hydrogenation of an unsubstituted alkene yields the greatest energy (according to theheats of reaction listed in Table 2). A less stable reactant yields a greater amount of heat upon reaction, so then-bond must be weakest in an unsubstituted alkene. The energy generated from oxidative cleavage, or anyreaction that breaks the n-bond, is greatest when the alkene is unsubstituted. The best answer is choice A.

Choice D is correct. The difference in reactivity between the cis and trans geometrical isomers of an alkene isattributecl to i.ntramolecular steric hindrance in the cis compound, because the substituents are on the same sideof the molecule. The resonance, hybridization, and electronegativity of carbon are the same in bothgeometrical isomers of the alkene. This eliminates choices A, B, and C and makes the best answer choice D.

Choice D is correct. This question should be a lvelcome freebie, relative to other questions in the passage.Alkene carbons have sp2-hybridization, and alkane carbons have sp3-hybridization. Two of the three carbonsin the compound are alkene carbons, while tire other carbon is an alkane carbon. The best answer is choice D,two sp2-hybriclized carbons and one sp3-hybtidized carbon. Pick choice D to get your point for correctness.

Choice B is correct. Table 1 lists the bond dissociation energy for various bonds, which is the energy required tobreak the bond in a homoiytic fashion. This in essence means that Table 1 lists the bond strength. The strongestbond, according to that data listed in Table 1, is the bond with the highest bond energy. The highest valueamong the answer choices is the bond between the methyl carbon and chlorine. The best answer is choice B.

Choice D is correct. Double bonds are stronger than single bonds, meaning that more energy is required to breaka double bond than a single bond. Choices A and B are thus eliminated. The lower heat of hydrogenation inthe second chart implies that the reactant alkene molecuie is more stable. The more stable the alkenecompound, the stronger its n-bond. This means that the double bond in the tetrasubstituted aikene is strongerthan the double bond in the unsubstituted alkene. The best answer is choice D.

L Choice D is correct. To convert the acid form of a nitrogen-containilg compound (in this case, a protonated iminespecies) into a neutral species, you must add a strong base (strong enough to deprotonate the iminium cation).Choice A is eliminated, because it is a strong acid, which protonates rather than deprotonates the compound.Choice B is eliminated, because it is inert and has no effect on caffeile. Choice C and choice D are both bases,

but the stronger base is NaOH, so choice D is the better choice. The carbonate base is not strong enough to fu111'

deprotonate ihu i*ir-ri.rrrr cation. For best results in a case like this, pick choice D.

76Copyright C bv The Berkcley Reviewo Section I Detailed Dxplanations

Er thetW''bon

izedrices

role.:theIhis

Choice B is correct. We know that the nitrogen in questions has sp2-hybridization for three reasons: It isinvolved in one rc-bond, it has three substituents attached and no lone pairs, and it is pianar with respect to theadjacent atoms bonded to it, with bond angles of approximately 120'. For so many good chemistry reasons, thenitrogen in question is sp2-hybridized,, so why choose anything but choice B?

Choice B is correct. Because of the delocalization of electron density throughout the n-network of the caffeinemolecule (achieved by the resonance between all adjacent non-hydrogen atoms), the compound must be planar toallow the p-orbitals to overlap correctly. This means that all of the atoms in a caffeine molecule, and thus allof the nitrogens, must be coplanar with respect to one another. Pick choice B for a whopping good answer.

Choice A is correct. Because of the resonance donation of the lone pair of electrons on nitrogen to the carbonylcarbon, the C=O bond takes on some single-bond properti"es, which makes it longer than a standard spr-hybridized C=O bond that lacks any resonance effect. The best choice is therefore choice A. |ust as a side note,the C=O bond in formaldehyde is longer than the C=O bond in carbon dioxide because of the varyinghybridization of carbon in the two compounds. In formaldehyde the hybridization of carbon is sp2,while incarbon dioxide it is sp. The more s-character in the hybrid orbital, the shorter the orbital and thus the shorterthe bond. The stable resonance structures for an amide are shown below:

rakevior-t,1,

andllger:nd-

r ther theany

reakrn in(ene

nger

nrneion).und.ases/

rully

lt

..o:9: longer than a stanc{ard

lF C=Obond

q

\r.-/t\ <----->

\gZL\N- N-tlChoice D is correct. Ionic forces are stronger than polar forces, so the ionic compound (acid salt form) shouldhave both a higher melting and a higher boiling point than the polar uncharged compound (free-base form).This makes choice D correct, and you want that which is conect. Follow society's infiuence and pick D.

Choice A is correct. If the N-CH3 group is replaced by an oxygen atom, the compound that remains has anoxygen between two carbonyls. This is referred to as an acid nnhydride (from the fact that the compounds formswhen two acids combine in a dehydration reaction). The best answer is choice A. There is no such term as an"acid ester", so choice B is elin-rinated. An ester inl,olves just one carbonyl, so choice C is eliminated. A lactoneis a cyclic ester, and given that the compound is not an ester at all, it can't be a lactone. Choice D is eliminated.

-5. Choice C is correct. A good soivent for dissolving a dye to form an ink is one that is a liquid at roorntemperature, evaporates quickly, and exhibits a high degree of dye solubility. Having a high vapor pressureimpiies that it evaporates readily, so choice A is eliminated. If it has functional groups that are similar tothe dye, then the dye is like1y to be highly soluble in the solvent, so choice B is eiiminated. If the boilingpoint is slightly above room temperature, then it is a volatile liquid, so choice D is eliminated. The solventshould not react with paper (celluiose), so the correct answer is choice C.

-:" Choice B is correct. Becar-rse like dissolves like, the best solvent for dissolving the dye should also havehydroxyl groups attached to it, just as the dye does. The best choice is therefore the alcohol, The carboxylicacid is not a good choice, because carboxylic acids are not as volatile as alcohols. If you have melting pointsmemorized, then yor-l may be aware that carboxylic acids that are three carbons or greater in length are solidsat room temperature. If yor-r don't have them memori.zed, like 99.999"/0 of us, that's okay too. Aldehydes andketones may work, but not as well as the aicohol. The best choice is B.

:-. Choice B is correct. Determining the IUPAC name for the compound requires that you count the longestcontinuous chain of carbon atoms in this straight chain compound, which yields a total of ten. Next, you mustidentify all functional groups on the molecu1e, including alkyl groups that are not a part of the carbon backbone.The only functional group on this compound is a carboxylic acid. If there were multiple functional groups, themore oxidized functior-ral gror,rp gets higher priority in the name of the compound. For instance, if there werealso an OH group, then it would be namecl a hydroxy substituent, rather than an alcohol. Having ten carbonsand a carboxylic acid group makes the compound decanoic acid. Choice B is the best answer.

-:ryright @ by The Berkeley Review@ 77 Section I Detaited Explanations

18.

19.

20.

21.

Choice C is correct. When water is spilled on paper, it diffuses across the surface of the paper. If the ink boundto the paper is soluble in water, it dissolves into the water and spreads out, or runs. So if an ink runs, it must besoluble in water. The colored portion of the ink is the dye, not the solvent. Running ink means ihat the ink dyeis water soluble. Pick choice C for best results.

Choice D is correct. As stated in the last paragraph of the passage, a compound must contain at least eightcarbons in its chain to be a good soap. The best soap has a polar and nonpolar end associated with the molecule.The negatively charged carboxylate is at one end and an organic tail is at the other end. Molecules withcharged and organic ends are optimal for making a soap. Choice D has both eight carbons and a charged end.

Choice B is correct. To be soluble in water, a compound must be either charged or polar. Because choices B and Dare ionic, they are better in this regard than choices A or C. The organic tail is smaller in choice B, so itdissolves into water more readily than choice D. Pick choice B, and feel the sensation of correctness.

Choice C is correct. Carboxylates are formed when a carboxylic acid is deprotonated. The Grignard reagent inchoice A deprotonates the carboxylic acid to form the carboxylate, but the cation is not sodium, so choice A iseliminated. Choice B is invalid, because the Grignard reagent deprotonates the carboxylic acid (formic acid),and the carboxylic acid does not have enough carbons to make sodium acetate. Choice D is invalid, because ithas too many carbons (propanoic acid has three carbons). The H3CCO2Na molecule results from thedeprotonation of acetic acid by a base with a counterion of Na+. The best choice is therefore choice C.

23.

Choice A is correct. Because of the resonance donation from nitrogen, the carbonyl bond (C=O) of an amide hassome single-bond character. Since a single bond is longer than a double bond, the single-bond character of theamide carbonyi bond results in a longer carbonyl bond than the unconjr-rgated carbonyl (as observed with theketone). This makes statement I a false (not true) statement. Because of the previously mentioned resonance,the carbon-nitrogen bond has some double-bond character, making it shorter than a standard carbon-nitrogensingle bond (as seen with a primary amine). This makes statement II a true statement. Because of theresonance, the carbonyl oxygen carries a partial negative charge. This makes the oxygen more basic thantypical carbonyl oxygens (such as the one in an aldehyde). Statement III is also a true statement. Onlystatement I is ttot true, so choice A is the best answer.

Choice D is correct. As emphasized in the passage, nitrogen donates electron density to oxygen throughresonance. This places a partial negative charge on oxygen (increasing its basicity) and a partial positivecharge on nitrogen (decreasing its basicity). Choice C is thus eliminated, and choice D is correct. Choice Ashould be eliminated, because oxygen is more electronegative than nitrogen. Choice B should be eliminated,because oxygen is smaller than nitrogen. If you want to do what you should do, pick D and gain incrediblesatisfaction doing what you should do.

Choice D is correct. Because of the resonance donation from nitrogen, the nitrogen has sp2-hybridization.Having sp2-hybridization results in trigonal planar geometry. The carbonyl carbon aiso has trigonal planargeometry, so the central two atoms force the three hydrogens and one oxygen to assume a coplanar orientation.Choices A, B, and C therefore ail must be eliminated as incorrect geometric descriptions, making choice D thebest answer. The two resonance forms and the resonance hvbrid are drawn below:

oil

,.'C\H

=fll

o-I

I

H

N

I

H

o6I'

sp2-Hybridizati,n li

Trigonal planar -7 Cf. ./HH a]T---._ sp2-HybridizationTrigonal planar

H

/H n-----------* H/ \N*-H

25. Choice C is correct. From the passage, we know that amides are protonated at the carbonyl oxygen, so choice Dis eliminated. Because nitrogen is less electronegative than oxygen, it donates more electron density to thecarbonyl oxygen (through resonance) than the ester oxygen donates to the ester carbonyl oxygen. This places a

larger partial negative charge on the amide carbonyl oxygen than or-r the ester carbonyl oxygen. The largernegative charge makes Site c the most basic site. Choice C is therefore the best answer.

Copyright O by The BerkeJey Review@ 7a Section I Detailed Explanations

Choice A is correct. The most stable hydrogen bond forms between the best lone-pair donor (most basic site) andthe hydrogen with the greatest partial positive charge. Because amides are protonated at the carbonyl oxygen(as stated and drawn in the passage), the carbonyl oxygen is most basic and thus donates one lone pair of itselectrons. This eliminates both choice C and choice D, Of the answer choices remaining, the hydrogen onnitrogen carries the partial positive charge, not the hydrogen on carbon. This means that the hydrogen bondforms between the carbonyl oxygen and the hydrogen bonded to nitrogen, thus the best answer is choice A. If youare unsure, think of the hydrogen-bonding in B-pleated sheets.

Choice D is correct. Choice A is the most stable resonance structure of the amide (all octet and no formalcharges are present). When nitrogen donates electron density to oxygen, choice B becomes the resonancestructure. This is a minor contributor due to the formation of charges on the molecule. The fact that it is an all-octet resonance structure is favorable. If the nitrogen were to pull its n-electrons back from the carbon in answerchoice B, the resonance structure represented by answer choice C would be formed. Because carbon does not havea complete octet in this resonance structure, it is a very minor contributor, but it is none-the-less a resonancestructure of the amide. It is not possible to form a double bond to the R-group, because that would require fivebonds to carbon. In order for carbon to donate in that manner (and have only four bonds), it must have had a lonepair (and thus a negative charge) in the original structure. R was not drawn as having a lone pair, so it isassumed that the R represents a standard alkyl group. The best answer is therefore choice D.

/o\il,. c5-R'

r\ 1\ ..-

I

HChoice A (major)

o-I

Choice B (minor)

o-I

o.r\*,R' -* o-tlN.RltHH

Choice C (very minor)

Choice A is correct. Given that the molecular masses of the three compounds are roughly equal (59 g/mol, 58g/llrol, and 44 g/mol), the top consideration for determining their boiling points is the intermolecular forces. Anamide has hydrogen-bonding, while a ketone and a hydrocarbon do not. At room temperature, most amides aresolids, acetone is a volatile liquid, and propane is a gas. Based strictly on the phases, the best answer (and theonly answer that lists acetamide as the highest) is choice A, BP'sslnmide > BPacetor.,u > BPpropane. Acetone hasa higher boiiing point than propane, because it is polar and more massive.

Choice C is correct. Chemist 1 considers molecular mass to determine the relative boiling points of compounds.We are looking for the exception, so the correct answer is the choice where the lighter compound has thehigher boiling point (or lower vapor pressure). A higher vapor pressure at room temperature corresponds witha lower boiling point. In choice A, the heavier compound of the two has the higher boiling point, so choice A isnot an exception to Chemist 1's general rule. In choice B, the heavier compound of the two has the lower vaporpressure (and thus higher boiling point), so choice B also follows the rule. In choice D, the heavier compound ofthe two has the lower vapor pressure (and thus higher boiling point), so choice D follows the rule too. In choiceC, the heavier compound of the two has the lower boiling point, so choice C contradicts Chemist 1's theory.

Choice A is correct. Chemist 2 considers intermolecular forces to determine the boiling point of a compound.The strongest intermolecular forces correspond to the highest boiling point. The passage states that alcoholshave stronger hydrogen-bonding than amines. This means that the alcohol (I) has the highest boiling point(and thus is listed first), because ethanol has the strongest H-bonds of the three compounds. The ether (II) hasthe lowest boiling point, because it cannot form hydrogen bonds with itself (due to its lack of an electropositivehydrogen). The order of the boiling points is therefore: I > III > II, making choice A the best answer.

Choice A is correct. Hydrogenation is the addition of H2 gas to an alkene to break the n-bond and reduce thecompound to an alkane. For every n-bond that is lost by the alkene molecule, two hydrogens are gained. Thisincreases both the molecular weight of the compound and the molecular flexibility of the compound (theproduct is both more massive and more flexible than the reactant). Both of these effects increase the meltingpoint of the compound, making the melting point of the product greater than the melting point of the reactant.This makes choice A the correct answer. Pick A and you'il be an MCAT supernova.

- :vright O by The Berkeley Review@ 79 Section I Detailed Explanations

34.

32.

33.

J5.

36.

J/.

Choice B is correct. According to Chemist 2 (and as a general rule), the stronger the intermolecular forces, thegreater the boiling point for a compound. The greater the boiling point for a compound, the less of it there isthat evaporates, thus the lower its vapor pressure. Pick choice B, to correctly interpret the logic of Chemist 2.

Choice D is correct. Higher elevation means fewer molecules of gas per volume of air, and thus a loweratmospheric pressure. The elevation and atmospheric pressure have no effect on the intermolecular forcesbetween molecules. Flowever, the lower atmospheric pressure means that less energy (heat) is required torgach a temperature at which the vapor pressure (Prrupo.) is equal to the atmospheric p."rs.tte (Patmosph"ric),the definition of the boiling point. The boiling point is therefore lowered as elevation increases. This'makeschoice D correct.

Choice A is correct. The boiling points of Compounds I and II are directly comparable, because they aregeometrical isomers. Compound I (the cis isomer) is poiar, while Compound II (the trans isomer) is nonpolar.This means that the boiling point of Compound I is greater than the boiling point of Compound II, whicheliminates choice C. Because Compound III is an alkane, it is flexible (whereas Compounds i and II are rigid,due to the n-bond), so Compound III is able to rotate between conformers. The most stable conformation ofCompound III is nonpolar, but because it can assume polar conformations on occasion, the compound is slightlypolar. The boiling point of Compound III is less than the boiling point of Compound L Compound III shbuldhave the second highest boiling point, because it is slightly polar, while Compound II is non-polar. Thus, thecorrect order is I > IiI > II, making choice A the best answer.

Choice A is correct. The greatest reduction in voltage is caused by the compound with the greatest dielectricconstant. The greatest dielectric constant is associated with the most polar compound. Choices B, C, and D areall symmetric, so they are a1l nonpolar. This eliminates choices B, C, and D. Only ArHF (choice A) is polar,meaning that ArHF has the greatest dielectric constant. Choice A is a fine choice in a situation like this.

Choice C is correct. The hydrolysis of an alkene forms an alcohol. An alcohol is polar, so choice A iseliminated. The halogenation of an alkane forms an alkyl halide. An alkyl halide is polar, so choice B iseliminated. The hydrogenation of an alkene forms an alkane. An alkane is most often nonpolar, so the bestanswer is choice C. Reduction of an amide forms a primary amine. An amine is polar, so choice D is eliminated.

Choice C is correct. Assuming that an alkyl iodide is polar to begin with, then replacing iodine with bromineresults in a more polar compound, because bromine is more electronegative than iodine, so that the difference inelectronegativity between the halogen and carbon has increased. A carboxylic acid is more polar than aprimary alcohol (or any alcohol, for that matter), so choice B results in a more polar compound. Alkenes aretypically nonpolar, so the addition of HBr forms an alkyl bromide, which increases the polarity, so choice D iseliminated. Because fluorine is more electronegative than chlorine, replacing a fh,rorine substituent with achiorine substituent results in a compound that is less polar, making choice C the choice that does nof result inincreased poiarity. Pick choice C to be a star of chemistry.

Choice C is correct. The dipole moment changes only when a compound's bonds are either stretched or bent, ifthe compound is asymmetric. This makes choices A and B less likely to exhibit the least change in dipolemoment. The dipole moment does not change drastically (if at al1), when the chemical bonds of a symmetriccompound are either bent or stretched. Therefore, the least change in dipole is observed in a symmetricmolecule. Stretching a symmetric molecule often balances out, meaning that the electron density is shifteduniformly in opposing directions, The result is that the dipole of the molecule does not change. The best answeris choice C. Bending a symmetric molecule can make it asymmetric, so choice D is not as good as choice C.

Choice B is correct. To be nonpolar, all of the ligands must pu1l in such a way that the vectors of eachindividual bond cancel out. Tetrahedral structures are not possible with six ligands, so choices C and D areeliminated. It is only when the two chlorine ligands are trans to one another that they cancel out one anotherin terms of polarity. The best answer is therefore choice B.

38.

39.

40. Choice C is correct. A dielectric constant of zero results from a nonpolar molecule. The only nonpolar moleculeamong the answer choices is trans dichloroethene. Cis alkenes are polar, so choice B is eliminated. Choice D istrans, but there are different substituents on each carbon, so it is polar. The best answer is choice C.

Copyright @ by The Berkeley Review@ 80 Section I Detailed Dxplanations

1e

15

:1, Choice D is correct. Choice A is nonpolar, because the vectors expressing the electron withdrawal of thefiuorines cancel out, so choice A is eliminated. Fluorine is more electronegative than chlorine, so theasymrnetric electron distribution is found n'ith choices C and D, which rules out choice B. In choices C and D,the fluorine atoms withdraw the electron der-rsity, making the molecule asymmetric. In the propane molecule(choice D), the methyl substituent donates eiectron density to the electron-poor central carbon, placing a

partial positive charge on the methyl groupi therefore, it increases the dipole moment. This makes thepropane molecule more polar than the ethane molecule. The structures are shown below:

All vectors cancel .'. nonpolar C-F vectors cancel, br-rt methyl donates.'. slightly polar

Vectors almost cancel .'. slightly polar C-F vectors almost cancel C-Cl vectors,but methyl donates .'. polar

Choice B is correct. For a tetrahedral structure, if the four ligands are not all equivalent, then the structure isasymrnetric. If the compound is asymmetric, then it must be polar (have an asymmetric distribution of electronder-rsity). This makes statement I true. Figure 1 shows an example of a 1,4-disubstituted cyclohexane molecuiethat is tlot polat, which means that statement Ii is rzol true. A11 optically active compounds must beasymmetric in order to be optically active, so they must be at least slightly polar, This makes statement IIItrtre. Only statement lI is not true, so choice B is the best answer.

Choice B is correct. A dication carries a +2 charge, so it must be coupled with an anion that is organicallysoluble. The only organic anion among the answer choices is choice B.

Choice A is correct. A micelle turns inside out from its aqueous structure rt'hen it is added to an organic solvent(as stated in the passage). Figure 1 shows a micelle as it appears in water, where the polar heads are exposedto the liquid, and the organic tails are protected in the core. In a hydrophobic (organic) solvent, the organictails are exposecl, and the polar heads form a protective core. This is best illustrated in choice A. Choice Cmay look familiar, in tl-Lat cell membranes arrange themselves in such a manner. Choice D is a "throw-away"answer, because the tail and the head of the compound exhibit no attractive forces.

Choice A is correct. To be absorbed through respiration, a cornpouncl must be a gas or a vapor, because only gases

are absorbeci through respiration. This means that any compound intended to be taken into the lung must be

either a gas or a liquid r,r,ith a low boiiing (one with a high vapor pressure). There is hydrogen-bonding inchoices B, C, ancl D, but not in choice A. Hydrogen-bonding increases the boiling point and thus lowers thevapor pressure. All of the compounds have roughly comparable masses (either 73 or 74 grams per mole). Theonly factor to consider in approximating the relative boiting points is hydroger-r-bonding. The best answer is the

ether, choice A. As a point of trivia, it is estimated ihat the average human adult takes in approximately3500 gallons of air a day. Just thought you might like to know.

Choice B is correct. To be water-soluble, the compound should be able to form hydrogen bonds. Choices C and Dare eliminated immediately, because they are hydrophobic. Although choice A has an alcohol group, it is

primaril;z organic. Choice B has two hydroxyl groups and an amide group. All of these functional groups formhvdrogen bonds, so choice B exhibits the greatest amount of hydrogen-bonding. The best answer is choice B.

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E47. Choice D is correct. The best micelle has an ionic (charged) head and a long carbon chain for its organic tail.Choices A and C are eliminated, because they have uncharged heads. Choice D is a better answer than choiceB, because it has a longer organic tail.

Choice D is correct. A micelle enhances the water solubility of a compound that is normally insoluble in water.This question therefore is asking for the least watersoluble compound. An alcohol, a carboxylic acid, and anamine all exhibit hydrogen-bonding (although within a tertiary amine, there is no protic hydrogen forhydrogen-bonding), so they should all be water-soluble to some degree. Because an alkene has no hydrogen-bonding (it has neither a lone pair nor an electropositive hydrogen), it is unlikely that it would be water-soluble at all. The best answer of the given choices is therefore choice D.

Choice A is correct. The organic tails of micelles are held together by the weak attraction associated with vander Waals forces (choice A). The organic tails are alkyl-based, so they are nonpolar, and they contain neithernitrogen, oxygen/ nor fluorine. This means that choices B and C are both eliminated, because to form hydrogenbonds, a compound must have an electropositive hydrogen bonded to either nitrogen, oxygen, or fluorine. ChoiceD is eliminated, because covalent bonds are formed in chemicai reactions, and the organic tails in micellesexhibit only attractive forces, nothing as strong as covalent bonding. The best answer, and thus choice to make,is choice A.

Choice D is correct. Hydrogen-bonding weakens the covalent bond to hydrogen and thus makes the bond easierto vibrate. This means that as the degree of hydrogen-bonding to a protic hydrogen increases, the IRabsorbance for the bond decreases in energy (in terms of wave numbers) and the peak broadens (showing a

variety of strengths associated with the hydrogen-oxygen covalent bond). The greatest amount of hydrogen-bonding is found with the carboxylic acid, as shown by the smallest wave number and broadest absorbance inthe IR. The best answer is choice D.

Choice A is correct. The broadest peak is associated with the compound having the greatest amount ofhydrogen-bonding. As is observed in alcohols, the amine with the least steric hindrance exhibits the greatestamount of hydrogen-bonding. The least steric hindrance is found in ammonia. The best answer is choice A. As a

point of interest, the tertiary amine has no N-H covalent bonds, so it has no hydrogen-bonding.

Choice A is correct. The relationship between bond. strength and IR absorbance is that the lower the absorbancevalue in the IR (as measured in cm-1), the lower the energy associated with the stretching vibration of thebond. The lower the energy necessary to stretch a bond, the lower the energy necessary to break the bond, andthus the weaker the bond. Longer bonds are usually weaker bonds. Thus, as bond length increases, the wavenumber of IR absorbance decreases. This eliminates choices B and D. An increase in the degree of hydrogen-bonding weakens and thus lengthens the bond. This eliminates choice C and makes choice A the best answer.

Choice D is correct. Of the choices, only primary amines have a protic hydrogen, which means that onlyprimary amines exhibit hydrogen-bonding. The best answer is choice D.

Choice B is correct. The strongest hydrogen bond comes from the more basic lone-pair donor (found on thenitrogen atom, which is less electronegative than oxygen) being donated to the most protic hydrogen (foundcovalently bonded to the oxygen). This makes choice B the best choice.

Choice D is correct. The addition of dimethyl sulfide to solution reduces the degree of hydrogen bondingexhibited by the alcohol, because less alcohols will be adjacent to one another to form hydrogen bonds. Theabsorbance associated with a hydroxyl peak sharpens with the reduced hydrogen bonding. Associated withreduced hydrogen bonding is a stronger covalent bond and thus an IR absorbance with a higher wave number.Pick choice D for optimum correctness satisfaction.

56. Choice A is correct. The IR absorbance of a covalent bond is affected by hydrogen-bonding as stated in thepassage/ so Statementl is not true. As hydrogen-bonding increases, the covalent bond lengthens, so Statement IIis true. The acidity of a proton increases with hydrogen-bonding, because the covalent bond to hydrogen isweakened. This is why acidity is higher in water than in other solvents. This makes Statement III true. Theonly not true statement is Statement I. The best answer is thus choice A.

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Choice D is correct. From the data in Table 1, it can be seen that branching in a hydrocarbon increases its octanerating. The greatest amount of branching is observed with 2,2,4-trtnethylpentane, choice D, which can be

inferred from reading that 2,2,4-trimethylpentane has an octane rating of 100, higher than the straight-chainhydrocarbons. Don't be a dodo, pick D.

Choice D is correct. According to Figure 1, kerosene has a boiling-point range of 175"C to 280"C, so thecomponent most likely to be found in kerosene should have a boiling point in that range. The four answerchoices are saturated hydrocarbons of eight, ten, ten, and fourteen carbons. The eight-carbon compound shouldbe found in the petroleum range, and the ten-carbon compounds are probably found in the petroleum-to-naphtharange. You should use test-taking iogic to eliminate choices B and C, because their boiling points are simiiar,given that their molecular masses are identical and their structures are similar. The best answer is choice D,with the highest molecular mass (and thus the higher boiling point) of the choices. To make an estimate ofthe boiling points for both n-octane and n-decane, you can use the trend in other straight-chain hydrocarbons,where n-hexane has a boiling point of 69"C and n-heptane has a boiling point of 9B'C. Following this trendpredicts that n-octane has an approximate boiling point of I25"C - 130"C and n-decane an approximate boilingpoint of 775'C - 180"C. The branching of Z,2-dimethyloctane reduces the boiling point from that of n-decane(the straight-chain, ten-carbon alkane) to somewhere around 165"C to I70'C. The2,2,4,4-tetramethyldecane is

most likely to have a boiling point in the 175"C to 280"C range. Your job, should you accept it, is to pick D,

Choice D is correct. From the data in Table 1, it can be seen that branching increases the octane rating of ahydrocarbon. For example, as branching increases, so does octane rating for the seven-carbon aliphatichydrocarbons 2,2,3-trtmethylbutane > 2-methylhexane > n-heptane. This makes choice A a valid statement,thus eliminating choice A. From the data in Table 1, it can be seen that as branching increases, the boilingpoint decreases (for hydrocarbons of comparable mass). This can also be seen with the boiling points of theseven carbon aliphatic hydrocarbons, which have relative boiling points of n-heptane > 2-methyihexane >

2,2,3-trrrnethyibutane. Choice B is a valid statement, so it is also eliminated. Due to branching, thehydrocarbon with the greatest number of alkyl substituents has the greatest mass of compound occupying the

smallest volurne. This results in an increase in density with branching. Choice C is a valid staternent, so it is

also eliminated. Hydrocarbons have no hydrogen-bonding, so regardless of the arnount of branching, hydrogen-bonding neither increases nor decreases from hydrocarbon to hydrocarbon, This makes choice D an invalidstatement as to the effect of branching. You should smile brightly when you pick choice D.

60. Choice B is correct. As density increases for a hydrocarbon (or any gas), it does not rise as easily. This meansthat as density decreases, the ability of the vapor to rise (ascend the cracking column) increases. Thiseliminates choice A. Choice D is eliminated, because as shown in the apparatus in Figure 1, the aromatichydrocarbons are not collected in the highest chamber of the cracking column. You could have immediatelydeduced that the correct answer is either choice B or C, because they are opposites and the boiling point is

listed in the diagram. As indicated by the picture in Figure 1, the hydrocarbons with the lower boiling pointsare collected towards the top of the cracking column, which makes choice B the best choice. You'd be sad if youwere to choose anything except choice B.

61. Choice D is correct. Octane rating is based on the ability of a compound to distribute heat uniformly as itcombusts. This ability is found in compounds that are capable of reieasing their heat energy steadily over an

extentied period of tirne. The best answer is therefore choice D. Do what is best, and pick choice D. The octane

rating does not depend on the enthalpy or entropy of combustion, although the favorability of the combustionreaction does. The ratio of carbon dioxide to water depends only on the number of carbons and hydrogens in the

fuel"

62. Choice D is correct. Because toluene and benzene have octane ratings higher than the other six- and seven-

carbon saturated hydrocarbons, it can be inferred that aromaticity increases octane rating. Statement I is

therefore a true statement. Because toluene (methylbenzene) has an octane rating of 120 and benzene has an

octane rating of 106, it is assumed that ethylbenzene should also have an octane rating in excess of 100.

Statement II is therefore a false statement. Because of the brar-rching associated wrth 2,2,3-trimethylbutane, ithas a high octane rating. A high octane rating is a quality associated with a good fuel additive, so a branchedhydrocarbon such as 2,2,3-trirnethylbutane is a good fuel additive. Statement IIi is therefore a true statement.

Because statements I and III are both true statements, the best answer is choice D.

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63. Choice D is correct. Because 2,2,4-ttimelhylpentane is a saturated hydrocarbon, all of its carbons have ahybridization of sp3. In the final product, the carbons are all pr"r"r-,i in the form of carbon dioxide. Thehybridization of^ carbon in carbon dioxide (CO2) is sp. This means that in this reaction, the hybridizationchanges from sp3 to sp. The best answer, and on- we highly recommend to all parties interested in success, ischoice D.

Choice A is correct. The start of the third paragraph states that the closer the electrons within a carbon-hydrogen bond are to the carbon nucleus, the more acidic the compound is. To determine the relative acidity,you must make a decision about how close the electrons are to the nucleus. The passage also states that themore p-character there is in the hybrid, the longer the bond is. Connecting the two concepts, you should reachthe conclusion that the shorter the bond, the closer the electrons are to the nucleus. This means that the less p-character there is in the hybrid, the more acidic the hydrogen. The most acidic hydrogen is thus found on ansp-carbon. In 3-methyl-1-pentyne, carbons 1 and 2 are sp-hybridized, but only carbon t has a hydrogenattached. Pick choice A for optimal results. Make a note from the conclusions that sp > sp2 > sp3 for uiiaity.

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55. Choice D is correct. The largest Ku is associated with the strongest acid. All of the choices are hydrocarbons,so the most acidic proton is the one on an sp-hybridized carbon, as opposed to either an sp2-I.rybridized o, sp3-hybridized carbon. Of the four answer choices, only choice D has a hydrogen bonded to an sp-hybridizedcarbon, so choice D is the best answer.

Choice A is correct. Al1 of the compounds are depro^tonated hydrocarbons (with a lone pair on carbon), so thestrongest base is the one with the lone pair on an spJ-hybridized carbon. The only choice with a lone pair ofelectrons ot-t an sp3-hybridized carbon is choice A. The iation is irrelevant to the pioblem, because it is sodiumin each answer choice.

Choice D is correct. NaNH2 is a base strong enough to deprotonate a hydrogen on an sp-hybridized carbon.Although this is true, it is not critical information in solving this question. Only one answer choice can becorrect, so the correct choice must be the compound with the most acidic hydrogen. This means that thisquestion is reduced to asking "Which compound, of the choices listed, has the most acidic proton?" The mostacidic hydrogen is attached to the oxygen, so you had better pick D.

Choice C is correct. The longest carbon-carbol bond is a single bond between the two largest orbitals. Thelargest of the three possible hybrid orbitals is sp3 , so the longest carbon-carbon bond is formed between an sp3-hybridized carbon and an spJ-hybridized carbon. Choices B and D are eliminated immed-iately, because theycontain no C-C single bond. Choice A is a bond between an sp-hybridized carbon and an sp3-ltybridized carbon,while choice C is between an sp2-hybtidized carbon and an sp3-hybridlzed carbon. An sp2-6t6rid orbital islonger than an sp-hybrid orbitai, so choice C is the best answer.

Choice C is correct. The weakest carbon-carbon bond is associated with the longest carbon-carbon bond. ChoiceA is eliminated, because the C-C bond is a triple bond, and triple bonds are the shortest of carbon-carbon bonds.Choice B is between an sp-hybridized carbon and an sp3-hybridized carbon, choice C is between an sp2-hybridized carbon and an sp3-hybtidi"ed carbon, and choice D is between an sp2-hybridized carbon and an sp2-hybridized carbon. Choice C is the longest, so it would terrific if you would pick choice C.

Choice A is correct. The iowest pK6 is associated with the strongest base. Because the most acidic proton isfound on an sp-hybridized atom, the strongest base must be a lone pair on un sp3-hybtidrzed atom. Choices Cand D are spz-hybridized nitrogens, so they are both eliminated. Choice A is better than choice B, because thelone pair of electrons on nitrogen in choice B is tied into resonance r,r'ith the adjacent alkene n-bond. Electron-withdrawing resonance reduces a compound's basicity.

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Choice D is correct. For a reaction to have an equiiibrium constant greater than 1.0, the reaction must be

favorable in the forward direction as written. A favorable acid-base reaction proceeds from stronger acid toweaker acid in the forward direction as written. The larger the equilibrium constant, the more favorable thereaction, so the strength of each acid can be inferred from the K"O values in the table. In choice A, the reactionproceeds from the weaker acid (C13CH with KuO of 8.0 x 10-9) to the stronger acid (H3CCH2SH with K"O of 5.2

x t0a). This means that this reaction is unfavorable and thus has u Keg < 1. Choice A is therefore eliminated'In choice B, the reaction proceeds from the weaker acid (H3COH witfi K"O of g.g x 10-1) to the stronger acid(H3CNO2 with K"O of 8.2 x 104). This means that this reaction is unfavorable and th.y_s \a1 l Sgq < 1. ChoiceB is therefore eliminated. In choice C, the reaction proceeds Jrom the weaker acid (H3CCOCH3 with Kg* of2.0 x 10-4) to the stronger acid (H3COH with K"O of e.g x 10-1). This means that this reaction is unfavorableand thus has a K"O < 1. Choice C is therefore elimlnated, It is only in choice D that the reaction proceeds froma stronger acia 6IiCCOCH2COCH3 with Keq = 1.2 x 106) to a weaker acid (CI3CH with Kuo = 8.0 x tg-v;. The

correct answer is choice D.

Choice C is correct. Reaction 2, the experimental reaction from the passage, involves the protonation of CSHS-

to form CSHO. This question asks for the reverse reaction. This means that any acid that shows an equilibriumconstant less than 1.0 has a conjugate base that is strong enough to deprotonate C5H6. Choices A and B are

eliminated, because they are acids, not bases. Because only methanol (CH3OH) shows an equilibrium constantless than 1.0, only methoxide anion (CH3O-) is strong enough to deprotonate C5H6. The best answer is choice C.

Choice D is correct. The reaction as drawn proceeds from the stronger acid to the weaker acid, therefore theequilibrium constant is greater than 1.0. When the equilibrium constant is greater than 1.0, the products are inhigher concentration at equilibrium than the reactants. This means that H3CCH2S- is in higher concentrationthan H3CO-. This eliminates choices A and C. To distinguish choice B from choice D, the initialconcentrations must be known. Because H3CCH2SH and H3CO- are mixed equally initially, they must be

equally concentrated at equilibrium, The best answer therefore is choice D'

Choice B is correct. The conjugate base of the 1,3-cyclopentadiene species has six n-electrons in a continuouscyclic planat affay of p-orbitals. These conditions result in aromatic stability. The best explanation for the

relative ease with which the 1,3-cyclopentadiene loses its proton is the aromaticity associated with the

conjugate base ( 1,3-cyclopentadienyl anion). The more stable that the conjugate base is, the stronger the acid

is. Pick choice B and be satisfied.

Choice A is correct. The weaker of the two acids has the larger of the two pKn values. This question is askingfor the weakest acid relative to the strongest acid. As the acid gets weaker, the reaction with CSHS- becomesless favorable, so the equilibrium constant for the reaction gets smaller. C13CH shows the Iowest equilibriumconstant of the answer choices, so it is the weakest acid and thus has the highest pKo value. It is only in choice

A that CI3CH is listed as having the highest pK3 value, which makes choice A correct.

Choice C is correct. A Lewis acid is an electron-pair acceptor. The classic example of a Lewis acid is choice D,

BF3, with highly electronegative fluorine atoms and an empty p-orbital that can readily accept electrons.

This makes the boron severely eiectron-deficient. Both CH3CI and CH3N 02 are listed as acids in the table, so

choices A and B are not good choices. NaCH3 cannot accept a lone pair, but instead readily donates a lone pair.

This means that choice C is not a Lewis acid, and in fact is a Lewis base. Pick C to be terrific.

Choice D is correct. Using Equation 1, the K"q for a reaction is found by taking L0 to the power of the productacid pKu minus the reactant acid pKu. In the itandard reaction, C5H6 is the product acid andits^pKuvalue is

given as fS.0" If the pKu of the reactant acid is 10.0, then the equilibri.tm cor-rsiar-,t wouid be 10(i5'0-- 10 0) = 195'

ihe question is thereiore: "Which acid in Table t has an equilibrium constant of roughly 105?" The best answer

is choice D, CH3NO2, with an equilibrium constant of 8.2. x 104 when it reacts with C5H5-.

-i. Choice A is correct. Reading from Table 1, the fastest reaction rate corresponds to the lowest negative logvalue. Of the answer choices, the slowest reaction is observed with an alkoxide in an aprotic solvent. The 1.44

makes it the fastest reaction rate of the choices offered to you. The best answer is therefore choice A.

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Choice B is correct. Ethyl sulfide would react with 2-iodopropane just slightly more slowly than methylsulfide, so the negative log of the reaction rate would be slightly higher tnai lsir for the ethyl sulfide. Thebest answer is choice B, 2.04. Following the trend for ether s6lvent silows a negative 1og value for HS- of I.76and for H3CS- of 7.97. It can be conciuded from this that the negative log of-rate for

"H3CCH2S- should be

greater than 7.97, but no iarger than about 2.2. The only choice in this .utlgJi, 2.04. The tact of linearity in thetrend is due to the fact that the change in steric hindrance from H3CS- to u3CCu2S- is less drastic than thechange from HS- to H3CS-. The difference il the change in steric hindrance can be seen in the smaller change i1negative log value for the reaction rate.

Choice C is correct. Because bromine is smaller than iodine, bromine forms a stronger bond with carbon thaniodine, so bromine is a worse ieaving group than iodine. Tl-re rate of the reaction dJpends on the electrophile.The worse the leaving group, the worse the electrophile, and the slower the nucleophilic substitution reaction.The bromine leaving group wotild yield a slower reaction (a decrease in the ,"u.tior-, rate) than the iodineleaving group, and consequer-rtly a larger negative log of the reaction rate. The best answer is thus choice C.

Choice B is correct. The weakest bond is the one that would be broken in a nucleophilic substitution reaction,such as Reaction 1. In Reaction 1, the bond that is broken is the one between carbon und iodi.,", thus the C-I bondmust be the weakest of the choices. Pick choice B. Choice A should be eliminated, because the H is not theleaving group in Reaction 1. Because a C-O bond and C-S bond are formed when a C-I bond is broken inseparate reactions listed in Table 1, it can be inferred that both the C-O and C-S bonds are stronger than theC-I bond. This eliminates choices C and D and further supports choice B as the best answer.

Choice A is correct- As the solvent is changed from ether to ethanol and finally water, the degree of hydrogenbonding in solution increases. It can be observed from the data in Table 1 that the reaction rate decreases. Theconclusion must be that hydrogen bonding decreases the reaction rate. Choices C and D are eliminated. Therate must decrease due to hindrance of the nucleophile, not enhancement. Your answer is choice A"

Choice D is correct. According to the data in Table 1, regardless of the solvent, the negative 1og of the reactionrate is greater with H3CS- than HS-. This means that changing the nucleophile from HS- to H3CS- decreasesthe reaction rate. Choice A is therefore eliminated. the negitive log values of the rate are lower with theether solvent than the water solvent, therefore water solvent must decrease the reaction rate. This eliminateschoice B. Decreasing the temperature always produces a clecrease in the reaction rate, so choice C is eliminatedtoo. The negative log values of the rate are greater with the water solvent than the ethanol solvent, soethanol solvent must increase the reaction rate. That makes choice D the best answer. This question requiresdetermining the relationship between the reaction rate ancl negative 1og of the rate.

Choice C is correct. In ether, aikoxides have lower negative 1og values for their reaction rates than alkylsulfides, so the alkoxides must react faster. This eliminates choices B and D. The passage states that thedifference in reactivity can be attributed to alkoxides being better bases than alkyl sulfides. The stronger thebase, the lower the pK6 value. Pick C and feel jovial for just a moment, at least until the next question stirts.

85. Choice C is correct. The most basic species is the compound containi-ng nitrogen. In general, nitrogen compoundsare more basic than oxygell- and sulfur-containing co-mpounds of eqtr"at hybiidization. This eliminates choices Aand D. The degree of substitution is irrelevant. Esters have no lone pair of electrons that can be readih-donated to a proton, so choice B is not correct. The best answer is an amine, independent of whether it isprimary, secondary, or tertiary. This means that you really should pick C for the sensation of correctness.

86' Choice C is correct. The rate referred to in Figure 2 is for a nucleop}rilic substitution reaction at a carbonyl site.There is a direct corelation between nucleophilicity and the rate of reaction. The graphs show that above apH of 7, as the pH increases, so does the reaction rate. This mea11s that ihe nucleoptritcity increases. As pHincteases, compounds are no longer cationic. This eliminates choices A ancl B. Size ls noi applicable here, sochoice D is eliminated. Afier eliminating the wrong choices, vou shor,rld settle for choice C as the best answer.

87. Choice D is correct. At pH = 9.0, all of the compounds should be neutral (although the amine in choice D mar-have a small fraciion that rernains protonated). Tl-re most reactive compound is the best nucleophile. Fornucleophilicity, an arnine is better than an ether, an ester, or an arnide. Foi this reason, pick D.

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Choice C is correct. The chlorine atoms are electron-withdrawing by the inductive effect (chlorine is moreelectronegative than carbon). Electron-withdrawing groups make the compound more acidic and thus lower itspKu value. Both choice C and choice D are lower than 10.3. Choice C is the better choice, because theinductive effect is not so substantial that it will make the ammonium cation that acidic. For the pKu to dropdown to 1.0 would mean that the three chlorine atoms on the methyl group increased the acidity by a factor of109'3 = 2 x 709 = 2,000,000,000 times. That is too much. Be conservative and pick C.

Choice D is correct. The sum of pKu + pKU for a conjugate pair in water is equai to 14.0 al25"C. This means thatpK6 for NH3 (the conjugate base of NH+*) is equal to 14.0 - 9.25 = 4.75. Pick D and score with the best of them(whoever they are).

Choice A is correct. The hybridization of carbon in a carbonyl compound, such as an ester (which contains one 7[-

bond), must be sp2 1th" n-bond requires one p-orbital, so only two p-orbitals remain for hybridization). This canalso be deduced from the trigonal planar structure of the carbonyl compound. The hybridization of carbon in thetetrahedral intermediate (which contains no lr-bonds) is spJ. The final product again has the carbonylfunctionaiity, only now with the nucleophile attached. Tl." carbonyl product still has trigonal planargeometry. The hybridization therefore changes from spt to spJ and back to spt in the overall reaction, makingchoice A your choice. Make that choice today!

Choice A is correct. In each case, there is a nitrogen with one alkyl group and two hydrogens. This defines aprimary amine , so both compounds are primary amines. It just so happens that the R-groups are aromatic rings/but they are not aromatic amines per se, because the nitrogen atom is not a part of the aromatic system. The bestanswer is choice A. Of the two amines, only one is conjugated (aniline), so choice C does not describe bothstructures.

Choice D is correct. Alcohois exhibit hydrogen-bonding, which increases their intermolecular forces. Thestronger forces make it harder to move a molecule from the liquid phase into the gas phase. This raises theboiling point of an alcohol compared to a molecule of comparable size, so choice D has the highest boilingpoint. Molecular mass is of concern as well, but choice D is also the heaviest of the choices.

Choice B is correct. The longest chain is eight carbons, so based on that alone, you know that the best answer ischoice B (octane). All you need to do is find the longest chain to decipher the correct answer choice.

Choice C is correct. Conjugation is defined as consecutive, alternating n-bonds. The structures are drawn below.OnIy Compound III has conjugation, so choice C is correct.

1,,4-cyclohexadiene 3-methylcyclohexene 2-methyl-1,3-cyclopentadiene

Choice C is correct. Choice C is the most stable compound, because it is the only diene that has conjugation.Note that the structures are straight chains and not rings. It is easy to insert the word "cyclo" inadvertentlyinto the name. Avoid careless mistakes and choose C. Drawn below are the structures of all four choices:

2-methyl-1,4- 3-methyl-1,4-pentadiene 2-methyl-1,3-pentadiene 1,5-hexadiene

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Choice D is correct. Saying that a compound has a dipole that is not equal to zero is equivalent to saying thatthe compound is polar. Cis compounds (both alkenes and cyclic structures) are aln'ays polar. This makes bothCompound I and Compound III polar. You need not even examine Compound II, because no answer choice includesall three compounds. Pick choice D to score more MCAT points.

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LOns - :vright @ by The Berkeley Review@ Section I Detailed Explanations

97. Choice D is correct. All alcohols have hydrogen-bonding, which increases their intermolecular forces and thusincreases their boiling points, so choice D has the highest boiling point. Note that all of the compounds haveexactly the same formula (and thus the same molecular mass). This eliminates the need to accbunt for anydifferences in molecular mass (which would also affect the boiling point). Because of the linear nature ofchoice B and its asymmetry (which makes it more polar than the two remaining choices), it should have thesecond-highest boiling point.

Choice C is correct. For a compound not to be polar, it must be s).'rnmetric. Cis compounds are asynunetric about apoint (although they may have mirror-plane symmetry, rather than point symmetry), and thus are alwayspolar, so choice A is eliminated. This leaves choices B, C and D as possible answers. An odd-numbered ringmust be polar when it has two substituents, so choice B is polar and thus ruled out. In choice D, the middlecarbon has varying substituents attached (not all four groups are identical), so it cannot be symmetric, thus it ispolar, too. By eliminating three choices, choice C must be the correct answer. Drawing choice C out shows thatthe individual dipoles for the bonds cancel each other out, making the compound nonpolar.

The individual vectors cancel out, so there isno net vector. The compound is nonpolar.

Choice B is correct. The most heat is generated by the least stable compound; thus finding the least stablecompound is the task at hand. All of the choices have the same formula (C6Hg) so it comes down to structuralfeatures. The four-membered ring is unstable, so choices A and B are good. Choice B has no conjugation, whilechoice A does (conjugation is a stabilizing feature), so this makes B the least stable compound a*ot g the answerchoices. The bond angles are not the optimal 109.5", whether the n-bond is in the ring or not.

Choice C is correct. Translating from the chemical formula into the structure yields the compound below:

H3CH2C

H3CO H H CFI3

There is no aldehyde group (which would have been represented as CHO), so choice A is eliminated. There isno ester group (which would have been represented as CO2R), so choice B is eliminated. There is a carbonyladjacent to two alkyl groups, so the compound has a ketone functionality. This eliminates choice D and makeschoice C the best answer.

98.

99.

100.

E-1,4-dichloro-2-butene

Copyright @ by The Berkeley Review@ Section I Detailed Bxplanations

Section IIStructure

Elucidationby Todd Bennelt

H:C

lsomerisma) Isomers

i. Constitutional Isomersii. Stereoisomersiii. Configurational lsomersiv. Optical and Geometrical Isomers

b) Conformational Isomersi. Eclipsed vs. Staggered Conformationii. Oauche vs. Anti position

!ii. Newmanprojectionsiv. Cycloalkanesv. Cyclohexanevi. Chair and Boat Conformationsvii. trquatorial vs. Axial Orientation

Structural trnsightsa) Structural Symmetryb) Units of Unsaturation

Spectroscopy and Analysisa) Qeneral Spectroscopyb) Infrared Spectroscopy

i. Theclry and Key Signalsii. CommonAppplicationsiil Hydrogen-Bonding

c) Ultraviolet Spectroscopyl. I heoryii" Common Appplications

d) NMR Spectroscopyi. Theoryii. Structural Symmetryiii. Proton NIVIRiv. Shift Valuesv. Splitting Patternoi: Signal Integrationvii. Structural Features

RERI{EIEYL)K.E'v.r.-b.wt

Speci ahzing in MCAT Preparation

Structure ElucidationSection Goals

Be able to identifv isomers from both their structure and their name.

@3

There are several types of isomers. Be familiar with structural isomers (identified by differentconnectivity), stereoisomers (identified bv asymmetry), optical isome-{s (and their ability to rotateplane-polaijzed lieht), geometrical isom'ers (found #ith rings and alkenes), and conformationalisomers (identified"by ro-tation about bonds or ring-flips). You should know how the different typesof isomers are related to one another.

Be able to identify the more stable chair conformation for six-membered rings.Cvclohexane and pyranose sugars involve three-dimensional ring structures. The most stableconformation result3 in the leaEt steric hindrance. As a general rule, axial orientation results ingreater steric hindrance than equatorial orientation. Know the difference in stability between axialSrientation and equatorial oriehtation. Recognize the steric repulsion associated *ith 7,2-diaxial,1,3-diaxial, and 1,4-diaxial orientation.

Be able to identify the more stable Newmann proiection for a compound.Structures can orient themselves in a staggered conformation, an eclipsed conformation, or someconformation between eclipsed and staggered. You must be able to identify the most stable orientationfor a structure and distinguish between gauche and anti positions.

Be able to use the molecular formula to determine the units of unsaturation.Some auestions reouire vou to determine the potential functional groups of a molecule. The presenceof eithbr a n-bond'and 1 rine within a strucfure results in a uniFof dnsaturation, which manifestsitself as two fewer hvdroeefrs in the formula. A fullv saturated hvdrocarbon or carbohydrate hasa total of 2n + 2 hydrdgen itoms in the compouLnd, where n is the nurhber of carbons in the fompouLnd.

Be able to translate structures from two dimensions into three dimensions.Know what the terms staggered, eclipsed, gauche, and anti mean, and be able to draw structuresin the Newmann projectidi to show ihe orlentation of substituents in these structures. Be able torotate about o-bonds.

@3

t? Be able to deduce structural features using IK spectroscopy.You should have a basic understanding of the operations of an infrared spectrophotometer. Youmust know the IR stretches for a carboiyl and a hydroxyl bond. You shou-ld be able to determinewhich structural features correspond to #hich IR absorbairces. You must be able to eliminate and/orconfirm possible structures, using IR data. You must be able to decipher IR spectroscopy graphsand identify the key peaks.

Be able to deduce structural features using NMR sPectroscopy.You must know the NMR shifts for carbonyl compor-rnds, alkene compounds, and aromaLic compounds.You should be able to determine the struch-re of in unknown compor.rnd using the spechal informationfrom the NMR. Most structures you will encounter on the MeAT are sm"all and symmetrical, sothev are easilv solved. You mustbe able to eliminate incorrect structures based onNMR data. Bes'.rri: to underitand. what the shift value (measured in ppm) tells you, what the integration tells you,and what the peak shaoe and couplins constants tell vou. Each piece of information can be usedto help deterniine the s'tructure of in u"nknowt-, compound. Use these data in coniunction with theunits of unsaturation.

l

*t

,-11-

*I-

-:iirM':lT

:tr:I,

Jtll

I'L

[ht*

arLi:r:

ll l:-i,

f, JIl

Jllli.ru

l8,itr

lll [.i ri

slruItI l[

ffir*

;r Ii|]U

ryr*ruxrllietu1ll[ir'u

]LLJ u

4$rW,ot!,r

$mfigll

Organic Chemistry Structure Elucidation Introduction

structure elucidation involves applying ail available information, fromspectroscopic data to chemical reactivity, to ascertain the three-dimensionalshape of a molecule. It entails determining the atoms within the molecule, thefunctional groups present on the molecule, and in advancecl cases, the three-dimensional folding of the structure. Structure elucidation involves determiningthe number of isomers that fits a molecular formula and then systematicallyeliminati.g isomers that do not fit the data untit, only one structure remains.In this section, lve shall address the concept of isomerism and the many classes ofisomers. Isomers have the same atoms within the molecule, but they differ insome manner, so that the molecules are not superimposable on one another. Thedifference could result from different bonding *il-,ir-, the molecules, similarbonding but different three-dimensional distribution about a stereogenic center,or the same bonds and stereogenic syrnmetry with different conformationalorientation. A significant part of structure elucidation is determining the exactisomers that are formed in a chemical reaction.

Other structure elucidation tools shall be discussed. Questions that involvestructure elucidation are often made easier by first determining the units ofunsaturation from the molecular formula of a compound. Thii informationprovides hints as to the presence of n-bonds and./or iings within the structure.Chemical tests can be carried out to determine the nuriber of rc-bonds, whichh'hen combined with the units of unsaturation, can specify the exact number ofrings and ru-bonds within a molecule.

In this section we shall also address spectroscopy a.d the information aboutstructure it can provide. I'frared (IR) spectros.opy ir typically used to determinethe functional groups within u .o*por'rd. It iin atso give some informationabout the _symmetry of the molecule, the hybridizatiJn of carbon, and thepresence of groups capable of forming hydrogen bonds. Ultraviolet-visible (uv-Yis) sPectroscopy tells us information about the n-bonds and conjugation withina molecule' Although all molectiles absorb ultraviolet radiation, for practicalirurposes/ we use it only to detect n-bonds. Nuclear magnetic ,"ror-rur.,.u (NMR)spectroscopy describes the connectivity of a molecule and its specific structuralreatures. In its simplest application, NMR can show the carbon skeleton of amolecule. In its more sophisticated application, NMR can show the presence ofstereoisomers and the exact positions of functional groups. we shall address:roth carbon-13 and proton NMR. combining xvn data with uv-visspectroscopy and IR spectroscopy data allows for precise determination of three-limensional molecular structure.

-t is best to review NMR with symmetry as your focus. The question ',How can' ou distinguish compounds by NMR?" .ur-r b" reduced to ,,How many different:r'pes of hydroge^s are there in each compound?" Multiple-choice NMRquestions can be answered easilv by predicting the spectra from possible:tructures. For instance, if you can narrow down the potential struciures to(etones, then it's just a matter of systematically eliminating ketones that do not fit:he spectral data. This is the perspective from which w.-e will approach NMR.i he abiiity to predict spectra from structures is best attained thiough practice.-\s you do the multiple-choice qtiestions il the spectroscopy sections,-pr"di.t th"spectra for the structures in this same manner. the cliiference belween the:pectra in each answer choice (A,8, c, or D) is what often answers the question.

Copyright O by The Berkeley Review Exclusive MCAT Preparation

Organic Chemistry Structure Elucidation Isomerism

ISOMERSStructures made of the same atoms

StereoisomersDiffer in spatial arrangement of atoms

Constitutional (Structural)Differ in connectivity (bonds)

ConfigurationalDiffer by orientation in space

Can't rotate to become identicai

ConformersDi f fer by orieniaion in space

Identical after rotation about o-bond

GeometricalDiffer by orientation in space

Can't rotate to become identical due tothe presence of a ring or n-bond

OpticalDiffer by orientation in space

Can't rotate to become identical due toasymmetry in the structure

EnantiomersNonsuperimposable mirror images

DiastereomersNonsuperimposable and not

mirror images

IsomerismIsomersIsomers are structures with the same formula, meaning they are made of theexact same atoms, but they differ in the location of each atom. The difference inposition can be the result of different connectivity (bonds), different spatiaiarrangement because of asymmetry in the structure, or different orientationabout a bond. The result is that there are several different types of isomers.Figure 2-1 shows a flow chart for determining the type of isomers.

Figure 2-1

Constitutional isonrcrs, which have different bonding, ate more commonlyreferred to as structtLrsl isomers. Structural isomers are most easily recognized bytheir difference in IUPAC name. The difference may arise from the functionalgroups (like an alcohol versus an ether, or a ketone versus an aldehyde) or it mayarise from the connectivity of the carbon backbone (like 2-methyihexane versus

3-methylhexane). Structural isomers can be further divided tnto functional group

islnrcrs, positionnl isonrcrs, and skeletal isonters.

Stereoisorners have exactly the same bonds (and therefore the same connectivity),but they differ in the spatial arlangement of their atoms. On a more general note,

stereoisomers can be categorized as either configurationnl isomers (which differ inspatial arrangement and cannot be converted into the other isomer withoutbreaking a bond) or conformational isomers (which differ in spatial arrangementbut can i:e converted into the other isomer by rotation without breaking a bond.)Within configurational isomers, there are optical isomers (isomers that rotateplane-poiarized light differently) , geometrical isomers (isomers that vary inorientalion about a n-bond), ensntiorners (nonsuperimposable mirror images), and

diastereomers (nonsuperimposable and not mirror images). Configurationalisomers are most easily distinguished by their IUPAC prefix. The IUPAC prefixcontains either R or S, if the isomers differ in chirality at a stereocenter, or E and

Z, if the isomers differ in their arrangement about a n-bond. We shall address

stereoisomers in detaii in later sectrons'

((mqn

rtm&ffi@

ffim

Copyright @ by The Berkeley Review The Berkeley Review92


\vhat types of isomers are 2-methyl-3-pentanol and 3-methyl-2-pentanol?A. Conformational isomersB. Geometrical isomersC. Structural isomersD. Stereoisomers

S ol uti onThe two compounds have different IUPAC names, so they are structural isomers.The two structures vary in the position of their alcohol and side chain methyl, so:hey are also positional isomers. The question was not that specific, so the best.rnswer is choice C. The two structures are drawn below:

-Constitutional Isomersconstitutional isomers (also referred to as structural isomers) are uniquemolecules that have the same formula, but different connectivity. In otherrgords, they have the same atoms, but the atoms have different bonding. Fortlstance, 3-metllylhexanal and 3-methyl-2-hexanone are constitutional isomers.They each have the formr-rla c7H140, but they have a different sequence ofoonds. They can also be referred to as positional isomers. Using nomer-rclaturel'relps to determine whether two structures are consiitutional isomers, because:onstitutional isomers must have different IUPAC names. Figure 2-2 showsthree pairs of structural isomers, one set of functional group isomers, one set ofpositional isomers, and one set of skeletal isomers.

stntcturnl: Different arrangement of atoms (i.e. different bonds)

HoM Ho- cH2cH2cH2cH3 & H3cH2c- o- cH2cH3 , o .1-butanol diethyl ether

LII, =/\

CHq

-l-.-^-.

I{1C- CH- CH2CH2CH3'lcl &

2-chloropentane

H3C- CH- CH2CH2CH3

CHa &2-methyipentane

HICCH2 - CH CH2CHTI

LI

3-chloropentane

H:C- CH- CH- CHIltCHs CHa

2,3-dimethylbutane

mC1

CHrI-

CHs

Structural isomers have different IUPAC names.

Figure 2-2

3-methyl-2-pentanol 2-methyl-3-pentanol


Organic Chemistry Structure Elucidation Isomerism (

Example 2.2How many possible constitutional isomers exist for a molecule with themolecular formula CaHlg?

A. 1

8.2c.3D.4

SolutionThe maximum number of hydrogen atoms possible on a four-carbon alkane isten, so there are no units of unsaturation in C4H1g. This means that there are non-bonds or rings in the molecule. To solve this question, chains of varyingcarbon connectivity (skeletons) must be considered. There is always one longestchain structure (C-C-C-C). There is also the possibility of a three-carbonchain with a methyl group ofi of the second carbon (if the methyl were on thefirst carbon, it is still butane). This means that there are two constitutionalisomers for C4H1g, butane and 2-methyipropane. Pick choice B for the smile thata correct answer brings.

Example 2.3

Which of the following pairs of molecules is NOT a set of constitutional isomers?

A. 2-Methylpentane and 3-methyipentaneB. Cyclobutanol and tetrahydrofuranC. 1-Chlorobutane and 2-chiorobutaneD. 4-Ethylchlorocyclohexane and 3-methylchlorocyclopentane

SolutionIn choice A, both compounds have the formula COH1+ and different IUPACnames, so they are constitutional isomers. In choice B, both compounds have theformula CaHgO and different IUPAC names, so they are constifutional isomers.In choice C, both compounds have the formula CaHgCl and different IUPACnames, so they are constitutional isomers. In choice D, the first compound hasthe formula C3H15C1, while the second compound has the formula C6H11Cl, so

they are not even isomers, let alone constitutional isomers. This makes choice Dthe correct answer.

For a given formula, tirere is a finite number of possible structural isomers. Thenumber of possible structural isomers depends on the molecular formula.Saturated aliphatic compounds (linear alkanes) are the simplest case. For each

extra carbon, the number of structural isomers increases. For instance, C3Hg has

only one structural isomer, while C6H14 has five different structural isomers. Itis important to realize that both formulae (CgHs and C6H1a) are for structuresthat are fully saturated (have no units of unsaturation). There is no easy formulafor determining the number of structural isomers possibie for a given formula,but there is a systematic way to determine the number. Figure 2-3 shows all ofthe structural isomers for C3Hg, C+HfO, CSHfZ, and C6H14, and lists them interms of charn length and substituent location.

Ttu$Shfls

0m!um

dnr:mrln

lrMffi

htw un

Will-fieffiD

mfrems

mr{dtfimd

ru[-;.*-@!-'0-

illifirrfntJ;-ut.E -

nery:lpirffiii* lllcmr

au]l I$riffwe

nUemr:

i[r_,MpliluCopyright @ by The Berkeley Review The Berkeley Review

Organic Chemistry Structure Dlucidation Isomerism

C3H8 (1 total):

CH../ '\

H:rC' CHg

CsHtz (3 total):

CHr -CHzHrC/

-\.f '-.ur,

S-Carbon chain

CaHro (2 total):

Hr./cHt\ ,r{"'4-Carbon chain

"rf_CH -CH3

Hrc/ -."i4-Carbon chain

ur./cH'\.t'ttt'' .r{"t

HsCI

"ra/tt-a",3-Carbon chain

-zc\HSC CH:3-Carbon chain

HqC CHr"\/

CeHr+ (5 total):

5-Carbon chain

CHr -CHzHrC/

-- an- - cH,"l

HsCS-Carbon chain

"rf-zcH\ .zCH'

Hzc' -cH, -c",S-Carbon chain

"rl

"..-t"- ,r('^u"lCHs

4-Carbon chain

t"\ ,t"'

*C/t-.t'tt'4-Carbon chain

e

C

IS

;o

D

\e

a.

*rAS

ItES

1a

,4,

ofin

Figure 2-3

This procedure of determining the number of structural isomers is systematic.First, start with the longest continuous chain of carbons (equal to the totalnumber of carbons in the formula). In the case of CoHt+, the longest possiblechain is six carbons. After drawing the longest chain, draw a carbon chain of oneIess carbon (five carbons) and systematically deduce all of the possible isomersby moving the methyl group across the chain one carbon at a time. In the case ofC6HI4, the next chain down from six carbons is five carbons and the extra (sixth)carbon is attached to one of the interior carbons in the chain. If the extra carbonwere attached to a terminal carbon, then the longest chain would be six carbons,not five. In the case of C6H14, it is not possible to have 1-methylpentane, becausethat is really n-hexane. A guideline to follow as you deduce isomers is thatstructural isomers must have different IUPAC names. If you are ever in doubtabout whether or not two compounds are structural isomers of one another,name them using IUPAC conventions. To complete the process of determiningthe isomers, systematically count isomers for each possible chain length,reducing the length by one carbon each time. When you are finished with eichpossible chain length, sum all of the structures and that's your answer. Foralkanes with functional groups attached, the procedure is the same except onceall of the skeletal structures are determined, there is an additional itep ofsystematically placing the functional group at all unique carbons. Example 2.4demonstrates this procedure.

:w Copyright @ by The Berkeley Review 95 Exclusive MCAT Preparation


Example 2.4How many structural isomers are possible for the formula CaH9C1?

A.38.4c.5D.6

SolutionFor a problem of this type, the possibilities for the carbon skeleton must be

determined first. The four carbons can either be aligned four in a row or three ina row with the fourth carbon coming off of the second carbon of the three-carbonchain.

STEP 1: 2 possible carbon skeletons

(

=

IIII(T

tTuft

The second step is to determine how many unique carbons each chain contains.

STEP 2: Each skeleton has two unique carbons

C-C-C-C

4-Carbon chain

Co- Cu- Cu- Co

2 Unique carbons

The last step requires placing a chlorine on eachtime and verifying your answer by checkingdifferent IUPAC name.

STEP 3: 4 structural isomers total

C-C-CI

C3-Carbon chain

Co- Cu -

Cu

I

ca2 Unique carbons

unique carbon one structure at a

to see if each structure has a

fiinio!ffiq

I

/i

I

C_C-C-CI

C1

1-chlorobutane

C_C-CtlCIC

C-C-C-C

C-C-C/\

CCl

I

C1

2-chlorobutane

1-chloro-2-methylpropane 2-chloro-2-methylpropane

The best choice is answer B, because there are four possible structural isomers.

This systematic procedure works every time. It is assumed that isomer problems

much beyond this example in terms of difficulty will be avoided on the MCATbecause of time constraints. The skills employed when deducing the number ofstructural isomers can also be used when deducing structure from spectral data,

such as IR and NMR information.


Organic Chemistry Structure Dlucidation Isomerism

Example 2.5How many possible constitutional isomers have the molecular formula CaHg?

A.48.5c.6D.7

SolutionThese questions are time-consuming, but unfortunately, there is not a convenientrvay around it. To start, you must determine the units of unsaturation (alsol<rrown as degrees of unsafuration).

Degreesof unsaturation =z(+)+?-(8) =8+2-8=10-8 -2 - 12222\Vith one unit of unsaturation, the structure must contain either a n-bond or aring. To get the correct answer, you must systematically consider each linearconnectivity and each cyclic connectivity. Be sure not to count stereoisomersi'including geometrical isomers). The alkenes are listed first, followed by thecvclic structures.

Possible alkene structures:

4-carbon chain (2 total)

HzC: CHCH2CH3 H3CHC: CHCH3

1-Butene 2-ButeneNote that there are two possiblegeometrical isomers for 2-butene.

3-carbon chain (1 total)

H. ,CH',C= C,,

H CHs

Methylpropene

Possible cyclic alkane structures:

3-carbon ring (1 total)

CH"/\'

H?C- CH-\CHg

Methylcyclopropane

Because there are five total constitutional isomers in all, the best answer is choiceB.

4-carbon ring (1 total)

H?C- CH,-ttH2C- CH2

Cyclobutane

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Organic Chemistry Structure Elucidation Isomerism {

StereoisomersStereoisomers are molecules of the same formula that have the same bonds(connectivity), but a different spatial arrangement of the atoms. included instereoisomers are configurational isomers (molecules that cannot be convertedinto one another through rotation about a o'-bond) and conformational isomers(caused by rotation and ring-flipping). Configurational isomers can be brokendown further into either geometrical isomers (associated with nonrotatingstructures, such as rings and alkenes) and optical isomers (isomers that rotatepiane-polarized light differently). Not all configurational isomers rotate plane-polarized light, as you have seen with meso compounds, but opticai isomersdiffer in the magnitude and possible direction in which they rotate incidentplane-polarized light. Stereoisomers of all types are not superimposable (theycannot be superposed onto one another.) Figure 2-4 shows the types ofstereoisomers. Rotamers are conformational isomers that vary in orientation inspace because of rotation about a sigma bond.

Stereoisomers; Same bonds, but a different spatial arrangement of the atoms

HOH\s &(-

H^ccHz/ -- at,

(S)-2-butanol

Configurational isomers(Optical isomers)

CH: & .'rtl CH3

(cis)-4-methylcyclohexanol (trans)-4-methylcyclohexanol

Configurational isomers(Geometrical isomers)

Note: Configurational isomers have different prefixes in their IUPAC names

I(firu

&

d[

m[

,0u

HOH\s

-z C:.

H3CCH2- CH2OH

1-butanol

.z C:.H:C CH2CHl

(R)-2-butanol

H"CCH' H& is-z

C:.H CH2OH

1-butanol

,Q@,{F

HH\rConformational isomers

(Rotamers)

Figure 2-4

Configurational IsomersConfigurational isomers are a subgroup of stereoisomers that have the samebonds, but a different arrangement of their atoms in space, no matter how thestructures are twisted and rotated. Common examples with which you arefamiliar include optical isomers and geometrical isomers.

Copyright @ by The Berkeley Review 9a The Berkeley Review


Optical Isomersoptical isomers are molecules of the same formula and the exact same bonds'connectivity), but a different spatial arrangement of the atoms due to asymmetry-'r'ithin the structure. An optical isomer cannot be rotated or manipulated tbassume the structure of another isomer. They cannot be converted into anothercptical isomer without breaking a bond. Optical isomers rotate plane-polarizedlight differently from one another. Figure 2-5 shows an example of a pair of..ptical isomers.

optical Isomers: Identical bonds with a different spatial arrangement aboutan asymmetric carbon that rotate plane-polarized light differently.

Optical

H\& Ctrrrtt- CH2CH2CHI

HaC-

(S)-2-chloropentane

Isomers

Figure 2-5

Jptical isomers are a class of configuratisnal isomers. Configurational isomers:an also be classified as enantiomers and diastereomers, so some optical isomers::l also be referred to as enantiomers or diastereomers.

E.rample 2.6:{orv can the relationship between the following two molecular structures BEST:: described?

H

",?l7t- cH2cH2cH3

(R)-2-chloropentane

H\

CH2C\IZ - CH2CH2CH2CI &

.4,. IdenticalmoleculesB. Optical isomersLl. Skeletal isomersf. Structural isomers

CIH,CH?CH?C..\

.rut7t- CH2CI

H

Solution',-e can start by naming each of the structures. Both have three chlorine atoms on. iive-carbon chain. The chlorine atoms are on carbons 7, 2, and 5, so each: rlecule has the IUPAC name 1,2,5-trichloropentane. To be skeletal or structural-:-rmers requires that the two compounds have different IUPAC names, so-:-oices C and D are eliminated. No matter how you rotate the first structure, it:::rnot be superposed onto the second structure. This implies that they are not:entical molecules, so choice A is eliminated. The left structure has s-

':ereochemistry at the second carbon, while the right structure has R-:::reochemistry at the second carbon. This confirms that the two structures are.:rical isomers, so the best answer is choice B.

-".pyright @ by The Berkeley Review 99 Exclusive MCAT Preparation


Geometrical IsomersGeometrical isomers, simply put, are the cis and trans forms of a rigid compound(where rigid implies that it is not free to rotate between conformations). They aresometimes referred to as cis/trans isomers, which applies to rings and alkenes.They are nonsuperimposable, because they are locked into an orientation by thecyclic structure or n-bond. If two substituents on a cyclic compound are on thesame side of the plane or if two substituents on alkene are on the same side of thecarbon-carbon n-bond, then they are said to be cls to one another. If twosubstituents on a cyclic compound are on opposite sides of the plane or if twosubstituents on alkene are on opposite sides of the carbon-carbon n-bond, thenthey are said to be trans to one another. IUPAC convention does not use cis ortrans in the naming of alkenes; instead, the letters E and Z are employed. Ingeneral nomenclature, the terms cis and trans are common. For geometricalisomers, which have different spatial arrangement about a n-bond, the prefix of Eis given for trans orientation of the two highest priority groups, while Z is givenfor cis orientation of the two highest priority groups. Figure 2-6 shows severalexamples of pairs of geometrical isomers, both alkenes and cyclic structures.

Geometrical Isomers: Identical bonds with a different spatialarrangement (found with double bonds and rings).

Alkenes

H CH,CHTCH3\/LLc:c &/\

HsC H

Trans (E)

HsC

Z ("Zis")

CHg

H

Cyclic Systems

L-L/\

HsC CH2CH2CH3

Cis (Z)

H?CO H

CHe

(down)

Trans

HH\/

&

cH" (uP)

(up) -)HrC-H &

Cis

(up) (t.tp)

Hrc\rA.rCH"\J&

(up) (down)

H3C\.n,.,rt\CH3UCis

(up) /YcHsHzC-_/--/Trans

Copyright @ by The Berkeley Review loo

Figure 2-6

The Berkeley Kevieu


The term E is derived from the German word entgegen meaning "across from"and refers to a compound where the two highest priority groups on eachrespective alkene carbon are trans to one another. The term Z is derived from thecerman word zusnmmen meanine "together" and refers to a compound where thetwo highest priority groups on each respective alkene carbon are cis to oneanother. Think of cis as "Zis", and you will always remember which is which.

Example 2.7All of the following are true about geometrical isomers EXCEpT that:A. the E designation for an alkene refers to the highest priority groups on each

alkene carbon in trans orientation.B. in both the E and Z isomers of an alkene, the atoms directly bonded to the

alkene carbons are all coplanar.c. molecuies capable of forming geometrical isomers have greater entropy than

linear alkanes of equal carbon chain lengthD. geometrical isomers have relatively static structural features, such as polarity

and solubility.

SolutionTo determine the geometry of an alkene, first locate the two carbons thatconstitute the alkene. Determine the highest priority substituent on each of thetwo alkene carbons, using the Cahn-Ingold-Prelog rules for assigning priorities tosubstituents by sequentialiy looking at the atoms attached to the site of interest.If the two highest priority groups are across from one another with respect to thelouble bond, then the compound is trans and thus is assigned the letter E. If the:rvo highest priority groups are both on the same side of the double bond, then'&e compound is cis and thus is assigned the letter Z. This makes choice A a true,qtatement, so it is eliminated. Because the two p-orbitals of the rc-bond arecoplanar with an orientation perpendicular to the substituents on the alkenecarbons, the four atoms bonded to the two carbons of the alkene must be.oplanar. It is not possible to rotate around the doubie bond, because the p-.rrbitals would no longer be coplanar, breaking the n-bond. This makes choice Ba true statement, which eliminates it. There are only two geometrical isomerspossible for an alkene. Because it is not possible to rotate about a double bondthe ru-bond would have to be broken), it is not possible to convert between the

cis and trans geometrical isomers without adding a great deal of energy. The.onsequence is that alkenes are rigid and thus have less entropy than alkanes of alomparable carbon chain length. This makes choice C a false statement, and thuslhe best answer. The structures of both an aikene and cyclic molecule arerelatively static and do not change drastically. The ring may flip-flop a littie, butrotation is observed only for the substituents on the ring, and not observed forJee bonds in the ring. Because the structure is static, the molecular features areconstant. This makes choice D a true statement, eliminating it.

lVe shall discuss stereoisomers oniy at this superficial 1evel for the time being. Inthe stereochemistry and carbohydrate sections, stereoisomers will be discussed ingreater detail.

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Organic Chemistry Structure Elucidation lsomerism

Conf ormational IsomersConformational isomer s, ot conformers, are molecules with identical connectivity

(bonds) that are nonsuperimposable because of rotation about a bond or

iontortion (often referr"d to u, iing-flipping) of the molecular structure' The most

stable conformation of tne motlculai structure is predictable based on

hybridization,stericrepulsion,andVSEPR(ValenceShellElectronPairnlpurcion) theory. Understanding conformational isomerism is vital, because

strlcture ii.tut", the reactivity anl stability of a compound. Figure 2-7 shows

two pairs of conformational isomers'

Conformationnl Isomers: Identical bonds with different spatial orientations

carrs"d by either rotation about sigma bonds or contortion of ring structures'

OH

H3

CH:

CHa

CHs

V^H

HHeC CHe HsC

H\ul vH

H\9)H

HsC

H:C

Figure 2-7

Conformational isomers are different orientations of the same molecule' Figure

2-8 shor"s six possible conformations (three-dimensional orientations) for butane'

Conformational isomers are in dynamic equilibrium at room temperature,

because rotation about sigma bonds is possible with the energy available'

Because molecules are consiantly rotating, butane does not remain in jusi one of

the conformations, but rather urrn*", all confirmations at one time or another'

It assumes the most stable conformation most frequently' This means that there

is a most favored and least favored conformation'

|

-| .._H-4

Hzc-=L--,/

CH"t" oHt\

OH

"--/--"H\/t/cwzcI

CHa

nT1.,,nrr/ Y;"

Structure #1

Structure #4 Structure #5 btructure ffo

Figure 2-8

Rotation about a carbon-carbon bond converts butane from one conformer into

another.Forinstance,al80"rotationabouttheCz-CzbondofbutaneconvertsStructure #1 in Figure 2-8 into Structure #4. The six structures represent 60'

incrementalrotationsabouttheC2_.Cgbond,acrossafull360"rotation.

OH

cH2cH3

Structure #2 Structure #3

Structure #6Structure #4

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Of the conformers in Figure 2-8, Structure #4 is the least stable, due to sterichindrance between the two methyl substituents (the largest groups). Structure#4 has hydrogens on carbon 1 and carbon 4 colliding with one another. Structure#1 is the most stable, because repulsion is minimized. To maximize stability, thelargest groups on each carbon (CH3, in this case) need to be as far apart from oneanother as possible (tn anti orientation), if they repel. This minimizes the stericrepulsion between the two groups. The least stable conformation (eclipsed, as it iscailed) has the two largest groups colliding into one another. The differentorientations are referred to as confonnational isomers. Figure 2-9 shows a side-r.iew of steric repulsion for the fully eclipsed butane molecule in Structure #4.

'H '.i. H./\ Steric repulsion between

hydrogens makes thisstructure less stable.

Figure 2-9

A major part of learning about conformational isomers is learning theterminology. Listed below is a giossary of terms that apply to conformationalisomers based on the structures shown in Figure 2-8.

Eclipsed: Eclipsed refers to the orientation where from a front view of themolecule, you are prevented from seeing all of the substituents on the reartarbon, because the substituents on the front carbon block the view of the rear.Structures #2, #4, and #6 are exampies of the eclipsed orientation of butane,Because of steric hindrance, the eclipsed conformations are less stable than thestaggered conformations. Figure 2-10 shows the eclipsed conformation of ethane.

Eclipsed Drawings of Ethane

HH

HzC.. ,CHz\-/

H\YZ VNHH

"Jy'-'Y';,.Figure 2-10

Staggered: Staggered refers to the orientation where from a front view of therrolecule, you can see all of the substituents on the front and rear carbons. There,s a 60" dihedral angle between neighboring substituents. Structures #1, #3, and=5 are examples of the staggered orientation of butane. Staggered is the most.:able conformation. Figure 2-11 shows the staggered conformation of ethane.

Staggered Drawings of Ethane

H

c-c

Figure 2-11

Hn

H v"H

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Organic Chemistry Structure Elucidation Isomerism(

Within staggered conformation, there are two terms used to describe the relativeposition of substituents on adjacent atoms. These terms arc gauche and anti, andthey refer to the position of a substituent on one atom relative to the position of a

substituent on an adjacent atom. In the case of ethane, where each carbon hasthree hydrogen atoms attached, each hydrogen has two hydrogens gauche to itand one hvdrogen anti to it. Butane, having a methyl group on both carbon-2and carbon-3, is a simplistic molecule to consider gauche and anti orientation.Rotation about the bond between carbon-2 and carbon-3 of butane generatesconformational isomers where there is a methyl group on each carbon. We shalldemonstrate gauche and anti orientation by referencing the methyl substituenton carbon-2 of butane relative to the methyl substituent on carbon-3 of butane.The terms are defined belorv.

Gsttclrc: Gauche refers to a staggered conformation of the molecule, where thetwo groups of interest (often the iargest groups) have a dihedral angle of 60".Structures #3 and #5 have the CH3 groups gauche to one another. Threedifferent perspectives of the staggered conformation of butane, with carbons 1

and 4 gauche to one another, are shown in Figure 2-12.

Staggered Conformation of Butane with CH, groups Gauche

m

L0

jj

il\

:[5

T

CH, H.C

Hfi!-cHr

'€>(" n'/H

,/tn.C- C..v"H

H:C

l:

:.

t-!,

Er

^1

arIC

-{B.C.

Sc

B€

L<

B"

br,

arst:

Figure 2-12

Anti: Anti refers to a staggered conformation of the molecule, where the twogroups of interest have a dihedral angle of 180". Structure #1 has the CH3 groupsanti to one another. This is the most stable structurel Three differentperspectives of the staggered conformation of butane, with carbons 1 and 4 antito one another, are shown in Figure 2-13.

Staggered Conformation of Butane with CH, groups Anti

+CH:

H

c-c.\,"

H

,/'n'

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Figure 2-13

The Berkeley Review Cr


io

:!

Because atoms within a molecule are in constant rotational motion, a structureloes not exist in one fixed conformation. Figure 2-14 shows the energy diagramthat corresponds to a complete 360' rotation about the CZ-CS carbon-carbon bondrf butane, starting and finishing with Structure #L from Figure 2-8. The energy:iagram starts with the lowest energy structure at 0' (which is the conformation

".-ith carbons 1 and 4 anti to one another) and rotates 360" about the C2-C3 bond:o return to the same orientation. Note that the highest energy structure iseractly 180" apart from the lowest energy structure, and that a 60" rotation takes-"-ou from an apex (a localized energy maximum) to a nadir (a localized energyninimum) on the energy diagram.

0" 60" 1.20' 180' 240'

Degrees of rotation

Figure 2-14

ihe energy diagram for butane is symmetrical, because butane is a symmetricalinolecule. When the compound is chiral (contains stereogenic centers), theenergy diagram is not symmetric. The energy axis of the diagram is not;uantified in Figure 2-74, so it demonstrates only a conceptual relationship. Ifany calculations are required on the MCAT, values will be provided for theenergy of the eclipsed and staggered conformations, and for the gauche and anti::'rteractions of various substituents.

Example 2.8Ilow does the energy diagram for the complete rotation about a central bond ofan asymmetric (chiral) compound compare to the energy diagram for itscomplete rotation about the CZ-CS bond of butane?

A. The energy diagram is symmetric like butane, but with higher energy values.B. The energy diagram is symmetric like butane, but with lower energy values.C. The energy diagram is asymmetric like butane, but wiih greater differences

in energy values.D. The energy diagram is asymmetric, with different energy values than butane.

SolutionBecause the structure is asymmetric, the energy diagram depicting the rotationassociated with the molecule must also be asymmetric, eliminating choices A andB. The energy diagram for the complete rotation about the C2--4.g bond ofbutane is symmetric, so choice C is eliminated. Choice D must be the correctanswer. The three staggered conformations are of unequal energy. Thestaggered conformational isomers of l1-2,3-dimethylpentane are:

360'300'



H,. .cH2cH3H3c<.^ ^/L -(. V"

CH:

Hz9 .cH.cH, H"C.r&.- _/tn"n' r-r.'c'32.^ ^,/tn,t',n,c/'- \:,f n)'- \:,t

HgC

CHs

1 methyl/methyl andI ethyl/methyl interactions

Lowest Energy Conformation

HgC\. CH"J

HH:C

H1 methyl,/methyl and

2 ethyl/methyl interactions

$CHs

HsC

HsC

HsC

2 methyl/methyl and1 ethyl/methyl interactions

I

=

I

=i

CHa

H

Figure 2-15 shows the exact orientation of the atoms at various times during thefirst 120" of rotation of ethane. The graph shows that at 20" and 100" thestructures are symmetric and thus of equal energy. This is also observed for theconformers at 40" and 80".

0 20 40 60 80 100 720Degree of rotational displacement from staggered conformation

Figure 2-15

If the substituents repel one another, then the anti orientation is lowest on theenergy diagram. However, when the force between neighboring substituents isattractive in nature, tl-re gauche orientation is most favorable. A good example isa vicinal diol, where the two hydroxyls on adjacent carbons exhibit hydrogen-bonding with one another.

CH2CH?

Rotation about theC-C bond of ethane

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Erample 2.9.he best explanation for why the most stable orientation about the C2-C3 bond of:i.e amino group and the carboxylic acid group in 3-aminopropanoic acid is:auche is which of the following?\. Gauche is the most stable, because it minimizes the steric hindrance between

the amino group and the carboxylic acid group.B. The gauche conformation is always more stable than the eclipsed

conformation.C. The carboxylic acid and amino groups form a hydrogen bond best from

gauche orientation.D. The carboxylic acid and amino groups form a hydrogen bond best from anti

orientation.

Solutionil the substituents attract one another, then the most stable conformation iss:aggered with the two attracting groups gauche with respect to one another.I-re best choice is C, because gauche has the amino and hydroxyl groups close.nough to form a hydrogen bond. Hydrogen-bonding is a stabilizing force. The:lrongest hydrogen bond is formed by the lone pair of nitrogen with the acidic:,r'drogen of the carboxylic acid group.

Hn&J*n,,o,-./ \:r:"

ll"o

Anti orientation

Gauche orientation

Hydrogen-bonding is possible when the two groupsare only 60' apart, but not when they are 180" apart.

o?c -

zHvdrogen bond------- - ,,H-o- ^. ta u^xr.I '.4

U- (. \Hz"&_J,'/ Y3"

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Organic Chemistry Structure Elucidation Isomerism TNewman ProjectionsUnderstanding the nuances of conformational isomers requires good three-dimensional viewing skills, so you may wish to dig out your molecular models ifyou still have them. Be able to recognize structures from the stick figure view(with dashed and bold wedges), as well as from Newman projections. Newmanprojections are front views of a molecuie. In drafting, three views are given tosee the whole. It is no different in organic chemistry. The side view is a dashed-and-bold wedge representation, the front view is a Newman projection, and atop view is a Fischer projection. Figure 2-16 shows a pictorial explanation of theconversion from a dashed-and-bold wedge drawing to a Newman projection,while Figure 2-77 shows a pictorial explanation of the conversion from aNewman projection to a dashed-and-bold wedge drawing.

When viewed from the right, substituents pointing out of the plane in a dashed-and-bold wedge drawing are on the left side in a Newman projection.Substituents behind the plane in a dashed-and-bold wedge drawing are on theright side in a Newman projection. Substituents on the left side in the Newmanprojection end up pointing out of the plane in the dashed-and-bold wedgedrawing. Substituents on the right side in the Newman projection end up behindthe plane in the dashed-and-bold wedge drawrng.

Conversion from dashed-and-bold wedge drawing to Newman projection:

r,n

strerydrit

wTtrmr

I[tuP

MW

od

Imft

H

,/ Y;,.",

q,)

d9.6

-o

CJ

o.o

oL

HsC

H

H

CHs

Figure 2-16

Conversion from Newman projection to dashed-and-bold wedge drawing:

behind

HgC :f&1"'OH

behindH +)CHs

frontHpiane

Figure 2-17

CycloalkanesCyclic alkanes which contain only one ring have the chemical formula C.,H2.,and contain no n-bonds. The stability of a given cycloalkane is rooted in itsability to form bond angies of approximately 109.5', the norm for sp3-hybridizedcarbons. The farther from 109.5" the angle is, the greater the reactivity of thecycloalkane. For this reason, three- and four-membered rings are reactive, whilefive- and six-membered rings are stable. When treated with hydrogen gas (H2),cyclopropane and cyclobutane readily form straight chain alkanes (propane andbutane). Cyclopentane and cyclohexane do not undergo hydrogenation.

in front

HH

L-\,of plane ,/,/ Y,::*,"

lnof

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T

:I

o

t-

a

e

I,

a

The reactivity of three- and four-membered rings is attributed to ring strain. Rtngstrain is defined as the energy difference between the linear and cyclic alkanes ofequal carbon iength. Because the bond angle in cyclopropane is 60", vastlydifferent than 109.5' associated with a normal spr-hybridrzed carbon, there is agreat deal of ring strain. To relieve this angle problem, cyclopropane forms whatare referred to as bent bonds (sigma bonds in which the electron density does notlie between the two nuclei). Because the electron density is not between the twonuclei, the bond is much weaker and thus easier to break. Figure 2-18 shows theorbital and bonding pictures for various cyclic alkanes.

Cyclopropane

Ring strain 27.4 kcal / mole

The carbon-carbon bonds are bent (not coilinear), makingthem weaker than standard carbon-carbon singie bonds.

Cyclobutane

Ring strain 26.2 kcal / nole

Cyclopentane

t-I.en

e

d

H2C- cH2

uri-."r-)rr,Rrng strain 6.8 kcai/mole

Cyclohexane

H

HH

Figure 2-18

lecause of their stability, most cyclic organic and bio-organic molecules are

=rther five-membered or six-membered rings. Six-membered rings are slightlyrore stable than five-membered rings. Given their frequent presence in:iological molecules, five-membered and six-membered rings have a high:robability of appearing on the MCAT.

HHHH

lnitsed

:le

t!.-J

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Organic Chemistry Structure Elucidation Isomerism jCyclopentaneCyclopentane does not require much^distortion of its bonds and shape toaccommodate the 109.5" angle for the sf -hybrid. A perfect pentagon has anglesof 108', so there is only a small discrepancy from 109.5". This angle differencedoes not account for the small ring strain of 6.8 kcals per mo1e. To achieve thecorrect angle and alleviate this torsional strain, cyclopentane forms what isreferred to as an enaelope shape, where one of the carbons is not coplanar with theother four. The major problem with cyclopentane is not the ring bond angles, butthe substituents on the ring that are in an eclipsed conformation as a result of thenear-planar ring structure. This eclipsing of hydrogen atoms causes furthercontortion of the structure, which accounts for the ring strain energy. Still,cyclopentane structures are relatively stable, Their stability makes them commonin such biological structures as ribose, deoxyribose and the purine ring of theDNA bases adenine and guanine. Figure 2-19 shows a few other common five-membered rings frequently encountered in the biological sciences.

OHHB-D-Fructofuranose

\(

m;ui"

lh

LU

cg

lllln

m

il

rilm

OH HO

B-D-Ribofuranose

COr-

t-*H"N-"lCH"t'nHN\---zN

L-Histidine

OHH2-Deoxy-B-D-ribofuranose

Figure 2-19

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CyclohexaneCyclohexane has the most stable ring structure of all of the cycloalkanes. This isevident in biological molecules, which have multiple rings of six atoms. Themost stable form of cyclohexane is the chsir conformation. There are twodifferent chair conformations for cyclohexane. The two conformational isomerscan interconvert through a process referred to as ring-flipping. In interconverting,the structure passes through the boat conformation. Because the two chairconformations are equally stable, AGr* - 0. The interconversion between the twochair conformations of cyclohexane requires 10.8 kcals/mole in activationenergy. The chair conformation offers two substituent positions: equatorial(named for its orientation around the equatorial plane of the ring) and axial(named for its vertical alignment like an axis). Equatorial is more stable thanaxial, so the most stable conformation of a cyclohexane compound has the largestsubstituents in the equatorial positions. Figure 2-20 shows chair conformationsof cyclohexane with detailed positions.

Cyclohexane showing equatorial HsH

Cyclohexane showing axial Hs

axial (up) axiai (up)

equatorial (up)

(up) eq (down)

eq (down)eq (up)

eq (up) axial (down)

equatorial (down)axial (down)

axial (down)

Cyclohexane showing all substituents

Figure 2-20

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Interconversion between Cyclohexane ConformersThe interconversion between the two chair conformations of cyclohexanerequires activation energy, but the free energy (AGr*) is zero, because there areno substituents on the cyclohexane ring. As cyclohexane undergoes ring-flipbetween chair conformations, it passes through intermediate structures. The twomost significant intermediate structures are tLle twist form and the boat form. Thetwist form is of slightly lower energy than the boat form, because the hydrogensare not eclipsed in the twist form. The ring-flip conversion of cyclohexane fromchair to boat and on to the other chair conformation is shown in Figure 2-21.

H +

ilm

rydffi

turoffimh

tu@ry

@ffiffim

il[m

@M

HH

H

p1 HTwist

Figure 2-21

Because molecules can readiiy rotate and contort at room temperature, anequilibrium exists between the two chair conformations of the cyclohexanemolecule. Either at low temperatures or'in cases where a substituent is too bulkyto orient itself in an axiai manner, one conformer is present exclusiveiy. Theproton NMR can be used to determine the more stable structure by focusing onthe different signals for the two orientations of hydrogen. You should be able todetermine the most stable conformation. The boat conformation is not as stableas the chair conformation, because its substituents encounter high energyeclipsed interactions its substituents encounter, while the substituents in thechair conformation are staggered with lower energy gauche and anti interactions.

Monosubstituted CyclohexaneThe ring-flip process is the same when there is a substituent on the ring as it is forcyclohexane, but the energetics are different. When a substituent is present, theactivation energy for interconversion increases, and the two chair conformationsdiffer in stability. Because the two chair conforrnations are no longer of equalenergy, the two twist forrns are also no longer of equal energy. The most stablechair conformation is the structure r,r'ith the least steric repulsion. Substituentswith axial orientation on the same side of the ring are close enough to repel(known as 1,3-diaxinl interactions), so axial orientation is less favorable thanequatorial orientation. This is indicatecl in Figure 2-22, where the two chairconformations of both methylcyclohexane ar-rd cvclohexanol are shown. The twochair conformations of methylcyclohexane differ in stabiliiy by 7.69 kcals,/mole.Because a hydroxyl grorlp is smaller than a methyl group, the difference inenergy for the two chair conformations of cr-clohexanol is only 1.04 kcals/mole.

HH

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oIH

MethylcyclohexaneEquatorial > Axial by 7.69 kcals/mole

Figwre 2-22

Disubstituted CyclohexaneDisubstituted cyclohexane exhibits different dynamics than monosubstitutedcyclohexane. The greatest energy difference is observed when comparingT,3-diaxial to 1,3-diequatorial, because the steric repulsion of 1,3-diaxial substituentsis the strongest repulsion encountered in cyclohexane. This is shown in Figure 2-22 for dimethylcyclohexane. When comparing 7,2-diaxial to 1,2-diequatorial,there are no diaxial interactions of the bulkiest substituents, because the twobulky substituents are trans to one another. This makes the energy differencebetween chair conformations less than what would be observed with the 1,3-cyclohexane. The energy difference between the 1,2-diaxial and 1,2-diequatorialorientations is also less than the energy difference between 1,4-diaxial and 1.,4-

diequatorial (which is another compound having the two bulkiest substituentstrans to one another), because the 1,2-diequatorialcyclohexane species has gaucheinteractions between the bulkiest substituents. The energetics of theconformational isomers of dimethylcyclohexane are shown in Figure 2-23.

CHs Cis-1,3-dimethylcyclohexane1,3-Diequatorial > 1,3-Diaxial by 8.4 kcals/mo1e

9H:

CyclohexanolEquatorial > Axial by 7.04 kcals/mole

H

CH"l"I

/-=--/4 H.c/114L;-/ : "'[.\1-\I

CH. Trans-7,2-dimethylcyclohexane" 1,z-Diequatorial > 1,2-Dtaxialby 2.8 kcals/mole

N :- "''H*"| .r,

CH, Traus-7,4-dimethylcyclohexane" t,+-Diequatorial > 1,4-Diaxialby 3.4 kcals/mo1e

Figure 2-23

F/Fcrr.H:c-\/-\ r

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Possible Orientations for Disubstituted Cyclohexanes

7,2-trans (upldown) or (down/up)1,2-cis (uplup) or (down/down)7,3-cis (uplup) or (down/down)7,3-trans (upldown) or (down/up)7, -trans (upldown) or (down/up)\, -cis (up/up) or (down/down)

€ C-1 axial C-2 axial or C-1 equatorial C-2 equatorial<+ C-1 axial C-2 equatoriai or C-1 equatorial C-2 axiale; C-1 axial C-3 axial or C,1 equatorial C-3 equatoriale> C-1 axial C-3 equatorial or C-1 equatorial C-3 axiale C-L axial C-4 axial or C-1 equatorial C-4 equatorial++ C-1 axial C-4 equatorial or C-1 equatorial C-4 axial

Table 2-1 lists the possible orientations for disubstituted cyclohexane. Transrefers to substituents on opposite sides of the ring, equating to either up/downor down/up. Cis refers to substituents on the same side of the ring, equating toeither up/up or down/down. Cis and trans do not refer to equatoriil or axialorientations, but rather to whether the substituents are above or below the ring.

(

:

:si

J:,

t.

Table 2-1

It is essential that you be able to translate from nomenclature to the most stableconformation. For instance, trnns-3-methylethylcyclohexane is a 1,,3-transcompound. A compound with 7,3-trans orientation has both an axial and anequatorial substituent. The ethyl group is larger than the methyl group, so theethyl group occupies the equatorial orientation in the most stable conformationof trans-3-methylethylcyclohexane. This is shown in Figwe 2-24.

-:rr.x

llhr

f__4b--_LcH:I

tl!I"}ff

:lhu

ilIJI,

IIIIIII

:ilh,r

dlilrru

[]rgft

I[!r!

cH2cH3

Figure 2-24

You should take note that when you convert from one chair conformer to theother, the axial substituents become equatorial (as seen with the ethyl group),and the equatoriai substituents become axial (as seen with the methyl group).The most stable conformation has the least steric repulsion.

Example 2.10The most stable conformation of cis-L,2,4-trimethylcyclohexane has which of thefollowing orientations for the three methyl groups?

CHg

A.B.C.D.

The chair conformation witl-r 3 equatorial methyls and 0 axial methylsThe chair conformation with 2 equatorial methyls and 1 axial methylThe chair conformation with 1 equatorial methyl and 2 axiai methylsThe chair conformation with 0 equatorial methyls and 3 axial methyls

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SolutionThe most stable conformation has as many methyls equatorial as possible. 7,2-cishas one axial and one equatorial substituent, so there must be at least one axialsubstituent. Answer B is best. Drawn below are the two chair conformations ofcis-7,2,4-trimethylcyclohexane (or (1R,2S,4R)-1,2,4-trimethylcyclohexane as

correctly name using IUPAC nomenclature rules.)

CH. CH, CH.l" l' 1"".r-) \-\4/ /

-\

\H:.cd -

WcH.,2 equatorial; 1 axial1 equatorial;2axial

This covers the topics associated with isomerism. These topics are applicable inboth the physical and biological sciences areas, so know them well. To implantthis section as a working knowledge base, there are many passages with which towork. From the very beginning, you want to emphasize the logic behind youranswers. The MCAT may not have passages that are verbatim duplicates ofwhat you see in here, but if you answer these questions using sound logic andfundamental concepts, then you will slowly get acclimated to the MCAT way ofthinking. At this point, passages may seem like an absurd form of askingquestions, but hopefully you wiil take a liking to the style. Passages presentinformation that you must incorporate into your background knowledge, andthen using all the information you have, you must reach a conclusion concerningtheir questions. Multiple-choice tests require that you find the best, mostreasonable answer. You are not required to solve detailed questions or derivefundamental concepts. Just find the best answer, as fast as you can.

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Organic Chemistry Structure Elucidation Structural Insights

# ffi#ffiilffitStructural Symmetry\A/hen deducing the molecular structure for an organic molecule, it helps to knowsomething about the symmetry of the compound and its units of unsaturation.symmetry can be broken into plane symmetry and point symmetry. In planesymmetry, the compound has two halves that are evenly displaced about animaginary mirror in the middle of the molecule. In a structure with pointsymmetry, there is an inversion point at the center of the molecule such that iftwo lines are drawn in opposite directions along the same axis, then both linesegments intercept identical atoms at the same diJtance from the inversion point.This may not seem clear in words, but looking at a structure helps illustrate theconcept. Figure 2-25 shows one compound with mirror plane symmetry andanother with inversion symmetry. Molecules with inversion points are nonpolar,because all of the individual bond dipole vectors cancel each other out.

cH20H

HgC

Molecule with mirror symmetry

HOH"C H'\r.-o1rY*'

*--*oof'- "\..' -,H CH2Q.H

Molecule with an inversion point

Figure 2-25

Symmetry within a molecule affects its NMR and IR spectra. As symmetryincreases, the number of signals in a spectroscopic study decreases. couplingsymmetry information with units of unsaturation helps to deduce the structuralfeafures and connectivity of a molecule.

Units of UnsaturationUnits of unssturation are calculated from the molecular formula. The units ofunsaturation give us information about the number of rings and/or n-bondspresent within a molecule. There is some minimum number of bonds needed tohold the atoms in a molecule together, and any additional bonds beyond theminimum are the units of unsaturation. To hold two atoms together, it takes onebond (Atom1-Atom2). To hold three atoms together, it takes two bonds(Atom1-Atom2-Atom3). The minimum number of bonds required to hold amolecule together is always one less than the number of atoms. The minimumnumber of bonding electrons is two times the minimum number of bonds. Anyelectrons beyond the bare minimum needed to hold the molecule together can beused to form additional bonds. For every extra pair of electrons, there is a unit ofunsaturation. To determine the units of unsaturation, the strategy is todetermine the number of excess bonding electrons. There are a few differentmethods for doing this.

1) C3H6 contains eleven atoms, which requires at minimum ten bonds (andthus twenty bonding electrons). There are three carbons with four bondingelectrons each. There are eight hydrogens with one bonding electron each.This means that propane (C3Hg) has exactly the twenty bonding electronsneeded. There are no extra bonding electrons, so propane has a linearstructure with no n-bonds. This is to say that propane has no units ofunsaturation.

Copyright O by The Berkeley Review rr6 The Berkeley Review


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The units of unsaturation for hydrocarbons and carbohydrates can bederived from the formula for aliphatic alkanes, Cr,H2', 1 2. "Aliphatic" refersto a structure that has no rings or n-bonds. An aliphatic alkane has the bareminimum number of bonds, so there are no units of unsaturation. For everyur-rit of unsaturation, there are two fewer hydrogen atoms than themaximum. Thus, the units of unsaturation can be obtained by comparing theactual formula to the fully saturated formula. For instance, C5H6 has fourhydrogens less than the fully saturated formula for five carbons, CSH1Z.Because it has four fewer hydrogens, it has two units of unsaturation.

The units of unsaturation depend on the surplus of bonding electrons. Tokeep any chain propa gated, every member of it must make two connections.ln a molecule, each atom must make two bonds to keep it intact. This meansthat every atom needs a minimum of two bonding electrons. Using thisperspective, we can determine the number of excess electrons per atom.Hydrogen makes just one bond, so you subtract one for each hydrogen in themolecule. Oxygen atoms are ignored, because they make the minimum twobonds that are needed. Carbons are multiplied by two, because carbonsmake four bonds, two beyond the minimum to propagate the chain. Thereare two ends to every chain, so two is added to the total. The units ofunsaturation refer to bonds, rather than bonding electrons, so the sum ofexcess electrons must be divided by two. This is summarized in Equation2.7.

Units of unsaturation =2(#C)+2-(#H) (2.1)

2

l"lethod 3 works with other atoms, too. Nitrogen makes three bonds, which is::-e more than the minimum needed, so you add 1 per nitrogen atom. Halogens:-':ke one bond, which is one less than the minimum, so you subtract 1 per. -:-,de. Equation 2.2 includes nitrogen and halogens.

Units of unsaturation =2(#c) + (#N) - (#H) - (#x) + 2 (2.2)

2

Erample 2.11:::",r' many units of unsaturation are present in a compound with the molecular::::nula C7H9N3O2Cl2?

t rlution-:,s question is solved by applying Equation 2.2.

-. )+ (3)-(9)-(2)+2 =r4+3-9-2+2 =72 -9 - B - 4Unitsof unsaturation2222

:::ause there are four units of unsaturation, choice D is the best answer. With: -rr units of unsatuiation, the compound could contain three n-bonds and one::ng, meaning it is potentially a benzene derivative. When there are four units of,:.saturation, you should immediately consider the possibility that the. --mpound contains an aromatic ring. While there are other combinations of four-::ts of unsaturation, there is a high probability of having an aromatic ring.

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Spectro3bop$ an AndysisSpectroscopyThe MCAT togics include infrared absorption spectroscopy, uitravioiet-visiblespectroscopy, 1H (proton) nuclear rnugtl"ti. ,"ronun." spectroscopy, and 13C

(carbon) nuclear magnetic resonance spectroscopy. There is nothing for you tofear, however. At just over one minute per problem, if you have to deduce astructure from spectroscopic information, then it will likely be easy or symmetric.For example, on a previous exam, there was a proton NMR of ethanol that manystudents said was very easy. Besides just having to determine the structure of anunknown, you may also have to assign signals and peaks to an existingcompound. The spectra should be interpreted using the typical features youhave learned. To date, the test has emphasized only a few features that havebeen stressed in course work. We shall start by reviewing the basic operations ofthe IR and its applications to structure elucidation. From there we sha1l considerultraviolet,/visible spectroscopy, proton NMR, and carbon-13 NMR.

Infrared SpectroscopyEr,'ery molecule produces a unique IR spectrum. Infrared spectroscopy starts byadding a monochromatic beam of IR photons to either a thin oil suipension (ifthe compound is a solid) or a neat solution (if the compound is a liquid) betweensalt plates. The molecule absorbs electromagnetic radiation that causestransitions between vibrational energy levels within it, meaning that themolecule vibrates more or less frequently as energy is absorbed or emitted.when a molecule soaks up the EM radiation and vibrates at a higher energy.This change in stretching (vibrating) between atoms within the molecul" .u.6ei'achange in the dipole moment, which can then be monitored. An infraredspectrometer uses light of wavelength 2,500 nanometers to 17,000 nanometers(recorded as 4000 cm-1 to 600 cm-1 on the graph). \A/hat we record is the changein the intensity of the EM radiation from when it enters the molecule to when itexits the molecule. This is compared to a reference beam that traverses a path ofidenticai length but does not pass through the compound itself. If the compound.absorbs a given wavelength of light (corresponding to some transition), then weobserve an absence of light exiting the sample tube. This is known as absorptionspectroscopy. The graph records transmittance as a function of wave number (cm-r), so absorbances are represented by drops rn intensity.

The frequency at which light is absorbed is specific for each type of bond. Asyou may have learned in physics, the frequency of light is directly proportionalto the masses of the two atoms in the bond and the bond strength. This is to saythat the potential energy in a resonating system (such as a spring that obeysHooke's law) is described by Equation 2.3

P.E. = 1 kx22

The k-term is the spring constant, which we can say describes the bond strength.The x-term describes the distance from equillbrium that the bond has stretched.The absorbance can be thought of as increasing the potential energy of the bond,so the absorbance is proportional to the energy of the bond. As a resuit, the bonddissociation is directly proportional to the energy that is absorbed. This is notexactly true, but close enough to help approximate spectra. Because a wavenumber is measured in cm-l, it is an inverse of the wavelength. The inverse ofthe wavelength is directly proportional to the energy of the photon. This meansthat the higher the wave number is, the greater its energy.

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Organic Chemistry Structure Dlucidation Spectroscopy and Analysis

For instance, a C=O boncl absorbs around 1700 cm -1, while a C-O bond absorbsaround 1300 cm -1. this is because a C=O bond is stronger than a C-O bond.Carbonyl functional groups a-re common, so you should know the absorbancevalue for a C=O bond. An sp3-C-to-H bond absorbs just below 3000 cm-1, whilean sp2-C-to-H bond absorbs just above 3000 cm-l, because an sf -C-to-H bond isstronger than an sp3-C-to-H bond. This is because the sp2-hybrid, having more s-character, is smaller than the sp3-hybtid. The result is that an sp2-C-to-H bond isshorter and thus stronger than an sp3-C-to-H bond. The stronger bond, having ahigher bond dissociation energy, has a higher energy absorbance.

Although the molecule as a whole absorbs the EM radiation, we can use theabsorbances we measure to fingerprint particular functional groups and bondsrvithin the molecule. The skill needed to make IR useful is an active process.Scientists use IR not only to confirm the presence of certain functional groups,but to also to help decide which functional groups are not there. IR is mostuseful as a supplement to the moiecular formula and the NMR spectra formolecules. Table 2-2lists several useful IR absorbances. The values are listed interms of wave numbers. Note that the absorbance of a given bond varies withihe compound in which the bond exists.

Table 2-2

Bond type Stretching (cm-1) Bending (cm-1)

O-H alcohol (no H-bonding) 3640 - 3580 v)

O-H aicohol (H-bonding) 3600 - 3200 (s, broad) 1620 - 1590 (v)

N-H amides 3500 - 3350 m)

N-H amines 3450 - 3200 m)

C-H alkvnes 3300 - 3220 s)

C-H aromatic 3100 - 3000 v) 880 660 (v)

C-H alkenes 3060 - 3020 m) 1000 - 700 (s)

C-H alkanes 2980 - 2860 s) 1470 - 7320 (s)

C-H aldehyde 2900+,2700+ (m, 2 bands)

O-H acids (H-bonding) 3000 2500 (s, broad) 1655 - 1510 (s)

C=C alkvnes 2260 -2720 (v)

C=N nitrile 2260 - 2220 (v)

C=O ester 7750 - 1735 (s)

C=O aldehyde 7740 - 7720 (s)

C=O ketone 1725 - 1705 (s)

C=O acid 1725 - 7700 (s)

C=O aryl ketone 1700 - 1680 (s)

C=O amide 1690 - 1550 (s)

C=O cr,B-unsaturated ketone 1685 - 1665 (s)

C=C alkene 1680 - 1620 (v)

C=C aromatic 1600 - 1450 (v)

C-O alcohols, ethers, esters 1300 - 1000 (s)

C-N amines, alkyl 1220 - 1020 (w)

s = strong absorptionm = medium absorption

w = weak absorptionv = variable absorption

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Organic Chemistry Structure Elucidation Spectroscopy and Analysis

using Table 2-2, you can evaluate IR data given in spectrum form to identifystructures. As for memorizing peaks, according to the MCAT student Maru,Lal youare required to "know the important ones," which is open to interpretation. Youdon't necessarily have to memorize all of the values, but if you do enoughproblems, the values you repeatedly see should become second nature.

As a diagnostic tool, IR is used to detect certain functional groups. You havereached the pinnacle of utility when you use it to determine which functionalitiesare not present as well as which functional groups are present. Just as peaksconfirm the presence of a certain bond, the absence of a peak supports theabsence of that bond. Here is an example of how IR spectroscopy is used:

An unknown compound with formula C_aHgO is analyzed by IR spectroscopy.An intense band is detected at 1770 cm-1 (IR absorbances are listeh by energyaccording to the wave number as measured in cm-1). By comparing the value toa chart of IR absorbances, this peak can be attributed to a C=O. The compoundhas one degree of unsaturation attributable to a C=O, which makes it possible tonarrow it down to a small number of isomers. The structure cannot be cyclic andhas a carbonyl. Given that the longest chain is four carbons, the carbonyl can beonly on carbon 1 or carbon 2. This narrows it down to only two butanederivatives. The iongest chain could be only three carbons, with a methylsubstituent on carbon 2. In that particular structure, the carbonyl group has to beon carbon one. This leaves only three possibilities, and they are:

1. H3CCOCH2CH3 (butanone)

2. H3CCH2CH2CHO (butanal)

3. H3CCH(CH3)CHO (2-methylpropanal)

Thus, we can reduce the choices from many types of compounds having one unitof unsaturation and one oxygen (a cyclic ether, for example) to a few. Ketonesand aldehydes have different chemical reactivity and physical properties, sowhen we combine IR information with chemical tests and the melting point ofthe compound, we can eliminate two of the three structures. This is a structureelucidation technique you have done many times in the past. Through examplesar-rd practice, you can familiarize yourself with the peaks and become talented atsolving the problems using deductive reasoning.

Example 2.12Which of the following compounds with the formula C5H19O cannot have an IRabsorbance peak between 1700 cm-1 and 1750 cm-1?

A. An aldehydeB. A ketoneC. A cyclic etherD. A11 of the above have an IR absorbance between 1700 cm-1 and 1750 cm-1.

SolutionAn IR absorbance between 1700 cm-1 and 1750 cm-1 implies that the compoundhas a C=O in its structure. Because it has no absorbance between 1700 cm-1 and1750 cm-1, it does not have a C=O bond. Choices A and B have a C=O in theirstructure, so they can be eliminated. The best answer is choice C. The onedegree of unsaturation associated wlth the formula is used in the ring. The oneoxygen in the formula is in the ether, which contains carbon-oxygen singlebonds. Choice D is also eliminated, because choices A and B are eliminated.

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Example 2.13How many structural isomers of C4H3O2 are possible that have an IR absorbancepeak between 7735 cm-1 and 1750 cm-1, u p"ik around 1200 cm-1 and no broadpeaks above 2500 cm-1 ?

4.2B.3c.4D.5

SolutionThe absence of a broad peak above 2500 cm-1 indicates that the compound doesnot have an o-H bond, which eliminates the possibility of it being an alcohol orcarboxylic acid. It is most likely an ester, although it could have both a carbonyland ether functionality. According to Table 2-2, the peak betweenIT35 cm-1 and1750 cm-1 indicates ihat there ;s a c=o bond of an ester and not a ketone oraldehyde. The compound must be an ester, so the question now becomes, "Howmany esters are there that contain only four carbons in their structure?" Thereare only four four-carbon esters, as drawn below, so the best answer is choice C.

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The two aldehyde structures (1 and 3) could be confirmed or eliminated by thepresence or absence of two peaks at 2900 cm-1 and2700 cm-1. If this were a realiaboratory scenario, it would be far easier at this point to use proton NMR todeduce the structure of the unknown compound. Structure t has four types ofhvdrogenin a 1:2:2:3 ratio. Structure2hasthree types of hydrogenina3:2:i ratio. Structure 3 has three types of hydrogen in a 1 : 1 : 6 ratio. Structure 4 hasthree types of hydrogen in a 3 : 2 : 3 ratio. The integration would be enough todistinguish anything except Structure 2 from Structure 4. To distinguish thesetrvo structures requires identifying the ppm shift value of each type of hydrogen.iVe shall address NMR spectroscopy later in this chapter. IR spectroscopy

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Hydrogen-Bonding in Infrared SpectroscopyBecause the formation of hydrogen bonds affects the covalent bond between anatom and a partially positive hydrogen involved in hydrogen-bonding, anyspectroscopy techniques focusing on the covalent bond to hydrogen, or thehydrogen itself, is affected by hydrogen-bonding. The effect is a broadened peak(observed in both the IR and NMR techniques). In the case of IR, the broadeningof the hydroxyl absorbance associated with hydrogen-bonding is caused by theweakening of the covalent bond between the hydrogen and the atom (nitrogen,oxygen, or fluorine) to which it is bonded. This lowers the energy of the covalentbond and thus lowers the energy of absorption for the bond. As the hydrogenbond increases in strength, the covalent bond weakens. Because not all of thehydrogens have the same degree of hydrogen-bonding, their covalent bondsexhibit many different absorptions, ranging from unaffected and thereforestandard covalent bonds to covalent bonds that are highly weakened by thehydrogen bond. This range of covalent bonds gets grouped together into the one

broad peak. The same alcohol exhibiting two different degrees of hydrogen-bonding is shown inFigure2-26.

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Wave number (cm-1)

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Given that molecules are in continuous random motion within a liquid, some

alcohols have strong hydrogen bonds, while others have no hydrogen-bonding-This means that the solution has a random distribution of hydrogen bonds andtherefore a random distribution of covalent bonds. The result is a distribution ofsignals in infrared spectroscopy. To see each individual peak for each differentcovalent bond requires a high resolution IR spectrophotometer. It is unlikely youused such an instrument, so the signal with which your are familiar is a broadcomposite signal covering the range of the individual signals. Figute 2-27 s

a high-resolution IR signal and the standard-resolution equivalent.

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Copyright @ by The Berkeley Review t22

Figure 2-27

The Berkeley

0rganic Chemistry Structure Dlucidation Spectroscopy and Analysisl5

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Example 2.14',"hat is true for the compound associated with the following IR spectra?

1457 cm-l ll r:ii .,n '& 736'7 cm-l

3317 cm-l

2903 cm1

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-{. It exhibits no hydrogen-bonding.B. It has a carbonyl group.C, Ithas ahydroxyl group.D. It has a molecular mass that is less than 30 grams per mole.

Solution- he compound represe-nted by the lR spectrum above in the question has a broad:-eak around 3300 cm-l and no peak near 1700 cm-]. These are the first areas to:onsider when looking at IR spectra. The compound has a hydroxyl group, butro carbonyl group. This makes choice B incorrect and choice C correct. Because:t has the hydroxyl group, it can exhibit hydrogen-bonding. This eliminates;hoice A. Because of the peaks in the 1300-1400 cm-1 range, we know the:ompound has a carbon-carbon bond, so it must have at least two carbons. TheLightest compound with two carbons is ethyne (HC=CH), which has a molecularnass of 26 grams per mole. However, because there is an oxygen present, thecompound must have a molecular mass greater than 30 grams per mole. ChoiceD is eliminated.

Ultraviolet/Visible SpectroscopyIn addition to infrared spectroscop/, there is also ultraviolet/visiblespectroscopy. \A4rile infrared photons (in the 3-to-10 kcals/mole region) affectthe vibrational energies of a molecule, ultraviolet (in the 70-to-300 kcals/mo1eregion) and visible (in the 40-to-70 kcals/mole region) photons affect theelectronic energy levels. when uv or visible photons are absorbed by amolecule, an electron is said to be excited from the ground state to an excitedstate. Because o-bonds are so much stronger than n-bonds, the lowest energyabsorbance for alkanes is significantly higher than the lowest energy absorbancefor alkenes. To excite an electron from the o-level (sigma bonding orbital) to theo*-level (sigma anti bonding orbital), photons of approximately 140 nm to 120 nmare necessary. However, because molecules in the air readily absorb energy inthis region, the spectra must be obtained in a vacuum, Because this constraint israther impractical, uv-visible spectra typically range from 200 to 800 nm, whereair does not interfere. As a result, we typically use only UV-visible spectroscopyto analyze molecules with n-bonds, especially conjugated systems. uv-visiblespectroscopy in organic chemistry focuses on transitions between the n and n*energy levels. For systems with conjugation, there are several n-levels, but wecare about only the lowest energy transition.

3W Copyright @ by The Berkeley Review t23 Dxclusive MCAT Preparation

Organic Chemistry Structure Elucidation Spectroscopy and Analysis IThe transition of interest is from fi to n*. The wavelength of highest absorbance,known aslambda max (),^u*), changes with the amount of conjugation. The valueof L,,'.u* depends on the amount of conjugation. As the conjugation increases, so

does the tnaaelength of L*o*. Figure 2-28 shows the l"*u* values associated withthe lowest energy n-to-n* transition of various conjugated hydrocarbons.

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Because the energy gap between n and n* decreases as the conjugation increases,the wavelength of maximum absorbance increases. 1,3,5,7-octatetraene has a

I*u* of 304 nm and 1,3,5,7,9-decaquintene has d Imax of 353 nm. When moreconjugation is added, the absorbance shifts into the visible range. Color resulFfrom excessive conjugation within a molecule. For instance, B-carotene (with 1L

n-bonds) has absorbances at 483 nm and 453 nm. Substituted benzenes have a

number of peaks. Conjugated aldehydes and ketones have about the same n-:i*absorbances as conjugated alkenes of the same number of n-bonds. However'conjugated aldehydes and ketones have other, more intense absorbances (e >10,000) that are of longer wavelength than their hydrocarbon counterparts. ThL"

is attributed to the n-to-fi* transition associated with aldehydes and ketones,

possible because of the lone pair of electrons on the carbonyl oxygen. Figwe2-29shows the L,^u* values associated with the lowest energy n-to-fi* and n-to-n*transitions of various ketones.

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Unlike infrared spectroscopy, ultraviolet/visible spectroscopy can alsoapplied in a quantitative fashion. Ultraviolet/visible spectroscopy can be used

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Copyright @ by The Berkeley Review 124 The Berkeley


Nuclear Magnetic ResonanceThe fundamental principle behind nuclear magnetic resonance, NM& is the sameas for other forms of spectroscopy. Energy, in the form of electromagneticradiation in the radio frequency band, is added to the system and analyzed interms of what is absorbed. The energy levels that are affeited are for the spin of anucleus in the presence of an external magnetic field. Normally all of the nucleihave spins of the same energy. However, when an external magnetic field isapplied, spins can either align with the field or align against the fie[, so multipleenergy states are possible. In the case of 1H, theie or"u t*o energy levels: o lineone aligned with the external magnetic field), and G (the one aligned against theexternal magnetic field.) The B energy level is defined as higher tlian the cxenergy level. The energetics of the two levels depends on the strength of theerternal magnetic field and the magnetogyric ratio of a particular nucieus. This:neans that the energy gap between the two levels also clepends on the strength of:he external magnetic field. As the external tnug.r"ii. field increarui th"::equency of the EM radiation needed to flip the spin in"r"as"s proportionally.

\ny nucleus with an odd number of protons (Z number) or an odd number of:-ucleons (A number) has a net spin. what is meant by "spin" is that as the:-ucleus precesses, it generates a weak magnetic field (as etections do). Just as a:rarged particle in linear motion generates a radial magnetic field, a charged:article in rotational motion generates a linear magnetic field. when the atomic:,ucleus has an odd number of protons (or nucleoni), the spins cannot pair up to:ancel one another out. The result is that the nucleus has a net spin. In the cises-t 1H, 13c, and 19F, it happens that there are only two energy ievels associated, ',1th the spins, so they can be analyzed without complicationl-A nucleus such as-)-i has spin, but there are more than two energy states, so its NMR spectrum is-.o complicated to analyze conveniently.

-:. the absence of any surrounding electrons, all identical nuclei exhibit the same-:in and therefore require the same energy for excitation in an external magnetic:::ld. within a molecule, two identical nuclei may be in different eleclronic. r'ironments. As a result of the difference in their local magnetic fields, caused:-'' the moving electrons, they do not require exactly the same amount of energy:: excite the nucleus to a higher-energy spin state. This can also be viewed is

- cal magnetic fields altering the strength of the apptied external magnetic field- :eded to get excitation (spin-flip) at a set frequency for the EM radiation. NMR:-achines can be designed to vary the frequency of the radiation or vary the.::ength of the magnetic external field. The NMR graph we observe typically:=:ords changes in the magnetic field strength along the x-axis, so we tnint orl,\lR in terms of varying external magnetic field strength.

l:oton NMR (which uses the 1H nucleus) is the most common form of NMR andrkes advantage of the magnetic spin associated with the hydrogen nucleus. The

l.:cAT test-writers focus on analyztng the graphs produced uy lumran. There is:,lrimal 13cxuR on the MCAT. rn"e scaie rot tuNvn is set from 0 to 10 parts: =r million (pp*) of the total magnetic field of the machine. Just as an inchls an..;h because someone made it a standard unit of measurement, NMR is:easured in ppp of the external magnetic field, because that is the arbitrary.::.ndard. All lHNMR shift values are relative to a standard compounj,=:ramethylsilane ((H3c)4si). All twelve of the hydrogens on tetramethylsilane.:e equivalent, so they absorb at the same value. This value is arbitrarily';signed to be 0 ppm, and all shift values are referenced against it. Rather than.- - Lnto other intricacies of NMR, we shall concentrate on how to read the graphs.

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Organic Chemistry Structure Elucidation Spectroscopy and Analysis j

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Symmetry and NMR SignalsThe best place to begin NMR for the MCAT is with molecular symmetry. Based

on the symmetry of a molecule, you can determine the number of equivalenthydrogens that it contains. We will consider symmetry within different groupsof molecules, starting with the four six-carbon esters shown in Figure 2-30'

Figure 2-30

The first two esters, n-butyl acetate and sec-butyl acetate, each have six uniquecarbons, of which five hav-e hydrogens. gach exiribits five signals in its 1HNMR

spectrum and six signals in its 13CNMR spectrum. Isobutyl acetate has fiveunique carbons, but only four contain hydrogens. This means that isobutylacetate exhibits four signals in its I HNMR spectrum and five signals in its13CNUR spectrum. Tertbutyl acetate has four unique carbons, but only twocontain hydrogens. This means that tertbutyl^acetate exhibits only two signals inits 1HNMR spectrum and four signals in its IsCNMR spectrum. The presence o{

only two signals in an NMR spectrum makes it easy to identify tertbutyl acetate.

The comparison of symmetry between isomers is highly useful, particularly withbenzene derivatives. Figure 2-31 shows three structural isomers of methylanisole, each of which has different symmetry.

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Para-methylanisole has six unique carbons, of which only four containhydrogens. This means that para-methylanisole has four types of hydrogens and

Copyright O by The Berkeley Review t26 The Berkeley Revieu


therefore four signals in its proton NMR. Meta-methylanisole has eight uniquecarbons, of which six contain hydrogens. This means that meta-meiirylanlstlehas six types of hydrogens and therefore six signals in its proton NMR. ortho-methylanisole also has eight unique carbons, of wtrich six contain hydrogens.This means that ortho-methylanisole also has six types of hydrogens"andtherefore six signals in its proton NMR. The structures in Figur e-2-Br"are para-,meta-, and ortho-methylanisole, respectively.

You may have noticed that many problems are often just variations on a singletheme. In Figure 2-32, compounds with comparable i\MR readings are shoirnside by side to demonstrate similarities in the distribution of lheir uniquehydrogens. Figure 2-32 shows two sets of three isomers that can be distinguishedirom one another using IHNMR by simply looking at the number of

"signals.

Butanol and pentanal each have five unique types of hydrogens. The hydrigensare in exactly the same ratio on both compounds, so thelr lHNMR spectra

"r,f,6itstrong similaritie^s. Each has five signals in its 1HNMR spectrum, ilthough theyiave different 13cxMR spectra arie to a different nrr-L", of unique ciruor,r.rhe

.1 HNMR spectra of the two compounds can be distinguished from one

another by the shift values of the respective signals. Methylpiopyl ether and.2-lentanone each have four unique types of hydrogens. rhe hydrogens are inexactly the same ratio on both compounds, so their 1HNMR spectra exhibitstrong similarities. Each has four signals with the same relative area in their-HNMR spectra, but at different shift values. Diethyl ether and 3-pentanone eachhave mirror symmetry and thus have similar carbons and similar hydrogens dueio this symmetry. There are two unique types of hydrogens in both diettyl ether:ld 3-pentanone. This means that diethyl ether and 3-pentanone have only two'ignals in their 1HNVn.

a

Butanol

./o-- -CF:. -CHzH CH, . CH; e-.r b- d-

Fir-e unique hydrogens labeleda-e irr a 7 :2 :2 :2: 3 ratio.

Methylpropyl ether

H,C/ o-a*t,-tF'-

a".d r b. -;'j

Four unique hydrogens labeleda-dtna3 2:2:3ratio.

Diethyl ether

F'c-cni o-..,; tF'bb

- rvo unique hydrogens labeledaandbina3:2ntio.

Pentanaloil

T-t-.f {'!'-cu1cF:Five unique hydrogens labeled

a-e in a 1 : 2 : 2: 2 : 3 ratio.

2-Pentanone

oil

-C. CH.H'rCt - CHr' t '\ CH.a" b' d'

Four unique hydrogens labeleda-dina3:2:2:3ratio.

3-Pentanone

flT"- c+{c-ctlr-cYu

b

Two unique hydrogens labeledaandbina3:2rctto.

Figure 2-32

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The comparison of butanol to methylpropyl ether and diethyl ether is similar tothe comparison of pentanal to 2-pentano-ne and 3-pentanone. For instance, the

presence of only two signals in the t HNMR spectrum makes it easy to

distinguish 3-pentanone from 2-pentanone and pentanal in the same way it is

easy to distinguish diethyl ether from methylpropyl ether and 1-butanol. Youmay see this theme repeated several times, so it is better to know basic trendsrather than specific examples.

In the case of alcohols, such as butanol, the protic hydrogen can be distinguishedfrom other signals by its broadness. Broadening results from hydrogen-bondingin solution. Hence, alcohols are easily distinguished from ethers by the presence

of a broad. peak in their 1HNMR spectrum. In the case of NMR, the local

environment of equivalent hydrogens undergoing hydrogen-bonding is notequal, so they appear at slightly different shift values. The degree of hydrogen-bonding varies, so the effect is also varied, causing the signal to be a broadened.Hydrogen-bonding causes the broadening of peaks in all types of spectroscopy.Broad peaks are a dead give-away for protic hydrogens.

These examples were designed to look at symmetry within a molecule. You willdo this over and over throughout the spectroscopy section. The key to predictingan NMR pattern for a compound is to understand the symmetry of the molecule.You must be able to identify unique hydrogens and then determine theirrespective features. This is where we shall start our analysis. The features weshall focus on primarily are the integration of the peak, the splitting pattem(shape) of the peak, and the shift value (measured i. pp*) of the peak. Be sure

that you understand the importance of each of these features and the factors that

can produce changes in them.

lrer,-i illlf,*::

lnteg:-ra n'^'' t---- L

-^-! !:_.AL L

: ack-r,

_5OU:eithershou-lihvdro,

Copyright @ by The Berkeley Review The Berkeley Revieu Copr:


1H Nuclear Magnetic Resonance

P-":iy:: 3-pentanone has two unique types of hvdrogens in a B:2 ratio, itslHNMR spectrum has two signals with relative areas o{g,2. Figure 2-33 showsthe signals from the 1Flrultn spectrum of 3-pentanone. Each p"ui i, explained interms of its splitting, integral, and shift value. The unique hyhrogens aie labeledin the same fashion as they were in Figure 2-32.

tra\

r)

llzc\ .zCHz

CH. CH"a' h'Being next to a C=O group yields a shiftvalue between 2.0 and 2.5 ppm.Being,next to a CH3 groupyields a quartet.

Being next to a CH2 group yields a shiftvalue between 0.9 and 1.5 ppm.Being next to a CH2 group yields a triplet.

1 ppm o pp-

Figure 2-33

The zero reference is ignored for analytical purposes, because it is there just to setlhe scale correctly. The integral is not drawn on the spectrum in this example. Inntost cases, you will be provided with the relative areas of the peaks, or you will:e given a summation line to evaluate the relative areas. Either way, you must:e able to apply the relative areas of the peaks to the quantity of hydrogens thateach peak represents. This is the start of NMR analysis.

Spectrum AnalysisiVe shall start off with how to anaryze the three basic components of the graph::'r.tegral (determined by the number of hydrogens making up a signal), splitting:attern (derived from the coupling between hydrogen neighbors), and shift aaluedetermined by the local magnetic field caused by either lone pair electrons in

rrotion or the electronic density associated with electronegative atoms). Eachpiece is equally important. At times, one piece of information may be a littlemore enlightening than the rest, but on the whole, every bit of data counts.

IntegralThe peaks for a signal can be integrated, meaning that the area under the curve:an be summed up, and set directly proportional to the number of hydrogens:hat the signal represents. For instance, a CH3 group has a signal with a relitivearea of 3 compared to a CH2 group with a signal of relative area 2, workingrackwards from the integration to the structure, it is possible to deduce theSroup from the integration. For instance, a relative area of 5 can be attributed toeither 2 equivalent CH3 groups or 3 equivalent CH2 groups. Further inspectionshould reveal which of the two scenarios is responsible for the six equivalenthydrogens.

11 Copyright O by The Berkeley Review 129 Exclusive MCAT Preparation

Organic Chemistry Structure Elucidation Spectroscopy and Analysis (

Splitting PatternThe splitting pattern, also referred to as coupling, corresponds to the number ofhydrogens on a neighboring atom. Like electrons, nuclear particles have spinthat can be classified as either up or down. The magnetization caused by thenuclear spin of hydrogen can be felt by the hydrogens on a neighboring atom.Because the spin can be either of two ways, the magnetic field may be additive orsubtractive. The random distribution of spins is used to determine the number ofhydrogen neighbors a group has. For instance, the CH3 group in 3-pentanone(labeled with an a in Figure 2-33) is next to a C}{2 (labeled with a b in Figure 2-33). The two hydrogens of the CH2 have one of four possible spin combinations:up/up, up/down, down,/up, or down,/down. Every CH3 group next to anup / up CH2 group has a slightly higher signal, while every CH3 group next to adown/down CH2 group has a slightly lower signal. Every CH3 group next to anup/down or down/up CHZ group has a normal signal, because the oppositespins cancel each other. The result is that one out of every four times, the CH3signal is slightly higher, two out of every four times the signal is unaffected, andone out of every four times the signal is slightly lower. This is why the CH3signal in 3-pentanone occurs as a triplet (in a 1 : 2 : 1 ratio).

Likewise, the CH2 group of 3-pentanone is next to a CH3. The three hydrogensof the CH3 have eight possible spin combinations: up / up / up, down/down/ up,dow n / up / down, up / down/ down, up / up / down, up,/ down / up, down / up / ap,or down/down/down. If all three spins are up (up/up/ up), then the net spin is+3/2. If only two spins are up, then the net spin is +7/2. There are threecombinations where two spins are up and one is down (up/up/down,up / down/ up, down/up / up), so this is three times as frequent as the all spin upcombination. The same thing can be done for the one spin up combinations(down/down/up,down/up/down, upldown/down) and the all spin downcombination. The result is that a quartet is found to be in a 1 : 3 : 3 : 1 ratio.

This is why the CH2 signal in 3-pentanone occurs as a quartet (1 ; 3 : 3 : 1). Thereare eight outcomes, but three of them share one value and three of them shareanother value, so we see only four different outcomes. Working from a spectrumto a structure, it is possible to say that a 1 : 3 : 3 : 1 quartet is the result of thehydrogens on a carbon being next to three equivalent hydrogens, often due to thepresence of a CH3 group as the neighbor. To determine the ratio of the peakswithin an overall signal (like the 1 : 3 : 3 : 1 value for the quartet), you can usePascal's triangle for binomial expansion to get the relative area of each peakwithin the signal. Table 2-3 shows Pascal's triangle along with a briefexplanation of what the relative numbers are expressing about the shape of thepeak and the abundance of the signal. As the relative amount gets smaller, it isharder to distinguish a peak from noise in the baseline signal.

)r_

t'rl:

'1:

:-:lll,::

::,L

iir:

It"yril

rit,r

liIp'llri

:r:

-:

.- :.-

ir

-*

It

tl

rt

li'

fr,

:-.j'.

i'

Neighbors Signal Shape Pascal's Triangle Ratio of Peaks in Signal

0Hs Sinslet 1 1 peak

1H Doublet 11 2peaks:L:1ratio2Hs Triplet t27 3 peaks: 1 :2: 1 ratio

3Hs Quartet 1331 4peaks:1:3:3:lratio4Hs Quintet 74641 5 peaks: 1, : 4: 6: 4 : 1 ratio

5Hs Sextet 1 5 1010 5 1 6peaks: 1 : 5 : 10 : 10 : 5 : 1 ratio

6Hs Septet 1 6 152075 6 1 Tpeaks: 1 : 6 : 15 : 20 : 15 : 6 : 1 ratio


Table 2-3

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Shift ValueThe shift value is a diagnostic tool for assessing the locai electronic environment.it is measured in parts per million (ppm) relative to the magnetic field necessaryto detect a standard compound, tetramethylsilane ((H3C)aSi). It is used inconjunction wiih known shift values to determine what functional groups areneighboring. In the 3-pentanone example, the CH2 group adjacent to thecarbonyl group feels the magnetic field of the electrons on a neighboring oxygenand thus requires a stronger external magnetic field to energize the spin levelsthan a CH2 group that is next to an alkyl chain. A larger shift value in thespectrum is thus observed than is typically observed for CH2 groups inhvdrocarbons. This is referred to as being shifted downfield, which indicatesrigher ppm values for the shift. To verify the presence of a C=O (carbonylgroup), consult Table2-4 to find that a value of 2.1 to 2.5 ppm is expected for ar-rvdrogen on a carbon adjacent (alphn) to the carbonyl. Table 2-4 lists many.orrunon lUNVR shift values used for analyztng spectra. The bold hydrogen ineach compound in Table 2-4 is the one of interest (to which the shift vaue;orresponds) and the shift value is measured in units of ppm.

Hydrogen Atom D (ppm)

RCHc 0.8 - 1.0

RCH2R (acyclic) 1.3 - 1.5

RCH2R (cyclic) 1.5 - 1.8

R3CH r.5 - 2.0

R2C=C(R')CH3 1.8 - 2.2

RCOCH3 (ketone) ) 1 _) q

,\rCH3 2.2 - 2.5

RC=CH 2.5 - 2.6

R.OCH3 (ether) 3.5 - 4.0

RCH2X (X = Cl, Br,I) 3.0 - 3.8

RCO2CHa (ester) 3.5 - 4.0

Hydrogen Atom d (ppm)

R2C=CH2 5.0 - 5.8

RCH=CRz 5.2 - 6.4

RNH2 1-3RNHCH3 2.0 - 3.2

ArNH2 3.5 - 5.0

RCONHR (amide) 5-9ROH (alcohol) 1 - 5 (broad)

AIOH (phenol) 4 - 7 (broad)

ArH (benzene) 7.0 - 7.4

RCOH (aidehyde) 9.0 - 9.8

RCO2H (acid) 10 -72

Table 2-4

Erample 2.15lentanal can be distinguished from 2-pentanone by what lgNIrrtR feature?

\. A 3H triplet at 1.5 ppmB. A 2H multiplet at 1.8 ppmC. A 2H triplet at 2.3 ppmD. A 1H triplet at9.7 ppm

Solutionlentanal has an aidehyde hydrogen, while 2-pentanone does not. The shift value:-r an aldehyde hydrogen is found between 9.0 and 9.8 ppm. The peak shape is a:rp1et, because there are two equivalent alkyl hydrogens on carbon 2. The two- .'-drogen nerghbors couple with the aldehyde hydrogen to split it into a triplet in:e lHNMR spectrum. This makes choice D the best choice. 2-pentanone could:e distinguished by the singlet of relative integration 3 caused by the isolated:'.ethyl group adjacent to the carbonyl. Because the carbonyl carbon has no:.','drogens attached, the methyl group of carbon one is not coupled to any other

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hydrogens, which makes it appear as a singlet in the 1HNMR spectrum. The twostructures are drawn below, along with their proton NMR features.

Ilm&Afr(Il

f,lfinffin

GTgM

ffim

lffi

-ftWhlLnrfnlfrl[ffitfrragffir

n:l[rfi.nfl,n"

&MmM

ilhstuffifiF

o

il

o

lt(-

.-/*\ b c d e

H CH2CH2CH2CH3

Aldehyde hydrogens show signals

around 9.7 ppmin the 1HNMR.

(-

^ ,./-\ b c cl

H:C CH2CHzCH3

Because there are no hydrogens on

the neighboring carbon, the methyl

group is a singlet in the 1HNMR.

Example 2.16\A/hat signals are present in the lIlUVn spectrum of chloroethane?

A. A downfield doublet and an upfield tripletB. A downfield triplet and an upfield doubletC. A downfield triplet and an upfield quartetD. A downfield quartet and an upfield triplet

SolutionChloroethane has two unique types of hydrogens. This results in two signals inits 1HNMR spectrum. The two hydrogens on carbon 1 are split into a quartet bvthe three hydrogens on carbon 2. Equally, the three hydrogens on carbon 2 are

split into a triplet by the two hydrogens on carbon 1. The lines split according tothe neighboring hydrogens and project down to the spectra. The quartet is

farther downfield than an ordinary alkyl group due to the electron density on thechlorine atom. This means that the triplet is upfield, making choice D the bestanswer.

There are two types of H, so there are two 1HNMR signals.

CIH\^ -.Lu

nttuTt-t\

are coupied to the three F{t i i i I

on the carbon 2, resulting , I I

in a quartetdownfield. i i I i

l]lttttt

H

on carbon 1, resulting in atriplet upfield.

Copyright @ by The Berkeley Review 132

Actual srectrum1

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Erample 2.17lentanol can best be distinguished from ethyl propyl ether by which of the:,rllowing features in the lHNMR spectrum?

\. A 3H triplet at 1.2 ppmB. A 2H triplet at 3.5 ppmC. A broad 1H peak between 1.0 ppm. and 5.0 ppmD. The total number of signals in the ether is substantially less

Solutionin alcohol is distinguishable from an ether by its broad peak between 1.0 ppm.:rd 5.0 ppm, so pick C for best results. The peak is broad due to the hydrogen-:onding within the alcohol. The broadening of the peak makes it difficult to=-,-aluate the integration of the alcohol hydrogen peak. Both structures contain a:;{ iriplet at1.2ppm, so choice A is eliminated. Both structures contain a 2H:rplet at 3.5 ppm, so choice B is eliminated. Choice D is eliminated, because--:.ere is a small difference in the number of peaks between the two compor-rnds.

E-rample 2.1.8',',hich of the following is a common feature in the 1HNMR spectra of all methylretones?

\, A triplet at 1.5 ppm (3H)B. A quartet at 2.3 ppm (2H)C. A doublet at 2.3 ppm (3H)D, A singlet at2.7 ppm (3H)

Soiution- methyl ketone has an isolated methyl group neighboring the carbonyl carbon'.;hich has no hydrogens). Having no neighboring hydrogens makes the peak a.nglet. The protons on a carbon alpha to a carbonyl are found between 2.0 andI 5 ppm. The signal has a relative integration of 3 hydrogens. The only answer:hat fits this is choice D, a 3H singlet at 2.1 ppm. The best choice is choice D.

Example 2.19

-- monosubstituted benzene has which of the following in its IUXVIR spectrum?

\. A peak at 1.2 ppm (5H)B. A peak at 5.3 ppm (5H)C, A peak at 7.2 ppm (5H)D. A peak at 8.1 ppm (5H)

Solution\lonosubstituted benzenes have a single peak around 7.0 ppm. The aromaticl'drogens appear as one singlet, despite the fact that they are not all equivalent:r'symmetry. The key to this question is not the integration or the peak shape,:ut the shift value. Choice D is just a little too high, so it is eliminated along with:hoices A and B, which are far too low" To do this problem quickly, you should:e familiar with the common peaks. Choice C is the best answer.

Be aware of certain peaks and features that occur over and over. For instance,,vhenever you see a triplet and quartet in a 3: 2ratio, you should conclude thatthere is an isolated ethyl group (H3CCH2-) in the molecuie somewhere.



tr\lhenever you see a doublet and a septet in a 6 : 1 ratio, you should concludethat there is an isolated isopropyl group ((H3C)2CH-) somewhere in themolecule. Rather than looking at molecules to determine the spectra (going fromstructure to spectrum), it is important to work problems from the spectrum to thestructure. By recognizing the combination of peaks, you will save time indetermining the unknown structure. This is very conunon in NMR spectroscopy.

We will use symmetric structures at first and then move on to more difficultexamples. The MCAT has traditionally asked simple questions about this topic,but it's better to be safe than sorry, so we will present examples that are harderthan the questions they have given on previous exams. In the following fewquestions are sample spectra from which you must determine the correspondingstructures. It helps to solve for the units of unsaturation first. Once these areknown, deduce possible functional groups that fit both the heteroatoms in theformula and the calculated units of unsaturation. For instancet zero units ofunsaturation and one oxygen can be an aliphatic ether or an aliphatic alcohol.Take advantage of the multiple-choice format by eliminating wrong answers asyou come across them. In the case of a compound with zero units ofunsaturation and one oxygen, an answer choice of a ketone is eliminatedimmediately. Any structures with rings or n-bonds should be eliminated. Thisability to eliminate wrong answers can be very useful in the multiple-choiceformat. To gain both insight and experience, try the following spectral problems:

Example 2.20\A/hat is the name of the compound that has the following 1I-nuraR spectrum, andwhose formula is C7H14O?

!

I1

ft

J

I(t

$W

Tffi

ffi&&

-frmil"l.c.D.

&Wtuffi:-frmm

ffiF

A. 2,4-dimethyl-3-pentanoneB. 2,2,4,4-tettamethyl-3-pentanoneC. 1,1,3,3-tetramethyl-2-propanoneD. 2,2- dimethyl-3-pentarone

SolutionThe septet and doublet in a 1 : 5 ratio are a dead give-away for an isopropdgroup. Choice D is eliminated, because it does not have an isopropyl group"Choice B is eliminated, because it contains too many carbons (nine instead od

seven). Choice C is eliminated, because the structure is misnamed. The bes,t

answer and only remaining choice is A.

Copyright O by The Berkeley Review 134 The Berkeley Revieu


Example 2.21

.The s-hape-of the signal at 2.3 ppm in the luNvtR spectrum in Example 2.20 is

best described as:

A. a quartet.B. a sextet.C. a septet.D. an octet.

Solutionlve counting seven apexes within the signal at2.3 ppm, and seven apexes (peaks)is referred to as a septet. check to see whether the ratio is 1 : 6 :rs :20: 15 : 6 : 1to be sure. This seems reasonable, so the best answer is choice C. If you alreadyknow the structure, then you can see that the septet results from six equivalenthvdrogen neighbors on the methyl groups neighboring the alpha carbon.

Example 2.22The ratio of the areas under the peaks within a quartet is:

-{.7:2:2:1..B. 1 :3:3: 1.

C. 2:5:5:2.D. 1:4:4:'J..

SolutionBv using Pascal's triangle, you can easily determine the ratio. It is a good idea toknow the ratios of the more common peaks such as a doublet, a triplet, and in--}'ris case a quartet. A quartethas a ratio of 1 : 3 : 3 : 1. Choice B is correct.

Example 2.23'r\'hat is the IUPAC name of the compound represented by the following lHNltRspectrum, whose molecular formula is C4H3O?

c4H8o

-{.. ButanalB. ButanoneC. Ethyl ethanoateD. Methyl propanoate

SolutionBecause the formula has only one oxygen, the two esters (choices C and D) areimmediately eliminated. The question is now reduced to determining whetherJre compound is an aldehyde or ketone. An aldehyde would have a peak in the

I Copyright O by The Berkeley Review 135 Exclusive MCAT Preparation


proton NMR between 9.0 and 9.8 ppm. There is no peak in that range, so choiceA is eliminated. Only choice B remains. To be certain, butanone(CH3COCHZCHS) has three types of hydrogens and thus three peaks in itsproton NMR spectrum. The peaks are a singlet (3H), a quartet (2H), and a triplet(3H). This is what the spectrum shows, so choice B is correct.

Example 2.24\iVhat is the IUPAC name of the compound represented by the following lHirll,tRspectrum, whose molecular formula is C4H3O2?

H

2 ppm 1 ppm

A. ButanalB. ButanoneC. Ethyl ethanoateD. Methyl propanoate

SolutionThis question is similar to the previous question, except the ketone and aldehydeare eliminated, because there are two oxygen atoms in the molecular formula.Choices C and D show identical peak shapes and integrals in their 1HNMR

spectra" The distinguishing feature is the shift value of each signal. The ethylethanoate (structure shown on the left below) exhibits a quartet near 4.0 ppm,making choice C correct. The structure of methyl propanoate is shown on theright below.

IE"r

-.:n

"4tffi

C.lL-il

50-.][jr:

tat

l,/{1/,1

- :-:

i

"0,

ffiC

OO

,.4.,/tT'---,, T,t-.-A o/'\,T" o l'T' 1'Tt o I

3H Singlet 2H Quartet 3H Triplet 3H Triplet 2H Quartet 3H Singletat - 2.1 ppm at - 3.8 ppm at - 1.0 ppm at - 1.0 ppm at - 2.3 ppm at - 3.5 ppm

Srntr

-_E

fri j

trr0g:

*Ltxn:

I.EI:I:-*l_: a.

=! JIlCopyright O by The Berkeley Review The Berkeley Review


Example 2.25Hydrogens on a carbon adjacent to two equivalent CH2 groups show which typeof signal in a IHNMR spectrum?

A. A1:3:3:lquartetB. A1:4:4:lquartetC. A1:3:5:3:lquintetD. A1:4:6:4:Lquintet

SolutionHaving two equivalent CH2 groups adjacent to the site of interest results in atotal of four equivalent hydrogen neighbors. Four equivalent hydrogens split asignal into a total of five (4 + 1,) peaks. This makes the signal a quintet, whichaccording to Pascal's triangle (or binomial expansion of any sort) has a ratio of L :

1:6:4: 1. The best answer is choice D.

Example 2.26\Vhat is the common name of the compound represented by the followingHtWtR spectrum, whose molecular formula is C6H16O2?

-A.. Para-ethoxy phenol (H3CH2COC6HaOH)B. Ortho-ethoxy phenol (H3CH2COC6HaOH)C. Para-methoxy anisole (H3COC6HaOCH3)D. Ortho-methoxy anisole (H3COC6HaOCH3)

SolutionThe symmetry in the 1HNVR spectrum is associated with a structure that is also>r'mmetric. The only way to get two types of hydrogens on a disubstitutedlenzene is to have two equal substituents on benzene para to one another. Thiseliminates choices B and D. Based on the formula, this molecule has tworLethoxy groups para to one another on the benzene. All of the benzenehvdrogens are equivalent, which explarns why only a singlet is observed. Thebest choice is thus answer C. Choice A would exhibit more than two peaks, so itLs eliminated.



Recognizing Special Structural FeaturesRecognizing special structural features requires knowing some general shiftvalues (6-values) from memory. You should know that a carboxylic acidhydrogen falls in the 6 = 1.0 - 12 ppm range and that the signal is broad. Analdehyde hydrogen falls in the 6 = 9 - 10 ppm range, aromatic hydrogens fal inthe 6 = 7 - 8 ppm range, vinylic hydrogens fall in the 5 = 5 - 6 pp* range, alkoxyhydrogens fall in the 6 = 3.5 - 4pp- range, and alphahydrogens fallin the E=2-?.5 pp* range. Figure 2-34 shows a molecular structure and its correspondinglHNMR spectrum that includes many of these key peaks.

Figure 2-34

Be certain that you can match the peaks in the spectrum to the hydrogens in thestructure drawn above it in Figure 2-34. This can be done on the exam using achart of values if one is given, but it is not a bad idea to know the values frommemory.

Example2.27\A/hat is the IUPAC name of an unknown compound with the molecular formulaC3H6O, an IR absorption at 1722 cm-l, and three 1HNVR peak,; one at 9.7 ppm(1H), one at 2.3 ppm (2H), and one at 1.4 ppm (3H)?

A. Propanoic acidB. PropanalC. PropanoneD. Methyloxyrane

SolutionThere is an excess of information in this question beyond what is needed toaswer it. The one and only piece of information you need is the peak at9.7 ppm,which makes the compound an aldehyde. Pick choice B and move on quickly.

IDftm

@mm

ffio

.f'dualfu@

ffi,fillfl

m.

W

o 3H 3H 3H

oHa

b 11H 2Hc

o

2Hd

1H

A A1H

12 ppm 10 ppm 8 ppm 6 PPm 4 ppm 2 PP* o ppm

Copyright @ by The Berkeley Review The Berkeley Revier


Distinguishing Disubstituted BenzenesIntegrals teil us the number of equivalent hydrogens in a signal and are oftenemployed to determine the position of substituents on disubstituted benzenerings. Structures that are highly symmetrical have more equivalent hydrogensthan asymmetrical structures. A 1,4-disubstituted benzene ring (referred to aspara") shows the fewest peaks in the aromatic region of the spectrum of all

disubstituted benzenes, due to its mirror symmetry. Both a 1,2-disubstituted anda 1,3-disubstituted benzene ring (referred to as "ortho" and "meta" respectively)have four unique hydrogens in the aromatic region of the spectrum. Figure 2-35shows two sets of disubstituted benzenes, one set of three with identical groupsand another set of tlrree with two different groups on the benzene.

Case 1: The two substituents on benzene are equal:

X

;f.Hb

3 different Hsin a 1 :2:lratio

Case 2: The two substituents on benzene are not equal:

Ha

Hb

X

Ha

Ha

Ha

Ha

HaUrlb

2 different Hsinal:lratio

Y

Hd

{ different Hs ina1:1:1:lratio

X

All Hs areequivalent

2 different Hsinal:lratio

4 different Hs ina1:1:1:lratio

Figure 2-35

Figure 2-36

Hb

Hc

Ha

Y

:rara substitution is the easiest arrangement to distinguish of the three possible-.:uctural isomers, because it has a doublet of doublets. The dissimilar heights of-:s peaks can be attributed to a mathematical phenomenon whereby peaks, as--:.ev near one another, begin to coalesce. Figure 2-36 shows the aromatic region: a l HNMR spectrum if u puru substituted benzene ring, where the two

.:bstituents are nonequivalent. Para coupling is a highly recognizable feature.

Enlargement of the aromatic region shows that thesplitting is a doublet of doublets, corresponding to a

para-substituted benzene ring. X

Ha

Hb

7.8ppm 7.6ppm 7.4ppm 7.2ppm 7.0ppm

-;.pyright @ by The Berkeley Review

Ha

Exclusive MCAT Preparation


Example 2.28what is the common name of the compound that has the formula C6Hgo, an IRabsorption at7722 cm-], and three notable lHNMR peaks at9.7 ppm (1H, s), 2.3ppm (4H. dd), and 2.2ppm (3H, s)?

A. Ortho-methylbenzoic acidB. Para-hydroxyacetophenoneC. Ortho-methylbenzaldehydeD. Para-methylbenzaldehyde

SolutionThe compound has only one oxygen, so neither a carboxylic acid (methylbenzoicacid) nor a hydroxy ketone (hydroxy acetophenone) is possible. Choices A and Bare eliminated. we know that the compound must be an aldehyde from thechoices that remain, so the lriNvtR peak at 9.7 ppm and the lR absorption at1722 crn-ldo not help our efforts to identify the compound. The trqrulan signalat 7.3 ppm is a doublet of doublet (dd), which indicates para-substitution. Thismakes choice D the best answer.

Deuterated Solvents for 1HNMRBecause the solvent is in substantially higher concentration than the solute, it isimperative that the solvent not have any hydrogens. If the solvent has 1H nuclei.then it would produce the largest signal in the spectrum, eliminating integrationand causing the other peaks to disappear into the baseline. To avoid thisproblem, solvents are chosen that have deuterium (2H) instead of the standardisotope of hydrogen (1u;. one potential problem occurs when protic compoundsare dissolved into deuterated protic solvents, such as D2O. protic hydrogens canundergo exchange with the protons of the solvent, if the solvent is protic.Although the dissociation constant (Ku) may be small for compounds such asalcohols, over enough time all of the hydrogens can be released and then are ableto reform their bonds. rf D2o is present in the solution, then deuterium willgradually replace protic hydrogens capable of undergoing exchange. This causesthe signal for the protic hydrogen to disappear gradually.

Example 2.29which of the following compounds does Nor lose a luNIvtR signal after D2ohas been added to a solution containing it?A. Carboxylic acidB. Cyclic etherC. Primary amineD. Secondary alcohol

SolutionIf a compound contains a protic hydrogen, then it loses a peak from its 1HNMRspectrum when D2o is added to the solution. Primary and secondary amineshave a hydrogen bonded to nitrogen, so thev are protic. This eliminates choice C.All alcohols have a hydrogen bonded ,o s11:gen, so all alcohols are protic. Thiseliminates choice D. A cirboxylic acid hasl dissociable proton, so it readilyexchanges with deuterated water. choice A is eliminated. An ether, whethercyclic or not, has all of its hycirogens bonded to carbon, so it is aprotic. WhenD2O is added to an ether, no exchange transpires. The best answer is choice B.

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rli J&e"r

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Copyright @ by The Berkeley Review The Berkeley Review -:r:


13CNMR SpectroscopyCarbon-13 NMR focuses on the carbons of a compound rather than thehydrogens. The NMR cannot detect carbon-12, because it has no nuclear spin.However, carbon-13 has an odd number of nucleons (particles in the nucleus), soii has a nuclear spin. For that reason, NMR can be used to detect carbon-13.Because carbon-13 constitutes only about 1"/' of all carbon atoms, a 13CNMR

requires maly more scans to obtain a spectrum than 1HNMR. This is why the'raseline in 13CNMR spectra is noisy (scattered and messy). l3CNltR is used tordentify the number of unique carbons in a compound. A typical application oft'hrs technique is to distinguish the substitution of disubstituted benzenes. Figurel-37 shows the proton-decoupled (ali singlet spectrum) C-13 NMR spectra for-,2-dimethylbenzene, 1,3-dimethylbenzene, and 1,4-dimethyibenzene. What is:reant by "proton-decoupled" is that the compound is irradiated constantlysimilar in concept to Dolby noise^reduction) with the coupling energy, so thatiere is no coupling between the IJC and IH atoms. This makes all of the peaks

=inglets, as drawn.

ppm 140 120 100 BO 60

Figwe 2-37

llhe shift values for 13CNMR are simple to recall. They come in blocks of 50rpm. An sp3-hybridized carbon next to a carbon has a shift value between 0 and50 ppm. An sp3-hybridized carbon next to an electronegative atom has a shift

80

#CH"

2040

Copyright @ by The Berkeiey Review t4t Exclusive MCAT Preparation

Organic Chemistry Structure Elucidation Spectroscopy and Analysis jvalue between 50 and 100 ppm (usually less than 75 ppm). An sp2-hybidized,carbon next to a carbon has a shift value between 100 and 150 ppm. A carbonylcarbon with sp2-hybridization has a shift value between 180 u"a zso ppm. Theshift value increases as the substitution increases, meaning that 3' > 2" > f inshift value. Quaternary carbons do not show up on the 13cruvR very well, dueto their long relaxation times. \44-rat is meant by a "long relaxation time" is thatthe excited energy state of carbon takes longer than a few seconds to dissipate theenergy into solution and relax back to a lower energy level. We will not focus onrelaxation times, but instead will focus only on the application of informationextracted from the spectra.

In the three spectra shown in Figure 2-37, you'lrnote that the number of peaks ineach spectrum corresponds to the number of unique carbons in the compoundthe spectrum represents. For the ortho compound, there are four carbons in a 1 :

1:1:1ratio. Thespectrumshows fourpeaks ina 1:1:1:1ratio. Three of thepeaks are from sp2-hyb.idized carbons and one is from an sp3-hybridizedcarbon. For the meta-substituted and para-substituted compounds, a similarrelationship between spectrum and unique carbons in the structure is observed.

using spectroscopy data, you should be able to solve any problem following asystematic procedure. It is best to use l3ctlVR to get-the symmetry oI astructure and to identify selected functional groups (in a manner similar to IRspectroscopy). Don't make much more out of i3cNun spectroscopy than this.Be sure to use your degrees of unsaturation in structure elucidation problems!

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coplCopyright @ by The Berkeley Review t42 The Berkeley Review

Organic Chemistry Structure Elucidation Section Summary

Key Points for Structure Elucidation (Section 2)

lsomerism

1. Isomers (compounds with the same type and number of atoms but differentspatial arrangement due to bonding, connectivity, or molecular contortion)a) Structural isomers (isomers with different connectivitv because of

different bonding)i. Have different IUPAC namesii. Can be ciassified as skeletal isomers, positional isomers, or functional

isomersiii. Their number can be determined by evaluating possible chain

lengths and connectivity

b) stereoisomers (compounds with the same bonding, but different spatialarrangement)

i. Can be classified as configurational isomers (geometrical and optical)or conformational isomers

ii. Have same IUPAC root, but a different prefixiii. Conformational isomers are formed by rotating or contorting a

structure (leading to eclipsed and staggered conformations, withgroups gauche and anti to one another)

iv. Maximum number of possible optical isomers is 2n, where n is thenumber of chirai centers in the compound

Nlewman projections (front view of moiecule)

Cyclic moleculesi. Three- and four-membered rings have ring strain that makes them

highly reactiveii. Five- and six-membered rings are stable, with six being the more

stable of the twoiii. Cyclohexane (and

conformation, withaxial

Structural Insights

1. Structural symmetrya) Plane symmetry (mirror plane in molecule splits molecule into equal

halves)

Point symmetry (molecule has an inversion pointUnits of unsaturation

at its center of mass)

i. Determined from excess bonding electrons divided by 2

ii. Unirs of unsaturation - 2(#C) + (#N) - (#H)- (#X) + 2

iii. Describes the number of n-bond., ur-ra .i-rg, in a molecule

Spectroscopy

1. IR spectroscopy (used for vibrational excitation)a) Ranges from 1000 cm-1 to 4000 cm-1 (about 3 kcai/mole to 10 kcal/mole)b) Correlates bond-stretching and bond-bending to absorbancec) Used to identify functional groupsd) Key peaks: C=O around 1700 cm-1, O-H around 3400 cm-1 (broad), and

C-H around 3000+ cm-1 (varies with hybridization-- ,p > ,p2 > ,p3j

c)

d)

six-membered rings in general) assume chairgroups equatoriai (more stable position) and

b)

c)


Organic Chemistry Structure Dlucidation Spectroscopy and Analysis

e) spectrophotometer uses salt plates to hold sample, because salt plateshave ionic bonds and therefore do not interfere rn'ith the simptemolecule's absorbances

2. Ultraviolet-visible (uv-vis) spectroscopy (used for electronic excitation)a) Ranges from 200 nm to 800 nm, increasing in wavelength as conjugation

increases

Typicallv used for analyzing compounds with n-bonds, especiallyconjugated systems

Peak intensity and wavelength increase as the amount of conjugationincreases

3. NMR spectroscopy (the basics of I HNMR analysis)a) IUXVIR Integration (Quantitative analysis using relative area under the

curves)i. Area under the curve for each signal is proportional to the number of

hydrogens responsible for the signalii. Connectivity can often be deduced from the integration ratio

b) lUXUR peak shape (coupling and J-values)i. The number of peaks within a signal equals the number of

neighboring hydrogens plus 1

ii. Hydrogens coupled to one another have the same j-valuesiii. The ratio of the area of the peaks within a signai can be determined

using Pascal's triangle

c) lUNVtn shift value (eiectron-rich environments affect shift values byexerting a magnetic field)i. Common signals include 9-10 ppm for an aldehyde and around 7

ppm for hydrogens on benzeneii. "Upfield" refers to shifts at lower ppm valuesiii. All shifts are referenced against si(CH3)a, which is assigned a value

of 0 ppm

d) 1HrultR special features (effects of deuterium and structural symmetry)i. Exchanging of deuterium for protons (peak disappearance)ii. Para substitution pattern (symmetric benzenes have unique spectra)iii. Solvent choice (solvent must be invisible)

e) 13CXUR and symmetryi. The number of signals in 13CNVR spectrum is equal to the number

of unique carbons in the moleculeii. Alkene carbons have signals between 100 ppm and 150 ppm, while

carbonyl carbons have signals between 180 ppm and 230 ppmiii. 13C is a rare isotope, so 13CNMR spectra require more scans and

have more noise in their baseline than IUXVn spectra

b)

c)

Copyright @ by The Berkeley Review r44 The Berkeley Review

IStnrctureElucidation

Passages14 Passages

l OO Questions

Suggested schedule:I: After reading this section and attending lecture: Passages I, V, VI, VIII, & XI

Qrade passages immediately after completion and log your mistakes.

II: Following Task I: Passages III, IV XII, & Xlll (27 questions in 55 minutes)Time yourself accurately, grade your answers, and review mistakes.

III: Review: Passages II, VII, IX, X, XlV, & Questions 98 - lOOFocus on reviewing the concepts. Do not worry about timing.

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Speci altztng in MCAT Preparation

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I. Chair Conformation

II. Rotational Energy Diagrams

III. Deuterated Cyclohexane

IV. Infrared Spectroscopy

V. Unknown Alkene Determination

VI. Structure Elucidation

VII. NMR and IR Spectroscopy

VIII. NMR Data Table

IX. Alkene-Coupling Experiment

X. Unknown Compound Identification

XI. Carbon-l5 NMR

XII. Distinguishing Isomers Using lnnmR

XIII. Structure Elucidation Using lHnMn and IsCNMR

XIV. Proton NMR of an Unknown


Structure Elucidation Scoring Scale

Raw Score MCAT Score

84 - 100 15-1566 B5 lo-1247 65 7 -9

34-46 4-6t-33 t-3

(r -7)

(B - 14)

(r5 - 22)

(23 - 2e)

(50 - 56)

(37 - 43)

(44 - so)

(5r - 57)

(58 - 64)

(65 - 71)

(72 - 78)

(7e - 84)

(B5 , eO)

(er - e7)

(eB - l oo)

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Passage I (Question 1 - 7)

Cyclohexane is not a planar molecule, but in its moslstable conformation, fbur of the six carbons that make up thering are coplanar. Studies using I HNMR and X-ray:rystallography demonstrate that the most stable:onformation of the molecule has carbon-carbon-carbon bond

-rngles of approximately 107.5' and 1l l' in the ring and that.nere are two types of hydrogens present, axiaL and equatorial.The axial hydrogens are bonded directly above and directly:elow the ring carbons. The equatorial hydrogens lie away:rom the cyclohexane ring.

A low-temperature lHNVR stucly was conducted to::termine the equilibrium constant for the conversion from:i axial- 1,2-dimethylcyclohexane to diequatorial- 1,2-dimethyl-:..clohexane by way ol a ring-f'lip plocess (Ksq-1,2). This.llue is directly comparable to the equilibrium constant fbr::e conversion fiom diaxial-1,4-dimetl-rylcyclohexane to

-.equatorial-1,4-dimethylcyclohexane by way of a ring-f1ip:rocess (Keq-l,a), given that they are both trcns-substituted:,,clohexanes. The difference between their equilibrium, rnstants is attributed to the gauche and anti orientations: -,ssible with the methyl substitucnts at the 1,2 positions.j:udies have shown that lalger substituents prefer the.:uatorial orientation of the so-called chair conformation.

.qure I summarizes the findings of the study.

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H H,c{{-/--_/ \

t\eq = ll'J I

*-H.

9H..

s,cG4,' \--l-{/'cH\

K"o+ +.Zl

, .l:Figure 1 Ring-flipping of trans-dimethylcyclohexane

The values fbr the equilibrium constants can be appliedJetermine the relative steric hindrance of one substituent

, rpared to anolher. A bulkier group exhibits greatet' steric" rdrance, so the equilibrium lies more towards the more.rle of the two possible chair conl'irrnations. Hence, a

.-3ater equilibrium constant irnplies that there is a greater

---ee of steric hindrance in the less stable confblmation.

I . Which of the following is the most stable orientation ofa substituent on a cyclohexane molecule?

A , Axial orientation ol a chair conlormation

B. Equatorial orientation of a cl.rair conlotmation

C . Bridge orientation of a boat conlormation

D . Oar orientation of a boat conlormation

-:pyright @ by The Berkeley Review@ 147 GO ON TO THE NEXT PAGE

t A carbon-deuterium bond is shorter than a carbon-l-rydrogen bond. Using this idea, how many deuteriumatoms assume axial orientation in the most stableconformation of the following molecule?

A. 0

B. 1

c.2D. 3

3 . What is the value of K.O- 1,4 for the conversion oftrans-1,4-dimethylcyclohexane from its diaxialconformation to its diequatorial conformation?

A. 0.0029

B. 2.16

c. 4.31

D. 345

4, Cis-i,4-dimethylcyclohexane, in its most stable chairconformation, CANNOT have which of theseinteractions?

A. CH3/H gauche

B. CH3/H anti

C. CH3/CH3 gauche

D. H/H anti

5 . Which of the following accurately describes the value ofK"O for the conversion from one chair conformation [o

the other chair conformation for the compound cis-t,2-dimethylcyclohexane?

A. Less than or equal to 0

B. Greater than 0 and less than IC. Equal to 1

D. Greater than 1

6. The value of K"O for the conversion of cls-1,3-dirnethylcyclohexane from its diaxial conformation to

its diequatorial conformation is:

A . less than 0.22.

B . between 0.22 and 1.

C . between I and 4.3 i.D . greater than 4.31.

7. Which three-dimensional conformation corresponds tothe 3-hydroxy-cls-decalin, shown below?

H

qlr"'

&Fffi*f=,"

OH

D.

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Passage ll (Questions 8 - 14)

Alkanes are hydrocarbons that contain all sigma bonds.Sigma bonds have linear electron density (electron densitythat is localized between the two nuclei of the bondingatoms). This allows for free rotation about a sigma bond.Rotation about sigma bonds is continually occurring attemperatures above absolute zero, although the rate of the

rotation varies. However, the rotation does not necessarilycomplete a full 360' cycle about the sigma bond.

Some conformations encountered during rotation are ofhigh energy (due to steric repulsion) and others are of lowenergy (due to minimal steric interactions). The most stableconformation occurs when the largest groups are as far apartas possible. When two groups are as far apart as possible,the conformation is referred to as staggered, and the bulkiestsubstituents are said to be anti to one another. The leaststable conformation occurs when the largest groups interferewith one another. This is known asfully ecLipsed. Drawn inFigure I is an energy diagram for the counterclockwiserotation about the C2-C3 bond fbr R-2-methyl-1-butanol.

g l8o 360

Degrees displaced from initial state

Figure 1 Energy during rotation about the C2-C3 bond

The three apexes occurring at 60', 180' and 300' on thegraph are not of equal energy. In 2-methyl-1-butanol, carbon2 is a stereocenter. Because of this asymmetry, none of the

eclipsed or staggered conformations are equal in energy. Al1

visual projections show asymmetric steric interactions.Although the molecule is constantly rotating about its bonds,it assumes its most stable conformation most of the time.

8 . Which of the following structures represents themolecule at the 240" point on the graph in Figure 1?

bob

+.1

C.

H:C. \CH2OH

sr,'HHJ\H CHr

B.

H.C CH.

Hf{,,cH:oH

D.

TF{::""

9. The 60'point on the graph in Figure 1 represents thestructure when it is:

A. eclipsed and the methyl substituent on carbon-2interferes with carbon-4.

B. eclipsed and carbon-l interferes with carbon-4.

C. staggered and the methyl substituent on carbon-2interferes with carbon-4.

D . staggered and carbon-l interferes with carbon-4.

10. Which of the following structures represents themolecule at the 330" point on the graph in Figure 1?

A.9Hr

H

CHzOH HOH2C

CH2OH

H

B.

CHc

I l. Which of the following statements BEST explains whythe MOST stable conformation of 2-amino-1-ethanol isgauche?

A. Hydrogen bonds are strongest when the twosubstituents have gauche orientation.

B. Hydrogen bonds are weakest when the twosubstituents have gauche orientation.

C. H is bulkier than NH2, due to the inductive effect.

D. H is bulkier than NH2, due to resonance.

[ 2. The strongest hydrogen bond occurs between which ofthe following?

A. A lone pair on nitrogen bonded to an H on nitrogenB. A lone pair on oxygen bonded to an H on nitrogen

C . A lone pair on oxygen bonded to an H on oxygenD. A lone pair on nitrogen bonded to an H on oxygen

H:C

-opyright @ by The Berkeley Review@ 149 GO ON TO THE NEXT PAGE

1 3. The reason there is no rotation diagram for trans-2-butene is that:

A. gauche is not favorable for alkenes.

B. anti is not favorable for alkenes.

C. steric hindrance does not affect alkenes.

D. rotation about a n-bond requires energy in excess ofroom temperature.

14. Which of the following compounds contains a singlebond about which complete rotation is not possible?

A . Dipropyl ether

B. 2-butanone

C. 2,3-butanediol

D. Methylcyclopentane


A researcher wishes to determine the relative stability ofaxial orientation versus equatorial orientation for deuteriumand hydrogen on cyclohexane. To do so, she treats benzene(COHO) with D2S04/D2O at 100"C for rhirry minutes tosynthesize monodeuterobenzene (COHSD), which is thentreated with H2 gas and palladium metal under 90 psi ofpressure to yield monodeuterocyclohexane (C6H11D). Formonodeuterocyclohexane (C6H11D), there are two possible

chair conformations, one with the deuterium having axialorientation and the other with the deuterium having equatorialorientation. At room temperature, the interconversionbetween the two chair conformations is too rapid to study andall eleven hydrogens appear as a singlet 1.38 ppm in theIHUUR. At low temperatures, the interconversion throughring-flip from one chair-conformation to the other is slowedgreatly, so that axial and equatorial hydr-ogens generatedifferent signals in the IHNMR. As a resulr, the ring-flipprocess can be monitored using IHNMR spectroscopy.

A IHNMR was recordecl at -89'C in deuterochloroformsolvent, A hydrogen with axiai orientation shows a lUNIrrtR

shift of 6 = l.5l ppm, while a hydrogen with equatorialorienrarion shows a IHNMR shift of 5 = i.25 ppm.Integration shows that the relative area of lHNIr,lR signals is1.12: l in favor of the 6 = 1.25 ppm signal. Because alarger amount of equatorial hydrogen is obser-ved than axialhydrogen, the deuterium has axial orientation in the morefavorable chair confbrmation.

The researcher proposes that a diff'erence in bond lengthbetween the C-H bond and the C-D bond, rather than a

difference in atomic size between hydrogen and deuterium,accounts for the equatorial preference of hydrogen overdeuterium. The difference in bond length is attributecl to thegleater relative mass of deuterium compared to carbon versusthe lesser relative mass of hydrogen compared to carbon.Because the center of mass remains constant when a bond isstretched, the greater difference in mass between hydrogen andcarbon than deuterium and carbon makes the carbon-hydrogenbond stretch more asymmetrically than a carbon-deuteriumbond. A carbon-hydrogen bond stretches more than a carbon-deuterium bond, and thus occupies a greater amount of space.Despite the differences in bond length, the bond angles incyclohexane remain similar, between 107.5' and I 1 l'.

15, The researcher reached the ultimate conclusion that a

bond between carbon and deuterium is shorter than a

bond between carbon and hydrogen, based on the tactthat:

A. the deuterium favors the equatorial orientation.

B. the deuterium f'avols the axial orientation.

C. the interconversion between the two possible chairconformations of the deuterocyclohexane moleculethrough ring flip is rapid a[ room temperature.

D. the lgNVtR shift at 6 = 1.25 ppm is fbrrher uplieldthan the IHNVIR shil'r ar 6 = l.5l ppm.


1 6. Addition of D2 gas and palladium metal instead of H2gas and palladium metal to deuterobenzene would haveshown what ratio of equatorial hydrogens to axialhydrogens in its most stable chair conformation?

A.6:58.5:2C.3:2D. 2:3

17. The addition of H2 gas and platinum metal rochlorobenzene (C6H5CI) leads to a product whose moststable conformation is:

A . boat with chlorine anti.

B. boat with chlorine gauche.

C . chair with chlorine axial.

D. chair with chlorine equatorial.

18. The D-C-H bond angle about the deuterated carbon is

closest to which of the following values?

A. 90'

B. 109.5'

c. 120'

D. 180"

19. The most stable form of cis- 1,3,5-trimethylcyclohexanehas the chair conformation with:

Hsc ctl 3

A. three methyl groups in the equatorial position andno methyl groups in the axial position.

B. two methyl groups in the equatorial position andone methyl group in the axial position.

C . one methyl group in the equatorial position andtwo methyl groups in the axial position.

D . no methyl groups in the equatorial position andthree methyl groups in the axial position.

2 0. How many units of unsaturation are there in C6H5D?

A. 3

B. 4

c. 5

D. 6

11. The reason that the diaxial orientation for cis-3-hydroxycyclohexanol (a cis-1,3-diol) is preferred overthe diequatorial orientation is that the hydroxyl groups:

A. are smaller than hydrogens, so they exhibit nopref'erence for the less hindered equatorialorientation.

B. are larger than hydrogens, so they exhibit a

preference for the less hindered axial orientation.

C. are larger than hydrogens, so they exhibit a

pref'erence for the more hindered equatorialorientation.

D. can form an intramolecular hydrogen bond from a

1,3-diaxial orientation, while they cannot form an

intramolecular hydrogen bond from the 1,3-diequatorial orientation.

I L Which structure represents the MOST stable form ofcis- 1,4-ethylmethylcyclohexane?

A. cH: B.

H,CN' >'--L CH:CH-1

C

HCh CH2CH3

CH2CH3 CH:

D.

CH2CHj

The Berkeley Reviewo l5l GO ON TO THE NEXT PAGE


In most research laboratories, fourier transform infrared(FTIR) spectrophotographers are used to obtain infraredspectra. The FTIR spectrophotographer works by passing an

electromagnetic pulse of multiple frequencies through a

sample and then collecting and analyzing outgoing radiation.The difference between the output signal and a referencesignal is digitized by computer and broken down into a set ofcomponent sine waves (this is the Fourier transform process).The signals are processed and recorded to yield the samespectra as those obtained using outdated variable frequency IRspectrophotorneters.

One advantage of the FTIR machine is that the wave'number for each signal is given precisely. A disadvantage isthat it is not possible to focus on one peak by using a

monochromatic light pulse. Focusing on one peak with a

monochromatic beam can be done in rate studies, althoughthe IR has a rapid shutter speed, faster than any reaction(including diffusion controlled reactions).

The IR information is most useful if certain peaks areunderstood. For instance, an O-H bond in a compound can berecognized by the broad peak it displays around 3300 cm-1,although the exact value varies with the degree of hydrogen-bonding. A carbonyl bond is found around of 1700 cm-1. Ifa carbon has spj-hybridization, the bonds it forms tohydrogen are found just below 3000 cm-l. All of thisinformation combines into a nice packet of data used todeduce the structure of a compound. Figure 1 shows the IRspectrum for 2-heptanone:

4000 3600 3200 2800 2400 2000 1800 1600 1400 1200 1000

Figure I Infrared Spectra for 2-heptanone

The information extrapolated from the IR spectra can be

coupled with NMR (nuclear magnetic resonance) data to forma powerful combination. For instance, aromatic hydrogensare found in the 7 to 8 ppm range in an 1HNMR spectrum.

23, An absorbance between 1700 cm-l and 1740 cm-1would NOT be present in:

A. ethyl propanoate.

B. butanal.

C. 2-pentanone.

D. diethyl ether.

24. The IR spectrum for a straight chain monosaccharidehas all of the following absorbance values EXCEPT:

A. 3300 cm-l.B. 2980 cm-l.C. 2300 cm-1.

D. 1715 cm-1.

2 5. Which of the following isomers of C3H6O2 exhibits a

broad IR signal around 2850 cm-l?

A.oB.

"oA.H2cH3 H3co

o

AcHs

c.o

H,cnrcoAn

D' ,4..oo

2 6. Which of the following pairs of compounds could bedistinguished by their splitting patterns in the protonNMR region between 7 and 8 ppm?

A . Methylpropanoate from ethylethanoate

B . 3-methyl-2-hexanone from 2-methyl-3-hexanone

C. l,4-methylphenol from l,4-ethylphenol

D . 1,4-methylphenol from 1,3-methylphenol

27 . To distinguish a tertiary alcohol from a primary alcohol(the tertiary alcohol exhibits more steric hindrance tohydrogen-bonding than the primary alcohol does), itwould be best to focus on which of the following IRfeatures?

A . The width of the peaks near 3300 cm-lB. The length of the peaks near 3300 cm-lC . The width of the peaks near 1700 qm-l

D . The length of the peaks near 1700 cm-1

Copyright @ by The Berkeley Review@ 152 GO ON TO THE NEXT PAGE

2 8. Which of the following isomers of C4H6O would NOThave an IR signal at 1715 cm-1?

A. 2-methylpropanal

B. Butanal

C. Butanone

D. Tetrahydrofuran

29. Hydrolysis of an ester could be supported by which ofthese IR spectroscopic data?

A. The appearance of a signal around 1700 cm-lB. The disappearance of a signal around 1700 cm-lC . The appearance of a signal around 3300 cm- 1

D . The disappearance of a signal around 3300 cm-l

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Passage V (Question 30 - 36)

An unknown alkyne with a molecular mass of 122.2-: mole is treated with H2lPd and BaSO4 to convert it into a

..s-alkene. The cis-alkene is isolated in high purity. The cis..rene is then treated with high-pressure ozone (O3) gas and

nc metal to convert both of the alkene sp2-hybridized:.rbons into carbonyl carbons. Because the cis-alkene was

, -ned by hydlogenation of an alkyne, it is disubstituted with=.:h alkene carbon holding one hydrogen. This means that-: products of ozonolysis are both aldehydes. Two unique- rducts are isolated from the product mixture. The two--known products are designated as Compound A (C3H6O)

--J Compound B (C6H1gO). An lHNIrrtR spectrum is,.ained for Compound B and is shown in Figure I below.

. -: lHNMR is cariecl out using CDCI3 as the solvent.

Figure I 1HNMR spectrum of Compound B

For Compound A, spectral data were obtained from an IR:::r,rum using pure Compound A in liquid form between" . -:lates. Table 1 lists the key IR absorbances collected for

-rfound A

Shilt(cm-11 Inrensir.y

2962 strong2912 medium2106 mediurn1126 strong1212 strong

all other peaks are irrelevant

Table 1 IR absorbances of Compound A

The structures of Compounds A and B can be deducedqreat accuracy from the spectral data in the passage.

-r engaging in structure elucidation, some information is: useful than others. As a general rule, NMR data are-:d last, as they have the most information.

-\ compound with one degree of unsaturation and twor\ygens CANNOT be:

A. a cyclic ketone.

B. a cyclic ether.

C. acarboxylicacid.D. an ester.

: . right @ by The Berkeley Review@ 153 GO ON TO THE NEXT PAGE

3 1. Which of the following I UNUR shifts wouldobserved for Compound A?

A. 12.0 ppm

B. 9.7 ppm

C. 7.5 ppm

D. 5.5 ppm

32. Which of the following IR absorbances wouldobserved for Compound B?

A. 1125 cm-1

B . 2200 cm-l

C. 1620 cm-lD. 3550 cm-l

3 3. What is the molecular formula for the original alkyne?

A. C9H14

B. C9H16

C. C9H1s

D. C9H26

34. Which of the fbllowing IR absorbance values isindicative of an alkene?

A. 3550 cm-lB. 2200 cm-1

C. 1"725 cm-lD. 1620 cm-1

35. Which of the following structuresCompound A?

A.oB.

t,"A.r, HicH2c

CHr

corresponds

o

A,,

OC.

3 6. Which of theCompound B?

lollowing structures corresponds to

A. '{x:,.o

oAC

Copyright @ by The Berkeley Review@ GO ON TO THE NEXT PAGE


An unlabeled bottle containing an unknown compound isfound in a lab storage locker. The compound is an odorlessliquid that does not evaporate rapidly when the bottle is leftuncapped. A lab technician labeis the bottle Compound T.Compound T exhibits three signals in its proton magneticresonance spectrum. The three signals are listed in Table 1.

shift Integration Shape J-Value

6 = 2.5 ppm (broad) 1H triplet 3Hz

6 = 3.1 ppm 1H singlet NA

6 = 4.3 ppm 2H doublet 3Hz

Table 1 1HNUR signals of Compound T

Shorthand for describing nuclear magnetic resonancespectra describes the chemical shift value (6), the number of

hydrogen atoms (relative area under each signal), and the

coupling along with the respective coupling constant, J.

Signal shapes can sometimes be described by single letters.such as s = singlet, d = doublet, m = multiplet, q = quarter-

and t = triplet.

For the unknown compound, Compound T, three bands

at 6 50 (0, 73.8 (d), and 83.0 (s) appear in the carbon-limagnetic resonance spectrum. The splitting in the l3CNMR

is due to the hydrogens directly bonded to the carbon. The

important absorbances in the infrared spectrum of CompouncT are found at 3350 cm-l (broad),2988 cm-l, 2116 cm-1 , anC

1033 cm-1. The spectral data can be applied to determine the

symmetry of a compound as well as the functional groups ot

the compound. The IR peak at2116 cm-l is indicative of a

triple bond, indicating the presence of either an alkyne or a

nitrile in Compound T. There is no spectral evidence tc

suggest that a nitrogen atom is present in the compound, slthe most logical assumption is that there is a carbon-carboritriple bond present in the structure.

From the spectral data, the lab technician concludes tha:

the compound contains no carbonyl functionality. However"

the compound contains a functionality that is involved irhydrogen-bonding, explaining the relatively slow evaporatior-

of Compound T. Using chemical tests and a polarimeter, the

lab technician determines that there is no mirror symmetry i:the molecule. This implies that the number of signals in thel3CNtvtR spectrum is also the number of carbons iiCompound T.

37. The shift at 1033 cm-l in the IR is caused by the C-C

bond of:

A. an ether.

B. an ester.

C. a carboxylic acid.

D. an alcohol,

,#! "

n

(

I

38. The absorbance at2116 cm-l in the IR implies that thecompound is an:

A. alkane.

B. alkene.

C. alkyne.

D. alcohol.

39. The l3CNIrrtR doublet ati3.B ppm can be attributed towhich type of carbon?

A . The terminal C of a terminal alkyneB . The internal C of a terminal alkyne

C. CofanaldehydeD. Cofanketone

tt). The number of carbons in an asymmetric molecule thatshows seven peaks in the 13CNMR is:

A. 3.

B. 4.

c. 7.

D. 14.

4 1. Hydrogens that are coupled to one another have thesame:

A. shift value.

B. peak integration.

C . area under theirpeaks.

D. same J-value.

Compound T is which of the following compounds?

A.H- C: C- Ct-te\

OH

c.HO- C=

"- o\

- cH,

B.H- C= C- Chtz .H'\ ,/

cilo

D.H-C=C-cH^\o,

-:'pyright @ by The Berkeley Review@ GO ON TO THE NEXT PAGE

43. the l3CNIr,tR peak at 50 ppm can be attributed to the:

A. CofaC=O.B. CofaC-O.C. H next to C=O.

D. H next to C-O.


From the leaves of a tree that grows wild along the coast

ofEcuador, three esters were extracted using ether. The three

esters were separated by fractional distillation and purified by

column chromatography. Once isolated and purified, aIUNVIR spectrum was obtained for each compound. Thethree spectra are shown in Figure 1 below. All three spectrawere obtained using the same NMR spectrometer in CDCI3

solvent from the same bottle. A dead give-away for an ester

is the shift for the hydrogens attached to the carbon bound to

the oxygen of the alkoxy group. These hydrogens are found

between 3.5 ppm and 4.0 ppm. Compound I has a molecularformula of CgHgO2, Compound II has a molecular formula

of CgH1602, and Compound III has a molecular formula ofCaHs02.

Figure I 1HNMR spectra of Compounds I, II, and III

4 4. Which of the following structures is Compound I?

.odC. O

o*.

ctl 3

B. o

..A,,/ocH.vD.

o"oYo'4 5. It important that each sample use the same solvent in

order to:

A . ensure the same reactivity in the magnetic field.

B. increase the pH of the compounds in solution.

C. view any common impurities between samples.

D . allow fbr extraction of protic compounds.


Compound I

i'"lrr

6 pom 4 ppm 2 ppm

Compound II

lm

illl2Hilli

6 ppm 4 DDm 2 PPm

Compound III

lr"ilril,'"

I

1 ppmI

2 ppmI

3 pDm

4 6. According to IUPAC nomenclature convention, what isthe name of Compound III?

A. Methyl ethanoate

B. Methyl propanoate

C. Ethyl methanoate

D. Propyl methanoate

4 7. Which of the following IR absorbances would you NOTexpect for an ester?

A. 3500 cm-1

B. 2980 cm-1

C. 1685 cm-1

D . 1300 cm-1

Fa

I.

ill I

:1

llil{:

,T!*

llr'r t]rll rl-

gj,

*.!n-:'

I i,]"

.i[[,u:-

4 8. Which of the following structures

B.

cH2cH3

represents Compound

o

ocH 2cH 3

tr?

A. O

oaC. O

d:,.D.

49. For the spectrum of Compound III, the three 1HNMR

signals are which of the following, respectively?

A. Triplet, doublet, triplet

B. Triplet, quartet, triplet

C. Singlet, doublet, triplet

D. Singlet, quartet, triplet

5 0. Which of the following would NOT work as an NNfR

solvent for an ester?

A. CDC13 (chloroform)

B. C6D6 (benzene)

C. D3CD2COD (ethanol)

D. D2CO (formaldehyde)

Oclr*"

r56 GO ON TO THE NEXT PAGE

Passage Vlll (Question 5t - 57)

Nuclear magnetic resonance spectl'oscopy is a powerful:ool for distinguishing isomers. By cornparing shift valuesmeasured in ppm), coupling consrants (J-values), and

rntegration ofpeaks (the area under the curve), it is possibleio deduce a structure with great accuracy. The couplinglonstants help us deduce which hydlogens are neighbors of,lne another. For instance, a carbon con taini ng two:quivalent hydrogens influences the signal for hydrogens on::re neighboring carbon to foln.r a triplet. The integration is,re area under the curve that is directly proportional to the-.umber of hydrogens within that signal. Table 1 is a general,sting of proton NMR shift values.

Hydrogen Shift (ppm) Hydrogen Shift (ppm)

-RCH3 0.8 RCH2R, 0.9

-COCH3 2.1 -COCH2R 2.3-OCH3 3.5 -OCH2R-CH=CH2 5.5 -CH=CHr 5.3

\H 7-8 -CH2CH=CH2 2.3

CH=O 9.7 -CO2H to-12

Table 1 IHXVR shift values fbr common hyclrogen types

When IHNMR informarion is coupled with IR shifr.lles, it is possible to narrow the structure down quickly to: possibility. Important IR absorbances are 3500 crl-1 iO-

': . 1700 cm-l 1C=O;, anct 1600 cm-l 1C=C;. There are.:::r absorbances, but from these three values, it is possible

,, estimate many other values. As the strengLh of a bondr:reases, it exhibits a higher absorbance value in the IR.l:; heavier the atoms in the bond, the higher the absorbance: -:e IR. This intbrmation makes it possible to predict shift-- res lor other bonds. For instance, a C-C bond must be:is than 1600 cm-I in absorbance in the IR, because a

- i.,-:on-carbon single bond is weaker than a carbon-carboni " -cle bond. However, the irnportance o1 NMR:E.Iroscopy is greater than that of infrared spectroscopy.

i l. ]-Methyl-3-pentanoneconrains:

-\. 3 non-equivalenthyclrogens in a9 : 2 : I ratio.B. 4 non-equivalenthydrogens in a4 : 3 : 3 : 2 ratio.C, 4 non-equivalenthydrogens in a6 : 3 : 2 : 1 ratio"

D. 5 non-equivtrlent hydrogens in zi 3 : 3 : 3 : 2 : Iratio"

il" What in the proton NMR is a dead give-away for anisolated isopropyl group?

{. A 6H doublet and a lH septet

B. A 6H singlet and a lH sextet

C. A 6H septet and a lH doubletD. A 6H sextet and a lH singler

- ::', nght O by The Berkeley Revie',v@

53. Pentanal is BEST distinguished from 2-pentanone by apeak:

A. above 1700 cm-l in the IR.B. berween 2.0 to 2.3 ppm in rhe IHNMR.

C. near 3.5 ppm in the IHNMR.

D. near 9.7 ppm in the IHNMR.

5 4. To confirm the presence of an alcohol, it would be bestto add:

A. D2O with NaOD and observe whether the broadpeak grows.

B. D2O with NaOD and observe whether the broadpeak disappears.

C . H2O with HCI and observe whether the broad peakgrows.

D. H2O with HCI and observe whether the broad peakdisappears.

5 5. A methyl ketone always has which of the followingproton NMR absorbances?

A . A 3H singlet between 2.0 to 2.3 ppmB. A 3H triplet between 2.0 to 2.3 ppmC. A 3H singlet between 3.5 to 3.8 ppm

D. A 3H triplet between 3.5 to 3.8 ppm

5 6. What is the common name for the compound with thefbrmula CaHgO, an IR absorbance at Il21 cm-l , andthree NMR peaks at 9.7 ppm (lH),2.2 ppm (lH), and1.0 ppm (6H)?

A. Tetrahydrofuran

B. 2-methylpropanal

C . Butanal

D. Butanone

5 7. A compound with one oxygen in its formula, onedegree of unsaturation, and no IR absorbance between1600 cm-l and 1750 cm-I must be:

A. a ketone.

B. an aldehyde.

C. an alkene.

D . cyclic.

157 GO ON TO THE NEXT PAGE

Passage lX (Question 58 - 64)

A 1HNMR was run for the cis and trans isomers of the

molecule shown in Figure 1 below. The spectra obtained forthe two geometrical isomers are also shown in Figure 1. The

cis compound can be distinguished from the trans compound

using the coupling constants for the vinylic hydrogens. The

trans species has a larger coupling constant due to bettertransfer of magnetization in the trans orientation. The shiftvalues on the spectra are listed in parts per million (ppm) ofthe frequency of the spectrometer. The hydrogen on the

carbon adjacent to benzene can be tbund downfield from the

other vinylic hydrogen due to the presence of oxygen."Downfield" implies higher shiti values.

n,.-G cH- cH- ocHj

Spectrum I 7.25 ppn (4 H)5.37 pptn (l H)4.88 ppm (1 H)3.76 ppm (3 H)2.25 ppm (3 H)

]H

rllrilfl ]l]i llLl

PPmz65432loSpectrum II 3 H 3H

7.25 ppm (4 H)5.37 ppn (l H)4.88 ppm (l H)3.76 ppm (3 H)2.25 ppnt (3 H)

4H

rllriltiltil

IHIH

ililPPmz6s432lo

Figure 1 1HNMR spectra of the cis and trans isotners

The spectra as drawn are different enough to distinguishthe trans and cis compounds fiom one another. The shiltvalues does not help to distinguish the cis lrom the trans

compound, the coupling does.

5 8. Spectrum I is associated with the:

A. trans compound, because the J value is smaller.

B . cis compound, because the J value is smaller.

C. trans compound, because the J value is larger.

D . cis compound, because the J value is larger.

5 9. The methyl group on benzene appears in the IHNMR

at:

A " 7.25 ppm.

B. -5.37 ppm.

C. 3.76 ppni.

D. 2.25 ppm.

Copyright O by The Berkeley Review@

P@60. Which of the following compounds has the same

number of signals in their carbon-13 NMR as there are

carbons in the compound?

A. An ortho substituted benzene with two identicalsubstituents

B. A meta substituted benzene with two identicalsubstituents

C . A meta substituted benzene with two differentsubstituents

D . A para substituted benzene with two differentsubstituents

61. Where do the vinylic hydrogens appear in IHNMR?

A. Above 7.0 ppm

B. Between 4.5 and 7.0 ppm

C . Between 2.0 and 4.5 ppm

D. At less than 2.0 ppm

6 2. Which of the following absorbances wouid be found rn

the IR spectrum for the compound in Figure I ?

A. 3500 cm-l

B. 2220 cm-l

C. 1120 cm'lD. 1640 cm-l

63. Had the compound had meta substitution rather than

para substitution, the ratio of the benzene hydrogens

would be which of the following?

A. 1 :1B. 1 :3C" 1:2:lD. 1:1:1:1

64. Had the compound had an ethyl group on benzene rather

than a methvi group, which of the following would be

observed in the proton NMR?

A . Doubiet (2H) and triplet (3H)

B . Doublet (3H) and triplet (2H)

C . Triplet i2H) and quartet (3H)

D . Triplet (3H) and quartet (2H)

ill:'tn:1

FitE'ru

*"'"iu. -

I Jr lrl

intmr i:)illi;IlL,I .i

ll,il T1:.

[:ifL l[ 1.i"

l:iiL'tt:ti Liud t'

t,i l,nr;',il]fiI'rl',':r:

pr|trnTI- '

r58 GO ON TO THE NEXT PAGE

Passage X (Question 65 - 71)

An unknown compound, labeled Compound B, has only:u'o singlets in its proton magnetic resonance spectrum. Oneshows a shift of 6 1.42 ppm and the other shows a shift of 6-.96. ppm, with relative intensities (from the integration of:tre 1HNMR spectrum) of 3 : l. The decoupled carbon-l3:ruclear magnetic resonance spectrum for Compound B shows:lur signals (with one of greater intensity ihan the other'lree) with shifts at 6 22.3 ppm, 2g.l ppm,79.9 ppm, and- r0.2 ppm. The signal at ljo.2 ppm is atributed to a:rrbonyl carbon. For carbons of equal hybridization, the:rgher the shift value in the carbon-l3 NMR, the greater the:-ectronegativity of the atoms bonded directly to that carbon.:igure I shows both the 1HNMR ana l3CNUR spectra ofl mpound B.

The carbon-13 NMR information leads to the conclusionfibi.: there are four nonequivalent carbons tn the compound.Tre important bands in the infrared spectrum of Compound Biurir,: found at 1738 cm-l , 1256 cm-l, and ll73 cml1. Theinnr:d at 1738 cm-1 is attributed to the stretching of a carbonyl*n'r:d. Elemental analysis of Compound B shows that it

Figure t lHNltn and I3CI.Ur,IR spectra of Compound B

rr:alns two oxygen atoms. The spectral data, inr.:unction with mass percent values from elemental

umn-vsis, can be combined to determine the structure ofinpound B. Overlapping spectral data, such as an IR

,niur*orbance at 1738 cm-l and a l3CNUR signal at 170.2can be used to verify aspects of the structure.

::vright @ by The Berkeley Review@ GO ON TO THE NEXT PAGE

6 5. The shift at 1738 cm-l in rhe IR can be attributed to theC=O of:

A. an aldehyde.

B. a carboxylic acid.

C. an ester.

D. a ketone.

66. The molecular ratio of hydrogens in the structure iswhich of the following?

A. 3:1B. 6:2C. 9:3D. 12: 4

6 7. The l3CNUR peak at 79.9 ppmcan be attributed to:

A. an H bonded to C adjacent to C=O.B. an H bonded to C adjacent to C_O.

C. the C in rhe C=O bond.

D. the C in the C-O bond.

6 8. Which of the following compounds would NOT show al3cNnaR peak above 100 pfm?

A. Methyl benzoate

B. 3-methyl-2-hexanone

C. 3-methyl-2-penranol

D. 3-methylpentanal

69. The lgNVtR peak at 1.96 ppm can be artribured to ahydrogen bonded to a carbon:

A. adjacent to a C=O bond.

B. adjacent to a C-O bond.

C. ofaC-Obond.D . of a C=O bond.

70. in IHNMR, a singlet is explained as the evidence ofhydrogens of the signal being coupled to:

A . equivalent hydrogens on all adjacent carbons.

B . non-equivalent hydrogens on all adjacent carbons.

C . only one hydrogen on an adjacent carbon.

D. no hydrogens, because there are no hydrogens on

any of the adjacent carbons.

7 1. Compound B is which of the fbllowing?

A.oR'

CH:

HIC CH:

Hrc

C.

"''K)rnHsC CH: O

D.

HrC- - O- -CH-r.XYH:C CHr O

H:C CH:



Carbon-13 NMR can be used to determine the number olr,rnique carbons in a compound. Carbon-13 NMR is similar

to hydrogen-1 NMR in that it generates separate peaks for

each unique isotope, in this c"s. 13C. Many I3CNMR

spectra are recorded without coupling (known as decoupled

spectra), so all peaks appear as singlets. Table I lists the

approximate shift value for selected types of carbons. It is a

brief guide to determining the types of carbon represented bieach I3CNMR peak. The range for the ppm of each signal is

approxirnate and on occasion, a peak may fall outside of the

range.

Carbon ppm Carbon ppm

R2C=O 205 - 220 .C-OH 50-70

RHC=O 185 - 200 -C-CI 40-45R2C=CH2 t20 - 140 -C-NH2 35-45

R2C=CH2 r 10 - 120 -C-Br 25-35

RC=CH 12 85 F{3C-C=O 20-35RC=CH 65 -10 R2CH2 l0-30

Table 1 l3cxltR shilt values

It should be noted that the peaks fbr carbons with r:hydrogen directiy attached are less intense than other carbo:

peaks. Carbonyl carbons and quaternary carbons therefb:.

generate shorter peaks than those for other carbons in t1::

spectrum. For peaks that represent rnore than one carbon, th:

intensity increases, but not in a way that is easily integrate;

Integration is generally not carried out on 13CNMR spectra.

A researcher used I3CNMR to distinguish two structur--

isomcrs formed when toluene (methylbenzene) was acylate:

using acetyl chloride by comparing the I3CNMR spectra :t

the two compounds. The molecular tbrmula fbr both isome:i

is CgH160. Spectra fbr both compounds were recorded usi:'s

the same NMR instrument for the same period of time in lt:l':

same concentration. The only clifTerence between the t''n:

samples invoived the isomers themselves. The spectral d:-ltbr the two isomers are listed below:

Ison-ier I: 2-1,34, 110, ll1, 123, 130,206

lsonierll: 31, 35, 108, 115, 119,123,126,131,?t-';

The dift'erence in the number of 13CNUR signals lisl:rfor each isomer is cause<l by the lack of symmetry betse:rthe two isomers. Isomer I has nine carbons but shows ol"'sevcn peaks in the l3CNMR, so it must have some type :$i

symrnetry that equates carbons within the structure. B'-;tr

isomers ale benzene rings with a n-rethyl substituent anC ar

acetyl group attached to carbons on the benzene ring.

- 3. Which of the followingstructure of Compound I?

structures represents the

72. Which of the following reactions CANNOT bemonitored by a change in the shift values in thel3CNvtR spectra?

A . Oxidation of a secondary alcohol to a ketoneB . Nitration of ethyl benzene to para-nitroethylbenzene

C . Deprotonation of a carboxylic acidD . Reduction of an alkene to an alkane with hydrogen

gas and nickel catalyst

ov.+t

dil

cH2cH

D.

I

Irs

t!

D

f,

A. B.

Gl3

C.

Which of the following functionalities are NOT presentin either of the two isomers?

A. Aldehyde

B. Alkene

C. Ketone

D. Methyl

How can it be explained that the least intense peak inthe spectrum for Compound II is found at209 ppm?

A . Carbons with sp2-hybridization show low intensityl3CNMR peaks.

B. Carbons with spj-hybridization show low intensityl3cNuR peaks.

C. Chiral carbons show low intensity l3CNVtRpeaks.

D. Carbons with no hydrogen atoms attached showlow intensity l3CNUR peaks.

)pyright @ by The Berkeley Review@ l6l GO ON TO THE NEXT PAGE

76. How many signals would be seen in the carbon-l3NMR of para-methoxy benzaldehyde?

A. Four

B. Five

C. Six

D. Eight

7 7 . What is true about the units of unsaturation for the twoisomers?

A . Isomer I, because it is more symmetric than IsomerII, has more units of unsaturation than Isomer II.

B. Isomer I, because it has less unique carbons thanIsomer II, has fewer units of unsaturation thanIsomer II.Isomer I and Isomer II each have three units ofunsaturation.

Isomer I and Isomer II each have five units ofunsaturation.

78. For the compound 2,2-djmethylbutane, which carbonshows the peak of lowest intensity?

A. Carbon 1

B. Carbon 2

C. Carbon 3

D. Carbon 4

C.

D.


Coupling in proton NMR is used to determine therelative positioning of hydrogens within a compound.Hydrogens are considered to be coupled when they are onneighboring carbons. Their respective magnetic fieldsinfluence the signals for one another. The effect is mutual,so the coupling interaction is equal for all coupled hydrogens.This can be seen in terms of identical J-values. A J-value isalso known as the coupling constant and is the distancebetween adjacent peaks within a proton NMR signal.

the lftNMR spectrum for two isomers of C4H3O2 are

collected under identical conditions using the same NMRmachine. Figure 1 shows the spectrum for Isomer I, whileFigure 2 shows the spectrum for Isomer II.

Figure 1 1HNMR spectrum of Isomer I

Figure 2 IHNMR spectrum of Isomer II

Both of the isomers are esters. The exact connectivity ofthe esters can be deduced from the shift values of the singletand quartet. IHNVIR signals around 3.5 to 4.0 ppm are dueto alkyl groups bonded to an oxygen while alpha hydrogenstypically show shift values between 2.0 and 2,5 ppm. Theintegration information verifies the substitution of eachcarbon in the ester.

Spectrum I 13 H

4ppm 3ppm 2ppm lppm 0

3pp- 2ppm lppm 0

Copyright @ by The Berkeley Review@ 162 GO ON TO THE NEXT PA

7 9. Which of the following is the common name for thecompound represented by Spectrum I?

A. Ethyl acetate

B. Acetone

C. Methyl acerare

D. Isopropyl formate

80. A triplet in the proton NMR is associated with t-be

hydrogens:

A. of a CH2 group.

B. on a carbon next to a CH2 group.

C . of a CH3 group.

D. of a carbon next to a CH3 group.

8 I . What type of compound is represented by Spectrum H-t

A. An ethyl ester

B. A methyl ester

C. An ethyl ketone

D. A methyl ketone

82. A lgNirAR signal in the range between 6 and 8

indicates the presence of which of the following?

A. a hydroxyl proton.

B. an aldehyde hydrogen.

C. abenzenehydrogen.

D. a carboxylic acid proton.

83. Which of the following features in the protonspectrum CANNOT be used to distinguish an esterof a carboxylic acid when the two compounds havesame molecular formula?

A. A sharp peak between 2.0 and 2.5 ppm

B. A sharp peak between 3.5 and 4,0 ppm

C . A sharp peak between 10.0 and 12.0 ppm

D. The observation that no peak integrates to a re

ratio of 1.

84. How many unique proton NMR signals are expected4-heptanone?

A. Two

B. Three

C. Four

D. Five

[h,ct- .

il.0-,D-.

frdmmm

A-mf-p,lU- s

Fassage Xlll (Question 85 - 90)

A chemist sets out to determine the structural identity forfih;ee structural isomers, using data from NMR spectroscopy,Ltu.L :iviolet-visible spectroscopy, and mass spectroscopy to

:tify each cornpound. The following spectral data are:l:rved for three separate compounds, all with the forrnula: Ht oo.

:-"pound I:r-:CNMR' 211 ppm (1), 31 ppm (1), 22 ppm (i), 17

ppm (1), 1l ppm (1):HNMR: 2.32 ppm (triplet, 2H),2.08 ppm (singlet,

3H), 1 .12 ppm (multiplet, 2H), 0.96 ppm(triplet,3H)

--\'-Visible: 268 nm and 189 nm

,".iound II:rCNMR: 12 ppm (1),24 ppm (2), 13 ppm (2)

HNMR: 3.71ppm (multiplet, iH), 1.88 ppm (broad,1H), 1.33ppm (multiplet,4H), 1.14 ppm(triplet,4H)

,f--Visible: No intense peaks above 180 nm

:, :cLutd III:'CNMR: 68 ppm (2),21ppm (2), l0 ppm (1)

FI-vMR: 3.58 ppm (triplet, 4H), 1.28 ppm(multiplet, 4H), 0.92 ppm (multipl et, 2H)

-,-"'-Visible: No intense peaks above 180 nm

1.ll NMR spectra are obtained using deuterated.::form as solvent. The mass spectrometer show anl-'e peak at 86 amu for all three isomers, contirming their

':ular fbrmula. Elemental analysis shows that carbon:'" drogen are not the only atoms present.

ilompound II is what type of compound?

A, An alcohol

B. An ether

C . A ketone

D. An alkene

^arbon-13 NMR is useful for determining all of the'.llowing EXCEPT the:

A . number of unique carbons in a structure.

B . presence of a carbonyl group.

C . substitution of a benzene ring.

D . geometry about a double bond.

right @ by The Berkeley Review@ 163 GO ON TO THE NEXT PAGE

8 7. What can be concluded from the data obtained usingUV-visible spectroscopy?

A. Compound I is a conjugated diene.

B. Compounds II and III are carbonyl compounds.

C. Compound I is a ketone.

D. There is no n-bond in Compounds I and II.

8 8. Which of the fbllowing is NOT a valid conclusion fromthe spectral data for Compound III?

A. The absence of a broad peak in the 1HNMR means

the compound cannot be an alcohol.

B. The absence of a peak above 175 ppm in thel3CNltR means the compound cannot be a

carbonyl.

C. The absence of an absorbance above 180 nmconfirms that the structure must contain a ring forits unit of unsaturation.

D. The large number of peaks in the 13CNMR means

the compound is an ether with little symmetry.

8 9. What is the IUPAC name for Compound I?

A. Pentanal

B. 2-Pentanone

C. 3-Pentanone

D. 3-Methylbutanone

90. The integration of a signal in a proton NMR is usefulfor determining the:

A . local magnetic field experienced by a hydrogen"

B. neighboring hydrogen atoms.

C . presence of an atom other than hydrogen or carbon.

D. relative quantities of unique hydrogens in thecompound.

Passage XIV (Questions 91 - 97)

Proton NMR is a valuable tool used to deduce thestructure of unknown organic compounds. It helps onedistinguish between structural features of two similarcompounds. The three useful components of the spectral datafbr extracting structural information are the shift value (6,measured in ppm), the coupling and coupling constants (peakshape), and the integral (the area under the curve of a signal).From the shift value, infbrmation about the electroncgativityof adjacent atoms may be obtained. Coupling is used todetermine the number of hydrogens on any atoms bonded tothe atom bound to the hydrogens producing the signal. Theintegral is directly proportional to the number of hydrogensin the compound, so it can be used to find the ratios ofhydrogens in the cornpound. Figure I and Figure 2 show thel gNVtR spectra for two simple organic structures,Compound A and Compound B, along with the moleculartbrmula of the compounds they represent.

Spectrurl ACzHrc

5H3H

8 ppm 6 ppm 4 ppm 2 ppm 0 pprn

Figure I IHNMR spectrum of Compound A

Spectrum BC3H602

Ij

iL) ppm 5 pprn o ppm

Figure 2 lHNMR spectrum of Compound B

Structures can often be deduced usins onlv some of thelUlltR information, such as the coupliig information andthe integration. Often, the coupling infbrmation is the mostimportant of all the data. The coupling constants can giveintormation about the connectivity of the structure, as well as

irints about the three-dimensional orientation of atoms within:ne n.iolecuie.

f.-,pvright O by The Berkeley Review@ GO ON TO THE NEXT P.{

91. Which of the following is the IUPAC name for rhe

compound represented by Spectrum A?

A. Phenylmethane

B. Toluene

C. Methylbenzene

D . Orthoxylene

92. In Spectrum B, what is the ratio of the areas underthree peaks?

A. 3 :2:3B. 1:1C. 2:3D. 1:2

4

5

3

9 3. What peaks are expected for 2-bromopropane?

A. A 6H sextet and a 1H singlet

B. A lH sextet and a 6H singlet

C. A 6H seprer and a lH doublet

D. A 1H septet and a 6H doublet

9 4 . Which of the following compounds would NOT har;quartet in its proton NMR spectrum?

A.

(H3C)2HC

C. D.

CHe H3CH2CH2C(H3C)2HCH2C

Which of the following compounds would NOT ha"'e

doublet in its proton NMR spectrum?

A. 2-methyl-1-pentanol

B. 3-methyl-1-pentanol

C. 4-methyl-1-pentanol

D. A11 isomers of methyl-1-pentanol have double-;their proton NMR spectrum.

o

{

o

A

B.

cH2cH3 H3CH2C(H3C)2C

o

Ao

A

95.

'r 6. The iodoform test involves the addition of hydroxideanion and iodine to a carbonyl compound. If a carboncontains three alpha hydrogens, then the iodine willreact with the carbonyl compound to yield a yellowprecipitate. A compound with a positive iodofonn test

would likely have which of the following signals in itsproton NMR?

A. Singlet

B. Doublet

C. Triplet

D. Septet

The broadness ofthe signal around 10 ppm in SpectrumB is explained as a signal caused by hydrogens:

A. on a carbon involved in resonance.

B. coupled to more than eight hydrogens.

C. on a carbon involved in hydrogen bonding.

D. involved in hydrogen bonding.

- ::.t O by The Berkeley Reviewo ENOUGH CHEMISTRY... FOR NOW!

Questions 98 through 100 are NOT based on a

descriptive passage.

9 8. The following two molecules are best described as:

A . structural isomers.

B. geomelrical isomers.

C . optical isomers.

D. the same molecule with altered spatial orientation.

9 9. The following two molecules are best described as:

A . structural isomers.

B. geometrical isomers.

C . optical isomers.

D. the same molecule with altered spatial orientation.

1 0 0. How many structural isomers of C5H12 are possible?

A. 3

B. 4

c. 5

D. 6

l.B 2.C 3.D 4.C 5.C 6.D1.8 8.C 9.A 10.A 11.A 12.D

13. D 14. D 15. B 16. C 1',7. D 18. B19. A 20. B 2r. D 22. D 23. D 24. C25. A 26. D 21. A 28. D 29. C 30. A3r. B 32. A 33. A 34. D 35. B 36. C31. D 38. C 39. A 40. C 41. D 42. A43. B 44. B 45. C 46. B 41. A 48. B49. D 50. C 51. C s2. A 53. D 54. B55. A 56. B 57. D 58. C 59. D 60. C6i. B 62. D 63. D 64. D 65. C 66. C61. D 68. C 69. A 10. D 11. D 72. C13. C 74. A 15. D 16. C 77. D 18, B19. A 80. B 8i. B 82. C 83. A 84. B85. A 86. D 87. C 88. D 89. B 90. D91. c 92. D 93. D 94. C 95. D 96. A91. D 98. B 99. C 100. A

--a

Structure Elucidation Answers

L. Choice B is correct. The most stable form of the cyclohexane ring is the chair conformation. The most s

position on the chair form is referred to as equatorial. Combine these two facts and the result is choice B.

a Choice C is correct. Because the three deuterium atoms are cis with respect to one another, they cannot allaxial nor all be equatorial. The most stable orientation (most stable chair confirmation) has as many deuteatoms with axial orientation as possible. However, because the deuterium atoms are all mutually cis toanother, the structure must have at least one deuterium with equatorial orientation. The best choice (

consequently your choice) is C, which is drawn below.

Passage

DD

J.

1 axialdeuterium2 equatorial deuteri

Choice D is correct. As the reaction is written, the value of Kgt-1,4 must be greater than 1, because the prodmore stable than the reactant. The reaction is favorable in the forward direction as written. This elimichoice A, a value less tiran 1.

The question now focuses on whether the conformational change with Ksq-1,4 is more favorable thanconformational change with K"O-1,2, because the value of Kqq-1,2 is 4.31. In the case of the 1,2-disubstihcompound, there are both diequdtorial versus diaxial interactions as well as gauche versus anti interactionsconsider. The diequatorial orientation is better than the diaxial orientation, because with diaxial thereeclipsed interactions with the axial hydrogens. Howevet, the anti orientation of the methyl groups is

than the gauche orientation.

Overall, diequatorial preference over diaxial is a more important factor than a preference for gauche overso the value of Keq-l,Z is greater than 1. The 1,4-disubstituted compound has no gauche-versus-anti inibetween the methyi groups to consider, because the carbons are far apart. Thus, the conformational prepurely an effect of the diequatorial orientation being preferred over the diaxial orientation. This makes

2 axial deuteriums and1 equatorial deuterium 11

cF{3

-

Anti > Gauche

value of Keq-1,4 greater than the value of Keq-1,2, making choice D the best answer. Drawn below is one

worth (apprbximately equal to 1000 words worth) of explanation.

Methyl groups are apart from Methyl groups are close to one Methyl grouPs collide withone another when diaxial. another when diequatorial. axial Hs when diaxial.

9Hs

Methyl groups do not collideone another when diequatorial

Diaxial < Diequatorial

No anti vs. gauche factor

1,2-diaxial orientation results in anti orientation, while1,2-dieqatorial orientation results in gauche orientation.The equilibrium still favors product (the right), but notas much as the L,4-equilibrium does.

1,4-diaxial orientation results in steric hindrfrom diaxial interactions with hydrogens, wl,4-diequatorial has no eclipsing steric hindrEquilibrium favors the products more than with 1

4" Choice C is correct. In order for substituents to be gauche or anti to one another, they must be bonded to ca

that are connected to one another. In the case of 1-4-dimethylcyclohexane, the methyl groups are notadjacent carbons, so the two methyl groups cannot be gauche or anti to one another. This makes choice C

There are hydrogens on every carbon, so H can be gauche to methyl. For the H to be anti to a methyl, the

group must assume axial orientation. The oniy possible chair conformation of cis-1,4-dimethylcyclohexaneone methyl group axial and the other in an equatorial orientation. This makes choices A, B, and D valid.

CH.I,CH1

-H.C,-

rr3L

I oiulgl = !,"guubtlcFI3

Copyright @ by The Berkeley Review@ 166 Section II Detailed Exp

Choice C is correct. Both of the chair conformations possible for cis-1,2-dimethylcyclohexane are equivalent inenergy. In both of the chair conformations, one methyl substituent assumes axial orientation and the othermethyl substituent assumes equatorial orientation. The equilibrium constant for the ring-flip process is equal to1, because the energy levei of the product is equal to the energy level of the reactant. This makes choice Ccorrect. Do the correct thing and pick C. As a point of interest, the value of KuO can never be less than or equal to0 (products and reactants always have some positive quantity), so choice A is an absurd answer.

HsC

HsC

Both structures have one methyi equatorial and the other methyl axial

Choice D is correct. With cis-1,3-dimethylcyclohexane, the cis-1,3-diaxial interactions (steric repulsion)between the two methyl groups makes the diaxial orientation less stable than the trans-1,2-diaxial orientation.This decrease in stability in the conformation drawn on the reactant side results in a greater value for Ksq aswritten. This means that K"O-1,3 ,K"q_I,2, which is choice D.

Methyl groups are apart fromone another when diaxial.

CH.IJ

Methyl grouPs are close to one Methyl groups collide with Methyl groups do not collide withanother when diequatorial. one another when diaxial. one another when diequatorial.

.r, a, fn, In, S^cF{"

1,2-diaxial orientation results in anti orientation, while 1,3-diaxial orientation results in increased. steric1.2-dieqatorial orientation results in gauche orientation. hindrance. This has an effect on the equilibrium byI?re equilibrium still favors product (the right), but not as shifting it heavily to the right. 1,3-diequatorialmuch as the 1,3-diaxial-to-diequatorial equilibrium does. has no eclipsing sieric hindrance.

Choice B is correct. The key to this problem is drawing the two hydrogen atoms on the bridging carbons cis to oneanother. When the hydrogens are cis to one another on adjacent carbons, one hydrogen assumes axial orientation,while the other assumes equatorial orientation. As a consequence, the carbon-carbon bonds to the left ring mustalso be axiai for one and equatorial for the other. The structure is drawn below with the hydrogens noted.Choice B is the best answer.

Note that the two hydrogen atoms andthe hydroxyl group are all cis and up.

OH

EquatorialAxial

Equatorial--+11

Choice C is correct. Because 240' is at the nadir (low point) of the graph in Figure 1, it correlates with thestaggered structure. This eliminates choices A and B. The compound has R stereochemistry at carbon two whichmakes the correct answer choice C and eliminates choice D, which has S stereochemistry.

Choice A is correct. Because 60" is at a local apex (high point) of the graph in Figure 1, it correlates with aneclipsed structure. This eliminates choices C and D, which have the compound in its staggered confirmation.The 60' point is not the highest point on the energy diagram, so it does not involve the largest groups (carbon 1and carbon 4) interfering with one another. This eliminates choice B. The best answer is choice A, with thetwo methyl groups eclipsing one another.

-:-:vright O by The Berkeley Review@ t67 Section II Detailed Explanations

10. Choice A is correct. The 330" point on the graph is near (30" away from) the most stable conformation (nh:has anti orientation of the CH2OH group and the CH3 group of carbon 4). This eliminates choices C andBecause of R stereochemistry, the correct choice is A. Stereochemistry is difficult to see in the Newrrprojection and can be seen more easily in other projections. Drawn below is a way to convert the Neu,c:projection back to the stick-wedge drawing and a subsequent evaluation of stereochemistry.

CHs

H:9

HcH2oH H CH.OH H\\"t"'

11.

1'

Choice A is correct. Conditioning may cause yoll to respond automatically that the best orientation is the anewith the fewest repuisive interactions. This is often true, but it does not tell the entire story in this case. The,most stable orientation can also be the result of the strongest attractive interactions. Hydrogen-bonding betn'the hydroxyl and amine groups occurs only from gauche orientation, where the two gtorpi are close enoughbond. Hydrogen-bonding cannot occur between substituents with anti orientation. This *uk"r choice A correct.

Choice D is correct. Hydrogen bonds have some acid-base character to them, so the most favorable pro::transfer reaction is a good indicator of the strongest hydrogen bond. Because the amine is more basic than ihydroxyl group, the nitrogen is the lone pair donor. Likewise, the hydroxyl is more acidic than the aminethe hydroxyl is the hydrogen donor. This makes choice D correct.

13. Choice D is correct. Because of the planar nature of electron density in a n-bond, rotation about a double bc:requires that the n-bond be broken. This is not observed under thermal conditions. To convert a cis n-bond int;trans n-bond, UV light is needed. Pick D to tally big points. The drawing below shows that a 90' rotation abcthe C-C breaks the r-bond. It requires substantial energy to break a rc-bond.

R1

90" rotation----------------

R2

Choice D is correct. The only single bond about which rotation is not possible is a single bond between two ar

in a cyclic compound. The only cyciic compound of the answer choices is methylcyclopentane, choice D. Theanswer is therefore choice D. One item of notable interest is that both n-bonds and rings lower the entroplcompound by lowering its degrees of freedom (i.e., its ability to rotate freely).

Choice B is correct. The most stable form of cyclohexane is the chair conformation (as opposed to the i:conformation), with the smallest substituents (determined by bond length) in the axial position. The inteqfrom the proton NMR shows that the ratio of peaks for 1H is 1.12 : 1 in favor of the 6 = 7-.25 ppm shift, the =due to the equatorial hydrogen. This indicates that more hydrogens are located in the equatorial position i1.25 ppn) than the axial orientation (6 = 1.51 ppm). Consequently, the preferred conformation has deuteriu-crthe axial position. For this reason, choose B. Drawn below are the two chair structures and their equilibriurc"

6 = t.st ppm ----->H L{ D

6 = 1.25 ppm --->H

R,

R3

1.4.

1.5.

I0

m

-

{,0

t[

rU

rfi

T

,fl

H6axiaiH:5equatorialH (IHNMRintegral 7.2:7) 5 axialH:6equatorialH llUruVRintegral 1:Actual ratio is I.1.2:1, rvhich shows H prefers the equatorial position, so the right structure is more stable.

ru-bond

Copyright @ by The Berkeley Review@ l6a Section II Detailed Explanat

Choice C is correct. Using D2 gas rather than H2 gas would have produced C6H5D7. The addition of the D2 is

in syn addition for all 6 deuteriums that are added, producing two possible conformational isomers, one havingthree hydrogens equatorial and two hydrogens axial (choice C) and the other having two hydrogens equatorialand three hydrogens axial (choice D). Choice C is more stable, because hydrogens prefer the equatorialorientation. Choose C if you want to be a star. The structure is drawn below:

-

.-

D

3 axialH:2 equatoriaiH 2axiaIH:3 equatorialH

H prefers the equatorial position, confirming that the structure on the right ismore stable. The K"O for this ring flip is > 1. The ratio is 2 axial: 3 equatorial.

This question could have also been answered without knowing the exact chemistry. The passage infers thathvdrogen prefers the equatorial position over deuterium. This means that in the most stable orientation, there

are more hydrogens with equatorial orientation than axiai orientation. This eliminates choice D. Because

there are only five hydrogens present (you start with five on deuterobenzene), the sum of the two numbers mustbe five. This eliminates choices A and B and makes choice C the best answer. Learn to answer questions as

quickly as you can, whether you use organic chemistry knowledge or common sense.

Choice D is correct. The addition of H2 gas to chlorobenzene results in the hydrogenation of the benzene ringand the formation of CH11Cl (chlorocyclohexane). For chlorocyclohexane, there are two possible stable

conformational isomers, one having a ring structure in a chair conformation with the chlorine equatorial(choice D) and one iraving a ring structure in a chair conformation with the chlorine axial (choice C). Chlorineis larger than hydrogen, so it prefers to be in the equatorial orientation. Choice D is more stable so choose it.

Clroice B is correct. The hybridi zatron of carbon remains sp3, whether it is bonded to deuterium or to hydrogen'The angle therefore should be around 109.5". It is stated in the passage that the bond angles are between 107.5"

and 111", which are both nearest to 109.5" of all the choices. Pick B for best results on this question.

Choice A is correct. The cis form of 1,3,5-trimethylcyclohexane allows for all three methyl groups to assume

identical orientation in terms of axial or equatorial. This means either choice A or choice D is the best answer.

Equatorial orientation is more stable than axial orientation, so choice A is the best answer. The conformation is

shown below:

H

HHsc

H

Choice B is correct. Like benzene, C6H5D has three n-bonds and one ring. The best answer is choice B. If you

recall the formula for units of unsaturation, (2(#C) - (#H) + 2)/2, you can calcuiate the units of unsaturation,knowing that D behaves the same as H. This would also yield a total of 4.

Choice D is correct. A hydroxyl group, OH, is larger than a hydrogen, so choice A should be eliminated. The

axial orientation is -or" iritld"redin terms of sterics, so choice B is invalid. Choice C is a good explanation forwhy the two hydroxyl groups would be found in a diequatorial orientation, but the question is looking for an

expianation foi why diaxiai, rather than diequatorial, is the preferred conformation. This eliminates choice

C. When the two hydroxyl groups are both in axial orientation, they can exhibit 1,3-diaxial interactions.

This is normally considered to be steric repulsion; but in the case of two hydroxyl grouPs, there exists the

ability to form hydrogen bonds. This makes choice D the best answer.

3axialH:2equatoriaiH

::',right O by The Berkeley Revierv@ r69 Section II Detailed ExPlanations

,,

25.

23.

24.

Choice D is correct. Only in choice C and choice D do the rings have their substituents with cis orientation.Choices A and B are eliminated, because the substituents are trans to one another. The more stable conformerhas the larger substituent (the ethyl group) in the equatorial position. This describes choice D.

Choice D is correct. An IR absorbance between 1700 cm-1 and 7740 cm-1 is the result of a carbonyl (C=O) group, asstated in the passage. Of the choices, only the_ether doesn'tcontain a carbonyl group, so it is the ether thaidoes not have an IR absorbance between 1700 cm-1 and.7740 cm-1. Pick D for best t"rrttr.

Choice C is correct. A straight-chain monosaccharide has O-H groups, a C=O functionality and C-H bonds.These groups have IR absorbances for bond stretching of 3300 cm-r,l7I5 cm-1, and 2980 cm-1 respectively. Thismakes choice C the best choice. You are required to know common values for the IR absorbances. No common-value peak is found around 2300 cm-1, so thit should lead you to answer choice C.

Choice A is correct. A broad signal near 2850 cm-1 indicates the hydroxyl group of a carboxyiic acid (O-H witinhydrogen-bonding). A hydroxyl group exhibits hydrogen-bonding, so the signal is broad. Because the O-Hbond of a carboxylic acid is weak, its absorbance is lower than that of standard hydroxyl groups. Of the ansirerchoices, only choice A has a carboxylic acid functionality, let alone a hydroxyl gio,rp. plit a for happiness.

26. Choice D is correct. The region between 7.0 and 8.0 ppm in the proton NMR can be attributed to hydrogens onaromatic ring. This immediately eliminates choices A and B, which have no aromatic rings associated rrithem. The difference between ethylphenol and methylphenol can be demonstrated by eithei the total numbof hydrogens (as shown in the integration) or by the coupling of the alkyl portion of the compounds. ThegrouP exhibits a 2H quarter and a 3H triplet, while the methyl group exhibits a 3H singlet. The extragroup associated with the ethylphenol compared to methylphenol is found in the 2.0 to 2.5 ppm region, notaromatic region. This eliminates choice C. To decide between the two compounds in choice D, one can lookthe coupling of the hydrogens on the benzene ring, found in the range between 7.0 ppm and 8.0 ppm.splitting for the hydrogens on a benzene ring with para-substitution is symmetric, while the splitting forhydrogens on a benzene ring with meta-substitution is asymmetric. This is the distinguishing iactor betuthe para-substituted and meta-substituted phenols in choice D. The best choice is answer D.

,n Choice A is correct. Because the broadness of hydroxyl peaks is associated with hydrogen-bonding, r

decreased hydrogen-bonding of a tertiary alcohol is reflected as a narrower absorbance in the IR spectru-n:.compared to a primary alcohol. The alcohol peak is found weli above 3000 cm-1, so the best answer ii choice

28. Choice D is correct. An IR absorbance at 1775 cm-1 is rndicative of a carbonyl group (the stretching of a Cbond)' Aldehydes and ketones contain a carbonyl group, so choices A, B, and C are eliminated. Only the(tetrahydrofuran) does not have a carbonyl functionality. Pick D and bask in the glow of correctness.

29. Choice C is correct. The hydrolysis of an ester results in the formation of a carboxylic acid and an aBoth the ester and the carboxylic acid have carbonyl groups, so each has an IR absorbance around 1700 c

This means that both before and after hydrolysis, there is a signal around 1700 cm-1, eliminating choices AB. An ester has no hydroxyl group, so initially there is no signai around 3300 cm-1. Following hydrolysis,an alcohol and carboxylic acid are formed, so a signal for the hydroxyl group appears. In particular"hydroxyl group of an alcohol shows an absorbance around 3300 cm-1. This means that during the course cd-

hydrolysis of an ester, an IR signal appears around 3300 cm-1, making the best answer choice C.

30. Choice A is correct. Because a ring takes one degree of unsaturation and a carbonyl takes one dunsaturation, a cyclic ketone has two degrees of unsaturation. The compound has only oneunsaturation, so it cannot be a cyclic ketone. The correct answer is choice A. A cyclic ether has one ring andbonds, so it has one degree of unsaturation. This eliminates choice B. A carboxylic acid has no rings andbond, so it has one degree of unsaturation. This eliminates choice C. An ester has no rings and one n-bond.has one degree of unsaturation. This eliminates choice D.

I

:

1

I,ii

t

(iu

fr

(u'@

rffi

Copyright @ by The Berkeley Review@ t70 Section II Detailed Expl

Choice B is correct. Ozonolvsis cieaves the double bond of an aikene, converting it into two carbonyl compounds(either an aldehyde or a ketone). In this reaction, the products formed are aldehydes, as mentioned in thepassage. Aldehvdes can be identified by their peak around 9.7 ppm in the 1HNMR. Choose B for best results.

Choice A is correct. Compound B, like Compound A, is an aldehyde. Like Compound A, it shows an IRabsorbar-ice near 7725 cm-1 for the stretching mode of the C=O bond. This makes choice A correct.

Choice A is correct. Compound A has one degree of r-rnsaturation (due to the carbonyl), and Compound B has twodegrees of unsaturation (one due to the carbonyl and the other due to a ring). The second degree of unsaturationhas to be from a ring, because there can be no double bond in the product from ozonolysis (had there been a n-bond, the ozone would have reacted with it.) This means that the original alkyne must have had a ring and a

triple bond, which in turn means that the original alkyne had three degrees of unsaturation. The alkyne hadnine carbons (the sum of the carbons from Compounds A and B). A nine-carbon compound with no units ofunsaturation has the formula CSHZO. For every unit of unsaturation, two hydrogens are removed from theformula, so a hydrocarbon with three units of unsaturation must have the formula C9H1.4, choice A. You canverify tlris with the molecuiar mass, which is roughly 122 g/mole, as stated in the passage.

Choice D is correct. An alkene is distinguishable by an IR absorbance between7620 cm-1 and 1660 cm-1 for thestretching of the C=C bond. You should know that i carbonyl C=O bond absorbs at around 1700 cm-1, and beingthat a C=C bond is slightly weaker than a C=O bond (they are both double bonds, but a carbonyl bond isslightly shorter), it takes siightly iess energy to stretch the C=C bond than the C=O bond. This means that a

C=C bond has an absorbance slightly less ihar-r 1700 cm-1. Only choice D is less than 1700.*-1, so choice D istl-re best answer. The test requires that you have some iR peaks in your memory; but for ones you don't recali,estimate them by comparison to the values you know. Choose D in this question for the feeling of correctitude.

Choice ll is correct. The products from the ozonolysis of the alkene are both aldehydes, so Compound A must bean aldehvde, which makes choice B the best answer. All of the choices have the correct units of unsaturation,so the bcst answer rnust be determir-red from the presence of the aldehyde proton.

Choice C is correct. As stated in the passage, Compound B must be an aldehyde, which eliminates choice A.The con'rpound cannot contair a double bond (the product from ozonolysis cannot have a double bond between

carbons), so choice B is eliminated. The lHXVR integration shows a peak ratio of 7:1:4:4, which indicatesthat there are no methyl groups in the compound. A methyl group would have shown a relative ratio in theintegrai of 3. This eliminates choice D. Choice D can also be eliminated, because it contains too many carbons.This narrows it down to choice C, which does in fact show four nonequivalent hydrogens in a ratio of 1:1:4:4. Itis vital that you solve this question by a muitiple-choice elimination process rather than structural deduction,because in a multiple-choice format, eliminatir-rg invalid structures is faster than elucidating the correct one.

Choice D is correct. The question here is not what type of bond is causing the shift, because all four answerchoices are C-O bonds. The question is: "What type of molecule is Compound T?" Because of the broad IRabsorbance at 3350 cm-1, tire compound has H-bonding, so it must be an alcohol. The correct answer is therefore

choice D. Choices B and C should be elimilated, because the passage states there is no carbonyl. Choice A can

be eliminated, because ethers exhibit no hydrogen-bonding.

Choice C is correct. An absorbance at 2\76 cm-1 implies that the compound is an alkyne. This is trivialknowledge for the most part, but the passage provided the information, in case you don't have carbon-to-carbontriple-bond IR data committed to memory. You can deduce the IR absorbance vaiue for an alkyne knowing thata C-C boncl has an IR absorbance between 1100 cm-1 and 1300 cm-1, and that a C=C bond has an IR absorbancebetween 1620 cm-1 and 1680 cm-1. As the bor-id strength increases (or bond length decreases), the IR shift valueincre;rses. Choices C is the best answer. The answer is given at the end of the tl-rird paragraph in the passage.

Choice A is correct. A doublet in the 13CNMR is due to a carrbon with one hydrogen attached to it. Because a

ketone carbon and an internarl alkyne carbon do not have hydrogens attached to them, they would be singletsrather than doublets. This eli.mi.nates both choice B ancl choice D. There is no peak in the IR spectrum,inciicating that there is a C=O present (no peak around 1700 cm-1), so choice C is eliminated. The only choice

left is choice A, the best choice.

t7l,:vright O by The Berkelev Reviervo Section II Detailed Explanations

40.

47.

42.

43.

Choice C is correct. If a compound has no symmetry, then thereis a 13CNMR peak for every carbon in -*-*

compor-rnd. The number of carbons in a compound exhibiting seven 13CWR peaks is therefore r"rr"rr. pick C :.rbest results.

Choice D is correct. When hydrogens are coupled to one another, they have the same coupling constants :values)' This makes choice D the best answer. Two different hydrogen groups have different shift values, .,:choice A is eliminated. Choices B and C are the same answer, so they shouldboth be eliminated. This lear'=only choice D as the best answer.

Choice A is correct. From tl-re NMR data, we know that Compound T has three carbon atoms and four hydrog.:atoms, so the structure must contain only three carbons. Choices B, C, and D al1 have four carbonr, ,o thuy .-,=eliminated without involving the spectroscopic data. The only choice that fits the formula is choice A. Or^.::this question is resolved, it can help to solve some of the other questions. This happens often on the MCAT.

Choice B is correct. The 13CNMR peak at 50 ppm is due to a carbon, not a hydrogen. This eliminates choices -and D. A carbonyl shows a 13CNMR peak around 200 ppm, so choice A is eliminafed. Only choice B remains,

'.,:you sl-rould pick it. The absence of an IR peak around 1700 cm-1 should reaffirm that a carbonyl is not present -Compound T. This means that choice B is the best answer.

Choice B is correct. The total number of hydrogens in the compound, according to the NMR is eight, of whr::five are on the benzene ring. This eliminates choice C. The pisug" states thal the compound is an ester,:.choice A, a ketone, is eliminated. The best answer is choice B, because the methyl shift is around 4.0 ppm. Th,-is tough to know without an NMR chart. In the passage, it is mentioned that the alkoxy group hai a shift :the 1HNMR between 3.5 and 4.0 ppm. if you can'f recali a fact, search for it in the passage.

Choice C is correct. The three separate IUXMR spectra are to be compared to determine the compounds :solution. It is important that nothing varies between samples, so the same solvent should be used i1 each ca.s=Any impurity peak wouid be common to all of the spectra and thus could be eliminated. This makes choice Cthe best answer. A common solvent has no effect on the pH or the behavior of a protic species in the lF{Nlt:Although lHUltR invokes an externai magnetic field, the solvent has no bearing on the iielcl.

Choice B is correct' This question is typical of spectra-to-structure-to-name questions. There are four carbons :Compor,rnd III, so choices A and C are elj.minated, bec;ruse they have only three carbons, according to t:-nomcnclature. Translating the narrles of cl-roices B and D yields the following two structures:

H3CH2C ocH3 H:C OCH2CH3

Ethyl ethanoateMethyl propanoate

Both A anci C would have identical peaks of a 2H quartet, 3H triplet, and a 3H singlet. The key feature is ti.3H singlet at approximately 3.5 ppm. This indicates that the methyl group is attached to the ester oxygen :iin choice B. You should pick B to score.

47. Choice A is correct. The four answer choices are the result of the following bonds: O-H (3500 "^-t), sp3C--

(2980 cm-l), C=O (1685 cm-1;, and C-O (1300 cm-1). An ester contains an sp3-f-g bond, a C=O bond, and a C-ibond, but it does not contain an O-H bond. This means that there will be no O-H stretcl-r (thus no IR peak at i:near 3500 cm-1; associatecl with the IR spectra of an ester. Cl-roose A and be a stellar achiever at choosr-r.:correct answers^

48. Choice B is correct. Compound II has two oxygerls in its formula, so choices A ancl C are out immediately. T:'.question here is reciuced to placing the a1ky1 group either on the ox\1gen (as in choice B) or on the carbon., -

carbon (as in choice D)" Because the peak for the two hydrogens is around 4.0 ppm, the ethyl group must'r=attached to the oxygen, making choice B the best answer. You should note that when there is a jH quartet ar.:a 3H tripiet in the spectra, there is an isolatec-l ethyl group ir-i the n-rolecule.

mhllL.

45.

46.

o

Ao

A

Copyright @ by The Berkeley Reviewo 172 Section II Detailed Dxplanations

+Y. Choice D is correct. Taken sequentially, the signals are a singlet (one apex), a quartet (four apexes), and atriplet (three apexes). This makes choice D the correct choice. The shape of a peak, you should recall, is due

to the presence of hydrogens on the neighboring carbon.

Choice C is correct. An NMR solvent should be inert and show no peaks in the spectra. All of the compounds are

deuterated, (contains 2H rather than 1H) so they show no peaks in the 1HNMR spectrum. The most reactivesolvent listed is the alcohoi. The problem with an alcohol is that when it is heated or given enough time, itcan undergo a transesterification reaction and change the ester. This makes choice C the best choice. An ester

should be soluble in all four of the solvents. It is a general rule that protic solvents are bad choices for theNMR, because they exchange for hydrogens and are typically more reactive than aprotic solvents.

Choice C is correct. 2-methyl-3-pentanone has four nonequivalent hydrogens, according to the symmetry of the

molecule. The structure of 2-methyl-3-pentanone is drawn below:

H*C"l

bl c

A../tt\ ,/t'=-.ff,'llo

This resuits in a ratio of 6:3:2:l for the four nonequivalent hydrogens, which makes choice C the best answer.

Choice A is correct. An isolated isopropyl group has two equivalent methyls groups (six hydrogens total)adjacent to a CH group (one hydrogen). The six hydrogens of the methyl groups are expressed as a doublet(being adjacent to one hydrogen), and the one hydrogen of CH are expressed as a septet (being adjacent to sixequivalent hydrogens). This results in a proton NMR with a septet (1H) and a doublet (6H). This is choice A,so choice A is the best of all possible choices,

Choice D is correct. Pentanal is an aldehyde, while 2-pentanone is a ketone. An aldehyde is bestdistinguished from other compounds by a peak near 9.7 ppm for the aldehyde proton. Choices A and B shouldbe eliminated, because both pentanal and l-pentanone have a peak in the IR just above 1700 cm-1 for the C=Obond and a peak in the proton NMR between 2.0 and 2.3 ppm for the alpha hydrogens. Neither comPound has a

proton NMR peak between 3.5 ppm and 4.0 ppm, because that shift value is attributed to hydrogens on a carbon

bonded to oxygen. Only choice D, a peak near 9.7 ppm, is for a peak that is unique to the aldehyde, so the

presence of the peak confirms that the compound is pentanal, while the absence of the peak supports that the

compound is 2-pentanone.

Choice B is correct. An alcohol hydrogen is rnildly acidic. This means that when a base is added to an alcohol,

ihe protic H on oxygen can be removed. If the alkoxide formed is placed in deuterium-labeled water, then the

alkoxide can remove deuterium from water to form a deuterium-labeled alcohol. Deuteriums do not appear inthe 1HNMR, so the peak for the H on oxygen disappears, making choice B the best answer.

o+DO

Choice A is correct. A methyl ketone has an isolated CH3 group adjacent to a carbonyl. Because the methylgroup has no hydrogen neighbors, it has no coupling, and thus is a singlet between 2.0 and 2.5 ppm. Pick A.

R_d:\ e

-

r\ "\/ +oo1

-H----/

D-dl eu "\/ +on1

-D---/

6 a hydrogens1 b hydrogen2 c hydrogens3 d hydrogens

D-\ o+Ro

H

D

-::r-right @ by The Berkeley Review@ 173 Section II Detaited Explanations

rytr5.

56. Choice B is correct. A 1HNMR peak at 9.7 pprr- indicates that the compound is an aldehyde. This eliminat.schoice A (an ether) and choice D (a ketone). The ratio of 1,:7:6 lor theirea of the signali indicates that the::are two equivalent methyl groups (that accounts for six equivalent hydrogens) and two unique hydrogens on +;,:compound. Only choice B has two equivalent methyl groups, so choice B is the best answer. Theie are fc::unique types of hydrogens on butanal, so butanal would show four signals in its 1HNMR, not just three signals.

Choice D is correct. Having no IR absorbance between 1600 cm-1 and 1750 cm-1 indicates that the compound h.:-.neither a C=C bond nor a C=O bond in its structure, both of which have IR signals between 1600 cm-1 and 1li,cm-1' This eliminates all of the choices except choice D. The compound ,rr.rit b" cyclic to account for the o:*degree of unsaturation.

Choice C is correct. In Spectrum I, the distance between the peaks in the aikene region (coupling constant :;greater than it is in Spectrum II. A larger coupling between vinyiic hydrogens is attributed to the tra:=compound. This is choice C.

Choice D is correct. The methyl group has three hydrogens, so the signai for the methyl group on benze:.cannot be the peak at either 7.25 ppm and 5.37 ppm, because neither of those peaks contains three hydroge:--This eliminates choices A and B. The alkoxy methyl group on the oxygen is found farther downfield than f *methyl on benzene. Thus from the values on the spectra, the methyl on benzene (3H singlet) comes at 2.25 pp::which is choice D.

Choice C is correct. In order for the disubstituted benzene compound to have the same number of peaks in -:*carbon NMR as it has carbons, the compound must not have symmetry. Symmetry results in equivallnt carbc:-.resulting in fewer 13CXUR signals. the only asymmetric iompound u-orlg tire answer choices is the n.-:.lsubstituted benzene with two different substituents. Choice C is correct. The choices are drawn below:

B.X

57.

59.

60.

A.

6 Carbons

3 l3cltttR

peaks

cc

I

/Cb,'i-'

?Cr -CcX

6 Carbons

4 13CNMR peaks

C.XI

c,/cu.- auil 1"

C.\ ^ t Cr--.

udY

6 Carbons

e 13CNMR peaks

D{

'iiT"'f.c.1 lzC.

$'6 Carbons

4 13CNMR peaks

61. Choice B is correct. The 1H peaks (the vinylic hydrogens) in the spectrum fall between 4.88 and 5.37 py=".implying that the hydrogens on the alkene are found in this range. Hydrogens on an alkene are referred tc mvinylic hydrogens. This means that vinylic hydrogens are found in the range of 4.5 ppm to 7.0 ppm. This ma..:schoice B the best choice.

Choice D is correct. The compound contains a C=C double bond, which has an IR absorbance between 1620 c::'')and 1660 cm-1. The best choice is 1640 cm-l, choice D. An IR absorbance of 3500 cm-1 is due to a O-H bond, ar. Jiabsorbance of 2220 cm-1 is d"ue to a C=C bond, and an IR absorbance of 7720 cn-7 is due to a C=O bond.

Choice D is correct. With meta-substitution, all of the benzvlic hydrogens would be different, because --:*

benzene compound would be asymmetric. There would thus be four nonequivalent hydrogens in a 1:1:1:1 ra-r:,Picking D is a beautiful thing on this question.

64. Choice D is correct. An ethyl group on benzene is composed of a CH3 group next to two equivaient hydrogi--r,.,(making it a 3H triplet) and a CH2 group next to three equivalent hvdrogens (making it a 2H quartet). I:describes choice D.

63.

Copyright @ by The Berkeley Review@ t74 Section II Detailed Explanatio

,ffi.

Choice C is correct. Because the compound contains two oxygen atoms, choice A (an aldehyde) and choice D (aketone) are both eliminated. There is no peak in the IR spectrum that is indicative of an O-H bond, so thecompound is not a carboxylic acid. This eliminates choice B. The only choice left is choice C.

Choice C is correct. The molecular ratio of hydrogens cannot be 3:1 in that region of the proton NMR spectrum,given the shift values. In order for there to be only one hydrogen on a carbon, there must be other groupsittached to that carbon. If the other group was caibot'r-based, ihere would be more signals in the IHNVIRspectrum than two and more signals in the 13CNMR spectrum than four. If the groups contained no hydrogen,then they would affect the shift value, making it farther downfield than 2.0 ppm. They are alkyl hydrogens,so the ratio is most likely 9:3, caused by the presence of equivalent methyl groups. Any alkyl group besidesmethyl would show coupling and not singlets. The structure must therefore have a tertiary butyl group (nineequivalent hydrogens forming a singlet) and an isolated methyl group (three equivalent hydrogens forming a

singlet). Choose C for best results.

Choice D is correct. A 13CNMR peak is for carbon, not hydrogen, which eliminates choices A and B. Thepassage states that the carbon of the C=O bond is found at170.2 ppm., which eliminates choice C. The bestanswer is choice D. Carbons bonded to an electronegative atom that arc spJ-hybridized typically show a

signal between 60 and 90 ppm.

Choice C is correct. As stated in the passage, the peak at 170.2 ppm is attributed to a carbonyl carbon.Carbonyl carbons are found between 770 and220 ppm in the 13CNMR. Choice A is an ester, choice B is a ketone,and choice D is an aldehyde, of which all three contain a carbonyl group. This means that choices A, B, and Dcan be eliminated, because they all exhibit a signal in the l3CNltR that is greater than 100 ppm. An alcoholhas a peak around 70 ppm to 80 ppm for the alcohol carbon, but not greater than 100 ppm. The best answer ischoice C.

Choice A is correct. The shift value for hydrogens on an alpha carbon (the carbon adjacent to a carbonyl) isfound to be between 2.0 ppm. and 2.5 ppm. The alpha hydrogens are described in the answer choices as

hydrogens on a carbon adjacent to a carbonyl (C=O) bond. This is choice A; your choice for a question like this.

Choice D is correct. A singlet in a 1HNMR spectrum occurs when there is an isolated hydrogen (or group ofequivalent hydrogens), with no hydrogens on the neighboring atoms. This eliminates choices A, B, and C, andmakes choice D the best answer.

Choice D is correct. Compounds A and C are eliminated, because they do not contain two oxygen atoms, as is

stated in the passage. Choice B is eliminated, because the CH3 of the methoxy group would show a shift valuebetween 3.5 ppm and 4.0 ppm. The peak for the lone methyl group is found near 2.0 ppm, which indicates thatthe methyl group is adjacent to the carbonyl. The correct structure is choice D.

tr!"

:a Choice C is correct. The oxidation of a secondary alcohol to a ketone results in a carbon that goes from ai3CIrIptR signal of roughly 75 ppm to a signal of roughly 200 ppm. This conversion can be monitored.easily by13CNMR, so choice A is valid. Nitration of ethylbenzene results in a carbon that changes from a TTCNMR

signal of roughly 115 ppm to a signal of roughly 140 ppm,.lecause the nitro group changes the immediatechemical environment. This conversion can be monitored by 13CNVR, so choice B is valid. Deprotonation of acarboxylic acid. does not change the immediate environment (adjacent atoms) of any carb.oSr so no drastic change

in change in any l3CNltR sifral is observed. This conversion cannot be monitored Uy isg*tR, so choice C is

invalid, and thus is the best choice. Reduction of an alkene to an alkane results in a carbon that goes from a13CNMR signal of roughly 100 ppm to a signal of roughly 20 ppm. This conversion can be monitored byl3CNVln, so choice D is valid.

:i Choice C is correct. According to the reaction in the passage, the compound is an aromatic ketone. The peak at

206 pprn, according to the data in Table 1, confirms that Isomer I is a ketone. Choice D (the aldehyde) is thuseliminated.. Becau-se there are only seven peaks for the 13CNMR, the compound must have symmetry, so

choices A and B are eliminated. The best answer (one that contains only seven unique carbons) is choice C.

J:n-right O by The Berkeley Review@ 175 Section II Detailed Explanations

74. Choice A is correct. An aldehyde, according to Table 1, is found between 185 and 200 ppm, and no peak is fou_i.:in this range for either structural isomer. Choice A may be the corect choice. It stated in the pu5ug" that c,:occasion, the shift may not fall exactly in the range given. An alkene, according to Table 1, is found beiween 1- -and 140 ppm, and there are peaks found in this range for both structural isomers. Choice B can thus i.eliminated. A ketone, according to Table 1, is found betr.r'een 205 and 220 ppm, and peaks are found in this rans=for both compounds. Choice C is thus eliminated. A methyl group, according to Table 1, is found between 10 a-r;35 ppm, and there are peaks found in this range for both structural isomers. Choice D is thus eliminated. Tf.-best answer is choice A.

Choice D is correct. According to the passage, the least intense peak is caused by a carbon with no hydroge:,.attached. This makes choice D the best answer. The carbons with no hydrogens attached in a disubstitute:benzene derivative with a carbonyl are two of the benzene carbons (the beniene carbons with a substituer:attached) and the ketone carbon. It has sp2-hybtrdization, not sp3-hybridization, and it is achiral. Th-"eliminates choices B and C. While it has sp2-hybridization, that is not the cause of its low intensity. Chor.:D is a better answer than choice A. This is a question that rewarded the test taker who sifted through t:-information in the passage. This will happen on occasion, because even the science sections are reading u*u*..

Choice C is correct. Para-methoxybenzaldehyde has a total of eight carbons in its structure, but the molecu-.contains a mirror plane that reflects two pairs of equivalent carbons. Using symmetry, this means that the:=are only six unique carbons, so the best answer is six signals, choice C. The structure is shown below:

o\^-t!u

I

CD.c/-\ c.iltoC-..

-Z CdeClr

ocH3

Choice D is correct. Because the two compounds are isomers, they have the same molecular formula. Havrr:the same molecular formula results in having the same units of unsaturation. This eliminates choices A and E

Each compound has a benzene ring and a carbonyl group. The benzene ring has three n-bonds and the one rir-r:There are four ur-rits of unsaturation due to the benzene ring alone, so choice C is eliminated. When the carbor.-. -

n-bond is accounted for, there are five units of unsaturation in the molecule, so the best answer is choice D.

Choice B is correct. It is stated in the second paragraph of the passage that quaternary and carbonyl carbor..generate peaks of low intensity. There is no carbonyl in the compound, but carbon 2 of 2,2-dimethylbutane ha-.four other carbons attached, making it a quaternary carbon. The best answer is choice B, carbon 2.

Choice A is correct. Spectrum I contains a quartet (2H), a triplet (3H) (this combination is a dead give-an-a:for an isolated ethyl group), and a singlet (3H) (a dead give-away for an isolated methyl group)- The t',=question here is whether the ethyl group or methyl group is attached directly to the oxygen of the ester grou;The degrees of unsaturation (1) and number of oxygens (2) tell you the compound must be either an ester or :carboxylic acid. The lack of a broad peak between 10 - 12 ppm eliminates the possibility of the compound bef :a carboxylic acid, so the compound must be an ester. Because the quartet is so far ciownfield (at a higher sh,:-value), the ethyl group is attached to the oxygen. This makes the best choice an ethyl group on the meth'. -

ester, whose common name is ethyl acetate. Choose A for best results.

Choice B is correct. A triplet is the result of coupiing to the neigl-rboring hydrogens (there are two hydrogens c.the adjacent carbon in the case of a triplet). The integral (quantitr. of hvdrogens for the signal) has no effect c:the shape of the signal, meaning that the peak shape does not tell vou any information about the hydrogens c-the signal, only about the neighboring hydrogens. This eliminates choices A and C. A triplet is therefore th=result of neighboring a CH2 group. Choose B for a grade A, genuine, altogether correct, best answer.

/J.

76.

78.

79.

80.

E&F

Copyright @ by The Berkeley Review@ 176 Section II Detailed Explanations

Choice B is correct. Spectrum II contains a quartet (2H), a triplet (3H) (this combination is a dead give-awayfor an isolated ethyl group), and a singiet (3F{) (a dead give-away for an isolated methyl group). Because the

3H singiet is found around 4.0 ppm, the methyl group is attached to the oxygen. This makes the best choice a

methyfgroup on an ethyl ester. Choose B for the happiness of another correct answer. The drawing below lists

how the name and structure are determined for the compound.

Singlet in the 3.5 to 4.0 PPm range

/ilH;C-.

o cHzcHj

--Jmethyl propanoate

Nomenclature rules state that the alkyl group on oxygen is named first, followed by the ester chain' This

makes this compound methyl propanoate.

Choice C is correct. This is one of those trivial facts that you should know. A peak in the neighborho od ol 7ppm is indicative of aromatic hydrogens, which are found in benzene compounds. Pick C, to score bigl

Choice A is correct. A carboxylic acid has one proton that forms a broad peak between 10 ppm and 12 ppm in the1nruVtR. The hydrogen in question is the acidic proton of the carboxylic acid. Because there is only one proton,

the peak between 10 ppm and 12 ppm has an integration value of one hydrogen, so choices C and D are

eliminated. An ester nui at'r alkyl group attached to the noncarbonyl oxygen of the ester. Protons on the firstcarbon from the oxygen have a peak between 3.5 ppm and 4.0 ppm. This eliminates choice B, leaving onlychoice A as the possible answer. The peak between 2.0 ppm and 2.5 ppm is the result of alpha hydrogens,

which ot" preser-ri in both an ester and a carboxylic acid. 'lhis means that a peak between 2.0 ppm and 2.5 ppm

cannot be used to distinguish an ester from a carboxylic acid. The correct answer is choice A'

Choice B is correct. 4-Heptanone is a seven-carbon ketone with the carbonyl directly in the middle- The

structure is symmetric, so there are many equivalent carbons and hydrogens. There are four unique carbons, of

which tl-iree contain hydrogens. This resulti in three signals in the 1HNMR for 4-heptanone. The best answer

is choice B.

Choice A is correct. For Compound II, the absence of a 13CNMR signal between 180 ppm and 230 ppm supports

the idea that it has no C=O group. This eliminates choice C. No peak above 5 pp* in the rHNMR spectrum

confirms that there is no double bond. This eiiminates choice D. This means that the unit of unsaturation in the

compound must be the result of a ring. The presence of a broad peak in the 1HNMR spectrum suPPorts,the idea

that the compound is an alcohol, eliminating choice B and making choice A the best al1swer. The IrCNMR

spectrum shows that there is great symmetrv in the structure. The choices are either cyclopentanol or 2,3-

dlmethylcyclopropanol. The integral of the proton NMR says that there are mostly CH2 groups present,

which iurro., ryclopentanol over 2,3-dimethylcyclopropanol. Cyclopentanol is drawn below:

13cNMR,

There is a mirror planecutting throughthemolecule, so there are three unique carbons,resulting in three different signals.

Broad------>1tO H<- Multiplet(1H) \ / (1H)

,zt\H"C CH"

MuitiDiet42c- v (4FU

\,2Tripiet (4H)

Copyright @ by The Berkeley Review@ Section II Detailed ExPlanations

86. Choice D is correct. Each unique carbon within a molecule exhibits a unique signal in the 13CNMR spectrum, sochoice A is valid. This eliminates choice A. A carbonyl carbon has a slgnaiaround 180 ppm in a 13CNVRspectrum, which b d+lf:=llom other signals in the 13CNMR spectru-. ih" presence of a carbonyl group canbe identified using l3CNl'tR spectroscopy, so choice B is valid and thus elirninated. The substitution of abenzene ring affects the symmetry of the molecule. For instance, if the substitution is para, then there is amirror plane in the molecule. This results in fewer unique carbons, which is seen with 13CNMR spectroscopy.This makes choice C a valid statement, and therefore it is eliminated. The geometry about a doulle bond, cisversus trans, does not express itself in l3CNltR spectroscopy. Typicaily, geJmetry is determined by looking at

ll:::_"-p_t*g constants of the vinyiic hydrogens in 1HNMR spectroscopy. bhoice D cannot be determined usingIrCNMR spectroscopy, so choice D is the best answer.

Choice C is correct. According to the data in the passage, only Compound I has a UV absorbance above 175 nm.This means that only Compound I has a r-bond. There is only one unit of unsaturation, so Compound I can have,at most, one n-bond. This means that Compound I cannot be a conjugated diene, which eliminites choice A. If itwere a conjugated diene, there would be a UV absorbance above 200 nm for the n to rc* transition. Compounds IIand III cannot be carbonyl compounds, according to their UV data. If there was a carbonyl gro.rp on themolecule, there would be a n-to-n* transition around 190 nm and an n-to-n* transition around jOO n*. thiseliminates choice B. Because Compound I has two UV absorbances and only one n-bond, it must be a carbonylspecies of some sort, While it is not possible to decide between an aldehyde and ketone based on thisinformation, choice C is a solid answer. Because Compounds II and III show no UV absorbance above 1,75 nm,there is no n-bond present. However, choice D refers to Compounds I and II, not II and III, so choice D iseliminated. Choice C is the best answer.

Choice D is correct. For Compound III, the absence of a broad peak in its 1HNMR spectrum confirms that thereis no alcohol in the compound. This makes choice A a valid statement, and thui the incorrect choice. Theabsence of a peak around 180+ ppm in its 13CNMR spectrum confirms there is no C=O present. This makes choiceB a valid statement, which eliminates it. The absence of an absorbance above 180 nm in the UV-visiblespectrum implies that there is no n-bond ir-r the compound, confirming that the structure must be cyclic to accountfor the one unit of unsaturation. This makes choice C a valid statement, which eliminates it. The I3CNIrARshows very few signals (only three), which implies that there is great symmetry in the structure. It must be asymmetric cyclic ether. This makes choice D an invalid statement, making it the best answer.

Choice B is correct. Key features from each spectrum must be extracted. From the molecular formula, we knowthere is one unit of unsaturation and one oxygen. This means that the compound must contain either a ring, aC=C bond, or a C=O bond. All of the answer choices fit these criteria, so we must use the spectroscopic dita.the l3CNVtR data show a peak at 211 ppm (1), and that no two carbons are alike. This means that thecompound is a carbonyl, which does nothing to eliminate any choices. If the compound were 3-pentanone, itwould show only three signals in the 13CNVR due to its symmetry, which eliminates choice C. The fact thatno two carbons are alike also eliminates choice D, which has two equivalent methyl groups. the 1HN\4R datashows a 3H singlet at 2.08 ppm, and no peak between 9 and 10 ppm. This means that the iompound is a methylketone and not an aldehyde, which elj.minates choice A and choice C. The ratio of the hydrogen signals in thelHNltn (3 :2 :2: 3) supports choice B. The structure is shown below:

1HNMR,

13CNMR,

There can be no branching,that the structure is linear.

This implies

E91"

87.

88.

89.

oll :v-vis:

18e nm and 268 nm

n"r/'T-.r'/ tV',. CNMR: 21 1 PPm

t' I t' trrJptet v,,rlipturrrlpt"t silgr"t(3H) (2H) (2H) (3H)All five carbons are different, so there are five different signals.

because the proton NMR shows that the integral values are 3,2,2,3.

Copyright @ by The Berkeley Review@ t7a Section II Detailed Dxplanations

90. Choice D is correct. Integration is used to determine the reiative quantity of hydrogens within a signal bylooking at the area of the signal. Integration does not change with magnetic environment, so choice A iseliminated. The neighboring hydrogen atoms affect the splitting, not the integral, so choice B is eliminated.Integration does nothing to determine the presence of atoms other than hydrogen, so choice C is eliminated.The best answer is choice D.

Choice C is correct. Five hydrogens constituting a singlet with a shift value between 6.0 and 8.0 ppm indicatesthat the compound is a monosubstituted benzene. The three remaining hydrogens make up a methyl group. Thisnow becomes a nomenclature question, rather than a spectroscopy question. The correct name for a methyl groupattached to benzene is methylbenzene. The common name for methylbenzene is toluene. Choose C for optimalresults.

Choice D is correct. The formula contains six hydrogens in all, so the sum of the ratio values must equal 6. Thefirst peak is shortest, the middle peak is the second tallest, and the last peak is the tailest. This means thatthe values must be ascending. The only combination of ascending values adding to 6 is 1:2:3. Choice D is thebest answer.

Choice D is correct. The compound 2-bromopropane has two unique types of hydrogens, so it has two peaks in itsIUNUn spectrum. The two terminal methyl groups are equivalent, so they are seen as one signal with anintegration of 6. The middle carbon (carbon 2) has one hydrogen, so it has a signal with an integration of 1. Thepeak shape is determined by adding 1 to the number of hydrogens on the adjacent carbons. The six equivalenthydrogens have one hydrogen neighbor, so there is a doubiet of integration 6. The one hydrogen has sixhydrogen neighbors, so there is a septet of integration 1. Choice D is the best ansr.ver.

Choice C is correct. A quartet is the result of the observed hydrogens being coupled to three equivalenthydrogens. This is often the result of hydrogens that are adjacent to a methyl group on one side and no otherprotons on the other. The quartet hydrogens are in bold face, and the neighboring three hydrogens are boxed inthe drawing below. Choice C is the oniy structure that shows no quartet in its proton NMR spectrum.

91.

92.

93.

9'1.

o

AA.

cHr@ fnrburclHrcyrc CH:

D.

H3CH2CH2

o

AB.

Choice D is correct. A doublet is the result of hydrogens on a carbon that neighbors a carbon with only one

hydrogen attached (most easily recognized as a tertiary carbon). In each of the first three answer choices, themethyl group attached to the interior of the carbon chain is bonded to a carbon with only one hydrogen (a

tertiary carbon), which results in every compound having a doublet with an integration of three hydrogens.This leaves choice D as the best answer.

Choice A is correct. A positive iodoform test, as stated in the question, is caused by a compound with threeaipha hydrogens on one carbon. This means that the iodoform test is positive for a methyl ketone, whichwould have a CH3 group adjacent to a carbonyl (there are no hydrogens on a carbonyl). With no hydrogens onthe neighboring carbon (carbonyl), there is no coupling and thus a the peak is a singlet. Pick A for the pleasureof correctness. The iodoform test works by removing an alpha hydrogen to form an anion. The anionsubsequently attacks iodine, adding an iodide to the alpha carbon. This is repeated two more times, until thereare three iodides bonded to the alpha carbon. The CI3 group is a great leaving group, and it forms a yellow,oily compound when protonated.

(H3C)2HC

C.

(H3C)2HCH2C cHr@

o

AC

o

AcHg

95.

q5.

-opyright @ by The Berkeley Reviewo Section II Detailed Explanations

97. Choice D is correct- The hydrogen responsible for the broadness is the carboxylic acid proton. It is bondecl to anoxygen, making choice A an incorrect answer. The couphng to eight or more other hvdrogens would result in amultiplet made of many sharp peaks, not a broadened signal, which eliminates ciroice B. Choice C can beeliminated, because hydrogens on carbon do not form hydrogen bonds. the hydrogen forming the hydrogen bondin this molecule is bonded to oxygen. This makes choice D the best answer.

Choice B is correct. Because the compound on the left has cis orientation about ttrre internal double bond, whilethe compound on the right has trans orientation abor-rt the internal doubie bond, the two compouncls must begeometrical isomers of one another. Pick choice B, and you won't be sorry. In case you were considering choiceC, the two compounds have the same absolute configuration, so they cannot be optical isomers.

Choice C is correct. Both double bonds are trans, so choice B is immediately eliminated. Both have the sameconnectivity, so that eliminates choice A. The last thing to check is the chiral centers, and each has just onechiral center. The compound on the left has R chirality, while the compound on the right has S chiralitr.This makes the two structures optical isomers, choice C. If you flip the structure on the right to align with thestructure on the left, it can be seen that the chirai center has changed" A change in a chiral center results in anoptical isomer.

100. Choice A is correct. The best way to do this probiem is the systematic counting of carbon backbones, startingwith the iongest carbon chain possibie (five carbons). The tally for each possible carbon backbone is drar,r,rbeiow. There are only three possible structures: pentane, 2-methylbutane, and 2,2-dimethylpropane. Tlr-cstructures that are in fact structural isomers must have different IUPAC names. Pick A to be a correct anst'e:picker person.

CI

I

99.

I

I

C

C-CI

C

,ars".s

Copyright O by The Berkeley Review@ Section II Detailed Dxplanations

Molecule does not fitScenterilr>v>a>.J

template!

TIEBITEIEYl) g. p- y.1.-6. u/'

Section IIIStereochemistry

by Todd Bennett

Molecule fits template!Rcenlelill>Y>a>l

Speciahztng in MCAT Preparation

Stereochemistry Section Goals

oB A stereogenic center (often referred to as a chiral center) is most commonly made up of a centralatom (gsually carbon) wlth four unique substitutents aitached. The stere6centei ii'identified as

Be able tq identify stereocenters and chiral com

Be able to and stereoisomers.

Ntrt,,,- -i,n_^_^

ct/

:,it-h.* R (lf,t1l

: lec,tusl or S (Latin,. sinister) to define its orientatio" i.rpuce. g";;ti;;ilI t"qiiir"that you identify the number of chiral centers and often label them accordini to convention.'ThelSguentlY cite.d example of a,non-t1.pical situation involves allene, which has sf-hybridized carbonsthat can be chlral. An example of allene enantiomers is drawn below:

H

CI

H

cl

&

a? Youmustunderstandthedifferencesbetweenenantiomersu''da

9"*""*.'.fl_1":"_.::Tf?.yf:ji9T lh"to op.tica] inactivity and structure. iL" rnlJi't i-'o,' specialcasesinvol,vi"g :"g?.t: such as anomers anh epimers- M<ist importantly, when giu; irvo strucLures,be able to identify their relationship if they are'stereoisomers" of one another.

It should be as second nature to yog that sugars occur naturally in the D form, *fli.f-r " defined bvhaving the penultimate carbon wittr R sterJochemistry. The iypical exception ir ;;;;ith ;t;AVpes where one of the sugars in the antigenic determin-ant is L-fticose. lt shbuld be as second nahurel.o,y,o:1la1a1ino.acids oicur naturally "in the L form, which is defined by il;ing the;lpd;";il;ylti-::1"^r:o-:,1:iil]V Jexqep,t

for cysteine). A typical exception is seen wi[h transiriptidase enzymeswnere the acttve amtno acrd is D-alanine.

@? JustIiketheboilrngpointandthemeltingpoint,theopticalrotationi'upnyffiDeused to Identlty a molecule. Ihe optical rotation is a measurement of the rotation ol .r nlan,,dl"'""'; _"': :"'::.,6 rrrL rrrcrrrrrS PUrrrr, .rrE uyrrldr rurdr.rurr ls d pftyslcal pl.opeftv tnat can

::,T:9P,i1""l,ilt: i?],"jq" The opti"cal,rotation is a measurement df t'he rotalion o1, pti""J

Be familiar with common exam of chiral molecules.

Be able to use rotation data to i an unknorun

p,."hlt::1,!,qht bi a solution of the cipticatly active .o*po.rtld. A;.-;;;;;ppli;"ti;"'il;h;ldentrtlcatron an unknown susar.sugar.

Be able to disti ish ic substitution mechanisms.

ancl the second is the Sg2. They aredefined bv the number of leactants in the rate determining stepof the mechanism. You.must 6e able io predict the reaction from the l.ltiri.."ailio"s, t"co"g"ir'"the reaction from the intermediate or trahsition state, and identify the reaction it;;ih6 p;;f,;;r.The differences include solvent, strength.of leaving group, sterii hindrance, ability to slabilize acarbocation intermedia te, and stereoch"emistry.

strength of a leaving group is dconjugate acid of tlieleaving pthe basicitv of the nucleoohilel

There are two mechanisms for nucleophilic-substitution that you must know. fft" fitst ir tt-r" Slryfand the second is the S1x2. They are deTined bv the number of ieactants in the rate cletermininc, sfen

gtg*p: Equally, the strength of anuileopl-rile can be predictdd fromof the nucleophilel Agpin, dolvaiion and st&ic hindrance it"y;;;l";ih;;;"dil;l

llili:li

ilit i:

Be able to recognize typical nucleophiles and leaving groups.You must recognize what makes a go-od leaving grgup, and what effect this has on the reaction. Thestrength of a lehvi is dependent onthe solvent and can be qredicted from the acidity of the

)il[*

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Organic Chemistry Stereochemistry Introduction

stereochemistry involves the asymmetry of a molecule. we can consider theasymmetry as a whole or the asymmetry about specific atoms in the molecule,most often carbon. If there is asymmetry within a molecular structure, thecompound's reactivity, physical properties, and stability are all impacted. Thestudy of stereochemistry has direct implications in the biological applications ofmolecules.

In this particular section we shall address the concept of configurationalisomerism and the many different classifications of configurational isomers. Aswith ail isomers, configurational isomers have the same atoms within themolecule, but they differ in some manner so that the molecules are notsuperimposable on one another. No matter how a compound is rotated orcontorted, it is not superimposable on its configurational isomer. As aconsequence of their different configurations, one configurational isomer mayhave the correct arrangement of atoms to offer minimal steric hindrance in achemical reaction with another asymmetric molecule, while anotherconfigurational isomer proves to be too stericaily hindered on one side toundergo reaction. This is frequently seen with enzymatic chemistry, whereenzymes have several stereogenic centers and are highly specific about whichconfigurational isomer can bind and undergo a reaction.

Configurational isomers can be categorized as either optical isomers orgeometrical isomers. optical isomers rotate plane-polarized light whilegeometrical isomers are structures with limited rotation. In addition to thatcategofization, configurational isomers can also be categorized as eitherenantiomers or diastereomers. Enantiomers are nonsuperimposable mirrorimages while diastereomers are nonsuperimposabie structures that are notmirror images. The two categorizations are not mutually exclusive; meaning apair of configurational isomers could be enantiomeric optical isomers,diastereomeric optical isomers, enantiomeric geometrical isomers, ordiastereomeric geometrical isomers.

In this section we shall also address nucleophilic substitution. We will considerthe two mechanisms for nucleophilic substitution: the S511-mechanism and theSp2-mechanism. In a nucleophilic substitution reaction that proceeds by an SN1-mechanism, the leaving group leaves to form a carbocation intermediate beforethe nucleophile attacks. An S511-reaction has a unimolecuiar rate-determiningstep. In a nucleophilic substitution reaction that proceeds by an Sp2-mechanism,the nucleophile attacks the electrophile from the opposite side of the leavinggroup and forces the ieaving group off of the electrophile. An Sy2-reaction hasoniy one step, a bimolecular step. We will compare and contrast the conditionsand features of an Sp1-reaction with that of an S52-reaction to establish a set ofcriteria you can use when deciding which mechanism (Sp1 or Sp2) is applicablefor a given nucleophilic substitution reaction.

We will consider the impact of stereochemistry on reactant interactions,transition state formation, and product distribution. We will present the basictenant that if the reactants are optically active, then the product mixture is likelyoptically active, and at the very least enantiomerically rich in one configurationalisomer (possible a geometrical isomer, which is optically inactive.) We will alsoconsider enantiomeric distribution in a product mixture and discuss ways toincrease the optical purity.


Organic Chemistry Stereochemistry Configurational Isorners

Confi gurational trsornersStereochemistryStereochemistry centers around the formation, orientation, and reactivity ofmolecules with stereogenic centers, referred to as stereoisomers. The moleculeswe shall consider in this section are configurational isomers.

Configurational IsomersConfigurationai isomers have identical bonds, but they have a different spatialarrangement of their atoms, no matter how the structures are contorted.Common examples, with which you are familiar, are optical isomers. Opticatisomers, due to their asymmetry, rotate piane-polarized light. This is used as adiagnostic test to identify a specific configurational isomer. We shall first look atasymmetry and chirality, as configurational isomers are based on chirality.

AsymmetryA molecule with asymmetry has a site about which there is uneven distributionof the bonded atoms. Analyzing symmetry is critical, because at least one of twostereoisomers must be asymmetric in some manner if the two structures are notsuperimposabie. To understand stereoisomers, it helps to be familiar with mirrorplane symmetry and chirality (molecular asymmetry). Figure 3-1 shows theasymmetry of carvone and the symmetry of 2,2-dichloropropane.

-H5€Hsc CHs

C1

No plane of symmetry, sothe compound is chiral.

Plane of symmetry, sothe compound is achiral.

Figure 3-1

ChiralityChiral is the term assigned to a molecuie with no plane of symmetry, therefore a

chiral molecule has an asymmetric structure. Simply put, chirality is the "leftand right handedness" of a molecule. From our perspective (keeping at the leve1of this test), a chiral molecule has at least one stereogenic center present. Astereogenic center is an atom within the chiral compound^that has asymmetrr-abor.rt it. For our needs, chiral (asymmetric) carbons are sf -hybridized carbonswith four unique substituents attached. Within 2-chloropentane there is onestereocenter (chirai carbon), as emphasized in Figure 3-2 below. Figure 3-2shows the two configurational isomers (enantiomers) of 2-chloropentane in sucha manner that the two structures are mirror images of one another. The planemirror reflects a configurational isomer as its image.

-;---- -'CHTCH2CH3plane I 'mlrror I

-_____;;__.v,1:I-ug" cl

vith four different groups attached.

'tLl

:tl&n

-;sp3-hyb.idirotitor-t

Copyright O by The Berkeley Review

Figure 3-2

The Berkeley Revieu

Organic Chemistry Stereochemistry Configurational Isomers

What makes this important is that an atom with four unique substituentsattached has two possible ways that the substituents may be connected (whichate mirror images of one another). The two structures, mirror images that haveidentical bonds, are stereoisomers that may exhibit different chemical propertiesdespite identical physical properties as one another. The biological ramificationsof chirality are important. For instance, humans digest only D-sugars (D refers toone of the two possible stereogenic orientations associated with the penultimatecarbon within a sugar backbone), because enzymes bind and react only with D-sugars. If this seems unclear, the following everyday analogies may help:a) A soew with right-handed threads does not fit into a left-handed nut.b) Your right hand does not fit rnto a left glove.:) A key with a groove on only one side and its mirror image do not open the

same iock.

Exampie 3.1 shows some examples of pairs of butane moiecules that have four'l1ique substituents on carbon two. The goal of the question is to develop skills:or quickly recognizing when two structures represent conformational isomersciscussed in section II) versus when they represent configurational isomers.

Example 3.1,ihich of the following structurai pairs represents the same molecule and not a:air of configurational isomers?

\.

F{3CH2C

DD.

irLcH2cH3

&

&

F{rCHrC&IH

""-&rx,.,,,!Br

i

-,./YcHrcH3&

D. cH.cH"l'J

:ul2clY/H &

CH"Clt-H3cH2clY/H

cH2cH3CHg

! olution- :u must decide whether the compournds are either enantiomers or the identical,:r-Lpound with different spatial orientation. Rotate the molecules in your mind' - see if the atoms overlay. If you do this successfully, then you will see that they=:e identical only in choice C. However, if this is hard for you to visualize, try a.=: of models. A shortcut you may recognize is that when two of the substituents

':e interchanged, the chirality of that stereocenter changes. In choices A, B, andl' two substituents have interchanged, making them enantiomeric pairs.

OH

-:pyright O by The Berkeley Review r85 Exclusive MCAT Preparation

Organic Chemistry Stereochemistry Confi gurational Isomers

We will start with a traditional approach to stereochemistry probiems and thenslowly work our way into shortcuts and visualization tricks. The first pair ofmolecules in Example 3.1 are mirror images of one another, because if you rotateeither structure by 180" about its carbon 1 - carbon 2 bond, the structures are notsuperimposable. According to our shortcut, when tlvo of the substituentsinterchange their locati.ons (H and I in this case), because there is only one chiralcenter, the tr,r'o molecules are mirror images of one another. Mirror images thatare non-superimposable are defined as enantiomers. The substituent exchange

shortcut shoulcl make it easier to recognize enantiomers.

The molecules in choice C in Example 3.1 are identical. If the left structure is

rotated counterclockwise by 120" about the carbon 2 - carbon 3 bond (as shown inFigure 3-3), the identical structure and orientation are formed. Note that when a

structure is rotated by 720" about a bond, the three other substituents interchangetheir locations on the molecule. The conclusion from this is that when threesubstituents are different from one structure to another, those two structuresrepresent different conformational isomers (orientations) of the same compound.

C1

I

BrlYcH2cH3zGH:-

Figure 3-3

Rotating molecuies in your mind becomes easier with practice, although if theski11 is never fully developed, you can stil1 answer stereochemistry questions brfollowing a few simple rules. If two groups are interchanged and the rest of themolecr,rle remains in place, then the two structures are configurational isomers. Ifthere is a mirror plane between two molecules and no mirror plane within themolecule, then the two structures are configurational isomers.

In addition to the symmetry of a compound with respect to another molecule.there is also internal symmetry to consider. For internal symmetry, you can lookfor either a plane within the molecule that reflects equal halves, or an inversiorr.

point. An inversion point is a point at the molecule's center of mass throughwhich a line passing through that spot will intersect the same atom at the same

distance, no matter which direction you proceed (positive or negative directionon the 1ine. Within asymmetric molecules, each stereogenic center is assigned a

letter, R or S, to describe its stereochemistry.

Determining Absolute ConfigurationThe identification and naming of a chiral center is based on nornenclatureconvention. There is a set of guidelines, the Cahn-Ingoid-Prelog rules, to follolfor determining R and S for a stereocenter (chiral carbon). The Cahn-Ingold-Preiog rules to determine the stereochemical orientation (R or S) are as follows:

1) First, you must prioritize (from heaviest to lighiest) the substituents that are

attached to the carbon of the stereocenter according to the atomic mass of the

atom directly bonded to the chiral carbon. (Get it from the periodic table)

2) Second, you must orient the molecule in such a way that the substituent wil;.priority number four points behild the plane of the molecule'

3) Third, you must draw a circular arc from substituent 1 through 2 and on to lIf the arc is clockwise, it is referred to as R. If the arc is counterclockwise, it r.referred to as S.

lx

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llr

s

llr ;

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'l,rCopyright O by The Berkeley Revierv ra6 The Berkeley Revieu


R is from the word rechts, which is right in Latin, while S is from the word sinisterwhich is left in Latin. If you point your thumb in the direction of substituentnumber four on a compound with R-stereochemistry, the fingers of your righthand will curl from one to two and on to three. Thus, R-chirality can be thoughtof as right handedness. The same holds true for your left hand with an s-center.

Figure 3-4 shows a generic R molecuie and a generic s molecuie as orientedaccording to convention. The steps of rotation presented, take the structure froma standard view to a view with the fourth priority substituent eclipsed, to theNewman projection from which the stereochemical identity is derived.

1Reposition by

rotating by 90" aboutC.6irot-C1 bond

Redraw tosee chirality_______>

Reposition by 3

rotating by 90' aboutRedraw to

see chlrality---------->C.51.o1-C1 bond

Counterclockwise: S

Figure 3-4

Knowing the terminology is key; it is recognition, not recail, that is emphasizedon a multiple-choice exam. Here are some nndified definitions of common terms.

Chiral Center (stereochemical center):A carbon with four unique substituents attached. Any carbon with four uniquesubstituents has two different orientations that it can assume (R and S). What isneant by "unique substituents" is not four different atoms, but four unique$oups including the atoms attached to the four atoms bonded directly to carbon.For example, carbon two of 2-chloropentane (see Figure 3-2) is chiral, because itras a chlorine (priority 1), a propyl group (priority 2), a methyl group (priorityit, and a hydrogen (priority 4)bonded to it. These four substituents are different:rom one another, therefore they are four unique substituents.

R-center:-\ carbon center that when you look down the bond from the chiral carbon to the:ourth priority substituent (usually a C-H bond) in a way that you can't see theiourth priority substituent, the remaining substituents form a clockwise arc,r-hen moving from priority one to priority two and on to priority three accordingio the priority rules. This can be thought of as a right-handed molecule whenllacing your thurnb in the direction you're looking and curling your fingers tonatch your right hand to the structure.

S-center:-1, carbon center that when you look down the bond from the chiral carbon to the:ourth priority substituent (usually a C-H bond) in a way that you can't see the:ourth priority substituent, the remaining substituents form a counterclockwise:rc when moving from priority one to priority two and on to priority three:ccording to the priority rules. This can be thought of as a left-handed molecule'";hen placing your thumb in the direction you're looking and curling your:rngers to match your ieft hand to the structure.

Clockwise: R

Copyright @ by The Berkeley Review ta7 Exclusive MCAT Preparation


Prioritizing Substituents to Determine R and STo prioritize, first you must look at the four atoms directly bonded to theasymmetric carbon. You then rank those atoms according to their atomic masswith the heaviest atom taking ihe highest priority. if two atoms are equal (as is

often the case n,ith carbon) you must continue down the molecule following thebonds outward from the chiral center. Figure 3-5 shown below presentsexamples with the priorities labeled on the molecules.

3

f',Ir

n4$'r n,sI

1

23

Tt' !n' T"'

"r'*'it H,cH,tA';;, "'t-t. "tY:.d ::,"=I>Br>C>H O>C>D>H Br>C=C>H C-C-C-C

CH2CH3 t CHs C=O > C=C > C{ > C-H

Figure 3-5

Shortcut to Determine R and S

As with so many other topics in organic chemistry, such as nomenclature, R andS questior-rs become easy and redundant with time. Once they become easy,there are useful quick tricks to help you to identify chiral centers as being eitherR or S. For instance, when the fourth-priority substituent is sticking out from the

rnolecule, the molecule must be rotated. To save time, it is easiest to first solvefor the arc using the structure as it is, and then take the opposite chirality for the

center. In the interest of saving time, this works well for use on tl're MCAT.Substituent number four can either be behind the plane, in front of the plane orin the plane. In each case, there is a technique to apply to arrive at the chiralcenter easily. Many techniques shall be presented, so choose your favorite.Figure 3-6 shows how to get the chirality easily when the structure is drawn irrthe conventional manner.

,l,.i(f-},.t-,,'(')t4t, Ho H

H in back of plane .'.

Take whai you observe as is.Clockwise Arc = R

Take what you observe as is.Ctlunterclockwise Arc = S

H?CH2

H in front of plane .'.

Reverse what you observe.Clockwise Arc reverses to S

H3CH

1

H in plane close to back group .'.

Reverse what you observe.Clockwise Arc reverses to S

H"

iY)HO CHs


Figure 3-6

The Berkeley Review L{._


-Example 3.2The following molecule has what type of chiral orientation?

H"C,\\

Ho$tyc cH2cH3

HA.RB.SC. The molecule has no chiral center.D. The compound is meso.

SolutionThe compound has one chiral center, so it cannot be meso (to be meso requires aneven number of chiral centers). The compound is chiral, because carbon two hasfour different substituents attached it. The molecuie is therefore either R or S.The priorities are oH > CH2CH3 > CH3 > H. Correct arignment of thesubstituents shows that the compound has an R chiral center. Acounterclockwise arc connects priorities 1,,2 and 3. Because the H (priority 4) isin front, the arc should be reversed. Pick A for best resuits.

Priority +S HsC

Priority#1 HO\\)

Priority #2

CH2CH3 Priority #4 in frontCounterclockwise = R

HPriority #4

Example 3.3r'\4rat is the chirality of the triol below according to the Cahn-Ingold-preiog rules?

o CH"OH

OH{. 2R,35,45,6RB. 2R,3R,4RC. 2R,3R,43D. 2R,35,45

Solutionlarbon six is not a chirallhis eliminates choice A.

o o cH2oH4S

oH oH CHsClockwise Arc and #4

in back .'. take as R

OH

center, because there are two methyl groups presentThe chiral centers are 2R, 35, 45 (choice D), as shown.

o CH"OHV<

Counterclockwise Arc and#4 in back .'. take as S

Clockwise Arc and #4in front ,'. reverse to S

-,:pvright @ by The Berkelev Review ra9 Exclusive IvICAT Preparation


Example 3.4Which of the foilowing compounds have R orientation?

C1

I

nrcnrc4/;cn,Compound I

Compound I(2-chlorobutane)tcl

HgN*

H CH2OHCompound II

Compound II(Serine)

o

Clockwise Arc and #4in fiont .'. reverse to S

HOH2C

HOHCompound III

Compound III(D-Glyceraldehyde)

H

A. Compound T onlyB. Compound II onlyC. Compound III onlyD. Compounds I and III only

SolutionHydrogen points out in each of the compounds. Whichever arc is seen from thisview must be reversed to get the arc that would be seen from the correct view.The priorities in Compound I are: Ci > CH2CHg > CHe > H. Compound I has anR chiral center. The priorities in Compound iI are: NH3+ > COZ- > CH2OH > H.Compound iI has an S chiral center. The priorities in Compound III are: OH >CHO > CH2OH > H. Compound III has an R chiral center. Choice D is best.

2

H3CH2

3

H3

HOH2o-

Counterclockwise Arc and

#4 in fiont .'. reverse to R

Counterclockwise Arc and

#4 in front .'. reverse to R

Example 3.5What is the stereochemical orientation of the foilowing molecule?

H .CH,"\- Eilr,.J)y'-'a.,

A. 2R,3RB. 2R,35C. 23,3RD. 2S,3S

SolutionFor the first chirai center (carbon 2), the fourth priority (hydrogen) is in the planeclose to the group in back (a reversing position). An arc from priority one to twoand on to priority three is counterclockrvise. However, because H is in a reverseposition, the chirality is R. The second chiral center (carbon 3) has H in front ofthe plane, so it is in a reversing position too. An arc from priority one to two andon to priority three is clockwise. Horr-ever, because H is in a reverse position, thechirality is S. The best answer is thus 2R, 35 which makes choice B correct.

:i

-s

lt*

i,:Il

lr:

2

Copyright O by The Berkeley Review r90 The Berkeley Review


Figure 3-7 shows a summary of the tricks presented in Figure 3-6 and applied inExamples 3.2 through 3.5.

If Priority #4 is in front, If Priority #4 is in back,then take arc as is.reverse the arc to opposite.

Figure 3-7

An alternative short cut in determining the chirality of a compound involvesinterchanging substituents to generate an easy structure to solve, and thenassigning the opposite chirality to the original molecule. This method is based onthe idea that it is easiest to solve for chirality when the fourth priority substituentis in back and that when two substituents are interchar-rg"a, tn" chirality isrnverted. Figure 3-8 shows the application of this method.

HOH2C< cHs HoH2r< H.J

'@ , lHtz'HO H*

If Priority #4 is in the plane far fromgroup in back, then take arc as is

Priority #4 not in back ...

interchange #4 and group in back

CH"OH

Priority #4 not in back ...

lnterchange #4 and group in back

Hgcl

,HOH2C

Priority #4 not in back .'.

lnterchange #4 and group in back

If Priority #4 is in the plane neargroup in back, then reverse arc.

Priority #4 now in back of plane .'.

New structure has a ciockwise arc, so it is RThe original compound must be S

H2OH

Priority #4 now in back of plane .'.

New structure has a clockwise arc, so it is RThe original compound must be S

Hscj 1oH

2

HOH2cPriority #4 now in back of plane .'.

New structure has a clockwise arc, so it is RThe original compound must be S

Figure 3-8

These are two-dimensional tricks that may be done on paper. There are otherticks that involve contorting your hand to model the molecule. No one methodis more accurate than another is, so once you find the one you prefer, hone it inand use it. We will cover the three methods in class. One involves pointing inthe direction your eye should be looking, and forming an arc that goes fromoriority 1 to 2 to 3. Another involves using your fingers to represent the bonds inthe molecule. And as mentioned earlier, one method involves matching yourthumb and curling fingers to a molecule.

J

Copyright @ by The Berkeley Review l9r Exclusive MCAT Preparation


Solution of Chiral Compound

light enters light rotates rotated light exits

^ndz porarlzet

Optical RotationOptical rotation is a physical measurement of the rotation of plane polarized lightby a solution with a chiral molecule. A soiution containing a pure compound ofknown concentration (dissolved into solvent) in a standardized cuvette rotatespiane-polarized iight the same amount each time. Consequently, the specificrotation (optical rotation under specific condiiions) is a physical measurement(like melting point and boiiing point) that may be used as a diagnostic test for the

identity of a compound. This is common with sugars which have multiple chiralcenters. The direction of the rotation is signified by either (+) or (-) orientationfoliowed by the degrees of rotation. The (+) denotation describes clockwiserotation of light while the (-) denotation describes counterciockwise rotation oflight by the molecule. If the R-enantiomer of a compound lotates the light in a

positive direction by X", then the S-enantiomer of the cornpound rotates light byX' in the negative direction. The measurement is taken with a polarimeter whichis made from a sample tube sandwiched between two polarizing plates' The

plates are rotated in a manner to allow for the greatest amount of iight to pass

through. If the plates are rotated 90' away from the orientation with highestintensity, then the new orientation of the plates does not allow for any light topass through. A sample polarimeter is shown in Figure 3-9 below.

Figure 3-9

Chiral molecules may be assigned a "+" or "-" preceding their name to indicate

the direction that the compound r,r'il1 rotate plane polarized light. For instance.

(+)-2-butanol rotates tight in a clockwise direction while its mirror image, (-)-2-

butanol, rotates light in a counterclockr.r'ise direction. There are compounds withR-stereochemistry that rotate plane polarized light in a clockwise fashion anC

other compounds with R-stereochemistry that rotate plane polarized light in a

counterclockwise direction. Consequently, R stereochemistry does notnecessarily correspond to either (+) or (-) rotation of plane polarized light. Irbiochemistry, the designation of D and L is based on threose, where (+) was

originally assigned to D-threose and (-) rvas assigned to L-threose. Flowever.because of the vast multitude of sr-rgars, there is no correlation between D and L

designation and (+) or (-) rotation oi pl3ns polarized light.

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Having chiral centers does not always result in the rotation of plane polarizedlight. Meso compounds have opposing chiral centers that cancel one another out,resulting in no net rotation of plane polarized light. Meso compounds aretherefore opticnlly innctiue, meaning they have a specific rotation of zero.

Polarimeters measure the optical rotation of a soiution. The specific rotation of acompound, [a]p, is calculated from the optical rotation using Equation 3.1.

[o]5 =observed rotation in degrees

(3.1)

(length of sample in dm)(concentration of sample it-r ItuTt)mL

As seen in Equation 3.1, the standard cell length of a cuvette is 10 centimeters(one decimeter), and the standard concentration is 1 gram solute per millilitersolution. The superscript T refers to temperature as measured in Celsius and thesubscript D refers to monochromatic light from a sodium lamp, known as the D-band. Specific rotation is the optical rotation observed under specific conditions.If the solution is too concentrated, then the rotation of tight is greater than itshould be. If the solution is less concentrated than standard conditions, then therotation of light is less than it should be. For this reason, the observed opticalrotation is converted to specific rotation to determine purity.

Types of Configurational IsomersUntil now, we have focused on the chirality of molecules and their stereogeniccenters. A moiecule is a chiral molecule when it is asymmetric. However, oftentimes, a molecule can be asymmetric in more than one way. There are terms thatdescribe the relationship between two stereoisomers. The relationship requiresdetermining whether the structures are mirror images and whether they aresuperimposable. Configurational isomers can be classified as either enantiomersor diastereomers. Enantiomers and diastereomers have the same bonds, but adifferent spatial orientation of their atoms. Enantiomers are configurationalisomers that are nonsuperimposable mirror images (reflections that you can'toverlay). Diastereomers are configurational isomers that are nonsuperimposableand that are not mirror images. Thus, to classify a pair of configurational isomersas either enantiomers or diastereomers requires evaluating whether thestructures are mirror images and whether they are superimposable.

There is a wonderful short cut for determining whether two compounds areenantiomers or diastereomers. Consider a molecule that has two chiral centers, Rand R. To be mirror images, ail of the chirai centers must differ, because eachchiral center must switch when it is reflected (just as a left hand in the mirrorturns into a right hand). If you were to place an R, R molecule into the mirror itrvould reflect an S, S molecule, so the S, S molecule is the enantiomer of the R, Rmolecule. This means that to be enantiomers, all of the chiral centers mustdiffer between the two configurational isomers. If no chiral centers differ, thenthe two structures are identical (the same molecuie). If one of two stereocentersdiffers, then the two compounds are neither mirror images of one another nor thesame molecule. This makes the two compounds diastereomers. If only a few,but not all chiral centers differ, then the compounds are diastereomers. Listedbelow are the modified definitions of enantiomers and diastereomers.

Enantiomer; Enantiomers are configurational isomers in which all of the chiralcenters in each molecule are different from one another.

Diastereomer: Diastereomers are configurational isomers in which at least one,but not all of the chiral centers in each molecule is different from one another.



These modified definitions should prove to be easier to use than the traditionaldefinitions. Recall how these definitions were derived and they will be easy toremember and apply. Figure 3-10 shows two pairs of enantiomers and Figure 3-11" shows two pairs of diastereomers.

HsC cH2cH3

Figure 3-10

cH2cH3

Diastereomers, because only one of the chiral centers (the left one) is different

ocH2cH?

Diastereomers, because only one of the chiral centers (the right one) is different

Figure 3-1L

After practice (and thus on your exam), you should be able to just scan structuresto look for interchanged substituents (chiral centers that are different.) First, lookfor any chiral centers (sf-carbons with four unique substituents). From thatpoint, compare the comparabie chiral centers in the two structures. If ttrstructure is oriented in a similar fashion, but two substituents are in differsfrpositions, then the chiral center is different between the two compounds. If ttrcstructure is oriented in a similar fashion, but three substituents are in differerilpositions, then the chiral center is the same between the two compound.*Finally, it is a matter of deciding i-f a1l, some, or none of the chiral centers on tlutwo molecules are different and then determining their relationship.

Irl

(

u

M

M

ffi

{m

s@

lt

nm

&

&

A,n0rt

HgC. /cF{2cH3\J8,,'f Vott

Enantiomeric pair because both'of the chiral centers are different

H'v

Enantiomeric pair because both bf the chiral centers are different

OCH"CH"-.-- ^

Ht J

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Example 3.5Ihe following pair of compounds is best described as which of the following?

cl. c(cH3)3\-JH.,c$\'/ Yrort

HH

H1C C(CH.):" \___./

'

"v \r"Ci OH

A. DiastereomersB. EnantiomersC. Identical achiral compoundsD. Identical chirai compounds

Solution\Vhen the orientation of the molecule remains constant and three substituents:hange their location, this implies that the compound has been rotated about thatchiral center. The left chiral center is just rotated between the two compounds,thus it has the same chirality. When two substituents interchange their location,this implies that the chiral center changed. The right chiral center has changed,because the H and aldehyde group have interconverted. This means that onlyone out of the two chiral centers has changed its orientation between the twostructures. The two compounds are therefore diastereomers of one another,naking choice A the best answer.

Example 3.7F{ow can the relationship of the following pair of molecules be described?

CHs

-{. DiastereomersB. EnantiomersC. Identical achiral compoundsD. Identical chiral compounds

Solutionln this example, the mirror plane between the two molecules can be seen easily aslhey are drawn. So without rotating or counting chiral centers, the two:ompounds can be identified as enantiomers of one another. Enantiomers arenonsuperimposable mirror images. Choice B is the best answer.

Example 3.8The following pair of isomers is best described as which of the following?

HO HHO.H

\-,)cH, &Hrcrsf \ f",.",

H OCH3

A pair of anomersA pair of constitutional isomersA pair of diastereomersA pair of enantiomers

A.B.

C.D.

Copyright @ by The Berkeley Review r95

ocH3



A

J";

SolutionThe two structures are not aligned in an equivalentstructures must be rotated into a structure equivalent

fashion, so one of the trt'oto the other structure.

H\

#'H"C Hro'&_/-

&/ \ao.tt,H CHs

(I

I

u

u

r(

u

L

T

Hsc

Br

H

CHs

H

CI

HeC

Br

CHs

H

H

&

C1

CHc

A. diastereomers.B. enantiomers.C. identical achiral comPounds.D. identical chiral comPounds.

SolutionThe structures are drawn as Fischer projections, which represent the top vierr

the all eclipsed form of the molecule. In a Fischer projection, the side gloups

coming olrt ut you in the three dimensional perspective,. The two isomers h

two c(iral ."r-tiuru. Only the chiral center with the chlorine is different r'r'

comparing the two compounds. The chiral carbon with bromine has

changed, t".urrr" the subitituents have not moved' Only one out of the

chiraf centers di{fers, so the compounds are diastereomers. This makes choice

the best answer.

"t.&--J"Itwo 60' rotations)

-z

/ poc",OCH? H CHe

H ocH3

After rotating the structure, it is easier to see that the left chiral center has tn'c

substituents thut huu" interchanged between the two structures, thus it ha-'

changed chirality. The same is true for the right chiral center after rotation, thus

it toJhas changed chirality. This means that both chiral centers have changed, sc

the compoundi are e.rut iio*"ts. There is no need to determine the chiralitv c:

the stereocenters (R or S) within molecules to determine whether they are

enantiomers or diastereomers. Deciding whether two molecules are enantiomers

or diastereomers is as easy as asking whether aII of the stereocenters or just sonr

of the stereocenters have changed their orientation between the two compounds'

(All centers differ = enantiomers; Some centers differ = diastereomers)

Example 3.9The following molecules are best described as:

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Example 3.10Ihe physical properties of the follolving stereoisomers relate in what way?

CH:

CH:

\. Same boiling point; different melting pointsB. Same density; different boiling pointsC. Different boiling points; different melting pointsD. Different densities; same boiling point

SolutionThe chiral center on carbon 2 is different between the two structures, because theCH3 and OH are interchanged. The chiral center on carbon 3 is not different:etween the two structures, because the H, CH3 and the Br have all,nterconverted. When three substituents interconvert, the chiral center is not:hanged. This means that one out of two chirai centers differ, so the two:ompounds are diastereomers. Diastereomers have different physical propertiesrcluding melting point, boiling point, and density. Pick C for best results.

Example 3.11lhe following pair of isomers is best described by which of the following terms?

cH?oH

H.c{ouBr-+H

I

CHs

&

\.B.rl.

D.

H:cr'''1)''\$oH r,.r,"'Q ",tt aH

AnomersDiastereomersEnantiomersidentical meso compounds

Solution-ne two compounds are mirror images of one another, so they are enantiomers,:.::oice C. To see the mirror plane relationship, one of the compounds must be::tated 180'. The drawing below shows the right structure berng rotated.

nr.r,,O..rr\oH "r@

t""O"rr\oH I

nirror ima

- -.pyright @ by The Berkelev Review

"or'O".,,tCHg



MesoMeso compounds are individual structures which contain a mirror plane slicingthrough the middle of the compound and an eaen number of chiral centerssymmetrically displaced about the mirror plane. The net optical rotation of a

meso compound is 0". It is zero, because the opposing chiral centers on each haUof the molecule cancel one another out, Ieaving no net rotation of plane polarizedlight. Meso compounds are referred to as optically inactiae. Remember thephrase, "Me so inactive", a high-energy rap lyric that describes your physicaistate while studying for the MCAT. Figure 3-12 shows a meso compound (it has

been rotated into a side view to see the mirror plane more easily).

top view side view

Figure 3-12

A meso compound may be identified by an inversion center in the middle of themolecule. Figure 3-13 shows two conformational isomers of a meso compound.one in its most stable conformation (where it has an inversion point), and theother in its least stable conformation (where it has a mirror plane of symmetry.A meso compound has the same number of R-stereocenters as S-stereocenters.

Inversion point

CHa

Figure 3-13

Example 3.12The reflection of a meso compound can be classified as which of the following?

A. Identical to the original compoundB. An enantiomer of the original compoundC. A diastereomer of the original compoundD. An ameso compound from the fifth dimension where evil lurks and the sock

that disappear from laundry loads in our world gather.

SolutionA meso compound when viewed in a mirror reflects the identical compouncThis makes choice A the best choice, although choice D is a close second. A:example of a meso compound and its reflection is drawn below.

I!

:

m

,[

t5

'n

"aI

.i:

:lntr--t

.

HHOO

H\,/

plane

"u"lmirror

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StereoisomerismAs mentioned before, stereoisomers are compounds that have identical bondsbut their atoms differ in spatial orientation. \Mhen a molecule contains more thanone chiral center, the maximum number of stereoisomers increases exponentiallywith each new chiral center according to the equation 2n, where n is the numberof chiral carbons in the molecule. There are less than 2n stereoisomers, if one ofthe possible structures is meso. If there are an odd number of chiral centers, thestructure cannot be meso, so there is exactly 2n possible stereoisomers. Forexample, consider the compound 3,4,5-trimethyloctane, which has three chiralcenters and thus eight possible stereoisomers. 3R, 4R, SR-trimethyloctane is justone of the eight possible stereoisomers. Table 3-1 shows the possiblestereoisomers for compounds with a variable number of chiral centers.

Chiral Centers MaximumStereoisomers

Stereoisomers

1 2 RorS

2 4 R& RS, SR, or SS

3 8 RRR, SRR, RS& RRS, SSR, SRS, RSS, OT SSS

Table 3-1

lVe are most concerned with stereoisomers when a molecule contains more thanrne chiral center. If there is only one chiral center, there cannot be diastereomers.Stereoisomerism is important in biological sciences, because only a very fewriological compounds have just one chiral center (some amino acids beingamongst of these few) with some proteins having in excess of 200 chiral centers.

E-xample 3.13I{ow many stereoisomers are possible for the

OH

following structure?

OH

o CHs CH:

-\. 4ts.8c. 72

D. 16

Solutionlhe molecule is drawn in a way to make you mistakenly see four chiral centers if- ou don't pay close attention, but the number of chiral centers is only three.larbon five is not a chiral center, because there are two methyl substituents::'iached to it (one that is drawn as a methyl substituent and the other that isj:awn as carbon six of the longest chain). By having two methyl groups.::ached, it does not have four different substituents attached, thus it is not a::riral carbon. This means that only carbons two, three, and four are chiral. The:-aximum number of stereoisomers is derived using the equation 2n, which is::rs case is 23. Because there is an odd number of chiral centers, the compound::mot be meso, so there are eight stereoisomers. The correct answer is choice B.

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Example 3.14The relationship of the following pair of compounds is best described as:

CHs OHA. AnomersB. DiastereomersC. EnantiomersD. Structural isomers

SolutionThe hydroxyl group is attached to different carbons in the two structures. In theleft structure the hydroxyl group is on carbon 4 while in the right structure, thehydroxyl group is on carbon 2. This makes the two compounds structuralisomers, which makes choice D the best answer. you may recall that if the twocompounds have different IUPAC names, then they are structural isomers. Byvirtue of the hydroxyl being in a different position, the two compounds inquestion have different IUPAC names.

Example 3.15Which of the following compounds islare optically inactive?

L 2R,3S-dibromopentane

II. 2S,3R-dichlorobutane

III. 1R,2R-diiodocyclopentane

A. Compound I onlyB. Compound II onlyC. Compound III onlyD. Compounds I and III only

SolutionTo be optically inactive, the compound must either be achiral or meso. All of thecompounds iisted have stereocenters, so achiral is not a possibility. The questio:.is whether or not each structure is meso. To be meso, the compound must b=symmetric and have an even number of chiral centers equally displaced abou:the internal mirror plane (i.e., R on one side and s on the other). Compound I seliminated, because it is not symmetric about the a plane (the mirror plane wou.l;have to slice through carbon three to break the five carbon species into halvesTo be symmetric, carbons two and four would have to have the sam"substituents, which they do not. Compound I is chiral, so it is optically acti,,'=and therefore eliminated. Compound III is eliminated, because it does not havropposite chiral centers (they are both R). To be meso, carbons one and trt:would have to have opposite absolute configuration, which they do nc:Compound III is chiral, so it is optically active and therefore eliminated. Tl:only choice left is Compound II, which is symmetric, because it has a mirrirplane that slices through the bond between carbon two and carbon three. Th:best answer is choice B.

:

T]f

1l!18

mrt

,&-

il"C.l!i't

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Organic Chemistry Stereochemistry Stereochemistry in Reactions

Stereoisomer FormationThe utility of stereochemistry lies in the selectivity of chiral reactants for oneanother. \Mhen reactions involve chiral reactants, they are often selective for onestereoisomer over another. This is a staple of enzymatic selectivity. Howevet,when a reaction involves reactants without any chirality, the formation ofstereoisomers is random and follows basic probability. Most reactions in organicchemistry produce stereoisomers. The type of stereoisomers formed depends onthe chirality of the starting reagents. \44ren a symmetric nucleophile can attack a

planar species from either side, there usually are two enantiomers formed inequal proportion. When a symmetric nucleophile is hindered from attacking a

planar species from one side more than the other (due to a chiral center in theelectrophile that creates greater steric hindrance on one side than the other), thereare two diastereomers formed in unequal proportions. Stereoisomers can resultfrom electrophilic addition reactions with alkenes as well as substitutionreactions. Figure 3-14 shows an example of an electrophilic addition reactionthat forms two enantiomers.

oll o

,rrc'c\nrcr,, j*",Planar carbonyl

""\ ftz c-.

.. O O..t,to\ l"o

t t\ St",€

.rC-. & ,C-.HsC CH2CH3 HsC CH2CH3

I

I

l nocu"HO. OCH, I "\l'._-,& ,rC-.

HsC cH2cH3 HsC

Enantiomers

cH2cH3

Figure 3-14

Example 3.16The addition of alkyl magnesium bromide (RMgBr) to a carbonyl in ether adds a:iew alkyl substituent to the carbonyl carbon, resulting in conversion of the:arbonyl into an alcohol. The addition of H3CMgBr to R-2-methylcyclohexanonen diethyl ether yields which products?

-A.. One meso compoundB. Two diastereomersC. Two enantiomersD. Two epimers

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SolutionIn this reaction, the methyl group can add to either the top or bottom of theplanar carbonyl group. This results in a new chiral center that can be either R orS. However, there is already a chiral center present in the reactant that is notinvolved in the reaction, which retains its original chirality. The chiral centerpresent in the reactant does not change during the course of a reaction, so theproducts cannot be enantiomers. This eiiminates choice C. The compound is nota sugar, so choice D is eliminated. It is not meso, so choice A is eliminated. Oneof the two chiral centers differs between the two stereoisomer products, so thevare diastereomers. Choice B is correct. Due to steric hindrance from the ethllgroup on carbon two, the product mixture of the two diastereomers is not 50/50.

HOo

0When two enantiomers are formed, they are formed in equal quantity, and theproduct mixture is said to be rncemic. When the two enantiomers are equalh-present, there is no net rotation of plane poiarized light. When tu'odiastereomers are formed, they are formed unequally, so the product mixture hasa major and a minor product. When the two diastereomers are present ir.unequal amounts, there can be a net rotation of plane polarized light.Enantiomers can be formed in an unequal ratio if a chiral catalyst is present. Thisleads to the concept of enantiomeric excess, used to analyze product distributionsfrom reactions with a chiral catalyst (most often an enzyme).

Example 3.17\Atrich of the following reactions produces no optically active compounds?

A. 2-Butanone treated with NaBH4 in ether followed by acidic workup.B. (Z)-Z-butene with KMnO4 in base (sgr addition of two hydroxyl groups).C. Reduction of HN=C(CH3)CH2CH3 using LiAiH4 in thf solvent.D. S-2-Bromobutane treated with NaCN in ether solvent.

SolutionFor a compound to be optically inactive, it must either be meso or achiral. Inchoice A, NaBH4 adds a hydrogen to the carbonyl carbon from either side,resulting in a racemic mixture of alcohols. Choice A is eliminated. In choice B,

KMnO4 adds a pair of hydroxyl groups to ihe alkene carbons to from a synvicinal diol. Because the alkene is svmmetric to begin with, symmetric additionresults in a symmetric product. The product is a meso diol, so choice B is thebest answer. In choice C, LiAIH4 adds a hydrogen to the imine carbon fromeither side, resulting in a racemic mixture of ami.nes following workup. Choice Cis eliminated. In choice D, a good nucleophile attacks an alkyl halide, resulting ininversion of stereochemistry. One chiral species is formed, so choice D is out.

.*rCHe+ HrCMgBr -----> .'rrCHs .,tfCHe

:E

:i:

.4_

ffi

,ID

i::::i

Choice B \ /\-*/

Symmetric about plane

+ KMnOr -+Symmetric

additioi-r HO OHSymmetric product:

-t' nHO OH

meso vicinal diol



The only way to get a meso compound from an addition to an alkene, is to have asymmetric addition to a symmetric alkene or an asymmetric addition to anasymmetric alkene. The best answer is choice B. Choices A and C involveracemization which is common when a nucleophile attacks an sp2-hybridizedcarbon. Choice D is the result of an SN2-reaction, which inverts the chirality.Because the compound starts optically active, inversion generates an opticallyactive product.

Enantiomeric ExcessEnantiomers rotate light in opposite directions of one another, but with equalmagnitude. When both enantiomers are present in equal quantities in soiution (a50-50 mixture), the solution exhibits no net rotation of plane-polarized light.Based on this idea, when a mixture is not in a 50-50 ratio, then the net rotation oflight by the solution is not zero. The farther the value deviates from zero, thegreater the difference in concentration of the two enantiomers. From theobserved rotation of the solution, the percentage of the enantiomer in excess canbe derived. Equation 3.2 shows how to determine the enantiomeric excess fromthe observed specific rotation. The enantiomeric excess is the difference inpercentage between the more abundant enantiomer and the less abundantenantiomer.

'oee (enantiomeric excess) =measured specific rotation x 700% (3.2)

specific rotation of the pure enantiomer

Example 3.18i\4rat enantiomeric distribution would account for a specific rotation of +13.6' if',Jre pure enantiomers have specific rotations of +27.2" and -27.2' respectively?

-\. (+)-enantiomer =25ok and (-)-enantiomer =75%B. (+)-enantiomer = 33oh and (-)-enantiomer = 67.hC. (+)-enantiomer = 67oh and (-)-enantiomer = 75%D. (+)-enantiomer =75o/o and (-)-enantiomer =ZS%

SolutionBecause the net rotation is positive, the (+)-enantiomer must be in higherconcentration than the (-)-enantiomer. This eliminates choices A and B. Todetermine the exact quantity, Equation 3.2 canbe applied.

"/oee =+13'6 x 100% =!x 100% = 50'/oinfavorof the(+)-enantiomer27.2 2

The exact ratio is found using the following reiationship:

'i'(+)-enantiomer + %(-)-enantiomer = 100% and (+)-enantiomer - (-)-enantiomer = Bjok

(+)-enantiomer = 75o/o and (-)-enantiomer = 25"/o, choice D


Organic Chemistry Stereochemistry Nucleophilic Substitution

Nubledphiliru1i;$fib$$.flfi ffi,bnNucleophilic SubstitutionOne application of stereochemistry is in nucleophilic substitution reactions.Nucleophilic substitution involves the attack of an electropositive carbon by a

nucleophile (Lewis base) to dislodge an atom or functional group (referred to as

the leaving group). This is a recurring reaction in organic chemistry and itinvolves the substitution of one functional group for another. Nucleophilicsubstitution can proceed by more than one reaction pathway. It can proceed bvthe two-step 51111 mechanism, or it can proceed by the one-step 5512 mechanism.Nucleophilic substitution reactions are based on the fundamental chemistrvconcept that negative charge seeks positive charge. The electron pair of thenucleophile hunts for an electron deficient carbon to attach to. It will beimportant to understand the steps of the reaction for both mechanisms, as

reactants proceed to products. Figure 3-15 shows an example of a nucleophiJicsubstitution reaction.

j

fi

s

lli

ri[

rl

sM

M..-..o.1. + H.CO

,M

d

Figure 3-15

We will discuss the mechanism of this reaction shortly, but for now, there aresome fundamental definitions with which to be familiar. Listed below are themost important definitions. Each definition is followed by some genere-comments about the relative reactivity of the species and how to discern i5reactive strength.

Nucleophile

The species donating an electron pair in a nucleophilic substitution reactic'n(Lewis base). As its name implies, it loaes (philes) a positiae charge (nucieo'Nucleophiles must have an available pair of electrons to share. Nucleophilestrength is closely approximated by its base strength, although steric factors(nucleophile size) affect nucleophilicity. Small nucleophiles are generally betternucleophiles. This is to say that steric hindrance plays a larger role innucleophilic substitution reactions than proton transfer reactions. The strengthof the nucleophile does not perfectly correlate with base strength, but it is closeenough to say that it parallels. A short list of nucleophilic strength in watersolvent is as follows:

SH-> CN- > I-> OR- > OH-> Br-> NHg > C6H5O- > CH3CO2- > Cf > F- > ROH > HzOIt should be noted that if the base is too strong, an elimination reaction can occur(as is the case with OR- and OH-). The solvent also has an effect in thatnucleophiles that can hydrogen bond are hindered in protic solvents, becausethey are solvated. The solvation by water binds the electron pair of thenucleophiie and reduces its nucleophilicity. This phenomenon explains why SH-,CN- and I- are stronger nucleophiles in water than OH- despite being weakerbases than OH-. In aprotic solvents and the gas phase, nucleophilicity moreclosely parallels basicity. The big difference in nucleophilicity is that size of theanion is not as important as it is with basicity.

ffii

s

m

.Lq

dI(f,



Electrophile

The species accepting an electron pair in a nucleophilic substitution reaction(Lewis acid). The electrophile holds the reactive .urborl and the leaving group.The weaker the bond between the leavrng group and the carbon, the better theelectrophile. Electrophile strength can be ipproximated by the stability of theleaving group once it is off of the electrophile. Electrophiiic carbons typicallynave a partially positive charge.

Leaaing groupThe functionai group that dissociates from the eiectrophile in a nucleophilicsubstitution reaction. The more stabie the leaving group, the weaker it ii as arase. This means that the strength of a leaving gio"p can be predicted by the.trength (pKu) of its conjugate acid. The theory is that the more stable the-eaving group, the less basic the leaving group, and thus the more acidic the:onjugate acid of the leaving group. The strengih of a leaving group increases as:he pKu of its conjugate acid decreases. This is most true in wafer, but can also beseen in organic solvents. f-eaving group strength increases as the strength of the:ond between carbon and the leaving gro.tp J""r"ur"r. This is why iJdine is aretter leaving group than fluorine.

?.scemic mixtrn.e

'+ product mixture that has an even distribution of enantiomers, 507u of each=nantiomer, in the product mixture. A racemic mixture is the observed product-'""hen the mechanism involves an intermediate where the reactive site is'an sp2--^,., bridized carbon (like a carbonyl or carbocation) and the molecule is symmeiric-:ras r-lo other chiral centers). There is no such thing as a racemic mixture of..iastereomers, because diastereomers have at least two chiral centers associated''ith them, and a chirai center present in the reaction hinders attack of one side of-: e electrophiie relative to the other, causing the distribution to not be fifty-fifty.

--.1 nucleophilic reactions involve a nucleophile, an electrophiie, and a leaving;:oup. some, but not all, nucieophilic reactions generate a racemic mixture.",hether a racemic mixture is generated or not depends on the reaction pathway.lhe fundamental question in a nucleophilic subititution reaction is, ;does the- ;cleophiie come in first, or does the leaving group leave first?,, This is the basic-..Jference between the sry1 and the slrJ2 mechanisms. The nucleophile attacks::st in an S]N2 reaction mechanism, while in the leaving $oup leaves first in an:';1 reaction mechanism. we will look at these two scenarios in more detail.

Erample 3.19-, iavorable nucleophilic substitution reaction has all of the following EXCEpT:d. a good leaving group.l. a reactant with a weak bond to the leaving group.:. a strong Lewis base as the nucleophile.f . a weak Lewis base as the nucleophile.

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SolutionA favorable nucleophilic substitution reaction is one that forms a stronger bondthan the one broken. A good nucleophile is one that forms a stronger bJnd withcarbon than the bond between carbon and the leaving group. A good leavinggroup is one that forms a weak bond with carbon, thus minimal eneigy is needeJto break the bond between it and carbon. when the leaving group ii strong, ihereaction is said to be favorable, so choice A is a valid statemet-,i. choic" e iseliminated. A weak bond to the leaving group makes it a good leaving group, sochoice B is also a valid statement. This eliminates choiie B. The n.t"t"optl1t"should be a strong Lewis base, so choice C is a valid statement and choice Dls aninvalid statement. This eliminates choice C and makes choice D the best answer.You might note that to determine the favorability of a nucleophilic substitution

"".rt"", t"" """, , leophile and leaving group.

-

sru2In an sjr]2 reaction, the nucleophile attacks prior to the ieaving group leaving. lr-u:1"1.9, the nucleophile comes in from the backside and pushes ine teaving gio.rpoff of the electrophile. An important factor to consider is the transition state tha:forms during the reaction. Transition states cannot be viewed directly (thei:lifetimes are too short), but evidence in the product (inversion of configuration aia chirai carbon) infers they exist. Backside attack by a nucleophile causes thr-.inversion at the chiral carbon. Certain features in the reactants (nucleophile an;electrophile) and the product (if it is chiral) indicate that the reaction proceedecby an slr]2 mechanism. These are idiosyncrasies of the reaction, and they dictatrthe pathway the reaction chooses. Each property favors one of the trr lmechanisms. They can be used to distinguish an slr]2 reaction from an sl-reaction. Figure 3-16 shows a generic mechanism for an sl.I2 reaction.

I

,i

L

eerm

s,nGln"

Trigonal bipyramidalTransition state

L'G'--+ ------> Nuc L.G.-

H

Figure 3-16

The reaction takes place in one step, so the rate of an sry2 reaction depend.s .-mr

both the concentration of the nucleophile and the concentration of t:eelectrophile. The nucleophile initiates the reaction by attacking the electrophieand forcing the bond between the carbon and the leaving group to stretch a:rdweaken. At the same time that the nucleophile approaches the electroph:-rccarbon, the electron density of the nucleophile repels the substituents on twelectrophilic carbon and thus they form the trigonal bipyramidal transition sta:e-As the leaving group begins to leave, the substituents on the electrophilic carb,cnmrbegin to fold in the direction of the less hindered side of the mblecule (lhindered because the leaving group has left). The hybridization finishes as #-Table 3-2 lists some key features associated with an sp2 reaction, accordingobservation order. What is meant by observation order is that the first featr::(features of the reactants) are observed before the reaction begins, the serfeatures (features of the transition state) are observed during the reaction,the last features (features of the products) are observed after the reaction end-s-

II

&,&riil

Nuc:

I

R+I

INuc---f---'L

l'==HH

.G.

Copyright @ by The Berkeley Review 206 The Berkeley


Reactant Features Course of Reaction Features Product Features

rhe preference for an 51112

::,.echanism is 1' > 2" > 3" in:erms of electrophiles

An 51112 mechanism forms a five'ligand transition state duringthe middle of the reaction

A single enantiomeric productis formed(No racemic mixture)

-{n S11i2 mechanism is favoredi'.-ith a good nucleophile

The S-ligand transition statethe highest energy state andexists for just a split second

isit

5512 reactions exhibit secondorder kinetics(rate = klNucllElectl)

ln 5512 mechanism is favored-l poiar, aprotic solvents such

- ethers and ketones

Steric forces destabilize thetransition state by forcing bondangles to values less than 109,5'

Sl12 reactions are one-stepreactions, so they have fastrates of formation

Table 3-2

lie reaction in Figure 3-17 proceeds by an Sy2 mechanism, because the:-ectrophile is primary and it has a good nucleophile. With a primary=-ectrophile, the reaction must proceed by an 5512 mechanism. The ether solvent

-. polar and aprotic, which further favors the Sp2 reaction pathway.

NC:@ +

Good\ucleophile

H"C,\

H'''yD

+I@StableAnion

CH"

r --+ NC-J-Et2O Y,o"

1" Electrophile

D

Inversion Product

Figure 3-17

E"xample 3.20-ll of the following are associated with an 51112 reaction EXCEPT:

A. backside attack of the electrophile by the nucleophile.ts. inversion of stereochemistry.C. nucleophile concentration affecting the reaction rate.D. rearrangement of alkyl groups from reactant to product.

5olutionohis question focuses on the fundamentals of an S1.tr2 reaction. For an St12

:eaction to proceed, the nucleophile must attack the electrophile from the

--rpposite side as the leaving group in a collinear fashion reiative to the bond to:he leaving group. This is referred to as backside attack, so choice A is valid, and--nus eliminated. Backside attack results in inversion of stereochemistry if thetlectrophilic carbon is a chiral carbon. This makes choice B valid, whicheliminates it. Because there is just one step in an 51112 reaction, the rate dependsrn all of the reactants, including the nucleophile. This makes choice C valid, andthus eliminates choice C. Rearrangement can occur when there is a carbocationpresent, because carbocations lack a bond. Carbocations are associated with 51111

reactions, not 51112 reactions, so choice D is invalid and thus the correct answer to:his question.


Organic Chemistry Stereochemistry Nucleophilic Substitution jsp1:

Lr an s|J1 reaction, the leaving group leaves before the nucleophile attacks. Lressence/ the nucleophile waits until the leaving group has left, allowing it moreroom to attack. The slr]1 reaction rate does not depend on nucleophileconcentration. once the leaving group has dissociated, a planar caiionicintermediate forms. Evidence for the intermediate comes from kinetics data aswell as stereochemicai evidence provided by the products. The carbocationintermediate has a long enough lifetime to be detected using spectroscopy. Bothrearrangement (hydride shifts and alkyl shifts) and a mixture of stereoisomers(formed from the spl-intermediate) are observed with sNI1 reactions. Thenucleophiie is free to attack from either side of the carbocation intermediate, ifthe carbocation is symmetric. As a result, a racemic mixture of enantiomers isformed as the product mixture. Figure 3-18 shows a generic mechanism for an5111 reaction.

R Nuc attacks720'A,."(

I fromrhe,9r__

-> Vf ffi;=-.R,, R, from the right 109

Trigonal planarCarbocation Intermediate

R>.. -R; L.c.'[r

R109.5"

,onf"$G"

R"

Nuc

R"

Figure 3-18

The reaction takes place in two steps, where the first step is the slowest. In thefirst step, the bond between the carbon and the leaving group breaks. As theleaving group begins to leave, the substituents on the carbon foia in the directionof the less hindered side of the molecule, allowing the-bond-angles to increasefrom 109.5' to 120" as the carbon re-hybridizes from ry3 to sp2. ihis results in aslight increase in stability, accounting for the intermediate being at a lowerenergy level than the first transition state in the energy diagram. In addition tore-hybridization, the planar cation is solvated, which also increases the stabilityof the intermediate. The nucleophile can attack the carbocation intermediate anddisplace the solvent from either side of the carbocation intermediate. Thisdisplacement of solvent and the re-hybridization from sp2 back to sp3 causes adecrease in stability from the intermediate to the second transition state. Finally,as the new bond is formed, the energy level decreases until it reaches the level ofthe products. Bond formation is an exothermic process. The hybridization of thecentral carbon finishes atsp3. Table 3-3 lists features associited with the sx11reaction. Like in Table 3-2, the features are iisted according to observation order.

Ll.:-,':

.q,.

D

C

D

Reactant Features Course of Reaction Features Product Features

The preference for an 5511

mechanism is 3" > 2' > 7"in terms of electrophiles

Steric hindrance pushes theleaving group off of theelectrophile

A racemic mixture formswhen the electrophilehas chirality

An S11 1 mechanism isfavored in a protic solventsuch as alcohol

The intermediate is a planar,three-ligand carbocation withsp2-hybrrdtzatton

SN 1 reactions exhibitfirst order kinetics(rate = k[Elect])

An Sry1 reaction is seenwith a poor nucleophile

An intermediate is observed inaddition to transition states

Sry1 reactions are slowtwo-step reactions

Table 3-3

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The Sxyl reaction can be complicated by rearrangement, because of thecarbocation intermediate formed. If a secondary carbocation (R2CH+) is formed,it can rearrange to form a tertiary carbocation (RgC*), if a tertiary carbocation ispossible. For alkyl carbocations, the relative stability is 3" > 2' > 1", the samepreference that is observed with free radicals. The features in an S5i1 reaction areopposite of those in the Sp2 reaction. These features of a reaction can be used topredict whether a reaction will proceed by an 51111 or 51112 reaction mechanism.The reaction in Figure 3-19 proceeds by an Sp1 mechanism, because theelectrophile is tertiary and it has a good ieaving group.

H3CH2CHTC'\ o

H3c\f NH3

H3CH2CI+& *IoH3N: +

AverageNucleophile

H?CH2CH2C

\

tr.o'YFH3CH2C

3" Electrophile

gH2cH2cH. stable| ' ' Anionol,,*_\,,.,r,

cH2cH3

Racemic Mixture of Products

Figure 3-19

The reaction is favorable, because the leaving group is a

the nucleophile forms a stronger bond with carbon thangood leaving group andthe leaving group.

Example 3.21Ihe addition of ammonia to R-3-iodo-3-methylhexane at low temperature wouldvield:

A. one product with R configuration exclusively (retentionB. one product with S configuration exclusively (inversionC. two products in an enantiomer mixture.D. two products in a diastereomeric mixture.

of stereochemistry).of stereochemistry).

SolutionFirst, we must determine whether the reaction proceeds by an S1r11 or 51..12

mechanism. The electrophile (R-3-iodo-3-methylhexane) is tertiary, so thereaction proceeds by an S1'tr1 mechanism. The chiral center is lost with theformation of the carbocation intermediate, because the intermediate is pianarwith symmetric sides. This results in a racemic product mixture of twoenantiomers. Choice C is a swell answer for this question.


Organic Chemistry Stereochemistry Nucleophilic Sutistitution

In Figure 3-19, the electrophile is tertiary. In cases where the reactive carbon issecondary, an slr]1 reaction can be complicated by rearrangement. This is shownin Figure 3-20.

H.C"\

,,L--r

-

Hv \'(H3C)2HC

H,.C H"C"l hydride shift "l

-([t -

"'h"-c H c* H/\ /\

HsC CHs HsC CHe2'carbocation 3" carbocation

HlCH,C" 'lI HAN:\-

-li,-

HsC CHa

3" carbocation

H3CH2C

\

urcsf.NH'HsC

achiral product

Figure 3-20

Rearrangement is rapid, because it is an intramolecular process. L:r the examplein Figure 3-20, the secondary carbocation rearranges to form a more stabletertiary carbocation before the ammonia nucleophile attacks the carbocationintermediate. This results in a tertiary product. The halide leaving group is notbasic enough to deprotonate the ammonium cation formed from the substitutionreaction, so the product remains as a cation.

If the electrophile has a chiral center at a site other than the electrophilic carbon,an Str11 reaction will form both a major and minor product. The major productresults from the transition state with least steric hindrance.

CHICH? CH,,CH" CH,CH" CH.CH,YYYYOCI -.> e.'C*

ry.s O'"?:il; Ort:il;

Figure 3-21

In the example in Figure 3-27, the ethyl group in front of the plane interferes withthe attack by the nucleophile, which results in an uneven distribution ofdiastereomers as the product mixture. The major product is formed whenammonia attacks the less hindered face of the carbocation (backside attack in thisexample). The minor product is formed when ammonia attacks the morehindered face of the carbocation (front side attack in this example).

Attack from the backside ismore favorable than frontsideattack due to steric hindrance.

MajorProduct

MinorProduct

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Distinguishing an Sp2 reaction from an SpL reactionThe first thing to look for when determining the mechanism by which a

nucleophilic substitution reaction will proceed, is the substitution of theelectrophile. Tertiary and allylic (adjacent to a n-bond) electrophiles wiil proceedby an Sx11 mechanism while methyl and primary electrophiles will proceed by an5512 mechanism. This is the first factor to view. If the electrophile is secondary,then the reaction can proceed by either mechanism. After considering thesubstitution of the electrophile, the next feature to consider is the nucleophilicstrength. The stronger the nucleophile, the more likely the reaction will proceedby an 5512 mechanism. The better the leaving grosp, the more likely the reactionwill proceed by an 51111 mechanism. Lastly, you should consider the solvent. Ifthe solvent is protic (capable of forming hydrogen bonds), the reaction wili havea tendency to proceed by an 5111 mechanism. If the solvent is aprotic (notcapabie of forming hydrogen bonds), the reaction will have a tendency toproceed by an Sry2 mechanism. These factors can be applied when looking at thereactants.

If rate data are given, then the mechanism can be inferred without ambiguity.The rate law associated with an 5111 mechanism is shown in Equation 3.3, whilethe rate iaw associated with an Sp2 mechanism is shown in Equation 3.4.

S1.tr1 Rate = k [Electrophile]

5512 Rate = k [Nucleophile][Electrophile]

If the rate of a reaction changes as the nucleophile concentration is varied, thereaction is proceeding by an Sp2 mechanism. Conversely, if the rate of a reactiondoes not'change as the nucleophile concentration is varied, the reaction isproceeding by an 5511 mechanism. Because the solvent can affect the strength ofa nucleophile, solvent and nucleophile are often considered together. The ratesof both reactions vary with a change in electrophile concentration.

The energy diagrams for the two mechanisms also differ. There is nointermediate associated with an 51112 reaction, only a transition state. There is ar"t

intermediate and two transition states associated with an 5511 reaction. Figure 3-

22 shows the energy diagrams for the one-step S52 reaction (on the left) and thetwo-step Sp1 reaction.

sru1+

Intermediate

Reactant

Reaction co-ordinate -+

Figure 3-22

(3.3)

(3.4)

t

boHG)

F]

Transition state

Reaction co-ordinate ------->

Product

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The energy level increases at the start of each energy diagram, because a bond isbeing broken. In the case of the S1r11 reaction, the intermediate is of lower energythan the transition state, because the carbocation can rehybridize to the leilcrowded sp2-center rather than an sp3-center and the intermediate can besolvated in a protic solvent. The increase in energy from the intermediate to thesecond transition state is associated with rehybridization to the more crowdedsp3-center and the intermediate losing solvition to allow the nucleophile toattack. It is important to be able to recognize these diagrams and apply theinformation they contain to conceptuai questions.

Example 3.22The addition of sodium methoxide to s-2-bromohexane at low temperaturewould yield:

A. one product with R configuration exclusively (retention of stereochemistry).B. one product with S configuration exclusively (inversion of stereochemistry).C. two products in an enantiomeric mixture.D. two products in a diastereomeric mixture.

SolutionFirst, we must determine whether the reaction proceeds by an sIrI1 or sry2mechanism. The electrophile (S-2-bromohexane) is secondary so the reaction canproceed by either an sl.J1 or Sry2 mechanism. The nucleophile is a strongnucleophiie, so we can assume the reaction will proceed via an Sp2 mechanismlThis results in inversion of the chiral center and the final product having Rstereochemistry, so choice A is the best answer. The low temperature isimportant, so that there is little to no E2 product formed. An E2 reaition resultsr" r^" trt-",t"" t

Example 3.23The following reaction shows what relationship between nucleophileconcentration and reaction rate?

H?CCH2S- + H,CCHTBT -----+ H3CCH2SCHTCH, + Br-

A. The reaction rate increases in a linear fashion with increasing nucleophileconcentration.

B. The reaction rate increases in an exponential fashion with increasingnucleophile concentration.

C. The reaction rate does not change with increasing nucleophile concentration.D. The reaction rate decreases in a linear fashion with increasing nucleophile

concentration.

SolutionThe reaction has a primary electrophile and a good nucleophile, which favors ans512 mechanism. The rate equation associated with a reaction proceeding by ans}..J2 mechanism is rate = k lElectrophile][Nucleophile]. The equation shows thatthe reaction rate is directly proportional to the nucleophile concentration. Therate increases in a linear fashion wiih increasing nucleophile concentration, asstated in choice A. The best answer is choice A. Choice B should be eliminated,because the rate of a nucleophilic substitution reaction does not depend on theconcentration of any species in an exponential fashion. Choice D should also beeliminated, because the rate will not decrease with additional nucleophile. It willeither increase in a linear fashion or not change.

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Example 3.24A transition state with no intermediate is associated with which of the followingreactions?

A. H3CCHZO- + H3CCH2BT ----t> H3CCH2OCH2CH3 + Br-

B. H3CCH2OH + (H3C)3CBr ------> H3CCH2OC(CHa)3 + HBr

C. (H3C)3CSH + (H3C)2CHOHz* + @3C)3CS+HCH(CH3)2 + H2o

D. NH3 + (H3CH2C)3CBr --.> (H3CCH2)3NH3+ + Br-

SolutionAs shown in Figure 3-27, no intermediate is associated with an sx12 mechanism,so we must find the reaction most likely to proceed by an sN2 mechanism. Thereaction most likely to proceed by an Sl.]2 mechanism should have a goodnucleophile and ideally a primary electrophile. A low temperature is importanthere so that there will be little to no E2 product formed. Choice C is the ieactionof a secondary electrophile with a poor nucleophile, so it will likely proceed byan S51 reaction mechanism. This eliminates choice C. Choices B and D involvetertiary electrophiles, therefore they definitely will proceed by an SlrJ1 reactionmechanism. This eliminates both choice B and choice D. Choice A has a primaryelectrophile and a good nucleophile, which makes it the most likely to proceedby an sg2 mechanism, and therefore makes it the best answer. The ethoxideanion is also a strong base, so elimination is possible in choice A. Despite thecompetition with the E2 reaction, choice A is still the best answer.

Reaction KineticsThe rate'of a reaction depends on several factors. The rate depends on theavailable energy for the molecules to collide, orient, and break the necessarybonds. The rate depends on the likelihood of the molecules colliding. For ansirl2 reaction, the rate depends on the availability of nucleophile, while it doesnot depend on the nucleophile concentration in an Sry1 reaction. Consider the5512 reaction shown in Figure 3-23 with an ethoxide concentration, [CH3CH2O-],of 0.01 M a 2-bromopropane concentration, [CH3CHBTCH3], of 0.01 M and a k1*of 2.53 x 10-2 L.mol-1.s1 at 2gBK.

HqCH?CO:

Figure 3-23

The concentrations are low, so the reaction is very slow. Plugging the values intoEquation 3.4 yields a rate of 2.53 x 70-6 M per second. The reaction rate may beincreased by increasing the reactant concentrations, increasing the temperature,or by adding a catalyst. A catalyst stabilizes the transjtion state complex andlowers Eu.1. Transition states are short-lived complexes. In the course of thereaction, reactants collide with the correction orientation (from backside attack)to form the transition state complex, when eventually splits io generate theproducts. Figure 3-24 represents the species of the S5tr2 reaction in Figure 3-23 atdifferent stages in chronological order over the duration of the reaction.



-/'n'Yg"

CH"..4- | -

..5H,cH"co ----+----Bri

H CH3Transition State

Reactants draw close to startbond formation and form the

Early Complex (C"u.1r).

H.C..6- '\

..6-H3cH2co.

;f --- Br!

HsCCeurly

Transition State starts to splitas the bond breaks to form theLate Complex (C1u1s).

..5-H3CH2CO.- - - /::'

\"CHs

Cl"t.

..6-Bri

H.CH"COJ L ..

Figure 3-24

Physical Properties of StereoisomersEnantiomers have identical physical properties (boiling point, melting point, anddensity to name a few), while diastereomers have slightly different physicalproperties. Because they have slightly different physical properties,diastereomers are easier to separate than enantiomers. Enantiomeric mixturesare difficult to purify, because a racemic mixture often has strongerintermolecular forces than the pure enantiomer. Table 3-4 lists the physicalproperties of the two enantiomers and the racemic mixture of 2-butanol.

Form Chirality CTD Boiling Point Density Index of Refraction

(+) S + 13.5" 99.4'C 0.808 1.398

(-) R - -t-J.5 99.4'C 0.808 1.398

(r) R/S 0' t01.2"c 0.840 1..442

Table 3-4

lrom the data in TabIe 3-4, it can be seen that a racemic mixture allows themolecules to get closer together. This can be thought of when considering yourhands, where a left and right hand fit together nicely. It is common that a

racemic solid mixture has a higher melting point and greater density than eitherenantiomer.

Reaction co-ordinate



Separating StereoisomersOne of the most chailenging tasks a synthetic organic chemist faces is theseparation of stereoisomers. If a reaction generates a new chiral center in theproduct, then it will be complicated by stereoisomerism. To generate a purestereoisomer as a product, chirality must be invoked at some point. Frombiochemical examples, we know that enzymes (chiral polypeptides) orientmolecules in a specific fashion, allowing just one stereoisomer to form, Thechirality of the enzyme helps to select for the desired product. In organicchemistry, there are compounds known as chiral nuxiliaries, which introducechiraiity to, or exaggerate existing chirality within, a reactant molecule. Chiralauxiliaries serve in a similar fashion to an enzyme. When aiming for one specificstereoisomer, it is often easiest to select for it in the reaction. If not, a mixture ofstereoisomers is formed and chirally selective separation techniques must beapplied.

Chirally selective separation techniques come in two types. The first involvesemploying an enzyme (or chirally selective molecule) to react specifically withone stereoisomer within the mixture. By reacting and therefore introducing anew functional group to only one stereoisomer, the two enantiomers now havedifferent physical properties and can easily be separated. Once separated, thesame enzyme can be employeci to return the compound back to its original form.An example of such an enzyme ts porcine renal acylase, which selectively acylatesthe N-terminal of L-amino acids. By acylating the L-amino acid, it is no longer a

zwitterion at neutral pH, while the D-amino acid is a zwitterion. Because onecarries a net charge, it is easily separated from the other.

The second chiraily selective separation technique involves invoking chirality inan existing separation technique. For instance, a column chromatography gel canbe made from a pure stereoisomer. If the column is made with an R-alcohol forinstance, then when a racemic mixture of alcohols is added, the S-enantiomer hasa greater affinity for the column and thus has a greater elution time. This is thebasic principle behind affinity chromatography in biochemistry, where anantibody is bound to the column so that it can selectiveiy bind an antigen. Chiralcolumns in organic chemistry are not as specific as enzyme columns and theyhinder solutes, but do not actualiy bind them. In theory, chiraiity could beinvoked in any organic chemistry separation technique, including distillation,but only it is chiral columns that are commonly used.


Organic Chemistry Stereochemistry Summary

SummaryThis section involved a few basic concepts. From this particular section, youshould be able to identify R and S chirality, determine the relationship betweenstereoisomers, understand the nuances of nucleophilic substitution, distinguishbetween the slri1 and sIN2 reactions, and apply stereochemistry to organic labtechniques and biochemistry.

Determining R and S: When a structure is drawn in dash-and-wedge style, ifpriority number four is in back or two substituents away from the substituent inback, then take the arc as is and assign the corresponding chirality (Clockwise =R; Counterclockwise = S). If priority number four is one substituent away fromthe substituent in back, then determine the arc and assign the opposite chiralityof the arc.

Enantiomer: Enantiomers are stereoisomers which are nonsuperimposablemirror irnages (reflections that you can't overlay). They have the same bonds,but they have a different orientation of atoms in space, as do all stereoisomers.Enantiomers can be thought of as stereoisomers in which all of the chiral centershave different orientation between the two molecules. Enantiomers haveidentical physical properties.

Diastereomer: Diastereomers are stereoisomers which are nonsuperimposableand that are not mirror images. They too have the same bonds and a differentorientation of atoms in space. To be diastereomers, the compounds must containa minimum of two chiral centers. Diastereomers are stereoisomers in which atleast one but not all of the chiral centers have different orientation between thetwo molecules. Diastereomers have close, but not identical, physical properties,

sy1 Reactions: Preferred when an electrophile is tertiary, when the solvent ispolar and protic, and when the electrophlle has u good leaving group, Sx1reactions form a planar carbocation intermediate that can undergorearrangelnent to form a more stable carbocation. Sru1 reactions result in a

racemic mixture when the reactive carbon is the only chiral center in the reactant.The rate of an SX11 reaction depends on only the electrophile and not thenucleophile (rate = k[Electrophile]).

5672 Reactio;,zs: Preferred when an electrophile is primary, when the solvent ispolar and aprotic, and when the nucleophile is good. 5112 reactions form a

transition state complex as the nucleophile forces the leaving group off from theelectrophile. SN2 reactions result in inversion when the reactive carbon is a

chiral center. The rate of an 5112 reaction depends on both the electrophile anCnucleophile (rate = k[Nucleophiie][Electrophile]).

Stereoisomer Mixtures: A fifty-fifty mixture of enantiomers is said to be a racemr:mixture. The diastereomers in a product mixture formed from a chemica,reaction are referred to as major and minor, because they do not occur in a fiftv-fifty ratio. Separating one diastereomer from another is easier than separatirrgone enantiomer from another. To separate enantiomers, chirality must b.incorporated into the separation technique.

Copyright O by The Berkeiey Review 216 The Berkeley Revien


Key Points for Stereochemistry (Section 3)

Chirality and Asymmetry

1. Chiral Moleculesa) Have asymmetry within their structure due to atoms that are unevenly

substitutedi. Stereogenic carbons are sp3-hybridized carbons with four unique

substituentsii. Chiral moiecules are predominant in many organic reactions

b) Stereogenic carbons are assigned an absolute configuration of either R orS to describe their chiralityi. Priorities are assigned according to atomic mass of the atoms

attached to the stereogenic center. If two atoms are identical, thenyou proceed along its connectivity until there is a difference

ii. when priority #4 is in back, a clockwise arc connects priorities #1,#2, and #3 in R-stereogenic centers. When priority #4 is in back, acounterclockwise arc connects priorities #I, #2, and #3 in S -stereogenic centers.

iii. To determine whether a center is R or S, you can place your thumb inthe direction of substituent #4 and curl your fingers from priority #1,through priority #2, and on to priority #3. Only one of your handscan do this. If it is a right hand that does this, the stereogenic carbonhas R-chirality. If it is a left hand that does this, the stereogeniccarbon has S-chirality.

c) short cuts for determining R and s involve the positioning of priority #4i. If priority #4 is in back, then the arc determines the chirality

(clockwise for R and counterclockwise for S). If priority #4 is infront, then the arc must be reversed to determine the chirality (aclockwise arc is reversed to represent S and a counterclockwise arc isreversed to represent R). If priority #4 is drawn in the plane close tothe group going back, then the arc is reversed to determine thechirality. If priority #4 is drawn in the plane far away from thegroup going back, then the arc as is determines the chirality.

ii. \Ahenever two groups are switched, the chirality reverses

Configurational Isomers

1. Same connectivity, but different spatial arrangement of atomsa) Can be categorized as either optical isomers or geometrical isomers

i. Optical isomers rotate plane-polarized lightii. Geometrical isomers differ about a feature in the molecule about

which rotation is not possible (n-bond or ring)iii. Optical isomers are identified by a standard rotation value

b) Can be categorized as enantiomers or diastereomersi. Enantiomers are nonsuperimposable stereoisomers that are mirror

imagesii. Diastereomers are nonsuperimposable stereoisomers that are not

mirror imagesiii. Enantiomeric optical isomers are better thought of as stereoisomers

where all of the chiral centers differiv. Diastereomeric optical isomers are better thought of as stereoisomers

where some, but not ali, of the chiral centers differ


Organic Chemistry Stereochemistry Section Summary

Stereochemistry in Reactions

7. Stereoisomers are formed when a nucleophile attacks an asymmetricmoiecule in multiple ways

a) Racemic mixtures form when there is no preexisting chiralityb) Diastereomers are formed in a major/minor distribution when one of the

reactants has a chiral center at a non-reactive sitei. Mixtures are resolved by using chiral reagents or lab techniques that

invoke chirality.ii. Enzymatic reactions use a chirai catalyst to cause the reaction to

drastically favor the formation of one stereoisomer over all otherpossible stereoisomer products

iii. When two enantiomers are present in unequal amount, there is saidto be an enantiomeric excess. Enantiomeric excess is used to describethe success in a stereoselective synthesis

c) Enantiomers have identical physical properties as one another whilediastereomers have different physical properties.

Nucleophilic Substitution

1. Proceeds by either an Sp1-mechanism or Sp2-mechanismIn an Sg1-mechanism, the electrophile is highly substituted, the solventis protic, and a carbocation intermediate is formed because the leavinggroup leaves in ihe first step. There is potentially rearrangement and theproduct mixture is often racemic. The reaction rate only depends on theelectrophile.

In an S52-mechanism, the electrophile is minimally substituted, thesolvent is polar and aprotic, and a transition state is formed because thenucleophile attacks to force the leaving group off in the only step. Thereis inversion of chirality so the product mixture is often the oppositechirality of the reactant and the reaction is fast. The reaction ratedepends on both the nucleophile and the electrophile.

a)

b)

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51

StereochemistryPassages

15 Passages

I OO Questions

Suggested schedule:I: After reading this section and attending lecture: passages I, II, vll & x

Grade passages immediately after completion and log your mistakes.

II: Following Task I: Passages IV V IX, & XIII (28 questions in 56 minutes)Time yourself accurately, grade your answers, and review mistaKes.

III: Keview: Passages III, VI, VIII, XI, XII, & Questions 92 - IOOFocus on reviewing the concepts. Do not worry about timing.

BSpeciaLtztng in MCAT Preparation

#fi il rtffi-iffit :lffiffing€s.

I. Isoleucine and Threonine

II. Unknown Stereochemically Active Compound

III. Classification of Isomers

IV. Stereoisomers and Optical Activity

V. Enantiomers from Diels-Alder Reaction

VI. Proposed S51 SYnthesis

VII. Nucleophilic Substitution

vlil. Nucleophilic substitution of chlorocyclohexane

IX. Leaving GrouP Strength

X. Reaction Rates of Nucleophilic Substitution

XL Elimination versus Substitution Experiment

XII. Enantiomeric Excess

XIII. Stereochemistry in Synthesis


(r -7)

(B - 14)

(15 - 21)

(22 - 28)

(2e - 35)

(36 - 42)

(43 - 4e)

(5O - s6)

(s7 - 63)

(64 - 70)

(7r -77)

(78 - 84)

(Bs - el)

(e2 - lOO)

lI:r

lll.r:

.lr-:

rlltut

illl:,"

rLllJ

Structure, Bonding, and Reactivity Scoring Scale

84 - 100

Passage I (Questions '1 - 7)

Isoleucine is one of eight essen.tirrl amino acids. The::n essential is applied to amino acids that humans cannot::rduce, and therefore must take in through diet. Isoleucine:, .sts as a zwitterion in aqueous solution. The natural form

isoleucine has the same chirality as other naturally-:urring amino acids at carbon number two. Isoleucine is

- lnd naturally in the L form. Naturally occurring amino.,-ds have an S chiral center on carbon two (except cysteine).-..rleucine has a second chiral center in addition to the one at-'rbon two. Seventeen of the amino acids we code for have: ,.rctly one chiral center, with glycine (which has no chiral- 'rter) and threonine (which has two chiral centers) being the::er exceptions. The second chiral center of L-isoleucine

-:s fixed chirality, so a diastereomer of isoleucine varying at. : side chain is not a biological substitute for L-isoleucine.. .sure 1 shows the zwitterion form of L-isoleucine.

CH 2CH r

o-L-Isoleucine

Figure 1 Zwitterion form of L-Isoleucine

The only other amino acid coded fbr by human beings.:at has a chiral side chain attached is threonine. The.ireonine side chain contains an alcohol functionality. Like.oleucine, the side chain chiral center must be specific for.ie amino acid to be biologically incorporated. Threonine is..so an essential amino acid. The diaslereomer of threonine.:at varies at the side chain is known as "allo-threonine."

1 . What is the stereoconfiguration for isoleucine?

A. 2R. 3RB. 2R, 35

C. 25,3RD. 25, 35

2. If the side chain chiral center were changed, the newstructure would be which of the fbllowing?

A . An enantiomer of isoleucine.B . A diastereomer of isoleucine.

C . An epimer of isoleucine.D . ldentical lo isolcucine.

3. Most naturally occurring amino acids have whichstereochemical orientation ?

A. RB. S

C. E

D. Z

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4 . The amino acid glycine has an H as its side chain. Whatwould you predict for the optical rotation for naturallyoccuning glycine?

A. (.+)32'

B. (-) 32'c. 0'D . More information is needed. (Never choose this!)

5 . If L-isoleucine were found to have an optical rotation of-62' from plane-polarized light studies, what would youpredict for the optical rotation of its enantiomer?

A. -62'

B. +62'c. -31'D. +31'

6 . Which of the following is the side chain of threonine?

A. -CH(CH3)2

B. -CH(oH)CH3C. -CH2CH2OH

D. -CH2OH

7 . Only twenty-five percent of synthesized isoleucine can beused biologically. This is best explained by which ofthe following explanations?

A. Only 25Vo exists as a zwitterion in the body.

B. 15Vo of synthetic isoleucine does not have thecorrect side chain.

C. In synthesizing isoleucine, the two chiral centersresult in four stereoisomers being formed. Only oneof the fbur is biologically usable.

D. In synthesizing isoleucine, the two chiral centersresult in eight stereoisomers being formed. Onlytwo of the eight are biologically usable.



In two controversial laboratory studies, Compound P,

shown in Figure I below, has been determined to reduceconstipation in the Southern European Red-Eared JumpingLizard during mating season. The effects are less significantduring courting periods and does nol occur at all during thefirst three days following the New Moon. The two studiescompared this particular stereoisomer, along with otherstereoisomers and structural isomers of Compound P, todetermine the effects of the drugs. The dosages used wereheld constant between compounds. The disagreementbetween researchers came in determining the binding activityofeach stereoisomer and its subsequent reactivity.

HOH

cH 2cH 3

Figure 1 Compound P

The exact mechanism for the constipation reducingbehavior is not known, but it is speculated to work inconjunction with sex hormones to induce muscle relaxation.In other studies, the compound has been applied to theabdomen of the Saskatchewan Green-Nosed Squatting Frog totest for similar effects. To date, no solid conclusions have

been formed as to the effect of Compound P on constipationin other organisms. Being such an important chemical to the

bowel process of reptiles and amphibians everywhere, a great

deal ofresearch is destined to be under way. Researchers havecontinued to develop other structural isomers that willhopefully show similar effects in creatures such as the veryrare Yellow-Tongued Sabertooth Crotch Cricket.

8 . If treated with PBr3, the OH groups can be converted

into Br groups through a reaction which proceeds by an

5512 mechanism. The dibromo product formed fromCompound P would show which of the following chiralassignments?

A. 35, 45

B. 3R, 45

C. 35, 4R

D. 3R., 4R

9 . Assuming Compound P is made by treating an alkenewith KMnO4 in basic water, what is the geometrical

orientation of the alkene precursor to Compound P?(KMnOa adds two hydroxyl groups with syn orientation)

A. E.

B. Z.C. Either, the reaction has no stereoselectivity.

D . Neither, the reaction has no stereoselectivity.

OH

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10. What is the stereochemistry of Compound P?

A. 35, 45B. 3R, 45C. 35, 4RD. 3R, 4R

1 1. The followingCompound P,

Jumping Lizard'

structure relates in what way tothe Southern European Red-Eared

s constipation medication?

H H HO Ct-12CHs

A. It is an enantiomer of Compound P

B . It is a diastereomer of Compound P

C. It is identical to Compound P

D . It is a meso structural isomer of Compound P

12. What is the stereochemical orientation of the followinscompound?

A. 15, 25

B. lR, 25

C. 15,2RD. lR, 2R

13. Enzymatic active sites are all of the followingEXCEPT:

A. chiral speciiic.

B. size specific.

C. functional group specific.D . isotope specific.

14. The enantiomer of Compound P has:

A. the same boiling point as Compound P

B. a higher melting point than Compound P.

C . A lower density than Compound P.

D . The same specific rotation as Compound P.

;"

-!:

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Isomers are compounds with identical formulas but adifferent arrangement of atoms in space. There are two basictypes of isomers: structuraL isomers (also known asconstitutional isomers) and stereoisonters (which can furtherbe categorized as either configurational isomers orconformational isonters). Structural isomers are defined asraving different connectivity of atoms, also referred to asdifferent bonds. Ethanol and dimethyl ether are an example ofstructural isomers.

Stereoisomers are isomers which have the exact same'londs, but diff'er by the position of their atoms (substituents)'.rithin space. Stereoisomers can be classified as either-tptical isomers or geometrical isomers. Optical isomersrave different spatial arrangement due to asymmetry about an.tom with in the molecule. A good example of optical.somers are the R and S enantiomers of a given molecule.3eometrical isomers have a different spatial arrangement of:ioms with fespect to a molecular plane. Cis and trans:utene are a good example of geometrical isomers.

A common, and often challenging, task in an organic:remistry laboratory is to separate and distinguish the)omers formed fiom a chemical reaction. Of all isomeric:ixtures, it is easiest to separate and distinguish structuralsomers, which have varying physical and chemical

::operties. Geometrical isomers also have different physical::operties, and can be separated readily. They can often be::stinguished from one another by their melting or boiling: rints. Of isomers, optical isomers are the most difficult to,:parate. To separate optical isomers, a pure chiral material-"n be used in a chemical reaction, which is then followed by::ecipitation or extraction. Once separated, the specific::tical rotation value may be used to identify the enantiomer'-.3t was isolated.

Other traditional laboratory techniques can be modified to,:ploy chirality to help separate optical isomers from one..other. As a general rule, the distinguishing features: -'lween isomers can be used to separate them in a laboratory::ocedure, although some are considerably easier than others.-:r identifying a purified isomer, spectroscopy and physicalr: lperties are most often employed.

- 5. The pure R enantiomer of some compound has aspecific rotation of +32.2'. A sample you make in labhas a specific rotation of +16. l'. This is best explainedby which of the following mixtures?

A. 100Vo R enantiomer present.B . 7 5Vo R and 25Vo S enantiomers present.C . 50Vo R and 50% S enantiomers present.D . 25Va R and l5Vo S enantiomers present.

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16. Which of the following explanations does NOT accountfor an observed optical rotation greater than that of apositive literature value for the pure species?

A . Both enantiomers are present in an unequal amount.B. The solution is too concentrated.C . The cell for the polarimeter is too long.D . The wavelength of light being used is greater than

that of the standard sodium light used.

17. How many degrees of unsaturation are present in acompound with formula C11H21NO2?

4.4B. 3

c.2D. I

18. How many structural isomers are possible for theformula C6Hla?

A. rl

B. 1

c. 6

D. 5

19. The chiral centers for the following molecule are:(note: OH is located on carbon 2)

A. 2R, 3R

B. 2R, 35

C. 25, 35

D. 25, 3R

20. Which of the following CANNOT form opticalisomers?

A . A four carbon gem diolB. A four carbon vicinal diolC. A fourcarbon secondary amineD. A four carbon secondarv alcohol

2 L. How many stereoisomers are possible

structure?

4.32B. 64

c. 128

D.256

for the following

OH

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Stereocenters are important features in chiral organiccompounds. Stereocenters are responsible for physicalproperties, chemical reactivity, and biological function. Anexample of the correlation between physical properties and

chirality is shown with the stereoisomers of tartaric acid.Drawn in Figure 1 is the generic structure of tartaric acid. and

a data table of the physical properties of the stereoisomers:

HO

(oH)

(oH)

o

Figure

Table 1 shows thestereoisomers of tartaricthe enantiomeric mixture.

1 Tartaric Acid

physical properties of the thre;acid and the physical properties c:

Form m.p. 0D Density H2O sol. @ 20'C

(+) 168-170 + 12" 1.7598 139 g/l00ml(-) 168-170 12' 1.7598 139 g/100ml

meso t46-148 0" r.5996 125 g/l}0mL

(t) 205-207 0' 1.7880 2l g/l}0mL

Table 1

The differences in physical properties can be attributed ::lattice formation in the solid phase. The compound can pa;r.

most easily when it is symmetric. This can be seen in t:*density and the melting point of each stereoisome:Examples of biological activity are numerous. A comni:,'r

example involves the digestion of D-sugars. Our bod,v car

metabolize D-glucose yet it cannot metabolize L-glucose i::eenantiomer of D-glucose). A recent example involves ne

compound L-Dopa. L-Dopa is used as an anti-Parkinson dr-r4

while D-Dopa has no effect and can in fact be toxic in largenough doses. L-Dopa is drawn in Figure 2 below.

HO NHz

Figure 2 L-Dopa

2 2 . Which of the following physical properties would hi'*u

the same value for morphine and a diastereomer ui{

morphine?

A. Melting point.

B. Density.C . Molecular mass.

D. Specific rotation.

{ts

fD

-{"E"ID"

How can a compound with an optical rotation of+233.0'be discerned from a compound with an opticalrotation of -121 .0"2

A . The intensity of the light is greater with theposilive optical rolation.

B. The sample with +233.0" oprical rotarion whendiluted to half of its original concentration wouldshow an optical rotation of +116.5".

C. The larger the absolute value of the opticalrotation, the greater the density of the compound.

D. It is not possible to distinguish the two compoundsfiom one another.

Given that the specific rotation of D-Glucose is +52.6.,what can be said about the specific rotation of D_mannose (the C-2 epimer of glucose)? Note: Anepimer is a diastereomer that varies at onlv onestereocenter.

^voIu-f oHIso-FHIH-1- oH

u -1- oHI

CH2OH

D_Glucose

A . It cannotbe +52.6', -52.6', or 0'.B. It must be either +52.6', -52.6', or 0'.C. It is 0'.D. It is -52.6 ".

How many stereoisomers are possible for penicillin V?

HIH'coc6Hsl/ t\a.r,

o ,h.y -CHr

co2H

A. X

8,2c" 4

D. 8

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2 6. What can be concluded about the packing of moleculesin the crystal lattice of the stereoisomers of tartaric acid?A. The meso compound packs most tightly while the

(+) enantiomer and (-) enantiomer pack the same.B. The meso compound packs least tightly while the

(+) enantiomer and (-) enantiomer pack the same"C . The meso compound packs most tightly of all of

the stereoisomers. The (+) enantiomir packs moretightly than the (-) enantiomer.

D. The meso compound packs most tightly of all ofthe stereoisomers. The (+) enantiomer packs lesstightly than the (-) enantiomer.

28.

27 . How many stereocenters are present in the moleculecamphor which shows an optical rotation of +44.3"?

cHs

CH:

A. 0

B. I

c.2D. 3

Which of the following statements best explains whyan R/S enantiomeric mixture has a higher melting pointthan pure samples of either the R enantiomer or the Senantiomer?

A . It requires more energy to break the hydrogen bondswithin a pure compound than within a mixture ofcompounds.

B. Enantiomers readily form covalent bonds with oneanother.

C . The covalent bonds are weaker when the material isone pure stereoisomer then when it is a mixture oftwo or more stereoisomers.

D. The R-stereoisomer packs more tightly with itsenantiomer than it does with itself.


The Diels-Alder reaction of isoprene upon itself at 100'Cyields two enantiomers of limonene in a fifty-fifty ratio. Thediene of one isoprene molecule reacts with the less hindereddouble bond of another isoprene molecule to form theproduct. Because isoprene is a planar molecule, there is anequal chance fbr the reaction to occur on the top face as thebottom face of isoprene. The Diels-Alder reaction thatsynthesizes limonene is shown in Figure 1 below.

Figure 1 Diels-Alder Condensation Reaction of Isoprene

The two enantiomers that are formed have similarphysical properties, but have different applications in theflavoring agent business. One enantiomer has a lemon scentwhile the other has an orange scent. The two enantiomers oflimonene are shown in Figure 2 below.

lemon odor orange odor

Figure 2 Enantiomers of Limonene

The percentage of each enantiomer in the product mixturecan be altered only with the addition of another chiral reagentinvolved in the transition state. To isolate either the lemonflavoring or the lirne flavoring, the chosen enantiomer mustbe separated from the product mixture by one of severalpossible laborator"y techniques that involve chirality.

29. Which of the following techniques does NOT work roisolate one enantiorner from the plesence of anenantiomeric mixture?

A. Adding the mixture to a chiral gel in a columnchromatography.

B. Distillation of the product mixture.

C . Crystallization of the mixrure with the addition ofan enantiomerically pure compound.

D . Filtering through an enzymatic membrane.

30. Which of the following will lead to a product mixturecomposed of more than fifty percent of one of theenantiomers (the product mixture not being racemic)'/

A. Carrying the reaction out with a chiral solvent.B. Carrying the reaction out with a chiral catalyst.C . Carrying the reaction out at a lower temperature.D. Carrying the reaction out at a higher concentration.

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31. The lemon flavored isomer has what stereochemisr,associated with it?

A . One chiral center with R stereochemistryB. One chiral center with S stereochemistryC. Two chiral centers with R, R stereochemistryD. Two chiral centers with S, S stereochemistry

32. Which of the following physical properties is MOSIlikely different for the two enantiomers?

A. Boiling pointB. DensityC. Alcohol solubiiityD. Optical Rotation

3 3. When hydroborane (BH3) reacts with limonene, ir a.rjboron to the less hindered carbon of both alkenes ::nhydrogen to the more hindered carbon of both alken::;,What is the major product when hydroborane reacts $:17

R-limonene?^"-&""&ca p,"-u\.4 n-u-=A

3 4 . What can be concluded about the olfactory receptors

A. They are symmetric (achiral) because thel ;:anr

distinguish between enantiomers.

B. They are asymmetric (chiral) because ther :md istinguish between enantiomers.

C . They are symmetric (achiral) because they c.::nm.distinguish between enantiomers.

D. They are asymmetric (chiral) because they c-rmum

distinguish between enantiomers.

3 5. If the following molecule were the reactantisoprene, how would the results differ?

rather than

A . The product mixture would no longer be racemic.

B. The product mixture would still be racemic.

C . The products would be achiral.

D . The products would have four stereocenters each.

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A chemist intended to study the effect of peripherychirality on a nucleophilic substitution reaction. To 50 mLof methanol at 45'C, the chemist added 0.10 moles of(l S,2R)-2-methylbromocyclopentane in the hopes of carryingout Reaction 1, which is shown below.

HrC Br HrC OCHrIJ LiO

CHroH'

O &

majorproduct

"'u''minor

product

Reaction I

The reaction was carried out for one hour at 45"C in an

aqueous solution buffered at a pH of 5.0. Excess methanolwas removed from the product mixture by fractionaldistillation under reduced atmospheric pressure. Theatmospheric pressure was reduced to lower the boiling pointof methanol, in the hopes that additional reactivity would be

minimized at a lower temperature. The optical rotation forthe crude product mixture was found to be 0', which is

contrary to what the chemist had expected. The chemist had

expected that the crude product mixture would exhibit opticalactivity, although the exact value would be different than that

of the reactant.

The chemist reevaluated the proposed reaction, Reaction1, and decided that the temperature and the pH should be

changed. Under the reaction conditions used, the proposed

reaction proceeds by a mechanism that is susceptible torearrangement. The chemist also failed to consider otherreactions that compete with nucleophilic substitution at

elevated temperatures, such as elimination. Under differentconditions, the chemist found that optical activity could be

retained.

36. The chemist attempted to carry out what type ofreaction?

A. SylB. S52

c. Er

D. Ez

3 7. Which of the fbllowing reaction mechanisms wouidexplain a 0' optical rotation?

A . An Spl reaction with inversion of configuration

B. An Sy2 reaction with rearrangement

C . An Sy2 reaction with inversion of configuration

D . An E1 reaction with rearrangement

3 8. The two products from the proposed reaction are relatedin what manner?

A . They are enantiomers of one another.

B. They are diastereomers ofone another.

C. They are identical to one another.

D. They are different orientations of the same mesocompound.

3 9. To increase the amount of substitution reactionobserved, the chemist would likely change the

temperature and pH in what way?

A . Increase the temperature and decrease the pH

B . Decrease the temperature and increase the pH

C . Increase both the temperature and the pH

D . Decrease both the temperature and the pH

4 0. The small amount of substitution product isolated wasfound to have both the OCH3 group and the CH3 group

both on the same carbon. This can best be explained inwhat way?

A. First an elimination reaction took place followedby a Markovnikov addition reaction.

B. First a Markovnikov addition reaction took place

lbllowed by an elimination reaction.

C . A hydride shift occuned.

D. A methyl shift occured.

41. PBr3, when added to an alcohol, converts an OH group

into a Br substituent with inversion of configuration at

the carbon. The mechanism is an Sy2 substitution ofthe Br for the OH. What alcohol can be used in thisreaction to produce the starting reagent in the proposed

synthesis in Reaction 1?

A . (1R,2S)-2-Methylcyclopentanol

B. (1R,2R)-2-Methylcyclopentanol

C . (1S,2R)-2-Methylcyclopentanol

D . (1S,2S)-2-Methylcyclopentanol

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42. The following distribution of products can best L,'e

explained by which ofthe explanations?

HrC CHr

t- su'/ \ cHroH

\,)

.->

",u,s.",

Pas

_:_

::[::rtu:

"::::::- ":ira

-u J:i'fui:

.F--

"tr ."

MI;:

\,

\r

llllus .,:

unit;urrTI

ill"ri-ullJ

maJor

"'ry,minor

A . The intermediate undergoes a hydride shift.

B. The methyl group on the carbon adjacent to tr:carbocation influences the attack of methanol.

C . The methyl group on the carbocation influences r,:attack of methanol.

D. The intermediate undergoes a methyl shift.

f,assage Vll (Questions 43 - 49)

Nucleophilic substitution is the process by which-nctional groups on an alkane can be exchanged. It is

,:mmonly viewed as a reaction which can proceed by one of'.o paihways. The first of the pathways is named S51 I for. rate dependence on one reactant. The second pathway is

-.-erred to as Sp2 1br its rate dependence on two reactants..:.e data in the tables were gathered for reactions in which the- -:leophile was varied in two different experiments involving,'r different electrophiles (one primary and the other.::iary). The reactions were monitored using UV:3rtroscopy wl-rere the magnitude of the rate of disappearance: the peak corresponding to the electrophile varies directly,h the reaction rate.

\ucleophile Reaction rate w/ CH3CH2CH2CI

H3CNH2 4.7 x 10-2 M/sec

NH: 4.2 x l0-2 M/sec

H3CSH 2.1 x 10-2 M/sec

H3COH 8.2 x 10-3 M/sec

Table 1 Reactions with n-Propylchloride

\ ucleophile Reaction rate w/ (Hf C)f CCl

H3CNH2 3.3 x l0-4 M/sec

NH: 3.6 x 10-4 M/sec

H3CSH 3.3 x 10-4 M/sec

H3COH 3.5 x 10-4 M/sec

Table 2 Reactions with t-Butylchloride

The reaction rates listed in Table I and Table 2 represent: initial rate ol'cach reaction. All other variables that can:ct the reaction rate besides the nucleophile, such as

-Derature and concentration, were held constant between.;S .

l From the data presented in the passage, which of thefbllowing is thc best nucleophile?

A. H3CNH2

B. NH3

C. H3CSH

D. H3COH

1 . A nucleophilic substitution reaction proceeds MOSTrapidly with the leaving group on what type of carbon?

A. 1'

B. 2"

c. 3'

D. The reaction rate is independent of the degree ofsubstitution.

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Time -->

45. Reaction of (2R,3S) 2-bromo-3-methylpentane withammonia yields which of the following products?

A . (2S,3S) 2-amino-3-methylpentane

B. (2R,3S) 2-amino-3-methylpentane

C. (2S,3R) 2-amino-3-methylpentane

D . A racemic mixture of A and C.

46. A reaction in which the specific rotation of the startingmaterial is + 32" and the product (which still contains a

chiral center) is 0" is which of the following?

A. Sp1

B. SNr2

C. S5E1

D. SpE2

4 7 . Which of the following electrophiles is the best choiceto react with NaOCH3 to yield the following ether?

H

{r'"""'

48.

H:C-o

A . (R)-2-chlorobutane

B. (S)-2-chlorobutane

C. (R)-2-aminobutane

D. (S)-2-aminobutane

Which of the following graphs BEST represents the Cl-concentration as a function of time for the reaction ofammonia with 1-chloropropane?

Time+

Time ---------_->

Time ---+

tz--/CH:

49. Monitoring the following reaction by optical rorationwould yield which of the fbllowing graphs?

HjCH2C

+ HqCSH -+

HjCH2CH2C

Time

Time

Br

Time ------->

Time _-_=>

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It is possible to exchange one functional group or. Isubstituted alkane for another by performing a nucleoplu.,:substitution reaction. There are two versions of nucleophi,r:substitution, known as S,y1 and S,rr'2. The number in ea:rreaction describes the rate dependence. The rate of an S.;

reaction depends only on the electrophile concentration ;-;not on the nucleophile concentration. The rate of an S.,lreaction depends on both the concentration of nucleophile cr;the concentration of electrophile. The difference between ;*two mechanisms boils down to the sequence of the steps. -:Sp1 reactions, a leaving group first leaves followed :'nucleophilic attack of the carbocation intermediate. In S.,-reactions, the nucleophile attacks the electrophilic carb':iforcing the leaving group off of the molecule. I: ,mechanistic study, a secondary alkyl chloride w4s rsa1ei r,iltwo different nucleophiles to get the same ether product. ; j(rate data were collected for each.

NaOEtt - t ----------------

HOEr.0'

Reaction 1

HOEIL-l--->pH=5,0"

Reaction 2

OEt

Tables I and2 show the initial rate data for threetrials lbr each of the two recctions.

Table I Initial Rates for Reaction 1

Reaction 2

Rate (M/s) ICHTCHzOHI lCoHrrCll1.93 x l0-3 0.20 M 0.05 M

1.95 x l0-3 0.40 M 0,0.5 M

3.83 x 10-3 0.40 M 0.10 M

Table 2 Initial Rates for Reaction 2

Based on the data presented, the nature of the mechan,n:(whether it follows Sy1 or Sp2 kinetics) can be determl:.:rThe key to the analysis is to observe the change in rate a! 'rrw

nucleophile concentration changes. As a rule, Sy2 reaci,,rn,are fil\tcr than S5 l reactions.

Reaction 1

Rate (M/s) ICH3CH2ONa] IC6H11Cl]

1.32 x 10-2 0.05 M 0.0-5 M

2.63 x 10-2 0.10 M 0.05 M

5.25 x l0-2 0.10 M 0.10 M

5 0. Reaction 1 and Reaction 2 are best described as whattype of reactions?

A. Reaction I is an Syl; Reaction 2 is an Sp2B. Reaction I is an Sp2; Reaction 2 is an SplC. Reaction I is an El; Reaction 2 is an Sp2D . Reaction I is an Spl; Reaction 2 is anE2

5 1. All of the following are associated with Reaction 2EXCEPT:

A . inversion of the chiral center.

B. a carbocation intermediate.

C . the rate depending on the electrophile.

D. a greater rate when a protic solvent is used thanwhen an aprotic solvent is used.

5 2. A product mixture from a nucleophilic substitutionreaction on an enantiomerically pure compound thatyields a product distribution of 87Vo R and 13% S canbest be explained by which of the following?

A. The reaction goes purely by an 5112 mechanism.

B. The reaction goes mostly by an Sp2 mechanismwith some Sp1 mechanism transpiring.

C. The reaction goes mostly by an Syl mechanismwith some 51,12 mechanism transpiring.

D. The reaction goes purely by an SNI1 mechanism.

5 3. If bromine were used as the leaving group from thecyclohexane in lieu of chlorine, what effect would youexpect on the rate? (Note that a C-Cl bond is strongerthan a C-Br bond)

A . Both the S51l and Sy2 rates would increase.

B. The Spl rate would increase, while the Sp2 rate

would decrease.

C. The Sy1 rate would decrease, while the Sp2 rate

would increase.

D . Both the Spl and Sp2 rates would decrease.

5 -1. Which of the following is the BEST nucleophile?

A. tteCO-

B. H3COH

C. Cl-

D. HCI

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Rxn Coordinate

5 5. Which of the following reactions is most likely toproceed by an Sy2 mechanism?

A. H3CO- + (H3C)3CBr -+B. H3COH + (H3C)3CBr -+

C. H3CO- + (H3C)2CHCHBTCH3 -+D. H3COH + (H3C)2CHCHBTCH3 -+

5 6. Which of the following energy diagrams corresponds toan exothermic Sp2 reaction?

A. B.

Rxn Coordinate

Rxn Coordinate

Passage lX (Questions 5Z - 63)

- -ucleophilic substitution reactions, the reactivity ofa: ..::,rophile dictates the reaction rate. The reactivity ofthe..=::-rphile is correlated to the str.ength of the leaving group.,..: :lectrophile with the better leaving group is the more::::tive electrophile, and thus reacts faster and undergoes the::re favorabie nucleophilic substitution reaction. A good.::r ing group is stable once it has left the electrophile, so as:able leaving group does not readily donate its electron pair.Thts implies that a good leaving group is a weak base. Theueaker the compound is as a base, the stronger its conjugateacid is. As acid strength increases, the pKu of the icicldecreases. This ultimately means that the lower the pKu ofthe conjugate acid o1'the leaving group, the more reactive theelectrophile.

Based on the pKs values, it is possible to predict therelative reactiviry of various electrophiles. The favorabilityof a nucleophilic substitution reaction can be approximatedby comparing the pK2 value of the conjugate icid of thenucleophile with the pKu value of the conjugate acid of theleaving group. Equation 1 can be ernployed to approximate areactivity constilnt, C, for the reaction:

C = 10[pKr(H-Nucleophile) - pKo(H-Leaving group)]

Equation 1

A reaction can be classified anywhere front very fbvorableto very unfavorable. The C value can be used as follows toapproximate the tavorability of a reaction:

If C > 103, then the reaction is very favor.ableIf 103 > C > l, then the reaction is slightly favorableIf I > C > l0-3, then the r-eaction is slightly unfavorableIf 10-3 > C, then the reaction is very unfavorableTable 1 lists pKn values 1br the conjugate acids of sorne

common leaving groups.

Acid pKa Acid PKaHI -r0.5 HCN 9.1

HBr -8..5 C6H5OH 10.0

HCI -1.0 H3CCH2SH 10.5

HF 3.3 Hzo 15.1

Table IThe pK6 data given in Table I can be userJ to pr.edict the

favorability of a nucleophilic substitution reaction. Theaccuracy is within experimental error fbr substitution stuclies.Equation I does not hold well in protic solvents due tovariations in nucleophile strength from hydrogen bonding.

57. Which of rhe fbllowing compounds is the BESTelectrophile'l

A. (CH3)3CI

B. (CH3)jCBr

C. (CH3)3CCI

D. (CH3)3CF

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59.

58. Which of the following compounds is the MOSTreactive when treated with cyanide nucleophile?

A. CH3CH2F

B. CH3CH2OC6H5

C. CH3CH2SCH3

D. CH3CH2BT

Which of the following is the MOST srable leavin.group?

A. HCNB. CN.

C. H3CCH2S-

D. H3CCH2SH

The best explanation of why NaSCH3 is a bette:nucleophile than NaSCH(CH3)2 is which of ri_:fbllowing?

A . Inductive effect of methyl is weaker than isopropr .

B. Resonance affects only three carbon fragmentsC . Hybridization of carbon varies with substitutionD . Steric hindrance is less with the methyl gr<_rup

The reaction of sodium cyanide with 2-iodobutane is:

A. very favorable.

B. slightly favorable.C. slightly unfavorable.D. very unfavorable.

The reaction of methylthiol (CH3SH) wirh R-2-buta::,is which of the following?

A . Very favorable.

B. Slightly favorable.C . Slightly unfavorable.D. Very unfavorable.

How can it be explained that fluoride, F-, is a bei:.rnucleophile than iodide, I-, in eiher solvent but a uc:l;lnucleophile in alcohol solvent?

A . In a protic solvent such as alcohol, F- is hinde::,rrby hydrogen bonding, and cannot migrate as we..

B. In an aprotic solvent such as ether, F- is hinde,:rby hydrogen bonding, and cannot migrate as wel.

C . In a protic solvent such as alcohol, F- exhibits : I

hydrogen bonding, so it cannot migrate as we1l.

D. In an aprotic solvent such as ether, F- exhibirs - Ihydrogen bonding, so it cannot migrate as we11.

60.

61.

62.

li:

Lll a

t\,,

.

-

a_

63.

r iir"i:

LLtt]tri :

r.4


The strength of a leaving group can be deternined by:ramining the reaction rate of either the S51 or Sp2 reaction.

The rates fbr both reaction mechanisms show a linearjependence on the concentration of the electrophile. It can be

.nferred from kinetic data that an electrophile with a better

.eaving group undergoes nucleophilic substitution at a faster

.ate than an electrophile with a worse leaving group. A:esearcher designed a study that varies the leaving group on an

:lectrophile while keeping all other factors constant. Factors.hat influence the rate include temperature, nucleophile,trength, solvent, and concentration. In a valid study, all of.hese factors should remain constant between trials.

A difficulty that can arise with this experiment is:ompetition between substitution reactions and elimination:eactions. To alleviate the problem of the competing:limination reaction, a one-carbon electrophile is chosen.i\-hen the electrophile has only one carbon, the substitution:eaction must proceed via an S52 mechanism. Reaction I.hown below is a generic reaction representing each of the

:irteen triai runs.

Nuc + CH3-X -+ Nuc-CH3 + X

Reaction 1

There were fbur nucleophiles and fbur leaving groups

-sed in the experiment to account for sixteen combinations.lable I is a matrix showing the four nucleophiles ancl lbur:aving groups used in the generic reaction. The vaiue listedr each box in the table is the log of the reaction rate.

\x -+)JucJ \ OSO3CH3 I CI OCHr

H:N -3.82 -3.31 -4.11 No Rx

CN. 1.39 1.tl L92 No Rx

H3CS- -2.16 1.92 )Aa No Rx

H3COH -5.21 -4.80 -5.10 No Rx

Table 1

The pKu values fbr the conjugate acids of the leaving::oups can be used to estimate reactivity. The relative pKus

':e pKulg.6gCHr) t PKa(HCl) > PKa(HOSOTCH:) > PKa(HI)'

The rate for the reaction is measured in rnolar per second,

:erefore the less negative the log value of the initial reaction:.rte, the faster the reaction. Because the rate of an S52-:action depends on both the nucleophile and the electrophile,.ris experiment can be used to determine the strength of both- -rcleophiles and leaving groups.

: J . The relative strength of the nucleophiles is bestdescribed by which of the fbllowing relationships?

A. CN- > CH3S- > NHr > HOCH3

B. CN- > CH3S- > HOCH3 > NH:

C. CH3S- > CN- > NH3 > HOCH3

D. CH3S- > CN- > HOCH3 > NH3

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66.

6 5 . What difficulty arises if Reaction 1 is carried out usinga secondary propyl electrophile instead of the methylelectrophiie?

A. The electrophile exhibits more steric hindrance.

B. The electrophile exhibits less steric hindrance.

C . There is the chance of an elimination side reaction.

D . The chance for an elimination side reaction isreduced.

Generally, nucleophiiicity and basicity run parallel toone another. What can be said about the correlationbetween the leaving grouprs basicity and the strength ofthe leaving group?

A . The less basic the leaving group, the better it is as

a leaving group.

B . The less basic the leaving group, the worse it is as

a leaving group.

C . The more basic the ieaving group, the better it is as

a leaving group.

D. There is no observable correlation between basicityand leaving group strength.

Which of the following does NOT directly aff'ect thestrength of the nucleophile?

A . The nature of the solvent.

B . The basicity of the nucleophile.

C . The steric bulkiness of the nucleophile.

D. The quality of the leaving group on theelectrophile.

Which of the following could NOT be determined froma similar experiment carried out with an Sy1 reactioninstead of the Sy2 reaction'?

A . The strength of the leaving group.

B . The strength of the nucleophile.

C . The ellect of temperature.

D . The effect of varying solvent.

The best explanation tor the lack of any observedreaction when NH3 was added to H3COCH3 is that:

A. NH3 is a poor nucleophile.

B. NH3 is a poor leaving group.

C . -OCH3 is a poor nucleophile.

D . OCH3 is a poor leaving group.

A nucleophile can also be classifled as which of the

following?

A. ALewisacid.B. A Lewis base.

C . An oxidizing agent.

D. A reducing agent.

o/.

68.

69.

233

70.



It is theorized that under identical conditions, whilerarying only the temperature of a solution, it is possible toconvert a reaction that yields purely substitution product (asubstituted alkane) into a reaction that yields purelyelimination product (an alkene). Studies have shown thatfavorable conditions for an elimination reaction involvehigher temperatures (elimination is endothermic and oftenhas a greater activation energy than the competingsubstitution reaction). Elimination is carried out in thepresence of eithel a strong bulky base (82) or a strong acid(Et) in solution. This implies that weak bases at lowtemperature react as nucleophiles rather than as bases.Reaction i, drawn below, was designed to verify this theory.

BrD

+ Nuc ---> ProductH$\l

H:C CH:

Reaction IThe idea was ro monitor the reaction by the optical

rotation of plane polarized light. Only the SyZ reactionshows retention of some optical activity although the exactvalue of the specific rotation (tolp) is not predictable. TableI shows the final specific rotations for each of the six trialsof Reaction 1, where either the nucleophile or temperaturewere varied. The initial specific rotation for the deuteratedalkyl bromide is +24'.

Table 1

The two cornpeting reactions when a good nucleophileis present in solution, in the absence of an acid, are thi E2and 5512 reactions. The internal alkene is the predominantproduct when elimination is observed. It is fbund thatdeuterium is less acidic than a proton due to the cleuteriumisotope effect rooted in the shorter bond length associatedwith the deuterium-carbon bond. An E1 r.eaction cancompete if the leaving group is a good leaving group and itis situated on a tel'tiary carbon.

71. Based on the data listed in Tabie l. Trial I ispredominantly what type ol reaction?

A. SplB. 5512

c. Er

D. Ez

Trial Temperature Nucleophile [cr]p productsTrial I l0'c NH: -JO

Trial II 60"c NH: -28"

Trial III l0'c NaOH -38"

Trial IV 60'c NaOH 0"

Trial V r0"c NaNH2 0'Trial VI 60'c NaNH2 0'

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7 2. When a secondarybase at 60'C, what

A. Sp1

B. S1q2

c. Er

D. Ez

alkyl bromide is treated with stronltype of reaction occurs?

7 3. The chirality of the reactant can BEST be described a.

which of the following?

A. 2R, 3R

B. 2R, 35

C. 25, 3R

D. 25, 35

7 4. In Trial II, the -28" optical rotation for the prodr_:lmixture can BEST be explained by which of ::,:following explanations'/

A . The reaction goes purely by an 5612 mechanism.

B. The reaction goes purely by an E2 mechanism.

C . The reaction is an Sp 1 reaction with sor:r*competing elimination reaction side produ:::tpresent.

D. The reaction is an Sy2 reaction with so:rrcompeting elimination reaction side proc::1.present.

7 5. Which of the following statements is true regarding :uinterchanging of stereoisomers for the reactant?

A . When the enantiomer of the reactant is used. ::usame geometrical isomers are formed, so either:ncreactant or its enantiomer may be used.

B. When the enantiomer of the reactant is us:'Ldifferent geometrical isomers are formed, sc irenantiomer cannot be substituted for the reactan:

C . When a diastereomer of the reactant is used. ::r'c

same geometrical isomers are formed, so either :lrw

reactant or its diastereomer may be used.

D . When a diastereomer of the reactant is usacdifferent geometrical isomers are formed. s: ur

diastereomer cannot be substituted for the reacta:r-

An elimination product for thiswhat optical rotation?

A. +24'B. 0"

c. -24'D. -42'

reaction would have

To support the theory that an E,2 reaction mechanrsm rs

taking place, it would be best to use chiral centers onwhich carbons of a deuterated 2-bromobutane?

A. Carbon2onlyB. Carbon 3 onlyC. Both calbon 2 and carbon 3

D. Carbons l, 2 and 3

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78.


CH2CHj

[o]o = +48'


Not ali stereocenters are chemically reactive. When areaction is carried out on a molecule with an unreactivestereocenter present, there exists the possibility thatdiastereomers will be formed in unequal quantities, due to theasymmetry in the molecule. This influence is referred to as

stereochemical control. Reaction l, drawn below,demonstrates this principle.

CH2CHj

..t\OH1. BHr(etrO)

---->

2'fuozcHs oH (aq)

[cr]o = +48' [cr]l = +18"

Reaction 1

The two products, Compound A and Compound B, are

nonsuperimposable and are not mirror images. Inhydroboration, the hydroxyl group prefers to add to the less

hindered carbon of the n-bond, so the reaction is referred to as

anti-Markovnikov. The hydroborane prefers to add to the less

hindered fbce of the molecule, which means that the twoproducts are present in unequal amounts. Their percentages

can be fbund using Equation 1 below. The percenlages

determined using this equation can be referenced against thequantitative values obtained using GC analysis.

Oobs=xacxa+(l -xu)cx6

Equation 1

uo6, is the observed optical rotation for the mixture and

xu is the mole fraction of component a in the mixture.

The same phenomena can be observed any time a

nucleophile is attacking an ,p2-hybridized carbon of an

asymmetric molecule. This means that unequal amounts ofdiastereomers may be observed with Sy 1 reactions,electrophilic addition reactions, and carbonyl additionreactions. Reaction 2, shown below, is an Sp1 reaction

involving the forrnation of two diastereomers.

@

...'St':NH:___l-

Reaction 2

For Reaction 2, what is the observed specific rotationfbr the product mixture?

A . Greater than 62"

B . Between 46' and 62"

C . Between 30" and 46'D . Less than 30'

,'r/CH3

oNHr

.$\

[s]o = +62' lulo = +30"

7 9. What would be the specific rotation for Reaction 1 ifCompound A is 807o of the diastereomeric mixture?

A. 48'B. 42"

c. 33"

D. 18'

8 0. The two products in Reaction

A . enantiomers.

B. epimers.

C . diastereomers.

D. identical.

2 are best described as:

8 1. The products of Reaction 1 can be distinguished fromone another by all of the following methods EXCEPT:

A . specific rotation.B. melting point.C. retention time on a GC.

D. IR spectroscopy.

82. Which of the following structures best represents themost stable conformer of product A of Reaction 1?

A' ,GH.CH3

B'Hn ayM.o ""#

/>cHt HrcH2i

OHC.9Hr

8 3. What are the orientations of the three chiral centers inthe reactant in Reaction 2, starting with the chirai centeron which the iodine is attached and moving clockwisearound the cyclopentane?

A. S,R,SB. R,R,SC. S,S,SD. R,S,S

CH2CHj

cH2cH3

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84. How many stereogenic centers (chiral carbons)present in the alkene reactant in Reaction 1?

A. 0

B. 1

c.2D. 3

at: rPa

Tlll_t

L\.

Passage Xlll (Questions 85 - 91)

A chemist sets out to perform a multistep synthesis.lhe first step, Reaction 1, is a standard Diels Alder reaction.

oo

"..\. q"ri,.o

Compound 1 Compound 2o

Compound 3

Reaction I

Compound 3 represents a mixture of enantiomers. The

:irture undergoes a chirally specific laboratory technique to.oiate Compound 3a, shown below, from Compound 3b.

HsC

oCompound 3a

In the second step, Reaction 2, Compound 3a is treated,'. rth meta-chloro peroxybenzoic acid, mcpba, in ether to form!ompound 4.

o

o

HgC

Compound 4 o

Compound 4 then undergoes Reaction 3 to formCompound 5.

-.:e("HicNH2H3;iq"o

Compound 4

oo

oCompound 5

Reaction 3

In Reaction 4, Compound 5 is hydrolyzed using water at

90"C to form Compound 6.

H3CHN

HOOHOH

Hec

Compound 6


8 5 . All of the following reagents, when added to Compound3a, result in a product with more asymmetric carbonsthan compound 3a EXCEPT:

A. Br2(l)/CCla(l)B. Hz(e)/Pd(s)

C . cold KMnO4(aq) at pH = l0D. NH3(l)

8 6. Which of the following is Compound 3b?

oB.A. o

..rr(o

,,,,,(

o

..r(o

HgC

C.

HeC

D.

8 7 . What laboratory technique would be MOST effective inobtaining a pure enantiomer from a racemic mixture?

A. Adding the mixture to a chromatography columnfilled with a gel with both enantiomers bound to it.

B. Adding the mixture to a chromatography columnfilled with a gel with just one of the enantiomers

bound to it.

Distilling the mixture using a vertical columnfilled with beads that contained both enantiomers

bound to their surface.

Using a chirally pure carrier gas in a gas

chromatography experiment.

8 8. What is the major product, most abundant stereoisomer,formed in Reaction 2?

HrC"

C.

D.

oB.A.

o


89. All of the steps in the overallpassage generate an opticallyEXCEPT:

A. Reaction IB. Reaction 2

C. Reaction 3

D. Reaction 4

synthesis shown in theactive product mixture

90. What is a likely side product of Reaction 3, if excessamine is used?

A.

H3CN NCFI3

HsC

H?CHN

HOHsC

H"CHN

HONHCFLNHCr-t

HsC

91. The final product mixture following Reaction 4 can bestbe described as:

A. an enantiomeric mixture with up = Q'.B. a diastereomeric mixture with ap = Q".

C. a diastereomeric mixture with up * 0".D. a meso mixture with up * 0".

B.

C.

D.

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Questions 92 throughdescriptive passage.

9 2 . The following molecule has which of thestereocherhical orientations ?

lollou i:g

93.

ocH3

-oH

A. 2R, 3R

B. 2R, 35

C. 25,3RD. 25, 35

The following molecule has which of thestereochemical orientations ?

A. 2R, 3R

B. 2R, 35

C. 25, 3R

D. 25, 35

The following pair of molecules can best be described xwhich of the following?

CHr

CH2CH3

cH2cH3

CH:

A . Diastereomers

B . Enantiomers

C. Epimers

D. Anomers

94.

t?aHHO

and

HOH\ts

CHr

ocH3

9 5. How many stereoisomers are possibie for the molecule1,2,3-trifl uoropentane?

4.2B. 4

c. 6

D. 8

,6. Addition of KMnO4(aq) at pH = l0 generates a vicinaldiol with syn stereoselectivity. What does the additionof KMnO4(aq) at pH = l0 to E-2-butene would yieldwhich of the following?

A. Two enantiomers

B. Two diastereomers

C . Two meso compounds (not identical)

D. One meso compound

9 7. Which of the following compounds is optically active?

A. 2R,3s-dibromobutane

B. 2R,4s-dibromopentane

C. 2R,4s-dibromohexane

D . cis-1,3-dichlorocyclohexane

9 8. Which of the following compounds CANNOT beoptically active?

A. 2-chlorocyclopentanol

B. 2-chlorocyclohexanol

C. 3-chlorocyclohexanol

D. 4-chlorocyclohexanol

9 9. Only twenty-five percent of synthesized isoleucine canbe used biologically. This is best explained by whichof the following explanations?

A . Only 25Vo exists as a zwitterion in the body.

B. 15Vo of synthetic isoleucine does not have thecorrect side chain.

C. In synthesizing isoleucine, the two chiral centersresuit in four stereoisomers being formed. Onlyone of the four is biologically coffect.

D. In synthesizing isoleucine, the two chiral centersresult in eight stereoisomers being formed. Onlyone of the eight are biologically correct.

Copyright @ by The Berkeley Review@ LIFE BEYOND CHEM STARTS NOW

100. Which of the following compoundsany optical rotation of planar light?

A. 2R, 3R dibromobutane

B. 2R, 35 dibromobutane

C . 2R, 3R dibromohexane

D. 2R, 35 dibromohexane

does NOT show

1.D 2.8 3.B 4.C 5.8 6.81.C 8.C 9.B l0.B 11.B 12.4

13. D 14. A 15. B 16. A 17. C 18. D19. B 20. A 21. B 22. C 23. B 24. A25. D 26. B 21. C 28. D 29. B 30. B31. A 32. D 33. A 34. B 35. A 36. A31. D 38. B 39. B 40. C 41. B 42. B43. A 44. A 45. A 46. A 41. A 48. A49. A 50. B 51. A 52. B 53. A 54. A55. C 56. B 57. A 58. D 59. D 60. D61. A 62. D 63. A 64. A 65. C 66. A61. D 68. B 69. D 70. B 1r. B 72. D73. B 14. D 75. A 76. B 77. C 78. B19. B 80. C 81. D 82. A 83. C 84. B85. D 86. A 87. B 88. B 89. A 90. C91. C 92. A 93. D 94. A 95. B 96. A91. C 98. D 99. C 100. B

Stereochemistry & Nucleophilic Subst'n Passage Answers

H:9

Choice D is correct. Using the Cahn-Ingold-Prelog rules for substituent priority and drawing the appropria:.arcs, the chiraiity for the two stereogenic carbons of isoleucine is determined as iollows:

H Priority #4 in back(an as isposition)

Countercioskwise Arc

CH2CFI3 "' ChiralitY isS

*HrN

Priority #4 in back(an as lsposition)

Countercloskwise Arc.'. Chirality isS

Coz-

Isoleucine

2.

Both of the chiral carbons have S chirality, which makes the compound 2S,3S. It wouid be swell of you:-choose D. The chiral center on carbon two of an amino acid must be S according to the rules discussed in :: .Passage, so choices A and B could have been eliminated eariy. Regardless, the chiral center on the side cha-:of the amino acid must be solved for. Determining R and S ii actuiily rather simple when you get the hang .:it. The key is to find a method that works for you and hone it in by repeated .rr". '

Choice B is correct. If the side chain (carbon three of isoleucine) were to change the orientation of its chr.-center while carbon 2 retained the orientation of its chiral center, then only ot" o] the two chiral centers lyoL-:differ between the two compounds. The two stereoisomers (isoleucine and the other compound) would ::classified as diastereomers. The correct choice is B.

Choice B is correct. This question can be solved from straight memorization. Naturally occurring amino ac:-.,are "L" as in life and natural. It is also stated in the passage that naturally occurring amino acids har'- :stereoconfiguration and that S stereochemistry is associated with L-amino acids. The besi answer is choice B

4. Choice C is correct. With H as the side chain, carbon two of glycine (the alpha carbon) has two hydrog.. -,attached, thus there is no chiral center present on glycine. Neither of the two carbons in glycine have::-:different substituents attached. The absence of a chiral center in glycine results in an opticai rotation o: ,

Choose C and be happy.

5. Choice B is correct. Enantiomers are nonsuperimposable mirror images, thus ail of the chiral centers ,:.different between the two structures. If all of the chiral centers are reversed, then the specific rotation shc*-:be completeiy reversed, which would lead to a value of +62" rather than -62". It would be terrific if you \{e:- ::choose B. Enantiomers always have the same absolute value for the specific rotation, only the sign (direc-* :of rotation) differs.

6. Choice B is correct. As stated in the passage, the side chain of threonine is an alcohol (eliminating cholct ..and it is chiral (eliminating choices A, C, and D), all choices except answer B are eliminated. Only chor:t S

contains a carbon that is asymmetric (chiral center).

7' Choice C is correct. Isoleucine contains two chiral centers, one for the alpha carbon and one in the side ch,*-Plugging into the stereoisomer equation 2n where n is the number of chiral centers, there are four pos'-- -rstereoisomers for the isoleucine structure. Because isoleucine contains two chiral centers that must have s1.;- -,-

orientation, only one of the four stereoisomers wili have the correct chirality to be biologically usable. Ti.,-" ,.stated in the passage in two fragments. The best choice is C.

8. Choice C is correct. PBr3 converts the OH group of an alcohol ir-rto a Br group through an S1.;2 reaction. Bec: --,othe reaction is by way of an S52 mechanism, the chiral centers inverts. If you recall, chiral centers inr e :: -Ln

Sp2 reactions, but not S1l11 reactions. This means the product shows stereochemistry of 35, 4R. Choose C fc: --:':

sensation of correctness and satisfaction.

Copyright @ by The Berkeley Review@ 240 STEREOCHEMISTRY EXPLANATIC,i\]$

9. Choice B is correct. If you rotate the original structure so that the hydroxyl groups are syn, then the two alkylgroups of the molecule also have syn orientation. The original alkene must therefore have the two alkylgroups cis to one another, resulting in Z geometrical alignment. Pick B for best results.

HO

cH2cH3

h B{r R"t"t " H + cH2cH3

OH cH2cH3 HHydxroxyl groups are syn

Using the rules of priorities, the following is determined:

Prioritv #4 in backClockwise = R

74

H Priority #4 in frontClockwise = S

This makes the compound 3R,45. Choose B for optimal correctness and the satisfaction that goes with it.

Choice B is correct. On the third carbon, the OH and H groups have interchanged, so that chiral center haschanged. On the fourth carbon, the ethyl group, hydrogen and hydroxyl group have interchanged, so thatchiral center has not changed. When only one out of two (some, not all) chiral centers change, it is not a mirrorimage, nor identical (superimposable), making the two compounds diastereomers. Pick answer B.

cH2cH3

?.,

Compound

HH H

P

HO CH2CH3

12. Choice A is correct. Using the rules of priorities,

Mystery Compound

the following is determined.

H in front, so reversechirality from R toS

H in back, so chiralityis S as shown

This rnakes the compound 15, 25. Choose A.

[3. Choice D is correct. The active site of an enzyme carries out a highly specific function (reaction), so they mustbe highly selective in terms of reactivity. As implied by the passage, active sites are highly specific in termsof chirality. This eliminates choice A. Although it is not stated in the passage, you should know that theactive site has specific dimensions, so it is size specific. This eliminates choice B. Active sites are highlyspecific for the functional groups involved in a chemical reaction, so choice C is eliminated. Because isotopesshow the same chemical reactivity, enzymes are unable to distinguish isotopes. This means that enzymeactive sites are nof isotope specific, making choice D the best answer.

So, utt "rlu*ust

have been cis

10. Choice B is correct.

11.

sr o

eilM

mt

3CH

m!

hile

:hr

&forhc

cH2cH3

HOH

lopyright @ by The Berkeley Review@ 241 STEREOCHEMISTRY EXPI,ANATIONS

't4. Choice A is correct. Enantiomers have the exact same physical properties, such as boiling point, melting point,and density. This makes choice A correct and eliminates choices B and C. Enantiomers have the samemagnitude for specific rotation, but with the opposite sign (in the opposite direction). This makes choice D anincorrect answer, and leave choice A as the best choice ini sea of many choices (weil maybe not many, but four.)

16.

Choice B is correct. A specific rotation for an enantiomeric mixture that is positive means that there is anexcess of the enantiomer with the positive rotation (in this case the R enantitmer). If the two enantiomerswere present in a fifty-fifty ratio, then the specific rotation would be 0". This eliminates answer choice C.Choice D is eliminated because an excess of th" S stereoisomer would result in an overall negative opticairotation, making choice D invalid. If the mixture were in fact 100% of the R enantiomer, thei the specificrotation would be +32.2" ' This is not the case either, so choice A is eliminated. By the process of elimination.choice B must be the correct answer. The problem could have been solved mathematically as follows:

%R (+32.2) + 7"5 (-32.2) = +L6.1.

%R (32.2) - %S (32.2) = +16.1

32.2 (%R - "/oS) = 16.1

%R - %S = 0.5 = 50/o; and %R + %S = 700'/"

Thus2(%R-%S)=%R+%S2%F.-2"k5=%R+%S%R = 3 %S, so o/,R = 75oh and %S = 25"/o

Choice A is correct. Because one enantiomer aiways has the opposite optical rotation of the other enantiome:(one is negative and one is positive with the same magnitude), li Uotn enantiomers are present in solution, theabsolute value of the rotation must decrease from the absolute value of the pure enantiomer. The observecoptical rotation is an average of the enantiomers in solution. Therefore, choice A cannot account for an obser'ecspecific rotation greater than that of the pure species, because the mixture of the two enantiomers would causethe value to be less than the literature value (given that the literature value is positive). Choices B and Cboth lead to an increase in the observed rotation. The standard rotation is based orrmonochromatic light from;sodium iamp. If monochromatic light from another source is used, it will interact differentlyiwith thecompound, resulting in a different optical rotation. Whether the rotation is greater or lower is uniertain, bu:because it is possible that the rotation is greater, choice D is eliminated.

17. Choice C is correct. The formula for units of unsaturation is Units of unsat =2(#C)+1(#N)+2-1(#H)

For a2

compound with a formula of C11H21NO2, the calculation is: Unitsof unsat - 2(77)+7(7)+2-7(27) - 4 -'

22Pick C. Remember that oxygen is not included in the formula for degrees of unsaturation. Iflrou didn't rememte:the formula, then count the bonding electrons for the compound (44 for the eleven carbons, 21 for the hydrogerr.3 for the nitrogen, and 4 for the two oxygens). There are a total of 72 bonding electrons, so the compound has lrbonds total. There are 35 atoms in the molecule, therefore only 34 bonds (minimum) are necessary to connect r;,eatoms and form the molecule. There are 36 bonds when only 34 are needed, thus there are two extra bonds. Th-long-winded path still leads to answer choice C.

18. Choice D is correct. The only way to do this problem is to count isomers systematically. Start with the longe=:chain (6) and list all the possible isomers. Then look at chain lengths of one less carbon (5) and list all of tho-possible isomers. This continues until the possible backbones are depleted. For this question, the carbi:backbone is shown for all of the possible isomers. There are five isomers total, so pick D.

rongest chaln rnnnact ^h.i- _ < ^-, longest chain = 4 carbons

= 6 carbons longest chain = 5 carbons A

-

\lc-c-c-c-c-c c-c-c-c-c c-c-c-c_c c_c_c_c c_c_c_cllrttccccc

E

15.

Copyright @ by The Berkeley Review@ 242 STEREOCHEMISTRY EXPLANATIO\5

Li\,,F,,,Y

19. Choice B is correct. It is stated that the OH substituent is on carbon 2. The OH is attached to an R chiral centerwhile the Br is attached to an S chiral center. Pick B. The rationale is shown below:

IBr

If the Br and H were switched, thecompound would be R. Because theyare not switched, it must be an S.

Looks like an S center fromthe arc, but because H is infront, it reverses to R.

20. Choice A is correct. A geminal diol has two hydroxyl groups on the same carbon (think of 'geminal' as beingequivalent to gemini, meaning that the two OH groups are twins.) Because the two hydroxyl groups are on thesame carbon, that carbon cannot be asymmetric. As a result, the molecule is achiral, so it cannot form opticalisomers. This makes choice A the best answer. A vicinal diol has two hydroxyl groups on neighboring carbons(think of 'vicinal' as meaning vicinity or vicino, the Spanish word for neighbor). Because this ensures that thesecond carbon has a hydroxyl group, the second carbon must be asymmetric. As a result, the molecule is chiral,so it can form optical isomers. This eliminates choice B. Four carbon chains with a secondary functional group,whether it is an amine or alcohol, have a carbon with four different substituents (H, CH3, CH2CH3, and thegroup), so they are chiral. Being chiral, four carbon chains with a secondary functional group can form opticalisomers. This eliminates choices C and D. Choice A is the best answer.

Choice B is correct. In polycyclic systems, if there is no plane of symmetr1z, then all tertiary and quaternarycarbons are stereogenic centers. In addition, any secondary carbons with a functional group are also stereogeniccenters. The compound has two quaternary carbons, three tertiary carbons, and one secondary hydroxyl group.This means that there are six chiral centers. Given that there are two possible orientations at each chiralcenter and the compound is not meso, the total possible number of unique stereoisome rs ts 26, which is 64, Pick B,and feel the warmth of correctness.

Choice C is correct. As indicated in Table 1 in the passage, diastereomers show different physical properties.The melting points of diastereomers are different, because the molecules pack into their respective latticestructures differently. The density is different between isomers, because the two diastereomers have differentconformations that also pack into their respective lattice structures differently. The optical rotation ofdiastereorners must be different given the fact that they are identified by their differences in optical rotation.This eliminates choices A, B, and D. Because diastereomers are isomers, and they have the same molecularformula, thus they have the same molecular mass. The correct answer is therefore choice C.

Choice B is correct. When using a polarimeter, an observed optical rotation of +233.0" and -1.27.0'would yieidthe same reading (given that a full circle is 360"). To discern one optical rotation from the other, the sampleshould be diluted to reduce the rotation. If the actual optical rotation is in fact +233.0', then the lowerconcentration would show a rotation less than +233.0' (less clockwise). If the actual optical rotation is in fact-727.0", then the lower concentration wouid show a rotation less than -727.0" (less counterclockwise). If thesolution concentration were cut in half for instance, the rotation would be either +116.5" or -63.5". The change inrotation can therefore determine the original rotation value. The only answer that indicates changing theconcentration is choice B.

Choice A is correct. If D-glucose has an optical rotation of +52.6' , then the enantiomer of D-glucose (L-glucose)must have an optical rotation of -52.6'. Mannose, the C-2 epimer of glucose (the diastereomer of glucose thatonly differs at carbon two) is neither of these two structures (L or D glucose), thus it does not show an opticalrotation of either + 52.6' or -52.6". Mannose is chiral and not meso, so it cannot have an optical rotation of 0".The best answer is choice A.

25" Choice D is correct. The number of stereoisomers (assuming that there is no meso structure), can be determinedby raising 2 to the power of the number of chiral carbons (stereocenters). There are three chiral carbons(stereocenters) associated wiih penicillin V, thus there will be eight (23) stereoisomers for the structure ofpenicillin V. The 2n formula represents the maximum number of stereoisomers. For every meso structure, youmust subtract one frorn the total. Choice D is correct.

21.

22.

23.

24.

Copyright O by The Berkeley Review@ 243 STEREOCHEMISTRY EXPLANATIONS

26. Choice B is correct. The greater the density of a compound, the more tightly packed the compound is in itscrystal lattice. This means that this question is a read-the-chart-to-fitr"a-tnu-d"r-rsity question. The mesccompound is less dense than the other compounds according to the data in the table, thus it must pack leasttightly of all of the choices. The best answer is therefore choice B.

Choice C is correct. Stereocenters can be identified quickly as sp3 carbons with four different substituentsattached. In camphor, there are two carbons that fit this deicriptibn. The correct choice is answer C. Dral'nbelow is camphor with the two chiral carbons (stereocenters) labeled:

27.

Top Side

28.

CH:

Choice D is correct. If the hydrogen bonds in a pure compound were stronger than the hydrogen bonds in amixture of stereoisomers, then the intermolecular forces would be greate"r in the p,rr"- .orripound. As aconsequence, the pure compound would have the higher melting point, which is the exact opposite of thepremise. Choice A is an untrue statement, and thus it is eliminated. Physical properties, such as'melting pointresult from intermolecuiar forces, not covalent bonds. Enantiomers are no -ore lit ely to form covaleni bondswith one another than diastereomers. Choice B is an untrue statement that does not explain the observeimelting points. Choice B is eliminated. Covalent bonds are not affected when a .o*porr1d melts, so choice Cshould be eliminated. The best explanation is choice D, because when the moleculei pack more tightly, therexert stronger forces on each other. Because the enantiomeric mixture (R with S) has a higher -"itlng po*,than the pure enantiomer (R with itself), the forces are in fact stronger in the enantiomeric mixture than thepure compound. This makes choice D the best answer.

Choice B is correct. To separate enantiomers from one another, the medium must be chirally pure (be chiralwith only one enantiomer present). The best method is the use of chiral gei in colum. .hro^uiography. Thetwo enantiomers exhibit different migration rates down the column, because the two adhere to thJcoiumn to adifferent extent. Choice A is a valid methocl. Distiilation r,vill nof separate the two enantiomers, so choice Bis the correct answer choice. The mixture can be selectively crystallized with a pure R or pure S compound. ThLis often carried out with tartaric acid. Enzymes are chiral, so chiral .o-porrr.i, pass thiough an enzyme filterat different rates. This makes choice C and D vaiid.

Choice B is correct. To prevent the product mixture from being racemic, chirality must be present in thetransition state. A change in temperature does not affect the chirality of the products. Choice Cis eliminated.The concentration does_not affect ihe alignment of the molecules in the transition state, only the frequency witi.which the reactants collide to form the transition state. This eliminates choice D. The presence of a chira-center in the solvent cloes not affect the chirality of the transition state unless the soivent is involved in thetransition state. Choice A cannot be eliminated yet, but it is not a likely choice. The only change that vr'i--definitely affect the distribution of enantiomers is the addltion of a chiral catalyst which affects thetransition state. This is the whole idea behind the activity and specificity of enzymes (chiral catalysts) i-r.biological reactions. The best answer is therefore choice B.

Choice A is correct' Both enantiomers have oniy one stereocenter, therefore choices C and D are eliminatecThe lemon fiavored extract is the structure on the left, which has its lone stereogenic center in R configgrationThis eliminates choice B and makes choice A the correct answer.

]!

29.

30.

31"

Copyright @ by The Berkeley Review@ .)At STEREOCHEMISTRY EXPLANATIONS

Choice D is correct. The one physical property that definitely changes with the chiral center is the opticalrotation. The specific rotation of a product mixture measures its enantiomeric purity (percentage of eachenantiomer). The boiling point, density, and solubility in a given solvent does not vary between enantiomers.These physical properties can vary between diastereomers, but enantiomers are identical in their packing andintermolecular forces, unless the alcohol solvent is chiral and optically pure. The best answer is choice D.

Choice A is correct. The regio-chemistry is correct in all of the answer choices (the boron has attached to theless hindered carbon of both alkenes). Hydroboration, you may recall, proceeds with anti-Markovnikovregioselectivity. The reaction calls for the R enantiomer, which has the alkenyl group sticking out of theplane (the structure on the left in Figure 2, labeled "lemon odor", has R stereochemistry). This eliminateschoices C and D. Because the boron and the hydrogen add syn to one another, the methyl substituent on the ringmust be trans to the bridging boron. This eliminates choice B and makes choice A the best answer.

First addition

->

of BH3(Et2O)Second additionof BH"(EI2O)

H-B

R stereocenter Trans addition to methyl

Choice B is correct. The two enantiomers have a different flavor (and thus different smell), so they must bindthe olfactory receptors differently. This eliminates choices C and D. Because they recognize the differencebetween the two enantiomers, they too must be asymmetric (and thus chiral). The correct answer is choice B.

Choice A is correct. The isoprene molecule has no chiral centers, so the product is a racemic mixture. Themolecule listed in the question as an alternative reactant has a chiral center present that will influence theorientation in the transition state. The product would contain three chiral centers (one new chiral centerformed and one each present in the two reactant molecules). The product would be present in a mixture ofdiastereomers. Diastereomers cannot be present in a racemic mixture, therefore the product mixture would notbe racemic for the new reaction. The best answer is choice A.

Choice A is correct. The proposed product is a mixture of diastereomers formed from a nucleophilic substitutionreaction. Because two diastereomers are formed, the proposed reaction must have been predicted to proceed byan Sp1 mechanism. The best answer is choice A.

Choice D is correct. Any nucleophilic substitution reaction that proceeds with either inversion of a chiralcenter or retention of a chiral center remains chiral, and thus is optically active. This eliminates choices A, B,

and C. The product from elimination is an alkene, which has no chirality, and thus no optical activity. Thecorrect answer is elimination, choice D. The rearrangement clause, although true, has little bearing.

Choice B is correct. There are two chiral carbons present on each compound. The tertiary carbon retains itschirality between stereoisomers, but the methoxy carbon has different chirality in the two stereoisomers. Thismeans that one out of two of the chiral centers differs, making choice B the best answer.

39. Choice B is correct. To increase the amount of substitution product that forms, the amount of eliminationproduct that forms must be reduced. The elimination reaction is by way of an E1 mechanism, because no strongbase is present. To reduce the amount of E1 product, the amount of acid should be reduced, and the reactionshould be carried out at a lower temperature. This means that the temperature should decrease (elimination isfavored at higher temperatures) and the pH should increase. The best answer is choice B.

40. Choice C is correct. There is a secondary electrophile, protic solvent, and a poor nucleophile present, so thesubstitution reaction takes place by an SX11 mechanism. The presence of the methyl and methoxy groups on thesame carbon can be explained by rearrangement. There is a secondary carbocation formed when the leavinggroup leaves. When the hydride shifts, a tertiary carbocation forms. This is a more favorable intermediate, so

the reaction proceeds via a hydride shift before the nucleophile attacks. The best answer is choice C.

i8.

nrC4,

Choice A

Copyright @ by The Berkeley Review@ 245 STEREOCHEMISTRY EXPLANATIONS

41, Choice B is correct. Given that the conversion of the hydroxyl group into a bromine goes with inversion o;stereochemistry, the alcohol must have opposite chirality at carbon 1 as the bromoalkane in Reaction 1. Thepassage states that the reactant is (1S,2R)-2-methylbromocyclopentane, so the alcohol precursor must har.echirality of 1R and 2R. The best answer is B. The reaction and chirality is shown below.

H in back on C-2,so take as is: R .)'rrOU .,rrr\OH PBre______*

H in front on C-1,so reverse it: R

42. Choice B is correct. The preference for the major product can be attributed to steric hindrance in the transitionstate (the transition state is asymmetric). The carbocation is sp2-hybridized,, so it is planar. The only steri:hindrance comes from the adjacent methyl group (which is above the ring). The adjacent methyl, by beir.eabove the ring, influences the nucleophile to attack from below the ring. The best answer is choice B.

43. Choice A is correct. The first reaction in Table 1 proceeds by way of an 5512 mechanism, because theelectrophile is a primarv alkyl halide. The second reaction in Table 1 proceeds by way of an 51'11 mechanisrr.because the electrophile is a tertiary alkyl halide. Only the rate of an 51112 reaction depends on thenucleophile, therefore to determine the best nucleophiie, the data from the reaction with the primary aik-, -halide (first set of data in Table 1) should be used. The best nucleophile is the compound that has the fastes:reaction rate, which according to Table 1, is methylamine (H3CNH2). Choose A if you're a table believer.

44' Choice A is correct. From the data in Table 1, the faster reactions are observed with 1-chioropropane an othe:electrophiles, so it is a safe and valid assumption that the reaction with the primary aikyl halide proceed.most rapidly. Choose A.

45. Choice A is correct. The reaction can proceed by either an S511 or Sl112 mechanism with a secondary alk,,'-halide as the eiectrophile. If the reaction were to proceed by an 51112 mechanism, then the product wouid bethe 25, 35 stereoisomer. Only the stereocenter from which the bromine substituent left underwent a change r,..

chirality (inversion), thus only that stereocenter will show a change in its orientation. If the reaction were tcproceed by an S1r11 mechanism, then the intermediate would undergo rearrangement, and thus the ammoni"wouid attack the third carbon leaving an achiral product. Given the answer choices, choices B and C car.:form, and there is no achirai choice, therefore the reaction proceeded by an SNI2 mechanism. This means tha:inversion of the second carbon will transpire to yield 25,35. Choose A.

46. Choice A is correct. To have optical activity and lose it during the course of the reaction eliminates choice B

because the 51'12 reaction proceeds with inversion (thus an optically active product is formed). By definitiorthe product mixture as described in the question is racemic. A racemic product mixture is associated with th.51111 reaction. Pick A.

47, Choice A is correct. The ether product has S stereochemistry as drawn and was formed by a substitutic:.reaction using NaOCH3 as the nucleophile. Sodium methoxide (NaOCH3) is a strong base, and thus it is also a

good nucleophile. Because the nucieophile is good, the reaction must have proceeded by an 51112 mechanisn'.Because the final product has S stereochemistry, the starting material (electrophile) must have had F.

stereochemistry to form the S product from inversion. This eliminates choices B and D. For the substitutior.reaction to proceed, the electrophile must have had a good leaving group. Ammonia is not a leaving grouptherefore the chlorine leaving group is the better choice. This makes the correct answer choice A.

48. Choice A is correct. The reaction proceeds at the fastest rate during the first segment of the reaction becaus:initially the concentration of L-chloropropane (a reactant in the rate determining step) is greatest and bot:rreactants (nucleophile and electrophile) are depleted over time. Chloride anion is the leaving group, thus rlsconcentration will increase over time. A11 of the graphs show increasing concentration. Over time, theconcentration of 1-chloropropane gradually decreases, thus the reaction rate decreases graduallyi this resul-in a slower production of Cl- anion. Graph A best depicts this gradual decrease in reaction rate. Choose A ancbe a wunder sfudent.

CFL#d Br

Iilll'


49. Choice A is correct. Because the electrophile is a tertiary alkyt halide, the reaction is an Sry1 reaction, whichproceeds with racemization. The optical rotation of the product mixture following an 5511 reaction is 0". Thiseliminates choice B. Graph C (the schizophrenic graph) shows the correct final optical rotation, but noreaction will proceed with the erratic change in rotation. Choice D shows that the reaction proceeds at aconstant rate until the reaction is complete. This would be seen with a zero order reaction, not a first orderreaction' The SIrI1 reaction is first otder, so answer choice D is eliminated. Graph A best depicts the gradualloss of chirality. The optical rotation will never switch to the opposite sign unless there is inversion *hi"h itnot possible with an sI.I1, reaction. Choose A and make your support group proud.

Choice B is correct. The ether product shown in both Reaction 1, and 2 is the result of a substitution reaction, notan elimination reaction, so choices C and D are eliminated. The data in Table 1 correlates to Reaction 1.Because the rate of the reaction varies directly with both the concentration of the nucleophile and theconcentration of the electrophile, Reaction 1 must be an 5512 reaction. The data in Table 2 correlates to Reaction2' Because the rate of the reaction varies directly with the concentration of the electrophile, but does not varywith the concentration of the nucleophile, the reaction must be an 51111 reaction. The rate of an 51111 reactiononly depends on the concentration of the electrophile, and does not vary with the concentration of thenucleophile. This means that you must choose B to live up to your potential.

Choice A is correct. Because the rate of Reaction 2 varies directly with the concentration of the electrophile,but does not vary with the concentration of the nucleophile, Reaction 2 must be an St'tr1 reaction. an S51treaction undergoes racemization, not inversion, so choice A cannot apply to Reaction 2. In addition, Reaction 2has no chirality, so choice A is invalid. The best answer is choice A. A carbocation intermediate correspondswith an Sp1 reaction, so choice B is valid and thus eliminated. All nucleophilic substitution .eactions,whether it is an 5111 or S1rJ2 rnechanism, have a rate that depends on the electrophile. Choice C is valid, andthus eliminated. A protic solvent helps to stabilize the carbocation intermediate and the leaving group, so aprotic solvent increases the rate of an 5511 reaction. This makes choice D valid, and thus eiiminateJ it. ChoiceA is in fact the top dog of choices.

Choice B is correct. If the reaction were to proceed purely by an 5512 mechanism, the product would be 100% R,because the reactant is enantiomerically pure and the SNI2 reaction results in complete inversion. If the reactionwere to proceed purely by an S1r11 mechanism, the product mixture would be 50% R and 5Ao/" S, because thereaction goes through a planar carbocation intermediate resulting in a racemic mixture. The mixture is 87% Rand 13% S, which is closer to the products of an 51'12 mechanism than the products of an 51111 mechanism. It isnot a pure reaction so the best answer is choice B.

Choice A is correct. Because the carbon-bromine bond is weaker than the carbon-chlorine bond, it is more easilybroken. This makes the bromine a better leaving group than chlorine. An alkyl bromide is therefore a morereactive electrophile than an alkyl chloride. With a better electrophile, the reaction is faster for both theSp1 and the Sir12 mechanisms, because they both depend on the electrophile. This makes choice A correct.

Choice A is correct. The strongest nucleophile is most willing to donate its lone pair to carbon. The answerchoices include two conjugate pairs. The conjugate base is the better nucleophile of the pair, so choices B and Dare eliminated. HCI is a strong acid while methanol is a weak acid, so methoxide is a stronger base thanchloride. This means that methoxide is more willing to donate electrons than chloride, and thereforemethoxide is the better nucleophile. Pick A and see your score improve.

Choice C is correct. An SNJ2 reaction favors a primary electrophile over second aty ot tertiary electrophile, anda good nucleophile is required. Choices A and B can both be eliminated, because the electrophiles are tertiary(and tertiary electrophiles proceed via the Sry1 mechanism). Choice C is better than choice D, becausemethoxide is a better nucleophile than methanol. Pick C for optimal results.

56, Choice B is correct. An 5512 reaction proceeds by way of a one-step mechanism, which eliminates choices A andC. An exothermic reaction has the products at a lower energy level than the reactants, which eliminateschoice D and leaves choice B as the correct answer. The apex of the graph represents the transition state, andthe absence of a valley on the graph implies that there is no intermediate for the reaction. Choice A is anexothermic 5111 reaction, choice C is an endothermic Sp1 reaction, and choice D is an endothermic Sp2 reaction.

50.

52.

53.

i4.

f5.


57.

58.

59.

Choice A is correct. The best electrophile is the compound with the best leaving group. The best leaving groupis the leaving group with the strongest conjugate acid. In this case, iodide is the best leaving group, beciuse HI(hydroiodic acid) is the strongest conjugate acid of the choices listed. This makes choice A the best answer.

Choice D is correct. This question requires that you identify the best electrophile. Again, the best electrophileis the compound with the best leaving group. The best leaving group has the strongest conjugate acid, which rnthis question is the bromide leaving group. The ranking of the conjugate acids for the leaving groups are: HBr >HF > HOC6H5 > HSCH3. It is in your best interest to choose D.

Choice D is correct. A leaving group, once it has left an electrophile, must have at least one lone pair (as a

result of the heterolytic bond cleaving). This stipulation eliminates choice A, because the carbon of the cyanicacid has no lone pair, and the nitrogen does not interact with a carbon to be a leaving group. The most stableleaving #oup is the weakest base. Of the three choices left, CH3CH2SH has the strongest conjugate acic(CH3CH2SH2+ is a stronger acid than CH3CH2SH and HCN), thus CH3CH2SH is a weaker base thanCH3CH2S- and CN-. CH3CH2SH is the weakest base of the choices remaining, therefore it is the best leavinggroup. Choose D to score a point in this contest of point collecting.

Choice D is correct. The difference between the two molecules is the alkyl group. According to the question, thesmaller molecule is the better nucleophile. The inductive effect would predict that the electron donatingmethyl groups would make the larger alkyl group more electron donating and thus more nucleophilic. This isthe opposite of what is observed, so choice A is eliminated. Resonance is not a factor, because there is no r-system. Choice B is eliminated. The hybridization is sp3 in both cases, so the difference in nucleophilicit',-cannot be attributed to hybridization. This eliminates choice C. Steric hindrance predicts that the smallernucleophile has less interference in the transition state, thus iL is a better nucleophile. In this case, sterichindrance plays a larger role than the inductive effect in the reactivity of the nucleophile. The correct ansn'eris choice D.

Choice A is correct. Sodium cyanide is a good nucleophile and iodide is a great leaving group, therefore thi.reaction shouid be very favorable. The difference between their pKu values ts 79.6, which implies that theKsq for this reaction is near 1919'6 = 4v1919. This defines a reaction that goes completely to product, whic-i.supports the evaluation that the reaction is very favorable. Do the correct thing, choose A.

Choice D is correct. Hydroxyl groups are terrible leaving groups. Thiols are average to poor nucleophiles. Thereaction between a poor nucleophile and an electrophile with a poor leaving group should be very unfavorab:eby intuitio.n. The difference between their pKu values is -5.2, which implies that the K"O for this reaction'.near 10-5'2 = 6 x 10-6. This defines a reaction that stays predominanlly as reactant, rirhich supports theevaluation that the reaction is very unfavorable. Choose D,

Choice A is correct. In protic solvents, there is hydrogen bonding, so choice C is eliminated. Equally, in a:,aprotic solvent, there is no hydrogen bonding, so choice B is eliminated. Hydrogen bonding affects fluoride an:not iodide, so the fact that fluoride is a worse nucleophile than iodide in alcohol implies that hydroger,bonding reduces the nucieophilicity of fluoride. This can be attributed to hindrance to migration caused t'ihydrogen bonding. The best answer is choice A. Choice D can be eliminated, because if it were true, then theopposite relative nucleophilicity would be observed for fluoride and iodide.

64. Choice A is correct. The strength of a nucleophile can be measured by its reaction rate in a second-orde:nucleophilic substitution reaction (an SX12 reaction). The nucleophiles are listed in the first column of Table 1

Any other column can be used to determine the relative strength of the nucleophiles, because all other factors ithe reaction are constant. The less negative the value in ihe table, the faster the reaction, and therefore th.rbetter the nucleophile. The CN- has the lowest value in all three columns, so the best nucleophile is CN-. Thrieliminates choices C and D. The question now is to determine whether the ammonia (NH3) or methanc-(HOCH3) is the better nucleophile. The stronger nucleophile is ammonia, because in each column, the 1e..negative value is associated with ammonia. The best answer is choice A, CN- > CH3S- > NHg > HOCH3.

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Copyright @ by The Berkeley Review@ STEREOCHEMISTRY EXPLANATIO\,(

55.

66.

67-

59.

70.

68.

Choice C is correct. As mentioned in the passage, the methyl electrophile is chosen to avoid the complicationof the competing elimination reaction. It is not possible to form a dbuble bond with only one carbon in thereactant (at least two carbons are required for the formation of a double bond). This makes choice C the bestanswer. Steric hindrance increases when using the isopropyl electrophile in lieu of the methyl electrophile,but that is not necessarily a difficulty. The effect should be uniform u"ross the reaction chart, so choices A andB can be eliminated.

Choice A is correct. The best leaving group is the functional group that takes electrons from carbon and retainsthem to the greatest extent. Retaining electrons can also be viewed as not sharing electrons. By not sharingelectrons, an ion or molecule can be viewed as being a weak base. The strength of i leaving group is generallycorrelated (in a linear fashion) to the strength of the conjugate acid of the leaving group. aJ un icid becomesstronger, the conjugate base becomes weaker. This implies that it is valid to compare the strength of theieaving grouP in a linear fashion with weakening base strength. This makes choice A, "The less basic theleaving group, the better it is as a leaving group,' the best answer. Choices B and C are essentially the sameanswer, therefore they should both be eliminated.

Choice D is correct. The strength of a nucleophile can vary with many reaction features. Depending on itsnature, a solvent hinders a nucleophile to a varying degree. For instance, if a solvent is capable of fbrminghydrogen bonds, then it will hinder the attack of nucleophiles that are capable of forming hydrogen bonds-.This can be seen in the differing nucleophilic strength of halides as they are observed in apiotic and proticsolvents. The fluoride is the strongest nucleophile of the halides in aprotic solvents whilelt is the weakestnucleophile of the halides in protic solvents. This eliminates choice A. The nucleophilicity of a compound canbe correlated to its basicity in terms of a Lewis base. Generally, for a nucleopiile that is more basic thananother, it is the better nucleophile of the two, with steric hindrance responsible for most deviations from thatpattern. This eliminates choice B. The strength of a nucleophile reduces with increasing bulk. This impliesthat nucleophilicity can vary with steric hindrance, which eliminates choice C. The only answer choice left ischoice D. The leaving group is independent of the nucleophile in nucleophilic substitution reactions. Thismakes choice D the best answer.

Choice B is correct. Because the rate of an Sp1 reaction depends only on the leaving group breaking free fromthe electrophile in the rate-determining step, the nucleophile is irrelevant .to the reaction rate f-or an Sy1reaction. The strength of a nucleophile cannot be determined by a rate study in which the nucleophile does notinfluence the rate. The rate can vary with changes in the leaving group strength (which can be viewed aschanges in the electrophile), the temperature (temperature always affects the rate of a reaction), and solvent.It is only the strength of the nucleophile that cannot be determined from the reaction rate data of an S1q1reaction. Choose B for yet another chance to flash a happy "I just got another one right" smile.

Choice D is correct. Choice B can be eliminated immediately, because NH3 is the nucleophile and not theleaving group. Choice C can be eliminated immediately, because -OCH3 is the leaving group (if it were toreact) and not the nucleophile. The data in Table 1 shows that no reaction was observed each time that theelectrophile was dimethyl ether (CH3OCH3). O.t the other hand, the data in Table tr shows that ammonia(NHg) is a reactive nucleophile with the other three electrophiles used in the experiment (CH3OSO3CH3,CH3I, and CH3CI). This implies that the lack of reactivity can be attributed to the electrophile rather thanthe nucleophile. The leaving group in the cases where dimethyl ether is the electrophile is a methoxide anion(-OCH3). Pick D to prosper and score... well score at least.

Choice B is correct. By definition, a nucleophile is a lone pair donor, which by yet another definition is aLewis base. This makes this question a freebie and the correct answer choice B.

71. Choice B is correct. From the low temperature of the reaction and the retention of optical activity in theproduct, it can be inferred that the reaction proceeds by way of an 51112 mechanism. Elimination and S1g1reactions produce products that lose their optical activity. Choices C and D can be eliminated because thealkene products would show no optical rotation because they lose both stereocenters in the formation of thealkene. The products from a reaction proceeding by an Sry1 mechanism in this case would be a mixture ofdiastereomers (not enantiomers), which would lead to an optical rotation close to zero. Enantiomers areobtained if the reactant is symmetric. The best (although not perfect) answer is choice B.


72.

/5.

74.

75.

Choice D is correct. At high temperature, the predominant reaction is elimination. This eliminates choices Cand D. Because a strong base (NaOH) was used, the mechanism must have been an E2 rather than an E1

mechanism. The best answer is choice D. This can be verified by looking at Trial IV in Table 1, which shou'identical reaction conditions as the reaction in the question. The loss of optical rotation in the product impl,vthat the reaction was an elimination reaction. A substitution reaction would yield some sort of opticalactivity. You have to know that base infers that the mechanism is E2.

Choice B is correct. There are two chiral centers in the reactant, located at carbons two and three. This can be

inferred from the answer selections. The number four priority is hydrogen on both chiral centers andconveniently it is in the back in both cases. The first chiral center (carbon two) has priorities bromine > deuteroethyl > methyl which form a clockwise arc. The first chiral center is thus R. The second chiral center (carbonthree) has priorities ethylbromide > methyl > deuterium which form a counterclockwise arc. The seconichiral center is thus S. The solution is drawn below:

Iu'\.n"}VH

carbon 2 = R

The correct answer is 2R,35, which makes choice B correct.

carbon 3 = S

Choice D is correct. If the reaction proceeded purely by an SNI2 mechanism, then the optical rotation would b,e

the same as was observed for trial I(-36"), a purely 5512 reaction. This eliminates choice A. An eliminatior.reaction would yield an optically inactive product so the optical rotation observed for an elimination reactio:.would be zero. This eliminates choice B. The optical activity observed implies that an S52 reaction must har-e

been occurring to some degree. The reduction in optical activity must be attributed to the Presence of soneimpurities (from some side reaction). The best answer is choice D.

Choice A is correct. Because the enantiomer is a mirror image of the reactant, it forms a transition state wher, -:is eliminating that is a mirror image of the reactant's transition state. This means that the products are ali:mirror images, but without stereogenic centers, they can be rotated to match as identical compounds. Th;symmetry presents itself in the product as an identical geometric isomer. Because a diastereomer varies at on-ione chiral center, it is not a mirror image of the reactant when it is eliminating. This asymmetry presents itse''in the product as d"ifferent geometric isomers. This may not make sense in words so the drawing below shou s

the products.

BrH

H$FTHsC CHs

Reactant

"T,>+"HsC CHsDiastereomer

DHBr

rotation to correctalignment

CH"' Elimnation with -"

rtrr"g b"*

Elimnation with -

-

a strong base

H$tjlHsC HEC CH:

Cis Methyls

Br H CH.

><

rotation to correct

H CHsEnantiomer

aiignment

The enantiomer forms the same products whilechoice A.


-..,i';*"

H\tyH:C HaC D

Trans methyls

HeC

Cis Methyls

a diastereomer forms geometric isomers, making the best ansl,:s"

250 STEREOCHEMISTRY EXPLANATION$

rotation to correct---------=--+alignment

DH

2

76.

77.

Choice B is correct. The elimination product would be optically inactive, because it will lose both stereocenters(chiral carbons) upon forming the alkene. This means that the specific rotation of the alkene product is zeromaking choice B the best answer.

Choice C is correct. An E2 mechanism involves the removal of a proton alpha to the leaving group which thenshifts the electrons into a n-bond forcing the leaving group off. The mechanism is concerted, iiplying that bothevents occur simultaneously. This results in very predictable geometry with respect to the p.oirct a-lkene. Tosupport the operation of the E2 mechanism, both carbons that become involved rn the n-bond should be chirallylabeled to trace the reaction. If the mechanism is specific, then the final product is a specific geometricalisomer. This is why the reactant is deutero labeled at iarbon three. Therefore-, the best answer is choice C.

78' Choice B is correct. Because the electrophile is a tertiary alkyl halide, Reaction 2 proceeds by way of an S.u1mechanism. The two products have specific rotations of +62' and +30' respectiveiy as writtln. A fifty / fiftymixture of the two products would yield an observed specific rotation of +46'" (the average of the values ior thetwo diastereomers). Because of steric hindrance from the bonds to the six membereJring, the nucleophile(It{Hs) prefers to attack from the backside of the molecule. This makes the major product Lompound A, thefirst product in Reaction 1. The -ujgt product has a specific rotation of +62', thus'the specific rotution of themixture is closer to +62" than +30". The specific rotation is between 46' and 62', which -ik", choice B the bestanswer.

79' Choice B is correct. The specific rotation for the mixture is found by taking a weighted average of the specificrotations for the components in the mixture, which is essentially what Equition 1 does. Becarise there is moreof the component with an ct = +48" than the component with an o = +18', the averaged value should be closer to+48' than +18'. However, because it is not purely one component, the specific rotation for the mixture must beless than +48" (within the range of +33" to +48'). The best answer is therefore +42', choice B. The exact

'aluecan be determined mathematically as follows:

oobs = 80% (+a8") + 20'/" (+18') = 38.4 + 3.6 = 42'

80' Choice C is correct. The two products formed in Reaction 2 both have identical bonds to one another and threechiral centers each. In comparing the two sttuctures, only one of the three chiral centers differs, making thetwo structures diastereomers. If some, but not all, of the chiral centers differ between two stereoiso^"rr,1h"yare not superimposable nor are they mirror images. This, by definition, makes them diastereomers. The bestanswer is choice 9 Jh" term epimers describes diastereomers that differ at one chiral center, but it appliesspecifically to the backbones of sugars.

81' Choice D is correct. The two products for Reaction 1 are diastereomers of one another so they have differentspecific rotations' This means that they can be distinguished by their different specific rotation values. Thiseliminates choice A. Because they have different geometry (asymmetry), thef pack differently into theirrespective solid lattices and thus they exhibit different melting points. Tiis eliminates choice B. Because oftheir varying asymmetry, they bind a gas chromatography

"ot.,*n differently (due to a difference in sterichindrance) and thus they show different retention times on the gas chromatogiapher (GC). It can be inferredfrom the last sentence in paragraph two of the passage that the two products hurr" dlff"r"nt retention times onthe gas chromatographer (GC), since the concentration values can be d.etermined (and thus verified in thisexample) using the GC. This eliminates choice C. Infrared spectroscopy measures the type of bonds in themolecules, therefore it is difficult, if not impossible, to diitinguistr ihe two diaster"o*ers by infraredspectroscopy' Diastereomers have identical bonds as one another. The best answer is choice D even if vou haveno idea what infrared spectroscopy does.

82" Choice A is correct. The first product has the methyl and ethyl groups both up and the hydroxyl group down.To retain the chiral integrity shown in Figure 1, the chair confoimatior,, -.rrt have ethyi and mjhyl Lp withthe hydroxyl down. This eliminates choices B and C, which have the exact opposite geometry (ethyl andmethyi are down and the hydroxyi group is up). The most stable conformer hai the least steric repulsion,which idealiy occurs if the three substituents are in equatorial orientation rather than having the threesubstituents in axial orientation. Because the substituents are on adjacent carbons, they alternate upTdown/up,which allows all the substituents to be equatorial. This makes the best answer choice A.


83. Choice C is correct. The three chiral centers are determined as follows:

41

Substituent number four is in front,so the arc must be reversed fromclockwise to counterclockwise.This makes the chiral centerS.

Substituent number four is in back,so the arc is correct as drawn. Thecounterclockwise arc makes thechiral center S.

Substituent number four is in back,so the arc is correct as drawn. Thecounterclockwise arc makes thechiral center S.

84.

85.

86.

87.

The correct answer is S, S, S, which makes choice C the correct answer.

Choice B is correct. In the alkene reactant in Reaction 1, only one carbon has four different substituent=attached, therefore only one carbon is chiral. The best answer is choice B. The chirality is specified in th=reactant for the ring carbon with the ethyl substituent attached.

Choice D is correct. This question is asking to determine which reactions generate new chiral centers. Whertreating an alkene with bromine, two bromides are added, one to each of the double bond carbons. The result -that the two double bond carbons become new chiral centers. This eliminates choice A. When treating a:alkene with hydrogen gas in the presence of a catalytic metal such as palladium, two hydrogen atoms ar-added, one to each of the double bond carbons. The result is that the double bond carbon with the methyl grou:becomes a new chiral center. This eliminates choice B. When treating an alkene with permanganate in bas-:water, two hydroxyl groups are added, one to each of the double bond carbons. The result is that the two doub,i=bond carbons become new chiral centers. This eliminates choice C. When Compound 3a is treated with a goc:nucleophile like ammonia, the acid anhydride is cleaved. The result is that one carbonyl group becomes -amide and the other becomes a carboxylic acid. Neither of these groups are chiral, so no new chiral centers a::formed. The best answer is choice D.

Choice A is correct. It is stated in the passage that Compound 3 represents a mixture of enantiomers, -: -

Compound 3b is the enantiomer of Compound 3a. Enantiomers are mirror images of one another, therefore ail --:

their chiral centers differ. There are two chiral centers in Compound 3, so Compound 3b must have the opposr::chirality of Compound 3a at these two chiral centers. This is true in choice A, so choice A is the best answer.

Choice B is correct. To separate a compound from its enantiomer, the technique must select for one of the t',"":

enantiomers. This requires that there is chirality (asymmetry) in the technique. If the mixture tra\ = j

through a column with one enantiomer attached, the other enantiomer is likely to adhere to the column as -:travels. This means that one enantiomer r,r'ill migrate quickly while the other travels slowly. This ma-.-choice B a strong answer. If the column has both enantiomers bound, then both of the free enantiomer: :solution will be hindered by the column, slowing equally. This does not result in separation. Choice A 'eliminated. The same logic can be used to eliminate choice C . If the distilling column has beads with b - -:enantiomers, then both of the free enantiomers in solution will exhibit the same affinity for the beads, a::will not separate as well. Choice C is eliminated. The compounds do not interact with the carrier gas in ;*chromatography to any notable extent. This means that choice D will do nothing to help separate --- {

enantiomers. Choice D is eliminated.

Choice B is correct. Reaction 2 converts Compound 3a into Compound 4, using a peroxyacid. The oxirane ::L:

form from either side of the n-bond. Flowever, because of the steric hindrance above the ring with the aLk.- *(caused by the six-membered ring), it is preferential to form the epoxide on the back side. Both bonds to oxr-;*nlin the epoxide must be on the same side, so choices C and D are eliminated. Because of steric hindrance, ci^ - -:*B is a better structure than choice A. The best answer is choice B.

88.

Copyright O by Thc Berkeley Review@ 252 STEREOCHEMISTRY EXPLANATIO\$

89.

90.

Choice A is correct. It is stated in the passage that Reaction 1 generates a mixture of enantiomers. Whenenantiomers are formed, they are formed in a racemic mixture. The result is that the optical activity is zero,because the two isomers cancel one another out. This means that Reaction 1 generates an optically inactiveproduct mixture. Once the enantiomer mixture is separated, isolating Compound 3a, the subsequent reactions allstart with an optically active starting material, resulting in optical activity in the end. Reactions 2 and 3 formdiastereomers and Reaction 4 cleaves the anhydride to retain the same chirality as the starting material.Because the materiai is optically pure at the start, the product of reaction 4 is also optically purel The bestanswer is choice A.

Choice C is correct. Methyl antine, like all amines, is a good nucleophile. Tl-re first equivalent reacts with themost reactive electrophile, which in this case is the epoxide ring. However, an anhydride is also highlyreactive. In the event excess amine is added, it can easily add to the carbonyi, cleaving the anhydride andgenerating a carboxylic acid and amide. Choices A and B can be eliminated immediateiy, becar-rse the three-membered ring is highly unstable and will not reform once broken. The OH group is a bad ieaving group, so it isnot likely that an amine will substitute for the hydroxyl group, once the airhydride is cleaved, so choice D iselimrnated. The result is that choice C is the best answer.

Choice C is correct. In Reaction 2, a mixture of stereoisomers is formed that is never resolved. As a result, thereactant in Reaction 4, Compottnd 5, represents a mixture of stereoisomers. Two of the chiral centers are thesame in all of the stereoisomers, so they cannot be enantiomers (where all of the chirai centers differ). Thiseliminates choice A. The product is not meso, it does not have an internal mirror plane, so choice D iseliminated. Because the mixture starts with optical activity, and Reaction 4 does nothing to affect thechirality, the product mixture in Reaction 4, Compound 6, must be optically active. This makes choice C thebest answer.

92. Choice A is correct. If you are well versed ir-r r-rsing the thumb technique, then you can place your thumb in thedirection of the C-H bond and curl your fingers from priorities 1-2-3 using a right hand for both stereocenters.This makes both centers R. For some, it may be easier to rotate the molecule to the side view rather than usingthe Newman projection to deterrnine the chirality. From the side perspective, it is much easier to see the threedimensional orientation of the molecule. As shown in the determination below, the molecule has 2Ii., 3Rstereochemistry. Choose A; be a chem star.

CFI?

ocFi3 H in back; R stereocenter

OHOH

H in back; R stereocenter

93- Choice D is correct. With Fischer projections, you must remember that when an H is drawn on the side, itrepresents an H coming out at you ln a three dimensional perspective. Hence, whatever arc you determine fromthe Fischer projection must sutsequently be reversed to get the chirality of the stereocenter. in this example,both chiral centers generate ciockwise circles in a two-dimensional perspective. But after reversing iheclockwise circles to counterclockwise becanse of the hydrogen atoms are in fiont, both centers have S chirility.Choose D and be a scholar.

H in front;S stereocenter

H in front; S stereocenter

91.

CH"

| ' Clockwise arc with1/lrF+T- OH\y

-Fs.,,.\ Clockwise arc r,r'ith

ocF{3

Copyright O by The Berkeley Review@ 253 STBREOCHEMISTRY EXPLANATIONS

94. Choice A is correct. On the second carbon, the OH and H groups have interchanged, so that chiral center differsbetween the two stereoisomers. On the third carbon, the OH and H groups have not interchanged, so tha:chiral center is identical in the two stereoisomers. On the fourth carbon, the ethyl and methyl groups haleinterchanged, so that chiral center differs between the two stereoisomers. When two out of three (i.e., somebut not all) of the chiral centers differ, the two stereoisomers are neither superimposable, nor are they mirro:images of one another, which defines diastereomers. Pick A and be jovial.

Choice B is correct. In the molecule, carbons two and three are chiral centers. However, carbon one is not rchiral center, because it contains two equivalent hydrogen atoms. The maximum number of stereoisomers i.equal to 2n where n is the number of chiral centers in the molecule, which in this case is 2. This yields a total c:fot.l (22) stereoisomers. Be stellar and choose B.

Choice A is correct. Syn addition of equivalent to a trans double bond in an alkene results in the formation c:two enantiomers (specifically the R,R and S,S enantiomers) as shown below. Choose A for best results. A mes:compound can be obtained from syn addition to a cis alkene, where the aikyl groups are equal on the alkene.

HO OH

95.

96.

98.

99.

100.

H.C Fi\ / KMnOalaq).,-, __-+/ \ pH=10

H CFL" Syt-,additionTrans reactant (asymmetric) (symmetric)

H CFI"

. H:cr,l /-.,,,"Hc' ,,,Tt

HO OH

\_JHrcY \"'*,

H CH:

As).rnmetric products

97. Choice C is correct. If a compound has stereocenters but is not optically active, this implies that the compou:.:must be meso. To be meso, a compound must have a mirror plane in the molecule about which the chiral centef.are evenly displaced. This mirror plane slices the molecule into two equivalent halves. 2R,3S-dibromobuta:.=and 2R,4S-dibromopentane each have equivalent mirror halves (thus mirror symmetry) so they are both me>:compounds. Cis-1,3-dichlorocyclohexane has equivalent halves as wel1, thus it is a meso compound. Or-.2R,4S-dibromohexane does not have two equivaient halves. The correct answer is thus choice C.

Choice D is correct. 4-chlorocyclohexanol contains no stereocenters (chiral carbons), therefore it cannot rota.:=plane polarized light. The result is that 4-chlorocyclohexanol is not optically active. Choose wisely ar.:choose D. The other three compounds have two chiral carbons and therefore are all opticaily active. It:'possible for the compound (4-chlorocyclohexanol) to exhibit isomerism in the form of geometrical isomers r.-'and trans), but geometrical isomers do not rotate plane polarized light.

Choice C is correct. Isoleucine contains two chiral centers, one for the alpha carbon and one in the side cha-Plugging into the stereoisomer equation 2n, wl-rere n represents the number of chiral centers, there are tc..:possible stereoisomers for the isoleucine structure. Because isoleucine contains two chiral centers that mu-.have specific orientation, only one of the four stereoisomers has the correct chirality to be biologically use:...,Choose C to be a successful point collector.

Choice B is correct. For a compound with chiral centers to be optically inactive, it must be meso. and tt-.contain an internal mirror plane of symmetry. The molecule must have R, S chiraiity to be meso. T-,-:eliminates choices A ar-rd C. The mirror plane must be through the C2-C3 bond according to the chiral centers -:the ansll'er choices. This mirror plane is possible only with butane, so the correct answer must be choice B.

Copr ri-sht O by flie Berkeley Review@ STEREOCHEMISTRY EXPLANATIO\i

Section HVAlkanes

a) Structures and Physical Propertiesi. Aliphatic and Cyclic Alkanes

b) Alkane Reactivityi. Free Radical Llalogenationii. Mechanismiii. Energeticsiv. SelecLivity

llydrocarbon R.eactionsa) Elimination Reactions

i. Ez Mechanismii" E1 Mechanismiii. Carbocation Stabilityiv" Electrophilic Additionv. Regioselectivityvi. Steneoselectivityvii. Stereoisomer Formation

b) 1,Z-Additi<ln versus 1,4-Addition

HydrocarbonReactions

by Todd Bennett

Limonene

Me c) Pericyclic Reactionsi. Diels-AlderReactionii. Cope Rearrangementiii. Claisen Rearrangement

Terpenesa) Classification (Carbon Count)b) l'ormation and Synthesisc) Connectivityd) Spectroscopic Properties

Geraniol

Speciahzing in MCAT Preparation

Ilydroctrrbons and KeactionsSection Goals

Recoanize the nomenclature associated with the alkanes and dienes.

aB

The Greek prefixes and suffixes associated with the hydrocarbons must be common knowledge.Know the terminology so that when names are presented in passages, you can draw the structureor recognize the structure in an answer choice. Be able to recognize the positioning of the n-bondsin conjugated n-systems.

Know the general mechanism for free radical addition to an alkene.Free radical halogenation of an alkane follows a simple three-step process. A free radical halogenis generated by the initiation reaction, requiring heat or light. The free radical halogen starts thepropagation reactions which undergo a continuous chain reaction until it is terminated. Terminationi"uitidtlr involve two free radicals coming together to form a bond. The overall reaction is a sumof the propagation steps.

Know the mechanisms associated with elimination reactions.You must know the E1 and pZ mechanism in detail. Know which reactant, solvent, and catalystcombination will result in which mechanism. As a rule, E1 occurs in acid and E2 occurs in strongbase. Rearrangement can offer complications in an E1 reaction. E2 reactions require the acidic protonand leaving group to be anti to one another.

Know the general mechanism for electrophilic addition to an diene.There are several reactions which involve the addition of an electrophile to one of the n-bonds in adiene. You must recognize the diffrence between the 1,2-addition and 1,4-additions of haloacidsand water across a conilgated diene. Recognize the stereochemical results associated with electrophilicaddition reactions of allienes.

Know the general mechanism for electrocyclic reactions of polyenes.The most common electrocyclic in organic chemistry is the Diels-Alder reaction. You must recognizeit and be able to predict r6latir.e reictivity of diflerent reactions based on electronic factors"andsteric hindrance.'You must also recognize that the six-electron transition state in the Diels-Alderreaction is also found in sigmatropiclearrangements like the Cope and Claisen rearrangements.You should have a basic understanding of n-systems.

Recoqnize the isoorene subunits in natural terpenes.Terpenes are natural oils derived from a five-carbon species. Isoprene is 2-methyl-1,3-butadiene,altliough the biologically active molecule is a derivative. Terpenes can be extracted and/or distilledfrom natural products.'You should be able to identify the isoprene units in biologically commonterpenes and terpenoids.

@?

"?

Organic Chemistry tlydrocarbons and Reactions Introduction

The crux of organic.chemistry centers on reactivity. In organic chemistry thereare many classes of reactions. Th"y are organized accoiding to mechanisticfeatures. Common .reaction types incrude acid-base reactions (section 1),nucleophilic substitution (section 3), free radical reactions (typically halogenationof an alkane or polymerization of an alkene), elimination, electrophilic Jdditlon,electrophilic aromatic substitution, electrocyclic reactions, oxidation-reductionreactions, and carbonyl substitution reactions (section 5). There are other typesof reactions, name reactions, which we shall address by functiond grtlp.However, in this section, we shall address the reactions of alkaner ur,a tt-r"reactions of n-bonds, as classified by their mechanisms.

Free radical reactions follow three basic mechanistic steps: initiation,propagation, and termination. Whether we consider free radical halogenation ofan alkane or free radical polymerization of an alkene, the first step of ih" pro.u5is initiation. Initiation entails the homolytic cleavage of a weak'sigma bond inonly a small number of the molecules present in tne reaction mixture. Inhalogenation reactions, the halogen-to-hilogen bond is cleaved to initiate thereaction. In free radical polymerization, the oxygen-to-oxygen bond of aperoxide initiator is cleaved to generate a low concentratiot of fr"e radicalcompounds. The product of a free radical halogenation reaction is an alkylhalide. An alkyl halide can und"ergo nucleophilic substitution reactions (as wesaw in the stereochemistry section) and elimination reactions. Eliminationreactions result in the formation of a n-bond, which opens the door to a plethoraof new reactions. starting with an alkane, ,oori u.ry compound can besynthesized.

Although the MCAT test writers do not require that you memorize the reactionsof alkenes, as they have required in the past, you are expected to know thegeneral classes of reactions that involve n-bonds. when the n-bond of an alkeneac-ts- as a nucleophile by attacking an electrophile, this starts an electrophilicaddition reaction. The ru-bond is not limited to being part of an alkene, u, lt .u1also be part of a diene, triene, or any other conjugaied system of multiple rc-bonds. V\4ren there is more than one n-bond in the system, we must consider thepossibility of the addition reaction occurring at different reactive sites. Thenucleophiiic n-bond can also be found on benzene, although the aromaticity ofthe n-system drastically reduces the nucleophilicity of a r-bond. However, if theelectrophile is strong enough, the n-bond oi benrene can attack it and eventuallye>rqhange the electrophile for a hydrogen, retaining the aromaticity of the system.This is known as an electrophilic aromatic substitulion. The last reaction we shallconsider for n-bonds is the Diels Alder reaction, an electrocyclic reactioninvolving the addition of an alkene to a conjugated diene to form a cyclohexenemoiety.

From this section on, we shall focus on the reactivity of organic compound.s. Tobest prepare for the MCAT, your goal shourd not be to memorize every reaction,but instead, learn a few simple, common mechanisms and have a conceptualpicture of how they work. If you can summarize the contents of this secti,on interms of the nucleophilicity of a n-bond, no matter what compound contains therc-bond, then you have a solid grasp of the topic at the MCAr level. Reactiondetails will be provided in the passage, so from this point on, know the generalreaction and work on your information extraction skills by reading graphs,tables, and data charts,

Copyright O by The Berkeley Review 257 Dxclusive MCAT Preparation

Organic Chemistry llydrocarbons and Reactions Alkanes

AlkanesAlkane StructureAlkanes contain only carbon and hydrogen atoms, and all of the carbons havespr-hybridization. AII of the hydrogens in aikanes use s-orbitals to form bonds.Alkanes are held together exclusively by o-bonds. In any alkane, there are onlytwo types of bonds present: C-C bonds (which are osD3-s'3) and C-H bonds(which are oro3-r). Both types of bonds present in alkanis aie shown in Figure 4-1 as molecular orbitals.

cffic @.orp3-tp3 os-sP3

Figure 4-1

Alkanes can be classified as either aliphntic (straight chain form) or cyclic(containing a ring in its structure). Aliphatic alkanes hive a molecular formula ofCnH2r, 12 while cycioalkanes with one ring have a morecular formula of C,.,H2r.,.For each additional ring in a cyclic alkane, the molecule has two less hydrogenatoms. Table 4-1 lists some common linear alkanes and cyclic alkanes up to-tencarbons, It should be noted that at least three carbons are needed to form a ring.

C^H2', * 2 for linear alkanes (no rings)

Methane CHa Hexane CoHra

Ethane CzH.e Heptane CzHrcPropane csHs Octane CgHrs

Butane C+Hro Nonane CsHzo

Pentane CsHrz Decane CrcHzz

C^H2., for cycloalkanes (with one ring)

Cyclopropane CsHo Cycioheptane CrF{uCyclobutane C+Hs Cyclooctane CsHro

Cyclopentane CsHro Cyclononane CsHra

Cvclohexane CoHrz Cyclodecane CroHzo

Table 4-1

l

I

;

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Organic Chemistry tlydrocarbons and Reactions Alkanes

Alkane PropertiesThe physical properties of concern associated with an alkane are its solubilityfeatures, its density, its melting point, and its boiling point. Alkanes arehydrophobic, nonpolar molecules. They can also be defined as lipophilic, whichimplies that they are highly soluble in oils such as the lipid membrane of a cell.As with all compounds, their physical properties vary with mass and branching.As the molecular mass increases, both the boiling point and melting pointsincrease. As the branching increases, the boiling point decreases. Table 4-2shows the physical properties of several aliphatic alkanes. From the data inTable 4-2, the effects of mass and branching on the physicai properties can beobserved.

Table 4-2

Example 4.1What is the molecular weight of 2,2-dimethyl-4-propyl-5-cyclopentylnonane?

A, 228.21,grams,/moleB. 266.51, grams/moleC. 268.57 grams/moleD. 280.54 grams/mole

SolutionFirst, we must determine the number of carbon atoms and hydrogen atoms.Dimethyl accounts for two carbons, propyl accounts for another three, pentylaccounts for another five, and nonane accounts for nine. The compound contains19 carbon atoms totai. Because of the "cyclo" in the name, there is one unit ofunsaturation. The one unit of unsaturation implies that there are 38 hydrogens(two less then the 40 that would be present in an aliphatic, linear alkane). Themolecular mass is thus 19(12) + 38 = 266, choice B.

Isomer Name BoilingPoint

MeltingPoint

Density(g/mL)

Mass(g/mole)

CH+ Methane -762"C -183"C 76.043

H3CCHe Ethane -89'C -183'C 30.070

H3CCH2CH3 Propane -42"C -787'C 44.097

H3C(CH2)2CH3 Butane 0"c -138'C 58.124

H3C(CH2)3CH3 Pentane 36'C -130'C 0.557 72.151

H3C(CH2)aCH3 n-Hexane 69"C -95.3'C 0.659 86.778

(HrC)zCH(CHD2CH3 2-Methylpentane 60'c -754"C 0.654 86.1.78

H3CCH2CH(CH3)CH2CH3 3-Methylpentane 63'C 118"C 0.676 86.778

(H3C)3CCH2CH3 2,2-Dimethylbutane 50"c -98"C 0.649 86.178

(H3C)2CHCH(CH3)2 2,3-Dimethylbutane 58'C -729"C 0.668 86.778

H3C(CH2)5CH3 n-Heptane 98'C -90.5"C 0.684 100.205

H3C(CH2)6CH3 n-Octane 126"C -57"C 0.703 1.1.4.232

H3C(CH2)7CH3 n-Nonane 151"C -54'C 0.778 128.259

H3C(CH2)sCH3 n-Decane 174'C -30"c 0.7362 142.286


Organic Chemistry llydrocarbons and Reactions Alkanes

Alkane ReactivityAlkanes undergo a minimal number of reactions, and the few they do undergoinvolve free radical chemistry. The two reactions of concern are free radi&lhalogenation (more specifically bromination, using Br2, and chlorination, usingC12) and combustion. Reaction 4.1 is the free radical chlorination of methane.

CH+(g) + Cl2(g) -il* CH3CI(g) + CHrCtr(1) + CHC13(I) + CCla(t) + C2Cl6(s)

80% 70% minor minor minor

Reaction 4.1

Free Radical Halogenation of AlkanesA free radical halogenation reaction starts with the addition of activation energ\-to cleave a halogen-halogen bond (the weakest bond in the reactants) to form twofree radical halogen atoms. A free radical is an atom, such as a halogen orcarbon, with one unpaired electron. According to the octet rule, most atoms wishto have eight valence electrons. In a free radical, there are only seven, so freeradicals are highiy reactive. Figure 4-2 shows a 3-D perspective of four alkyl freeradicals with the p-orbital filled with one electron. The three substituents aredrawn slightly below the carbon atom, because the free radical molecule isslightly trigonal planar due the electrostatic repulsion from the single electron.Because the electron can exist in either lobe of the p-orbital, the substituents carrbe angied up or down, so the average of the two trigonal pyramidal forms isplanar' The stability of a free radical depends on its substitution. Tertiary freeradicals are more stable than secondary free radicals, which in turn are morestabie than primary free radicals. Figure 4-2 shows the relative stability of thealkyl free radicals.

>H.

3" Free radical 2' Free radical 1' Free radical Methyl free radical

Figure 4-2

Once a free radicai halogenation reaction has been initiated by the addition crlight (activation energy), a free radical halogen atom then attacks an alkane ancabstracts a hydrogen from the alkane to leave behind an alkyl free radical. Thehalogen free radical abstracts the first hydrogen it encounters, but because alkr.free radicals can react with other alkanes, over time the distribution favors theformation of the more stable tertiary free radicals. The conversion from a

primary free radical into a tertiary free radical is shown in Figure 4-3 below.

t'.\ \-^*i n,\ n,.\

H*f cH.cH,. JQ_ cH"cH1----* .,,,| cH"cH^ . .| cH^cH

H:C HsC HaC HsCPrimarv Free Radical Tertiarv Free Radical

Figure 4-3

The reaction in Figure 4-3 heavily favors the formation of product, because thetertiary free radical is substantially more stable than the primary free radical.

"Gx'+o-6Xii.* ' "-6*gF ' o-6*9,# '

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Organic Chemistry flydrocarbons and Reactions Alkanes

Free Radical MechanrsmThe mechanism for a free radical halogenation is a chain reaction processinvolving an initiation reaction, followed by propagation reactions, andultimately a termination reaction. An initiaiion reaction breaks a covaient bondin a homolytic fashion to form two free radicals, so an initiation reaction goes

from no free radicals to two free radicais. Homolytic cleavage is djfferent fromheterolytic cleavage in that each atom gets a single eiectron (resulting in freeradicals) as opposed to the more electronegative atom getting both bondingelectrons (resulting in a cation and an anion). In halogenation, it is the haiogen-halogen bond that is broken. Propagation steps involve the abstraction of an

atom from a neutral molecule by a free radical to form a new free radical.Propagation reactions include the consumption and formation of a free radical,so a propagation reaction goes from one free radical to one free radical. In freeradical halogenation, there are two propagation steps. Termination occurs whentwo free radicals combine to form a neutral, stable molecule. Termination steps

involve the consumption of two free radicals, so a termination reaction goes fromtwo free radicals to no free radicals. There are several possible terminationreactions in a free radical halogenation reaction, most of which form minor sideproducts. The steps for a generic halogenation of an alkane are shown below as

Reactions 4.2,4.3, and 4.4.

Tnitiation: --------->

Reaction 4.2

Propagation: tGi*ila X.+.X

Termination:

-c?a

n! * '.R

nli.x

-----> R. + H-XReaction 4.3a

-------> R-X + .X

Reaction 4.3b

-------> R-RReaction 4.4a

-------> R-XReaction 4.4b

--------> x-xReaction 4.4c

*li.*

Overall Reaction: R-H + X-X --------+ R-X + H-XReaction 4.5

The sum of the propagation steps for a free radical halogenation reaction givesthe overall reactiort, Reaction 4.5. The initiation step is brief, for just a splitsecond at the start of a free radical reaction. However, the propagation steps

continue until the reaction is quencired or the free radicals are completelyconsumed in termination reactions. There are always multiple termination steps

possible. One of the possible termination steps is the reverse of the initiationreaction. A termination step is any reaction that combines two free radicals togenerate a stable compound. Termination steps are responsible for several minorside products.

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Example 4.2Which of the following reactions represents an initiation step?

A. H3CCHzCHz. + Br. -) H3CCH2CH2BTB. H3CCH2CH2. + H3CCH2CH3 -+ H3CCH2CH3 + @3C)2CH.C. (H3C)2CH. + Br2 -) H3CCHBTCH3 + Br.D. HOBr --> HO. + Br.

SolutionIn an initiation reaction, free radicais are generated, so the product side has moreradicals than the reactant side. Choice A is eliminated, because there are fewerradicals on the product side, making it a termination reaction. In choices B andC, there is the same number of radicals on both sides of the reaction, so they arepropagation reactions. This elimrnates choices B and C. In choice D, the reactiongoes from zero free radicals on the reactant side to two free radicals on theproduct side, so it is an initiation reaction.

Example 4.3In the free radical chlorination of ethane, butane is a minor side product. Horr'can this best be explained?

A. An ethane molecule attacked an ethyl chloride in a nucleophilic substitutior,reaction to form butane.

B. An ethyl free radical removed a hydrogen from an ethane molecule.C. Two ethyl free radicals combined to form a new sigma bond.D. The carbon-carbon bond of ethane was cleaved during initiation to forrn

methyl free radicals, which rapidly combine to form iong-chain alkanes.

SolutionAn ethane molecule has neither a lone pair of electrons nor an available pair o:bonding electrons to share (like a n-bond), so it is definitely not going to act as a

nucleophile. This eliminates choice A. An ethyl free radical can definitel,,'remove a hydrogen from an ethane molecule. However, that does not resuit ir.the formation of butane, it simply regenerates the same molecules.

H3CCH2. + H3CCH3 -+ H3CCH3 + H3CCH2.

Choice B is eliminated. When two ethyl free radicals combine, they form a ne\^.'

sigma bond between the two free radical carbons. Combining the two two-carbon fragments results in the formation of a four-carbon fragment. The resuL:is that butane is formed from the termination reaction of two ethyl free radicals,The best answer is choice C. Carbon-carbon bonds are not easily cleaved, so theactivation energy added in the initiation step is not high enough to cleave a

carbon-carbon bond. Even if it could cleave the carbon-carbon bond to forr.methyl free radicals, there is iittle likelihood that four CH3 groups woulccombine to form C+H1O. The loss of two hydrogen atoms would not occur.

Copyright O by The Berkeley Review 262 The Berkeley Kevieu


Free Radical Halogenation Energeticswith each reaction, we consider how much product is formed (thermodynamics)and how fast the reaction proceeds (kinetics). AH and AG vaiues determine howmuch is formed while the activation energy (Eo.1) dictates the speed. For thechlorination of methane, Reaction 4.1, the foilowing enthalpy valuel apply:

Ci-Cl -) Cl.+ .Cl

H3C-H + .Cl -+ H3C. + H-CIH3C. + Cl-Cl -) H3C-CI + .Cl

AHi.itiatic,n = 58 kcal

AH1 = I ks.1

AH2 = -26kcaI

Figure 4-4 shows the energy diagram associated with propagation steps inReaction 4.1.

Reaction Coordinate

Figure 4-4

The energetics of each free radical halogenation reaction is different. Forinstance, bromine has a lower bond dissociation energy than chlorine, sobromination requires less activation energy than chlorination. This is whychlorination requires light for initiation, while either light or heat can initiate abromination reaction. F{owever, this does not mean that bromination proceedsfaster than chlorination. The reaction rate is determined from the aitivationenergy of the rate-determining step in propagation. Table 4-3 lists the energiesfor the halogenation of methane. From Table 4-3, you can determine whichreaction is fastest and which reaction generates the most heat. The data showsthat free radical iodination of methane is unfavorable and free radicalfluorination of methane is too favorable, generatrng enough energy to explode.

AH values {Kcal/*o1")

Halogenation Reaction X=F X=Cl X=Br X=Ikritiation; X2 -+ 2X. 38 58 46 36

X. + CHa -+ HX + H3C. -32 +1 +76 +33

H3C" + X2 --+ H3CX + X. - /t) -26 -z+ -20

CH4 + X2 -+ H3CX + HX -102 -25 -8 +13

Table 4-3

HCl + CH3

frCJ

tr.l

OJ

oA

overall AH = AHr

CH3C1 +


Organic Chemistry Ilydrocarbons and Reactions Alkanes

Free Radical Halogenation SelectivityBecause tertiary free radicals are more stable than other free radicals,halogenation occurs preferentially at tertiary carbons. When there is more thanone unique carbon in the alkane reactant, the product distribution is a result ofthe relative free radical intermediate stability, the relative abundance ofequivalent hydrogens, and the rate of the reaction. For the chlorination reaction,the relative reactivity of 3" :2' : 1" carbons is 5 : 3.g : 1 at room temperature. Attemperatures around 100"C, chlorination selects for tertiary over seiondary overprimary by a ratio of roughly 4 : 2.5 : r. As the temperature increases, thereaction proceeds faster, and is therefore less selective. For the brominationreaction, the relative reactivity of 3' : 2" : 7" carbons is 1600 : 62 : r at roomtemperature. This means that the product of free radical bromination is almostexclusively tertiary, while the free radical chlorination reaction gives a morebalanced product mixture. Figure 4-5 shows the product disiribution forcomparable chlorination and bromination reactions.

Br

I

.t\-/97.6"/o Seconda

Figure 4-5

As a general rule, slower reactions are more selective than faster reactions,because reactants have more time to select the best site for reacting. Chlorinationreactions are faster than bromination reactions, so bromination is more selectivethan chlorination. In the slower bromination reaction, if a primary free radicai jsformed, it has time to abstract a hydrogen from another alkane and form a ner!.,possibly more stable free radical. If the new free radical is tertiary, it is likely toonly react with a dihalogen molecule. Consider Reaction 4.6 and, the data inTable 4-4 corresponding to the distribution of mono-halogenated products.

>\+X2grb.hxReaction 4.6

c+Hro

cl"__>hv

C+Hro

77.7"/o Secondary$.,

28.3% Primary

$Br2.4"h Prirnary

Br,€hv

ty

Trial Halogen Temp (K) Product A Product B Product C Product DI Br2 298 03% 89.40/" 70.1% 0.2%II Clz 298 29.1% 243% 32.00/" t4.6%III Btz J/J 0.4% 88.3% 1,L.1% 0.2%IV Clz J/J JJ.J 70 aa ao/LL.L /O 27.8% T6]%V 12 373 No Rxn No Rxn No Rxn No Rxn


Table 4-4

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The data in Table 4-4 confirms that bromination is more selective thanchlorination for tertiary over secondary over primary. Bromination is slowerthan chlorination, but it should also be noted that bromination is more reversiblethan chlorination, so it is more likely undergo a reverse reaction from an unstableproduct to ultimately form the most stable product. The data in Table 4-4 alsoshows that temperature has an effect on the reaction rate and therefore on theselectivity. As the reaction temperature is increased, the reaction proceeds at afaster rate, resulting in the formation of products based more on randomprobability rather than selection for the most stable intermediate. In addition,Boltzmann's law states that as energy is added to a system, the distribution ofcompounds is shifted to the less stable compounds, to absorb the energy.Actually, who reaily knows if Boltzmann said it, or if it's even a law. The keything is that as energy is added, less stabie compounds are formed.

Example 4.4\44ry in Trial II of Table 4-4 is Product A formed to a greater extent than productB?

A. Product A results from the more stable free radical, thus it is selected for.B. Product B rearranges to form product A.c. Product B is more stable, but there are more hydrogens that lead to product

A, so overall less Product B is formed.D. Product A is more stable, but there are more hydrogens that lead to product

B, so overall less Product B is formed.

SolutionProduct A results from a reaction at a primary carbon, so it proceeds via aprimary free radical. Product B results from reaction at a tertiary carbon, so itproceeds via a tertiary free radical. This eliminates choice A. Rearrangement isseen with carbocations, but not with free radicals or carbanions, so choice B iseliminated. The best answer is choice C. It is often possible to answer a questionwithout fuil analysis. The reason for the substantial amount of Product A isbecause there are six hydrogens that lead to Product A while there is only onehydrogen that leads to Product B. Although tertiary reactivity with chlorinationis roughly four to five times greater than primary reactivity, the six-to-oneabundance ratio outweighs the four or five-to-one reactivity preference, makingthe probability of forming Product A greater than the probability of formingProduct B. This is makes choice C the best answer. Choice D can be eliminated,because there are more hydrogens available to form Product A than Product B.



Example 4.5Why is there no reaction observed when iodine is used?

A. Iodine cannot form a free radical.B. The iodine-iodine bond is too strong to cleave.C. Free-radical iodination of an alkane is too reactive.D. Free-radical iodination of an alkane is unfavorable.

SolutionBecause chlorine and bromine form free radicals, we can be assume that anotherhalogens, such as iodine, can also form a free radical. This eliminates choice A.Iodine is lower in the periodic table than chlorine and bromine, so the iodine-iodine bond is weaker than the chiorine-chlorine bond and the bromine-brominebond. Because Cl2 and Br2 are cleaved, it is safe to assume that 12 is even easierto cleave. This eiiminates choice B. Iodine forms weak bonds to carbon andhydrogen, so the products are less stable than the reactants. Because theproducts of free radical iodination are less stable than the reactants, the reactionis unfavorable, so there is no reaction observed with iodine. This makes choice Dthe best answer and eliminates choice C. Fluorine is not used for completelyopposite reasons. Fluorine forms strong bonds to carbon and hydrogen, and th!fluorine-fluorine bond is weak. The products are so much more stable than the

:j'"",,^" t'"" ",Example 4.6Why are there minimal di-halogenated products formed in the free-radicalchlorination of an alkane?

A. The acldition of the halogen makes the alkyl halide less acidic than thealkane, so it is less reactive to subsequent halogenation reactions.

B. The addition of the halogen makes the alkyl halide more acidic than thealkane, so it is more reactive to subsequent halogenation reactions.

c. After the first halogen is added to the alkane, the carbon-hydrogen bondsgrow weaker and thus more reactive.

D. After the first haiogen is added to the alkane, the weakest bond is a carbon-halogen bond and not the carbon-hydrogen bond. As a result, it is easier toremove the halogen rather than the hydrogen from the mono-substitutedalkyl halide.

SolutionHalogens are electron-withdrawrng, so their presence on a molecule increases itsacidity. This eliminates choice A. Choice B is invalid, because an increase inreactivity would imply that more poly-halogenated products would form, notless. Choice C can be eliminated for almost the same reason. If the carbon-hydrogen bond is weaker, and thus more reactive, then it would be easier to adda_second halogen than the first, making poly-haiogenation preferable. once analkane is haiogenated, the weakest bond is the carbon-halogen bond, not acarbon-hydrogen bond. If a second halogen free radical ruuit, with an alkvlhalide (rather than an alkane), it preferentiully r"rr,orr"s the halogen (breaking theweakest bond), forming a non-halogenated alkyi free radical. this is because thereverse halogenation reaction (in propagation) is more favorable than theremoval of a hydrogen and subsequent additional halogenation reaction. Choice

"'*.

j

-1.

-j

E.

-,1L

ts(-

D

5-c

:._

:_-:

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Organic Chemistry Ilydrocarbons and Reactions Alkanes

Example 4.7To synthesize a primary alkyl halide from an alkane in highest yield, whatshould be done?

A. Bromination at25"CB. Chlorination at25'CC. Bromination at 100"CD. Chlorination at 100'C

SolutionA primary alkyl halide is the least favorable product, so the best free radicalhalogenation reaction is the one with lowest selectivity. According to the data inTable 4-3, chlorination is less selective than bromination and selectivity isreduced at higher temperatures. This means that choice D, chlorination at thehighest listed temperature, is best.

Example 4.8Using the data listed in Table 4.3, what percent of the mono-halogenatedproducts is 2-bromo-2-methylbutane following the free-radical bromination of 2-methylbutane at75"C?

A. 88.1%B. 88.7%c. 89.1%D. 89.7%

SolutionTo answer this question, you need to read Table 4-3. In trials I and III, product Bis 2-bromo-2-methylbutane. Therefore, we need to estimate how much Product Bis formed at75"C. At25'C, there is 89.4% product B formed, while at 100"C thereis 88.3% product B formed. This means that the amount of product B formed at75'C should be between 88.3% and 89.4o/o, which eliminates choices A and D.The amount formed at75'C should be closer to 88.3% than89.4"h, so choice B isthe best answer.

Example 4.9How many mono-chlorinated structural isomer products are possible when 2,5-dimethylhexane undergoes free radical chlorination?

A.38.5c.6D.8

SolutionThis question is asking for how many structural isomers there are for chloro-2,S-dimethylhexane. Because of the mirror plane through the carbon 3-to-carbon 4,bond, there are three unique carbons on 2,5-dimethylhexane. This means thatthere are just three carbons that can be chlorinated, so there are only three mono-chlorinated structural isomers. If stereoisomers were included, the value wouldincrease to four, given that chlorination of the secondary carbon yields a

stereogenic center. The best answer is choice A.


Organic Chemistry tlydrocarbons and Reactions tlydrocarbon Keactions

Aydroiberbdh R€adtiddb , ,,-,,.,*,,,,,,r',i,,,t,,,,,,,f,,,,

Elimination ReactionsThe reaction that forms an alkene from a substituted alkane is elimination. It isnamed from the fact that a functionai group and a hydrogen on adjacent carbonsare eliminated in order to form a n-bond. The-reac-tion requires elevatedtemperatures to help overcome the activation energy and to p.rsh the reaction inthe forward direction. Like the nucleophilic substiiution ,*u.tior,r, there are tworeaction mechanisms, appropriately named E1 and E2. As with nucleophilicsubstitution, the two versions are named also for theiireaction orders (kineticrate dependence). E1 is similar to sry1 and E2 is similar to slr]2, except that theproduct is an alkene. In an E1 elimination, the leaving gto.rp first ieaves andthere is a carbocation formed. The empty p-orbital of thi c"arbocation eventuallr-becomes one of the two p-orbitais in the new n-bond. In an E2 elimination, a baseremoves an alpha hydrogen to force the leaving group off of the neighboringcarbon. Elimination converts a functionarized alkyigroup into an alkenJ.

E2 Reaction (Carried out under Basic Conditions at High Temperature)The E2 reaction is corrcerted (one-step) rike the sNI2 r;action, with one majorexception being that the E2 reaction occurs at higher temperatures (temperatuiesabove ambient temperature) than the S52 reaction. An E2 reaction also requiresthat the base be bulky. Because of the steric hindrance aisociated with a bulkybase, it is.less apt to act as a nucleophile and thereby minimlze in;;;;fi;;Sp2 reaction. An important feature of the E2 reacti-on is that the substituentibeing eliminated must be onti to one another (have a dihedral angle of 1g0.). Themechanism for an E2 reaction is shown in Figure 4-6.

(^H

! oR-

H3Ctt,,) (trrH

H

H-----+

AHt

R:o

I

I

HH

H,CV X-

H

HgCH

Figure 4-6

In an E2 reaction, the compound may have to rotate to proper conformationbefore the reaction can proceed to form the alkene. This is to siy that the leavinggroup must be oriented anti to the hydrogen being eliminated before the baseattacks the proton to start the reaction. whenever there are two alpha hydrogen,.that can be eliminated from the starting reagent, there are two posiible prodicts.The consequence is that when the starting material has a specific stereoChemistl'at one of the reactive carbons, a corresponding specific geometrical isomerproduct (either cis or trans) forms. Figure 4-7 shows an exampte where there aretwo alpha protons to choose between. One of the two protoni has been replaceclby deuterium, so that the structure can be monitorecl and the product can beused to support the idea that the elimination reaction o.c.rrr"d from the antiorientation. Figure 4-7 shows the rotation to anti and formation of both products.

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Organic Chemistry Ilydrocarbons and Reactions llydrocarbon Reactions

DHLINR^ \ /----1- EJ

^/\ H:C CH:

H.C H

F<H cHs

Figure 4-7

If the reaction were not carried out through a mechanism that required antiorientation, then there would have been four elimination products total, ratherthan just two. The four products wourd include the cis and trans alkenesfeaturing the deuterium, and the cis and trans alkenes with no deuterium. Figure4-B shows the two alkenes not formed,in the reaction in Figure 4_7.

HH

><HsC CHs

Not formed in the elimination reaction of (2s, 3R)-3d-2-bromobutane,therefore the reaction must have proceeded by the anti orientation.

Figure 4-8

Multiple products can result when their are multiple hydrogens available fordeprotonation. In Figure 4-7, onry the products with in iniernai n-bond aredrawn. The internal n-bond is more favorable then the terminal n-bond by about1.5 kcals, making 1-butene a minor side product in the reaction in Figure [-2. rn"relative stability of substituted alkenes ii shown in Figure 4_9.

\_-/nHRDisubstituted

R

\--/

H

Monosubstituted

Figure 4-9

Terms such as anti-periplanar orientatiott should sound familiar and immediatelymake you think of an F2 reaction. Any time you see a strong bulky base, ahydrocarbon with a leaving group, and high temperature, you should think of anE2 reaction.

Bt LiNR,

--+AH CH:

CH:

R R'''\_JR' R''

Tetrasubstituted

R' R''

Trisubstituted

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Organic Chemistry tlydrocarbons and Reactions tlydrocarbon Reactions

E1 Reaction (Carried out Under Acidic Conditions at High Temperature)

Tu E-t reaction also yields an alkene, but it goes through a different mechanismthan the E2 reaction. The mechanism for the t1 reactioi is quite similar to that ofthe sI.J1 reaction. In both the E1 and sNJ1 reactions, there is (1) a carbocationintermediate formed after the leaving group leaves in the rate determining stepof the reaction and (2) the possibility of rearrangement with the carbocationintermediate. A schematic for a typical E1 reaction is shown in Figure 4-10.

HsQ ooHzOH --+-n------>

A

CHs

l,,rt\\ CH, cH2cH3

,?_-H

CH:

,rg, + Hro

T'.';*1f\'" '. Hb\

-HA HJ

i

Hb CHs Ha H

Figure 4-10

To recognize E1, as opposed io E2, it is easiest to look at the reaction conditionsE1 reactions are best carried out under acidic conditions while E2 reactionsrequire basic conditions. Both reactions proceed under thermal conditions(elevated temperature) and yield the most substituted alkene as the majorproduct. There may be some of the less substituted alkene formed in smal1quantity. The reaction in Figure 4-11 shows the formation of both the 2-alkenemajor product and the 1-alkene side product.

.atll*HO:

\SH2CH3 7

H-'-'--.-H."O ..S"1:.t,*r"au"R

HsC H

'\) s- ::.

- H3c$\'7-l -:

HrC /, H

H"C CH,

><J

1t

r.3Urclrr,,rp-W, rrrttl CH2C H.H,c7\ /F.",t ts-{./


Figure 4-11

The Berkeley Review

Organic Chemistry Ilydrocarbons and Reactions Ilydrocarbon Reactions

Example 4.10What is the majorconcentrated sulfuric

A.

product whenacid at 50"C?

2-methylcyclopentanol is treated with

CHz

cHs

SolutionAt an elevated temperature in the presence of a strong acid, the reaction proceedsby an E1 mechanism, so the product is a highly substituted aikene. Choice A iseliminated, because it is a terminal alkene. Choice B is eliminated, because it isan alkane and not an alkene. Choice D is eliminated, because it is not the mostsubstituted alkene and it formed via a secondary carbocation, a less stableintermediate than the tertiary carbocation. The best answer is choice C, becausethe n-bond is highly substituted and it includes the tertiary carbon, implying thatit was formed via a tertiary intermediate.

Rearrangement is possible with carbocation intermediates, so E1 reactions aresusceptible to forming rearranged products. Hydride and alkyl shifts are rapid,intramolecular processes that occur before the soivent can remove theneighboring proton to form an alkene.

Carbocation StabilityBecause carbocations are common intermediates, it is important to know therelative stability of the various carbocations. The relative stability of alkylcarbocations is 3' > 2' > 1' > Methyl. For carbocations conjugated to a t-bond,vinylic and benzylic carbocations, there is additional stability because ofresonance. Carbocations can undergo rearrangement by having hydrides oralkyl groups shift, resulting in a different carbocation. For instance, if a

secondary carbocation (R2CH+) is formed, it can rearrange to a tertiarycarbocation (R2R'C+), if a tertiary carbocation is possible. Figure 4-12 shows therelative stabiiity of alkyl carbocations.

B.

".. €'slx: '"{+H: '"" €*= '" :6"'"o:.)@..,,,trtu

3" Carbocation 2'Carbocation 1'Carbocation MethylCarbocation

Figure 4-12

The conclusion that can be drawn from the relative stability of alkyl carbocationsis that methyl groups are electron donating to electron poor carbons, which canbe extrapolated to say that alkyl groups are electron donating. Figure 4-13 showsthree rearrangements where a less stable carbocation is converted into a more


Organic Chemistry tlydrocarbons and Reactions llydrocarbon Reactions

stable carbocation via a hydride shift. The third example forms an allyliccarbocation, which is more stable than alkyl carbocations, beiause of resonance.

HI

H

HI"\_ o_,/cylH-----> "Yt-t\r,

'H

H2")'w;*1'

CH"

H l''\ .1--.n,--------> nf-.vn J

3"H

l" 2" Allvlic

Figure 4-13

which of the following carbocations is apt to undergo rearrangement?A. (H3C)3C+B. (H3C)2CH+C. H3CH2C+D. (H3C)2CHCH2+

SolutionChoice A is already a tertiary carbocation, so it has no reason to rearrange, leastof all to a primary carbocation. Choice A is eliminated. Choice B is a secondarr-carbocation, but can only rearrange to form a primary carbocation, so it iseliminated. choice C is a primary carbocation that only has primary carbons, soit is eliminated. In choice D, a hydride shift can covert a primary carbocationinto a tertiary carbocation, making choice D the best answer.

H

HHHH\/\/c-c H c-c H/ \ o/ ------> / -\ o /'H "YJt"

H ,,f-'Y#.,

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Organic Chemistry tlydrocarbons and Reactions flydrocarbon Reactions

Electrophilic Addition ReactionsAlthough isolated alkenes are not a topic tested on the MCAT, polyenes are atopic, so generic electrophilic addition reactions are viable. Electrophilic additionreactions involve a n-bond actrng as nucleophile by attacking an electrophile toform a substituted alkane product. In the first step of electrophilic additionreactions, the n-bond of the alkene donates its eiectron density to an electrophile.The first step of a generic electrophilic addition reaction is shown in Figure 4-14.

----------t-

Step IE* represents any electrophile (lone pair acceptor)

Figure 4-14

After the electrons from the n-bond are donated to the electrophile, a positivecharge is situated on the most substituted carbon of the original alkene in thecarbocation intermediate. This is the first step in almost all electrophilic additionmechanisms. In the second step of the mechanism, a nucleophile attacks thecarbocation formed in the first step of the electrophilic addition reaction. Thesecond step of a generic electrophilic addition reaction is shown in Figure 4-15.

\-/'H-Nu:J \

-----------.r>

H-NuSteP II

H-Nuc represents any nucleophile (lone pair donor)

Figure 4-15

It is important to recognize that when there is a carbocation intermediate, there ispotential for rearrangement. If the carbocation is unstable, then prior to theattack by a nucleophile, the carbocation can rearrange by way of a hydride shiftor alkyl shift to form a more stable carbocation. Rearrangement is not shown inthis example, but it can occur between steps one and two of the mechanism. Thenucleophile should be a weak base, otherwise it can deprotonate the intermediateto carry out the reverse reaction (elimrnation) and regenerate an alkene. To avoidthis, eiectrophilic addition reactions are carried out under acidic conditions. Thefinal step is the neutralization of the cationic product, which is carried out by a

solvent basic enough to deprotonate the cationic species. The final step of a

generic electrophilic addition reaction, a workup step, is shown in Figure 4-15.

sorv ! -\ \-,/t ----------.>L-# \\-/ Step lll (workup)

Solv represents any polar/protic solvent

Figure 4-16

A protic solvent is capable of forming hydrogen bonds and transferring protons.


Organic Chemistry Ilydrocarbons and Reactions tlydrocarbon Reactions

It is a good idea to understand basic mechanism of electrophilic addition todienes and to recognize the class of reaction. Know how to diaw the reaction ifthey describe the mechanism for the reaction in words. The MCAT doesn't focusmuch on memorization. The focus is on conceptual understanding, so be certainthat the corresponding terms such as Markovnikov, syn and anti ire completeiyunderstood. Later in this section we will look at a variations on this samemechanism with 1,2 versus 1,4 addition reactions of conjugated alkenes.

Regioselectivity (Markovnikov versus Anti-Markovnikov Addition)The concept of regioselectivity occurs when the reactant electrophile and thealkene both lack mirror plane symmetry. This is to say that the double-bondcarbons are not equaily substituted. Because the two carbons are not equallysubstituted, one is less sterically hindered than the other. As a mechanistii rulein electrophilic addition reactions, the first substituent attacks the less hinderedcarbon of the alkene, leaving the second substituent to add to the other carbon.Cenerally, the less hindered carbon of the intermediate is the aikene carbon thatwasn't attacked by the first substituent. A Marlcoanikoa addition product resultsfrom the addition of the electrophile to the less substituted carbon of the alkeneand the nucleophile to the more substituted carbon of the alkene. In the casewhere a strong acid reacts with an alkene the electrophile is a proton, you cansimply say, "H goes where H is." An onti-Markoanikoa ad,dition prod,rct is theopposite of a Markovnikov addition.

Stereoselectivity (Syn versus Anti Addition)The concept of stereoselectivity occurs when the alkene reactant or intermediatehas asymmetric faces, and thus non-uniform steric hindrance. syn additiolrefers to a reaction where the two new substituents add to the same side (face) ofthe alkene reactant or intermedrate. Anti addition refers to a reaction where thetwo new substituents add to opposite sides (face) of the alkene reactant orintermediate. As a mechanistic rule, a substituent attacks at (and adds to) theless hindered face of the alkene (or intermediate). If the two substituents add atthe same time, they add in a syn fashion to the least crowded face of the alkene.If the two substituents add at different times, they add in an anti fashion toopposite faces of the alkene. once the first substituent adds to an alkene, itmakes one side of the intermediate more crowded than the other side. This isreferred to as stereoselectivity, because the face at which the substituent attacksdictates the stereclchemistry of any newly formed chiral centers.

Stereoisomer FormationYou should always consider if stereochemistry is involved in a reaction; whetherit forms a racemic mixture of enantiomers or a major/minor mixture ofdiastereomers. When no chiral center is present on an alkene reactant, there is noasymmetry to influence the reaction. There is an equally likeiy chance to attackthe alkene from either face. The result is the formation of a racemic mixture ofenantiomers, assuming that new chiral centers are formed. When a chiral centeris present on an alkene reactant, its asymmetrv influences the stereoselectivity ofthe reaction. There is a greater chance to attack the alkene from the less hincleredface than the more hindered face. The result is the formation of a major/minormixture of diastereomers. A good rule to follow is that if there is no opticalactivity in the reactants, then there can be no optical activity in the productmixture. This means that the presence of chirality influences further chirality.


Organic Chemistry Hydrocarbons and Reactions llydrocarbon Reactions

1'2-Addition to a Conjugated Diene vs. 1,4-Addition to a Conjugated DieneElectrophilic addition reactions are not iimited to alkenes with isolated n-bonds,as these reactions also take_place on conjugated dienes. with conjugated ciienes,there are multiple potential products. t,z--aaaition refers to an addition reactionthat adds substituents to the first and second atoms (usually carbon) in theconjugated n-network. 1,4-addition refers to an addition reaction that addssubstituents to the first and fourth atoms in the n-network. As a general rule, r,4_addition is favored at higher temperatures, because the more s-table product isformed (thermodynamic control). At lower temperatures, the more stableintermediate dictates the product, so r,z-ad.d.Ltion is favored. This is known askinetic control. There is always the option to add either 7,2 or 1,4 when thefysJem has conjugation. Figure 4-17 shows the product distribution for thel-rydration of 1,3-butadiene at two different temperatures.

H

H,c-.')ot"r.- \ ttr.-

\H2C: CH Lz- -\ H:SOq{aq)-

HC_ CH2 o"c

H.C- CH.\ H2SO,1(aq)

HC- CH2

62%

HC- CH2

16%

HC- CH2OH38%

HC- CH2OH84%

50"c

1,2-AdditonProduct 1,4-Additonproduct

H

)otl-hL- c H,c- cH' \

'r-\

HC- CHz

1,2-Additonproduct 1,4_Additonproduct

Figure 4-12

In the example in Figure 4-77, there are two possible carbocation intermediates.The mechanism and both intermediates ur" ,ho*r, in Figure 4-1g. To start thereaction, the proton adds to the terminal carbon (least hiidered carbon) leavi'gthe carbocation on carbon two (a secondary carbocation). The carbocation isallylic meaning it can resonate through tie n-network. It is in essence apropylene cation that can have cationic density at either end of the rc-network.What this means is that the carbocation can resonate to the terminal carbon(carbon number four). This forms a primary carbocation, whicl-r is not as stableas a secondary carbocation. According to the Boltzmann distribution law, astemperature increases, the higher energy levels become more populated toabsorb this increase in energy. This means that at higher temperatures there aremore primary carbocations than at lower temperatuies. This is why we see theproduct distribution favoring substitution of the alcohol at carbon two at lowertemperatures while we at higher temperatures we observe the substitution of thealcohol at the fourth carbon. Figure 4-18 shows the energy diagram andmechanism associated with this reaction, which includes t*o poisiute iuth*uyr.

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Organic Chemistry Ilydrocarbons and Reactions Ilydrocarbon Reactions

1" carbocation

+2" carbocation is most stable

s1" carbocation is least stable

HOI H H3C-CH H

-J o/, \ /rirC=Cri :HrC-C^\ .e HC-C@' \ )^)^.. ' \

HC:CH2 HC:CH2 H

1f.""H@

HsC-.i""'

Kinetic Pathway Thermodynamic Pathway(favorable at lower temperatures) (favorable at higher temperatures)

H?C- CH"\oHC- CH2OH2HC: CH2

H*

1f.""

1f".1rH

>oHHeC- C-"\ \HC: CH2

Monosubstituted alkeneis least stable

H"C- CH"\HC- CH2OH

Disubstituted alkeneis most stable

Lower energy intermediateleads to the kinetic product.

2' tion Higher energy intermediate leadsto the thermodynamic product.

Figure 4-18

The energy diagram shows the two possible pathways for the reaction and theirrelative energetics. The dashed pathway represents L,2-addition to theconjugated alkene, while the solid pathway represents 1,4-addition to theconjugated alkene. From this data, you should be able to predict the moreabundant product at a given reaction temperature.

Icarboca

Reaction Coordinate *--*-+


Organic Chemistry flydrocarbons and Reactions Ilydrocarbon Reactions

Pericyclic ReactionsPericyclic reactions involve the repositioning of both sigma-bonds and pi-bondsthrough a cyclic transition state. These reactions are believed to be concerted,meaning that the formation and breaking of all bonds occur simultaneously.Pericyclic reactions include cycloaddition reictions,which most notably include theDiels-Alder reaction, sigmatropic rearrnngement, and electrocyclic reactiors. We willaddress only cycloadditions and sigrnatropic rearrangement. The significantdifference between a cycloaddition reaction and sigi-Latropic rearrangementinvolves the number of molecules. In cycloaddition, two separate compoundscome together, resulting in a single new compound. In sigmitropicrearrangement reactions, it is an intramolecular rearrangement that takes place.

Paramount to understanding these reactions is having a good idea about orbitaloverlap in both sigma-bonds and pi-bonds. In everything we'll address in termsof the MCAT, we shall only consider the positioning oith" atoms and not thespin of the electrons within the moleculai orbitals. the first reaction we shallconsider is the Diels-Alder reaction.

Diels-Alder ReactionThe Diels-Alder reaction, an electrocyclic addition reaction, involves the additionof a conjugated diene (4 n-electrons) to an alkene (2 n-electrons) to from a sixmembered cyclohexene ring. The transition state for a Diels-Alder reaction issimilar to the resonance of benzene, as shown in Figure 4-19.

_->

Figure 4-19

six n-electrons in a cyclic n-network make benzene aromatic, so we refer to thetransition state of a Diels-Alder reaction as aromatic (containing 6 n-electrons in aring). Diels-Alder reactions involve the addition of a 1,3-diene to a dieneophile.The diene must have cis orientation about the central sigma bond to undergo aDiels-Alder reaction. A sample Diels-Alder reaction is driwn in Figure 4-20.

dieneophileo

.A( CHstcH2cH2

49Diels-Alder Reaction

diene

1

22"

,(+4

1

2//\6

^*ll l+3\-/s

4

First: draw thecyclohexene ring

Figure 4-20

cyclohexene derivativeo

CHg

cH2cH3

Second: draw the rest ofthe molecule connectedto the numbered carbons


Organic Chemistry llydrocarbons and Reactions Hydrocarbon Reactions

The six membered ring that is formed is cyclohexene. The carbons are numberedto help identify the product, which will make large, polycyclic products easier toevaluate. Figure 4-21 shows a more complex Diels-Alder reaction. Because boththe diene and dienophile have substituents, there is a chance that stereoisomerscan form. The stereoselectivity is driven by orbital overlap in the transition state.The two stereoisomers products, diastereomers, are drawn for the reaction.

1

o4

EWG

/)

\

<:+

zl.H?co .s- 6*

Figure 4-21.

The last thing for us to consider is regioselectivity. when the diene anddienophile have substituents, there exists the potential for different structuralisomers. Regioselectivity can be predicted using resonance, where the mostelectron rich terminal carbon of the diene attacks the electron poor carbon of thedienophile. The reaction is optimized when the dienophile has electronwithdrawing groups. Figure 4-22 shows the effect of electron donating andelectron withdrawing groups on the diene through resonance.

G EDG

EWG = electron withdrawing group EDG = electron donating group

Figure 4-22

when it comes to predicting regiochemistry, it's as simple as plus attracts minus.Figure 4-23 shows a Diels-Alder reaction where regioselectivity is an issue.

o

\o

ro

^*:o:::;1"no

Endo Product

oEW

)

l_-.-.

o

*

oEDG

t)4l-/ €<)

'a:

H --+A

H3CO CHsMajor

Exo Product


Figure 4-23

The Berkeley Review

Organic Chemistry Ilydrocarbons and Reactions flydrocarbon Reactions

Example 4.12\A/hich of the following Diels-Aider reactions is fastest?

A.

B.

C.

D.

2l*HrC\

2l*nrao\

(+

H.c^

2I,"r.o\

o

4.",ll --+\A

CHa

o

4.".tl- -^-CHa

CHe

CHa

o

,4.',,Ar-H:C -CHs

H3CO

-^>H3CO

CH:

SolutionThe rate of a Diels-Arder reaction is increaseci by the presence of an electrondonating_ group on the diene and an erectron withdrawing group on thedienophile. The rate of a Diels-Alder reaction is decreased by the presence ofbulky groups in the transition state. Choices C and D are eliminated because ofthe two methyl groups on the dienophile. No matter how the molecure alignsentering the transition state, one of the methyl groups wiil be in the way. Inchoice B, there is a methoxy group on the diene, *-hil" ,^ choice A is just a methyr,grou.p.

Methoxy groups donate electron density through resonance, so choice Bhas the more electron rich diene, resulting in a faster reaction than choice A.

Cope RearrangementThe Cope rearrangement is a sigmatropic rearrangement i.vorving two pi-bondsand one sigma-bond. Figure 4-24 shows a simpl"e Cope rearrangement carriedout on a 1,S-diene. In more comprex examples, stereo.i-rerr-ristry rnay be an issue,because stereocenters can be both formed and rost as hybridizaiior-, changes.

HaC

4,-,,.4,

'G-\g --+A

Figure 4-24

cHs

CHs

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Organic Chemistry Hydrocarbons and Reactions Hydrocarbon Reactions

The Cope rearrangement involves two pi-bonds and one sigma-bond aligned insuch a way that the terminal p-orbitals of the two pi-bonds are close enough tooverlap. The reaction requires the input of energy to overcome the activationbarrier. The structure of the product is dictated by the orbital overlap in thetransition state.

Claisen RearrangementThe Claisen rearrangement is similar to Cope rearrangement, except that in aClaisen rearrangement, the reactant is a vinylic allyi ether. The rearrangementinvolves two pi-bonds and one sigma-bond and has a transition state that issimilar to the one observed with the Cope rearrangement. The difference is thepresence of an oxygen. Figure 4-25 shows a Claisen rearrangement.

o^\,

Figure -l-25

when the ether is benzylic instead of vinylic, the cyclic ketone can quicklytautomerize to form a phenol. The preference of a phenol over the cyclic ketoneis due to the aromaticity of the benzene ring.

Example 4.13\z\4rich statement is valid in terms of the sigmatropic rearrangement reactions?

A. Aldehydes are formed from a Claisen rearrangement of a vinylic allyl etherwhen the allylic ether carbon is unsubstituted.

B. Aldehydes are formed from a Claisen rearrangement of a vinylic allyl etherwhen the vinylic ether carbon is unsubstituted.

c. Aldehydes are formed from a Cope rearrangement of a vinylic allyl etherwhen the allylic ether carbon is unsubstituted.

D. Aldehydes are formed from a Cope rearrangement of a vinylic aliyl etherwhen the vinyiic ether carbon is unsubstituted.

SolutionChoices C and D are eliminated immediately, because Cope rearrangementresults in the conversion of one 1,5-diene into another 1,5-diene, not an aidehyde.The question is reduced to determining which carbon in the reactant forms thecarbonyl group following Claisen rearrangement. The reaction is shown be1ow.

asw ---l>A

HIvinylicT\

---l>altvtic etner

l

aidehyde

It is the vinylic ether carbon that becomes the carbonyl carbon, not the allylicether carbon. This means that the vinylic ether carbon must only have a

hydrogen, and no carbons, in order to form an aldehyde and not a ketone. Thebest answer is choice B.


Organic Chemistry tlydrocarbons and Reactions Terpenes

Terp,€he$Classification

Terpenes and terpenoids, btological molecules derived from terpenes, arenatural hydrocarbons found in plants and animals that are made from S-carbonisoprene (2-methyl-1,3-butadiene) units. The five-carbon skeleton of isoprene canbe found in terpenes. Terpenes are classified by their number of carbon atoms.Monoterpenes have ten carbon atoms, sesquiterpenes have fifteen carbons,diterpenes have twenty carbons, sesterterpenes have twenty-five carbons and soon, The in aitro synthesis of terpenes and terpenoids is called nstural productsynthesis. Some naturally occurring monoterpenes are shown in Figure 4-26.

Myrcene(Oil of Bay)

o,-Terpinene(Oil of Coriander)

I

^ttY\oHi. -Menthol

(Oil of Peppermint)

Figure 4-26

Citronellol

y-Terpinene(Oil of Coriander)

CH"t'u? lS."-\1./

CH:cr-Pinene

(Oil of Turpentine)

Dimethylallyl

ooililo- P- o- P- oH

lloo-Isopentyl pyrophosphate

Figure 4-27

These molecuies add to one another in a way where the n-bond of isopentylpyrophosphate is the nucleophile and pyrophosphate of another molecule is theleaving group. A proton is lost from the nucleophilic moiety to regenerate a ru-

bond. The reaction involves head-to-taii addition. When cyclizing, the bond thatis formed to complete the ring is rarely connected head-to-tai1. Figure 4-28 showsthe reaction of isopentenyl pyrophosphate and dimethylallyl pyrophosphate.

Geraniol(Oil of Germanium)

Limonene(or Limin)o

Carvone(Oil of Spearmint)

L,Isoprene

Studies in biogenesis show that the large terpenes are synthesized usingisopentenyi pyrophosphate rather than isoprene. Pyrophosphate adds across thediene of isoprene to form either isopentenyl pyrophosphate or dimethylallylpyrophosphate, which are interconverted by isomerization. Figure 4-27 showsisoprene, isopentenyl pyrophosphate, and dimethylallyl pyrophosphate.

ooillto- P- o- P- oHtlo- o-pyrophosphate

r CH,OgHCitronellal

r CH.ON"Citral

(Oil of Lemongrass)

H"C. -CH""XJ

LRHeC o

Camphor

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Organic Chemistry Hydrocarbons and Reactions Terpenes

Isopentyl pyrophosphate

ooll ll - r

HO- P- O- P- O- Y

ooll il Geranyl pyrophosphate

ooililHO- P- O- P- Otto- o-

Figure 4-28

Both plants and animals synthesize terpenes. Larger terpenes are built frommultiple additions of isoprene units, including both isopentenyl pyrophosphateand dimethylallyl pyrophosphate. Geranyl pyrophosphate (a C-10 terpenederived from the head-to-tail connection of two isopentenyl pyrophosphatemolecules) is the first monoterpene in many natural synthetic pathways.Another isoprene unit can be added to geranyl pyrophosphate to form farnesylpyrophosphate (a C-15 terpene). These molecules can undergo further addition,dimerization, or modification into other terpenes and terpenoids. Figure 4-29shows a generic pathway for the biosynthesis of larger terpenes.

Iloo4./\o-li-o-ll-o.,

tto- o-Isopentyl pyrophosphate

(3-Methyl-3-butenyl pyrophosphate)

Monoterpenes (C1s)

l"o,.^^o-ll_o_ll_on

or Dimethylallr, Orr3pn"rf,i",", (3-Methyl-2-butenyl pyrophosphate)

I Isopentyl

{PFophosthate

ooillto- P- o- P- oHtt

+_ Geranyl pyrophosphate O- O-(C1o-pyrophosphate)

ttsesquiterpenes (C15)

- Farnesyl pyrophosphlt" \ o- o-

(C,5-pyrophosphate) \ ,

I tsoPentYl squalene (c36)

I pyrophosphate I

j rr-"r--""r-."".- tDiterpenes(C2s) # C2g-pyrophosphate

I

VTetraterpenes (Ca6)

Figure 4-29

Lanesterol (C36)

{Chloesterol (C27)

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Organic Chemistry ffydrocarbons and Reactions Terpenes

As Figure 4-29 shows, the triterpene squarene can undergo further reactivityto generate cholesterol, which does not have a number of carb'ons that is divisibleby five. so while cholesterol may not be a terpene, its synthesis involves terpenesand terpenoids. The basic schematic for the biosynthesis of cholesterol startinjfrom isopentenyl pyrophosphate is shown in Figure 4_30.

ooililo -P- O- P- OHj. f"

Isopentenyl pyrophosphate

Figure 4-30

, As a general rule, smaller terpenes are found primarily in plants, while somelarger terpenes, such as lanesterol (a C-30 precursor to steroid hormones) and B-carotene (a C-40 source of vitamin A), aie found in plants and animals. Forinstance, the monoterpene pinene is found only in plants while Vitamin 41, aditerpene, is found in both plants and animals. Frg.rr" 4-31 shows some selectedlarger terpenes.

Elemene(oiiofCoral)

Z+

?orFarnesol

(Oil of Lemon)

"

Vitamin .A,1G-Selinene

(Oil of Celery) WHeC CHg

Figure 4-31


CH: CHs OH

Exclusive MCAT preparation

Organic Chemistry flydrocarbons and Reactions Terpenes

One of the skills that you must deveiop to do well on terpene relatedquestions on the MCAT is to recognize terpenes and be able to identify theisoprene subunits in the carbon skeleton. Figure 4-32 shows the analysis of someterpenes for the isoprene units in the skeletal fragments.

+Me

o,-Terpinene

Me

Me

y-Bisabolene

^4{- ^'r ,-

ll t

,ll

t,,-=A

^4;

'

Me

"*AM

+ixPatchouli Oil

Figure 4-32

A common lab technique employed to isolate terpenes is steam distillation,where the pulp of some natural material is placed into water and boiled so thatthe natural oiis are distilled from the pulp. Steam distillation allows the essentialoils to vaporize at a temperature lower than their boiling point, so they do notdegrade. The distillate is a mixture of water and terpenes, which are easilyseparated using extraction techniques. Terpenes can also be extracted from pulp.When isolating terpenes, it is a mixture of geometrical isomers that is collected.The different geometrical isomers of a terpene are given the prefixes a"-, B-, y- andso on. The different geometrical isomers have similar physical properties, butbecause of differences in conjugation, they exhibit differences in the absorption ofphotons in the ultraviolet (UV) and visible range of the EM spectrum.

Terpenes are UV active, because of their n-bonds. An isolated alkene has a

UV absorbance around 180 nm. A conjugated diene has a UV absorbance around225 nrrl, which is significantly more intense than the absorbance of an isolatedalkene. As the conjugation of a n-network increases, the wavelength ofmaximum absorbance, l,*u", and the intensity of absorbance, t, increase. Someterpenes contain oxygen, which is added in a way that does not alter the carbonskeleton. Carbonyls exhibit absorbances of greater wavelength than alkenes ofthe same conjugation, For instance, carvone (shown in Figure 4-26) is evident b,v

a carbonyl absorption at 7744 cm-1 in its IR spectrum and a strong UV (e >10,000) absorption at L,p31 = 242 nrrr. Terpenes are often isolated in educationallaboratory experiments. Because of their biological significance and the fact thevare isolated in lab experiments, they are highly represented on the MCAT. If youhave a fundamental understandrng of terpenes, then you should be fine.

Zingerberine


Organic Chemistry flydrocarbons and Reactions Section Summary

Key Points for Hydrocarbons and Reactions (Section 4)Alkanes

1' Hydrocarbon compounds with only carbons, hydrogens, and sigma bondsa) Only contain C-C and C-H single bonds

i. Can be aliphatic (straight chain) or cyclicii. Low water soiubility, row boiling point, and row melting pointiii. Relatively inert compounds that are used as solvents

b) Undergo free radical halogenation reactions with chlorine and brominei. Involves initiation, propagation, and termination in that orderii. Bromination is more selective than chlorinationiii. The relative stability for alkyl free radicals is: 3" > 2' > l" > methyl

Hydrocarbon Reactions (Reactions involving ru-bonds)1. Elimination

a) Loss of an H and a leaving group to form a ri-bondi. Requires high temperatureii. Competes with nucleophilic substitution reactionsiii. Goes by way of one of two mechanisms: E1 or E2

b) E1 reactions are similar to 5511 reactionsi. Requires a strong acid (Bronsted-Lowry or Lewis)ii. Forms carbocation intermediate so rearrangement is possibleiii' Forms most substituted and least sterically hindered alkene

c) E2 reactions are similar to 5512 reactionsi. Concerted reaction that requires a strong, bulky baseii. Proton to be lost and the leaving group must be anti to one anotheriii. No intermediate formed, only a transition state

2. ElectrophilicAddition

a) An electrophile can be added to a n-bond followed by nucleophilic attacki. The n,bond is a weak nucleophileii. The reaction is driven by the strength of the electrophile

b) Electrophilic addition reactions exhibit regioselectivityi' Steric hindrance and carbocation stability influence the site of attackii. when the electrophile attacks the ress substituted carbon, the

reaction is said to be a Markovnikov additioniii. when the electrophile attacks the more substituted carbon, the

reaction is said to be an anti-Markovnikov additionc) Electrophilic addition reactions exhibit stereoselectivity

i. Steric hindrance is the most influential factor in stereoselectivityii. when substituents add one at a time (the first is added before the

second on attacks) the product exhibits anti addition stereochemistryiii. when substituents add simultaneously (both add at the same time)

the product exhibits syn addition stereochemistry


Organic Chemistry Ilydrocarbons and Reactions Section Summary

3. 1,2-Additionversus 1,4-Addition

a) Electrophilic addition reaction can proceed in multiple ways withconjugated n-systems

i. Typically, \,2-addition is favored at lower temperatures (kineticcontrol)

ii. Typically, 1,4-addition is favored at higher temperatures(thermodynamic controi)

4. Diels-AlderReaction

a) Reaction of a diene and a dienophile (alkene)

i. An electrocyclic reaction carried out with either light or heat

ii. Forms a cyclohexene productiii. Stereoselectivity: Endo product is preferred over exo productiv. Regioselectivity: Depends on the resonance nature of the groups on

the reactants

5. Sigmatropic Rearrangement (Cope and Claisen Rearrangements)

a) Both sigma-bonds and pi-bonds are broken and formed via a cyclictransition state

i. Cope rearrangement converts a y,8-unsaturated alkene into anothery,8-unsaturated alkene via a realignment of molecular orbitals

ii. Claisen rearrangement converts a vinylic allyl ether into another y,5-

unsaturated carbonyl via a realignment of molecular orbitalsiii. Sigmatropic rearrangement requires heat

Terpenes

1. Natural products derived from the connecting of five-carbon units

a) Derived via biosynthesis involving either isopentyl pyrophosphate ordimethylallyl pyrophosphate

b) Can be cleaved into isoprene subunits

i. Terpenes are named for their carbon count: 10 C = monoterpene, etc.

ii. Isolated by steam distillation or extraction as natural oilsiii. Presence of n-bonds results in UV absorbances. As conjugation

increases, intensity and l"*u* both increase.


Alkanes : and,, ftr5rdroctrrbon' Keaction:s

I. Free Radical Halogenation Selectivity

II. Free Radical Reactions

III. Elimination and Stereochemistry

IV. Elimination Study

V. Phermones

VI. Creen Synthesis

VII. Conjugated n-Networks

VIII. Diels-Alder Reaction Rate Study

IX. Diels-Alder Reaction

X. Claisen and Cope Rearrangements

XI. Isoprene Units

XII. Terpenes

XIII. Fatty Acids and Oils

XIV. Occidentalol Synthesis


Alkanes and Hydrocarbon Reactions Scoring Scale

Kaw Score MCAT Score

84 - 100 l5-1566-85 10-1247 -65 7 -9

34-46 4-6t-33 t-5

(r -7)

(B - 14)

(r5 - 2t)

(22 - 28)

(2e - 35)

(36 - 4t)

(42 - 48)

(4e - 54)

(s5 - 60)

(6r - 67)

(68 - 73)

(74 - BO)

(81 - 86)

(87 - e2)

(e5 - loo)

Passage I (Questions 1 - 7)

Alkyl halides can be lormed tiom a halogen reacting withan alkane. Alkanes can be treated riith either chlorine gas andUV radiation to form chloroalkanes. or rvith bromine liquidand UV radiation to form bromoalkanes. The bromination ofan alkane is a slower and more selective reaction than thechlorination of an alkane. Because of this greater selectivity,bromination is preferred over chlorination in the synthesis ofhighly substituted haloalkanes.

For the chlorination of an alkane, the reactivitypreference for carbon substitution through liee radicalsfollows the trend 3'carbon > 2" carbon > 1'carbon by a

factor of 4 : 2.5 : I at a given temperature. This means thatfor a compound like butane with four secondary hydrogensand six primary hydrogens, the ratio of chlorination productsis not based on random probability aione. Randomprobability predicts the formation of two 2-chlorobutanes forevery three 1-chlorobutanes. Because of the reactivitypreference of secondary carbons over primary carbons by aratio of 2.5 : l, the product distribution instead is five 2-chlorobutanes to three 1-chlorobutanes. This implies that thepercentage of secondary products is 62.5Vo (rather than 40Vowhich is expected when there is no site preference).

The calculation of the number of products is based on thereactivity coefficient times the number of unique hydrogens.In a molecule like pentane, there are three types of hydrogensin a 3 : 2'. I ratio. Six hydrogens are secondary and six areprimary. The abundance of the primary hydrogens is 6hydrogens x 1 reactivity for 6. The abundance for carbon twoand carbon four is 4 hydrogens x 2.5 reactivity for 10. Theabundance for carbon three is 2 hydrogens x 2.5 reactivity for5. This means that the chlorination product ratio is six l-chloropentane to ten 2-chloropentane to five 3-chloropentane.

1 . How many degrees of unsaturation are there in thecompound C5H9C1?

A. 0

B. 1

C,2D. 3

2 . How many structural isomers of C5H11Cl are there?

A. 5

B. 6

c.1D. 8


3 . The chlorination of methylcyclopentane would yieldhow many different structural isomers?

4.2B. 3

c. 4

D. 5

4. What is the most abundant product in the brominationof 2-methylbutane?

A. 1-bromo-2-methylbutane

B. 2-bromo,2-methylbutane

C . 2-bromo-3-methylbutane

D. 1-bromo-3-methvlbutane

5 . Following the free radical monochlorination of pentanewhat is the ratio of2-chloropentane to 1-chloropentane?

A. 2:3B. 5:3c.2.5:1D. 3: 1

6 . In the monochlorination of n-hexane, how can a ratio of1.07 2-chlorohexane to one 3-chlorohexane beexplarned?

A. Based on the relative reactivity of the carbons andthe abundance of hydrogens, a ratio of 1.07:1 isexpected.

B. The second carbon of n-hexane is less stericallyhindered than the third carbon.

C. The second carbon ofn-hexane can better stabilize afree radical due to resonance.

D. The third carbon of n-hexane can better stabilize a

free radical due to the inductive effect.

7 . If the alkane reactant exhibits ring strain, the instabilityof the free radical intermediate may cause the ring tobreak. Which free radical alkane is LEAST stable?

A . Propyl free radical (C:HZ.)

B. Cyclopropyl free radical (CSHS.)

C . Butyl free radical (C+HS.)

D. Cyclobutyl free radical (CqHl.)


A halide can be substituted onto an alkane by way of afree radical mechanism. Halogenation of an alkane isinitiated by the homolytic cleavage of a diatomic halogenmolecule into free radical halogen atoms. During subsequentsteps in the reaction, an alkane reacts with the halogen freeradicals to form an alkyl halide. The reaction requires somesource of activation energy to cleave the halogen-halogenbond. Depending on the halogen-halogen bond strength, theamount of activation energy necessary varies from thermal toultraviolet radiation for the initiation step.

The mechanism is a sequence broken down into stepsthat fit into one of three categories: initiation, propagation,and termination, in that order. The initiation step involveshomolytic cleavage of a halogen-halogen bond to form twofree radicals. The second phase of the reaction sequence ispropagation where the free radical is transferred through a setof abstraction reactions. The last phase of the sequence is thetermination step where two free radicals combine to fbrm asigma bond. The reaction involves two transition states. inwhich the second is of higher energy than the first. Table 1

shows the bond energies of the halogens and the reactionenthalpies for the various halogenation reactions:

Compound B.D.E. AH.t

F2 154 kJ

mole-483 kJ

mole

Clz n9 kImole

114 kJmole

Br2 193 kJ

mole-33 kJ

mole

I2 149 kJ

molet1'.. kJ

mole

Table I

The enthalpy of a chemical reaction can be found byusing Equation l.

AHr^n = I Energy6onds broken - Energy6on6s fbrmed

Equation 1

The average bond dissociation energy {br a sigma bondbetween an spJ-hybridized carbon and ahydrogen is 413 kJper mole. A hydrogen free radical cannot be formed in thisreaction mechanism.

8 . The MOST stable type of carbon fr-ee radical fonned inthe monobromination of (R)-3-methylhexane is bestdescribed as:

A. primary.

B. seconclary.

C. tertiary.

D. quatefnary.


Reaction Coordinate

9. Which of the following energy diagrams corresponds toreaction of 12 with an alkane?

1 0. The first propagation step in a free radical reaction iswhich of the following?

A.XZ + R. -> RX + X.

B. RH + X. -> HX + R.C.Xz -> 2X.D. X. + R. -> RX

1 1. The strongest halogen-halogen bond corresponds towhich of the following?

A. The shortest halogen-halogen bond.

B. The second shortest halogen-halogen bond.

C . The longest halogen-halogen bond.

D. The second longest halogen-halogen bond.

12. Which of the following steps is NOT found in a freeradical halogenation reaction ?

A. RH + X. -> RX + H.B. XZ + R. -> RX + X.

C. RH + X.-> HX + R.

D. X. + R. -> RX

B.

Ibo

EnReaction Coordinate

Reaction Coordinate Reaction Coordinate

13. Which of the following conclusions can be inferredfrom the observation that usually only one halide reactswith the alkane and minimai multiple halogenatedproducts ar.e isolated from the product mixture?I. Halogens do not help to stabilize free radical

intermediates.

II. Halogens, once on an alkane, increase the C-Hbond strength.

III. Halogens, once on an alkane, decrease the C_Hbond strength.

A. II onlyB. III only

C. I and II only

D. I and III only

1 4. Given that a C-F bond energy js 462 kJ per mole and aH-F bond energy is 588 kJ per mole, what is the heat ofreaction associated with fluorination of an alkane?

A. - 483 kJmole

B. -lt6 kJrnole

C. +116 kJmole

D. + +83 kJmole


Passage lll (Questions 1S- 21)

Due to the usefulness of alkenes in synthesis, they areoften a starting material in many synthetic transformations.This can be attributed to the numerous addition reactionsalkenes undergo. Because of their synthetic usefulness, it isimportant to be able to synthesize alkenes in a geometricallyspecific manner. This is to say that it is beneficial to be ableto create predominantly the trans (E) geometrical isomer orpredominantly the cis (Z) geometrical isomer. Reactionsproceeding by the E1 mechanism result in the formation ofmostly the trans geometrical isomer with some cisgeometrical isomer formed. The E2 mechanism, on the otherhand, allows for the formation of either the trans or cisgeometrical isomers in high purity, if there are chiral centerspresent in the reactant. Reaction 1 is an E2 reaction carriedout in the hopes of synthesizing Z-3-methylpentene, to beused in subsequent steps ofg total synthesis process:

BrCH"' rbutOK-->

t-butOH

HqC CHrCH'

><H CHrH:C CH2CHj

Reaction I

Reaction 1 proceeds by an E2 mechanism at elevatedtemperature, so the bromine leaving group and the hydrogenon carbon 3 must be aligned in the anti orientation. Asdrawn in Reaction 1, the bromine leaving group and thehydrogen on carbon 3 are not correctly-al-igned for f2elimination, nor is the structure drawn in its most stableconformation. Counting the conformer shown in Reaction l,there are three staggered conformations total for the reactant.Only one of the three orientations has the anti orientationnecessary for the E2 reaction.

15. What conclusion can be made about the eliminationreaction that generates the following data?

Trial (H3C)3CBr t-ButOK Rate

I 0.25 M 0.25M 4.61 x t0-3 M/s

il 0.50 M 0.25M 9.29 x t0-3 Mls

III 0.40 M 0.50M 1.47 x 10-2 Mts

The reaction proceeds by an E1-mechanism,because the data indicate that the rate-determiningstep is unimolecular.The reaction proceeds by an E1-mechanism,because the data indicate that the rate-determiningstep is bimolecular.

C , The reaction proceeds by an E2-mechanism,because the data indicate that the rate-determiningstep is unimolecular.

D. The reaction proceeds by an E2-mechanism,because the data indicate that the rate-determiningstep is bimolecular.

A.

B.

16. Which of the staggered complexes is MOST stable forthe reactant?

A. The conformer drawn in the example.

B. The conformer fiom which the E2 eliminationtakes place.

C . The conformer with carbons I and 4 anti to one

another.

D. All three staggered conformers all equal in stability.

17. The loss of optical rotation in the reaction can be

explained by which of the following statements?

A. Ez eliminates the carbon 2 chiral center.

B. EZeliminates the carbon 3 chiral center.

C . Ez eliminates both the carbon 2 chiral center and

the carbon 3 chiral center.

D . The product is a pair of enantiomers.

I 8. Which of the following structures (from the perspective

of the eye) is the Newman projection for the reactant?

CH:

H$t),H:C CHzCHT

B.

Br

A.Br",$.".CH2CHj

c.Br

;"4;CH2CH3

CH2CH3

D.Br

;"V:CH2CH3

CH:

H

19. The ENANTIOMER of the reactant has which of the

followin g stereochemical orientations ?

A. 2R, 3R

B. 2R, 35

C. 25,3RD. 25, 35


20. What is the role of concentrated sulfuric acid in an E1

elimination reaction?

A. Sulfuric acid protonates the leaving group makingit a better leaving group.

B. Sulfuric acid serves to dehydrate the solventpreventing back reaction.

C. Sulfuric acid dissociates into sulfate which helps

remove the proton allowing the leaving group toleave.

D . Sulfuric acid stabilizes the carbocation intermediateby protonating the cationic carbon.

21 . How many units of unsaturation are present in the

product of chlorocyclohexane and strong base and heat?

A. 0

B. Ic.2D.3

292 GO ON TO THE NEXT PAGE copl:

prurea{reag

)'t


Elimination results in the formation of a new n-bondfollowing the loss of two groups from the molecule. Thereaction can proceed by one of two possible mechanisms: E1

or 82.

In E1 reactions, the leaving group first leaves to form a

carbocation intermediate. The carbocation has the potentialto undergo rearrangement. The solvent serves as a base and

deprotonates a hydrogen off of the carbon adjacent to thecationic carbon. The result is the formation of a new fi-bond.

In E2 reactions, the leaving group leaves simultaneouslyas the proton is removed by a strong base. The proton and

leaving group must be anti to one another in a staggeredconformation to undergo an E2 reaction. An E2 reactionrequires that the base be strong, to remove a weakly acidichydrogen, and bulky to reduce the amount of side productsformed from nucleophilic substitution. Figure I shows threeelimination reactions carried out concurrently.

Reaction I:CI

Yg

@Reaction II:

CI&Reaction III:

OH&

t-butOK-----'+t-butOH,50"C

t-butOK-..+rbutOH,50'C

conc. HrSOa_->

60'c

Figure L. Three Elimination Reactions

The product shown in each of the reactions is the majorproduct. Among the minor side products for each of the threereactions is the alkene products from one of the other tworeactions.

2 2. Which of the elimination reactions in Figure 1 involvesrearrangement?

A. Reaction I only

B. Reactions I and II only

C . Reactions II and III only

D. Reaction III only


23. Which of the reactions results in an optically activeproduct mixture?

A. Reaction I only


C . Reactions II and III only

D . Reaction III only

24. When the following reaction is carried out, why does

the optical activity disappear?

conc. HrSOa Major ProductcH3 60- (crp=6';

CHr

A. After the sulfate group substitutes for the hydroxylgroup, the chiral centers cancel one another.

B. The product is meso.

C . The major product is an achiral alkene, resultingfrom rearrangement.

D. The product is an achiral alcohol.

2 5 . Which of the following observations are consistent withthe mechanisms discussed in the passage?

I. Increasing the base concentration in Reaction Iincreases the reaction rate.

II. The amount of alkene product is maximized atlower temperatures.

III. Reaction II has a competing an Sp2 reaction.

A. I onlyB. II onlyC. I andll onlyD. I and III only

26. Why do Reactions I and II yield different majorproducts?

A. Et reactions can undergo rearrangement if the

hydrogen is on the correct side of the plane.

B. Et reactions are influenced by steric hindrance.

C . EZ reactions require that the leaving group and

proton are anti to one another.

D . EZ reactions require that the leaving group and

proton are syn to one another.

27 . How does the hybridization change fbr the carbonscommon to both rings in Reaction III?

A, sp2 to sp3

B. spz to sp

C. sp3 to sp2

D. spi to sp

28. If diethyl amine, (H3CCH2)2NH, is used in Reacrion Iinstead of t-butOK, what is the major product?

a'E,rN



Phermones are chemicals secreted by animals (mostcommonly insects) that elicit a specific behavioral reaction inother members of their same species. They are effective inlow concentration in sending signals between members of thesame species for such things as reproduction, dangerwarnings, and aggregation (in the case of a food supply.)Many phermones are simple hydrocarbons. For instance,when in danger, ants secrete undecane (.Cy1H2$ or tridecane(Cf :HZS) to inform other ants of the trouble. Many of thetraps and sprays we use to capture and kill insects takeadvantage of sex attractants. The structures of four sexphermones are shown in Figure 1.

Tiger Moth sex attractant(2-Methyl heptadecane)

o

o4.",Oriental Fruit Moth sex attractant

[(E)-8-Dodecen- I -yl acetare]

HpCa .CttHzt\J/\HH

House Fly sex attractant(Muscalure [(9Z)-Tricosenel)

Silkworm Moth sex attractant(Bombykol)

Figure 1 Four random phermones

Phermones are specific to each species, because receptorproteins are highly selective in what they bind. In one of therare cases where two geometrical isomers both elicit the sameresponse, the Oriental Fruit Moth responds to both the E-isomer, shown in Figure 1, and the Z-isomer. There arecases where two similar species to a phermone that is similarin structure, but not exactly the same. For instance, theGrape Berry Moth uses (Z)-9-dodecen-1-yl acetate as a sexattractant in roughly the same concentration that the OrientalFruit Moth uses (Z)-8-dodecen-1-yl acatate.

29. The Silkworm Moth sex phermone has all of thefollowing structural features EXCEPT:

A. one stereogenic center.

B. one cis double bond.

C. no tertiary carbons.

D. conjugation.

30. Relative to the Oriental Fruit Moth sex phermoneshown in Figure 1, the follorving compound is:

oil

oA cH.

A. a confbrmational isomer.

B. a geometrical isomer.

C . an optical isomer.

D. a structlu'al isomer.

31.. Which of the following statements accuraiely relates the

four structures shown in Figure 1?

I. Muscalure has a shorter wavelength of rnaximumabsorbance in UV-visible spectroscopy bombykol.

II. The Tiger Moth sex phelmone has the more unitsof unsaturation than undecane.

m. The Oriental Fruit Moth sex phermone can beclassifled as a terpene.

A. I only

B. iII only

C. I and II only

D. i and III only

3 2. What physical property is NOT expected for muscalure?

A. Low miscibility in water

B. A boiling point above room temperature

C. High lipid solubilityD. The ability to rotate plane-polarized light

33. Which of the phermones in Figure t has the greatestnumber of primary carbons?

A . The Tiger Moth sex phermone

B. The House Fly sex phermone

C. The Oriental Fruit Moth sex phermones

D. The Silkworm Moth sex phermone

34. The Green Peach Aphid def'ense phermone is shownbelow.

A. It can be synthesized from isoprene units.

B . It has *o.e sp2-hybridized carbons than spj-hybridized carbons.

C . It can potentially undergo 1,4-addition.

D. lt is highly flexible.

What is NOT true of the structure?

Copyright O by The Berkeley Review@ 295 GO ON TO THE NEXT PAGE

3 5. Which spectroscopic observation does NOT correlatewith the conesponding compound?

A. The Oriental Fruit Moth sex phermone: an IRabsorbance at 1141 cm-l

B. Bombykol: a UV-visible absorbance at22'l nm

C . Muscalure: two signals around 5.00 ppm in itsluNIrvlR

D . The Tiger Moth sex phermone: 14 signals in itsl3cNltR spectrum

Passage VI (Questions 36 - 41)

In the recent years, many chemists around the world haveshifted their focus to so called green chemistry. Greenchemistry, also called sustainable chemistry, aims to developchemical reactions and processes that are environmentallysafe. The goal is to reduce waste generation rather thanernploying waste management--the "end of the pipe"solution. The most significant alteration to traditionalchemistry is the recycling of solvent, or in optimal cases, theelimination of solvent. This is achieved in many ways,including doing reactions under high pressure to make thesystem act like a supercritical f'luid.

Areas of current research in green chemistry include theuse of renewable raw materials, direct oxidations usingoxygen, improved separation during the course of a reaction,and all forms of catalysts. The aim is to maximize atont-economy, the tracking of how many atoms used in thereaction end up in the product, by not using solvents orprotecting groups. Figure I lists three atom-economicalreactions used in green synthesis.

Reaction I:

H:CCOzCHT

.+ i.sd'u.":.:.co2cHj

98%o vterd

Reaction II:

H2C= CHCHj

o

AH CH:CHzCHr+CO+Ho-€- cat. Rh O

JtH CH(CHr)r

Pd, H2

Supercritical CO2

cHr lHrC Hrc

> 997o yield

Figure 1. Three Atom-economical Green Syntheses

Reaction I is a Diels-Alder reaction. Reaction 2 is a

hydrofbrmylation reaction, and Reaction 3 is a hydrogenationreaction. All o1' the reactions in Figure i start with alkenes,a common starting material in the production ol plastics andpolymers. Green chemistry is ideal fbr polymerizationreactions, which by design aims to minimize the materialneed to carry out the propagation reaction. Green synthesistechniques can be applied to any reaction.


-l 6 . In Reaction I, the alkyne is besr described as:

A . a dienophile.

B. an electrophile.

C. a nucleophile.

D. an oxidant.

3 7. Which of the following intermediates is consistent withthe two structural isomers formed in Reaction II?

A.oilC

/\HzC- CHCH3

c' .cH,/HrC- C-tt

H2C- O

38. Which of the reactions in Figure I involves thefbrmation of new stereocenters?

A . Reaction III only


C . Reactions I and III only

D . Reactions II and III only

3 9. Which of the following changes does NOT fit with th;philosophy of green chemistry?

A . Using supercritical fluid as a solvent.

B. Using protecting groups and not removing theruntil the very last step of the reaction.

C . Using solid-state catalysts built into the 1ar

equipment.

D. Running a constant stream of oxygen gas througithe reaction vessel fbr oxidation reactions.

40. Reaction III can be described by all of the follorvin_.terms EXCEPT:

A. reduction.

B. hydrogenation.

C. stereoselective.

D. regioselective.

B.oil

,zt\CHz

D' CH.,/

HC-CttH2C-O

e(+

uoee("q"

4 1 . What is the major product of the reaction below?

A.

olt

la" -:+,\A

o

oB'o

o

D.

o

oC.

oo



A chemist set out to synthesize a series of conjugateddienes. Starting with an allylic alcohol, generating aconjugated diene involves an acid catalyzed eliminationreaction. Elimination by way of an E1 mechanism to form aconjugated diene is shown in Reaction 1.

OH

conc. HrSO,

A

Compound II

Reaction 1

Reaction 1 is monitored using UV spectroscopy. Overthe course of the reaction an intense UV absorban ce at 179nm diminishes as a new peak at 222 nm appears.

When the product of Reaction I is treated with acidicwater, two products of detectable quantity are formed. FigureI shows the distribution of the two hydration products at35'C, labeled Compound 3a and Compound 3b.

OH

Compound IV60.\Vo

Figure I

The percentage ofthe secondary alcohol formed increasesas the temperature ofthe hydration reaction increases. This isattributed to a shift from kinetic control to thermodvnamiccontrol.

It is found that if the allylic alcohol in Reaction I isreplaced by a new compound containing both an alcoholgroup and a carbon-carbon n-bond, with the exception of avinylic alcohol, a conjugated diene is formed upon treatmentwith concentrated strong acid at elevated temperatures.

4 2. Which of the following starements accurately reflectReaction I ?

I. Rearrangement is possible during the reaction.

II. A vinylic carbocation is formed as an intermediatein the reaction.

m. The first step of the reaction is the protonation ofthe hydroxyl oxygen.

A. I onlyB. II onlyC. I and II onlyD . I and III only

tuCompound I

OH

I

@Compound III

39.27n

ln43. Which spectroscopy technique is MOST effectivedistinguishing Compound III from Compound IV?

a. IHNlaR

B. Infraled

C. Ultraviolet

D. Visible

Cornpound IV is best described as:

A. a single, achiral molecule

B. a single, chiral molecule.

C . a pair of diastereomers.

D . a pair of enantiomers.

4 7. Which of the following species is NOT an inrermediateformed during the hydration of Compound II?

A' tgr, t;,,\--

A^ 'eD\-ry

44.

45. Which of the following compounds is an allylicalcohol?^er* "ff*'er-* 'er*

46. 1,2- and 1,4-addition is possible in all of the followingcompounds EXCEPT:

A. 2,4-Hexadiene

B. 2-methylcyclopentadiene

C. 1,4-cycloheptadiene

D. 1,3-cyclohexadiene


4 8. What is the major product of the reaction below?

O[.",H2O/H2SOa

-..>

75'C

B' oHA.

d.,, d..,"G;


The Diels-Alder reaction is quicklv becoming the premierway to synthesize polycyclic compounds. New carbon_carbon bonds are formed when a conju_sated diene reacts witha compound containing a n-bond (dienophile) at elevatedtemperatures.

The reaction is fastest when the conjugated diene has anelectron-donating substituent and the dienophile has anelectron-withdrawing substituent. Reaction 1 was carried outfor a total of eight trials to determine the effects of electronicsand steric hindrance on the reaction rate. Table I lists thedata fbr the eight trials.

Reaction I

Table 1

Copyright @ by The Berkeley Revierv@

Trial A B C D X Relativerate

H H H H H t.0

2 CH: H H H H 5.3

J CH: H H cHr H 8.1

4 CH: CH: H H H 0.00s0

5 H H H H '\'. .

o82

6 cHr H H H 'Y'"',o

4t1

cHr H H CHr T,"'o

613

8 CH: CH: H H '1('"'o

0.34

The product of a Diels-Alder reaction is a cyclohexenederivative. The Diels-Alder reaction is classified as anelectrocyclic reaction and it is believed to proceed by aconcerted mechanism that goes through a so called aromatictransition state, where the six n-electrons in the reactantinteract to form new bonds.

In the presence of a Lewis acid, the rate of the reactionincreases substantially, which implies that the Diels-Alderreaction has an alternative mechanism, where the conjugateddiene acts as a nucleophile by attacking the dienophile. Thisinfbrmation is in agreement with the rate shift associatedwith the addition of electron-withdrawing groups to thedienophile.

49 . The product shown in Reaction I is one of twoenantiomers that is formed. Which of the followingpairs of molecules represents the product mixtureformed in Trial 6?

A. CH:

CH: CH:

CH:

cHs

CHr

&

""1(o

o

"J(

CHr

CH:

cHr

&CHq

"ff::, &

o

CH:

CH:

cHr

cHr

'$f;: &

o


50. How would the relative rate change if ethyl groups wereused instead of methyl groups in Trial 4?

A. The relative rate would be significantly lower than0.005.

B. The relative rate would be about 0.005.

C . The relative rate would be significantly higher than0.005.

D. The relative rate would be 200, the reciprocal of0.005.

Addition of which of the following species to Trial 7would increase the rate?

A. AICI3

B. CCla

C. KHD. LiAlHa

If Trial 2 proceeds by a nucleophilic mechanism, ratherthan a concerted mechanism, the intermediate carriescharges. Which of the following structures best showsthe structure of the intermediate in Trial 2?

51.

(t

c' cHr oo D' cHq ooy*,(/*,

53. How can the significantly lower reaction rate in Trial 4than Trial 2 best be explained?

A. The methyl groups on the conjugated diene exhibitsteric hindrance when the conjugated diene anddienophile form the transition state.

B. The methyl group on the conjugated dienewithdrawals electron density from the conjugateddiene.

C. The two methyl groups are acting likeintramolecular Lewis acids on the dienophile.

D . The carbonyl group is more electron-withdrawingon the dienophile in Trial 2 than it is in Trial4.


s4. Which of the followingdienophile?

A.NB.

compounds is the WORST

Io'""''COCHzCHT

c.o

("

D' cHrCHe/

()

Passage lX (Questions 55 - 60)

In a Diels-Alder reaction, the alignment of the diene anddieneophile determines the structural orientation of thesubstituents in the final product. If both the diene and thedieneophile are asymmetric, then there are two differentorientations that the two reactants can assume when theyalign to form the transition state. The preferred alignmentcan be predicted using resonance theory. The two reactantsalign in a manner so as to have a partially positive siteattacking a partially negative site. A generic Diels-Alderreaction of an asymmetric diene with an asymmetricdieneophile is shown in Figure 1.

o

.+.X XoX*ff..q.

Compound Ao

CompoundB

Figure I Asymmetric Diels-Alder reaction

If the reactants are asymmetric, the product distributionof Product A to Product B is never 50-to-50. When X iselectron donating and Y is electron donating, Product A is themajor product. When X is electron withdrawing and y iselectron donating, Product B is the major product. Table Ilists the product distributions for a series of reactions wherethe X and Y groups are varied. In both Product A andProduct B, the Y group is always cis to the carbonyl group.

X Y A B

OCH3 NHCH3 94Vo 6Vo

OCH3 CH: 88Vo l2%o

CH: NHCH3 87Vo l3Vo

cHr CH: 63Vo 37Vo

COCH3 NHCH3 lSVo 82Vo

COCH3 CH: 3lVo 69Vo

Table 1

The product distribution in Table 1 supports theprediction about the electron donating and withdrawing effectsbased on resonance theory. A methyl substituent isconsidered to be mildly electron donating.

55. If the Y-substituent is a second carbonyl functionalgroup (-CR=O), making the alkene reactantsymmetric, what would be predicted for the distributionbetween Product A and Product B?

A. >50Vo Product A;<S}VoProductBB. <50Vo Product A:>50Vo ProductBC. 507o Product A:50Vo Product B

D. Product A and Product B are the same compound

Copyright @ by The Berkeley Review@ 301

c.


5 6. Which of the following conclusions can be drawn fromthe data in Table IA. OCH3 is more electron donating than CH3 because

OCH3 in the X position yields more producr A.B. OCH3 is more electron donating than CH3 because

OCH3 in the X position yields more producr B,C. CH3 is more electron donating than OCH3 because

CH3 in the X position yields more producr A.D. CH3 is more electron donating than OCH3 because

CH3 in the X position yields more product B.

5 7. Predict the major product for the following reaction:

H3CH2CO O

)^.l+L^*\ cHrcr

A.

''""44.\,4 cH2cl

c. D.

oo

B.

CH2CI

____>150'C

,o^O:")

H3CH2CO H3CH2CO

eq: Oq:

5 8. What is the major product for the following reaction?

59. Two structural isomers are formed fr.om Diels-Alderreactions that involve:

A. a symmetric diene with a symmetric dieneophile.

B. an asymmetric diene with symmetric dieneophile.

C. a symmetric diene with asymmetric dieneophile.

D. an asymmetric diene with asymmetric dieneophile.

60. Counting stereoisomers, how many possible productscan be formed from the followine reaction?

HrC

CH.'----*A

H:Co

4.2B.4c.8D. 16

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Pericyclic reactions are single-step reactions that involvethe movement of electrons through cyclic transition statesthat involve pi and sigma orbitals. One class of pericyclicreactions is the sigmatropic rearrangement, which involvesthe migration of a sigma-bonded group across a pi-electronsystem. The two most common sigmatropic rearrangementsare the Cope rearrangement and the Claisen reatangement. Inthe Cope rearrangement, one 1,5-hexadiene yields a new 1,5-hexadiene. In the Claisen rearrangement, an allyl vinylicether yields an unsaturated carbonyl compound. Bothreactions are shown in Figure 1 below.

Cope rearrangement

A new 1,5-diene

JA y,&unsaturated

carbonyl compound

rearrangements

Step II I av

o

t3,31 _

An allyl vinylic ether

Figure 1 Cope and Claisen

A 1,5-diene

Claisen rearrangement

o/Nl)\v

When the reactant in a Claisen rearrangement includes a

benzene ring, the ketone formed from the allyl phenylic etherundergoes tautomerization and converts into a phenol. Theultimate product is the one that is most stable. Figure 2shows a synthesis pathway that involves a Claisenrearrangement, a Cope rearrangement, and tautomerization.

CH:

HrcStep I__>

A

HrcStep III<-

CH: cHrFigure 2 Synthesis using Claisen and Cope rearrangements


64"

65. rnA.B.C.D.

Copyright

OH

61. How can it be supported that the Cope rearrangement isconcerted rather than a multistep process involvingsubstitution?

A . All of the stereocenters are retainedB . All of the stereocenters are invertedC . Several cross products are formed

D. No cross products are formed

62. Step III in the synthesis shown in Figure 2 occursbecause the product:

A. loses steric hindrance after oxidation.B. gains aromaticity after reduction.

C . loses resonance after tautomerization.D. gains aromaticity after tautomerization.

6 3 . Heat serves what role in the Claisen rearrangement?

A. To provide energy to overcome the activationbarrier

B. To drive the exothermic reaction

C. To generate pi-bonds

D. To fbrm sigma-bonds

6 4 . Which of the following orbital arrangements representsthe transition state of a Cope rearrangement?

A.

6 5. The reactant in Figure 2 is best described as a:

A. allyl benzylic erher.

B. allyl phenylic ether.

C. vinyl benzylic ether.D. vinyl phenylic ether.


B.

6 6. What spectroscopic evidence supports the formation of aproduct in a Claisen rearangement?

A . Appearance of a signal at 9.5 ppm in the 1HNMR

B. Appearance ofa broad absorbance around 3400 cm-tin infrared spectroscopy

C. Disappearance of an absorbance around 1700 cm-lin infrared spectroscopy

D . Disappearance of two signals between 5 and 6 ppmin rhe IHNMR

6 7. What is true of the Claisen and Cope rearrangements inFigure 2?

A . Step I is Claisen rearrangement and step II is Coperearrangement; the units of unsaturation decreasefrom 5 to 4 during the Claisen rearrangement.

B. Step I is Cope rearrangement and step II is Clasienrearranggment; the units of unsaturation decreasefrom 5 to 4 during the Cope rearrangement.

C . Step I is Claisen rearrangement and step II is Coperearrangement; the units of unsaturation remain 5during the Claisen rearrangement.

D. Step I is Cope rearrangement and step II is Claisenrearrangement; the units of unsaturation remain 5during the Cope rearrangement.



The isoprene unit is one of nature's favorite buildingblocks. Isoprene (2-methyl-1,3-butadiene) reacts at carbonsone and two or one and four with other isoprene molecules toibrm terpenes, a class of bio-organic molecules. Terpenes arefound in such natural products as rubber and essential oils.Nearly all of the naturally occurring terpenes result from thehead-to-tail connectivity of isoprene units. They connect byundergoing either nucleophilic substitution or electrocyclicaddition, such as a Diels Alder reaction. Figure 1 showssome common terpenes.

cHrH:C

Caryophyllene

i"'

-sCH:

cr-Pinene

HSC CH: CH:

Vitamin.\

cHr OH

t CHr

*o^Citronellol

Figure 1. Common Terpenes

Both plants and animals synthesize terpenes. Pinene, a

monoterpene, is fbund in plants and Vitamin A1, a diterpene,is found in both plants and animals. Smaller terpenes arefound primarily in plants, while some larger terpenes, such as

lanesterol (a C-30 precursor to steroid hormones) and B-

carotene (a C-40 source of vitamin A), are found in plants andanimals. Terpenes can be modified into other compounds,known as terpenoids.

Studies in biogenesis show that the large terpenes aresynthesized starting from isopentenyl pyrophosphate. Thepyrophosphate adds across the diene of isoprene to form eitherisopentenyl pyrophosphate or dimethylallyl pyrophosphate.These molecules then add to one another by way ofnucleophilic substitution reactions, where the pyrophosphateacts as a leaving group. Geranyl pyrophosphate (a namegiven to C-10 terpenes) is the first monoterpene in manynatural synthetic pathways. More isoprene units are added tothe monoterpene to forfir other terpenes and terpenoids.Figure 2 shows a basic schematic for the biosynthesis ofcholesterol starting from isopentenyl pyrophosphate.


i

il1l

1i

HOCholesterol

Figure 2. Biosynthesis of Cholesterol

6 8. Which of the following compounds is NOT a terpene?

A. Limonine (C16H16)

B. Geraniol (C16H1gO)

C. Patchouli alcohol (C15H26O)

D. Stearol (C1gH3gO)

69. What irregularity in a sample of a sesquiterpene (1-<

carbon terpene) would indicate that the compound wa:synthesized in lab as opposed to extracted from a plant.'

A . The compound is not enantiomerically pure.

B. The compound is not a racemic mixture.

C. The compound had impurities with 16 carbons.

D . The compound had impurities with 20 carbons.

70. Which carbon is most susceptible to nucleophilic attackin isopentyl pyrophosphate?

oil

oilP

I

()

o- P- o- P- oHI

o-

A. Carbon 1

B. Carbon 2

C . Carbon 3

D. Carbon 4

7 1. Isoprene units are believed to be formed from threeacetyl coenzyrne A molecules. What is a likely sideproduct from the reaction?

oilI

--.\HrC- - SCoAAcetyl Coenzyme A

A . Carbon dioxide gas

B. Ethanol

C . Acetic acid

D. Isopropanol

The biosynthesis of which of the following moleculeslikely involved a Diels Alder reaction with isopreneunits?

A . Caryophyllene

B. Citronellol

C. a-PineneD. Vitamin A1

Which of the labeled bonds in y-terpinene was formedin the biological synthesis fiom isoprene units?

Bond d

CH:

A. Bonda

B. Bond b

C. Bondc

D. Bondd

1''

73.

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Passage Xll (Questions 74 - B0)

Terpenes are natural organic molecules found in plantsand animals. They are formed from the basic subunit ofisoprene, a five-carbon conjugated diene. Terpenes andterpenoids, biological molecules derived from terpenes, have atotal carbon count that is divisible by five. Terpenes areclassified according to the number of carbon atoms theycontain. Monoterpenes have ten carbon atom, sesquiterpeneshave filteen carbon atoms, diterpenes have twenty carbonatoms, sesterterpene has twenty-five carbons and so on.

Because terpenes are natural products, they are commonin many household items, such as flavoring agents in various{bods and the active molecule in many drugs. Much oforganic chemistry research involves the development of inuilro synthesis of terpenes and terpenoids. Some naturallyoccurring monoterpenes are shown in Figure 1.

MyrceneA"

Camphor

Figure 1. Four Common Monoterpenes

Some terpenes contain oxygen, which is added in a waythat does not alter the carbon skeleton. Carvone is evident bya carbonyl absorption at 1750 cm-1 in the IR spectrum and astrong UV (e > 10,000) absorption at L.'nu* = 242 nm.

7 4. When limin is converted into carvone, what type ofreaction has to transpire?

A . Oxidation of carbon

B. Reduction ofcarbon

C . Hydrolysis of a n-bond

D. Nucleophilic substitution

7 5. Which compound in Figure 1 is LEAST likely tor.rndergo ozonolysis when treated with 03?

A. Camphor

B. Carvone

C. I-imin

D. Myrcene

Limin

Carvone

76. How many singlets does camphor show in its proronNMR spectrum?

A. Two

B. Three

C. Six

D. Nine

7 7 . If myrcene reacts with another isoprene unit, what kindof terpene is fbrmed?

A. Diterpene

B. Monoterpene

C . Sesquiterpene

D. Sesterterpene

78. Camphor is likely to show which of rhe followingphysical and chemical proper ties'/

L High water solubility

II. A boiling point above 298K

m. No specific rotation of plane polarized light

A. I onlyB. II only

C. I and II only

D. I and III only

79. Does limin display a stlong (log e > 4) UV absorption?

A. Yes, because the n-bonds are spaced fal apart.

B . Yes, because of the six-membered ring,

C. No, because there is no carbonyl group.

D. No, because the n-bonds are not conjugated.

8 0. Which compound is the direct product of a Diels Aldercondensation of two isoprene units?

A. Carnphor

B. Carvone

C " LiminD. Myrcene


Passage Xlll (Questions B1- BG)

Many processed fbod products often contain partiallyhydrogenated vegetable oil as one oftheir ingredients. Partialhydrogenation serves to reduce some of the n-bonds found innatural oils. Naturally occurring fatty acids, such as

vegetable oil, often have long carbon chains. They can behydrogenated to convert the alkyl chain, which may containmultiple double bonds, into to an aliphatic R group.Hydrogenation raises the compound's rnelting point, andofien converts a naturally occurring liquid into a solid. Thefatty acids can be found as either the carboxylic acid or as panof a fatty acid triglyceride. Figure I shows the enzymaticalll'controlled conversion of a fatty acid triglyceride into glyceroland three fatty acids.

o

-oA*,o

- oA R2 + H2o Li'ases

o

-oA*. [I

o

soAn,o

* HoAnuo

HoAn,Figure 1 Enzymatic Hydrolysis of a Triglyceride

The fatty acid is isoiated when a tatty acid triglyceride ishydrolyzed. Three carboxylic acids are formed from the fattyacid triglyceride. The R in Figure 1, represents any alkylgroup. In naturally occurring fatty acids, the R has an oddnumber of carbons. Including the carbon of the carboxylicacid lunctional group, naturally occurring fatty acids have an

even number of carbons. This is attributed to the fact tharfatty acid biosynthesis occurs two carbons at a time, viaacetyi coenzyme A. Natural fats can be distinguished fiomsynthetic lats by their carbon chain length. Table 1 listssome common fatty acids that are naturally found in animals:

Table 1 Common Fatty Acids

Acid Formula n

Arachidic CH3(CH2)1sCO2H 0

Arachidonic CH g (C Hz)+( CH =C HC H2) +(fH) zCO zH 4

Behenic CH3(CH2)2sCO2H 0

Lauric CH:(CHz)roCO2H 0

Lignocaric CH3(CH2)22CO2H 0

Linoleic CH j (CH 2 )a(C H=CHC Hz) z(CH) oCO zH 2

Linolenic CH3CH2(CH=CHCHz): (CH2)6CO2H 3

Myristic CHj(CH2)12CO2H 0

C)leic CH 3 (C H2 )7 CH =C H ( CHz ) 7 CO 2H I

Palmitic CH3(CH2)1aCO2H 0

Palmitoleic CH 3 (C H 2 ).5 CH =C H (CHz ) 1 CO 2H I

Stearic CH3(CH2)16CO2H 0

Vegetable oils generally have more unsaturation thananimal fats. For instance, corn oil is 63Vo linoleic acid and26Vo oleic acid, with the rest being made of other saturatedfatty acids. Safflower o1I is'75Vc linoleic acid, l4To oleicacid, and 4Vo linolenic acid with the rest being made of othersaturated fatty acids. As the amount of unsaturationincreases, the melting point of the fatty acid decreases,assuming that the number of carbons remains constant. Forthis reason, many animal fats are solids while manyvegetable oils are liquids at room temperature. Fatty acidscan play one of three roles in biological systems. They arefound as the building blocks of cell walls as phospholipidsand glycolipids. Fatty acids form derivatives that serve as

hormones (intercellular messengers). Fatty acids are alsoused for fuel through fany acid metabolism.

81. What is the structure for the MOST abundant fatty acidfound in corn oil?

B.o

HO

HO

HO

The n-bond of a fatty acid can be reduced viahydrogenation when treated with hydrogen gas and acatalytic metal or by FADH2. Treatment of linoleicacid with FADH2 yields a product:

C.

with lower molecular mass and a lower meltingpoint than the reactant.

with higher molecular mass but a lower meltingpoint than the reactant.

with a lower molecular mass but a higher meltingpoint than the reactant.

with a higher molecular mass and a higher meltingpoint than the reactant.

D.

82.

A.

B.

D.


8 3. Addition of D2 with Pd catalysr reduces n-bonds byadding deuterium to each n-bond carbon. Treatment ofoleic acid with D2 and palladium yields a compoundwith how many chiral centers?

A,7nroB. One

C. Two

D. Four

8 4. Treatment of an alkene with potassium permanganateyields a vicinal diol at the alkene carbons. Where do thehydroxyl groups add when the most unsaturated fattyacid in safflower oil is treated with KMnO4 under basicconditions?

A. Carbons 8, 9, 11, 12,14, and 15

B. Carbons 1,9, ll, 13, 15, and 17

C. Carbons 9,10,12, and 13

D. Carbons 9,10, 12,13, 15, and 16

8 5 . Bromine liquid is used as a quantitative test reagent todetermine the amount of n-bonds per molecule of a

compound. Which of the following acids consumes theMOST Br2 per molecule?

A. Arachidic

B. Arachidonic

C. LinoleicD. Linolenic

86. Complete hydrogenation of palmitoleic acid yieldswhich of the following acids?

A. MyristicB. PalmiticC. Stearic

D. Arachidic

Passage XIV (Questions 87 - 92)

Terpenes and terpenoids are natural compounds found inplants and animals that are built fiom 5-carbon reactants.Figure I shows the sesquitetpenoid (+)-occidentalol.

Figure 1 (t)-Occidentalol

A multistep synthesis leading to occidentalol beginswith the conversion of Compound 1 into Compound 6,which can further react to form occidentalol. Fisure 2 showsthe synthesis of Compound 6.

'rr4<

loH

CO2Me

Compound 2

I l' HoaoHStep 2 | rrou. phH. A

Y z. LeH, eroEt

CO2Me

Compound 1

la"150'C, -Cq---+

Step I

SOj.Py, THF,0"C;<-

Compound 4

I

Step 4ll NHCI. HOAcI

V

EtO\ -j-Eto- fr SMe

_oo ---_->HMPA, DME,62'C

Compound 5 Step 5 Compound 6

Figure 2 Synthesis of Compound 6

Step 1 involves a Diels-Alder cycloaddition followed bydecarboxylation of the polycyclic system. Step 2 involvesthe conversion of the ketone group of Compound 2 into a

ketal followed by the reduction of the ester into a primaryalcohol" Further reduction of the vinylic alcohol group inCompound 3 forms a methyl group in Compound 4. Theprotecting group is removed in Step 4 to relorm the ketone.Compound 5 undergoes a variation on the Wittig reaction toform an alkene in Compound 6.


88.

8 7 . How many chiral canters are present in occidentalol?

A. IB. 2

c. 3

D.4

Terpenes containing flfteen carbons are best describedAS:

A. monoterpenes.

B. diterpenes.

C . triterpenes.

D. sesquiterpenes.

Which compound has the longest maximum wavelengthof absorbance, Imax, in ultraviolet spectroscopy?

A. Compound IB. Compound 2

C. Compound 3

D. Compound 4

What intermediate compound forms in Step I before thedecarboxyl ation takes place'?

A . A cyclohexadiene

B. A cyclohexene

C . An cr,B-unsaturated ketone

D. Alactam

To synthesize occidentalol, Compound 6 is firstconverted into a methyl ketone, which is then treatedwith MeLi in Et2O at *70'C. What is the role ofMeLi?

A. To act as an electrophile and accept electron densityform the alpha carbon.

B. To act as a nucleophile and donate electron densityto the carbonyl carbon.

C . To act as a base and deprotonate the alpha proton.

D. To act as an acid and protonate the carbonyloxygen.

If Compound 4 were treated with strong acid, at whichcarbon in the n-network is it most likely to gain H+?

A . Carbon a

B. Carbon b

C " Carbon c

D. Carbon d

89.

90.

91.

92.

Questions 93 through 100 are NOT based on adescriptive passage.

93. Leukotriene A4, LTAa, is derived from arachadonicacid. Its structure is shown below:

corakWhat is NOT true of LTA4?

A. LTAa has six units of unsaturation.

B. LTAa has six n-electrons in a conjugated system.

C . LTAa is capable of undergoin C 1,2-, 1,4-, 1,6-, or1,8-addition when treated with an electrophile andnucleophile.

D. LTAa has more sp2-hybridized carbons than it hasspj-hyb.idir"d carbons.

94. What is the major organic product of the reactionbelow?

n't\.0$1, Br.__=+

t1cbHcH., hv

o."*,.:,'rlT,

I{TCCHCH3 HTCCHCH3

D' nrc ..,noTr

I

I!CCBTCHT

B.H:C

"'"r"

95. Which of the following reactions is a propagationreaction?

A. H3C. + H3CCH2CH3 -+ 2 H3C. + H3CCH2.B. H3C. + H3CCH2. -+ H3CCH2CH3

C . H3CH2C. + H3CCH3 -+ 2 H3CCH 2, + | t2H2D. H3C' + H3CCH2CH3 -+ CH4 + (H3C)2CH.

9 6. The terpene (+)-B-trans-Bergamontene shown below:

A. is generated from three isoprene units.B. is a terpenoid and not a terpene.

C . has three units of unsaturation.D . has sixteen possible stereoisomers.


FIlCCHCH3

97, A conjugated diene is necessary in which of thefollowing reactions?

A . Claisen rearrangement

B. ClemmensenreductionC . Cope rearangementD . Diels-Alder cycloaddition

9 8. All of the following observations are associated with anE2 reaction EXCEPT:

A. the base must be bulky and strong.

B . whether the major product has cis or trans geometrydepends on the stereochemistry of the reactant.

C. that rearrangement is observed.

D. heat is required to drive the reaction.

9 9. All of the following are physical properties of a terpeneEXCEPT:

A. High lipid solubiliryB. High boiling pointC. Low volatilityD. High specific rotation

I 0 0. Which of the followingequal to zero?

L E-butene

II. CzHq

il. Z-butene

A. I only

B. II only

C . iII only

D. I and III only

molecules have dipoles NOT

1.B 2.D 3.C 4.8 5.8 6.B1.8 8.C 9.D l0.B ll.B 12.A

13. c 14. A 15. D 16. C 1'.7. C 18. D19. c 20. A 21. c 22. D 23. B 24. C2s. D 26. C 21. C 28. B 29. A 30. D31. A 32. D 33. A 34. D 35. D 36. A37. A 38. A 39. B 40. D 41. C 42. D43. A 44. D 45. A 46. C 47. B 48. A49. D 50. A 51. A 52. D 53. A 54. D55. D 56. A 51. A 58. A 59. D 60. C61. D 62. D 63. A 64. B 65. B 66. A61. C 68. D 69. C 70. D ',71. A 12. C13. B 14. A "/5. A 16. B 11. C 78. B19. D 80. C 81. B 82. D 83. C 84. D8s. B 86. B 87. C 88. D 89. A 90. B91. B 92. A 93. D 94. D 95. D 96. A91. D 98. C 99. D 100. C

ALL DONE, NO MORE!

Alkanes and Hydrocarbon Reactions passage Answers

Choice B is correct. Hahdes are like hydrogens when considering degrees of unsaturation, because they, likehydrogen, make only one bond with carbon. Plugging into the ronJ*itl[ formula for units of unsaturation yieldsan answer of 1. Pick B for best results.

Units of Unsturation =2(#C) + 2- (#H) - (#cl) _ 2(s)+2-(e)-(1) _

2

Choice D is correct. According to the molecular formula, there are no units of unsaturation within the molecule,so there are three linear carbon backbones to be considered. No other carbon backbones are possible, becausethey would be either on9_of the following structures, but drawn differently, violaie the units of unsaturation, ornot be structurally possible.

C

I

C-CI

C

70 +2-9 -72

_12-10 _2_,___ r22)

(-C-CC-CI

I

C

on each of the carbon backbones, the chlorine must be systematically placed to deduce the total number ofstructural isomers that are possible (stereoisomers do not count in this question). There are three non-equivalentcarbons in the five carbon siraight chain, therefore there are three monochloro isomers of pentane. There arefour non-equivalent carbons in the four carbon chain, therefore there are four monochloro isomers of 2-methylbutane. There are two non-equivalent carbons in the three carbon chain, but one of the carbons (thecentral carbon) already has for-rr bonds, therefore there is only one monochloro isomer of 2,2-dimethylpropane.This means that there are eight structural isomers total, so pict< o for greatest success.

C_C-C_C-C C_C-C-C-C C_C-C_C-Ctti1-chloropentane

C-C-C-C

C12-chloropentane

C-C-C_C

C13-chloropentane

rlclc

1 -chloro-2-methylbutane

/\CCI

2-chloro-2-methylbutaneCCi

2-chloro-3-methylbutane

C-C-C_ClrCCI

1-chloro-3-methylbutane

C_ CII

C- C- C 7-chloro-2,2-dimethylpropaneI

C

8 total

J. Choice C is correct. By symmetly, there are four unique carbons on methylcyclopentane, therefore chlorinationcan occur at a total of four different sites (four different carbons). This yields u tot^1 of four structural isomers.The correct answer is choice C.

C-C-C_C

&.{C1

HYDROCARI]ONS EXPLANATIONSCopyright @ by The Berkeley Review@

4.

5.

6.

7.

Choice B is correct. Bromination is highly selective for tertiary carbons over secondary and primary carbons, sothe exact quantity of each type of hydrogen need not be accounted for. The most abundant product frombromination results from the bromination of a tertiary carbon. Carbon number two is tertiary, so the mostabundant product is 2-bromo-2-methylbutane. Pick B, and be a chemistry master.

Choice B is correct. 1-chloropentane and 2-chloropentane are formed from chlorination of a primary andsecondary carbon, respectively. There is a difference in reactivity between primary and secondary carbons infree radical chlorination reactions. Secondary carbons are 2.5 times as reactive as primary carbons. This meansthat the ratio of hydrogens (abundance) and the relative reactivity must both be accounted for whenestimating the final ratio. Drawn below is the application of quantity and reactivity. Hydrogens symbolizedby Hu lead to 1-chloropentane and hydrogens symbolized by HU lead to 2-chloropentane. They are multipliedby their respective reactivity factor.

"\ /to "\ l'4-.^-c-.^-C\^,, H,

Hu-l L \- Ha

Ha H,6Hux7=6:4H6x2.5=10

@'ffin*'"irr{--

i----2-chloropentane

The ratio is 10: 6, which reduces to 5 : 3. This makes choice B the best answer.

Choice B is correct. Four secondary hydrogens lead to 2-chlorohexane and four secondary hydrogens lead to 3-chlorohexane, sotheproductratio shouldbe 1:1. Becausethe ratio is7.07: l andnot 1: l,choiceAiseliminated. There is no resonance in the molecules (because there are no lone pairs and no n-bonds), so choice Cis eliminated. The inductive effect wouid favor the formation of 3-chloropentane, so choice D is eliminated.Steric hindrance is the best explanation why the carbon-2 is selected over carbon-3, so choice B is best.

Choice B is correct. The question addresses ring strain destabilizing not only the alkane, but the free radicalintermediate. Based on that, choices A and C are eliminated, because they are not ring structures. A three-membered ring has more ring strain than a four membered ring, so the best answer is choice B.

8. Choice C is correct. The stability of the carbon free radical is attributed to the donation of electron densityfrom the alkyl substituents through hyperconjugation. Because hydrogens cannot donate electron densitythrough hyperconjugation, the stability of a free radical depends on the number of alkyl substituents attached.As a result, the stability of free radicals is tertiary > secondary > primary (3' > 2" > 1"). Quaternary freeradicals cannot exist, because the presence of four bonds and a free electron on carbon would exceed the octet rulefor carbon. The best answer is therefore choice C.

9. Choice D is correct. According to Table 1, the enthalpy change is positive for the reaction of iodine with analkene, so it is an endothermic reaction. This means that the products are in a higher energy state than thereactants, which eliminates choices A and B. In the next to the last sentence of the first paragraph, it is statedthat the second transition state is of higher energy than the first transition state, so the best answer is choiceD. If you find yourself asking "do they really expect me to know this?", it's probably in the passage... find it!

10. Choice B is correct. Initiation forms the halogen free radical (X.), so the first propagation step involves thehalogen free radical as a reactant. This eliminates choices A and C. In propagation step 1, the free radicalabstracts a hydrogen from an alkane, which is choice B. Eliminate choice D, because it is a termination step.

1.L" Choice B is correct. According to the data in Table 1, the strongest halogen-halogen bond is formed between twochlorine atoms. Chlorine is the second halogen from the top in the column VII of the periodic table. Atomicradius increases as you descend a column of the periodic table, so chlorine is the second smallest halogen.Therefore, the strongest halogen-halogen bond is formed between the second smallest halogen. The fluorine-fluorine bond is an exception to the shorter bond equals stronger bond rule, because of inter-nuclear repulsion andthe odd fact that the single bond is actually a pi-bond, rather than a sigma-bond. The best answer is choice B.

Copyright O by The Berkeley Review@ HYDROCARBONS EXPLANATIONS

12.

13.

Choice A is correct. Hydrogen free radical is unstable, and thus it cannot be formed in the reaction mechanism.Choice A produces a hydrogen free radical, therefore the best answer is choice A.

Cl'roice C is correct. If the halogen were to stabilize the free radical when attached, it would lower thetransition state and make the reaction pathway for adding a second halogen more favorable. The halogen willnot stabilize the free radical, because halogens are election withdrawiig by the inductive effect. Hilogenstherefore do not heip stabilize the free radical intermediate so statement I is true. Because a second halidedoes not add to an alkyl halide, but a halide does add to an alkane, we assume that the alkane is morereactive' The difference is either in the bond broken or the bond formed. It is safe to assume that the reactionwili choose a pathway of lowest energy so by breaking the C-H bond of the alkane preferentially over the C-Hof the alkyl halide, it can be concluded that the bond broken is of lower energy. rhi, makes statement II true.If statement II is true, then statement III cannot be true. The best answer is therefore choice C. As a note, an R-H bond is stronger than an R-X bond, so if a halogen free radical (X.) were to react with an alkyl haiide, itwould abstract the halide, not a hydrogen. This is the more logical explanation for the absence of multiplesubstitution products with free radical halogenation.

Choice A is correct. The enthalpy of reaction for a reaction between an alkane and fluorine is listed in Table 1.By simply reading the chart, it can be seen that best answer is choice A. Sometimes answers are this easy toget, so don't be fooied into thinking every question on the MCAT will be difficult.

Choice D is correct'-. Wl"l comparing Trial I with Trial II, both the electrophile concentration and reactionrate have doubled. This behavior is expected no matter what mechanism is employed. When comparing TrialI with Trial III, the base concentration has doubled, the electrophiie concent.ation has increasea (bi7 sixtypercent), and the rate has more than doubled. If the reaction depended just on the electrophile, the raie"wouldbe less thang'2 x 10-3 M/s. Because the rate in Trial III is morl than tire rate in Trial I, ihe rate-determiningstep depends on both reagents, making it bimolecular. E2-reactions are bimolecular, so choice D is correct.

Choice C is correct. The most stable conformation has the largest substituents with anti orientation to oneanother' The bulkiest group on carbon two is the methyl group comprised of carbon 1 and the bulkiest group oncarbon three is the ethyl group comprised of carbons + anJ s, so choiie C is correct. The Newman projections forthe three staggered conformations are drawn below:

14.

15

t6.

"-;,-4i'"'HsC cH2cH3

Conformer fromthe example

CHs

cHs

Br. .CH"

\->cir.cH"H\Y \HsC H

Conformer from whichelimination takes place

cH2cH3

CHs

Br. .*CH"CH3

"t,><HsC CHs

Conformer withC1 and Cn anti

HsC-r H3CH2C H

CH:

CH2CH3 H CHs

'l'7' Choice C is correct" Carbons in the reactanthave sp3-hybridization while two carbons i1 the alkene producthave sp2-hybridization, and thus those carbor-r, .ur-rr-roi have chirality. There are two stereocenters in thereactants (on carbon two and carbon three), and both stereocenters are lost in the elimination reaction. Becausethe alkene product has no chirai centers, it can have no optical activity. The best answer for the question ischoice C.

18' Choice D is correct. From the perspective of the eye as shown, the bromine atom is bonded to the rear (eclipsed)carbon, therefore choices A and C are eliminated. From the perspective of the eye, the methyl would stick outto the left and the hydrogen would stick out to the right. The-besi choice is thus answer choice D.

CHs

Copyright @ by The Berkeley Review@ 312 HYDROCARBONS EXPLANATIONS

1

Carbon 2 = R

't9. Choice C is correct. The enantiomer of the reactant is its mirror image, therefore both chiral centers must beopposite that of the reactant's chiral centers. The determination oiboth of the reactant's chiral centers isshown below. The chirai centers of the enantiomer are 23 and 3R. The best answer is choice C.

Compound is (2R,3S), so theenantiomer must be (2S,3R)

H$)cH2cH3 HsC

Carbon 3 = S

Choice A is correct. The role of a strong acid in an elimination reaction is to protonate the leaving group andincrease its tendency to leave. Protonation occurs in the E1 reaction rather thin the E2 reaction. T;hI acii willnot dehydrate the solvent, so choice B is eliminated. The sulfate ion is not a strong enough base to remove aproton, so choice C is both wrong and impossible. Choice D should be eliminatia i*J"aiutely, because acarbocation cannot be protonated. The best answer is choice A.

Choice C is correct. Chlorocyclohexane when treated with a strong base und.ergoes an elimination reaction byway of an E2 mechani-sm to yield cyclohexene. There are two units of unsaturatiion in cyclohexene, one unit ofunsaturation for the n-bond and one unit of nnsaturation for the ring. The correct answer is choice C.

Choice D is correct. This question is another way of asking, "Which reaction proceeds via a carbocationintermediate in its mechanism?" The first two reactions proceed by an E2-mechanism, which is a one-stepprocess having no intermediate. The E2-mechanism is predictable, becaur" of th" presence of the strong, bulkybase. The base must be strong enough to remove a proton from carbon and it must te bulky enough to #inimizecompetition with an Sp2-reaction. Reaction III proceeds by an E1-mechanism, because of the preJence of strongacid, which protonates the hydroxyl group. This makes a better leaving group, and ultimaiely facilitates theformation of a carbocation. In fact, the location of the doubie bond in theiiial product can only be explained by

- a hydride shift. The hydride shift results in the conversion from a ,".orliury carbocation into a tertiarycarbocation. Only Reaction iii exhibits rearrangement, so the best answer is choice D.

Choice B is correct. This question involves counting the chiral centers in each product in Figure 1 to see whichmolecules are chiral, and then analyzing to see if chirality is involved in the reaction. The m-ajor products fromReaction I and Reaction II both have chiral centers present. In Reaction I, two of the original three chiralcenters are lost , but that still leaves one remaining chiral center, so the mixture is optically uJtirr". In ReactionII, one of the original three chiral centers is lost , which leaves two remaining chiral centers, so the mixture isoptically active' In Reaction III, all three chiral centers are lost, so the fiial product mixture exhibits nooptical activity. The best answer is choice B.

Choice C is correct. The reaction involves the use of concentrated strong acid at high temperature, so thereaction is apt to proceed via an E1-mechanism, An E1-mechanism entails the hydroxyl [roup being protonatedand then leaving, producing a secondary carbocation. A hydride shift resuits in ihe conversion fropf secondarycarbocation into a tertiary carbocation, which is the intermediate from which deprotonation to form thealkene product occurs. A11 three chiral centers are lost, which is also seen in Reaction III. This is bestexplained in choice C. On the MCAT, test-takers are expected to be able to see the analogy between thereaction in a question and a reaction in the passage.

Choice D is correct. Reaction I is an E2-elimination reaction. One of the characteristics of an E2-reaction isthat the reaction is concerted, so the rate of the reaction depends on both reactants. If you increase theconcentration of either reactant, in statement I the base, the reaction rate increases. This makes Statement I avalid statement. Elimination reactions require heat, so a decrease in temperature decreases the amount ofelimination product that forms. You may recall that to maximize the subsiitution product and minimize thecompeting elimination product, temperatures are reduced. This makes Statement Il invalid, which eliminateschoices B and C. Reaction II is an E2-reaction, so it has a competing Sp2-reaction occurring in the same flask.This makes Statement IiI valid, and makes choice D the best u.,r*"r.

BrcHg

20.

21..

))

23.

,t

25.

Copyright @ by The Berkeley Review@ JIJ HYDROCARBONS EXPLANATIONS

26. Choice C is correct' First and foremost, with,a strong, bulky base, the reaction proceeds by an E2-mechanism.This eliminates choices A and B' The E2-mechanism-requires that the hydrogeir being lost be oriented anti tothe- leavrng group, in this case chlorine. In Reaction I, there are two anti irydrigens, and the one that is chosen,is the one that leads to the more substituted alkene product. In Reaction Ii, the"re is only one anti hydrogen, soonly one-product may be. formed. The hydrog"n o., ii-," more substituted alpha carbon has gauche orientation, soit is not deprotonated. The best answer is choice C.

27 ' Choice C is correct. The two carbons common to both rings finish as doubie bond carbons in Reaction III, so theyboth finish with sp2-hybridization. Only choice C finish"es witn spi 1-rrAr1ii")ution, so it is the best answer.

Choice B is correct. Diethyl amine is a weak base, while potassium tert-butoxide, t-butoK, is a strong bulkybase' For elimination by an E2-mechanism, a strong, bulky base is necessary. This means that when usingdiethyl amine, there can be no elimination reaction, which eliminates choices C and D. The reaction withdiethyl amine is nucleophilic substitution. The electrophiie is a secondary alkyl chioride, which could go byeither an 5511 or S52 mechanism. Because the nucleophile is a good ,,rr.l"ophil", the best choice is aJ sNzreaction' An 5512 reaction results in the inversion of the reactive iite, which puts tne amine group behind theplane of the molecule and makes choice B the best answer.

28.

29. Choice A is correct' The Silkworm Moth sex phermone has two double bonds, one that is cis substituted and theother that is trans substituted. This eliminates choice B. All of the carbons are part of a straight chain with nobranching, so it has two terminal carbons that are primary and all the internal .uiborr, are secondary. There are

l"^:"1:::I_:11,,""?,:: t:tu: C.is eliminar:d 91y:ne sinsle bond separates the two doubie bonds, so they are inl1::^t"T1q111,_lllt

eliminates choice D. Ali of the carbons have at least two identical substituents br theynave sp'-hybridization, so there are no stereogenic centers. This makes choice A the best answer.

fact coni

30' Choice D is correct. The Oriental Fruit Moth sex phermone differs from the structure in the question at theposition of the double bond. The structure in the question ha9 cis geometry, while the phermone in Figure t hastrans geometry. This can be misleadingand tempt you to pick choice g. gut, the conneitivity is not the same, sothe structures are structural isomers. The n-boncl is between carbons B ancl 9 in the Oriental Fruit Moth sexphermone, but it is between carbons 9 and 10 in the compound that is shown. This makes choice D the bestanswer' The two structures are not interchangeable by a rotation about a bond, so they are not conformationalisomers' This eliminates choice A. There is no itereogenic carbon, so the two structures cannot be optical isomers.This eliminates choice C. The two structures are structural isomers, making choice D the best answer.

31' Choice A is correct. UV-visible spectroscopy is used to detect n-bonds. Bombykol has conjugated n-bonds, whilemuscalure just one.n-bond. Conjugation tednces the transition energy, so the #avelength oi ria*i^.rm absorbanceincreases with conjugation. This means, that bombykol has a greatJi l,*u* than

^1rr."ulrrr", making Statement I avalid statement. Choice B is eliminated. The Tiger Moth r"* ho.-on"1s an aliphatic alkane, roit hu, no units

of unsaturation. Undecane is an 11-carbon aliphalic hydrocarbon, so it too has no units of unsaturation. The twocompounds have the same units of unsaturation, zero, so Statement II is invalid. This eliminates choice C.Terpenes and terpenoids contain a number of carbons that is divisible by five and a predictable connectivity thatcan be partitioned into isoprene subunits. The oriental Fruit Moth sex ih"r-o1" has twelve carbons in its chainand two more for the acetate group. Fourteen is not divisible by five nor is the structure one that can be brokeninto isoprene subunits. This makes statement III invalid and makes the choice A the best answer.

32' Choice D is correct. Muscalure is a cis alkene made of 23 carbons and 46 hydrogens. It is a long chainhydrocarbon, so it has low water miscibitity. Choice A is a valid statement, and ttereby eliminated. It isexcreted, so it must be a liquid under ambient conditions. This makes choice B a valid statement, whicheliminates it. Because it is a long chain-hydrocarbon, lipids can dissolve into it. It has high lipid solubility,making choice C a valid statement, thereby eliminating it. There are no stereogenic centers on muscalure, so itwill not rotate plane-poiarized light. This makes choiJe D an ir-rvalid statement] so choice D is the best answer.

33' Choice A is correct. All of the compounds in Figure t have straight chains (no rings), so each has at least twoprimary carbons' only 2-methylheptadecane his branching, ro lt hu, an extra primary carbon. No otherstr.ucture in Figure t has any branching, so 2-methylheptadeiane, the Tiger Moth sex phermone, has the rnostprimary carbons. Choice A is the best answer.

Copyright O by The Berkeley Review@ 314 HYDROCARBONS EXPLANATIONS

34. Choice D is correct. The Green Peach Aphid defense phermone has 15 carbons connected in such a way that it canbe partitioned into three isoprene subunits. Choice A is a valid statement, so it is eliminated. The Green peachAphid defense phermone has four double boncls, so it has eight sp2-hybridized carbons. With fifteen carbonstotal, more than half of the carbons are sp2-hybridized. Choiie B is a valid statement, so it is eliminated. TheGreen Peach Aphid defense phermone has a conjugated diene, so it can undergo 1,4-addition of an electrophile.Choice C is a valid statement, which eliminates it. Because of all of the n-donds in the Green peach Aphiddefense phermone, it is not very flexible. Choice D is an invalid statement, which makes it the best answer.

Choice D is correct. The Oriental Fruit Moth sex phermone has an ester group. A carbonyl has an IR absorbanceabove 1700 crn-l , so an ester accounts for the IR absorbance at1741 .i',-r.'Ch"i;; A is a valid correlation ofstructure to spectroscopic observation, so it is eliminated. Bombykol has two n-bonds in a conjugated network.The presence of n-bonds in a structure results in an absorbance in the ultraviolet-visible tuig" of the EMspectrum, so an absorbance of 227 nm seems viable for a conjugated diene. Choice B is a valid iorrelation ofstructure to spectroscopic observation, so it is eliminated. Muscalure has two carbons involved in double bonds,each of which has a hydrogen attached. This means that the hydrogens on those carbons will be founddownfield, resuiting in two signals with values around 5.00 ppm in-lUx"traR. Choice C is a valid correlation ofstructure to spectroscopic observation, so it is eliminated. The Tiger Moth sex phermone has eighteen carbons

?*.i:-{""e oj.srynmetry. There are two equivalent methyl gtorrp!, so there arJseventeen uniqui carbons. TheTTCNMR would show 17 signals if it were of high enough iesolution. The reality is that ,.uny of the signalswould overlap, so it would likely show less. It will not show fourteen signals in its 13CNMR spectrum, so choiceD is an invalid correlation of structure to spectroscopic observation, makLg it the best answer.

Choice A is correct. Reaction I is a Diels-Alder reaction. Diels-Alder reactions involve the reaction of aconjugated diene and a dienophile. Normally we think of the dienophile as an alkene, but the onlyrequirement is that it has a r-bond. The alkyne meets this requirement, so it is a dienophile. The best answeris choice A. There is no nucleophilic substitution gorng on, so choices B and C are eliminated. There is no changein oxidation state, so the compound is not an oxidant or reductant, eliminating choice D.

Choice A is correct. Each of the answer choices is lacking two hydrogens from the formula of the final product,CaHgO. This means that hydrogenation converts the intermediate to tne final product. The final product is analdehyde and hydrogenation cannot convert a cyclic ether into a carbonyl, so choices C and D can be eliminated.Hydrogenation adds two hydrogen atoms to neighboring carbons. Because an aldehyde is generated, we knowthat one of the hydrogen atoms is added to the carbonyl carbon. The ring is cleaved, so iydrogenation mustbreak the strained ring by adding a hydrogen to the carbonyl carbon and a h"ydrogen to the neighbJring atom. Inchoice A, the hydrogen atoms are correctly displaced on the intermediate-(CH2-CH-CH3) to lorm bothaidehyde products (CH2-CH2-CH3 and CH3-CU-CH3) when a single hydrogen is added to tne akyl chain.This is not the case in choice B (CH2-CHZ-CHZ), so it is eliminated.

Choice A is correct. On the product of Reaction I, ail of the functional groups are bonded to sp2-hybridizedcarbons, so there are no stereocenters. This eliminates choices B and C. Bised bn the remaini.,g ur,r*", choices,Reaction III must have formed a new stereocenter. The new stereocenter is located on the rin{ carbon with themethyl substituent. Hydrogenation can occur from above or below the ring, so the methyl is albove the plane infifty percent of the product mixture and below the plane in fifty percent of tnu mixture. Reaction III forms aracemic mixture. In Reaction II, the products have no chiral centers, so no new stereocenters were formed duringthe reaction. This eliminates choice D and makes choice A the best answer.

39' Choice B is correct. The basic tenet of green chemistry is to minimize waste and side products and maximizeatom-economy. Atom-economy aims to get every atom added to the reaction containe. et-rait-rg up in the product.Using a supercritical fluid as a solvent makes for easy recovery and reuse of the solventl Iieaction III usessupercritical CO2, so choice A is valid and thus eliminated. Using protecting groups adds extra atoms to thesolution that are not destined to be part of the product, so it violatei ine princfof, of atom-ecor-romy. protectinggroups are difficult to recycle without spending a great deal of solvent, so they do not fit the green chemistr!philosophy. Choice B is an exception. If catalysts are part of the lab equipment, such as cataljrtic beads, thenthey are easily recovered and reused. This makes choice C in philosopnicat agreement with ti-re principles ofgreen chemistry. This eliminates choice C. It is stated in the passage that direcl oxidation using oxygen fits inthe philosophy of green chemistry, so choice D is eliminated. Choose B and be on top of your gui".

35.

36.

J/.

38.

Copyright O by The Berkeley Review@ 315 HYDROCARBONS EXPLANATIONS

40. Choice D is correct. Reaction III is described in the passage as hydrogenation, the term applied to a reactionthat adds hydrogen atoms. This eliminates choice g. rne gain of Uoias to hydrogen, u i"r, electronegativeatom than carbon, is defined as reduction, so choice A is eliminated. Although it is not specified in thepassage, when hydrogenating with a metal catalyst, the process adds the hydrJgen atoms in a syn fashion,which means that the reaction is stereoselective. This eiiminates choice C.- Both carbons gain a hydrogenatom, so there is no regioselectivity. This makes choice D the best answer. Choose D for the safe of correctness.

Choice C is correct. The reaction is a Diels-Alder reaction, similar to Reaction I, except an alkene is serving asthe dienophile, rather than an alkyne. A Diels-Alder reaction results in the formation of a new six-memberedring, which is observed in all of the choices. When the reaction involves a conjugated cliene and an alkene, theproduct is a cyclohexene ring, so choices B and D are eliminated. To form choice"B, the dienophile would haveneeded to be an alkyne, like Reaction I. The n-bond in the product is located between the two internal carbonsof the-original conjugated diene, so they should be found otr the left side of the central ring. This makes choiceC the best arswer. Choose C for a brighter smile when scoring iike you did.

41..

eK-r+ _^* e[("42. Choice D is correct. Reaction 1 is an elimination reaction by way of an E1 mechanism. In an E1 reaction, a

carbocation is formed, so rearrangement is possible. Whether lt is observed iepends on the compo,ind, but it ispossible. Statement I is a valid statement. The intermediate is an allylic carbocation, where the cationiccarbon is bonded to one of the carbons in the double bond, not a vinylic carbocation, where the cation carbon isone of the carbons in the double bond. Statement II is invalid, which eliminates choices B and C. The first stepin acid catalyzed reactions is the protonation of some functional group on the reactant. In this particular case,the hydroxyl group is the most basic site, so it is protonated. Thii generates a good leaving grorrp, which thenieaves in the second step. Because the hydroxyl group is protonated to stait the reaction, Statement III isvaiid. This makes choice D the best answer.

Choice A is correct. Compound III and Compound IV are structural isomers of one another. They each have ahydroxyl

.group and an aikene functionality, so infrared spectroscopy yields the same key absorbances. Thismakes infrared spectroscopy ineffective at distinguishing the two iilyiic alcoho1s, so choice B is eliminated.Ultraviolet spectroscopy is great for determining the amount of conjugation in a system. However, bothcompounds have the same number of rc-bonds, one, so ultraviolet spectroscopy yields essentially the samespectrum for both compounds. Choice C is eliminated. Neither structure absorbs light in the visible range,given that neither structure has extensive conjugation. This can be inferred from the passage when thlymention that the peak at 779 nm disappears. That peak is associated with Compound I, which tappens to beone of the two enantiomers represented by Compound III. Choice D is eliminatei, because 1,79 nmis not in thevisible range of the EM spectrum. The best method is 1HNMR, which can distinguish structurai isomers bytheir equivalent hydrogens. Choice A is the best answer.

Choice D is correct' The tertiary carbon with the hydroxyl group has four unique substituents, so it is chiral.This eliminates choice A. There is only one chiral center, so diastereomers are not possible (there must be atleast two chiral centers for diastereomers to be possible). This eliminates choice b. Because the hydroxylgroup can be above the plane or below the plane, there is more than one structure possible for Compound IV.This eliminates choice B and makes choice D the best answer.

{).

44.

OHCompound IV


mOH

Enantiomers

HYDROCARBONS EXPLANATIONS

45.

46.

Choice A is correct. The first paragraph of the passage implies that Compound I is an allylic alcohol. If yourecall your general nomenclature, then you should know that when a functional group, in this case the hydroxylgroup, is on the carbon bonded to the alkene carbon, it is said to be allylic. Choice A fits this description.Choice D should have been eliminated early, because it does not contains a n-bond. Choice B is eliminated,because it is a vinylic alcohol (hydroxyl group directly bonded to the alkene carbon). Choice C has the doublebond too far from the alcohol group to be allylic, so it is eliminated as well.

Choice C is correct. The option for either 1,2-addition or 1,4-addition occurs when the reactant has conjugatedn-bonds. Choices A and D should be eliminated immediately, because when the two numbers describing the n-bonds differ by 2, then the n-bonds are conjugated. Cyclopentadiene has only five carbons, so one n-bond must bebetween carbons one and two. The second ru-bond must be between carbons three and four, because in a five-carbonring, no matter how you place two double bonds, for the ring to not be so strained it can't exist, they must beconjugated. Only in choice C are the n-bonds not conjugated, so choice C is the best answer.

Choice A Choice B Choice D

47.

Choice C

.A/ f'*Choice B is correct. The hydration of Compound II starts with the addition of a proton to the conjugated n-network. The easiest carbon to protonate, because of steric hindrance and resonance stability, is the secondary,terminal carbon of the system. This generates the structure in choice C, so choice C is eliminated. Thatstructure can undergo resonance to generate the structure shown in choice D. This eliminates choice D. If waterwere to attack the structure shown in choice C, the structure in choice A, a new intermediate, forms. Thiseliminates choice A. By default, the best answer is choice B. Choice B is not possible, because the structurewould have to gain a proton at the more sterically hindered terminal carbon of the n-system.

, .:--\Ha H,o. \) - ' lra Choice C

eI).\eDI

mChoice A is correct. At 35"C, the hydration of Compound II using sulfuric acid and water yields abottt 40"hsecondary alcohol and 60"h tertiary alcohol. It is stated in the passage that "the percentage of the secondaryalcohol formed increases as the temperature of the hydration reaction increases. This is attributed to a shiftfrom kinetic control to thermodynamic control." This means that at 75'C,it is reasonable to suspect that thesecondary, allylic alcohol is the major product. This eliminates choice C. The product is not a vinylic alcohol,so choices B and D are eliminated. In choice A, a secondary allylic alcohol is formed, so it is the best answer.Taking the information in the third paragraph of the passage and Figure 1, and erasing the cyclopentane ringcould aiso solve this question.

49. Choice D is correct. It is easiest to start by evaluating which pair represents enantiomers. Enantiomers, yourecall, are non-superimposable mirror images. In this case, it is easier to compare the chiral centers ratherthan reorient the structures to see if they are mirror images. If all of the chiral centers differ, then the twostructures are enantiomers. In choices A and C, only one of the three chiral centers differs between the pair, sothey are diastereomers, not enantiomers. The next factor to consider is the alignment of the carbonylsubstituents. They are cis to begin with (on the alkene reactant), so they should finish cis. In choice B, the twocarbonyl substituents are trans, so choice B is eliminated. In choice D, the groups are cis, so it is the best answer.

48.

Copyright @ by The Berkeley Revierv@ 31'/ HYDROCARBONS EXPLANATIONS

50. Choice A is correct. In Triai 4, with methyi groups in positions A and B of the conjugated diene, the rate isroughly 0'005 that of 1,3-butadiene. The reason for thli reduced rate is the steric hindrance associated withthe methyl qrgup pointing to the middle of the transition state. No matter how the molecule contorts, one ofthe two methyl groups is aiways pointing inwards, where the transition state forms. If ethyl groups were usedin lieu of methyl groups, then the steric hindrance would be even more substantial and the reaction rate wouldbe even slower. Only choice A presents a slower reaction rate, so choice A is the best answer.

Choice A is correct' According to the passage, the addition of a Lewis acid to the system increases the rate ofthe Diels-Alder reaction. Choice A, AlC13, is a Lewis acid, because the aluminum lacks a complete octet.Choice B, CCl4, is a common organic solvent where each atom has a satisfied octet. Because CCI4 acts like asolvent and not a Lewis acid, choice B is eliminated. Choice C, KH, is a strong base that reaaity donateselectrons. Because KH acts like a base and not a Lewis acid, choice C is eliminaied. Choice D, LiAiH4, is astrong reducing agent where each atom has a satisfied octet. Because LiAIH4 acts like a reducing agent and nota Lewis acid, choice D is eiiminated. The best answer is choice A.

Choice D is correct. If the reaction proceeds by a nucleophilic mechanism, then we must determine whichmolecule is acting as the nucleophile and which is acting as the electrophile. The reaction is best when thedienopirile has an electron-withdrawing substituent, so lei us assume that the diene is the nucleophile and thedienophile is the electrophile. Based on this mechanism, the part of the intermediate that originally camefrom the diene should carry a positive charge (because it donated electrons) and the part of the intermediatethat originally came from the dienophiie should carry a negative charge (because it accepted electrons). Thisis observed in each answer, choice eicept choice A, so choicJ A is eliminatecl. on the bright side, based on theanswer choices, we know that our assumption about the diene being the nucleophiie anJdienophiie being theelectrophiie is valid. The drawing belolt' shows the forination o? th" first intermediate after nucleophilica ttack.

51.

52.

The second intermediate drawn matches choice D, so choice D is the best answer.been elimi'ated by the incorrect location of the positive charge.

53' Choice A is correct. The dienophile is the same in Trial 2 and Trial 4, so the difference in reactivity must beattributed to the conjugated diene' This eiiminates choice D. Methyl groups, when bonded to a structure, do notact as Lewis acids, because their octets are complete. This eliminates choice C. The methyl groups are mildiyelectron-donating, not eiectron-withdrawing, to .hoi." B is incorrect. The most significant iacior in the reactionrate is that the extra methyl group on the diene in Trial 4 causes steric hindrance in the transition state. Nomatter how the diene contorts, one of the two methyl groLlps interferes with the incoming dienophiie. The bestanswer is choice A.

Choices B and C could have

&ilf*'--->U;Minimal steric hindlance Signi licant steric hindrance

Choice D is correct. It is stated in the passage that a dienophile is enhanced when it has an electron-withdrawing group conjugated to the alkene. Choice A is enhanied by the carbonyl groups conjugated to thealkene, so choice A is eliminated. Choice B, albeit an alkyne and not an alkene, has el-ectronlwlthdrawingester groups conjugated with the n-bond, so it is enhanced as a dienophile. Choice B is eliminated. Choice C isenhanced by the carhonyl groups conjugated to the aikene, so choice C is eliminated. In choice D, the aminegroup is an electron-donating group that lessens the reactivity of the dienophile. Choice D is the best answer.

cHe

a ,f.Ur.3 __>

Copyright @ by The Berkeley Review@ 3r8 HYDROCARBONS EXPLANATIONS

55.

5b.

57.

58.

Choice D is correct. If the Y-group is a carbonyl group, then it is the exact same substituent as the othercarbonyl group (on the adjacent carbon), making the compound symmetric and thus indistinguishabie. BothProduct A and Product B are the same compound, if the reactant is symmetric, so choice D is the best answer.

Choice A is correct. In the second paragraph of the passage it is stated that when X is an electron-donatinggroup/ product A is the major product. Because the OCH3 reactant yields more Product A than the CH3 reactantin comparable reactions, it can be concluded that an OCH3 group is more electron donating than a CH3 group.The best answer is choice A. Choices B and D should have been eliminated, because tie major prlJuct isproduct A, not product B.

Choice A is correct. By analogy, OCH2CH3 (ethoxy) is an electron donating group like OCH3 (methoxy). Thepresence of the eiectron-donating group makes Product A the more favorable product. Product A fiom thegeneric reaction of the passage is choice A. Be careful not to choose B without paying attention to the locationof the double bond. The double bond in choice B is on the side opposite from where it should be.

Choice A is correct. Two five-carbon species are combined, so the final product can have only ten carbonsaltogether. Choice D is eliminated for having twelve carbons total. One o] the double bonds is in the wronglocation in both choice B and choice C. The best answer is therefore choice A. The stereochemistry with thenew cyclopentyl ring trans to the bridging carbon is what is referred to as the "endo product." The arrow-pushing schematic from the reactant to the product is drawn below:

ryO->OrO ..rrrl\\

.,,tttlll

Choice D is correct. Structural isomers have different bonds (connectivity of atoms). Product A and Product B inthe sample reaction in Figure 1 are structural isomers. Structural isomers result when both reactants areasymmetric. The best answer is choice D.

Choice C is correct. For the reaction as drawn in the question, with two asymmetric reactants, there are twopossible structural isomers (corresponding to product A and product B in the generic reaction) that can form. Inboth structural isomers, there are two new chiral centers formed. For a compound with two chiral centers, thereare four (22) possible stereoisomers, meaning that there are four possible stereoisomers for each structuralisomer. The result is that there are eight possible isomers total, so the best answer is choice C. In reality, notall eight isomers are observed to any measurable level in a Diels-Alder reaction. The major product resultsfrom the transition state of least steric hindrance. In a typical Diels-Alder reaction suclt as thir, th" majorproducts are an enantiomeric pair of one of the two possible structural isomers. The less favorable structuralisomer may also be formed, resulting in an enantiomeric pair, but it is generally in much lower concentrationthan the more favorable structural isomer.

Choice D is correct. A concerted reaction occurs in one step. Given that a sigmatropic rearrangement involvesjust one molecule, if it occurs in just one step, then only one product can be formed. This eliminates choice C,because there are not multiple products, let alone cross products. The stereochemistry can be lost at centers thatgo from sp3-hybridizationl, sp2-hybridization and it can be gained at centers that go from sp2-hybridizationto spr-hybridization. Carbons that do not change hybridization cannot experience a change in stereochemistry.This means that there is no set rule about the complete retention or the complete inversion of all stereocenters.This eliminates choices A and B. The only possible answer is the one that supports no cross products beingformed, because the molecule only reacts one way. Choice D is the best answer.

62. Choice D is correct. Step III converts a cyclic ketone into a phenol, so the product has aromaticity that thereactant does not. The gain of aromaticity drives the reaction, so choices A and C are eliminated. Theconversion from a ketone to phenol shifts the n-bond from the carbonyl to the benzene ring, so it is the result oftautomerization, not reduction. The best answer is choice D.

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Choice A is correct. The role of heat in any pericyclic reaction is to provide energy for the reactant to realignits orbitals to achieve the transition state. The best answer is choice A. c-hoice B should have beeneliminated, because exothermic reactions generate heat, so no heat must be added to drive them. Heat isreleased when bonds are formed, sigma or pi, so choices C and D are elimilated.

Choice B is correct. The Cope rearrangement involves a 1,S-diene, so there are six carbons within the molecularorbital of the transition state. Choice A is elimrnated because it has only four carbons. Choice C is eliminatedbecause the orbitals show no fi-overiap between adjacent carbons. Choice D is eliminated because there is nooverlap across the complete cycle. The best overlap is choice B, where the sigma-bond is present o1 the left andthe terminal orbitals are aligned correctly to form i pi-bond.

Choice B is correct. The oxygen is directly bonded to the benzene ring in the reactant, so it is phenylic a'd notbenzylic. This eliminates choices A and C. The oxygen is also bonded to the carbon alpha to ihe aikene. Thismakes the carbon allylic, so choice B is the best answer.

Choice A is correct. The Claisen rearrangement converts an ether into a carbonyl, so the spectroscopic evidencemust depict either the loss of an ether or gain of a carbonyl group. Aldehyde protons shor,n, a signai around 9.5ppm in the 1HNMR, so the formation of in aldehyde wouliin fact correspond with the appearince of a signalaround 9.5 ppm. Choice A is the best answer. Infrared absorbances uro.r,-rd 1700 cm-l inaicate the presence of acarbonyl grouP and broad infrared absorbances around 3400 cm-1 indicate the presence of an alcohol group. Nohydroxyl group appears in either the reactant or product, so choice B is eliminated. The reaction would besupported by the appearance of an absorbance around 1700 cm-1 in infrared spectroscopy, not a disappearance, sochoice C is eliminated' Signals between 5 and 6 ppm in the 1HNMR co.respond to vinylic hydroge-ns bonded toaikene carbons, which are present before and afier the reaction, so choice D is eliminated.

Choice C is correct- The first reaction in the synthesis in Figure 2, Step I, invoives the oxygen. The Claisenrearrangement involves oxygen as the ether is converted into a ketone. This eliminates ihoices B and D.According to the remaining choices, Step II is a Cope rearrangement. To determine the best answer, we mustdecide if the units of unsaturation decrease by one cluring the Claisen rearrangement or r,r.hether they remailconstant at five. In all compounds in Figure 2, there are four n-bonds and one .lig, ,o there are aiways iive unitsof unsaturation. This makes choice C the best answer. You could also conclud"J thut the units of unsaturationdo not change by looking at the Claisen rearrangement in Figure 1, where there are two n-bonds in both theredctant and product.

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Choice D is correct. Terpenes are composed of isoprene subunits which are made of five carbons. To be a terpene,a molecule must have a nurnber of carbons that is divisible by five. Stearol has eighteen carbons, so it cannot bea terpene. The correct choice is D.

Choice C is correct- If the sesquiterpene were derived from a natural source (such as extraction or distillationfrom a plant), then any impurities would be naturally occurring impurities. If there were two enantiomerspresent, that would be explained by attack at a planar site from two sides. This can occur in nature althoughenzymes strongly favor synthesis of one enantiomer over another. Choices A and B are eliminated, becausechiral impurities can occur in nature. The dead give-away would be an impurity with sixteen carbons. Terpeneshave muitiples of five for their carbon values. Because sixteen carbons is not possible, choice C is the bestchoice. A twenty-carbon impurity is a terpene, thurs it is naturally occurring.

Choice D is correct. The carbon that is most susceptible to nucleophilic attack is the carbon with a leavinggroup attached" Carbon four, with the pyropl-rosphate leaving group; is the most electrophilic. Alkene carbonido,act as electrophiles oir occasion, but in this iornpound, carbon four is more electrophilic than an alkenecarbon" The best answer is choice D.

Choice A is correct. Combining three acetyl coenzyme A molecules result in six carbons total. Isoprene unitshave only five carbons, so one carbon must be in a side product. Carbon dioxide contains only one carbon, sochoice A is the best choice. Ethanol and acetic acid each contain two carbons ancl isopropanol contains threecarbons. Choices B, C, and D are all eliminated.

Copl,right @ by The Berkeley Review@ HYDROCARBONS EXPLANATIONS

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Choice C is correct. A Diels-Alder reaction forms cyclohexene, so caryophyllene and citronellol cannot havebeen formed from a Diels-Alder reaction. This eliminates choices A and B. Both c,-pinene and Vitamin 41have a cyclohexene moiety, so we must look closer. Diels-Alder reactions involve a di"ne and dienophile, sowe can look at the compounds in a retrosynthetic fashion. In Vitamin ,A.1, the retro Diels-Alder reaction doesnot generate terpene fragments, so choice D is eliminated. Choice C is the best answer by default.

Choice B is correct. Bond a can be eliminated immediately, because the fragments formed from the break arethree carbons and seven carbons. Bond c can be also eliminated immediately, because the fragments formed fromthe break are nine carbons and one carbon. These are not multiples of five, therefore the two fragments cannotbe involved in the synthesis. This eliminates choices A and C. Bond b and Bond d when broken can leave a tencarbon molecule, so neither can be eliminated. The trouble with bond d is that the fragment to the right of thebreak cannot form a 2-methylbutene, because it loses the tertiary carbon. Choice D is eliminated. Isopreneunits must be isopentenyl, not straight chain pentenyl, thus the break is not allowed. The two possible retrosynthesis pathways are shown below, and only Bond b is involved. Choice B is the best answer.

Bond d Bond d

CFI?

Bond b must have been formedto connect the isoprene units.

CIT

None of the labeled bonds wereformed to connect the isoprene units.

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Choice A is correct. Carvone differs from limin by a carbonyl group. To go from iimin to carvone, a carbon mustlose two bonds to hydrogen and gain a double bond to oxygen. This is oxid.ation, so choice A is the best answer.

Choice A is correct. Ozonolysis is the oxidative cleavage of a double bond between two carbons. The resultingproducts are carbonyl compounds that vary from aldehydes to ketones to carboxylic acids, depending on thework up step. To undergo ozonolysis, the reactant must contain an alkene functional group. All of the compoundshave an alkene functionality except for camphor. This makes choice A, camphor, the best answer.

Choice B is correct. Singlets in the proton NMR are caused by unique hydrogen atoms in an environment wherethe adjacent atoms have no bonds to hydrogen, and thus there are no neighboring hydrogens with whichcoupiing can take place. In camphor, ali of the methyl groups are bonded to quaternary carbons, so they all fitthis description. Because the cyciic structure is incapable of rotation, like an alkene, the two methyl groupsbonded to the bridge carbon are not equivalent, causing them to express different NMR signals. The result isthat each of the methyl groups are represented by a singlet in the proton NMR. All of the remaining hydrogenson camphor are on carbons adjacent to neighboring carbons with hydrogens, so there are no other singlets thanthe ones from the methyl groups. This generates three proton NMR singlets, so the best answer is choice B.

No Hs on neighbor, cannot berotated to be equivalentwithother bridge methyl group.

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No Hs on neighbor, cannot berotated to be equivalent withother bridge methyl group.(3H singlet)

H

No Hs on neighbor, isolatedg H3cs )g-- methyl grorrpl (3H singlet)

77. Choice C is correct. Myrcene contains ten carbon atoms, so the addition of another isoprene unit would result in aproduct with fifteen carbons total. According to the first paragraph of the passage, terpenes having fifteencarbons are referred to as sesquiterpene, making choice C the best answer.

Copyright @ by The Berkelel, Revie*'3 321 HYDROCARBONS EXPLANATIONS

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Choice B is correct. Can-rphor has a carbonyl group (water soluble) and a iarge alkyl ring system (not watersoluble). It is hard to decide based on the structure. It happens that the compound is watei soluble, which youmay know first hand from using camphor-containing cleaning agents for skin. The question is whether or not itis highly water soluble. Because there is some ambiguity, let's say for now thai it 1ikely not highly watersoluble, and consider statement I to be invalid. Camphor is a liquid at room temperature, as you ti-tignt n^rr"seen if you synthesized it in a lab experiment. Being a liquid at standard temperature, its boiling point"is above298 K. Statement II is valid. Camphor has two chiral carbons, so it rotates plane-polarized lighi. This makesstatement iII invalid. Choice B is the best answer, but not with one hundred percent certainty.

Choice D is correct. It is stated in the passage that carvone has a strong UV absorbance (e > 10,000). Carvonehas conjugation, which causes its intense UV absorbance. On the other hand, limin has no conjugation, so its UVabsorbance is not as intense as that of carvone. This means that the UV absorbance for limin his an e less than10,000 (therefore, log e < 1og 104 = 4). The best answer is choice D. This question required some backgroundinformation on UV spectroscopy. The minimum you should know is that n-bonds ire UV active, and withconjugation, the intensity of the absorbance increases and the energy of the absorbance decreases.

Choice C is correct. A Diels-Alder reaction is a cyclization reaction that involves the addition of a diene to adienophile (alkene) to form a cyclohexene product. Both limin and carvone are cyclohexene compounds,eliminating choices A and D, but carvone has a carbonyl group and isoprene contains only Cs and Hs. The bestanswer is limin, choice C.

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Choice B is correct. It is stated in the passage that corn oil is 63'/, iinoleic acid. Looking at table 1 shows thatlinoleic acid is made up of eighteen carbons and has two n-bonds. Choices C and D are eliminated, because theyonly have sixteen carbons. Choice A is eliminated, because it l-ras three n-bonds. The best answer is choice B.You can try to match the exact location of each n-bond from the formula in Table 1 to the drawing in the answerchoices, but doing so is not time efficient.

Choice D is correct. Linoleic acid contains two n-bonds, both with cis geometry. When linoleic acid is treatedwith FADH2, the result is hydrogenation of the diene and the formation of the aliphatic carboxylic acid (ofeighteen carbons) stearic acid. The gain of four hydrogen atoms increases the moleCular mass of tne acid (byfour), and the loss of unsaturation results in more molecular flexibility, which results in a higher melting poini.Unsaturated fats, with less flexibility and therefore less ability to engage in intermoleculaiinteractioni, havelower melting points than saturated fats of comparable mass. This is Co**orl organic chemistry knowledgethat you should have addressed when comparing vegetable and animal fats. The coirect answer is choice D.

Choice C is correct. Oleic acid is an eighteen-carbon acid with one n-bond between the eighth and ninthcarbons. The a-bond in oleic acid has cis orientation. Treating oleic acid with deuterium (D2) and a catalyticmetal like pailadium adds two deuterium atoms across the n-bond of the alkene molecule. The two deuteriumatoms add syn to one another at carbons eight and nine. The result is the formation of two new chiral centers.There are no chiral centers to begin with, so the product has tr.tro chiral centers. The best answer is choice C.

Choice D is correct. Potassium permanganate reacts with alkenes to form diols by adding two hydroxyl groupsis a syn addition fashion to the carbons of the n-bond. Linolenic acid has three n-bonds, located betweencarbons 9 and 10, carbons 12 and L3, and carbons 15 and 1,6 when the carboxylic acid carbon is considered to becarbon one (IUPAC convention). Hydroxyl groups form at all sp2-hybridized carbon sites. This results in aproduct with hydroxyl groups at carbons 9,70, 12,13, 15, and 16, as listed in choice D.

Choice B is correct. The most bromine per moiecule is consumed by the fatty acid with the greatest number of n-bonds present. For every n-bond present, one molecule of bromine tiquid will be consumed. Arachidonic acid hasfour n-bonds. Arachidic acid has no n-bonds present in its structure, linoleic acid has two n-bonds present in itsstructure, and linolenic acid has three n-bonds present in its structure. Arachidonic acid is the mosiunsaturatedof the choices. The correct answer is thus choice B.

Choice B is correct. Paimitoleic acid has sixteen carbons and one n-bond. When palmitoleic acid is fuliyhydrogenated, it forms the aliphatic acid of sixteen carbons (listed in Table t as pilmltic acid). The bestanswer choice is B.

Copyright O by The Berkeley Review@ 322 HYDROCAR.BONS EXPLANATIONS

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Choice C is correct' Figure 1 shows occidentaiol with three specified chiral centers, so choices A and B.ffi::,il;lXl*"1'fli,*:lr":::,-".11",::lE_::ly:lt1: "itn".iul" ,;a_i;;fi;ii;" or they are methyrene(cHz) sroups' rhe methvr group and tertiirv arcohol carbon o" ""inl.i#J:T:.ffi:ffil":J""f:tJ:ii::chiral centers on the structure. The best answer is choice C.

Choice D is correct' The first paragraph states that occidentalol is a sesquiterpenoid. occidentalol has fifteencarbons total' so it is a tuatotlibl" conclus.ion that sesquiterpenes hurr" iift""r-r carbons. Choice D is the bestanswer' Monoterpenes have ten carbons, diterpenes trave twenty carbons, and triterpenes have thirty carbons.

choice A is correct' The maximum wavelength of absorbance, r*u*, increases as the conjugation of the rc-network increases' All four choices are conjugaied dienes, but Compound 1 also has a carbonyl in the conjugatednetworks' As such, the compound with the ioigest L'u* is Compound 1, choice A.

Cl'roice B is correct' step 1 involves a Diels-Alder reaction followed by decarboxylation. The intermediatecompound is the Diels-Alder product' There is. no nitrogen present in either reactant, so the compound cannot bea lactam (cyclic amide)' Thii eliminates choice ? Tie p.oar.t of a Diels-Alder reaction involving a dieneand an alkene (dienophile) is cyclohexene, so choice s is the best answer. The diene in Compound 2 isregenerated from cyclohexene after decarboxylation, so choice A is eliminated. The reaction is shown below:

150'C____*>

9"t_.

92.

o Cyclohexene- Lactone (not lactam)

CO2Me Carbonyl is NOT conjugated to alkene

Compound 1.5 ( intermediate compound in Step 1)

The compound is a cyclohexene with a lactone that is not conjugated to the alkene.

Choice B is correct' occidentalol has one. more me-thyl group than compound 6, so the role of methyl lithium,MeLi' must be to add a methyl group to compouna o. int eliminates choices C and D. Methyt lithium has ananionic carbon' so it acts as a nucleophile rither than an electrophile. This eliminates choice A and makeschoice B the best answer' The methyi anion attacks the carbonyl carbon in the same fashion as an alkylmagnesium bromide anion attacks a caibonyl in a Grignard reaction.

Choice A is correct' A conjugated diene can be protonated at either terminal carbon of the ru-network, becausethe carbocation that results is resonance stabilized. This eiiminates choices B and C. Carbon a is a secondarycarbon while carbon d 1s l jgrtiary carbon. It is easier to protonate the less hindered site, so carbon a is the sitethat is most likely to gain H+. The best answer is choice A.

93' Choice D is correct' Leukotriene 44, LTA4, has four alkene n-bonds, one carbonyi n-bond, and an epoxide ring.Il':-:"::i:i::::i::^:l,T'"*'l.lion. choice,A is a valid sratement, which eliminates it. LrAa has threealkene n-bonds in conjugation, resulting in six n-electrons in a conjugated system. cnrr.?i:r,

" Jtffnri"1"""."",,which eliminates it' Because of the extensive conjugation, there are several sites at which a nucleophile andelectrophile may add' For instance, if the epoxide o*yg"n were protonated, a nucleophile could attack the ringor the left carbon of any n-bond to add u"ros the slstem, Tilis means that r,2-addition, r,4-addition, 1,6-addition, and 1,8-addition are all possible. Choice C is a valid statement, which eliminates it. In alllikelihood, this answer choice uurrr"d the coveted "huh?", meaning you can,t eliminate it, because you're justnot sure" on a multiple choice exam, this is not a problem. You just need to look at choice D and use your testinglogic' LTAa has four alkene n-bonds and one carbonyt n-bond, so tn"." ur"

'ri'r"-rp'r+yrrialzed carbons. Thereare twentv carbons total in LTA4, so eleven of them.are sp3-hybridlzed. There "r"

_.1",pl;;,[;;""urUo.*than sp2-iybridized carbons, ,o .hoi." D is an invalid statement, which makes it the best answer.

C02Me

Compound 1

CO2Me

Copyright @ by The Berkeley Revierv@ .11i HYDROCARBONS EXPLANATIONS

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Choice D is correct. ,Bromine, in the presence of light, adds to the most substituted carbon of an alkane by wayof a free radical mechanism. The most substituted carbon is tertiary, so the choice with bromine added to thetertiary carbon (most substituted) is the best answer. This makes choice D the correct choice. T'he productshown, as well as its enantiomer, are both formed.

Choice D is correct., Free radical propagation reactions keep the free radical reaction going, so to be apropagation reaction, there must be the same number of free radicals on each side of the equation. In choice A,there is one free radical on the reactant side and three free radicals on the product side, so it is not apropagation step. Choice A is eliminated. In choice B, there are two free radicals on the reactant side and nofree radicals on the product side, so it is a termination step and not a propagation step. Choice B is eliminated.In choice C, there is one free radical on the reactant side and two free radicals on the product side, so it is not apropagation step. Choice C is eliminated. In choice D, there is one free radical on the reactant side and onefree radical on the product side, so it is a propagation step. Choice D is the best answer. You may not recognizethe reaction from the overall mechanism, but it converts a less stable free radical into a more stable freeradical, which ultimately impacts the product distribution.

Choice A is correct. Bergamontene contains fifteen carbons, so it is likely made from three S-carbon isopreneunits. We can't be sure without analyzing the structtire to find the isoprene fragments, but that is not timeefficient. Choice A is the best answer so far, and shall remain our choice until a better one comes.Bergamontene is a hydrocarbon with no heteroatoms, so it is a terpene and not a terpenoid. This eliminateschoice B. Bergamontene has two n-bonds and a cyclohexane ring, so at first look choice C is tempting. But themolecule is bicyclic, meaning it has a second ring, the four-membered ring connected to the cycioheiane ring.Bergamontene has four units of unsaturation, not three. This eliminates choice C. To verify this, bergamontenehas 24 hydrogens and therefore a formula of C15H24. The units of unsaturation are 1z1iS; + Z -24\/2 = 4, sochoice C is eliminated. There are three chiral centers on bergamontene, so 8 (23) is the maximum number ofstereoisomers, not 16. This elimir-rates choice D and secures choice A as the best answer.

Choice D is correct. Both the Claisen rearrangement and the Cope rearrangement require dienes, but they neednot be conjugated. The two n-bonds must be separated by three sigma bonds, so choices A and C are eliminated.Clemmensen reduction converts a carbonyl into an alkane, so no diene of any kind is required. Choice B iseliminated. A Diels-Alder reaction involves the cyclization of a conjugated diene and a dienophile, so it mustl-rave a conjugated diene. This makes choice D the best answer.

Choice C is correct. For an E2 reaction, the base must be strong enough to remove a proton from carbon and bulkyenough to not undergo substitution. This makes choice A a valid statement and thereby eliminates it. For an E2reaction, the leaving group and proton being lost from carbon must be positior-red anti to one another, so thegeometry of the product is dependent upon the alignment of the reactant. Cis versus trans results from theorientation and stereochemistry, so choice B is a valid statement and thereby eliminated. For an E1 reaction, aleavilg group first leaves, resulting in a carbocation. With carbocations, rearrangement can be observed, so it iswith E1 reactions that we see rearrangement, not E2 reactions. Because E2 reactions are concerted, there is norearrangement, so choice C is an invalid statement and thereby the correct answer. Heat is required to driveboth El and E2 reactions, so choice D is a valid statement. It is eliminated, leaving choice C as our choice.

Choice D is correct. Terpenes are hydrocarbons of 70,15,20, etc... carbons, so they are somewhat massive lipids.Because they are hydrocarbons, they are lipid soluble, so choice A is a valid statement. Choice A isconsequently eliminated. Terpenes have molecular masses of about 140 g/mole, about 210 g/mole, about 280g/moie, etc..., so they have somewhat high boiling points. High is a relative term, so we ian't be certain ineliminating choice B. However, choices B and C are the same concept, so they mutually exclude one anotherfrom consideration. It is important that you use all of your test taking stcllls. The specific rotation of acompor-rnd is dictatecl by its chiral centers, which a terpene may or may not have. Given that there is nogeneral rule about the chirality of terpenes, we cannot conclude that they have high specific rotations" ChoiceD is the best answer.

Choice C is correct. To have a dipoie nol eqr-ral to zero is to have a dipole. To have a dipole is to be polar. Ciscompouuds are always polar so Compound III is polar. Ethvlene is perfectly symmetric, io choice Compound IIis nonpolar. The question comes down to: "Is Compound I polar?" Compound I is not polar, because thl methylgroups on t1-ie alkene cancel one another and sum to a resultant vector of 0. Choose C for best results.

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