the berkeley review mcat organic chemistry part 1

OrganicChemistry

Part ISections I-IV

Section IStructure, Bonding, and Reactivity

Section IIStructure Elucidation

Section IIIStereochemistry

Section IVHydrocarbon Reactions

,

BERKELEY

Specializing in MCAT Preparation

ERKELEYR • E • V • I • E • W

F.O. Box 40140, Berkeley, California 94704-0140Phone: (800) 622-8827Internet: [email protected]

(800) M C AT-TBRhttp://www.berkeleyreview.com

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Section I

Structure,Bonding,

and

Reactivity

ii

by Todd Bennett

H H

* /C C,

/ Y*HH H

BerkeleyUr-E-V-I^E'W®


Nomenclature

a) IUPAC Nomenclature

b) General nomenclature

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

d) Molecular Orbitals and Bonds

i. Single Bondsii. Double Bonds

iii. Triple Bondse) Molecular Structures

0 Octet Rule (HONC Shortcut)

9) Charged Structures

Hybridizationa) Hybridization of Atomic Orbitals

i. sp-Iiybridzationii. sp2-Hybridzationiii. sp3-Hybridzation

b) Common Shapes

Bond Energy

a) Bond Dissociation Energyb) Ionic Bonds

Intramolecular Features

a) Resonance

b) Inductive Effect

c) Steric Hindrance

d) Aromaticity

Fundamental Reactivity

a) Proton Transfer Reactions

b) Lewis Acid-Base Reactions

c) Acid and Base Strengthi. Primary Effectsii. Secondary Effectsiii. Values and Terminology

d) Electrophiles and Nucleophiles

Physical Propertiesa) Hydrogen-Bondingb) Polarityc) Van der Waals Forces

d)

,EYSolubility and Miscibility

o»

•>

Structure, Bonding, and ReactivitySection Goals

Be able to correlate structures with common and 1UPAC nomenclature.

It is expected that you can answer questionsabout organicmoleculeswhen given either the molecularstructure or the name of the compound. You will need to be able to rapidly convert from name tothe structure or from the structure to name. You are expected to be able to recognize commonstructural features like substitution and functional groups. Youmust be able to name small organicmolecules according to IUPAC conventions.

Be able to predict relative bond lengths, bond strengths, and structural angles.

Therewill be questions that require you to comparethe structural features of similar molecules. Youshould know the hybridization-to-bond angle correlation. Youshould also know what effect the s-character of a hybrid orbital has on bond length and strength. Youmust know the common moleculargeometries and shapes and how they correlate to hybridization. Youmust be able to read data tablesand explains trends in bonding features.

Be able to draw resonance structures and determine which is the most stable.

Some questions on the MCATrequire you either to count the resonance structures or determinewhichresonance structureis moststable. Stable resonance structureshave octetstabilityabout allatoms except hydrogen, haveminimal charges, andwhencharges arepresent, negative charge resideson the more highly electronegative atom and positive charge resides on the less electronegative

Know the structure of aromatic compounds and their unusual stability.Benzene isthetypical aromatic compound, because it iscompletely planar with a cyclic, conjugatedarrangement of ^-electrons that obey Huckel's rule. Huckefs rule states that aromatic compoundsmust have 4n + 2 n electrons in a cyclic array ofp orbitals, where n is any integer (orzero). Benzenehas aromatic stability due to its sixnelectrons in a continuous cyclic arrayofp orbitals. You shouldbe able to recognize aromaticity in structures other than benzene too.

Know the common organic acids and bases and their reactivity.Common organic acids include phenols, thiols, protonated amines, andcarboxylic acids. Commonorganic bases include amines and carboxylates. Youshould be able to determine the direction of aproton transferreaction fromthe pKa values. You shouldbe ableto giveapproximate ratiosof theconjugate acid and base in a buffered solution. You must have a solid understanidng on therelationship between pH and pKa.

Be able to predict the relative acidity and basicity of organic compounds.Acidity is determined bybothprimaryeffects (involving atomsthat aredirectly bonded to theacidicProton) and secondary effects (involving atoms that are not directly bonded to the acidic proton),

rimaryeffects include atomic size, electronegativity, and hybridization. Secondary effects includeresonance, theinductive effect, aromaticity, andsteric hindrance. You mustknow therelative impactof the various effects.

Be familiar with fundamental reactions, energetics, and mechanisms.Knowing that reactions in organic chemistry involve the interaction of electron-rich sites withelectron-poor sites, be able to identify the reactive sites of organic molecules. You must have afundamental understanding of electrophiles and nucleophiles and how they interact in transitionstates. It is alsoimportant that you be ableto correlate a reaction mechanism to an energydiagram,identifying reactants, transitionstates,intermediates, and products.

Be able to determine relative boiling and melting points.Physical properties like boiling and melting point are the result of intermolecular forces such ashydrogen-bonding, dipole-dipole interactions, and Van der Waals forces. You should be able topredict the effect ofintermolecular forces, molecular mass,and structuraldetails(like branchingandthe presence of rc-bonds) on the physical properties ofa compound. You shouldbe ableto predictrelative physical properties from functional groups.

Organic Chemistry Molecular Structure

Molecular StructureThe 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 of

atomic orbitals overlapping to form molecular orbitals. 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 electrons 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 acomplete valence electron shell. Bysharing electrons,each atom moves closer tocompleting its shell. This is the driving force behind the formation of stablecovalent bonds.

Having looked briefly at electron distribution, we can introduce the idea ofelectronic orbitals, which are three-dimensional probability maps of the locationof an electron. 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 onsimilarities in their bonding of particular atoms, known asfunctional groups. Eachfunctional group shall be considered in terms of its unique electron distribution,hybridization, 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 particular organic molecule. The review of nomenclature iscontinuous throughout all sections of this book.

Then, we shall consider the factors that affect the distribution of electron 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 aselectrophiles. If you can correctly label a molecule in terms of the region thatcarries 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 critical thinking skills, which will carry over into yourreview for other portions of the exam. Despite what you may have perceivedwas a girth of information when you initially studied organic chemistry, youdon't need to review that much material to prepare successfully for the MCAT.

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Introduction

Exclusive MCAT Preparation

Organic Chemistry Molecular Structure Nomenclature

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. Beaware that "R" stands for any genericalkyl group. When R is used, it indicates that the carbon chain size is irrelevantto the reaction. Table 1-3summarizes the nomenclature process by listing severalfour-carbon compounds.

Carbons Prefix

1 meth-

2 eth-

3 prop-

4 but-

Carbons Prefix

5 pent-

6 hex-

7 hept-

8 oct-

Carbons Prefix

9 non-

10 dec-

11 undec-

12 dodec-

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Table 1-1

Functionality Compound Name Bonding

R-CH3 Alkane C—C & C—H

R-O-R Ether C—O—C

R-CO-H AldehydeOII

C—C—H

R-CH2-OH Alcohol C—O—H

R-CO-R Ketone

O

IIC—C—C

R-CO-OH Carboxylic acid0II

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

The Berkeley Review


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

® l^.O

3-methylpentanoic acidLongest chain: 5 carbons

Carboxylic acid groupMethyl substituent at C-3

3-ethylcyclopentanoneRing of 5 carbons

Ketone groupEthyl substituent at C-3

Figure 1-1

4-chloro-5-methyl-3-heptanolLongest chain: 7 carbons

Alcohol groupChloro substituent at C-4

Methyl substituent at C-5

©

Br Br H

3,3-dibromobutanal

Longest chain: 4 carbonsAldehyde group

2 Bromo substituents at C-3

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 carbonatoms attached to thecentral carbon atom (carbon atom of interest). A carbon with one other carbonattachedis referred to as a primary (1°) carbon. A carbonwith two other carbonsattached is secondary (2°). A carbon with three other carbonsattached is tertiary(3°). Figure 1-2shows some sample structures.

H CH3\ ^ Tertiary carbon ^ ? Secondary carbon \ ? Primary carbon

H3CH2C OH H3CH2C CIIsobutane Sec-butanol n-Propyl chloride

(2-Methylpropane) (2-Butanol) (1-Chlropropane)

s<H,C QH,

H CH3 H H

Figure 1-2

Nomenclature is an area of organic chemistry best learned through practice andexperience. We will deal with nomenclature throughout the course, as weintroduceeachnew functional group. Understandingnomenclature is especiallyimportant in MCAT passages where names rather than structuresare given. Besure to know the Greek prefixes for carbon chain lengths up to twelve carbons.


nomenclature


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 1.1What is the Lewis dot structure for H2BF?

A.

:h—b-

1H

• •

• •

-f:• •

B.

:h-• •

-b—f:

1 -H

• •

C.• •

H—B-

1H

• •

-f:• •

D.

h--b—f:

1 -H

Solution

Boron has only three valence electrons, hence it can make only three bonds.There is no lone pairon theboron atom, elirninating choices A and C. Hydrogenhas onlyoneelectron, which is in thebond toboron,so thereis never a lonepairon a bonded hydrogen. This eliminates choices A (already eliminated) and B.Fluorine hasa completed octet, so it makes onebondandhas 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 themostprobable location foran electron. In mostcases, with the exception of ligandbonds (known also asLewis acid-base bonds), oneelectron from each atomgoes intoforming the bond. When electrons are shared evenly between two atoms, thebond is said to be a covalent bond. When electrons are transferred from one atomto another, the bond is said to be an ionic bond. The difference between a covalentand ionicbond 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 polarityof the bond. Therelativeelectronegativity of two atoms canbe determined by measuring the dipole of the bond they form. When thedifference in electronegativitybetween 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 1.5but less than 2.0, thenthe bond is said to be polar-covalent (or partially ionic).

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Example 1.2Which of the following bonds is MOST likely to be ionic?

A. C—O

B. N—F

C. Li—H

D. Li—F

Solution

A 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 Bonds

Bonds can be classified in one of three ways: ionic, polar-covalent, and covalent.A covalent bond occurs when electrons are shared between two atoms havinglittle to no difference in electronegativity. As the difference in electronegativitydecreases, the covalent nature of the bond increases. There are two types ofcovalent bonds: sigma bonds (a), defined as having electron density sharedbetween 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 exclusivelyofparallel p-orbitals. In almost all cases, the sigma bond is stronger than the pibond, with molecular fluorine (F2) being a notable exception. Figure 1-3shows ageneric sigma bond. You maynotice thatwithin a sigma bond, only about eightytoninety percentof the electron density liesbetween thenuclei, not allof it.

.Nuclei

o CD

Electron density

Figure 1-3

Example 1.3Which drawing depicts theelectron density ofa carbon-carbon sigma bond?

A. L~D

C D

Solution

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


Bonding and Orbitals


Organic Chemistry Molecular Structure Bonding and Orbitals

Figure 1-4 shows a generic 7C-bond. Within a rc-bond, there is no electron densitybetween the two nuclei. The electron density hi a rc-bond results from electronsbeing shared between the adjacent lobes of parallel p-orbitals.

Nuclei

Electron Density

Figure 1-4

In organic chemistry, covalent bonds are viewed in great detail, taking intoaccount hybridization and overlap. In alkanes, carbons have srAhybridizationand all of the bonds are sigma bonds. In alkanes, there are two types of bonds:C— H (GSp3-s bonds) and C— C (CJSp3-Sp3 bonds). In alkenes, there are sigmaand pi bonds present. The 7t-bond consists of p-orbitals side by side, and itscarbons have sp2-hybridization. The C=C bond is made up of a GSp2_Sp2 bondand a 7t2p-2p bond. Bond length varies with the size of the orbitals in the bond.For instance, a sigma bond composed of an s/?2-hybridized carbon and an sp^-hybridized carbon isshorter than a sigma bond comprised of two sp^-hybridizedcarbons. Hydrogens use s-orbitals to form bonds. Figure 1-5shows three sigmabonds with their relative bond lengths. The longer bond is associated with thelarger orbitals (bond radii: dz>dy >dx).

asp2-sp2 <V-sp3

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 lie between the two nuclei. 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 re-bond between two 2pz-orbitals, which is typical for nearly alln-bonds encountered in organic chemistry,because carbon, nitrogen, and oxygen have 2p-orbitals in their valence shells.

2pz-2pz

Figure 1-6

Copyright © by The Berkeley Review 8 The Berkeley Review


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 one7t-bond, withno o~-bond present. This is attributed to the smallsize 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 CI2, eventhoughchlorine 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 whencoupled with a bonding orbital. It is important to recognize theshape andlocation ofelectron density inmolecular orbitals. Figure 1-7 shows thecommon bonding and anti-bonding orbitals associated withorganic chemistry.

^>

sp sp

Sigma bonding molecular orbital

<V-sp3

p o3 3 CT*«n3 ,n3sp sp SP _SP

Sigma anti-bonding molecular orbital

P

P P 7C2p-2p

Pi bonding molecular orbital

$

P P 'L'2p-2p

Pi anti-bonding molecular orbital

Figure 1-7

The shading of the lobes in each orbital represents the direction of spin for theelectron. In 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 samedirection as traffic is flowing, youcan easily blend into traffic. This is afavorable interaction. If you join a freeway in theopposite direction as traffic isflowing, you cannot easily blend into traffic. This is an unfavorable interaction.





Molecular Bonds

Of 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 thesigma and pi nature of bonds. In organic molecules,there are only single, double, and triple bonds. Between like atoms, thedescending order of relative strengths of bonds is triple bond > double bond >single bond. Another rule to consider is that for bonds between like atoms, thelonger the bond, the less the electron density overlaps between nuclei, and thusthe weaker the bond. This is summarized as: longer bonds are weaker 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 fewer electrons are present. Ethane has sigma bonds onlyand is shown in Figure 1-8 in both stick figuresand with the relevant orbitals.

H H HH

IiH

H /OCc C, o

/ t*H /

/op

H

<V-sp3 H

Ii H Ii

Figure 1-8

Double Bonds: Double bonds are composed ofone sigma bond and one pibondbetween two adjacent atoms. Ethene (C2H4) has four sigma bonds betweencarbon and hydrogen, a sigma bond between the two carbons, and a pi bondpresent between the two carbons to complete the carbon-carbon double bond.Ethene isshown inFigure 1-9 inboth stick figures and withthe relevant orbitals.

H/,">... ..»tfH

rHH1C= C

^sp--sp-

Figure 1-9

Triple Bonds: Triple bonds are composed of a sigma bond and two pi bondsbetween two adjacent atoms. Triple bonds are shorter than either single ordouble bonds. Ethyne (C2H2) has two sigma bonds between carbon andhydrogen, asigma 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.

py-py

C c

G s-sp

Figure 1-10

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Example 1.4Whatis the relativebond strength of carbon-carbon bonds in the molecule shownbelow?

bondb

H3C I CH3bond a -*\ I /

c=tc

*/ \bond c

H,C

H^ VCH2CH3Sf

H | CH3bondd

A. Bond a > Bond b > Bond c> Bond dB. Bond b > Bond a > Bond c > Bond d

C. Bond d > Bond a > Bond c> Bond bD. Bond b > Bond c> Bond a > Bond 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-hybridizedcarbons. Bond a is stronger than bond c, despite both sharing ansp2-hybridizedand an sp3-hybridized carbon, because bond c contains the more highlysubstituted carbon. Choice B is the best answer.

Molecular Structures

We shall continue from the fundamental concept that a valence electron can beshared between two nuclei rather than being isolated to justone nucleus, becausethe attractive force of two positive sites is greater than the attractive force ofone.This is the basic, perhaps oversimplified, perspective ofa chemical bond. Thesharing ofelectrons 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 ofcovalent bonding like single, double, and triple bonds andtheir component a-bonds and rc-bonds). Table 1-4 shows the skeletal structuresof molecules that containcarbon, nitrogen, oxygen, and hydrogen.

AtomValence

Electrons

To CompleteShell

Number of Bonds inCompounds

Neutral

Carbon (C) 4 •£• 4 e" needed 41 _/ .—C^

Nitrogen (N) 5 -N- 3 e" needed 3N-. =N* :n=

Oxygen (O) 6 -O-• •

2 e" needed 2 JK -*Hydrogen (H) 1 H« 1 e" needed 1 H—

Table 1-4





Octet Rule and the HONC Shortcut:

Every molecular structure should have atoms that obey the octet rule (eightvalence electrons forC,N,and O). Thenumbers ofelectrons needed to completethe shellin the Table 1-4 are derivedfrom the electrons needed to obey the octetrule. All neutral structures have atomic arrangements as described in Table 1.4.To complete the octet valance shell, carbon requires four electronpairs in theform of bonds, nitrogen requires one lone pair in addition to the three bonds itmakes,and oxygenrequires 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). Inaneutral compound, hydrogen makes one bond,oxygen makestwo bonds, nitrogenmakes three bonds, and carbon makesfour bonds. Neutralstructures always obey this rule. Figure 1-11 shows examples of valid andinvalid structures anda brief description ofthebonding to thecomponent atoms.

H CH3\ / 3C=C

/ \H H

All carbons have 4 bonds.

All hydrogens have 1 bond.

Good Structure

H3C—C=C— CH2CH2CH3

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

Good Structure

H CHo\ / 3N=C

\CH3

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

Nitrogen has 3 bonds.Good Structure

H CH2

C=C/ \

H H

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

Bad Structure

CH:

H,C— c—o

CH2Cri3

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

but oxygen has 3 bonds.Bad Structure

H,C CH,\ /N=C

/ \H H

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

but nitrogen has 4 bonds.Bad Structure

Figure 1-11

You can validate molecular structures byseeing whether they satisfy bondingrules (HONC-1234) and conventions with regard to the number of bonds andlone pairs. Ifa 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 ananion, except with carbon. For instance, if oxygenmakes three bonds and has one lone pair, it carries a positive charge. Whennitrogen makes two bonds and has two lone pairs, it carries a negative charge.When carbon makes three bonds, the charge depends onthepresence or absenceofa lone pair(presence yields ananion, while absence yields a cation).

Copyright ©by TheBerkeley Review 12 The Berkeley Review


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 typicallyhas sixvalence electrons and wishes to have eight. This means that oxygen makes twobonds to complete its valence shell (and thus satisfy the octetrule). However, ifan oxygen atom had only five valence electrons, it would be short one electronfrom its original six and would consequently carry a positive charge. Havingonly five valence electrons, the positively charged oxygen wouldneed to makethree bonds (one more than its standard two) to complete its octet. We canconclude that oxygen with threebonds carries a positive charge. Table 1-5 showssome common organic ions to commit to memory:

Number of Bonds

to Neutral Atom

Number of Bonds to

Cationic Atom

Number of Bonds to

Anionic Atom

Carbon (C)

4 •&•

3 1C+ =C+—

3 • •_ •_, ^^

Nitrogen (N)• •

3 *N*•

4 1 /^N-™,, =N+ —N+S

Oxygen (O)

2 •(>• •

%o$™,, —o*' :o+= i -o-:• •

Table 1-5

Drawing Molecular StructuresDrawing molecular structures from a given formula requires following the octetrule for all atoms except hydrogen. On occasion, there 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.

Molecular and

Structural Formula

C2H60CH3CH2OH

C2H7NCH3NHCH3

C2HsO+CHgCHO+H

Lewis Structure

H HI I ..

H—C—C—O-HI I '•

H H

H H HI I I

H—C—N—C—HI •• I

H H

H HI I

H—C— C=0+-HI

H

Figure 1-12


3D Structure

H

.c—c

THH« '.O-H

13

HH

.C— N"

HH

HH

.C— H

H H\ /.c—C

sg| WH

H 0+-H



Organic Chemistry Molecular Structure Hybridization

*&&&*£•CM*.

Hybridization of Atomic Orbitals

Hybridization entails relocating electron density in atomic orbitals prior tobonding, in order to minimize the repulsion between electron pairs and therebyallow for bondingbetween atoms. There are three main types of hybrid orbitalstoconsider: sp, sp2, and sp3 hybrids. Hybrid orbitals are atomic orbitals that areinvolved inmaking bonds between atoms. Listed in Table 1-6 aresome pertinentfacts and structural features foreach ofthe three types ofhybridization. Table 1-6represents general trends that are observed in nearly all molecules withhybridized orbitals involved in their molecular orbitals.

Hybrid sp sp2 sp3Atomic Orbitals s + p s + p + p s+p+p+p

Angle 180° 120° 109.5°

Shape linear trigonal planar tetrahedral

a-bonds and e" pairs 2 3 4

rc-bonds 2 1 0

Table 1-6

The number ofrc-bonds in Table 1-6 is for typical molecules thatobey 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. Inboranes, the boron atom has only threevalence electrons, so a neutral boron cannot satisfy the octet rule. Theresult isthat boron has sp2-hybridization for its three sigma bonds, but no pi bond.

Structures with Orbitals

Lewis structures are usedas shorthand representations of molecules. However,in organic chemistry, molecules shouldbe visualized in threedimensions, whichhybridization helps to facilitate. 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 with orbitals and three-dimensionalorientation. Structures should be drawn with and correct bond lengths and bondangles should be basedon hybridization, sterichindrance,and VSEPR rules.

Copyright©by The Berkeley Review

sp -hybridized

BH „—?<H

sp -hybridized 0•N: "IIInh3 h-^ ayH

H

sp -hybridizedH

CH.H'

H

sp -hybridizedH

H2° H-^Sf?Figure 1-13

14 The Berkeley Review


The orbital shown for BH3 in Figure 1-13 is actually an empty p-orbital, while theother orbitals depicted in Figure 1-13 are hybrid orbitals. The pz-orbitalof BH3 isdevoid of electrons, so the hybridization is sp2. While an empty p-orbital doesnot actually exist, we consider the region where an electron pair could beaccepted. The threehybrid orbitals are detailed in Figures 1-14,1-15, and 1-16.

sp-Hybridization: s/>hybridization is the result of the mixing of thes-orbital andthe px-orbital.

♦OJY

Two atomic orbitals

>Hybridizesto become

Two sp-hybrid orbitals

Figure 1-14

sp2-Hybridization: sp2-hybridization is the result ofthe mixing of the s-orbital,the px-orbital, and the py orbital.

VThree atomic orbitals

Hybridizesto become

Three sp -hybrid orbitals

Figure 1-15

sp3-Hybridization: sp3-hybridization is the result of the mixing ofthe s-orbital,the px-orbital, the py-orbital, andthe pz-orbital.

.3

^\f ~~J~~ TVFour atomic orbitals t0 become Four sp -hybrid orbitals

' Hybridizes V.

Figure 1-16

Example 1.5What is thehybridization ofeachcarbon in propene (H2C=CH-CH3)?A. sp,sp, andsp3B. sp2, sp, and sp3C. sp2, sp2, and sp3D. sp3, sp3, and sp3

Solution

There are three carbons in propene. The first two carbons are involved in a it-bond, so they areeach sp2-hybridized. This makes the best answer choice C. Thelast carbon isnot involved inany 7i-bonds, so it has s/?3-hybridization.


Hybridization


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 explanation 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 explainbond lengths and bond angles. Figures1-17,1-18, and 1-19 showstructures with their corresponding geometry and structural features.

Tetrahedral and sp3-HybridizationA 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 1-17showsdifferent structures with tetrahedral geometry about the central atom, butdifferent molecular shapes.

Tetrahedral Structure

H3

sp -hybridization

4 atoms/0 lone pairsBecause of symmetry, all bondlengths and bondangles areequal

C—H:1.10A&<HCH: 109.5°

Trigonal Pyramidal Structure

H

3 atoms/1 lone pair

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

N—H: 1.00A & <HNH: 107.3°

2 atoms/2 lone pairs

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

O—H: 0.96A & <HOH: 104.5°

Figure 1-17

Trigonal Planar and sp2-HybridizationAcentral atom with three other atoms, twootheratoms andonelone pair,oroneother atom andtwo lone pairs attached has trigonal planar geometry ofthe threesubstiruents (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.

Copyright©by The Berkeley Review 16 The Berkeley Review


Planar Structure2

sp -hybridizationH A H

c=c

/ \H H

3 atoms/0 lone pairsEach carbon in ethene

has sp2-hybridization.

Spatial Representation

1.09A

Figure 1-18

117.5°

Linear and sp-HybridizationA central atom with two other atoms or one other atom and one lone pairattached haslinear geometry ofthe two substituents (or electron pairs) about thecentral atom. This does not mean that the shape is linear (although in most casesit is), but that the orientation about thecentral atom is linear. Figure 1-19 showsthe linear structure and spatial representation of the bonds in ethyne. The stickand ball representation shows thethree-dimensional perspective for ethyne.

Linear Structure

sp-hybridization

AH C^C H

2 atoms/0 lone pairs

Each carbon in ethynehas sp-hybridization.

Spatial Representation

180.0°

1.08A 1.21A

Figure 1-19

Example 1.6The hydrogen-carbon-hydrogen bond angle in formaldehyde (H2CO) is BESTapproximated by which of thefollowing values?A. 108.3°

B. 111.7°

C. 118.5°

D. 121.5°

SolutionThe first feature to look at is the hybridization ofcarbon. Carbon is involved inone 71-bond, so the hybridization is sp2. The bond angle about an sp2-hybridizedcarbon ispredicted tobe 120°. The question here is whether the angle isslightlygreater 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 should be slightly less than 120°. The bestanswer is thus choice C.

Copyright © by The BerkeleyReview 17

Hybridization


Organic Chemistry Molecular Structure Bond Energies

Bond Dissociation EnergyIn organicchemistry, theenergyrequired to cleave a bond in a homolytic fashionis commonly used to compare relative bond strengths. Homolytic cleavage refersto the breaking of a chemical bond into two free radical fragments. This istypically viewedin the gas phase or an aprotic,nonpolar solvent, where ions aretoo unstable to exist. It is important that you recall that energy is released whena bond is formed and that energymust be absorbed by the molecule to break abond. By subtracting the energy released upon forming new bonds from theenergy required to break bonds, the enthalpy of a reaction can be determined.This is shown in Equation 1.1.

AHReaction =lEnergy(bonds broken) - Energy(bonds formed) (1.1)

If the enthalpy ofa reaction is known, then the bond dissociation energies forbonds that are formed and broken during the course of a reaction can bedetermined. It is thismethod that allows for the comparison of bondsbetweenidentical atoms within different molecules. For instance, the theory ofaromaticity is supportedby theexcess energythat is released upon the formationof a rc-bond that completes the aromatic ring. The release of excess energyimplies that the molecule is morestable than expected from the standard bonddissociationenergies,so some other factor must be involved. Table 1-7lists somebond dissociation energies for typical bonds insome common organic molecules.

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

A B = H Me Et f-Pr t-Bu Ph OH NH2

Methyl 105 90 89 86 84 102 93 85

Ethyl 101 89 88 87 85 101 95 85

n-Propyl 101 89 88 86 85 101 95 85

Isopropyl 98 89 87 85 82 99 96 85

f-Butyl 96 87 95 82 77 99 96 85

Phenyl 111 102 100 99 96 115 111 102

Benzyl 88 76 75 74 73 90 81 71

Allyl 86 74 70 70 67 N/A 78 69

Acetyl 86 81 79 77 75 93.5 107 96

Ethoxy 104 83 85 N/A N/A 101 44 N/A

Vinyl 112 102 101 100 95 105 N/A N/A

H 104.2 105 101 98 96 111 119 | 107Table 1-7

Agreater value inTable 1-7 implies that the bond isstronger. You may note thatthe weakest bond listed inTable 1-7 isanO—O single bond within a peroxidemolecule (EtO-OH). Because this bond issoweak, peroxides arehighly reactivespecies, often used to oxidize other compounds. The data in Table1-7also revealthat the substitution of the carbon and the position of the bond within themolecule affect thebondenergy. The effect ofhybridization canalsobe extractedwhen comparing bond energies between vinyl andmethyl substituents.



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

A. An sp2-carbon toa primary sr^-carbonB. An sp2-carbon toasecondary sp^-carbonC. Asecondary sp3-carbon toa primary sp3-carbonD. Asecondary sp3-carbon to a secondary sp3-carbon

Solution

The most stable bond is the strongest bond. The strongest bond has the greatestbond dissociation energy, so to solve this question, the bond energiesfromTable1-7 must bereferenced. An sp2-hybridized carbon isfound inthe double bond ofan alkene. This is described as a vinylic carbon, so the vinyl entry in Table 1-7 isnecessary forchoices A and B. Considering thatwearelooking 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, asecondary carbon would come from a group such as ethyl or n-propyl.Considering only Et is listed as a substituent in Table 1-7, the value for Et isnecessary in choices B,C, and D.

Choice A is foundby looking at the entryforVinyl—Me, which is 102 kcal/mole.Choice Bis 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 entryfor 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 arecommonbetween metals and nonmetals. The strength of an ionicbond 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 chargeoneach ion, divided by the square of distance between the twocharges, which aretreated as point charges:

F = kqi q2 - 1 qiq2r2 4jce0 r2

where F = force, q = charge, r = distance, and e0= 8.85 x10r12_Cf_N-m2

(1.2)

The greater the charge on the ion, the stronger the bond; and the closer the ionsare to oneanother, the strongerthe 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 otherwords,despite the strength ofionic bonds, theyare 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.


Bond Energies


Organic ChemiStry Molecular Structure Intramolecular Features

Intramolecular FeaturesJ-tfW •O&h^O'.w


Intramolecular 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 electrondensitywithin a molecule. I liketo callthem the "five excuses" to explainorganicchemistry. They are resonance, the inductive effect, steric interactions, aromaticity,and hybridization. We have already examined hybridization and seen the effect ithas on the structure ofa molecule in termsof 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 resonancetheory.

Resonance

Resonance is an intramolecular phenomenon whereby electron density is shiftedthrough regions of the molecule via rc-bonds. Resonance is defined as thederealization of electrons througha continuous array of overlapping p-orbitalsfa-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(excludinghydrogen).

2. Thebeststructure minimizes the numberof formal charges throughoutthe molecule.

3a. If the molecule contains a negative charge, it is bestplacedon the mostelectronegative atom.

3b. If the molecule contains a positive charge, it is bestplaced 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 ofall the majorresonance contributors.

O O" OS-

X — A. => \tH3C NH2 H3C ^ +NH2 H3C NH2More stable form Less stable form Resonance hybrid

Figure 1-20

The resonancestructure farthest to the left in Figure 1-20is more stable than themiddle structure, because there is no separation of charge. Youmust 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 resonancetheoryto otherfeatures ofchemical 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 onmolecular structure in proteinfolding.

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OrganiC Chemistry Molecular Structure Intramolecular Features

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 rc-bond. The electrons from the adjacent rc-bond turn into a new lone pair.

® :o: h (^b: h :o:0 hII U — Nl I „ I I

H3C <V ~H H3C C NH H3C

©

H H H

Resonance Hybrid:

\W

:o^ o: Ao^-Po: cr o: o' ^o

OS" H

I: I

H3C C NH?H

.. © #..#© 5-o: *Q* O

v^%<

Resonance Hybrid

® CH3 CH3 CH3

HJCr CH, H9C ^CH, H2C *CH•2 "2*- ^-"2

Resonance Hybrid

® ?H3 CH3 CH3

H,C<->CH9 H2C ^CH2 H2C *cVCH,l2 n2\w ui2 1*2^ *-ri2

Resonance Hybrid

Figure 1-21

InExamples 1,2, and 3,shown in Figure 1-21, thenegative charge moves everyother atom between resonance forms. The lone pair becomes a rc-bond, and therc-bond becomes a lone pair two atoms away. This is true when the numberoftotal 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. All 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. Theresonance hybrid is acomposite of the individual resonance contributors. The most stable resonancestructures (major resonance contributors as they arecalled) have the greatest effecton reactivityand structure for a compound exhibitingresonance.

Copyright ©by The Berkeley Review 21 Exclusive MCAT Preparation

Organic ChemiStry Molecular Structure Intramolecular Features

Example 1.8The C-O bond length is LONGEST in the compound on the:

O O

A. left,becausenitrogendonateselectronsthrough resonance.B. right, because nitrogendonateselectronsthrough resonance.C. left,because nitrogen withdrawselectrons through resonance.D. right,because nitrogen withdrawselectrons through resonance.

Solution

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

QV O' r^-ii.. r~6"Double i

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 moreelectronegative than carbon.

Inductive Effect

The 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 inturnpulls electron density from itsneighbor. The effect dissipates over distance,butit can affect bonds between atoms up tothree orfour atoms away. We mostoften consider the inductive effect when a molecule has a halogen.The inductive effect increases withtheelectronegativity of theatom. 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, soit in turn pulls electrons from its neighbor. Ultimately,as weseewithpolarity, theelectron density in themolecule is pulled towards themostelectronegative atom in thecompound. An electronegative atom thereforewithdraws electron density andthus can increase a compound's acidity, increaseitselectrophilicity, decrease itsbasicity, ordecrease itsnucleophilicity.



For the relative electronegativity of common atoms in organic chemistry, thefollowing relationship holds: F>0>N>Cl>Br>I>S>C>H. Just recall"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 inductiveeffectas it applies to the nucleophilicity of amines reacting with an alkyl halide.

H H

"SA H

F N

^C-^ HJ "^ -log rate =4.311 's. -lograte = 1.44

H H F F

Methylamine Trifluoromethylamine

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

Figure 1-22

The withdrawal of electron density by the fluorine atoms decreases thenucleophilicity of the amine compound by pulling electrons away from thenitrogen atom. Aselectron densityis removed, thecompound becomes electron-poorand thus a worseelectron donor. This canbe verified by the reaction rate ina substitution reaction. As the negative log of the rate increases, the rate of thereaction decreases.

Example 1.9Which of the following compounds undergoes a nucleophilic substitutionreaction with ethyl chloride at the GREATEST rate?

A. H3CCHFNH2B. FH2CCH2NH2C. H3CCHCINH2D. CIH2CCH2NH2

Solution

The greatest reaction rate (the fastest reaction) is observed with the bestnucleophile. Each answer choice hasone halogen, soall the choices have slowerrates than ethyl amine. The question asks which experiences the least inductivewithdrawal. Chlorine is less electronegative than fluorine (Fonclbrisch), sochoices A and B are eliminated. The inductive effect diminishes with distance, sothe least electron withdrawal is observed with choice D. You must consider bothproximity and electronegativity whenlooking at theinductive effect.

Although not applicable in Example 1.9, you must also consider whether theinductive effect involves electron donation or electron withdrawal. For instance,methyl amine is more nucleophilic than ammonia (NH3), because the methylgroup is electron-donating. Varying the R-group changes theinductive effect. Italso changes the sizeof the molecule, so steric hindrance canaffect the reaction.For instance, trimethyl amine ((H3Q3N) is less nucleophilic thandimethyl amine((H3Q2NH), because the electron donation by theadditional methyl group doesnot compensate for the increase in molecular size.




Organic Chemistry Molecular Structure Intramolecular Features

Steric Hindrance

Sterichindrance occurs any time two atoms attempt to be in the same place at thesame 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.

hXhHoC CHj

/ \ / V1 C=C i

HoC CHo\ /hX«

Reduced bond angle in dimethylbutene,because themethylgroup hydrogens repel.

Figure 1-23

l3

Larger C-C-C bond angle Smaller C-C-C bond angle

Example 1.10Which of the following functional groups is MOST likely to be found in theequatorial position on cyclohexane?A. —OCH3B. —OCH2CH2CH2CH3C. —OCH(CH3)CH2CH3D. -OC(CH3)3

Solution

On cyclohexane, substituents with axial orientation experience greater sterichindrance than substituents with equatorial orientation. Because of sterichindrance, the substituent mostlikely to assumethe equatorialorientation 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.

Copyright ©byThe Berkeley Review 24 TheBerkeley Review


AromaticityAromaticity is stability generated from having 4n + 2 tc electrons in a continuous,overlapping ring of p-orbitals, where n is any integer including 0. This is knownas the HUckel rule. The stability is rooted in the molecular orbital model, wherean energy level is completely filled when there are 4n + 2 ^-electrons in the cyclicrc-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.

l!^ pd(s> * l^J AH =-28kcal/mole

0^-0 AH =-56kcal/mole

1!^ Pd(s) * l^J AH =-54kcal/mole

i^J Pd(s) i^J AH =-49kcal/m0le

Despite the presence of three rc-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 7t-bond. The third entry shows that conjugation results instability, reducing the amount of heat released upon hydrogenation, but onlybyabout 2 kcals/mole. Based on -28 kcal/mole for each 7t-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.

Not all cyclic, conjugated polyenes show such a large deviation from theexpected value for the enthalpy of reaction. 1,3,5,7-Cyclooctatetraene (CsHs)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 actualvalues. Because benzene has 6 rc-electrons in a continuous rc-cycle, it obeys theHiickel rule (it has 4n + 2 rc-electrons where n = 1), while 1,3,5,7-cyclooctatetraene(CsHg) has 8 rc-electrons in a continuous rc-cycle and does not obey the Hiickelrule. This lack of aromaticity results in a less stable reactant, so more heat isgenerated in the hydrogenation reaction.




Organic Chemistry Molecular Structure Intramolecular Features

Example 1.11The hydrogen-carbon-hydrogen bond angle about the terminal carbon in thefollowing alkene is BEST approximated by which of the following values?

H3CH2C H

\C=

/ •0H3C H

A. 108.3°

B. 111.7°

C. 118.5°

D. 121.5°

Solution

First we must consider the hybridization of carbon. It is involved in one rc-bond,so thehybridization is sp2. The bond angle about an sp2-hybridized carbon ispredicted to be 120°. The question here 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 1.12Which of thefollowing explanations accounts for thepKa of1,3-cyclopentadienebeing only 15, while the pKa for hydrogen on other sp^-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 srP-carbon weakens the C-H bond on that carbon.

Solution

The acidity of a compoundcapableof losing a proton (H+) can be determined bythe stability ofitsconjugate base. When an ordinary sp3-hybridized carbon (onethat is not stabilizedby resonance or the inductive effect) is deprotonated, thecarbanion that is formed is unstable. Carbon is not electropositive, so it does notreadily lose a proton. With 1,3-cyclopentadiene, however, the carbanion that isformed 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

\ /H*

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

Becauseit includes the word resonance, choice Bmay 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.



Fundamental ReactivityFundamental 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 samefundamental 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 Bronsted-Lowry acid-base chemistry.

Proton Transfer Reactions (Bronsted-Lowry Acid-Base Reactions)Bronsted-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 Bronsted-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 termprotonation to describe the gain of an H+. A hydrogen atom has one proton in itsnucleus and one orbiting electron (in a ls-orbital). When hydrogen loses anelectron to become H+, all that remains is a proton. This is to say that H+ is aproton, and thus the gain of H+ can be referred to as protonation. Deprotonation isthe loss of H+.

To be a Bronsted-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.

H,C

H3CH

Base

(Proton-acceptor)

Bond forming

Bond breaking

Acid

(Proton-donor)

H3C

\© ..©n—h + :ci:

Bond "

formed broken\\»V Bond

H

H3C

ConjugateAcid

ConjugateBase

Figure 1-25

In the reaction in Figure 1-25, you should note that an arrow going from a lonepair to an atombecomes 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. Thefavorability isattributed to the fact that HCl is a strong acid. Proton-transfer reactions proceedfavorably (AG < 0) from the side with the strongeracidand strongerbase 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, HN03, HCl, HBr, and HI. 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:

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




Organic Chemistry Molecular Structure Fundamental Reactivity

r

o

PK3=

OH

3-5

Carboxylic acid

Lewis Acid-Base Reactions

Lewis 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, A1C13, FeBr3, and SOCI2. Figure 1-26 shows a Lewisacid-base reaction, where ammonia is the Lewis base and BF3 is the Lewis acid.

H,C

H3C*7H

Base Acid

(Electron Pair Donor) (Electron Pair Acceptor)

Bond forming

....»»^.F.:

HoC\ ^ ••\0 BiMl:.N B

\\»V BondH3C q formed

H f:

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 Bronsted-Lowry acid or a Lewis acid. In organic chemistry,the terminologyvaries,and Lewisbases 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 toorganicchemistryis that negativecharges seek out and bond to positive charges.

AcidityAcidity is defined by three definitions: the Arrhenius definition, the Bronsted-Lowry definition, and the Lewis definition. The Arrhenius definition is that anacid yields H30+ 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 amolecule. They are responsible for the distribution of valance electrons, whichaccounts for the chemicalbehavior (such as acidity) of the molecule. An acid isstronger 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 worse electron-pairacceptor. The primary task associated with evaluating organic acid strength is todecide which groups are electron-donating and which are electron-withdrawing.Figure 1-27shows some common organic acids and their pKa values.

OHpKa«10

Phenol Alkyl ammonium cation Alkoxide Carbonyl a-proton

R"

HpK.-9-lO

H

pKa~15

R OH

Figure 1-27

R

O

A pKa = 17-20

CH,

Copyright © by The BerkeleyReview 28 The Berkeley Review


BasicityBasicity is most easily thought of as the opposite of acidity. Basicity is alsodefined by three definitions: the Arrhenius definition, the Bronsted-Lowrydefinition, and the Lewis definition. The Arrhenius definition is that a baseyields OH" when added to water. The Bronsted-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 stronger therespective compound (either acid or base). Figure 1-28 shows some commonorganic bases.

A-R

Carboxylate

pKb-9-11

O"

0-pKb =4

Phenoxide

R

pKb = 4

•"VhH

Alkyl amine

pKj,-l

R— Q":

Alkoxide

Figure 1-28

Althoughyou are not required to memorize exactpKa values, it is a good idea toknow the "-5-10-15-20 general rule" for organic acids. The pKa for a 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 1.13What is the pKa for p-nitrophenol and the pKbfor its conjugatebase?A. O2NC6H4OH has pKa = 7.2;02NC6H40- has pKb = 12.8B. O2NC6H4OH has pKa = 11.6;O2NC6H4O- has pKb = 8.4C. O2NC6H4OH has pKa = 7.2;02NC6H40- has pKb = 6.8D. 02NC6H4OH has pKa = 11.6;O2NC6H4O- has pKb = 2.4

Solution

The nitro group is electron-withdrawing, which makes nitrophenol a strongeracid than phenol. As an acid becomes stronger, its pKa value decreases. Thismeans that the pKa for nitrophenol is likely to be 7.2 rather than 11.6. Thiseliminates choices B and D. The pKa and pKbsum to 14,so choiceC is the bestanswer. This question could alsobe answered froma base perspective. The nitrogroup is electron-withdrawing, whichmakes nitrophenoxide a weaker base thanphenoxide. As a base becomes weaker, its pKb value increases. Thismeans thatthe pKb for nitrophenoxide is greater than 4.0. This eliminates choice D. Again,the pKa and pKb sum to 14, so choice C is the best answer. Pick C and gainincredible satisfaction doing what you should do.

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



O pKb«-6--3

R' CH2"

Anionic oc-carbon



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 1-29,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 that 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.

o . u . Oxygen a o H Oxygen a:o" HProtonation at // k /Oxygen a 'U '

5 **H,C C -* • H3C C.

Oxygen a o: mm >T@// All-octetresonance form, making

H3C C this the more stable product.

Oxygen bio H pi :"6:©//> /

^ H3C— c -*—• H3C— ceProtonation at \ © \ ©

Oxygen b Oxygen b:o H Oxygen b:o H/ /

H 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 canbe concluded that carboxylic acids are protonatedat the carboxyl oxygen (C=0), not the hydroxyl oxygen (OH). This means thatthe carbonyl oxygen is morebasicthan the alcohol oxygen. In reality, we rarelysee carboxylic acids acting as bases, but amides are similar in structure, and it iscriticalin biochemistry that we know where they are protonated and thus wherethey exhibithydrogen-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 oxygenis 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 carries a partial negative charge due to resonance donation from thenitrogen.



Solution

As 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 Bis eliminated, because oxygen is smallerthan nitrogen, not larger. Becauseof resonance, oxygen carries a partial negativecharge, while nitrogen carries a partial positive charge. This means that choice Cis false and choice D is true. This explains the basicity of amides.

:o:®.*'. more.basic due to negative charge.

less basic due to absence of electrons.

R

Acid Strength FactorsFactors affecting the strength of an acid or base can be broken down into primaryeffects and secondary effects. Primary effects depend on the bond directly to theacidic proton. The weaker the bond to the acidic proton, the more readily itbreaks, and consequently more acidic the acid. With acids, the bond breaks in aheterolytic fashion, forming ions. Primary effects include size, electronegativity,and hybridization of the atom directly attached to the acidic proton. Secondaryeffects depend on the effect of the molecule on the atom bonded to the acidicproton. The more electron-rich that atom, the less acidic the proton. The moreelectron-poor that atom, the more acidic the proton. Secondary effects includeresonance (the delocalization of electrons within a molecule through rc-bonds),the inductive effect (the delocalization of electrons within a molecule through a-bonds), and electron cloud repulsion (deformations of a molecule and elongationof bonds within the molecule). There is also aromaticity to consider, but that willbe addressed as a special case. Secondary effects involve intramolecular forces,which dictate where the electron density within a molecule lies, and thus theydictate the reactivity of a compound. We shall look at how these features affectthe electron distribution within a molecule, starting with the primary effects.Table 1-8 shows common acids and bases in organic chemistry, listed accordingto relative strength (in both the acids and the bases).

Strong Acids

Weak Acids

Strong Bases

Weak Bases

H2SQ4 > HI > HBr > HCl > HNO3

HF > HC02H > H3CCO2H > H2CO3 > H3CSH > H3CNH3CI >H5C6OH > H3COH

CH3(CH2)3Li > NaNH2 > KH > NaOCH2CH3 > NaOH « KOH

H3CNH2 > NaHC03 > H3CC02Na > HC02Na > H3COH

Table 1-8

Primary Acid Strength FactorsPrimary factors directly affect the strength of the bond to the protic hydrogen.As a bond weakens (homolytic and heterolytic bond dissociation energy lessens),acid strength increases and conjugate base strength decreases. The three primaryfactors to consider are atomic size (when comparing acids involving atomswithin the same column of the periodic table), electronegativity (whencomparing acids involving atoms within the same row of the periodic table), andhybridization (when comparing acids where hydrogen is on the same atom).





Strength Dependence onAtomic 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 weakerbonds (in both a homolytic and heterolytic fashion), theacidity of a compound directly depends on the length and strength of the bond.A significant differencein 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 becausesulfur (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 baseperspective. To apply this atomic size theory to bases, compare the distributionof electron density about a small atom versus a larger atom. A conjugate base ismore 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. If 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.

H H H H

PKa =10.4HS CH3 pKa =15.7H° CH3Stronger Acid Weaker Acid

Figure 1-30

You should understand that atomic size (polarizability of the conjugate base) canbe 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 haloacidstrength for haloacids is: HI > HBr > HCl > HF, corvfirming thatsize is more important than electronegativity for elements within the samecolumn 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 getslarger, because of the greater space over which the negative charge is distributed.

In the halogen case, the less concentrated (more diffuse) negative charge oniodide (T) is not as readily shared as the negative charge on bromide (Br"),chloride (CI"), 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 atthe relative reactivity of halogen containing organic compounds in reactionswhere the halide is a leaving group. The relative acidity of haloacids should befamiliar to you. The mathematics of acidity and basicity is important in generalchemistry, but we shall consider only values such as pKa to compare the relativestrengths of acids. In organic chemistry, by quantifying acidity with pKa values,we can support or disprove relationships.



Strength Dependenceon ElectronegativityWithin 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 pulled 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 1-31 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 moreelectronegativethan nitrogen, the following relationshipholds true:

H—F >

H H

X >H H

X »H H

XHO CH3 H2N CH3 H CH3

Strongest Acid Amphoteric Very Weak Acid Weakest Acid

PKa = 3.2 pKa = 15.7 PKa= 33 PKa = 49

Figureil-31

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

• •©HNj^— CH3 >pKb = -19

..©:o — ch3

pKb = -1.5

> :f:upKb = 10.8

Figure 1-32

Amides (R2N") 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 pKavalues above 17.




Organic ChemiStry Molecular Structure Fundamental Reactivity

Strength Dependence 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 1-33shows the comparison of acids where hybridization explainsthe difference in acid strength.

spCpKa =26 H—C=C—CH3

Stronger Acid

H CH,\ / 3

sp2CC=C/ \

pKa~36 H CH3Weaker Acid

pKa-26 H pKa«33 H|

•N^CHgsp2N |

H

Stronger Acid

sp3V*N^CH3A

H H

Weaker Acid

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 electronegativeatom 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.


OrganiC CnemiStry Molecular Structure Fundamental Reactivity

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 willnot 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 Dependenceon ResonanceResonance, being the delocalization of electrons through an overlapping string of7t-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 7t-bond is electron-withdrawing, while an adjacentatom with a lone pair is electron-donating.

Typicalexamples of organic acids that are affectedby resonanceinclude 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 thepKa only when it can delocalize Tt-electrons.

O OWith no it-bond, there •'' \_ '• \_ Methoxy lone pair cancan be no donation rt3*-u^ / un > "3*-^ / wn donate to acid groupthrough resonance \ / \ / through the 7t-bond

pKa = 4.7 pKa= 5.6

Figure 1-34

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

HO

pKa = 8.4 pKa = 10.0

Figure 1-35

It is important to know that electron-withdrawing groups make the compound abetter electron-pair acceptor (Lewis acid), so the acidity increases with electron-withdrawing substituents. Electron-donating groups make the compound aworse 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 thisis not always true. Use the electron-donating or electron-withdrawingcharacteristics to determine the effect on the acidity or basicity.

Copyright ©byTheBerkeley Review 35 Exclusive MCAT Preparation


H H

pKa = 1.68

HO

Strength Dependence ontheInductive EffectThe inductive effect is the delocalization of electrons through the sigma bonds ofa 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 carbonpulls electron density from its neighbor. Ultimately the electron density in themolecule (as with polarity) is pulled towards the most electronegative atom andaway 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 electrondensity and thus increases an acidic compound's acidity or decreases a basiccompound's basicity. For common atoms in organic chemistry, the order ofelectronegativity you should recall is:F>0>N>Cl>Br>I>S>C>H. Thishelps you predict the relative acidity due to the inductive effect. The carboxylicacids in Figure 1-36show the effectof electronegativity.

HO

pKa = 2.92

Figure 1-36

O

HO

H H

pKa = 3.17

HO

The inductive effect depends not only on the electronegativity of thewithdrawing atoms, but also on theirproximity. In Figure 1-37, the proximity ofthe chlorine atom to the acidic group affects the acidity of the carboxylic acid.Theeffect on the compound's aciditydrops with distance and is negligible whenthey are over four atomsapart. Thecloserthe electron-withdrawing substituentto the acid functionality, the greater the increasein the acidity.

H H

CI H

pKa = 2.88

CH, HO

O CI H

H H

pKa = 4.07

CH, HO

H H

CH2C1 HO

Figure 1-37

H H

CH,

pKa = 4.82

It may seem strange,but alkyl groups are electron-donating. This is attributed tothe fact that hydrogen is less electronegative than carbon. When the substituentis electron-donating, the inductive effectdecreases the acidity. As a result, alkylgroups reduce the acidity of a compound. In Figure 1-38, the compoundsbecome less acidicdue to the electron-donating methyl group.

HO

o

K >

H H

O

JL^CH3HO jC

H H

s=3.78 pKa = 4.74

Figure 1-38

pKa = 4.82



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

Ka of the acid increases acid dissociation increases

pKa of the acid decreases pH of a 1.0 M solution decreasesAG for reaction with a base decreases conjugate base strength decreasespKb of the conjugate base increases the stability of the conjugate base increases

You must also be able to use the qualitative concepts. A proton transfer reactiongoes favorably from the side with the stronger acid to the side with the weakeracid; a favorable reaction proceeds from stronger acid to weaker acid. Todetermine the Kgq for a proton-transfer reaction, first decide which of theacids(or conjugate bases) is stronger. From there, it is a matter of deterrnining thedirection of the equilibrium. If the equilibrium lies in favor of the reactants, theKgq <1. If theequilibrium lies infavor oftheproducts, the Kgq >1.

Example 1.16Which of the following compounds is the STRONGEST acid?

A. H3CCH2OHB. H3CCO2HC. F3CCH2OHD. F3CCO2H

Solution

Carboxylic acids are stronger acids than alcohols, due to the withdrawing ofelectron density by the carbonyl oxygen through resonance. This eliminateschoices A and C. Fluorine is highly electronegative, so it withdraws electronsfrom the acidic hydrogen via the inductive effect, thus increasing the acidity.Thestrongest acid is the carboxylicacid with fluorinesattached, choiceD.

Example 1.17How can the difference in acidity between trifluoroacetic acid and trichloroaceticacid be explained?

A. Fluorine is larger than chlorine, so trifluoroacetic acid is a stronger acid.B. Chlorine is larger than fluorine, so trichloroacetic acid is a stronger acid.C. Trichloroacetic acid is a stronger acid, because chlorine is more

electronegative than fluorine.D. Trifluoroacetic acid is a stronger acid, because fluorine is more

electronegative than chlorine.

Solution

Fluorine is smaller than chlorine, so choice A is eliminated. The acidic proton isnot bonded to the halogen, so atomic size does not dictate the acid strength. Thiseliminates choice B. Fluorine is more electronegative than chlorine, so Fwithdraws electron density more than CI, making trifluoroacetic acid moreelectron poor and a stronger acid (better electron pair acceptor). The pKa oftrifluoroacetic acid (F3CCO2H) is 0.18, while the pKa of trichloroacetic acid(CI3CCO2H) is 0.64. A lower pKa value confirms that trifluoroacetic acid(F3CCO2H) is the stronger of the two acids. Choice D is the best answer.




OrgailiC Chemistry Molecular Structure Fundamental Reactivity

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

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

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

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

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

Solution

Regardless of the effect (whether it is resonance or the inductive effect), electron-withdrawing groups increase acidity and lower the pKa while electron-donatinggroups decrease acidity and raise the pKa. This eliminates choices A, C, and Dand leaves choiceB as the best answer. The effect of an electron-withdrawinggroup can be seen in the following trend: H3CCO2H(pKa = 4.74) is less acid thanH2C1CC02H (pKa = 2.85), which is less acidic than HC12CC02H (pKa = 1.26),which in turn is less acidic than CI3CCO2H (pKa = 0.64). As the amount ofelectron-withdrawal increases, the acidity increases.

Example 1.19Which nitrogen atom in the following molecule is the MOSTbasic?

a

NH2

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

Solution

The 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 delocalizedanywhere within the molecule.) The best answer is choice C.



Example 1.20What 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 lesselectronegative than theamine nitrogen.

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

Solution

Bothnitrogen atoms are equally electronegative, because neither carries a charge.This eliminates choices A and B. One could argue that having differenthybridization makes the electronegativity different, but the goal here is to findthe best answer, and if it is necessary to stretch the definition of terms, you'rebetter off finding a better answer choice. The histidine is protonated on the iminenitrogen, because the lone pair on the amine nitrogen is being shared with thecyclic 7C-system through resonance. This makes the ring aromatic, so the lonepair on the amine nitrogen is not available to be donated as a base. The bestanswer is choice D, as shown below.

H,N

.N

Imine nitrogen

H"

NH

I Amine nitrogen

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


NH V' NH

HN^ HN "Resonance is still possible, sothe 7t-system remains aromatic.



OrganiC Chemistry Molecular Structure Fundamental Reactivity

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

pH = -log[H30+] (1.3) [H30+] = 10"PH (1.4)

pOH = -log[OH"l (1.5) pH + pOH = 14(at25°C) (1.6)

pKa = -log Ka (1.7) Ka = lO-pKa (1.8)

pKb = -log Kb (1.9) pKa + pKb = 14(at25°C) (1.10)

Given pKa values, the AG0 and Keq 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 =pKa +log-^J- (1.11)[HA]

KjtpH-pKa^i^l (1.12)[HA]

For HA +H20 -^=^ H30+ +A" Ka =[H3°+][A][HA]

(1.13)

Keq =10P^a(product acid)" pKa(reactant acid) (1.14)

Equation 1.14 is derived from the fundamental reaction:

HA +B" -^-^-A" +HB, where K^ =I™1L^_1q [HA][B1

By definition:

Ka(HA) =[H30l[A"] =10-pKa(HA) and Ka(HB) =[H3°1[B"] =10"PKa(HB)K ' [HA] K ' [HB]

Using equilibrium:

K _[HB][A-] _ [A'] ^ [HB] _ [H3Q+][A-] ^ [HB]eq [HA][B1 [HA] [B"] [HA] [H30+][B"]

~ 1 Ka(HA)= Ka(HA) x —-i = vKa(HB) Ka(HB)

Thus:

KoQ =Ka(HA) =10'pKa(HA) =10-pKa(HA) xi0+pKa(HB) =i0(pKa(HB) -pKa(HA))q Ka(HB) 10"PKa(HB)

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

Kp = l0pKa(Pr°ductacid) - pKa(reactant acid)



Reaction TypesIn organic chemistry, there are a few fundamental reactions that describe themajority of reactivity in organic chemistry. In addition to acid-base chemistry,there are substitution and addition reactions. All of these reactions involve theaddition of an electron pair to an electron-poor site. If you are able to identify theelectron pair that can be shared and the electron-poor site within the reactants,you will be successful at predicting chemical reactivity. Addition andsubstitution reactions are often further classified according to their nucleophileor electrophile, but for now, we shall consider only a genericaddition reaction tointroduce reaction pathways, mechanistic logic, transition states, intermediates,and energy diagrams.

Electrophilic Addition ReactionsElectrophilic addition reactions are common in alkene chemistry. The alkenenucleophile donates its mostavailable electron pair (the electrons of the rc-bond)to an electrophile. The reaction involves a strong acid, often a haloacid, as theelectrophile. If the alkene is asymmetric, then you need to consider orientationfactors in the reaction in terms of minimizing steric hindrance in the transitionstate and stability of the intermediate. When looking at an alkene, the termasymmetric refers to a state where the two carbons have an unequal types ofcarbons bonded to each. When the reactant alkene is asymmetric, the reactionfollows Markovnikov addition rules. The electrophilic substituent adds to theless hindered carbon of the rc-bond, according to the rules associated withMarkovnikov addition.

Mechanisms (Reaction Pathways)Mechanisms are the step-by-step account of all species thought to exist asreactants proceed to form products. Mechanisms are commonly referred to asreaction pathways. Mechanisms focus on intermediates, propose transition states,and break reactions into steps. Fewreactions in organic chemistry occur in onestep, somechanisms often involve many steps before reaching a product. Youmay recall from general chemistry that the rate ofa reaction depends onthe rate-determining step. Whenever there is more than one step in a reactionmechanism, there exists a rate-determining step.

The mechanism for a reaction isproposed based onthe products thatareformed,how fast thereaction proceeds, and which changes inconditions alter the rate. Itcan be supported by monitoring isotopically labeled atoms on a molecule. Amechanism can never be proven, only disproved. Because reaction data arebased on the rate-determining steps, rate data are critical in supporting amechanism. Mechanisms are the most probable pathway that accounts for thedata presented. Most reactions in organic chemistry involve a nucleophileattacking an electrophile, which is often referred to as nucleophilic substitution.Before any mechanistic studies areencountered, besure that you have a stronggrasp of the definitions of the terms.The reason for thoroughly addressing a reaction using arrow-pushing techniqueis to learn themes in reactions that are repeated in other reactions. In otherwords, understanding mechanisms allows you to predict products ofunknownreactions. For instance, if a carbocation is formed during one reactionmechanism, an analogous reaction is likely to form a carbocation as well. Fromthis perspective, almost all organic chemistry reactions can be classified asLewisacid-base reactions. These include all reactions that involve a nucleophileattacking an electrophile.





Example 1.21A carbocation can be classified as:

A. an intermediate.

B. a transition state.

C. a product.D. a catalyst.

Solution

A carbocation is an intermediate, because it is not stable enough to be presentatthecompletion ofa reaction, eliminating choice C. It is not presentat thestart 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 presentfor afinite period oftime, soit isnota transition state, meaning choice Biseliminated.It has a finite lifetime and is semi-stable, so the best answer is choice A.

Arrow-Pushing in MechanismsWhen drawing mechanisms, the arrow starts with theelectron pair destined toform a bond. Ifa bond isbroken, then the arrow starts at thebond being brokenand finishes as either a new bond being formed, or as a lone pair on anotheratom. Figure 1-39 explains the arrow-pushing associated withthe electrophilicadditionofhydrobromic acid, HBr, to a symmetric alkene, E-2-butene.

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

H3C / H\ / /C=C

/7i-Bond\H breaking CH3

Nucleophile(e* Pair Donor)

8+ ••a-

Bond breaking

H3C Bond H\ formed£- „

/ \H CH3

Planar carbocation(Electron poor)

..0:Br:

Electrophile(e"Pair Acceptor)

Bromide anion(Electron rich)

H,C

Step 2: Alone paironbromine goes to thepositive carbon ofthecarbocation intermediate to form a sigma bondbetween C andBr. The product isanalkane withbromine (alkyl bromide).

H,C

\v.c-

TBr

H

\CH3

H

Planar carbocation Bromide anion(Carbonis electron-poor) (Bromine is electron-rich) Alkyl bromide product

Figure 1-39



Competing MechanismsWhen an asymmetric alkene reacts with hydrobromic acid, there are multiplepotential products. Thereaction canproduce either a secondary or tertiary alkylbromide product. Thepathwayleadingto 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 thethermodynamic product. Thepathway with the lower activation energy yieldsthe 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 alkenewith hydrobromic acid.

Observed Reaction

(Leads to theMore Stable Productby way oftheMore Stable Intermediate)

H3C\ //C=C

/ \H,C

H

CH,

Nucleophile(Alkene 7t-bond)

H3C,\/ £*H©c— C^/ \

CH3

3° Carbocation

(Electron-poor)

H3C

H

8+ • • 8"

Electrophile(Strong haloacid)

Bromide anion(Electron-rich)

H3C H

©c—c^/ \

H3C CH338 Carbocation Bromide anion

(3* > 2° in stability) (Leavinggroup)

H ..©:Br:

H3C\

H

:b7 ch3

3"Alkyl bromide product(Morestable than 2° product)

Minor Product Reaction

(Leads totheLess Stable Product byway ofthe Less Stable Intermediate)

H3C

H3C

H3C\..c-

A**'H^VH3C

H

CH3

2° Carbocation(Electron-poor)

c=c + H^f* -*-CH3

Nucleophile Electrophile(Alkene7t-bond) (Stronghaloacid)

Bromide anion(Electron-rich)

H,C

H3C CH32° Carbocation Bromide anion

(2° < 3° in stability) (Leaving group)

A*v

Figure 1-40


H

HoC

..©:Br:

H ..

\ Brl

H3C

..c—cr"

\CH,

2" Alkylbromide product(Less stable than 3"product)

43




I

Energy DiagramsAnenergy diagram accounts for theenergy of thesystem 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. Asenergy is released from thesystem(upon making bonds or reorienting to a more stable structure), theenergy goesdown. The overall difference from start to finish represents theenergy changefor thereaction (either AG orAH.) The diagram reaches an apex witha transitionstate, and a localized nadir in the middle with an intermediate. The startrepresents theenergy of the reactants, and the end represents the energy of theproducts. Eachapex represents a transition state in the reaction, which results ina step in the reaction. Figure 1-41 shows the two energy diagrams associatedwiththe twopossible reactions whenaddingHBr toan asymmetric alkene.

t i'l

; tn

Reactants

(H3C)2C=CHCH3 + HBr

Reaction coordinate

/ t \ AG*2* Cation

2(2')

Products

2°:(H3C)2CHCHBrCH33°:(H3C)2CBrCH2CH3

AGr(2.} =-7.63 ** /molekcalrAGr(3.) =-8.77icca7mole

Figure 1-41

Energetics and KineticsWill a reaction proceed or not? This question deals with thermodynamics(overall energy of the process) and kinetics (overall rate of the process).Reactivity can be analyzedas follows:

1. Identify and label the reactants (in most cases as nucleophile andelectrophile).

2. Predict the products of the reaction. If the reaction occurs inmultiple steps,predictproductsfor eachstep.Show the flow of electrons ineach ofthe steps. This isknown asdrawing themechanism.

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, Keq (theequilibrium constant measuring the ratio ofproducts to reactants) isgreaterthan 1 (Keq >1). Because AG0 = AH° - TAS°, andmost organic reactions takeplace in solution where entropy is relatively unchanging, AH" can beapproximated from AG0. In conclusion, to determine whether a reaction isthermodynamically feasible, look at AGe, Keq, and AH°.

3.



5. Evaluate whether the reaction can proceed at the given temperature. Thereare favorable reactions that never take place, because the activation energy(Eact) is too great. When looking at reactions, there isAG* toconsider, 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 (krx), 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 F^ct equals fast reaction), and intermediates(stable intermediates equal fast reaction).

Effect of Temperature on Reaction RateThe rate of a reaction increases as temperature increases, because there is morefree energy available in solution for reaction. Temperature is part the rateconstant, krx, mathematically expressed Equation 1.15.

kr* = A eTX

•Eact/RT (1.15)

where A is the Arrhenius constant and Eact is the activation energy.

The energy diagrams in Figure 1-42 show the change in energy level as thereaction proceeds and Figure 1-43 shows the molecular energy distributionthroughout theentire system at two different temperatures (Ti andT2).

SB

Cw

Transitim state

Eact\' \

AGrx

f

(t)

h 1act Reaction rate

}

Reaction coordiante

Figure 1-42

Minium Eactfor reaction

Kinetic energy

Figure 1-43

AtT2, the average kinetic energy ofthemolecules isgreater than it isatTi. Thus,at T2, more molecules have enough energy to overcome the activation energybarrier of the reaction.




OrgaiUC ChemiStiy Molecular Structure Physical Properties

Wf^^H^B^", "" -*";• '; ?-*£};.; "'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 willtakeplace in a chemical reaction or physical change. Common MCAT questionsinvolving intermolecular forces include determining the boiling points of tworelated compounds, as well as a compound's solubility properties or meltingpoint. The ruleis simple: thegreater theforces, thehighertheboiling point. Wewill use relative boiling points to verify the relative intermolecular forcesbetween compounds. The following are intermolecular forces that affect theboiling points, listedin descending orderofstrength.

Hydrogen bondingHydrogen bondingis a weakbond (approximately 4 to 8 kcals/mole) that existsbetween a lone pair ofelectrons anda hydrogen thatcarries a substantial partialpositive charge. Ahydrogen hasa substantial partialpositive charge whenit isbonded to a small electronegative atom such as nitrogen, oxygen, or fluorine.There arenohydrogen bonds involving hydrogens thatarecovalently 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 pairbinds theprotic hydrogen ofanamine. The strength of a hydrogen bond can be estimated from the baseproperties ofthelone pairdonor and theacid properties ofthehydrogen donor.Polar Interactions

Polar interactions are the Coulombic interaction between partially chargedparticles (approximately 1 to 3 kcals/mole). Negatively charged sites attractpositively charged sites. The greater the partial charge on the site of themolecule, the stronger the force between opposite charges. The strength of theforce also increases as the distance between oppositely charged sites decreases.A typical example of a polar interaction is the dissolving of ions and polarspecies into water.

Van der Waals

Van der Waals forces exist between all compounds. Van der Waals forces areconsidered only when noother forces exist toanyextent. They are theresult 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 intermolecular forces are the primary consideration when approximatingphysical properties. When forces are not enough to determine the 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 meant by"molecular flexibility" is the ability to twist and conform to allow for moresurface area, and thus more intermolecular interactions. Van der Waals forcesarerarely used toexplain anything except whythere isnot a complete absence ofintermolecular force. Figure 1-44shows the different forces.



Hydrogen-bonding:Hydrogen bonding involves the sharing of a

pj jlone pair with an electropositive hydrogen.

r^O\ H

3 \ J.N HCH,

HoC CH,

Dipole-Dipole Interaction:

<hq2

Coulombic attraction exists here. Thelarger the charge, the greater the force.

F= k

H3C

8•rH,C

H

Van der Waals interactions:

Coulombic attraction existshere temporarily.Charges are momentary dipoles.

H^ **CH38f 6t+~CH,

CH, H

H

Figure 1-44

Example 1.22Why is methanol (CH3OH) a liquid at room temperature, while ethane(CH3CH3)isagas?

A. Ethane is more polar than methanol.B. Methanol is significantly heavier than ethane.C. Methanol hasstronger vanderWaals interactions thanethane.D. Methanol has hydrogen-bonding, while ethane doesnot.

SolutionMethanol (CH3OH) isa liquid at room temperature, while ethane (CH3CH3) isagas, somethanol has the greater boiling point. Ethane isa nonpolar molecule, sochoiceA is eliminated. Methanol has a molecular mass of 32 grams/mole, whileethane has a molecular mass of30 grams/mole, somethanol is only slightly (notsignificantly) heavier than ethane, so choice Bis eliminated. The van der Waalsinteractions are roughly equal for all molecules, so choice C is eliminated.Methanol has hydrogen-bonding, while ethane has no protic hydrogen andtherefore no hydrogen-bonding. The higher boiling point is due to thehydrogen-bonding ofCH3OH, so choice D is all yours!


Physical Properties


OrganiC ChemiStry Molecular Structure Physical Properties

Hydrogen-BondingHydrogen-bonding is the strongest of the common intermolecular forces. It canbe thought of as a weakcovalent bond between a hydrogen that carries a partialpositivechargeand the lonepair on a nearby atom. The strength of a hydrogenbond variesbetween4 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 toeither fluorine, oxygen, ornitrogen (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 onfluorine anda 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 firstlookforhydrogen-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 areofroughly equal mass.

OH O

H

Butanol (b.p. =117.4'C) Butanal (b.p. =76.1°C)

Figure 1-45

Example 1.23Which of the following compounds has the HIGHEST boiling point?A. (H3Q3NB. (H3Q2NHC. (H3Q3CHD. H3COCH3

Solution

This 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 ifacompound has hydrogen-bonding, it also has dipole-dipole interactions,so thecompound with hydrogen-bonding has thegreatest intermolecular forces.To form a hydrogen bond, both a lone pair ofelectrons and a hydrogen on ahighly electronegative atom (N, O, or F) arerequired. Choice Cdoes notcontaina lone pair, so it is eliminated. Choices Aand Dhave all of their hydrogens oncarbon, so they are both eliminated. This leaves choice Bas the only moleculethat forms hydrogen bonds.



PolarityPolarity is defined as an asymmetric distribution of electron density within amolecule. Perhaps an easier way to think of this is as a tug-o-war for theelectrons between atoms. The more electronegative atoms pull the electronsmore tightly. If the molecule has more of one atom on one side of the moleculethan another—that is, if it is asymmetric about a central point- then it is polar.Figure 1-46shows a series of chlorinated methane derivatives that demonstratepolarity (or lack of polarity) based on structure.

H CI H H CI

/Vh h/°S^h Ad n/C;i,,/cl n/C;V'//cl" H H H CI U CI u CI

'-1- ™_i__ r»_i__ NonpolarH H

Nonpolar Polar Polar Polar

Figure 1-46

A polar compound has a dipole, which for all intents and purposes is a linedrawn form the positive sideof themolecule to the negative side ofthe moleculeina way that sums up thepolarity vectors ofeach bond in the molecule. The lineindicates the directionin which the electrondensityshifts. Figure1-47 shows themolecules from Figure1-46, with dipolesnow drawn in.

H CI H H CI

HH

Nonpolar

./H

c.V'H:#f//

H

Polar

CIh-"^ a^G. C

;^"a n/ s?"aCI C1 ci

NonpolarPolar Polar

Figure 1-47

The interaction ofthe dipoles between two nearby molecules accounts for a weakforce, asshown inFigure 1-44. The alignment of dipoles isbest when the partialpositive of one molecule aligns near the partial negative of another molecule.

Example 1.24Which of the following molecules has a dipole moment ofzero?A. Carbon monoxideB. Dichloromethane

C. 2,2-DichloropropaneD. Trans-l,4-dichlorocyclohexane

SolutionTo have a dipole moment of zero, the molecule must be symmetric. Carbonmonoxide (CO), dichloromethane (see Figure 1-47), and2,2-dichloropropane areasymmetric and thus polar. The two carbon-chlorine bonds of trans-1,4-dichlorocyclohexane are on opposite sides of the molecule, symmetricallydisplaced about a central point in the molecule, so their dipoles cancel out.Choice D is the best answer.


Physical Properties


OrganiC Chemistry Molecular Structure Physical Properties

Van der Waals Interactions

Van 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 factvan 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 lard are moreflexible, they are better able to interactwith another molecule (tie knots aroundanother molecule, ifyou will) than isvegetable oil. Perhaps it is easier topicturelard asa pile ofstrings that can tie knots around themselves, while vegetable 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 removeastraw from the straw pile, because the straws are nottangled up. Their rigidityprevents interactions between the straws. This is why saturated fats (whosemolecules are flexible like strings) are solid at room temperature, whileunsaturated fats (whose molecules are rigid like straws) are liquid at roomtemperature. The greater thenumber of Tt-bonds in a compound, the lower itsmelting point, and the greater the odds it isa liquid at room temperature. The n-bonds in the fatty acids in a phospholipid bilayer affect the fluidity of a cellmembrane.

Vegetable Oil(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 (HOCH2CH(OH)CH20H),plus two fatty acids, and aphosphate attached to the oxygen atoms of glycerol.Acell membrane would be most rigid ifboth ofits fatty acids were:A. completely saturated and short molecules.B. completely saturated and longmolecules.C. unsaturated and short molecules.D. unsaturated and longmolecules.

Solution

For the membrane to be rigid, there must be many interactions between the lipidchains. The maximum degree of interaction occurs with long, saturated fattyacids. Pick Bandfeel good yetagain.



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 dissolves like. This means that a polar moleculedissolvesmost readily in a polar solvent, and a nonpolar molecule dissolves most readilyin a nonpolar solvent. Miscibility and solubility canbe used as diagnostic tests tohelp determine the identity of an unknown substance. There are threecombinationsof properties that a solvent may have. It may be polar and protic(capable ofhydrogen-bonding), polarand aprotic (no hydrogen-bonding, but hasdipole-dipole interactions), and nonpolar and aprotic (weak intermolecularforces). Types ofsolvents and theirproperties aredescribed in theTable 1-9.

Type Intermolecular Forces Examples

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

Water, Alcohols

Polar, AproticDipole-dipole and vander Waals

Ketones, Ethers, Alkyl halides

Nonpolar, Aprotic van der Waals Oils, Petroleum

Table 1-9

The following 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 dissolvesinto oil, because it is entropically favorable to randomize.Individual van der Waals interactions are weak; but over a longmolecule, they quickly becomesignificant.

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 chromatography, the greater the solubility ofa solute in the solvent, the greater theaffinity ofthe particle for the mobile phase,meaning itcan travel farther and faster. In extraction, solutes are separated fromone another by their relative solubility (or miscibility) in two solvents. Thedifference insolubility isattributed to functional groups on both the solvent andsolute. Micelles can beemployed toincrease the apparent solubility ofa 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 asoap (surfactant) to do this, itmust be both hydrophilic (water-soluble) and hydrophobic (water-insoluble)simultaneously. Such molecules contain a polar (or charged) end (referred toasthe head) and a alkyl chain (referred toas the tail). Originally, soaps were madeby treating animal fats with strong base to convert the ester to acarboxylic acid,and thenfurther convert the carboxylic acid to carboxylate (its conjugate base), a

Copyright©by The Berkeley Review 51

Physical Properties


Organic Chemistry Molecular Structure Physical Properties

species 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 is greater than roomtemperature, but less than the temperature of the flame. The oil and vapor thatform burnwhen exposed toheatfrom theflame and to oxygen. Only theoilnearthe flame bums, soheat isnot evenly distributed in the system. The workings ofa candlecan be explained using fundamental principles in science. Quite often,the simplest things in life are organicchemistryin action.

Example 1.26Which of the following compounds would make the BEST micelle?A. H3C(CH2)3C02HB. H3C(CH2)3C02-C. H3C(CH2)i4C02HD. H3C(CH2)14C02-

Solution

The 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 ithas a longer organic tail. Pick D to score big on this question.

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Key Points for Molecular Structure (Section 1)

Nomenclature

1. IUPAC Nomenclature (Names are assigned systematically based on functionalgroups 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 6)d) Common Nomenclature (Prefixes based on substitution and relative

positions of functional groups; i.e., geminal diol and secondary alcohol)


1. Bonding (An attractive interaction between neighboring atoms)a) Covalent Bond (The sharing of electrons between atoms; carbon makes

covalent bonds)i. Single bonds are made of a sigma-bond; they are weaker than both

double and triple bonds,ii. Double bonds are made of one sigma-bond and one n-bond; they are

weaker than triple bonds, but stringer than singlebonds.Triplebonds are made of one sigma-bond and two rc-bonds; they arestronger than both single and double bonds.Sigma-bonds share electron density between nuclei while rc-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

therearesigmaand pi orbitals to describe molecular bonds.Sigma bonding orbitals aremore stable than pibonding orbitals, whilesigma anti-bonding orbitals are less stable than pi anti-bondingorbitals. They fill aZjtVV2Anti-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 theirvalence 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 ofatomic orbitals (s and p in organic chemistry) to form hybrid

orbitals capable ofcombining to make molecular orbitals.a) sp-hybridization results in linear compounds, often with two rc-bonds, a

180° bond angle, and the shortest ofallhybrid orbitals.b) sp2-hybridization results in trigonal planar compounds, often with one n-

bond, a 120* bond angle, and a mediumsizedhybridorbital.c) sp3-hybridization results in tetrahedral compounds, with no 7t-bonds, a

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

ui.

IV.

u.

m.

Bond Dissociation Energy

1. The energy required tobreak a bond ina homolytic fashion (into free radicals)is the bond dissociation energy.Higher BE refersto a strongerbond.a) BDE depends on theatoms, thesubstitution, andelectron distributionb) Ionic bonds,rare in organic chemistry, breakin a heterolytic fashion.


Section Summary


Organic Chemistry Molecular Structure Section Summary


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

i. Most stable resonance structure has an octet for all atoms but

hydrogen, 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 rc-bondsiii. Atoms with lone pairs are generally electron-donating while atoms

with a Jt-bond and no lone pair are generally electron-withdrawing.b) InductiveEffect (The pull of electrondensity through sigma bonds)

i. Depends on electronegativity of atomsii. Diininishes over distance,becomingnegligibleafter four carbons.

c) StericHindrance (Repulsionof two atoms at the same location)d) Aromaticity (Stability for cyclic systems with a set number of Tt-electrons)

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


1. Organic Chemistry at it foundation is Lewis Acid-Base Chemistrya) Acid-Base Chemistry

i. Bronsted-Lowry deals with proton transfer while Lewis deals with theacceptingand donating of electronpairs,

ii. Electron pair donors are Lewis bases and nucleophiles, while electronpair acceptors are Lewis acids and electrophiles.

b) Detenriining Acid Strength (stronger acids have lower pKa 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 acidstrengthare resonance and induction,iii. Base strength goesin the opposite fashionas an acid.

c) Acidand Base Terminology and Factsi. Acids: pH =-log[H30+], [H30+] =10-PH, pKa =-log Ka, Ka =10'PKaii. As acid strength increases: 1) acid dissociation increases, 2) Ka

increases, 3) pKa decreases, 4) pH in an aqueous solution decreases,and5)conjugate base strength decreases andstability increases

iii. Bases: pOH =-log [OH"], [OH"] =10-POH, pKb =-log Kb, Kb =10"PKbiv. As base strength increases: 1) base hydrolysis increases, 2) Kb

increases, 3)pKb decreases, 4)pH in aqueoussolutionincreases, and 5)conjugate acidstrengthdecreases and stabilityincreases

d) Mechanisms involve tracking thepathway ofelectron transfer

Physical Properties

1. The physical properties ofa compound are affected by intermolecular forcesa) Hydrogen Bonding (Occurs between H onN,O,or Fand a lone pair)

i. Increases boiling pointand melting pointby increasing attraction.b) Polarity (Interaction between dipoles ofadjacent compounds

i. Polarinteractions are weakerthan a hydrogen bondsc) Van derWaals Forces (Weak interaction between temporary dipoles)

i. Small impact on physicalpropertiesd) Solubility and Miscibility (Ability to dissolve into a solvent)

i. Based on the idea that "Like dissolves like."

ii. Solid intosolvent is solubility while liquidintosolventis miscibility


Structure, Bonding,and Reactivity

Passages

13 Passages

100 Questions

Suggested Structure, Bonding, and Reactivity Passage Schedule:

I: After reading this section and attending lecture: Passages I - III & VI - VIIIGrade 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.

Ill: Review: Passages X - XIII & Questions 92 - 100Focus on reviewing the concepts. Do not worry about timing.


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 3* Alcohols

IX. Petroleum Distillation

X. Acidity and Hybridization

XI. Keq and Acidity

XII. Alkoxides and Alkyl Sulfides

XIII. nucleophilicity and Basicity

Questions not Based on a Descriptive Passage

Structure, Bonding, and Reactivity Scoring Scale

Raw Score MCAT Score

84 - 100 13- 15

66-83 10- 12

47 -65 7 -9

34-46 4-6

1 -33 1 -3

(1-8)

(9 - 14)

(15-21)

(22 - 28)

(29 - 34)

(35 - 42)

(43 - 49)

(50 - 56)

(57 - 63)

(64 - 70)

(71 -77)

(78 - 84)

(85-91)

(92 - 100)

Passage I (Questions 1 - 8)

As an approximation when determining the enthalpy ofreaction from bond energies, it is assumed that a bondbetween two atoms has a fixed value for its bond dissociation

energy, regardless of the substituents on the molecule. Thisis to say that one assumes a covalent bond between carbonand iodine always has the same bond energy, whether thecarbon is a tertiary or primary carbon. A more critical viewof bond energies, however, shows this assumption to beinaccurate. Table 1 lists a series of energies for commonbonds in a wide range of organic molecules.

Bond BE(kcal)mole

H3C—I 56

H3C—CI 84

H3C—OH 91

H5C2-I 53

H5C2—CI 81

H5C2—OH 91

(H3Q2CH—I 52

(H3Q2CH—CI 80

(H3Q2CH-OH 90

(H3Q3C-I 50

(H3O3C—CI 79

(H3Q3C-OH 89

Bond BE(kcal)mole

H3C—CH3 88

H5C2—CH3 85

(H3Q2CH-CH3 84

(H3Q3C-CH3 81

H2C=CH—CH3 97

H3C—H 104

H5C2—H 98

(H3C)2CH-H 95

(H3Q3C-H 91

H2C=CH—H 108

H3CO—H 102

H5C2O—H 103

Table 1

The values in Table 1 demonstrate the effect of alkylgroups on neighboring atoms and the bonds that they form.There is a correlation between carbon substitution and bond

strength, as well as between atomic size and bond strength.There is also a correlation between hybridization and bondstrength, but it is not substantiated by the limited data inTable 1, whichpresents too few examplesof varying degreesof hybridization to reach a solid conclusion about theeffect ofhybridization on bondstrength. The effectof substitution onan alkeneon bond strength can also be evaluatedusing bondenergetics. Table 2 lists the enthalpy of reaction for thehydrogenation reactions of various alkenes. The differencesin AH values for the various hydrogenation reactions are dueto the effect of alkyl groups on the strength of a 71-bond.

AlkeneAH(kcal}

mole

H2C=CH2 -32.6

RHC=CH2 -30.2

cis-RHC=CHR -28.5

R2C=CH2 -28.3

trans-RHC=CHR -27.4

R2C=CHR -26.7

R2C=CR2 -26.4

Table 2

Copyright © by The Berkeley Review® 57

The presence of an alkyl group on an alkene strengthensits 7t-bond. Alkyl groups on vinylic carbons are consideredto be electron donating, so xc-bonds must be electronacceptors.

1. What bond dissociation energy would you expect for thebond between carbon-1 and hydrogen and the onlycarbon-carbon single bond in H—C=C—CH3?

A. Ci—H 92 kcal/mole; C2—C3 86 kcal/moleB. Ci—H 116 kcal/mole; C2—C3 86 kcal/moleC. Ci—H 92 kcal/mole; C2—C3 110 kcal/moleD. Ci—H 116 kcal/mole; C2—C3 110 kcal/mole

2. Bromine would make the STRONGEST bond with

which type of carbon?

A. MethylB. PrimaryC. SecondaryD. Tertiary

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

A. Smaller atoms form longer, stronger bonds thanlarger atoms.

B. Smaller atoms form longer, weaker bonds thanlarger atoms.

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

D. Smaller atoms form shorter, weaker bonds than

larger atoms.

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

A. unsubstituted.

B. monosubstituted.

C. disubstituted.

D. trisubstituted.

5 . Thedifference in enthalpy of hydrogenation between thecis and trans alkenes can be attributed to a difference in:

A. resonance.

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

GO ON TO THE NEXT PAGE

6.

8.

The hybridization of the carbons in H2C=CH—CH3can best be described as:

A. sp, sp, and sp2.B. sp,sp, and sp3.C. sp3, sp3, and sp2.D. sp2, sp2, and sp3.

Which of the following single bonds is the strongest?

A. H3C—IB. H3C—CIC. (H3O2CH—ID. (H3C)2CH—CI

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. (H3C)2C=C(CH3)2


Passage II (Questions 9-14)

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 extracted

from tea leaves by employing acid-base extraction usingstandardextraction 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 their

crystalline solid state and in solution, but they exhibitdifferent physical properties due to the ionic nature of the acidsalt form. The acid salt form is more water-soluble and has a

higher 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 formcan be converted into the acid salt form by treatingthe free base with a strong acid. Drawn in Figure 1 are thefree-baseand hydrochloric acid salt forms of caffeine.

Freebase Acid salt

Figure 1 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 conjugateacid form of caffeine has a pKavalue that is slightly higher than 6, making caffeine a weakacid. It existsprimarily in its deprotonated form at a pH of 7in an aqueous solution. The extraction of caffeine from tealeavescan be carriedout using vinegar to lower the pH of theaqueous layerand using an ether solvent (organic layer)as thesecond layer of the biphasic system. The caffeine cationdissolves into the aqueous layer, while the other organiccomponents of the leaves dissolve preferentially into theorganic layer.

9. The acid salt form of caffeine can be converted to the

free-base form most readily by adding which of thefollowing reagents?

A. HCl

B. NaCl

C. CaC03

D. NaOH


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

A. sp

B. sp2C. sp3D. Nitrogen atoms do not exhibit hybridization.

11. Which of the following structural descriptions BESTdescribes the relationship between the four nitrogens incaffeine?

A. Perpendicular planarB. Coplanar

C. Tetrahedral

D. Inverted planar

12. The length of the carbonyl bonds (C=0) in the caffeinemolecule are BEST described by which of the followingstatements?

A. Both C=0 bonds in caffeine are longer than theC=0 bond in cyclohexanone.

B. Both C=0 bonds in caffeine are shorter than theOO bond in formaldehyde.

C. Both C=0 bonds in caffeine are shorter than theC=0 bond in carbon dioxide.

D. All C=0 bonds are of equal length, regardlessof thecompound.

13. 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.

Copyright ©by The Berkeley Review® 59

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

A. acid anhydride.B. acid ester.

C. ester.

D. lactone.


Passage III (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 asolvent. When ink is applied to a porous surface such aspaper, the pores of the material absorb the solution. Then asthe volatile organic solvent evaporates away, the solid dye isleft bound to the pores of the material. This is why ink cansmear when initially applied, but once it has dried (oncethesolvent 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 runningand 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.

Anothercommon householdproductderived fromorganiccompounds is soap. Each soap molecule has a hydrophilic(water-loving) 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 micelle (thespherical cell formed by several aligned andcoagulated soapmolecules) is water-soluble, because the hydrophilic end ofeach molecule composing it solvates in the water. A micelleis removed 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)CH2OH) andcarboxylate anions (fatty-acid carboxylates) by a reactionreferred to as saponification. Carboxylic acids oncedeprotonated form carboxylates (the conjugate base of theacid). The organic chain of a soap molecule is most usefulwhen it containsat least eight carbons. Longercarbon chainsare common in soaps that are used to remove oils havinglonger carbon chains.

15. All of the following would be ideal properties for asolvent used to dissolve a dye within an ink EXCEPT:

A. exerting a highvaporpressure at room temperature.B. containing functional groups similar to the dye.C. being highly reactive with cellulose.D. having a boiling point slightly above room

temperature.


16. 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. PropanolC. PropanalD. Propanoic acid

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

CH3CH2CH2CH2CH2CH2CH2CH2CH2CO2H

A. Nonionicacid

B. Decanoicacid

C. Undecanoic acid

D. Dodecanoic acid

18. Some kinds if ink run when water is spilled on thepaper to which they adhere. This can best be explainedby 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 in

water.

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

19. Whichof the followingwould be the BEST soap?

A. CH3CH2CO2H

B. CH3CH2C02Na

C. CH3CH2CH2CH2CH2CH2CH2CO2HD. CH3CH2CH2CH2CH2CH2CH2C02Na

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

A. CH3CH2CO2H

B. CH3CH2CO2K

C. CH3CH2CH2CH2CH2CO2H

D. CH3CH2CH2CH2CH2CO2K

21. Which of the following reactions forms CH3C02Na?

A. CH3CO2H + CH3MgCl

B. HCO2H + CH3MgCl

C. Ethanoic acid + NaOH

D. Propanoic acid + NaOH


Passage IV (Questions 22 - 28)

For years, chemists pondered whether amides wereprotonated on the nitrogen or oxygen atom. The amide is ananalog to peptide linkages, so the root of this question isfounded in the chemistry of proteins. By determining the siteof protonation, conclusions about hydrogen-bonding in thesecondary structure of proteins can be made. Before theadvent and advancement of x-ray crystallography, proteinstructure could only be hypothesized. Due to the importanceof hydrogen-bonding in protein structure, determination of theprotonation site was critical. Figure 1 shows the structuraleffects of protonation on the oxygen atom of the amide.

R'

OIIC,

N*I

H

R'

VH

XR Sr-R"

IH

OI

.H

/C^©XR'R N

IH

Figure 1 Protonation of amide on oxygen

Figure 2 shows the structural effects of protonation on thenitrogen atom of the amide.

O

II

R NI

H

O

IINo resonance

R N/\

H H

Figure 2 Protonation of amide on nitrogen

Protonation at the oxygen site is favored because of theresonance stabilization of the protonated product, similar towhatis observedwhen protonatingesters. Despite the greaterbasicity of nitrogen relative to oxygen (oxygen is less basic,because it is more electronegative), the resonance stability isgreat enough to favor O-protonation. This manifests itself inprotein structure through the formation of hydrogen-bondsfrom the carbonyl oxygen (lone-pair donor) to the nitrogenproton (partially positive proton). Support for thisconclusion is found in the planar 6-pleated sheets and helicesobserved in the secondary structure of proteins.

Because of the resonance structures with O-protonation,the six atoms of the amide are all coplanar. This is due tothe^-hybridization of carbon, oxygen, and nitrogen.

2 2. Which of the following statements CANNOT be true?

I. The C=0 bond of an amide is shorter than the C=0bond of a ketone.

n. The C-N bond of an amide is shorter than the C-Nbond of a primary amine.

HI. Amides are more basic that aldehydes.

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


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 TC-electrons by the nitrogen.

D. carries a partial negative charge due to resonancedonation of 7t-electrons from the nitrogen.

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

O

H

H 'N'

H

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 outsideof the planecreatedby theotherfive atoms.

D. The six atoms are coplanar.

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

b

0II

c

0||

RO'K^OL*NHR'

A. Site a

B. Siteb

C. Sitec

D. Sited

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.


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

B. 0-I

.R'

A.O

c.

R N1 1I

H1

H

0"1

D. 0.1

R N1

1

R+^n'1

i

H1

H

R'

2 8. Which arrangement accurately relates the boiling pointsof acetamide (H3CCONH2),acetone (H3CCOCH3), andpropane to each other in descending order?

A. BPacetamide > BPacetone > BPpropaneB. BPacetone > BPacetamide > BPpropaneC. BPpr0pane > BPacetamide > BPacetoneD. BPpr0pane > BPacetone > BPacetamide

Copyright© by The Berkeley Review® 62

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 pointcorrelates with the heat energy required to vaporize a moleculeof it from solution. As the heat energy of vaporization(AHvap0rjzation) increases, the boiling point for thecompound increases.

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

Chemist 1

Chemist 1 proposes that the difference in boiling pointsfor two similar organic compounds is related to thedifferences in their molecular masses. The heavier the

molecule, the more energy that must be required to liberatethe compound into the gas phase from the liquid phase. Toliberate the molecules into the gas phase, heat energy mustbe added to the solution, which increases the temperature ofthe solution. Chemist 1 concludes that heavier molecules

have higher boiling points than lighter molecules.

Chemist 2

Chemist 2 proposes that the boiling point of acompound depends primarily on the strength of the attractiveintermolecular forces between molecules in solution. The

stronger the attractive intermolecular forces betweenmolecules, the harder it must be to remove a molecule fromthe solution to the gas phase. As it becomes more difficultto liberatea moleculefrom its liquid phase into its gas phase,more heat energy is required to carry the process out. Theresult is that the boiling point of a compound increases as themolecules bind to each other more tightly in solution. Whenthe intermolecular forces are greater, fewer moleculesvaporize, so the boiling point of the compound increases, andthe vaporpressureof the compounddecreases.

The hierarchy in attractive intermolecular forces ishydrogen-bonding first, polarity is second, and van der Waalsforces rank third in strengths. Hydrogen bonds contain thegreatest amount of energy of these three forces, but not allhydrogen-bonds are equal in strength. For instance, alcoholshave stronger hydrogen bonds than amines, because thehydrogen of the alcohol is more acidic than the hydrogen ofthe amine. The greater acidity allows the hydrogen to acceptelectron density more readily. Any compound capable offorming hydrogen bonds is also polar. Polar attractions arestronger than van der Waals forces, the weakest of theintermolecular forces.


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

A. (H3C)2CHOH has a higher boiling point thanH3CCH2OH.

B. H3CCH2OH has a higher vapor pressure thanH3CCO2H.

C. H3CCH2CH2OH has a higher boiling point thanH3CCH2OCH2CH3.

D. H3COH has a higher vapor pressure thanH3C(CH2)6CH3.

3 0. How would Chemist 2 rank the following compoundsaccording to their boiling points?

I. H3CCH2OH

H. H3COCH3

m. H3CCH2NH2

A. H3CCH2OH> H3CCH2NH2 > H3COCH3

B. H3CCH2OH> H3COCH3 > H3CCH2NH2

C. H3CCH2NH2 > H3CCH2OH> H3COCH3

D. H3CCH2NH2 > H3COCH3 > H3CCH2OH

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

A. Both the molecular mass and the melting pointincrease.

B. The molecular mass increases, while the meltingpoint decreases.

C. The molecular mass decreases, while the meltingpoint increases.

D. Both the molecular mass and the melting pointdecrease.

32. 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.


33. 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 of hydrogen-bondingdecreases.

B. decreases, because the amount of hydrogen-bondingincreases.

C. decreases, because the atmospheric pressureincreases.

D. decreases, because the atmospheric pressuredecreases.

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

I.

n.

m.

H3C CH;

Hci a

a

H"H3C a

h3c CH;

h*t y*hci a

A. Compound I > CompoundHI > CompoundIIB. Compound I > CompoundII > CompoundHIC. Compound II > Compound I > Compound HID. Compound III > Compound I > Compound II


Passage VI (Questions 35 - 42)

The dipole of a bond can be found 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 isdetermined by the symmetry of the structure. Each bond istreated individually, and the sum of the component vectors isthe approximate dipole. The estimated dipole is good enoughto predict chemical behavior. Figure 1 shows examples of apolar and a nonpolar cyclohexane derivative.

XIH'<ox

Nonpolar Polar

Figure 1 Polar and nonpolar disubstituted cyclohexanes

The magnitude of a dipole is measured by placing thecompound 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 dipole. This technique works because ofthe ability of a neutral polar compound when added to anelectric field to align with the field. If there is a net chargeon the molecule, it migrates toward the capacitor plate withthe opposite charge.

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

A. ArHF

B. C6H6C. N2D. C02

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

A. Hydrolysis of an 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 bromine on an alkyl halideB. Oxidizing a primary alcohol into a carboxylic acidC. Replacing fluorine with chlorine on an alkyl halideD. Adding HBr anti-Markovnikov to an alkene


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 an asymmetricmolecule

C. Stretching a carbonyl bond in a symmetricmolecule

D. Bending a carbonyl bond in a symmetric molecule

3 9. Which is the BEST description of the nonpolar structureofFe(NH3)4Cl2?

A. Octahedral shape with the two CI ligands cisB. Octahedral shape with the two CI ligands transC. Tetrahedral shape with the two CI ligands cisD. Tetrahedral shape with the two CI ligands trans

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

A. 1,1-dichloroethaneB. cis-1,2-dichloroethene

C. trans-1,2-dichloroetheneD. E-l-chloro-2-fluoroethene

41. Which compound has the LARGEST dipole moment?

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


I. For a tetrahedral structure, if any of the four ligandsare not equivalent to the others, the molecule ispolar.

n. All 1,4-disubstituted cyclohexane molecules arepolar.

HI. All optically active molecules are polar.

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


Passage VII (Questions 43 - 49)

A common problem facing pharmacists is developingdrugs in a form that can be easily ingested by human beings,particularly the problem of getting organic compounds todissolve in water. As a general rule, organic compounds arenot water-soluble, so it is difficult for them to migratethrough the bloodstream. Because most organic compoundsexhibit little to no hydrogen-bonding, they are referred to ashydrophobic (Greek for "water fearing"). To overcome thehydrophobic nature of organic compounds, one of twotechniques can be employed.

Technique 1

Technique 1 involves the use of micelles, three-dimensional bulbs composed of compounds that are partlyionic and partly hydrophobic (organic). A prime example ofa compound that forms a micelle in water is the conjugatebase of a fatty acid (H3C(CH2)nC02* Na+). The micelle is aspherical membrane that forms when the organic tailsaggregate as shown in Figure 1.

Polar heads Hydrophobic tails

Figure 1 Aqueous arrangement of molecules in a micelle

An organiccompound(such as an antibiotic)prefers thecore of the micelle over the aqueous solution. Overall, themicelle is water-soluble due to the polar heads of theindividual fatty-acid carboxylate anions. After migratingfrom an aqueous environment to a hydrophobic environment(lipid), a micelle turns itself inside out and releases theorganic compound in its core. This mechanism is whatenables water-insoluble drugs to be transported through thebloodstream(an aqueous environment) to hydrophobic targetregionsof the body (such as lipid membranes).

Technique 2

Technique 2 involves converting the compound into awater-soluble derivative that decomposes into its active,water-insoluble form once inside the body. This ofteninvolves converting neutral organic compounds into ions byeither protonation or deprotonation or altering a functionalgroup, such as converting an alcohol into an ester by reactingit with an acyl group. The drug returns to its original activewhen it is exposed to physiological conditions.


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

A. H3C(CH2)nC02H

B. H3C(CH2)nC02-

C. H3C(CH2)nNH2

D. H3C(CH2)nNH3+

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

A. * B.

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

A. (H3Q2CHOCH3B. (H3C)2CHCH2OHC. (H3Q2CHNHCH3D. (H3Q2CHCH2NH2

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

B.H

6h °yS>0HOH

D. CH;

OCT


47. Which of the following compounds would make theBEST micelle?

A. H3C(CH2)3C02H

B. H3C(CH2)3C02-C. H3C(CH2)i3C02H

D. H3C(CH2)i3C02-

48. Which of the following compounds would require amicelle to make it water-soluble?

A. An alcohol (RCH2OH)

B. A carboxylic acid (RCO2H)

C. An amine (RCH2NH2)

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-bondingD. Covalent bonding


Passage VIII (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 1 shows the IRabsorbances associated with four different hydroxyl groups.

3° Alcohol

t3471cm"1

2° Alcohol

t3423cm"1

Carboxylic acid

1° Alcohol

t3396cm.-1

2812cm-1

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 areequivalent, 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.

50. According to the IR absorbances in Figure 1, which ofthe following compounds exhibits the GREATESTamount of hydrogen-bonding?

A. The tertiary alcoholB. The secondary alcoholC. The primary alcoholD. The carboxylic acid


51. Which of the following aminecompounds should showthe BROADEST signalabove3000cm"1?

A. Ammonia

B. Propylamine

C. Dipropylamine

D. Tripropylamine

52. How is the absorbance value in the IR for a covalent

bond between oxygen and hydrogen affected by the bondlength and hydrogen-bonding to other atoms?

A. As the bond length increases, the wave number(cm*1) of theabsorbance decreases; so as thedegreeof hydrogen-bonding increases, the bond lengthincreases and the wave number (cm-1) of theabsorbance decreases.

B. As the bond length increases, the wave number(cm"1) of theabsorbance increases; so as thedegreeof hydrogen-bonding increases, the bond lengthincreases and the wave number (cnr1) of theabsorbance increases.

C. As the bond length increases, the wave number(cm-1) of the absorbance decreases; so as thedegreeof hydrogen-bonding increases, the bond lengthdecreases and the wave number (cm"1) of theabsorbance increases.

D. As the bond length increases, the wave number(cm-1) of the absorbance increases; so as thedegreeof hydrogen-bonding increases, the bond lengthdecreases and the wave number (cm*1) of theabsorbance decreases.

5 3. Hydrogen-bonding occurs within which of the followingcompounds?

A. Aldehydes

B. Esters

C. Ketones

D. Primary amines

5 4. 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 of N and a hydrogen bonded to N.


55. 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 wave

number 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.


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

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

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

A. I onlyB. IlonlyC. I and II onlyD. Hand HI only


Passage IX (Questions 57 - 63)

The petroleum industry provides roughly forty percent ofthe annual energy needs of the United States. Crude oil is amixture of hydrocarbons that is refined to produce fuels,including heating oil and petroleum. Many lightweight,alkene by-products from the refinement of crude oil are usedas raw materials in making polymers. The industrial processfor refining crude oil into useful components is referred to ascracking and is similar to fractional distillation. Figure 1shows a schematic representation of the cracking column usedto refine crude oil and the fragments collected at differentlevels of refinement.

held at 110°C

iiltiiiiiil liiiiiiiliiL^| held at 1754C

JLheld at 250°C

held at 300'C

Vapor

Petroleum

258C-175°C

Naptha110°C-190°C

Kerosene

175°C-280eC

Heating oil250°C-350°C

Lubricating oil"*~ 300'C-375°C

Figure 1 Cracking column used to refine crude oil

Petroleum distillate is sold as gasoline, the fuel mostcommonly used in internal combustion engines. The bestair-petroleum mixture for such engines is the one thatproduces the most uniform distribution of heat over theperiod of time that the piston is doing work. This allows foran even expansion of the gas in the piston, which results inmore useful work. The result is a smooth lifting of thepiston, rather than an explosive jerk. Engine efficiencydepends on the uniformity of heat distribution within it, sothe choice of fuel influences engine efficiency.

Gasoline is given an octane rating that is based on itscombustion rate. An octane rating is a measure of a fuel'stendency to cause knocking (non-uniform combustion.) Thescale is set using 2,2,4-trimethylpentane, which is assignedan octane rating of 100, and n-heptane, which is assigned anoctane rating of zero. A higher octane rating implies a betterfuel. Table 1 lists the octane ratings and boiling points forsome components of petroleum distillate.


Hydrocarbon Octane Rating Boiling Point

2-Methylbutane 93 28*C

Benzene 106 80°C

n-Hexane 25 69°C

Toluene 120 104°C

n-Heptane 0 98°C

2-Methylhexane 42 88eC

2,2,3-Trimethylbutane 125 82eC

2,2,4-Trimethylpentane 100 104°C

Table 1

57. Which of the following eight-carbon hydrocarbons hasthe GREATEST octane rating?

A. 2-MethylheptaneB. n-Octane

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

5 8. Which of the following components is MOST likely acomponent of kerosene?

A. n-Octane

B. n-Decane

C. 2,2-DimethyloctaneD. 2,2,4,4-Tetramethyldecane

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

A. An increase in octane ratingB. A decrease in boiling pointC. A increase in densityD. An increase in hydrogen-bonding

6 0. The cracking (refining) column operates according to theprinciple that:

A. more dense hydrocarbons rise higher than less densehydrocarbons.

B. hydrocarbons with lower boiling points rise higherthan 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.


61. The efficiency (octane rating) of a fuel depends on the:

A. enthalpy of combustion.B. entropy of combustion.C. ratio of CO2 to water in the exhaust.

D. rate of combustion.

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

I. Aromaticity increases octane rating.

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

HI. 2,2,3-Trimethylbutane is a good fuel additive toincrease fuel efficiency.

A. I onlyB. Ill onlyC. I and II only

D. I and in only

63. The hybridization of carbon in the aerobic combustionof 2,2,4-trimethylpentane changes from:

A. sp3 to sp2.B. sp2 to sp3.C. sp to sp3.D. sp3 to sp.


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 form new hybrid orbitals that arecorrectly aligned to make up the covalent bonds. Hybridorbitals are oriented to align the atoms within the moleculeinto the least sterically hindered position for bonding. Theorientation of electrons allows the molecule to form

structures with a central carbon that has either tetrahedral

(jp-^-hybridized), trigonal planar (.s/>2-hybridized), or linear(s/?-hybridized) geometry. Although the geometry dictates thehybridization, the hybridization of a carbon within a moleculecan be used to predict the structure of the molecule.

The more p-character there is in the hybrid, the longerthe hybrid orbital is, and thus the further away the electronsare from the nucleus. This variation in length can be used toexplain differences in chemical reactivity and physicalproperties. When estimating properties of a bond, one mustconsider that acidity results from heterolytic cleavage, whilebond energies are determined from homolytic cleavage.Figure 1 shows both heterolytic and homolytic cleavage forthe C—H bond of a terminal alkyne.

Hetreolytic Cleavage (into ions)

I C=C -H© ©

c=c: + h

Homolytic Cleavage (into free radicals)

H ^= + H«

Figure 1 Heterolytic and homolytic cleavage of a C-H bond

Acidity can be explained in termsof heterolytic cleavage.The closer the electrons of a carbon-hydrogen bond are to thecarbon nucleus, the more acidic the hydrogen on that carbon.This is to say that as the electrons in the bond get closer tothe carbon nucleus, the bond is easier to break heterolytically,and thus the acidity of the hydrogen increases. Electrons getcloser 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 difficultfor 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 thelength of the bond between carbon and the atom to which itis bonded. As the bond becomes longer, it becomes weakerin a homolytic sense.


64. The MOST acidic hydrogen on 3-methyl-l-pentyne ison which carbon?

A. Carbon-1

B. Carbon-2

C. Carbon-3

D. Carbon-4

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

"o. BaCH.

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

A. CH3CH2CH2CH2Na

B. CH3CH2CH=CHNa

C. CH3CH2CNa=CH2C. CH3CH2C=CNa

67. NaNH2 is a base strong enough to deprotonate the firsthydrogen on a terminal alkyne. Which of the followinghydrogenscould it also deprotonate?

A. H on carbon-1 of 2-methyl-l-buteneB. H on carbon-2 of 2-methy1-1-buteneC. H on carbon-1 of 2-methyl-l-butanolD. H on oxygen of 2-methy1-1-butanol

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

A. H—C=C CH3 B. H—C=C H

C.

h/--<ch:c=CH 3 * H>^ ^ H

C=CC

vf H


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

A-H—C=C—H

C.

rf H

H3C^— C^s: C-~ CH3

CH~ D-"ic-«H

H;>c=cs

70. The LOWEST pKb is associated with which of thefollowing nitrogen containing compounds?

"aB. 1 NH

NH2 CH.

C. D.

O a NH


Passage XI (Questions 71 - 77)

Most reactions using organic reagents require a solventother than water, so acid-base chemistry must be viewed fromeither the Br0nsted-Lowry definition or the Lewis definition.The Br0nsted-Lowry definition of an acid is a compound thatacts as a proton donor, while the Lewis definition of an acidis a compound that accepts electron pairs. In an organicsolvent, acid-base reactions involve the transfer of a protonfrom one reactant to another. There is an equilibriumconstant associated with this process that is predictable basedon the pKa values of the two acids in the reaction (one acid isa reactant, and the other acid is a product).

To determine the Ka value for organic acids, an organicacid is added quantitatively to an organic base. Theequilibrium constant (Keq) is determined from theconcentration of each species, once equilibrium is reached.Reaction 1 is a generic reaction between an organic acid andthe conjugatebase of a second organic acid

HA + B" ^ » A' + HB

Reaction 1

Equation 1 can be used to determine the equilibriumconstant for Reaction 1.

j^ =[A'][HB] =Ka (acid HA) =1Q(pKa (HB) -pKa (HA))[HA][B'] Ka(acidHB)

Equation 1

A series of six different organic acids are treated with 1,3-cyclopentadienyl anion, as shown in Reaction2.

HA + C5H5- „ *• A" + C5H6

Reaction 2

The structure of 1,3,-cyclopentadienyl anion is shown inFigure 1.

..©

Figure 1 1,3-Cyclopentadienyl anion

The concentration of each organic species at equilibriumis determined using UV-visible spectroscopy wheneverpossible. In cases where no 7i-bond is present in both thereactant acid and the product acid, the concentrations aredetermined using gas chromatography. The conjugate baseand reactant base concentrations are determined by thedifference between initial acid concentration and equilibriumacid concentration. The concentrations are used to determine

equilibrium constants. The calculated values are comparedtovalues found using pKa numbers in Equation 1. It is foundthat the error is greatest when Kgq is greater than 104.


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

Organic Acid (HA) Equilibrium Constant (Keq)

H3CCOCH3 2.0 x 10-4

H3CCOCH2COCH3 1.2 x 106

H3COH 3.9 x 10-l

H3CCH2SH 5.2 x 104

CI3CH 8.0 x 10"9

H3CNO2 8.2 x 104

Table 1

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

A. Cl3CH +H3CCH2S-^:fcsCl3C- +H3CCH2SH

B. H3COH + H2CN02- ^^ H3CO- + H3CNO2C. HsCCOCHs +HsCO-^^*

H3CCOCH2- + H3COH

D. H3CCOCH2COCH3 +Cl3C--i5S^sH3CCOCHCOCH3- + CI3CH

72. Which of the following compounds can deprotonateC5H6?

A. H3CCOCH3

B. H3CCH2SH

C. H3CO-

D. H2CNO2'

7 3. For the following reaction:

H3CCH2SH +H3CO-^^H3CCH2S- +H3COHwhat is true about the relative concentrations of eachspecies at equilibrium, if the reactants are mixed inequal molar portions?

A. [H3CO-] > [H3CCH2S-]; [H3COI > [H3COH][H3CCH2SH] > [H3CO-]

B. [H3CCH2S-] > [H3CO-]; [H3COH] > [H3CO-][H3CCH2SH] > [H3CO-]

C. [H3CO-] > [H3CCH2S-]; [H3CO-] > [H3COH][H3CCH2SH] = [H3CO-]

D. [H3CCH2S-] > [H3CO-I; [H3COH] > [H3CO-][H3CCH2SH] = [H3CO-]


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 the

inductive 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. pKa(ci3CH) >pKa(CH3OH) >pKa^CNC^)B. pKa(H3CN02) >PKa(CH3OH) >pKa(d3CH)

C. pKa(CH3OH)>pKa(d3CH)>pKa(H3CN02)D. pKa(CH3OH) > pKa(H3CN02) >P^C^CH)

7 6. Which of the following compounds is NOTan exampleof a Lewis acid?

A. CI3CH

B. H3CNO2

C. NaCH3D. BF3

77. Which of the following acids has a pKa value close to10.0, given that the pKa for C5H6 is 15.0?

A. H3CCOCH3

B. CI3CH

C. H3COH

D. H3CNO2


Passage XII (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.

H3C H3C

H3C

Nuc

^/H SolventI H3C

Reaction 1

+ r

Table 1 lists the negative logs of the reaction rates forReaction 1 observed in a series of solvents reacting with aseries of nucleophiles. In each case, the reaction is carriedout at 30°C, and with an initial concentration of 0.10 M forthe nucleophileand of 0.11 M for the electrophile.

Nucleophilic Solvent -Log rate

H3CO- Ether 1.44

H3CS- Ether 1.97

HO- Ether 1.03

HS- Ether 1.76

H3CO- Ethanol 3.19

H3CS- Ethanol 2.12

HO- Ethanol 3.35

HS- Ethanol 1.96

H3CO- Water 4.22

H3CS- Water 2.47

HO- Water 5.62

HS" Water 2.14

Table 1

Because the value in Table 1 is the negative log of therate, the magnitude of the effect of nucleophile and solvent onthe reaction rate is not immediately apparent. The smallerthe negative log of the rate, the greater the rate. Thedifference between the rate of SH_ and the rate of RS" is

attributable to differences in their molecular size. The

difference in reaction rates between the various solvents isattributable to a change in the mechanism from S^-like toSNl-like (as the solvent changes from ether to water). Themore the solvent binds the nucleophile, the less rapidly thenucleophile can attack the electrophile and thus the slower therate of the nucleophilic substitution reaction. This affects thereaction rate of nucleophilic substitution in a protic solvent.


78. 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 alkyl sulfide in an aprotic solventD. An alkyl sulfide in a protic solvent

7 9. What value should be expected for the negative log ofthe reaction rate, if ethyl sulfide (CH3CH2S-) wereaddedto 2-iodopropane in ether solvent?

A. 1.05

B. 2.04

C. 3.09

D. 4.52

80. If iodinewas replaced with brominein Reaction 1, oneshould expect that the negative log of the reaction ratewould:

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 the reactionrate woulddecrease.

81. According to the data in Table 1, which of thefollowing bondsis the WEAKEST?

A. C—H

B. C—I

C. C—O

D. C—S

82. From the data in Table 1, what can be concluded abouttheeffect of hydrogen-bonding on reaction rate?A. Hydrogen bonding hinders nucleophilic attack and

thus lowers the reaction rate.

B. Hydrogen bonding enhances nucleophilic attack andthus lowers the reaction rate.

C. Hydrogen bonding hinders nucleophilic attack andthus increases the reaction rate.

D. Hydrogen bonding enhances nucleophilic attack andthus increases the reaction rate.

Copyright ©byThe Berkeley Review® 73

83. 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 waterC. Changing the reaction temperature from 30°C to

25°C

D. Changing the solvent from water to ethanol

84. In comparing the properties of alkoxides and alkylsulfides dissolved in ether, one notes that alkoxides havea:

A. higherpKband react fasterthan alkylsulfides.B. higher pKbandreactslower thanalkylsulfides.C. lowerpKband react fasterthanalkylsulfides.D. lowerpKbandreactslowerthanalkyl sulfides.


Passage XIII (Questions 85-91)

Esters are semi-reactive carbonyl compounds thatundergo substitution chemistry 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 different nucleophiles in astandard substitution reactionusingan ester electrophile. Forthe reaction in Figure 1, three different compounds(CompoundA, Compound B, and CompoundC) were used.

CH3 CH3

Figure 1 Deacylation of an ester

Compound A has the formula C13H24S, Compound Bhas the formula C9H18O, and Compound C has the formulaC7H9N. Figure 2 shows three graphs depicting the changein reaction rate of thedeacylation reaction as a function of thesolution pHforeachof thethree separate compounds.

8.0 6.0

PH pH pH

Figure 2 Reaction rate as a function of solution pH

Each reaction obeys standard mechanistic behavior forcarbonyl substitution. They are believed to proceed through amechanism where the nucleophile attacks the carbonylcarbon, breaking the C=0 rc-bond and forming a tetrahedralintermediate. A lone pair of electrons on oxygen thenreforms the 7t-bond, ejecting the leaving group. Althoughthe nucleophilicity of the different compounds is not equal,the similar mechanisms make the reactions comparable. Atlow pH, the carbonyl compound can be protonated, making ita better electrophile. This negates the effect of decreasingnucleophilicity of alcohols and thiols, because they remainunchargedat low pH values.

85. Which of the following types of compounds is theMOST basic?

A. Primary alcoholsB. Esters

C. SecondaryaminesD. Tertiary thiols


Compound C

8 6. The nucleophilicity of each reagent in aqueoussolution:

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.

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

A. H3CCH2CH2OCH3B. H3CCH2CCO2CH3

C. H3CCH2CH2CONH2

D. H3CCH2CH2CH2NH2

88. If the pKa for H3CNH3+ is 10.3, which of thefollowing is the BEST approximation for the pKa forC13CNH3+?

A. 17.3

B. 12.3

C. 8.3

D. 1.0

89. If the pKa for NH4+ is 9.25, then pKb for NH3 mustbe equal to whichof the following?A. 9.25

B. 7.00

C. 5.75

D. 4.75

90. How does the hybridization of the carbonyl carbonchange during the reaction?

A. Itchanges from sp2 to sp3 and back to sp2.B. Itchanges from sp3 to sp2 and back to sp3.C. It remains sp2 throughout the reaction.D. It remains sp3 throughout the reaction.

91. Anilineand benzylamine, drawn below,are both:

NH2^NH2

Aniline

A. primary amines.B. aromatic amines.

C. conjugated amines.D. nonalkyl amines.

Benzylamine


Questions 92 through 100 are NOT based on adescriptive passage.

92. Which of the following compounds has the HIGHESTboiling point?

A. 2-penteneB. Diethyl etherC. HeptanalD. Cyclohexanol

93. The correct RJPAC 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

94. Which of the following compounds exhibitsconjugation?

I. 1,4-cyclohexadiene

n. 3-ethylcyclohexene

HI. 2-methyl-1,3-cyclopentadiene

A. I onlyB. IlonlyC. monlyD. I and monly

95. Whichof the following compounds is MOST stable?

A. 2-methyl-1,4-pentadieneB. 3-methyl-1,4-pentadieneC. 2-methy1-1,3-pentadieneD. 1,5-hexadiene

96. Which of the following molecules wouldhave a dipolemoment NOT equal to zero?

I. Z-l,4-dichloro-2-butene

n. E-l,4-dichloro-2-butene

m. cis-l,2-dichlorocyclopentane

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


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

A. H3CCH2OCH2CH3B. H3CCH2CH2OCH3C. (H3Q2CHOCH3D. H3CCH2CH2CH2OH

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

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

99. Which of the following compounds releases theGREATEST amount of heat upon combustion?

A.

\B.

\

\ \D.

100. Which of the following functional groups is found inC2H5CH(OCH3)C(0)CH(CH3)2?

A. AldehydeB. Ester

C. Ketone

D. Oxirane

1. D 2. A 3. C 4. A 5. D 6. D

7. B 8 D 9. D 10. B 11. B 12. A

13. D 14. A 15. C 16. B 17. B 18. C

19. D 20. B 21. C 22. A 23. D 24. D

25. C 26. A 27. D 28. A 29. C 30. A

31. A 32. B 33. D 34. A 35. A 36. C

37. C 38. C 39. B 40. C 41. D 42. B

43. B 44. A 45. A 46. B 47. D 48. D

49. A 50. D 51. A 52. A 53. D 54. B

55. D 56. A 57. D 58. D 59. D 60. B

61. D 62. D 63. D 64. A 65. D 66. A

67. D 68. C 69. C 70. A 71. D 72. C

73. D 74. B 75. A 76. C 77. D 78. A

79. B 80. C 81. B 82. A 83. D 84. C

85. C 86. C 87. D 88. C 89. D 90. A

91. A 92. D 93. B 94. C 95. C 96. D

97. D 98. C 99. B 100. c

iCHEMICA ES FIN!

Structure, Bonding, and Reactivity Passage Answers

Passage I (Questions 1-8) Bond Dissociation Energies

5.

6.

7.

Choice D is correct. Table 1 does not listany alkynes, so thebond energies mustbe estimated from trends in thedata. According to data in Table 1, a bond between two s/?3-hybridized carbons has a bond dissociation energy,BDE, between 81 and 88 kcals/mole. Abond between an sp2-hybridized carbon and an s/?3-hybridized carbonhas a BDE of97 kcals/mole. This means that the bond between C2 and C3, a bond between ansp-hybridizedcarbon and an sp3-hybridized carbon, should have a BDE greater than 97 kcals/mole. This eliminates choicesAand B. Abond between a hydrogen and an sp3-hybridized carbon has a BDE between 91 and 104 kcals/mole.A bond between a hydrogen and ans/?2-hybridized carbon has a BDE of 108 kcals/mole. This means that thebond between hydrogen and an s/?-hybridized carbon should have a BDE greater than 108 kcals/mole. Thiseliminates choice C and makes choice D the best answer.

Choice Ais correct. According to the bond dissociation energies listed in Table 1,iodine and chlorine bothmaketheir strongest bonds to methyl 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 Ais best.

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

Choice Ais correct. The hydrogenation of an unsubstituted alkene yields the greatest energy (according to theheats of reaction listed in Table 2). Aless stable reactant yields a greater amount of heat upon reaction, so theTi-bond must be weakest in an unsubstituted alkene. The energy generated from oxidative cleavage, or anyreaction that breaks the 71-bond, is greatest when the alkene is unsubstituted. Thebest answer is choice A.

Choice Dis correct. The difference in reactivity between the cis and trans geometrical isomers of an alkene isattributed to intramolecular 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 welcome freebie, relative to other questions in the passage.Alkene carbons have s/?2-hybridization, and alkane carbons have sp3-hybridization. Two of the three carbonsin the compound are alkene carbons, while the other carbon is an alkane carbon. The best answer is choice D,two sp -hybridized carbons and one s/73-hybridized carbon. Pick choice Dto get your point for correctness.

Choice Bis correct. Table 1lists the bond dissociation energy for various bonds, which is the energy required tobreak the bond in ahomolytic fashion. This in essence means that Table 1lists the bond strength. The strongestbond, according to that data listed in Table 1, is the bond with the highest bond energy. The highest valueamong theanswer choices is thebond between themethyl carbon and chlorine. The bestanswer is choice B.

Choice Dis correct. Double bonds are stronger than single bonds, meaning that more energy is required to breaka double bond than a single bond. Choices Aand Bare thus eliminated. The lower heat of hydrogenation inthe second chart implies that the reactant alkene molecule is more stable. The more stable the alkenecompound, the stronger its 71-bond. This means that the double bond in the tetrasubstituted alkene is strongerthan the double bond in the unsubstituted alkene. The best answer is choice D.

Passage II (Questions 9 -14) Structure of Caffeine

9. Choice Dis correct. To convert the acid form of anitrogen-containing compound (in this case, aprotonated iminespecies) into a neutral species, you must add a strong base (strong enough to deprotonate the iminium cation).Choice Ais eliminated, because it is a strong acid, which protonates rather than deprotonates the compound.Choice Bis eliminated, because it is inert and has no effect on caffeine. Choice Cand choice D are both bases,but the stronger base is NaOH, so choice Dis the better choice. The carbonate base is not strong enough to fullydeprotonate the iminium cation. For best results ina case like this, pick choice D.

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10. 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 planar with respect to theadjacent atomsbonded to it, with bond angles of approximately 120°. For so many good chemistry reasons, thenitrogen inquestion issp2-hybridized, sowhy choose anything but choice B?

11. Choice B is correct. Because of the delocalization of electron density throughout the rc-network of the caffeinemolecule (achieved by the resonance between alladjacent 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 thenitrogens, must becoplanar with respect tooneanother. Pick choice Bfor a whoppinggoodanswer.

12.

13.

14.

Choice A is correct. Because of the resonance donation of the lone pair of electrons on nitrogen to the carbonylcarbon, the C=0 bond takes on some single-bond properties, which makes it longer than a standard sp2-hybridized C=0bond that lacks any resonance effect. The best choice is therefore choice A. Just asa side note,the C=0 bond in formaldehyde is longer than the C=0 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:

:o:

N

I

•OI longer than a standard*""* C=Obond

Choice 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 Dcorrect, and you want that which is correct. Follow society's influence 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 anhydride (from the fact that the compounds formswhen two acids combine in a dehydration reaction). The best answer ischoice A. There isnosuch term as an"acid ester", so choice Biseliminated. An ester involves just one carbonyl, so choice Ciseliminated. Alactoneis acyclic ester, and given that the compound is not an ester at all, it can't be alactone. Choice Dis eliminated.

Passage III (Questions 15 - 21) Solubility of Dyes and Soap

15. Choice C is correct. A good solvent for dissolving a dye to form an ink is one that is a liquid at roomtemperature, evaporates quickly, and exhibits a high degree of dye solubility. Having a high vapor pressureimplies that it evaporates readily, so choice Ais eliminated. If it has functional groups that are similar tothe dye, then the dye is likely to be highly soluble in the solvent, so choice Bis eliminated. If the boilingpoint is slightly above room temperature, then it is a volatile liquid, so choice Dis eliminated. The solventshouldnot react with paper (cellulose), so thecorrect answer is choice C.

16. Choice B is correct. Because 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 you may be aware that carboxylic acids that are three carbons or greater in length are solidsat room temperature. If you don't have them memorized, like 99.999% of us, that's okay too. Aldehydes andketonesmay work, but not as well as the alcohol. Thebest choice is B.

17. Choice B is correct. Determining the IUPAC name for the compound requires that you count the longestcontinuous chain ofcarbon atoms in this straight chain compound, which yields a total of ten. Next, you mustidentify all functional groups on the molecule, including alkyl groups that are not apart of the carbon backbone.The only functional group on this compound is acarboxylic acid. If there were multiple functional groups, themore oxidized functional group gets higher priority in the name of the compound. For instance, if there werealso an OH group, then itwould be named a hydroxy substituent, rather than an alcohol. Having ten carbonsanda carboxylic acid group makes the compound decanoic acid. Choice Bis the best answer.


18. Choice C is correct. Whenwater is spilled on paper,it diffuses across the surface of the paper. If the ink boundto the paper is solublein water, it dissolves into the waterand spreads out, or runs. So if an ink runs, it must besoluble in water. The colored portion ofthe ink is the dye, not the solvent. Running inkmeans that theinkdyeis water soluble. Pick choice C for best results.

19. Choice D is correct. As stated in the last paragraph of the passage, a compound must contain at least eightcarbons in its chain tobe a good soap. The best soap hasa polar andnonpolar end associated with themolecule.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 Dhas both eight carbons anda charged end.

20. Choice Bis correct. To besoluble inwater, a compound must beeither charged orpolar. Because choices BandDare 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 choiceD. PickchoiceB, and feel the sensation of correctness.

21. Choice Ciscorrect. Carboxylates are formed when acarboxylic acid is deprotonated. The Grignard reagent inchoice A deprotonates thecarboxylic acid to form the carboxylate, but the cation is not sodium, so choice A iseliminated. Choice Bis 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 H3CC02Na molecule results from thedeprotonation of acetic acid by a base with a counterion of Na+. Thebest choice is thereforechoiceC.

Passage IV (Questions 22- 28) Amide Protonation

22. Choice Aiscorrect. Because of the resonance donation from nitrogen, the carbonyl bond (C=0) ofanamide hassome single-bond character. Since a single bond is longer than a double bond, the single-bond character of theamide carbonyl bond results ina longer carbonyl bond than the unconjugated 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 itshorter than astandard 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 not true, so choice A is the best answer.

23. 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 Bshould 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.

24. 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 also has trigonal planargeometry, so the central two atoms force the three hydrogens and one oxygen to assume a coplanar orientation.Choices A, B, and Ctherefore all must be eliminated as incorrect geometric descriptions, making choice Dthebest answer. The two resonance forms and the resonance hybrid are drawn below:

O O-II i

M ' sp -HybridizationH .C .H .. . , _^^ ^sp2-Hybridization

^^ ^.' Trigonal planarrT ^N' -* • H^ ^N+

H pj ' Trigonal planar

25. Choice Cis correct. From the passage, we know that amides are protonated atthe 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 alarger partial negative charge on the amide carbonyl oxygen than on the ester carbonyl oxygen. The largernegative charge makes Site c the most basic site. Choice C is therefore the best answer.

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26. Choice A is correct. Themoststablehydrogen bond forms between thebest lone-pair donor (mostbasic 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 andthe hydrogen bonded tonitrogen, thus the best answer ischoice A. Ifyouare unsure, think of the hydrogen-bonding in 6-pleated sheets.

27. 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 Bbecomes the resonancestructure. This is a minor contributor due to the formation of charges on the molecule. The fact that it is an all-octetresonance structure is favorable. If the nitrogen were to pull its 7i-electrons back from the carbon in answerchoice B, the resonance structure represented by answer choice Cwould 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. Inorderforcarbon todonate in thatmanner (and haveonly four bonds), it musthavehad a lonepair (and thus a negative charge) in the original structure. Rwas not drawn as having a lone pair, so it isassumed that the Rrepresents a standard alkyl group. The best answer is therefore choice D.

r rxr>,R' -ct .R',R'

R

H

Choice A (major)

R" N1

H

Choice B (minor)

R N

IH

Choice C (very minor)

28. Choice A is correct. Given that the molecular masses ofthe three compounds are roughly equal (59 g/mol, 58g/mol, and 44 g/mol), the top consideration for determining their boiling points is the intermolecular forces. Anamide has hydrogen-bonding, while aketone and ahydrocarbon 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, BPacetamide >BPaCetone >BPpropane- Acetone hasa higher boiling point than propane, because it is polar andmore massive.

Passage V (Questions 29 - 34) Physical Properties and Intermolecular Forces

29. Choice Cis correct. Chemist 1considers molecular mass todetermine the relative boiling points ofcompounds.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). Ahigher 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 Aisnot an exception to Chemist l's general rule. In choice B, the heavier compound of the two has the lower vaporpressure (and thus higher boiling point), so choice Balso follows the rule. In choice D, the heavier compound ofthe two has the lower vapor pressure (and thus higher boiling point), so choice Dfollows the rule too. In choiceC, the heavier compound of the two has the lower boiling point, so choice Ccontradicts Chemist l's theory.

30. 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 Athe best answer.

31. Choice Ais correct. Hydrogenation is the addition of H2 gas to an alkene to break the 7U-bond and reduce thecompound to an alkane. For every 7i-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'll be an MCAT supernova.


32.

33.

34.

Choice Bis correct. According toChemist 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 ofit 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. However, the lower atmospheric pressure means that less energy (heat) is required toreach a temperature at which the vapor pressure (Pvapor) is equal to the atmospheric pressure (Patmospheric)/the definition of theboiling point. The boiling point is therefore lowered as elevation increases. This makeschoice D correct.

Choice Ais correct. The boiling points of Compounds I and II are directly comparable, because they aregeometrical isomers. Compound I (the cis isomer) is polar, 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 rc-bond), so Compound III is able to rotate between conformers. The most stable conformation ofCompound III is nonpolar, but because itcan assume polar conformations on occasion, the compound is slightlypolar. The boiling point of Compound III is less than the boiling point of Compound I. Compound III shouldhave the second highest boiling point, because it is slightly polar, while Compound II is non-polar. Thus, thecorrect order is I > in > II, making choiceA the best answer.

Passage VI (Questions 35 - 42) Molecular Structure and Polarity

35.

36.

37.

38.

39.

40.

Choice Ais correct. The greatest reduction in voltage is caused by the compound with the greatest dielectricconstant. The greatest dielectric constant isassociated with the most polar compound. Choices B, C, and Dareall symmetric, so they are all 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 Biseliminated. The hydrogenation of an alkene forms an alkane. An alkane is most often nonpolar, so the bestanswer ischoice C. Reduction ofanamide forms aprimary amine. An amine ispolar, sochoice Diseliminated.

Choice Cis correct. Assuming that an alkyl iodide is polar to begin with, then replacing iodine with bromineresults ina more polar compound, because bromine ismore electronegative than iodine, sothat 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 Bresults in a more polar compound. Alkenes aretypically nonpolar, so the addition of HBr forms an alkyl bromide, which increases the polarity, so choice Diseliminated. Because fluorine is more electronegative than chlorine, replacing a fluorine substituent with achlorine substituent results in a compound that is less polar, making choice C the choice that does not result inincreased polarity. Pick choice C to be a star ofchemistry.

Choice Cis correct. The dipole moment changes only when a compound's bonds are either stretched orbent, ifthe compound is asymmetric. This makes choices Aand Bless likely to exhibit the least change in dipolemoment. The dipole moment does not change drastically (if at all), 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 inopposing directions. The result is that the dipole of the molecule does not change. The best answerischoice C. Bending asymmetric molecule can make itasymmetric, so choice Disnot as good as choice C.

Choice B is correct. To be nonpolar, all of the ligands must pull 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.

Choice Ciscorrect. Adielectric constant of zero results from anonpolar molecule. The only nonpolar moleculeamong theanswer choices is trans dichloroethene. Cis alkenes are polar, sochoice Bis eliminated. Choice Distrans, but there aredifferent substituents oneach carbon, soit ispolar. The bestanswer is choice C.

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41.

42.

Choice D is correct. Choice A is nonpolar, because the vectors expressing the electron withdrawal of thefluorines cancel out, so choice A is eliminated. Fluorine is more electronegative than chlorine, so theasymmetric electron distribution is found with choices C and D, which rules out choice B. In choices C and D,the fluorine atoms withdraw the electron density, making the molecule asymmetric. In the propane molecule(choice D), the methyl substituent donates electron density to the electron-poor central carbon, placing apartial positive charge on the methyl group; therefore, it increases the dipole moment. This makes thepropane moleculemore polar than the ethane molecule. The structures are shown below:

All vectors cancel ,\ nonpolar

f4/F h

Vectors almost cancel.-. slightly polar

C-F vectors cancel, but methyl donates.•. 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 isasymmetric. If the compound is asymmetric, then itmust be polar (have an asymmetric distribution of electrondensity). This makes statement I true. Figure 1shows an example of a 1,4-disubstituted cyclohexane moleculethat is not polar, which means that statement II is not true. All optically active compounds must beasymmetric in order to be optically active, so they must be at least slightly polar. This makes statement IIItrue. Only statement II is not true, so choice B is the best answer.

Passage VII (Questions 43- 49) Micelles

43.

44.

45.

46.

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 outfrom itsaqueous structure when it is added to an organic solvent(as stated in the passage). Figure 1shows 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 exposed, and the polar heads form a protective core. This is best illustrated in choice^ A. Choice Cmay look familiar, in that cell membranes arrange themselves in such a manner. Choice Dis a "throw-away"answer, because the tail and the head of the compound exhibit no attractive forces.

Choice Ais correct. To be absorbed through respiration, acompound must be agas or a vapor, because only gasesare absorbed through respiration. This means that any compound intended to be taken into the lung must beeither a gas or a liquid with a low boiling (one with a high vapor pressure). There is hydrogen-bonding inchoices B, C, and 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 boiling points is hydrogen-bonding. The best answer is theether, choice A. As a point of trivia, it is estimated that the average human adult takes in approximately3500 gallons of air a day. Just thought you might like to know.

Choice Biscorrect. To be water-soluble, the compound should be able to form hydrogen bonds. Choices Cand Dare eliminated immediately, because they are hydrophobic. Although choice A has an alcohol group, it isprimarily organic. Choice Bhas two hydroxyl groups and an amide group. All of these functional groups formhydrogen bonds, so choice Bexhibits the greatest amount of hydrogen-bonding. The best answer is choice B.


47. 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.

48. 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 water-soluble 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.

49. 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 Band C are both eliminated, because to form hydrogenbonds,a compound musthave an electropositive hydrogen bonded toeithernitrogen, oxygen, or fluorine. ChoiceD is eliminated, because covalent bonds are formed in chemical reactions, and the organic tails in micellesexhibit only attractive forces, nothing as strong as covalentbonding. The best answer, and thus choice to make,is choice A.

Passage VIII (Questions 50- 56) IR Determination of 1°,2°,and 3°Alcohols

50. Choice D is correct. Hydrogen-bonding weakens thecovalent bond to hydrogenand 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 avariety 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.

51. 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 apoint ofinterest, the tertiary amine has no N-H covalent bonds, so it has no hydrogen-bonding.

52. ChoiceA is correct. The relationship between bond strength and IRabsorbance 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 tobreak 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 Band D. An increase in the degree of hydrogen-bonding weakens and thus lengthens the bond. Thiseliminates choice C and makes choice A the best answer.

53. 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.

54. Choice Bis 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.

55. 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 Statement I is not true. As hydrogen-bonding increases, thecovalent 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.


PassageIX (Questions 57 - 63) Petroleum Distillation

57. 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-trimethylpentane, choice D, which can beinferred 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.

58. 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 inthe petroleum-to-naphtharange. You should use test-taking logic to eliminate choices Band C, because their boiling points are similar,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 98°C. Following this trendpredicts that n-octane has an approximate boiling point of 125°C - 130°C and n-decane an approximate boilingpoint of 175°C - 180°C. The branching of 2,2-dimethyloctane reduces the boiling point from that of n-decane(the straight-chain, ten-carbon alkane) to somewhere around 165°C to 170°C. The 2,2,4,4-tetramethyldecane ismost likely to have a boiling point in the 175°C to 280°C range. Your job, should you accept it, is to pick D.

59. 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-trimethylbutane > 2-methylhexane > n-heptane. This makes choice Aa 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-methylhexane >2,2,3-trimethylbutane. 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 thesmallest volume. This results in an increase in density with branching. Choice C is a valid statement, so it isalso eliminated. Hydrocarbons have no hydrogen-bonding, so regardless of the amount of branching, hydrogen-bonding neither increases nor decreases from hydrocarbon to hydrocarbon. This makes choice Dan invalidstatement as to the effect of branching. You shouldsmile brightly when you pick choice D.

60. Choice Bis correct. As density increases for a hydrocarbon (or any gas), it does not rise aseasily. 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 islisted 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 Bthe 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 releasing their heat energy steadily over anextended period of time. The best answer is therefore choice D. Do what is best, and pick choice D. The octanerating 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 mthefuel.

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 istherefore a true statement. Because toluene (methylbenzene) has an octane rating of 120 and benzene has anoctane 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 branching associated with 2,2,3-trimethylbutane, ithas ahigh octane rating. Ahigh octane rating is a quality associated with a good fuel additive, so a branchedhydrocarbon such as 2,2,3-trimethylbutane 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.

Copyright ©by The Berkeley Review® 83 Section I Detailed Explanations

63. Choice D is correct. Because 2,2,4-trimethylpentane is a saturated hydrocarbon, all of its carbons have ahybridization of sp3. In the final product, the carbons are all present 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 one we highly recommend to all parties interested in success, ischoice D.

Passage X (Questions 64 - 70) Acidity and Hybridization

64. 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 twoconcepts, 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-l-pentyne, carbons 1 and 2 are sp-hybridized, but only carbon 1 has a hydrogenattached. Pick choice A for optimal results. Make a note from the conclusions that sp >sp2 >sp3 for acidity.

sp-hybridized

65.

66.

67.

68.

69.

70.

Choice D is correct. The largest Ka is associated with the strongest acid. All of the choices are hydrocarbons,so the most acidic proton is the one on ansp-hybridized carbon, as opposed to either an sp2-hybridized or 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. All of the compounds are deprotonated hydrocarbons (with a lone pair on carbon), so thestrongest base is the one with the lone pair on an sp3-hybridized carbon. The only choice with a lone pair ofelectrons on an sp3-hybridized carbon is choice A. The cation is irrelevant to the problem, because itis sodiumin each answer choice.

Choice D is correct. NaNH2 is a base strong enough to deprotonate a hydrogen on ansp-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, soyou hadbetter pick D.

Choice C is correct. The longest carbon-carbon 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 sp3-hybridized carbon. Choices Band Dare eliminated immediately, because theycontain noC-C single bond. Choice Aisa bond between ansp-hybridized carbon andansp3-hybridized carbon,while choice C is between an sp2-hybridized carbon and an sp3-hybridized carbon. An sp2-hybrid orbital islonger than an sp-hybrid orbital, so choice C is the best answer.

Choice C is correct. The weakest carbon-carbon bond isassociated with the longest carbon-carbon bond. ChoiceAis eliminated, because the C-C bondis a triple bond, and triple bonds are theshortest of carbon-carbon bonds.Choice Bis between an sp-hybridized carbon and an sp3-hybridized carbon, choice C is between an sp2-hybridized carbon and an sp3-hybridized carbon, and choice Dis between an sp2-hybridized carbon and an sp2-hybridized carbon. Choice Cis the longest, so itwould terrific ifyou would pick choice C.

Choice A is correct. The lowest pKb is associated with the strongest base. Because the most acidic proton isfound onansp-hybridized atom, the strongest base must be a lone pair onan sp3-hybridized atom. Choices Cand Dare sp2-hybridized nitrogens, so they are both eliminated. Choice A is better than choice B, because thelone pairofelectrons on nitrogen in choice Bis tied into resonance with the adjacent alkene 7C-bond. Electron-withdrawing resonance reduces a compound's basicity.


Passage XI (Questions 71 - 77) Keq and Acidity

71. Choice D is correct. For a reaction to have an equilibrium constant greater than 1.0, the reaction must befavorable 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 Keq values in the table. Inchoice A, the reactionproceeds from the weaker acid (CI3CH with Keq of 8.0 x10"9) to the stronger acid (H3CCH2SH with Keq of 5.2x 104). This means that this reaction is unfavorable and thus has a Keq < 1. Choice A is therefore eliminated.In choice B, the reaction proceeds from the weaker acid (H3COH with Keq of 3.9 x 10_1) to the stronger acid(H3CNO2 with Keq of 8.2 x104). This means that this reaction is unfavorable and thus has a Keq <1. ChoiceB is therefore eliminated. In choice C, the reaction proceeds from the weaker acid (H3CCOCH3 with Keq of2.0 x10-4) to the stronger acid (H3COH with Keq of 3.9 x10'1). This means that this reaction is unfavorableand thus has a Keq <1. Choice Cis therefore eliminated. It is only in choice Dthat the reaction proceeds fromastronger acid (H3CCOCH2COCH3 with Keq =1.2 x106) to aweaker acid (CI3CH with Keq =8.0 x10"9). Thecorrect answer is choice D.

72. Choice C is correct. Reaction 2, the experimental reaction from the passage, involves the protonation ofC5H5"to form C5H6. 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 CsH6. Choices A and Bareeliminated, because they are acids, notbases. Because only methanol (CH3OH) shows an equilibrium constantless than 1.0, only methoxide anion (CH3O-) is strong enough to deprotonate CsH6. The best answer is choice C.

73. 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 beequally concentrated at equilibrium. The best answer therefore ischoice D.

74. Choice Bis correct. The conjugate base of the 1,3-cyclopentadiene species has six rc-electrons in a continuouscyclic planar array of p-orbitals. These conditions result in aromatic stability. The best explanation for therelative ease with which the 1,3-cyclopentadiene loses its proton is the aromaticity associated with theconjugate base (1,3-cyclopentadienyl anion). The more stable that the conjugate base is, the stronger the acidis. Pick choice B and be satisfied.

75. Choice A is correct. The weaker of the two acids has the larger ofthe two pKa values. This question is askingfor the weakest acid relative to the strongest acid. As the acid gets weaker, the reaction with C5H5- becomesless favorable, so the equilibrium constant for the reaction gets smaller. CI3CH shows the lowest equilibriumconstant of the answer choices, so it is the weakest acid and thus has the highest pKa value. It is only in choiceA that CI3CH is listed as having the highest pKa value, which makes choice A correct.

76. Choice Cis correct. ALewis 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 electron-deficient. Both CH3CI and CH3NO2 are listed as acids in the table, sochoices Aand Bare 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.

77. Choice Dis correct. Using Equation 1, the Keq for areaction is found by taking 10 to the power of the productacid pKa minus the reactant acid pKa. In the standard reaction, C5H6 is the product acid and its pKa value: isgiven as 15.0. If the pKa of the reactant acid is 10.0, then the equilibrium constant would be 10^• " • >- 10 .The question is therefore: "Which acid in Table 1has an equilibrium constant of roughly 10^ ?" The best answeris choice D, CH3NO2, with an equilibrium constant of 8.2. x104 when it reacts with C5H5".

Passage XII (Questions 78 - 84) Alkoxides and Alkyl Sulfides

78. Choice Ais 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 1.44makes it the fastest reaction rate of the choices offered to you. Thebest answer is therefore choice A.

Copyright ©by The Berkeley Review® 85 Section 1 Detailed Explanations

79.

80.

81.

82.

83.

84.

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 than 1.97 for the ethyl sulfide. Thebest answer is choice B, 2.04. Following the trend for ether solvent shows a negative log value for HS" of 1.76and for H3CS" of 1.97. It can be concluded from this that the negative log of rate for H3CCH2S" should begreater than 1.97, but no larger than about 2.2. The only choice in this range is 2.04. The lack of linearity in thetrend is due to the fact that the change in steric hindrance from H3CS" to H3CCH2S' is less drastic than thechange from HS"to H3CS". The difference in the change in steric hindrance can be seen in the smaller change innegative log value for the reaction rate.

Choice C is correct. Becausebromine is smaller than iodine, bromine forms a stronger bond with carbon thaniodine, so bromine is a worse leaving group than iodine. The rate of the reaction depends on the electrophile.The worse the leaving group, the worse the electrophile, and the slower the nucleophilic substitution reaction.The bromine leaving group would yield a slower reaction (a decrease in the reaction rate) than the iodineleaving group, and consequently a larger negative log of the reaction rate. The best answer is thus choiceC.

Choice B is correct. Theweakest bond is the one that wouldbe broken in a nucleophilic substitution reaction,such as Reaction 1. In Reaction 1, the bond that is broken is the one between carbon and iodine, 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. Thiseliminates choices C and D and furthersupports choice B as the best answer.

Choice A is correct. As thesolvent is changed from ether to ethanol and finally water, the degree ofhydrogenbonding 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 choiceA.

Choice D is correct. According to the data inTable 1, regardless ofthe solvent, the negative log ofthereactionrate 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 negative 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 decrease 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 andnegative logof the rate.

Choice C is correct. In ether, alkoxides have lower negative log values for their reaction rates than alkylsulfides, so the alkoxides must react faster. This eliminates choices Band 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 pKb value. Pick Cand feel jovial for just amoment, at least until the next question starts.

Passage XIII (Questions 85 - 91) Nucleophilicity and Basicity

85.

86.

87.

Choice Ciscorrect. The most basic species is the compound containing nitrogen. In general, nitrogen compoundsaremore basic than oxygen- andsulfur-containing compounds ofequal hybridization. This eliminates choices Aand D. The degree of substitution is irrelevant. Esters have no lone pair of electrons that can be readilydonated to a proton, so choice Bis not correct. The best answer is an amine, independent of whether it isprimary, secondary,or tertiary. Thismeans that you really should pick C for the sensation of correctness.

Choice C is correct. The rate referred toin Figure 2 isfor a nucleophilic substitution reaction at a carbonyl site.There is a direct correlation between nucleophilicity and the rate of reaction. The graphs show that above apHof 7, as the pH increases, so does the reaction rate. This means that the nucleophilicity increases. As pHincreases, compounds are no longer cationic. This eliminates choices A and B. Size is not applicable here, sochoice D is eliminated. After eliminating thewrong choices, youshould settle forchoice C as thebestanswer.

Choice D is correct. AtpH =9.0, all of the compounds should beneutral (although the amine in choice D mayhave a small fraction that remains protonated). The most reactive compound is the best nucleophile. Fornucleophilicity, an amine isbetter than an ether, anester, or anamide. For this reason, pick D.


88. 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 itspKa 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 pKa 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 109 = 2,000,000,000 times. That is too much. Be conservative and pick C

89. Choice D is correct. The sum of pKa + pKb for a conjugate pair in water is equal to 14.0at 25°C. This means thatpKb for NH3 (the conjugate base of NH4"1") is equal to 14.0 - 9.25 = 4.75. Pick D and score with the best of them(whoever they are).

90. Choice A is correct. The hybridization of carbon in a carbonyl compound, such as an ester (which contains one k-bond), mustbe sp2 (the 71-bond requires onep-orbital, so only two p-orbitals remain for hybridization). This canalso be deduced from the trigonal planar structure of the carbonylcompound. The hybridization of carbon in thetetrahedral intermediate (which contains no 71-bonds) is sp3. The final product again has the carbonylfunctionality, only now with the nucleophile attached. The carbonyl product still has trigonal planargeometry. The hybridization therefore changes from sp2 to sp3 and back to sp2 in the overall reaction, makingchoice A your choice. Make that choice today!

91. Choice A is correct. In each case, there is a nitrogen with one alkyl group and two hydrogens. This defines aprimary amine , soboth compounds areprimary amines. It just so happens that the R-groups arearomatic rings,but they are not aromatic amines per se, because thenitrogen atom is not a part of thearomatic system. The bestanswer is choice A. Of the two amines, only one is conjugated (aniline), so choice C does not describe bothstructures.

Questions 92-100 Questions Not Based on a Descriptive Passage

92.

93.

94.

95.

96.

Choice D is correct. Alcohols 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 ofconcern as well, butchoice D is also the heaviest of the choices.

Choice Bis correct. The longest chain is eight carbons, so based on thatalone, you know that the best answer ischoice B(octane). All you need to do is find thelongest chain to decipher the correct answer choice.

Choice Cis correct. Conjugation isdefined asconsecutive, alternating 7t-bonds. The structures are drawn below.Only Compound IIIhas conjugation, so choiceC is correct.

<?

1,4-cyclohexadiene 3-methylcyclohexene 2-methyl-l,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-l,4-pentadiene 3-methyl-l,4-pentadiene 2-methyl-l,3-pentadiene 1,5-hexadiene

Choice D is correct. Saying that a compound has a dipole that isnotequal to zero isequivalent to saying thatthe compound is polar. Cis compounds (both alkenes and cyclic structures) are always polar. This makes bothCompound Iand 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.

Copyright © by The Berkeley Review13 87 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 account 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.

98. Choice C is correct. For a compound not to be polar, it must be symmetric. Cis compounds are asymmetric 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 middlecarbonhas varyingsubstituentsattached (notall four groupsare 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.

E-l,4-dichloro-2-butene

The individual vectors cancel out, so there isno net vector. The compound is nonpolar.

99. 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 (C6Hs) so it comes down to structuralfeatures. The four-membered ring is unstable, sochoices Aand Baregood. Choice Bhasno conjugation, whilechoice A does (conjugation isa stabilizing feature), so this makes Bthe least stable compound among the answerchoices. The bond angles are not theoptimal 109.5°, whether the7i-bond is in the ringor not.

100. Choice C is correct. Translating from the chemical formula into the structure yields thecompound below:O

H3CH2C CH,

H3CO CH,

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 Bis 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.

"Chem is part of a nutritious breakfast!"


Section II

Structure

Elucidationby Todd Bennett

Isomerism

a) Isomers

i. Constitutional Isomers

ii. Stereoisomers

iii. Conflgurational Isomersiv. Optical and Geometrical Isomers

b) Conformational Isomers

i. Eclipsed vs. Staggered Conformationii. Gauche vs. Anti Position

iii. Newman Projectionsiv. Cycloalkanesv. Cyclohexanevi. Chair and Boat Conformations

vii. Equatorial vs. Axial Orientation

Structural Insights

a) Structural Symmetryb) Units of Unsaturation

Spectroscopy and Analysisa) General Spectroscopy

b) Infrared Spectroscopy

i. Theory and Key Signalsii. Common Appplicationsiii. Hydrogen-Bonding

c) Ultraviolet Spectroscopy

i. Theoryii. Common Appplications

d) nMR Spectroscopy

i. Theoryii. Structural Symmetryiii. Proton NMR

iv. Shift Values

v. Splitting Patternvi. Signal Integrationvii. Structural Features

BerkeleyJL#R-E.V-KE>W®


Structure ElucidationSection Goals

Be able to identify isomers from both their structure and their name.There are several types of isomers. Be familiar with structural isomers (identified by differentconnectivity), stereoisomers (identified by asymmetry), optical isomers (and their ability to rotateplane-polarized light), geometrical isomers (found with rings and alkenes), and conformationalisomers (identified by rotation about bonds or ring-flips). Youshould know how the different typesof isomers are related to one another.

Be able to identify the more stable chair conformation for six-membered rings.

Cyclohexane and pyranose sugars involve three-dimensional ring structures. The most stableconformation results in the least steric hindrance. As a general rule, axial orientation results ingreater sterichindrance than equatorialorientation. Knowthe difference in stabilitybetween axialorientation and equatorial orientation. Recognizethe steric repulsion associated with 1,2-diaxial,1,3-diaxial, and 1,4-diaxial orientation.

Be able to identify the more stable Wewmann projection for a compound.Structures canorient themselves in a staggered conformation, an eclipsed conformation, or someconformation betweeneclipsed and staggered. You mustbe ableto identifythe moststableorientationfor a structure and distinguish betweengaucheand anti positions.

Be able to use the molecular formula to determine the units of unsaturation.Some questions require youtodetermine thepotential functional groupsofa molecule. Thepresenceof either a rc-bond and a ring within a structure results in a unit of unsaturation, which manifestsitselfas two fewer hydrogens in the formula. Afully saturatedhydrocarbon or carbohydrate hasa total of2n+2hydrogen atoms inthecompound, where n isthenumber ofcarbons in thecompound.

Be able to translate structures from two dimensions into three dimensions.

Know whatthe terms staggered, eclipsed, gauche, andantimean, andbe able to drawstructuresin the Newmann projection to show the orientation of substituents in these structures. Be able torotate about a-bonds.

Be able to deduce structural features using IR spectroscopy.

You shouldhavea basic understanding of theoperations ofan infrared spectrophotometer. Youmust know the IRstretches for a carbonyl and a hydroxylbond. You should be able to determinewhich structural features correspond towhich IR absorbances. You must beable toeliminate and/orconfirm possible structures, usmg IR data. You must beable to decipher IR spectroscopy graphsand identify the key peaks.

Be able to deduce structural features using NMR spectroscopy.You must know the NMRshifts for carbonyl compounds, alkene compounds, andaromatic compounds.You should beable todetermine thestructure ofanunknown compound using thespectral informationfrom the NMR. Moststructures you will encounter on the MCAT are smalland symmetrical, sothey are easily solved. You must be able to eliminate incorrect structures based on NMR data. Besuretounderstand whattheshift value (measured inppm) tells you, whattheintegration tells you,and what the peak shape and coupling constants tell you. Each piece ofinformation can beusedto help determine the structure ofan unknown compound. Usethese data in conjunction with theunits of unsaturation.

Organic Chemistry Structure Elucidation

Structure ElucidationStructure elucidation involves applying all 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 advanced cases, the three-dimensional folding of the structure. Structure elucidation involves determiningthe number of isomers that fits a molecular formula and then systematicallyeliminating isomers that do not fit the data until, only one structure remains.

In this section, we 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 within the molecules, similarbonding but different three-dimensional distribution about a stereogenic center,or the same bonds and stereogenic symmetry 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. This informationprovides hints as to the presence of rc-bonds and/or rings within the structure.Chemical tests can be carried out to determine the number of rc-bonds, whichwhen combined with the units of unsaturation, can specify the exact number ofrings and rc-bonds within a molecule.

In this section we shall also address spectroscopy and the information aboutstructure it can provide. Infrared (IR)spectroscopy is typically used to determinethe functional groups within a compound. It can also give some informationabout the symmetry of the molecule, the hybridization of carbon, and thepresence of groups capable of forming hydrogen bonds. Ultraviolet-visible (UV-vis) spectroscopy tells us information about the rc-bonds and conjugation withina molecule. Although all molecules absorb ultraviolet radiation, for practicalpurposes, we use it only to detect rc-bonds. Nuclear magnetic resonance (NMR)spectroscopy describes the connectivity of a molecule and its specific structuralfeatures. 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 addressboth carbon-13 and proton NMR. Combining NMR data with UV-visspectroscopy and IR spectroscopy data allows for precise determination of three-dimensional molecular structure.

It is best to review NMR with symmetry as your focus. The question "How canyou distinguish compounds by NMR?" can be reduced to "How many differenttypes of hydrogens are there in each compound?" Multiple-choice NMRquestions can be answered easily by predicting the spectra from possiblestructures. For instance, if you can narrow down the potential structures toketones, then it's just a matter of systematically eliminating ketones that do not fitthe spectral data. This is the perspective from which we will approach NMR.The ability to predict spectra from structures is best attained through practice.As you do the multiple-choice questions in the spectroscopy sections, predict thespectra for the structures in this same manner. The difference between thespectra in each answer choice (A, B, C, or D) is what often answers the question.

Introduction

Copyright © by The Berkeley Review 91 Exclusive MCAT Preparation

Organic Chemistry Structure Elucidation Isomerism

Isomerism **-"~.^

Isomers

Isomers 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 spatialarrangement 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.

ISOMERS

Structures made of the same atoms

Constitutional (Structural)Differ in connectivity (bonds)

Stereoisomers

Differ in spatial arrangement of atomsI ZZ

Conformers

Differ by orientaion in spaceIdentical after rotation about a-bond

OpticalDiffer by orientation in space

Can't rotate to become identical due to

asymmetry in the structure

Diastereomers

Nonsuperimposable and notmirror images

T_

ConfigurationalDiffer by orientation in space

Can't rotate to become identical

Geometrical

Differ by orientation in spaceCan't rotate to become identical due to

the presence of a ring or rc-bond

Enantiomers

Nonsuperimposable mirror images

Figure 2-1

Constitutional isomers, which have different bonding, are more commonlyreferred to as structural 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-methylhexane versus3-methylhexane). Structural isomers can be further divided intofunctional groupisomers, positional isomers, and skeletal isomers.

Stereoisomers have exactly the same bonds (and therefore the same connectivity),but they differ in the spatial arrangement of their atoms. On a more general note,stereoisomers can be categorized as either configurational 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 be converted into the other isomer by rotation without breaking a bond.)Within configurational isomers, there are optical isomers (isomers that rotateplane-polarized light differently), geometrical isomers (isomers that vary inorientation about a 7t-bond), enantiomers (nonsuperimposable mirror images), anddiastereomers (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 andZ, if the isomers differ in their arrangement about a 7t-bond. We shall addressstereoisomers in detail in later sections.



Example 2.1What types of isomers are 2-methyl-3-pentanol and 3-methyl-2-pentanol?

A. Conformational isomers

B. Geometrical isomers

C. Structural isomers

D. Stereoisomers

Solution

The 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, sothey are also positional isomers. The question was not that specific, so the bestanswer is choice C. The two structures are drawn below:

OH

CH33-methyl-2-pentanol

OH

2-methyl-3-pentanol

Constitutional Isomers

Constitutional isomers (also referred to as structural isomers) are uniquemolecules that have the same formula, but different connectivity. In otherwords, they have the same atoms, but the atoms have different bonding. Forinstance, 3-methylhexanal and 3-methyl-2-hexanone are constitutional isomers.They each have the formula C7H14O, but they have a different sequence ofbonds. They can also be referred to as positional isomers. Using nomenclaturehelps to determine whether two structures are constitutional isomers, becauseconstitutional 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.

HO

Structural: Different arrangement of atoms (i.e.different bonds)

& H3CH2C-0-CH2CH3 S qHO- CH2CH2CH2CH3

1-butanol

H3C- CH- CH2CH2CH3

CI &

2-chloropentane

H3CH2C-0-CH2CH3

diethyl ether

H3CCH2-CH CH2CH3

CI

3-chloropentane

H3C- CH- CH2CH2CH3

CH3

2-methylpentane

H3C— CH— CH— CH3

& CH3 CH32,3-dimethylbutane

Structural isomers have different IUPAC names.

Figure 2-2

CH3

CH,

Isomerism



Example 2.2How many possible constitutional isomers exist for a molecule with themolecular formula C4H10?

A. 1

B. 2

C. 3

D. 4

Solution

The maximum number of hydrogen atoms possible on a four-carbon alkane isten, so there are no units of unsaturation in C4H10. This means that there are norc-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 off of the second carbon (if the methyl were on thefirst carbon, it is still butane). This means that there are two constitutionalisomers for C4H10, butane and 2-methylpropane. Pick choice B for the smile thata correct answer brings.

Example 2.3Which of the following pairs of molecules is NOT a set of constitutional isomers?

A. 2-Methylpentane and 3-methylpentaneB. Cyclobutanol and tetrahydrofuranC. 1-Chlorobutane and 2-chlorobutane

D. 4-Ethylchlorocyclohexane and 3-methylchlorocyclopentane

Solution

In choice A, both compounds have the formula C6H14 and different IUPACnames, so they are constitutional isomers. In choice B,both compounds have theformula C4H.8O and different IUPAC names, so they are constitutional isomers.In choice C, both compounds have the formula C4H9CI and different IUPACnames, so they are constitutional isomers. In choice D, the first compound hasthe formula C8H15CI, while the second compound has the formula CgHnCl, sothey are not even isomers, let alone constitutional isomers. This makes choice Dthe correct answer.

For a given formula, there 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 eachextra carbon, the number of structural isomers increases. For instance, C3H8 hasonly one structural isomer, while C6H14 has five different structural isomers. Itis important to realize that both formulae (C3H8 and C6H14) are for structuresthat are fully saturated (have no units of unsaturation). There is no easy formulafor determining the number of structural isomers possible for a given formula,but there is a systematic way to determine the number. Figure 2-3 shows all ofthe structural isomers for C3H8, C4H10, C5H12, and C6H14, and lists them interms of chain length and substituent location.



C3H8 (1 total): C4H10 (2 total): H3C3|CH2

H3C CH3

CH2 .CH3

H3C CH2

CH

H3C ^CHa4-Carbon chain 3-Carbon chain

C5H12 (3 total):n3y

CH2 .CH2 1

H3C CH2

H3C^ CHg

H3C CH35-Carbon chain 4-Carbon chain 3-Carbon chain

C6H14(5total): CH2 ^CH2 ^»CH3

H3C^ CH2 ^CH26-Carbon chain

CH, CH2 H3CH3C^ CH CH3 1

| ^H ^\H3C H3r CH2 CH3

5-Carbon chain 5-Carbon chain

H3C CH3\ /C. .CH3

H3C CH24-Carbon chain

H,C

H3C

IXH /CH3

CH

CH34-Carbon chain

Figure 2-3

This procedure of deteimining 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 C6Hi4, the longest possiblechain is six carbons. After drawing the longest chain, draw a carbon chain of oneless 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 ofQ>Hi4, 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 CgHu, 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 eachpossible 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 step ofsystematically placing the functional group at all unique carbons. Example 2.4demonstrates this procedure.

Isomerism



Example 2.4How many structural isomers are possible for the formula C4H9CI?

A. 3

B. 4

C. 5

D. 6

Solution

For a problem of this type, the possibiUties for the carbon skeleton must bedetermined 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

C—C—C—C C—C—C

IC

4-Carbon chain 3-Carbon chain

The second step is to determine how many unique carbons each chain contains.

STEP 2: Each skeleton has two unique carbons

*-a— Q> Qj —Ca Ca Cb Ca

Ca2 Unique carbons 2 Unique carbons

The last step requires placing a chlorine on each unique carbon one structure at atime and verifying your answer by checking to see if each structure has adifferent IUPAC name.

STEP 3: 4 structural isomers total

c—c—c—c1

c-—c—c—c1

1

CIl

CI

1-chlorobutane 2-chlorobutane

c—c—c

1 1CI c

c—c—c

/\c ci

l-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 problemsmuch 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 Elucidation Isomerism

Example 2.5How many possible constitutional isomers have the molecular formula C4H8?A. 4

B. 5

C. 6

D. 7

Solution

These questions are time-consuming, but unfortunately, there is not a convenientway around it. To start, you must determine the units of unsaturation (alsoknown as degrees of unsaturation).

Degrees of Unsaturation =2(4)+2"(8) =8+2^8 =1CL8 =2 =12 2 2 2

With one unit of unsaturation, the structure must contain either a 71-bond or aring. To get the correct answer, you must systematically consider each linearconnectivity and each cyclic connectivity. Be sure not to count stereoisomers(including geometrical isomers). The alkenes are listed first, followed by thecyclic structures.

Possible alkene structures:

4-carbon chain (2 total) 3-carbon chain (1 total)

H2C=CHCH2CH3 H3CHC=CHCH3 H% ,CH31-Butene 2-Butene /C=Cn

Note that there are two possible " CH3geometrical isomers for 2-butene. ., , ,

Possible cyclic alkane structures:

4-carbon ring (1 total) 3-carbonring (1 total)

HoC*^- CH9 CHo

II /\H2C—CH2 H2C CH

Cyclobutane CH3Methylcyclopropane

Because there are five total constitutional isomers in all, the best answer is choiceB.

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Stereoisomers

Stereoisomers 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 a-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 rotateplane-polarized light differently). Not all configurational isomers rotate plane-polarized light, as you have seen with meso compounds, but optical 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

Ha h

&

HO, H

H3CCH2 CH3 H3C CH2CH3

(S)-2-butanol (R)-2-butanol

Configurational isomers(Optical isomers)

HO^-/ V^CH3 & HO^-/ y«iilCH3

(cis)-4-methylcyclohexanol (trans)-4-methylcyclohexanol

Configurational isomers(Geometrical isomers)

Note: Configurational isomers have different prefixes in their IUPAC names

H H H3CCH2 H

H3CCH2^ CH2OH H CH2OH

1-butanol 1-butanol

Conformational 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.



Optical IsomersOptical isomers are molecules of the same formula and the exact same bonds(connectivity), but a different spatial arrangement oftheatoms due to asymmetrywithin the structure. An optical isomer cannot be rotated or manipulated toassume the structure of another isomer. They cannot be converted into anotheroptical isomer without breaking a bond. Optical isomers rotate plane-polarizedlight differently from one another. Figure 2-5 shows an example of a pair ofoptical isomers.

Optical Isomers: Identical bonds with a different spatial arrangement aboutan asymmetric carbon that rotate plane-polarized light differently.

H H

\ \HC^!!/*" CH2CH2CH3 & ClUu*!/C CH2CH2CH3

cr H3C

(R)-2-chloropentane (S)-2-chloropentane

Optical Isomers

Figure 2-5

Optical isomers are a class of configurational isomers. Configurational isomerscan also be classified as enantiomers and diastereomers, so some optical isomerscan also be referred to as enantiomers or diastereomers.

Example 2.6How can the relationship between the following two molecular structures BESTbe described?

H C1H2CH2CH2C

C1H CVVUJ/C CH2CH2CH2C1 & qW1!)/C CH2C1CV H

A. Identical molecules

B. Optical isomersC. Skeletal isomers

D. Structural isomers

Solution

Wecan start by naming each of the structures. Bothhave three chlorine atoms ona five-carbon chain. The chlorine atoms are on carbons 1, 2, and 5, so eachmoleculehas the IUPACname 1,2,5-trichloropentane. Tobe skeletalor structuralisomers requires that the two compounds have different IUPAC names, sochoices C and D are eliminated. No matter how you rotate the first structure, itcannot be superposed onto the second structure. This implies that they are notidentical molecules, so choice A is eliminated. The left structure has S-stereochemistry at the second carbon, while the right structure has R-stereochemistry at the second carbon. This confirms that the two structures areoptical isomers, so the best answer is choice B.



Geometrical Isomers

Geometrical 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 7t-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 7t-bond, then they are said to be cis 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 it-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 7t-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 geometricalisomers, both alkenes and cyclicstructures.

Geometrical Isomers: Identical bonds with a different spatialarrangement (found with double bonds and rings).

Alkenes

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H CH0CH9CH0\ / 2 2 3C=C &

/ \H3C H

Trans (E)

H3CO—\ (

H3C H

Z ("Zis")

&

H H\ /C=C

/ \H3C CH2CH2CH3

Cis(Z)

H3CO

Cyclic Systems

CH>P)

(up)H,C

Cis

(up) (up)

H3C^/\^CH3

Cis

&

Figure 2-6

100

Trans

(up) (down)

h3c*y'/''nN-»^ch3

Trans

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OrganiC Chemistry Structure Elucidation Isomerism

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 theGerman word zusammen meaning "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 eachalkene carbon in trans orientation.

B. in both the E and Z isomers of an alkene, the atoms directly bonded to thealkene carbons are all coplanar.

C. molecules capable of forming geometrical isomers have greater entropy thanlinear alkanes of equal carbon chain length

D. geometrical isomers have relatively static structural features, such as polarityand solubility.

Solution

To 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 sequentially looking at the atoms attached to the site of interest.If the two highest priority groups are across from one another with respect to thedouble bond, then the compound is trans and thus is assigned the letter E. If thetwo highest priority groups are both on the same side of the double bond, thenthe compound is cis and thus is assigned the letter Z. This makes choiceA a truestatement, so it is eliminated. Because the two p-orbitals of the it-bond arecoplanar with an orientation perpendicular to the substituents on the alkenecarbons, the four atoms bonded to the two carbons of the alkene must becoplanar. It is not possible to rotate around the double bond, because the p-orbitals would no longer be coplanar, breaking the rc-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 bond(the n-bond would have to be broken), it is not possible to convert between thecis and trans geometrical isomers without adding a great deal of energy. Theconsequence is that alkenes are rigid and thus have lessentropy than alkanes of acomparable carbonchain length. Thismakeschoice C a false statement, and thusthe best answer. The structures of both an alkene and cyclic molecule arerelatively static and do not change drastically. The ring may flip-flop a little, butrotation is observed only for the substituents on the ring, and not observed forthe bonds in the ring. Because the structure is static, the molecular features areconstant. This makes choice D a true statement, eliminating it.

We shall discuss stereoisomers only at this superficial level for the time being. Inthe stereochemistry and carbohydrate sections, stereoisomers will be discussed ingreater detail.

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Conformational Isomers

Conformational isomers, or conformers, are molecules with identical connectivity(bonds) that are nonsuperimposable because of rotation about a bond orcontortion (often referred to as ring-flipping) of the molecular structure. The moststable conformation of the molecular structure is predictable based onhybridization, steric repulsion, and VSEPR (Valence Shell Electron PairRepulsion) theory. Understanding conformational isomerism is vital, becausestructure dictates the reactivity and stability of a compound. Figure 2-7 showstwo pairs of conformational isomers.

Conformational Isomers: Identical bonds with different spatial orientationscaused by either rotation about sigma bonds or contortion of ring structures.

OH

H3C

OH

H-^T^-HH3C^/^.H

CH2CH3

CHqOH

OH

Figure 2-7

Conformational isomers are different orientations of the same molecule. Figure2-8shows 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 constantly rotating, butane does not remain in just one ofthe conformations, but rather assumes all confirmations at one time or another.It assumes the most stable conformation most frequently. This means that thereis a most favored and least favored conformation.

H3C HStructure #1

H H

Structure #4

H CH3

H h

Structure #2

HaC*&

Structure #5

Figure 2-8

Structure #3

H^^HH3C^ H

Structure #6

Rotation about a carbon-carbon bond converts butane from one conformer into

another. For instance, a 180° rotation about the C2—C3 bond of butane convertsStructure #1 in Figure 2-8 into Structure #4. The six structures represent 60"incremental rotations about the C2—C3 bond, across a full 360° rotation.



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 (in anti orientation), if they repel. This minimizes the stericrepulsion between the two groups. The least stable conformation (eclipsed, as it iscalled) has the two largest groups colHding into one another. The differentorientations are referred to as conformational isomers. Figure 2-9 shows a side-view of steric repulsion for the fully eclipsed butane molecule in Structure #4.

HoC CH,Steric repulsion betweenhydrogens makes thisstructure less stable.

Figure 2-9

A major part of learning about conformational isomers is learning theterminology. Listed below is a glossary 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 rearcarbon, because the substituents on the front carbon block the view of the rear.Structures #2, #4, and #6 are examples of the eclipsed orientation of butane.Because of steric hindrance, the eclipsed conformations are less stable than thestaggered conformations. Figure 2-10 shows theeclipsed conformation ofethane.

Eclipsed Drawings of Ethane

H H

\ /

H

Q.

H

Figure 2-10

Staggered: Staggered refers to the orientation where from a front view of themolecule, you can seeall of the substituents on the front and rearcarbons. Thereis a 60' dihedral anglebetween neighboring substituents. Structures #1, #3, and#5 are examples of the staggered orientation of butane. Staggered is the moststableconformation. Figure2-11 shows the staggeredconformation ofethane.

HStaggered Drawings of Ethane

H H

/WH hAH^„

HH/ \ '«,

HH

Figure 2-11

Isomerism



Within staggered conformation, there are two terms used to describe the relativeposition of substituents on adjacent atoms. These terms are gauche and anti, andthey refer to the position of a substituent on one atom relative to the position of asubstituent 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 hydrogen 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 below.

Gauche: Gauche refers to a staggered conformation of the molecule, where thetwo groups of interest (often the largest 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 1and 4 gauche to one another, are shown in Figure 2-12.

Staggered Conformation of Butane with CH3groups Gauche

CH, H3<=> .CH,

i^c H

C.

> HH

H

Figure 2-12

Anti: Anti refers to a staggered conformation of the molecule, where the twogroups ofinterest have a dihedral angle of180°. Structure #1 has theCH3 groupsanti to one another. This is the most stable structure! Three differentperspectives of the staggered conformation of butane, with carbons 1 and 4 antito one another, are shown in Figure 2-13.

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Staggered Conformation of Butanewith CH3 groups Anti

CH,

hA

CH,

104

.CH,

> HH

Figure 2-13

The Berkeley Review


Because atoms within a molecule are in constant rotational motion, a structuredoes not exist in one fixed conformation. Figure 2-14 shows the energy diagramthat corresponds to a complete 360° rotation about the C2-C3carbon-carbon bondof butane, starting and finishing with Structure #1 from Figure 2-8. The energydiagram starts with the lowest energy structure at 0° (which is the conformationwith carbons 1 and 4 anti to one another) and rotates 360° about the C2-C3 bondto return to the same orientation. Note that the highest energy structure isexactly 180° apart from the lowest energy structure, and that a 60° rotation takesyou from an apex (a localized energy maximum) to a nadir (a localized energyrrtinimum) on the energy diagram.

180°

Degrees of rotation

Figure 2-14

The energy diagram for butane is symmetrical, because butane isa symmetricalmolecule. When the compound is chiral (contains stereogenic centers), theenergy diagram is not symmetric. The energy axis of the diagram is notquantified in Figure 2-14, so it demonstrates only a conceptual relationship. Ifany calculations are required on the MCAT, values will be provided for theenergy ofthe eclipsed andstaggered conformations, andfor the gauche andantiinteractions of various substituents.

Example 2.8How 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 C2—C3 bond ofbutane?A. The energy diagram issymmetric like butane, butwith higher energy values.B. The energy diagram issymmetric like butane, butwith lower energy values.C. The energy diagram is asymmetric like butane, butwith greater differences

in energy values.D. The energy diagram isasymmetric, with different energy values than butane.

SolutionBecause the structure is asymmetric, the energy diagram depicting the rotationassociated with the molecule must alsobe asymmetric, eliminating choices A andB. The energy diagram for the complete rotation about the C2—C3 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 ofR-2,3-dimethylpentane are:

Isomerism



H. VCH2CH3 H3C ,CH2CH3

/ V'*H

CH3

2 methyl/methyl and1 ethyl/methyl interactions

H3C CH,

CH2CH3

CH,

H,C

1 methyl/methyl and1 ethyl/methyl interactions

Lowest Energy Conformation

H3<=>H,C**<

H

H3C

Isomerism

CH2CH3

CH,

CH2CH3

CH,

1 methyl/methyl and2 ethyl /methyl interactions

Figure 2-15 shows the exact orientation of theatoms at various times duringthefirst 120° of rotation of ethane. The graph shows that at 20° and 100° thestructures are symmetricand thus of equal energy. This is also observed for theconformers at 40° and 80°.

Rotation about the

C—C bond of ethane

T

60 80 100

Degree ofrotational displacement from staggered conformation120

Figure 2-15

If the substituents repel one another, then the anti orientation is lowest on theenergy diagram. However, when theforce between neighboring substituents isattractive in nature, thegauche orientation is mostfavorable. A goodexample isa vicinal diol, where the twohydroxyls on adjacent carbons exhibit hydrogen-bonding with one another.



Example 2.9Thebest explanation for why the most stable orientation about the C2-C3 bond ofthe amino group and the carboxylic acid group in 3-aminopropanoic acid isgauche is which of the following?A. 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 fromgauche orientation.

D. Thecarboxylic acid and aminogroups form a hydrogen bond best from antiorientation.

Solution

If the substituents attract one another, then the most stable conformation isstaggered with the two attracting groups gauche with respect to one another.Thebest choice is C, because gauche has the amino and hydroxyl groups closeenough toform a hydrogen bond. Hydrogen-bonding isa stabilizing force. Thestrongest hydrogen bond is formed by the lone pair ofnitrogen with the acidichydrogen of the carboxylic acid group.

HO

Hx NH2

fr H

O

Anti orientation

NH2

O^OH

O R

•Hydrogen bond

/0=C,

HjA

:®e-H H

Gauche orientation H

Hydrogen-bonding is possible when the twogroupsare only60° apart,but not whentheyare 180° apart.

Isomerism



Newman 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. Beable 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 molecule. 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 Fischerprojection. Figure 2-16 shows a pictorial explanation of theconversion from a dashed-and-bold wedge drawing to a Newman projection,while Figure 2-17 shows a pictorial explanation of the conversion from aNewman projection to a dashed-and-boldwedge drawing.

When viewedfrom the right, substituentspointing 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 ontheright sidein theNewman projection end up behindthe plane in the dashed-and-bold wedgedrawing.

Conversion from dashed-and-bold wedgedrawingtoNewmanprojection:

)H

$> c "a H3C^r ^yH JOh

3 "3

plane

Figure 2-16

Conversion from Newman projection to dashed-and-bold wedge drawing:behind

<z

OH

H3C^"^\H

•ksi CH,

H

H

HaC^Plane.OH

in front

of plane S

H

Figure 2-17

behind

S^,, planein front "•

of plane

CH,

CycloalkanesCyclic alkanes whichcontain only one ring have the chemical formula CnH2nand contain no 7t-bonds. The stability of a given cycloalkane is rooted in itsability to form bond angles ofapproximately 109.5°, thenormforsp3-hybridizedcarbons. The farther from 109.5° the angle is, the greater the reactivity of thecycloalkane. For this reason, three- andfour-membered rings arereactive, whilefive- and six-membered rings arestable. When treated withhydrogen gas(H2),cyclopropane and cyclobutane readily form straight chain alkanes (propane andbutane). Cyclopentane and cyclohexane do not undergo hydrogenation.



The reactivity of three- and four-membered rings is attributed to ring strain. Ringstrain is defined as the energy difference between the linear and cyclic alkanes ofequal carbon length. Because the bond angle in cyclopropane is 60°, vastlydifferent than 109.5° associated with a normal s/?3-hybridized 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 cyclicalkanes.

Cyclopropane M

HH

Ring strain 27.4 kcal/moleH

IIH

The carbon-carbon bonds are bent (not collinear), makingthem weaker than standard carbon-carbon single bonds.

Cyclobutane

Ring strain 26.2 kcal/mole

H\ H H

v4hH

CyclopentaneH

Ring strain 6.8 kcal/mole H pj

H

HH

H

H2C-

H2C

CH,

CH* \

"CH2

Figure 2-18

Because of their stability, most cyclic organic and bio-organic molecules areeither five-membered or six-membered rings. Six-membered rings are slightlymore stable than five-membered rings. Given their frequent presence inbiological molecules, five-membered and six-membered rings have a highprobability of appearing on the MCAT.

Isomerism



CyclopentaneCyclopentane does not require much distortion of its bonds and shape toaccommodate the 109.5° angle for the s/^-hybrid. Aperfect pentagon has anglesof 108°, so there is only a smalldiscrepancy from 109.5°. This angle differencedoes not accountfor the small ring strain of 6.8 kcals per mole. To achieve thecorrect angle and alleviate this torsional strain, cyclopentane forms what isreferred to as an envelope shape, whereoneof the carbons is not coplanar with theother four. The major problem withcyclopentane isnot theringbondangles, butthe substituents on the ring that are in an eclipsedconformationas 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 arerelatively stable. Theirstability makes themcommonin 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 ringsfrequently encountered in thebiological sciences.

CH2OH

CH2OH

OH

CH2OH

OH H

C-D-Fructofuranose

OH

OH HO

fi-D-Ribofuranose

Copyright© by The Berkeley Review 110

Figure 2-19

C02"

+H,N—C H

CH

CH2OH

H I H

OH H

2-Deoxy-C-D-ribofuranose

OH

The Berkeley Review


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 chair 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, AGrx = 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.

Cyclohexaneshowing equatorial Hs

axial (up)

Cyclohexane showing axial Hs

axial (up)

axial (down)

equatorial (up)

axial (up)

axial (down)

equatorial (down)axial (down)

Cyclohexane showing all substituents

Figure 2-20

Isomerism



H H H

Chair

Interconversion between Cyclohexane ConformersThe interconversion between the two chair conformations of cyclohexanerequires activation energy, but the free energy (AGrx) is zero, because there areno substituents on the cyclohexane ring. As cyclohexane undergoes ring-flipbetweenchairconformations, it passes through intermediatestructures. The twomost significant intermediate structures are the twist form and the boat form. Thetwistform is ofslightly lower energy than theboat form, because the hydrogensare not eclipsed in the twistform. Thering-flip conversion of cyclohexane fromchairto boatand on to theotherchairconformation is shownin Figure 2-21.

H ^=

HH

Twist

HH

Twist

H H^s^ H h

^ H

Figure 2-21

Because molecules can readily rotate and contort at room temperature, anequilibrium exists between the two chair conformations of the cyclohexanemolecule. Either atlow temperatures orincases where asubstituent istoo bulkyto orient itself in an axial manner, one conformer is present exclusively. Theproton NMR can be used todetermine the more stable structure byfocusing onthe different signals for the two orientations ofhydrogen. You should beable 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 arestaggered withlower energy gauche and antiinteractions.

Monosubstituted CyclohexaneThe ring-flip process isthe same when there isa substituent onthe ring asit isforcyclohexane, but the energetics aredifferent. When a substituent is present, theactivation energy for interconversion increases, and the two chair conformationsdiffer in stability. Because the two chair conformations are no longer of equalenergy, the two twist forms are also no longer ofequalenergy. Themoststablechair conformation is thestructure with the leaststeric repulsion. Substituentswith axial orientation on the same side of the ring are close enough to repel(known as 1,3-diaxial interactions), so axial orientation is less favorable thanequatorial orientation. This is indicated in Figure 2-22, where the two chairconformations ofbothmethylcyclohexane and cyclohexanol are shown. Thetwochair conformations ofmethylcyclohexane differ in stability by 1.69 kcals/mole.Because a hydroxyl group is smaller than a methyl group, the difference inenergy for the twochair conformations ofcyclohexanol is only1.04 kcals/mole.



CH, OH

p^ \^\,CH, t=Ci,OH

MethylcyclohexaneEquatorial > Axialby 1.69 kcals/mole

CyclohexanolEquatorial > Axial by 1.04kcals/mole

Figure 2-22

Disubstituted CyclohexaneDisubstituted cyclohexane exhibits different dynamics than monosubstitutedcyclohexane. The greatest energy difference is observed when comparing 1,3-diaxial to 1,3-diequatorial, because the steric repulsion of1,3-diaxial substituentsis thestrongest repulsion encountered in cyclohexane. This isshown in Figure 2-22 for dimethylcyclohexane. When comparing 1,2-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 tooneanother), because the1,2-diequatorialcyclohexane species has gaucheinteractions between the bulkiest substituents. The energetics of theconformational isomers of dimethylcyclohexane are shown in Figure2-23.

CH

HoCJHoC

Cjj Traws-l,2-dimethylcyclohexane3 1,2-Diequatorial >1,2-Diaxial by 2.8 kcals/mole

CH

CH3

3 1,3-Diequatorial >1,3-Diaxial by 8.4 kcals/mole

H,C

O's-l,3-dimethylcyclohexane

CH,

H,C

CH Trans-l,4-dimethylcyclohexane3 1,4-Diequatorial >1,4-Diaxial by 3.4 kcals/mole

Figure 2-23


CH,

CH,



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 equatorial or axialorientations, but rather to whether the substituentsare above or below the ring.

Possible Orientations for Disubstituted Cyclohexanes

1,2-trans (up/down) or (down/up)1,2-cfs(up/up) or (down/down)1,3-ris (up/up) or (down/down)1,3-trans (up/down) or (down/up)l,4rtrans (up/down) or (down/up)1,4-cz's (up/up) or (down/down)

C-1axial C-2axial or C-1 equatorial C-2equatorialC-1axial C-2equatorial or C-1 equatorial C-2 axialC-1 axialC-3axial or C-1equatorial C-3equatorialC-1 axialC-3equatorial or C-1equatorial C-3axialC-1 axialC-4axialor C-1 equatorial C-4equatorialC-1 axialC-4equatorial or C-1 equatorial C-4axial

Table 2-1

It is essential that you be able to translate from nomenclature to the most stableconformation. For instance, frans-3-methylethylcyclohexane is a 1,3-transcompound. A compound with 1,3-trans orientation has both an axial and anequatorial substituent. The ethyl group is larger than themethyl group, so theethyl group occupies the equatorial orientation in the most stable conformationof frans-3-methylethylcyclohexane. This is shownin Figure 2-24.

CH,

CH,H3CH2<

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 equatorial substituents become axial (as seen with the methyl group).Themoststable conformation has the leaststeric repulsion.

Example 2.10Themost stableconformation of ris-l,2,4-trimethylcyclohexane has whichof thefollowing orientations for the threemethylgroups?

HoC

H3C

A. Thechairconformation with3equatorial methyls and 0 axial methylsB. Thechairconformation with2 equatorial methyls and 1 axialmethylC. Thechairconformation with1 equatorial methyland 2 axialmethylsD. Thechairconformation with0 equatorialmethylsand 3 axialmethyls


XT

The Berkeley Review


Solution

The most stable conformation has as many methyls equatorial as possible. 1,2-cfshas 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 ofris-l,2,4-trimethylcyclohexane (or (lR,2S,4R)-l,2,4-trimethylcyclohexane ascorrectly name using IUPAC nomenclature rules.)

CH,

1 equatorial; 2 axial 2 equatorial; 1 axial

This covers the topics associated with isomerism. These topics are applicable inboththephysical and biological sciences areas, so know them well. Toimplantthissection as a working knowledge base, there aremany passages withwhichtowork. 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, thenyou willslowly getacclimated to theMCAT wayofthinking. At this point, passages may seem like an absurd form of askingquestions, but hopefully you will take a liking to the style. Passages presentinformation that you must incorporate into your background knowledge, andthen using all theinformation youhave, youmust reach a conclusion concerningtheir questions. Multiple-choice tests require that you find the best, mostreasonable answer. You are not required to solvedetailed questions or derivefundamentalconcepts. Just find the best answer,as fast as you can.

Isomerism


Organic Chemistry Structure Elucidation Structural Insights

Structural InsightsStructural SymmetryWhen 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 distance 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 andanotherwithinversion symmetry. Molecules with inversion pointsare nonpolar,because all of the individual bond dipole vectors cancel each other out.

CH2OH

CH2OH

Molecule with mirror symmetry

HOH2C H

.•£&kcr

- 6 ,H CffcQH* •.

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 structuralfeatures and connectivityof a molecule.

Units of Unsaturation

Units of unsaturation are calculated from the molecular formula. The units ofunsaturation give us information about the number of rings and/or 7C-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. Tohold two atoms together, it takes onebond (Atomi—Atom2). To hold three atoms together, it takes two bonds(Atom]—Atom2—Aton-13). 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 minimumnumber of bonds. Anyelectrons beyond the bareminimumneeded 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) C3H8 contains eleven atoms, which requires at minimum ten bonds (andthus twenty bonding electrons). There are three carbons with four bondingelectronseach. There are eight hydrogens with one bonding electron each.This means that propane (C3H8) has exactly the twenty bonding electronsneeded. There are no extra bonding electrons, so propane has a linearstructure with no rc-bonds. This is to say that propane has no units ofunsaturation.


Organic Chemistry Structure Elucidation Structural Insights

2) The units of unsaturation for hydrocarbons and carbohydrates can bederived from the formula for aliphatic alkanes,CnH2n +2- "Aliphatic" refersto a structure that has no rings or 7t-bonds. An aliphatic alkane has the bareminimum number of bonds, so there are no units of unsaturation. For everyunit 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, C5H8 has fourhydrogens less than the fully saturated formula for five carbons, CsHi2-Because it has four fewer hydrogens, it has two units of unsaturation.

3) The units of unsaturation depend on the surplus of bonding electrons. Tokeep any chain propagated, every member of it must make two connections.In 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 eachhydrogen 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 Equation2.1.

tt •* c * *• 2(#C) + 2-(#H) ,91vUnits of unsaturation = ——- -—— (2.1)

2

Method 3 works with other atoms, too. Nitrogen makes three bonds, which isone more than the minimum needed, so you add 1 per nitrogenatom. Halogensmake one bond, which is one less than the minimum, so you subtract 1 perhalide. Equation2.2includesnitrogen and halogens.

tt •♦ < *, it 2(#C) + (#N)-(#H)-(#X) +2 ( .Units of unsaturation = ——-—i— U.ZJ

Example 2.11How many units ofunsaturation arepresent ina compound with the molecularformula C7H9N3O2CI2?

A. 1

B. 2

C. 3

D. 4

Solution

This question is solved by applying Equation 2.2.

2(7) + (3) -(9) -(2) +2 = 14+ 3-9-2 + 2 = V7^9 = 8 = 4Units ofunSaruration2 2 2 2

Because there are four units of unsaturation, choice D is the best answer. Withfour units of unsaturation, the compound could contain three 7t-bonds and onering, meaning it ispotentially a benzene derivative. When there are four units ofunsaturation, you should immediately consider the possibility that thecompound contains an aromatic ring. While there areothercombinations offourunits of unsaturation, there is a high probability of having an aromatic ring.


OrgSiniC Chemistry Structure Elucidation Structural Insights

SpectroscopyThe MCAT topics include infrared absorption spectroscopy, ultraviolet-visiblespectroscopy, 1H (proton) nuclear magnetic resonance 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 havebeenstressed in course work. We shallstart by reviewing the basicoperations ofthe IR and its applications to structure elucidation. From there we shall considerultraviolet/visible spectroscopy,proton NMR,and carbon-13NMR.

Infrared SpectroscopyEvery molecule produces a unique IRspectrum. Infrared spectroscopy starts byadding a monochromatic beam of IR photons to either a thin oil suspension (ifthe compound isa solid) or a neatsolution (ifthe 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.Thischangein stretching(vibrating) between atoms within the moleculecausesachange 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). What we record isthe changein the intensity of the EM radiation from when it enters the molecule to when itexits themolecule. This iscompared to a reference beam that traverses a pathofidentical length but does notpass through thecompound itself. If thecompoundabsorbs a given wavelength of light (corresponding to some transition), then weobserve an absence oflight exiting thesample tube. This is known asabsorptionspectroscopy. The graph records transmittance as a function of wave number (cm"1), so absorbances are represented by drops in 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 directlyproportionalto the masses of the twoatoms in the bondand the bond strength. Thisis to saythat the potential energy in a resonating system (such as a spring that obeysHooke's law) is describedby Equation2.3

P.E. =Ikx2 (2.3)2

Thek-termis thespring constant, whichwe can say describes the bond strength.The x-term describes the distance from equilibrium 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 result, 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"1, it isan inverse ofthe wavelength. The inverse ofthe wavelength is directlyproportional to the energy of the photon. This meansthat the higher the wavenumber is, the greater its energy.

Copyright ©byTheBerkeley Review 118 The Berkeley Review

Organic Chemistry Structure Elucidation Spectroscopy and Analysis

For instance, a C=0 bond absorbs around 1700 cm _1, while a C-O bond absorbsaround 1300 cm _1. This is because a C=0 bond is stronger than a C-O bond.Carbonyl functional groups are common, so you should know the absorbancevalue for a C=0 bond. Ansp3-C-to-H bond absorbs just below 3000 cm"1, whileansp2-C-to-H bond absorbs justabove 3000 cm"1, because ans/^-C-to-H bond isstronger than an sp3-C-to-H bond. This isbecause the sp2-hybrid, having more s-character, issmaller than the sp^-hybrid. The result is that ansp2-C-to-H bond isshorter andthus stronger than ansp^-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 bondswithin 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 molecular formula and the NMR spectra formolecules. Table 2-2 lists several useful IR absorbances. The values are listed in

terms of wave numbers. Note that the absorbance of a given bond varies withthe compound in which the bond exists.

Bond type Stretching (cm-1) Bending (cm"1)

O—H alcohol (no H-bonding) 3640-3580 (v)

O—H alcohol (H-bonding) 3600-3200 (s, broad) 1620-1590 (v)

N—H amides 3500-3350 (m)

N—H amines 3450-3200 (m)

C—H alkynes 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) 1470-1320 (s)

C—H aldehyde 2900+, 2700+ (m, 2 bands)

O—H acids (H-bonding) 3000-2500 (s, broad) 1655 -1510 (s)

C=C alkynes 2260-2120 (v)

C^Nnitrile 2260-2220 (v)

C=0 ester 1750-1735 (s)

C=0 aldehyde 1740-1720 (s)

C=0 ketone 1725-1705 (s)

C=0 acid 1725 -1700 (s)

C=0 aryl ketone 1700 -1680 (s)

C=0 amide 1690-1650 (s)

C=0 a,fi-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 absorption w = weak absorptionm = medium absorption v = variable absorption

Table 2-2


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 Manual 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 theabsenceof that bond. Here is an example of how IRspectroscopy is used:

An unknown compound with formula C4H8O is analyzed by IR spectroscopy.An intense band is detected at 1710 cm"1 (IR absorbances are listed by energyaccording to the wave number asmeasured in cm"1). By comparing thevalue toa chart of IR absorbances, this peak can be attributed to a C=0. The compoundhas one degree of unsaturationattributableto a C=0, which makes it possible tonarrow it down to a small number of isomers. The structure cannot be cyclicandhas 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 longest chain could be only three carbons, with a methylsubstituent on carbon2. 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 choicesfrom 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 techniqueyou have done many times in the past. Through examplesand 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 C5H10Ocannot have an IRabsorbance peakbetween 1700 cm"1 and 1750 cm"1?A. An aldehydeB. A ketone

C. A cyclic etherD. All of the above have an IR absorbance between 1700 cm"1 and 1750 cm"1.

Solution

An IR absorbance between 1700 cm'1 and 1750 cm"1 implies that the compoundhas a C=0 in its structure. Because it has no absorbance between 1700 cm"1 and1750cm"1, it does not have a C=0 bond. Choices A and Bhave a C=0 in theirstructure, so they can be eliminated. The best answer is choice C. The onedegree of unsaturation associated with 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.



Example 2.13How many structural isomers of C4H8O2 are possible that have an IR absorbancepeak between 1735 cm"1 and 1750 cm"1, a peak around 1200 cm"1 and no broadpeaks above 2500 cm"1?A. 2

B. 3

C. 4

D. 5

Solution

The absence ofa broad peakabove 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 carbonylandether functionality. According toTable 2-2, thepeak between 1735 cm"1 and1750 cm-1 indicates that there is a C=0 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.

1) O 2) O

H' OCH2CH2CH3 H3CH2C OCH,

3) Q 4) O

H OCH(CH3)2 H3C OCH2CH3

The two aldehyde structures (1 and 3) could be confirmed or eliminated by thepresence or absence oftwo peaks at 2900 cm"1 and2700 cm"1. Ifthis were a reallaboratory scenario, it would be far easier at this point to use proton NMR todeduce the structure of the unknown compound. Structure 1 has four types ofhydrogen in a 1: 2 : 2 : 3 ratio. Structure 2 has three types of hydrogen in a 3 : 2 :3 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 thesetwo structures requires identifying the ppm shift value of each type of hydrogen.We shall address NMR spectroscopy later in this chapter. IR spectroscopyshould be applied to identify functional groups.



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 effectis 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 onebroad peak. The same alcohol exhibiting two different degrees of hydrogen-bonding is shown in Figure 2-26.

Weaker hydrogen-bond H Stronger hydrogen-bond H

H

\H

O

Stronger covalent bond

e

R

H

R

H H

\O

°1H

Weaker covalent bond

H

R

Figure 2-26

Given that molecules are in continuous random motion within a liquid, somealcohols 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 of

signals 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. Figure 2-27 showsa high-resolution IR signal and the standard-resolution equivalent.

3420 3400 3380 3360 3340

Wave number (cm"1)High-resolution spectrum

Figure 2-27

3420 3400 3380

Wave number (cm"1)Low-resolution spectrum

1 1 r

3360 3340 3320



Example 2.14Whatis true for the compound associated with the following IRspectra?

3317 cm

2987 cm"

1457 cm" 1393 cm"1

& 1367 cm-'

A. It exhibits no hydrogen-bonding.B. It has a carbonyl group.C. It has a hydroxyl group.D. It has a molecular mass that is less than 30 grams per mole.

Solution

The compound represented by the IR spectrum above in the question has a broadpeak around 3300 cm"1 and no peaknear1700 cm"1. These are thefirst areas toconsider when looking at IR spectra. The compound has a hydroxyl group, butno carbonyl group. This makes choice B incorrect and choice C correct. Becauseit has the hydroxyl group, it can exhibit hydrogen-bonding. This eliminateschoice A. Because of the peaks in the 1300-1400 cm"1 range, we know thecompound has a carbon-carbon bond, so it must have at least two carbons. Thelightest compound with two carbons is ethyne (HCsCH), which has a molecularmass 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 spectroscopy, there is also ultraviolet/visiblespectroscopy. While infrared photons (in the 3-to-10 kcals/mole region) affectthe vibrational energies of a molecule, ultraviolet (in the 70-to-300 kcals/moleregion) 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 a-bonds are so much stronger than Jt-bonds, the lowest energyabsorbance for alkanes is significantly higher than the lowest energy absorbancefor alkenes. To excite an electron from the a-level (sigma bonding orbital) to thecrMevel (sigma anti bonding orbital), photons of approximately 140 nm to 170 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 200to 800nm, whereair does not interfere. As a result, we typically use only UV-visible spectroscopyto analyze molecules with 7C-bonds, especially conjugated systems. UV-visiblespectroscopy in organic chemistry focuses on transitions between the jcand n*energy levels. For systems with conjugation, there are several 7t-levels, but wecare about only the lowest energy transition.



The transition of interest is from n to 7t*. The wavelength of highest absorbance,known as lambda max (Xmax)> changes with the amount of conjugation. The valueof ^max depends on the amount of conjugation. As the conjugation increases, sodoes the wavelength of kmax- Figure 2-28 shows the Xmax values associated withthe lowest energy 7t-to-7t* transition of various conjugated hydrocarbons.

7t"

175 nm

n

H2C^— CH2

217 nm

H0C=CH

7C4%3

%2"1

\CH=CH2

Figure 2-28

258 nm

*5

7C3%2

H2C= CH\

CH

CH

H2C= CH

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^max of 304nm and 1,3,5,7,9-decaquintene has a kmax of 353 nm. When moreconjugation is added, the absorbance shifts into the visible range. Color resultsfrom excessive conjugation within a molecule. For instance, fi-carotene (with 11Jt-bonds) has absorbances at 483 nm and 453 nm. Substituted benzenes have anumber of peaks. Conjugated aldehydes and ketones have about the same n-n*absorbances as conjugated alkenes of the same number of 7C-bonds. However,conjugated aldehydes and ketones have other, more intense absorbances (e >10,000) that are of longer wavelength than their hydrocarbon counterparts. Thisis attributed to the n-to-n* transition associated with aldehydes and ketones,possible because of the lone pair of electrons on the carbonyl oxygen. Figure 2-29shows the A.max values associated with the lowest energy n-to-Jt* and 7t-to-7t*transitions of various ketones.

O o

CH, H,C

n-»Ji* A,max = 270nm n-»7t* A,max = 324nm n-»7t* X,max = 368nm

n -» jc* X^ax = 187nm n -»n* X^^ = 219nm %-» n* X^^ = 267nm

Figure 2-29

Unlike infrared spectroscopy, ultraviolet/visible spectroscopy can also beapplied in a quantitative fashion. Ultraviolet/visible spectroscopy can be used todetermine the yield of a reaction, if it involves a UV-visible active compound. Inorganic chemistry, a compound must have a 7t-bond to be UV-visible active.


Organic ChemiStry Structure Elucidation Spectroscopy and Analysis

Nuclear Magnetic ResonanceThe fundamental principle behind nuclear magnetic resonance, NMR, 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 affected 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 field, so multipleenergy states are possible. In the case of1H, there are two energy levels: a (theone aligned with the external magnetic field), and 6 (the one aligned against theexternal magnetic field.) The 6 energy level is defined as higher than the aenergy level. The energies of the two levels depend on the strength of theexternal magnetic field and the magnetogyric ratio of a particular nucleus. Thismeans that the energy gap between the two levels also depends on the strengthof the external magnetic field. As the external magnetic field increases, thefrequency of the EM radiation needed to flip the spin increases proportionally.

Any nucleus with an odd number of protons (Z number) or an odd number ofnucleons (A number) has a net spin. What is meant by "spin" is that as thenucleus precesses, it generates a weak magnetic field (as electrons do). Just as acharged particle in linear motion generates a radial magnetic field, a chargedparticle in rotational motion generates a linear magnetic field. When the atomicnucleus has an odd number of protons (or nucleons), the spins cannot pair up tocancel one another out. The result is that the nucleus has a net spin. In the casesof1H, 13C, and 19F, it happens that there are only two energy levels associatedwith the spins, so they can be analyzed without complication. A nucleus such as14N hasspin, but there are more than twoenergy states, so its NMR spectrum istoo complicated to analyze conveniently.

In the absence of any surrounding electrons, all identical nuclei exhibit the samespin and therefore require the same energy for excitation in an external magneticfield. Within a molecule, two identical nuclei may be in different electronicenvironments. As a result of the difference in their local magnetic fields, causedby the moving electrons, they do not require exactly the same amount of energyto excite the nucleus to a higher-energy spin state. This can also be viewed aslocal magnetic fields altering the strength of the applied external magnetic fieldneeded to get excitation (spin-flip) at a set frequency for the EMradiation. NMRmachines can be designed to vary the frequency of the radiation or vary thestrength of the external magnetic field. The NMR graph we observe typicallyrecords changes in the magnetic field strength along the x-axis, so we think ofNMR in terms of varying external magnetic field strength.

Proton NMR(which uses the *H nucleus) is the most common form of NMRandtakes advantage of the magnetic spin associated with the hydrogen nucleus. TheMCAT test-writers focus on analyzing the graphs produced by 1H NMR. Thereis minimal 13C NMR on the MCAT. The scale for XH NMR is set from 0 to 10parts per million (ppm) of the total magnetic field of the machine. Just as an inchis an inch because someone made it a standard unit of measurement, NMR ismeasured in ppm of the external magnetic field, because that is the arbitrarystandard. All 1H NMR shift values are relative to a standard compound,tetramethylsilane ((H3Q4SO. All twelve of the hydrogens on tetramethylsilaneare equivalent, so they absorb at the same value. This value is arbitrarilyassigned to be 0 ppm, and all shift values are referenced against it. Rather thango into other intricacies of NMR, we shall concentrate on how to read the graphs.

Copyright© by The Berkeley Review 125 Exclusive MCAT Preparation


Symmetry and NMR SignalsThe best place to begin NMR for the MCAT is with molecular symmetry. Basedon 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.

O

HqC OCH2CH2CH2CH3a beds

Five unique hydrogens labeleda-e in a 3:2:2:2:3 ratio.

O

H3C^ ^OCH2CH(CH3)2a bed

Four unique hydrogens labeleda-d in a 3:2:1:6 ratio.

O

H3C^ ^OCH(CH3)CH2CH3a b c d e

Five unique hydrogens labeleda-e in a 3:1:3:2:3 ratio.

H3C*a"*

O

OC(CH3)3b

Two unique hydrogens labeleda and b in a 1:3 ratio.

Figure 2-30

The first two esters, n-butyl acetate and sec-butyl acetate, each have six uniquecarbons, ofwhich five have hydrogens. Each exhibits five signals in its1HNMRspectrum 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 1HNMR spectrum and five signals in its13CNMR spectrum. Tertbutyl acetate has four unique carbons, but only twocontain hydrogens. This means that tertbutyl acetate exhibits only two signals inits1HNMR spectrum and four signals in its13CNMR spectrum. The presence ofonly two signals in an NMRspectrum 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.

OCH,

Four unique hydrogenslabeled a-d

in a 3:2:2:3 ratio.

OCH, OCH,

Sixunique hydrogens Sixunique hydrogenslabeled a-f labeled a-f

ina3:l:3:l:l:l ratio. ina3:3:l:l:l:l ratio.

Figure 2-31

Para-methylanisole has six unique carbons, of which only four containhydrogens. Thismeans that para-methylanisole has four types of hydrogens and



therefore four signals in its proton NMR. Meta-methylanisole has eight uniquecarbons, of which six contain hydrogens. This means that meta-methylanisolehas six types of hydrogens and therefore six signals in its proton NMR. Ortho-methylanisole also has eight unique carbons, of which six contain hydrogens.This means that ortho-methylanisole also has six types of hydrogens andtherefore sixsignals in its proton NMR. The structures in Figure 2-31 are para-,meta-, and ortho-methylanisole, respectively.

Youmay have noticed that many problems are often just variations on a singletheme. In Figure 2-32, compounds with comparable NMR readings are shownside by side to demonstrate similarities in the distribution of their uniquehydrogens. Figure2-32 shows two sets of three isomers that canbe distinguishedfrom one another using 1HNMR by simply looking at the number of signals.Butanol and pentanal each have five unique types of hydrogens. The hydrogensare inexactly thesame ratio onbothcompounds, sotheir 1HNMR spectra exhibitstrong similarities. Each has five signals in its 1HNMR spectrum, although theyhave different 13CNMR spectra due to a different number of unique carbons.The 1HNMR spectra of the two compounds can be distinguished from oneanother by the shift values of the respective signals. Methylpropyl ether and 2-pentanone each have four unique types of hydrogens. The 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 their1HNMR spectra, butatdifferent shift values. Diethyl ether and 3-pentanone eachhave mirror symmetry and thus have similar carbons and similar hydrogens dueto this symmetry. There are two unique types of hydrogens in both diethyl etherand 3-pentanone. This means that diethyl ether and 3-pentanone have only twosignals in their1HNMR.

Butanol

^ o^ . CH2H CB, c CH9'a b" d"

CH,

Five unique hydrogens labeleda-e in a 1: 2 :2 : 2 : 3 ratio.

Methylpropyl ether

^O^ CH9HoCa J

CH2 c CH3

Four unique hydrogens labeleda-d in a 3 : 2 :2 : 3 ratio.

Diethyl ether

H3C>. ^OCH2

b b

CH2 a3

Two unique hydrogens labeleda and b in a 3:2 ratio.

H'a

OPentanal

^CH2^ ^CH,CH2 c CH2 e

b~ d

Five unique hydrogens labeleda-e in a 1: 2 : 2 : 2 : 3 ratio.

2-Pentanone

O

CH,

HoCa05

CH2'b

CH,

Four unique hydrogens labeleda-d in a 3 :2 : 2: 3 ratio.

H3C.

3-Pentanone

O

CH2 CH2b b

CH,

Two unique hydrogens labeleda and b in a 3 :2 ratio.

Figure 2-32



The comparison of butanol to methylpropyl ether and diethyl ether is similar tothe comparison of pentanal to 2-pentanone and 3-pentanone. For instance, thepresence of only two signals in the 1HNMR spectrum makes it easy todistinguish 3-pentanone from 2-pentanone and pentanal in the same way it iseasy 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 presenceof a broad peak in their 1HNMR spectrum. In the case of NMR, the localenvironment 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 pattern(shape) of the peak, and the shift value (measured in ppm) of the peak. Be surethat you understand the importance of each of these features and the factors thatcan produce changes in them.



1H Nuclear Magnetic ResonanceBecause 3-pentanone has two unique types of hydrogens in a 3:2 ratio, its1HNMR spectrum has two signals with relative areas of3:2. Figure 2-33 showsthe signals from the 1HNMR spectrum of3-pentanone. Each peak isexplained interms of its splitting, integral, and shift value. The unique hydrogens are labeledin the same fashion as they were in Figure 2-32.

H^^CH2

b

\ 'CH2

b

CH,

Being next to a C=0 group yields a shiftvalue between 2.0 and 2.5 ppm.Being next to a CH3 group yields a quartet.

a:b = 3:2

2 ppm

Being next to a CH2 group yields a shiftvalue between 0.9 and 1.5 ppm.Beingnext to a CH2 group yields a triplet.

JL

lppm Oppm

Figure 2-33

The zero reference is ignored for analytical purposes, because it is there just to setthe scale correctly. The integral is not drawn on the spectrum in this example. Inmost cases, you will be provided with the relative areas of the peaks, or you willbe given a summation line to evaluate the relative areas. Either way, you mustbe 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 AnalysisWe shall start off with how to analyze the three basic components of the graph:integral (determined by the number of hydrogens making up a signal), splittingpattern (derived from the coupling between hydrogen neighbors), and shift value(determined by the local magnetic field caused by either lone pair electrons inmotion 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 curvecan be summed up, and set directly proportional to the number of hydrogensthat the signal represents. For instance, a CH3 group has a signal with a relativearea of 3 compared to a CH2 group with a signal of relative area 2. Workingbackwards from the integration to the structure, it is possible to deduce thegroup from the integration. For instance, a relative area of 6 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.



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 magnetic field resulting from 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 numberof hydrogen neighbors a group has. For instance, the CH3 group in 3-pentanone(labeled with an a in Figure 2-33) is next to a CH2 (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 CH3group next to adown /down CH2group has a slightly lower signal. Every CH3 group next to anup/down or down/up CH2 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,down/up/down, up/down/down, up/up/down, up/down/up, down/up/up,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 +1/2. 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, up/down/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 CH2signal 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 carbonbeing 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 baselinesignal.

Neighbors Signal Shape Pascal's Triangle Ratio of Peaks in Signal

OHs Singlet 1 1 peak

1H Doublet 1 1 2 peaks: 1:1 ratio

2Hs Triplet 12 1 3 peaks: 1:2:1 ratio

3Hs Quartet 13 3 1 4 peaks: 1:3:3:1 ratio

4Hs Quintet 14 6 4 1 5 peaks: 1:4:6:4:1 ratio

5Hs Sextet 1 5 1010 5 1 6 peaks: 1:5 :10 :10 :5 :1 ratio

6Hs Septet 1 6 152015 6 1 7 peaks: 1:6:15:20:15:6 :1 ratio

Table 2-3



Shift Value

The shift value is a diagnostic tool for assessing the local electronic environment.It is measured in parts per million (ppm) relative to the magnetic field necessaryto detect a standard compound, tetramethylsilane ((H3C)4Si). It is used inconjunction with 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 inhydrocarbons. This is referred to as being shifted downfield, which indicateshigher ppm values for the shift. To verify the presence of a C=0 (carbonylgroup), consult Table 2-4 to find that a value of 2.1 to 2.5 ppm is expected for ahydrogen on a carbon adjacent (alpha) to the carbonyl. Table 2-4 lists manycommon 1HNMR shift values used for analyzing spectra. The bold hydrogen ineach compound in Table 2-4 is the one of interest (to which the shift vauecorresponds) and the shift value is measured in units of ppm.

Hydrogen Atom d (ppm)

RCH3 0.8 -1.0

RCH2R (acyclic) 1.3 -1.5

RCH2R (cyclic) 1.5-1.8

R3CH 1.5-2.0

R2C=C(R')CH3 1.8-2.2

RCOCH3 (ketone) 2.1 - 2.5

ArCH3 2.2-2.5

RC^CH 2.5-2.6

ROCH3 (ether) 3.5-4.0

RCH2X (X = CI, Br, I) 3.0-3.8

RCO2CH3 (ester) 3.5-4.0

Hydrogen Atom 3 (ppm)

R2C=CH2 5.0-5.8

RCH=CR2 5.2-6.4

RNH2 1-3

RNHCH3 2.0-3.2

ArNH2 3.5-5.0

RCONHR (amide) 5-9

ROH (alcohol) 1 - 5 (broad)

ArOH (phenol) 4 - 7 (broad)

ArH (benzene) 7.0-7.4

RCOH (aldehyde) 9.0-9.8

RC02H (acid) 10-12

Table 2-4

Example 2.15Pentanal canbedistinguished from 2-pentanone bywhat^NMR feature?A. A 3H triplet at 1.5 ppmB. A 2H multiplet at 1.8 ppmC. A 2H triplet at 2.3 ppmD. A1H triplet at 9.7 ppm

Solution

Pentanal has an aldehyde hydrogen, while 2-pentanone does not. The shift valuefor an aldehyde hydrogen is found between 9.0 and 9.8 ppm. The peak shape is atriplet, because there are two equivalent alkyl hydrogens on carbon 2. The twohydrogen neighbors couple with the aldehyde hydrogen to split it into a triplet inthe ^NMR spectrum. This makes choice D the best choice. 2-pentanone couldbe distinguished by the singlet of relative integration 3 caused by the isolatedmethyl group adjacent to the carbonyl. Because the carbonyl carbon has nohydrogens attached, the methyl group of carbon one is not coupled to any other



hydrogens, which makes it appear asa singlet in the1HNMR spectrum. The twostructures are drawn below, along with their proton NMR features.

O

a^x^ "V^ b c d eH CH2CH2CH2CH3

Aldehyde hydrogens show signals

around 9.7 ppm inthe 1HNMR.

O

H3Cbed

CH2CH2CH3

Because there are no hydrogens on

the neighboring carbon, the methyl

group is a singlet in the HNMR.

Example 2.16What signals are present in the 1HNMR spectrum ofchloroethane?A. A downfield doublet and an upfield tripletB. A downheld triplet and an upheld doubletC. A downfield triplet and an upfield quartetD. A downfield quartet and an upfield triplet

Solution

Chloroethane has two unique types of hydrogens. This results in two signals inits1HNMR spectrum. The two hydrogens oncarbon 1aresplit into a quartet bythe three hydrogens on carbon 2. Equally, the three hydrogens on carbon 2 aresplit 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 isfarther 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 HNMR signals.


H

H H

| Predictedsplitting

Carbon 1 has two Hs thatare coupled to the three Hson the carbon 2, resultingin a quartet downfield.

I

Carbon 2 has three Hs that

are coupled to the two Hson carbon 1, resulting in atriplet upfield.

Actual spectrum



Example 2.17Pentanol can best be distinguished from ethyl propyl ether by which of thefollowing features in the1HNMR spectrum?A. 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

Solution

An alcohol is distinguishable from an ether by its broad peak between 1.0 ppm.and 5.0 ppm, so pick C for best results. The peak is broad due to the hydrogen-bonding within the alcohol. The broadening of the peak makes it difficult toevaluate the integration of the alcohol hydrogen peak. Both structures contain a3H triplet at 1.2 ppm, so choice A is eliminated. Both structures contain a 2Htriplet at 3.5 ppm, so choice B is eliminated. Choice D is eliminated, becausethere is a small difference in the number of peaks between the two compounds.

Example 2.18Which of thefollowing is a common feature in the1HNMR spectra ofallmethylketones?

A. 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 at 2.1 ppm (3H)

Solution

A methyl ketone has an isolated methyl group neighboring the carbonyl carbon(which has no hydrogens). Having no neighboring hydrogens makes the peak asinglet. The protons on a carbon alpha to a carbonyl are found between 2.0 and2.5 ppm. The signal has a relative integration of 3 hydrogens. The only answerthat fits this is choice D, a 3H singlet at 2.1 ppm. The best choice is choice D.

Example 2.19Amonosubstituted benzene haswhich ofthefollowing inits1HNMR spectrum?A. 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

Monosubstituted benzenes have a single peak around 7.0 ppm. The aromatichydrogens appear as one singlet, despite the fact that they are not all equivalentby symmetry. The key to this question is not the integration or the peak shape,but the shift value. Choice D is just a little too high, so it is eliminated along withchoices A and B, which are far too low. To do this problem quickly, you shouldbe 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,whenever you see a triplet and quartet in a 3 : 2 ratio, you should conclude thatthere is an isolated ethyl group (H3CCH2-) in the molecule somewhere.

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Whenever you see a doublet and a septet in a 6 :1 ratio, you should concludethat there is an isolated isopropyl group ((H3Q2CH-) 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 common 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 instance, 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 7C-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 2.20What is thename ofthe compound thathasthefollowing 1HNMR spectrum, andwhose formula is C7H14O?

'

J{

i

J, I'1 " ••"•

2 ppmI1 '

lppm1

Oppm

A. 2,4-dimethyl-3-pentanoneB. 2,2,4,4-tetramethyl-3-pentanoneC. l,l,3,3-tetramethyl-2-propanoneD. 2,2-dimethyl-3-pentanone

Solution

The septet and doublet in a 1 : 6 ratio are a dead give-away for an isopropylgroup. Choice D is eliminated, because it does not have an isopropyl group.Choice B is eliminated, because it contains too many carbons (nine instead ofseven). Choice C is eliminated, because the structure is misnamed. The bestanswer and only remaining choice is A.



Example 2.21The shape of thesignal at 2.3 ppm in the1HNMR spectrum in Example 2.20 isbest described as:

A. a quartet.B. a sextet.

C. a septet.D. an octet.

Solution

We counting seven apexes within the signal at 2.3 ppm, and seven apexes (peaks)is referred to as a septet. Check to see whether the ratio is 1: 6:15:20 :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 equivalenthydrogen neighbors on the methyl groups neighboring the alpha carbon.

Example 2.22The ratio of the areas under the peaks within a quartet is:

A. 1:2:2:1.

B. 1:3:3:1.

C 2:5:5:2.

D. 1:4:4:1.

Solution

Byusing 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 inthis case a quartet. A quartet has a ratio of 1:3:3:1. Choice B is correct.

Example 2.23What isthe IUPAC name oftihe compound represented bythe following 1HNMRspectrum, whose molecular formula is C4H8O?

A. Butanal

B. Butanone

C. Ethyl ethanoateD. Methyl propanoate

Solution

Because the formula has only one oxygen, the two esters (choices C and D) areimmediately eliminated. The question is now reduced to determining whetherthe compound is an aldehyde or ketone. An aldehydewould have a peak in the



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(CH3COCH2CH3) 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 2.24What is theIUPAC name ofthe compound represented by thefollowing 1HNMRspectrum, whose molecular formula is C4H8O2?

A. Butanal

B. Butanone

C. Ethyl ethanoateD. Methyl propanoate

Solution

This 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 1HNMRspectra. 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.

O

O

CH2A '

? ^J3H Singlet 2H Quartet 3H Tripletat ~ 2.1 ppm at ~ 3.8ppm at ~ 1.0 ppm


1

3H Tripletat ~ 1.0 ppm

O

CH

2H Quartetat ~ 2.3 ppm

OA

3H Singletat ~ 3.5 ppm

The Berkeley Review


Example 2.25Hydrogenson a carbonadjacent to two equivalentCH2groups show which typeofsignal in a 1HNMR spectrum?A. Al:3:3:l quartetB. A1:4:4:1 quartetC Al:3:5: 3:1 quintetD. A1:4:6:4:1 quintet

Solution

Having two equivalent CH2 groups adjacent to the site of interest results in atotal of four equivalent hydrogen neighbors. Fourequivalent hydrogens split asignal into a total of five (4 + 1)peaks. This makes thesignal a quintet, whichaccording to Pascal's triangle (orbinomial expansion ofanysort) hasa ratio of1:4:6:4:1. The best answer is choice D.

Example 2.26What is the common name of the compound represented by the following1HNMR spectrum, whose molecular formula is C8H10O2?

C8H10°2 3H

2H

. 1 i8 ppm

1-

6 ppm 4 ppm 2 ppm Oppm

A. Para-ethoxyphenol (H3CH2COC6H4OH)B. Ortho-ethoxyphenol (H3CH2COC6H4OH)C. Para-methoxy anisole (H3COC6H4OCH3)D. Ortho-methoxy anisole(H3COC6H4OCH3)

SolutionThe symmetry in the 1HNMR spectrum isassociated with a structure that isalsosymmetric. The only way to get two types of hydrogens on a disubstitutedbenzene is to have two equal substituents on benzene para to one another. Thiseliminates choices B and D. Based on the formula, this molecule has twomethoxy groups para to one another on the benzene. All of the benzenehydrogens areequivalent, which explains why only a singlet is observed. Thebest choice is thus answer C. Choice A would exhibit more than two peaks, so itis 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 8 = 10 -12 ppm range and that the signal is broad. Analdehyde hydrogen falls in the 8=9- 10 ppmrange, aromatic hydrogens fall inthe 8=7-8ppm range, vinylic hydrogens fall in the 8=5- 6ppm range, alkoxyhydrogens fall in the8=3.5 -4ppmrange, andalpha hydrogens fall in the8 =2-2.5 ppmrange. Figure 2-34 shows a molecular structure and its corresponding1HNMR spectrum that includes many of these key peaks.

yd c f

OCH3 bh H

H3C

r12 ppm 10 ppm

3H 3H 3H

1H

JL8 ppm 6 ppm 4 ppm 2 ppm Oppm

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 bedone on theexam using achart of values if one is given, but it is not a bad idea to know the values frommemory.

Example 2.27What is theIUPAC name ofan unknown compound with themolecular formulaC3H6O, an IR absorption at1722 cm-1, and three aHNMR peak,; one at9.7 ppm(1H), oneat 2.3 ppm(2H), and oneat 1.4 ppm (3H)?A. Propanoic acidB. PropanalC. PropanoneD. Methyloxyrane

Solution

There is an excess of information in this question beyond what is needed toaswerit. Theoneand onlypiece ofinformation youneed is the peak at 9.7ppm,which makes thecompound analdehyde. Pick choice Band move on quickly.

Copyright ©by The BerkeleyReview 138 The Berkeley Review


Distinguishing Disubstituted BenzenesIntegrals tell 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 as"para") shows the fewest peaks in the aromatic region of the spectrum of alldisubstituted 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 three with two different groups on the benzene.

Case 1: The two substituents on benzene are equal:XXX

2 different Hs

in a 1:1 ratio

3 different Hs

in a 1:2:1 ratio

*V J HaX

All Hs are

equivalent

Case 2: The two substituents on benzene are not equal:

XXX

Hc Hd Y

4 different Hs in 4 different Hs in 2 different Hs

a 1:1:1:1 ratio a 1:1:1:1 ratio in a 1:1 ratio

Figure 2-35

Para substitution is the easiest arrangement to distinguish of the three possiblestructural isomers, because it has a doublet of doublets. The dissimilar heights ofits peaks can be attributed to a mathematical phenomenon whereby peaks, asthey nearone another, begin to coalesce. Figure 2-36 shows thearomatic regionof a 1HNMR spectrum of a para substituted benzene ring, where the twosubstituents are nonequivalent. Paracoupling isa highly recognizable feature.

Enlargement of the aromatic region shows that thesplittingis a doubletofdoublets, corresponding to apara-substituted benzene ring. X

J1 1 1 T 1

7.8 ppm 7.6 ppm 7.4 ppm 7.2 ppm 7.0 ppm

Figure 2-36



Example 2.28What is the commonname of the compound that has the formula CgHsO, an IRabsorption at1722 cm-1, and three notable 1HNMR peaks at 9.7 ppm (1H, s), 7.3ppm (4H,dd), and 2.2ppm (3H, s)?

A. Ortho-methylbenzoic acidB. Para-hydroxyacetophenoneC. Ortho-methylbenzaldehydeD. Para-methylbenzaldehyde

Solution

The compound hasonly oneoxygen, soneither 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 aHNMR peak at 9.7 ppm and the IR absorption at1722 cm"1 do not help our efforts to identify the compound. The 1HNMR signalat 7.3 ppm is a doublet of doublet (dd), which indicates para-substitution. Thismakes choice D the best answer.

Deuterated Solvents for *HNMRBecause the solvent is in substantially higher concentration than the solute, it isimperative that thesolvent nothaveanyhydrogens. If thesolvent has 1Hnuclei,then it would produce the largest signal in thespectrum, 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 (1H). One potential problem occurs when protic compoundsaredissolved into deuterated protic solvents, such asD2O. Protic hydrogens canundergo exchange with the protons of the solvent, if the solvent is protic.Although the dissociation constant (Ka) may be small for compounds such asalcohols, overenough time allof thehydrogens canbe released and then are ableto reform their bonds. If D2O is present in the solution, then deuterium willgradually replace protic hydrogens capable ofundergoing exchange. This causesthesignal for theprotic hydrogen to disappear gradually.

Example 2.29Which of thefollowing compounds does NOT lose a 1HNMR signal after D2Ohasbeenaddedto a solution containing it?A. Carboxylic acidB. Cyclic etherC. Primary amineD. Secondary alcohol

Solution

Ifa compound contains a protic hydrogen, then it loses a peakfrom its 1HNMRspectrumwhen D2O is added to the solution. Primary and secondary amineshave a hydrogen bondedtonitrogen, so theyare protic. Thiseliminates choice C.All alcohols have a hydrogen bonded to oxygen, so all alcohols are protic. Thiseliminates choice D. A carboxylic acid has a dissociable proton, so it readilyexchanges with deuterated water. Choice A is eliminated. An ether, whethercyclic or not, has all of its hydrogens bonded to carbon, so it is aprotic. WhenD2Ois added to an ether,no exchange transpires. Thebest answer is choiceB.



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), soit 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 13CNMRrequires many more scans to obtain a spectrum than 1HNMR. This is why thebaseline in 13CNMR spectra isnoisy (scattered and messy). 13CNMR isused toidentify the number of unique carbons in a compound. A typical application ofthis technique is to distinguish the substitution of disubstituted benzenes. Figure2-37 shows the proton-decoupled (all singlet spectrum) C-13 NMR spectra for1,2-dimethylbenzene, 1,3-dimethylbenzene, and 1,4-dimethylbenzene. What ismeant by "proton-decoupled" is that the compound is irradiated constantly(similar in concept to Dolby noise reduction) with the coupling energy, so thatthere isnocoupling between the 13C and 1H atoms. This makes all ofthe peakssinglets, as drawn.

CH,

&CH,

t 1 1 1 1 1 1 Tppm 140 120 100 80 60 40 20 0

CH,

CH3vrryv?i>rtT'VwrTfvTVtW/rfV»V'dSyi^^

1 1 1 1 1 1 rppm 140 120 100 80 60 40 20

T

Figure 2-37

The shift values for 13CNMR are simple to recall. They come in blocks of 50ppm. Ansp3-hybridized carbon nextto a carbon hasa shift value between 0 and50 ppm. An sp3-hybridized carbon next to an electronegative atom has a shift



value between 50 and 100 ppm (usually less than 75 ppm). An sp2-hybridizedcarbon next to a carbon has a shiftvalue between 100 and 150 ppm. A carbonylcarbon with sp2-hybridization has a shift value between 180 and 230 ppm. Theshift value increases as the substitution increases, meaning that 3° > 2° > 1° inshift value. Quaternary carbons do not show up on the 13CNMR very well, dueto their long relaxation times. What is meant by a "long relaxation time" is thattheexcited energy state ofcarbon takes longer thana few seconds todissipate theenergy into solutionand 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'll note that thenumber ofpeaks ineach spectrum corresponds to the number of unique carbons in the compoundthe spectrum represents. For the ortho compound, there are four carbons in a 1:1:1:1 ratio. The spectrum shows four peaks in a 1 : 1 : 1 : 1 ratio. Three of thepeaks are from sp^-hybridized 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 13CNMR to get the symmetry of astructure and to identify selected functional groups (in a manner similar to IRspectroscopy). Don't make much more out of13CNMR spectroscopy than this.Be sure touse your degrees ofunsaturation instructure elucidation problems!

"Chemistry is good? You betcha Tiger!"


Organic Chemistry Structure Elucidation Section Summary

Key Points for Structure Elucidation (Section 2)

Isomerism

1. Isomers (compounds with the same type and number of atoms but differentspatial arrangement due to bonding, connectivity, or molecularcontortion)a) Structural isomers (isomers with different connectivity because of

different bonding)i. Have different IUPAC names

ii. Can be classified as skeletal isomers, positional isomers, or functionalisomers

iii. Their number can be determined by evaluating possible chainlengths 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 gaucheand anti to one another)

iv. Maximum number of possible optical isomers is 2n, where n is thenumber of chiral centers in the compound

c) Newman projections (frontview ofmolecule)

d) 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 two

iii. Cyclohexane (and six-membered rings in general) assume chairconformation, with groups equatorial (more stable position) andaxial

Structural Insights

1. Structural symmetrya) Plane symmetry (mirror plane in molecule splits molecule into equal

halves)b) Point symmetry (molecule hasaninversion point at itscenter ofmass)c) Units of unsaturation

i. Determined from excessbonding electronsdivided by 2.. TT . , L L. 2(#C) +(#N)-(#H)-(#X) +2ii. Units of unsaturation = -^—'-—-——-——-—-

2

iii. Describes the number of 7t-bonds and rings in a molecule

Spectroscopy

1. IRspectroscopy (used for vibrationalexcitation)a) Ranges from 1000 cm-1 to4000 cm"1 (about 3kcal/mole to10 kcal/mole)b) Correlatesbond-stretchingand bond-bendingto absorbancec) Used to identify functional groupsd) Key peaks: C=0 around 1700 cm"1, O-H around 3400 cm"1 (broad), and

C-H around 3000± cm"1 (varies with hybridization-- sp >sp2 >sp3)



e) Spectrophotometer uses salt plates to hold sample, because salt plateshave ionic bonds and therefore do not interfere with the samplemolecule'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

b) Typically used for analyzing compounds with rc-bonds, especiallyconjugated systems

c) Peak intensity and wavelength increase as the amount of conjugationincreases

3. NMR spectroscopy (the basics of1HNMR analysis)a) 1HNMR Integration (Quantitative analysis using relative area under the

curves)

i. Areaunder the curve for eachsignal is proportional to the number ofhydrogens responsiblefor the signal

ii. Connectivity canoften be deducedfromthe integration ratiob) 1HNMR peak shape (coupling and J-values)

i. The number of peaks within a signal equals the number ofneighboringhydrogens plus 1

ii. Hydrogens coupled to one another have the same J-valuesiii. Theratio of the area of the peaks within a signal can be determined

using Pascal's triangle

c) 1HNMR shift value (electron-rich environments affect shift values byexerting a magnetic field)

i. Common signals include 9-10 ppm for an aldehyde and around 7ppm for hydrogens on benzene

ii. "Upfield" refers toshifts at lowerppm valuesiii. All shifts are referenced against Si(CH3)4, which is assigned a value

of 0 ppm

d) 1HNMR special features (effects of deuterium and structural symmetry)i. Exchanging ofdeuterium for protons (peak disappearance)ii. Para substitution pattern (symmetric benzenes have unique spectra)iii. Solventchoice(solventmust be invisible)

e) 13CNMR and symmetryi. The number of signals in 13CNMR spectrum isequal 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

havemore noise in their baseline than1HNMR spectra

Copyright ©byThe Berkeley Review 144 The Berkeley Review

StructureElucidation

Passages

14 Passages

100 Questions

Suggested schedule:

I: After reading this section and attending lecture: Passages I, V, VI, VIII, & XIGrade passages immediately after completion and log your mistakes.

D: Following Task I: Passages III, IV, XII, & XIII (27 questions in 35 minutes)Time yourself accurately, grade your answers, and review mistakes.

DI: Review: Passages II, VII, IX, X, XIV, & Questions 98-100Focus on reviewing the concepts. Do not worry about timing.

•^Sr

J^R.E.V-I-E«W"Specializing in MCAT Preparation

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-13 NMR

XII. Distinguishing Isomers Using !HNMR

XIII. Structure Elucidation Using !HNMR and 13CNMR

XIV. Proton NMR of an Unknown


Structure Elucidation Scoring Scale


84 - 100 13- 15

66-83 10- 12

47 -65 7-9

34-46 4-6

1 -33 1 -3

(1-7)

(8 - 14)

(15-22)

(23 - 29)

(30 - 36)

(37 - 43)

(44 - 50)

(51 -57)

(58 - 64)

(65-71)

(72 - 78)

(79 - 84)

(85 - 90)

(91 -97)

(98 - 100)

Passage I (Question 1 - 7)

Cyclohexane is not a planar molecule, but in its moststable conformation, four of the six carbons that make up thering are coplanar. Studies using ^NMR and X-raycrystallography demonstrate that the most stableconformation of the molecule has carbon-carbon-carbon bond

angles of approximately 107.5° and 111" in the ring and thatthereare two typesof hydrogens present,axialand equatorial.The axial hydrogens are bonded directly above and directlybelow the ring carbons. The equatorial hydrogens lie awayfrom the cyclohexane ring.

A low-temperature ^HNMR study was conducted todetermine the equilibrium constant for the conversion fromdiaxial-1,2-dimethylcyclohexane to diequatorial-1,2-dimethyl-cyclohexane by way of a ring-flip process (Keq-i,2)- Thisvalue is directly comparable to the equilibrium constant forthe conversion from diaxial-1,4-dimethylcyclohexane todiequatorial-l,4-dimethylcyclohexane by way of a ring-flipprocess (Keq.1,4), given that they are both mws-substitutedcyclohexanes. The difference between their equilibriumconstants is attributed to the gauche and anti orientationspossible with the methyl substituents at the 1,2 positions.Studies have shown that larger substituents prefer theequatorial orientation of the so-called chair conformation.Figure 1summarizes the findings of the study.

Figure 1 Ring-flipping of fran.s-dimethylcyclohexane

The values for the equilibrium constants can be appliedto determine the relative steric hindrance of one substituentcompared to another. A bulkier group exhibits greater sterichindrance, so the equilibrium lies more towards the morestable of the two possible chair confirmations. Hence, agreater equilibrium constant implies that there is a greaterdegree of steric hindrance in the lessstable conformation.

1. Whichof the following is the most stable orientation ofa substituent on a cyclohexane molecule?

A. Axial orientation of a chair conformation

B. Equatorial orientation of a chair conformationC. Bridge orientation of a boat conformationD. Oar orientation of a boat conformation

®Copyright © by The Berkeley Review'

2.

3.

147

A carbon-deuterium bond is shorter than a carbon-

hydrogen bond. Using this idea, how many deuteriumatoms assume axial orientation in the most stable

conformation of the following molecule?

D

D

A. 0

B. 1

C. 2

D. 3

D

What is the value of Keq-i,4 for the conversion oftrans- 1,4-dimethylcyclohexane from its diaxialconformationto its diequatorial conformation?

A. 0.0029

B. 2.16

C. 4.31

D. 345

Cis-1,4-dimethylcyclohexane, in its most stable chairconformation, CANNOT have which of theseinteractions?

A. CH3/Hgauche

B. CH3/Hanti

C. CH3/CH3 gauche

D. H/Hanti

Which of thefollowing accurately describes thevalue ofKeq for the conversion from one chair conformation tothe other chair conformation for the compound cis-1,2-dimethylcyclohexane?

A . Less than or equal to 0B. Greater than 0 and less than 1

C. Equal to 1D. Greater than 1

The value of Keq for the conversion of cis-1,3-dimethylcyclohexane from its diaxial conformation toits diequatorial conformation is:

A. less than 0.22.

B. between 0.22 and 1.

C. between 1 and 4.31.

D. greater than 4.31.


7. Which three-dimensional conformation corresponds tothe 3-hydroxy-c/s-decalin, shown below?

H

or"H

A.OH

B.OH

C. D.

/^cp^t™ rp^ OH



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 therotation 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 least

stable conformation occurs when the largest groups interferewith one another. This is known asfully eclipsed. Drawn inFigure 1 is an energy diagram for the counterclockwiserotation about the C2-C3 bond for R-2-methyl-l-butanol.

c

W

180 360

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 theeclipsed or staggered conformations are equal in energy. Allvisual projections show asymmetric steric interactions.Although the molecule is constantlyrotating 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?

A. B.

H3C CH3

H\^ ^W HH

C.

H3C

H

CH2OH

H3C ,CH3

HWV ^"CH2OHH H

D.

^CH2OH H3C

H\^CH3 H

^CH2OH

CH3


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-1 interferes with carbon-4.C. staggered and the methyl substituent on carbon-2

interferes with carbon-4.

D. staggered and carbon-1 interferes with carbon-4.

10. Which of the following structures represents themolecule at the 330° point on the graph in Figure 1?

A. B.H3C CH3 H3C CH3

C.

H-

H^C^HCH3

H3C CH2OH

HOH2C

D.

HOH2C CH3

H3C

H

H

11. Which of the following statementsBEST explains whythe MOST stable conformation of 2-amino-l-ethanol is

gauche?

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.

12. The strongest hydrogen bondoccurs between which ofthe following?

A. A lone pair on nitrogen bonded to an H on nitrogenB. A lone pair on oxygenbonded to an H on nitrogenC. A lone pair on oxygen bonded to an H on oxygenD. A lone pair on nitrogen bonded to an H on oxygen

Copyright ©by The Berkeley Review®

13. 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 rc-bondrequires energy in excess ofroom temperature.

14. Which of the following compounds contains a singlebond about which complete rotation is not possible?

A. DipropyletherB. 2-butanone

C. 2,3-butanediol

D. Methylcyclopentane

149 GO ON TO THE NEXT PAGE


A researcher wishes to determine the relative stability ofaxial orientation versus equatorial orientation for deuteriumand hydrogen on cyclohexane. To do so, she treats benzene(C6H6) with D2SO4/D2O at 100*C for thirty minutes tosynthesize monodeuterobenzene (CgHsD), which is thentreated with H2 gas and palladium metal under 90 psi ofpressure to yield monodeuterocyclohexane (C6Hi]D). Formonodeuterocyclohexane (C^ii\\D), there are two possiblechair conformations, one with the deuterium having axialorientation and the other with the deuterium having equatorialorientation. At room temperature, the interconversionbetween the two chairconformations is too rapid to study andall eleven hydrogens appear as a singlet 1.38 ppm in the'HNMR. At low temperatures, the interconversion throughring-flip from one chairconformation to the other is slowedgreatly, so that axial and equatorial hydrogens generatedifferent signals in the 'HNMR. As a result, the ring-flipprocess can bemonitored using 'HNMR spectroscopy.

A lHNMR was recorded at -89*C in deuterochloroformsolvent. A hydrogen with axial orientation shows a 'HNMRshift of 8 = 1.51 ppm, while a hydrogen with equatorialorientation shows a 'HNMR shift of 8 = 1.25 ppm.Integration shows that the relative area of 1HNMR signals is1.12 : 1 in favor of the 8 = 1.25 ppm signal. Because alarger amount of equatorial hydrogen is observed than axialhydrogen, the deuterium has axial orientation in the morefavorable chair conformation.

The researcher proposes that a difference in bond lengthbetween the C-H bond and the C-D bond, rather than adifference in atomic size between hydrogen and deuterium,accounts for the equatorial preference of hydrogen overdeuterium. The difference in bond length is attributed to thegreater 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, thegreater difference inmass between hydrogen andcarbon than deuterium and carbon makes thecarbon-hydrogenbond stretch more asymmetrically than a carbon-deuteriumbond. A carbon-hydrogen bond stretches more thana carbon-deuterium bond, and thus occupies agreater amount of space.Despite the differences in bond length, the bond angles incyclohexane remain similar, between 107.5* and 111*.

15. The researcher reached the ultimate conclusion that abond between carbon and deuterium is shorter than abond between carbon and hydrogen, based on the factthat:

A. the deuterium favors the equatorial orientation.B. the deuterium favors the axial orientation.

C. the interconversion between the two possible chairconformations of the deuterocyclohexane moleculethrough ring flip is rapid at room temperature.

D. the 'HNMR shift at 8=1.25 ppm is farther upfieldthan the 'HNMR shift at 8= 1.51 ppm.


16. 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. 6:5

B. 5:2

C. 3:2

D. 2:3

17. The addition of H2 gas and platinum metal tochlorobenzene (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. chairwith chlorineequatorial.

18. The D-C-H bond angle about the deuterated carbon isclosest 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:

CH*

H3C^^ CH

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

21. The reason that the diaxial orientation for cis-3-

hydroxycyclohexanol (a cis-l,3-diol) is preferred overthe diequatorial orientation is that the hydroxyl groups:

A. are smaller than hydrogens, so they exhibit nopreference for the less hindered equatorialorientation.

B. are larger than hydrogens, so they exhibit apreference for the less hindered axialorientation.

C. are larger than hydrogens., so they exhibit apreference for the more hindered equatorialorientation.

D. can form an intramolecular hydrogen bond from a1,3-diaxial orientation, while they cannot form anintramolecular hydrogen bond from the 1,3-diequatorial orientation.

22. Which structure represents the MOST stable form ofcis-1,4-ethylmethyIcyclohexane?

B.

H*Z^Z- CH2CH3

CH2CH3

C. D.

CH3 CH2CH3

CH2CH3 CH3



In most research laboratories, fourier transform infrared(FTIR) spectrophotographers are used to obtain infraredspectra. The FTIR spectrophotographer works by passing anelectromagnetic pulse of multiple frequencies through asample 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 samespectraas those obtained using outdated variable frequencyIRspectrophotometers.

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 amonochromatic light pulse. Focusing on one peak with amonochromatic 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 itdisplays around 3300 cm"',although the exact value varies with the degree of hydrogen-bonding. Acarbonyl bond isfound around of1700 cm"'. Ifa carbon has ^-hybridization, the bonds it forms tohydrogen are found just below 3000 cm'1. All of thisinformation combines into a nice packet of data used todeduce the structure of a compound. Figure 1 shows the IRspectrumfor 2-heptanone:

1 1 1 1 1 r4000 3600 3200 2800 2400 2000 1800 1600 1400 1200

Figure 1 InfraredSpectrafor 2-heptanone

1000

The information extrapolated from the IR spectra can becoupled with NMR (nuclear magnetic resonance) data toforma powerful combination. For instance, aromatic hydrogensare found in the 7 to8 ppm range inan 'HNMR spectrum.

23. An absorbance between 1700 cm"1 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. 3300cm-1.

B. 2980 cm"'.

C. 2300 cm-'.

D. 1715 cm"1.

25. Which of the following isomers of C3H6O2 exhibits abroad IRsignal around 2850 cm*1?A. O

XHO CH2CH3

C.u

A.H3CH2CO H

B. O

AH3CO CH3

D.

26. Which of the following pairs of compounds could bedistinguished by their splitting patterns in the protonNMRregion between7 and 8 ppm?

A. Methylpropanoate from ethylethanoate

B. 3-methyl-2-hexanone from2-methyl-3-hexanoneC. 1,4-methylphenol from 1,4-ethylphenolD. 1,4-methylphenol from 1,3-methylphenol

27. To distinguish a tertiary alcohol froma 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. Thewidth of thepeaks near3300 cm-1B. Thelength of thepeaks near 3300 cm"'C. The width of thepeaks near 1700 cm"1D. Thelength of thepeaks near 1700 cm"1


2 8. Which of the following isomers of C4H8O would NOThave anIRsignal at 1715 cm-'?A. 2-methylpropanal

B. Butanal

C. Butanone

D. Tetrahydrofuran

29. Hydrolysis of an ester could be supported by which ofthese DR. spectroscopic data?

A. The appearance ofa signal around 1700 cm-1B. The disappearance ofa signal around 1700 cm"1C. The appearance ofa signal around 3300 cm"'D. The disappearance ofa signal around 3300 cm"'


Passage V (Question 30 - 36)

An unknown alkyne with a molecular mass of 122.2g/mole is treated with H2/Pd and BaS04 to convert it into acis-alkene. The cis-alkene is isolated in high purity. The cisalkene is then treated with high-pressure ozone (O3) gas andzinc metal to convert both of the alkene jp2-hybridizedcarbons into carbonyl carbons. Because the cis-alkene wasformed by hydrogenation of an alkyne, it is disubstituted witheach alkene carbon holding one hydrogen. This means thatthe products of ozonolysis are both aldehydes. Two uniqueproducts are isolated from the product mixture. The twounknown products are designated as Compound A (C3H6O)and Compound B (C6H10O). An 'HNMR spectrum isobtained for Compound B and is shown in Figure 1 below.The 'HNMR is carried out using CDCI3 as thesolvent.

1 H

1 4 H

1

1 19.8 ppm 9.6 ppm

4 H

1 1

1 H 1

ii 11 11 I1 1 1

2.5 ppm 2.0 ppm 1.5 ppm1

1.0 ppm1

0.5 ppm1

Oppm

Figure 1 'HNMR spectrum of Compound B

For Compound A, spectraldata wereobtained froman IRspectrum using pure Compound A in liquid form betweensaltplates. Table 1 lists the key IR absorbances collected forCompound A

Shift fern- '} Intensitv

strong2962

2912 medium

2706 medium

1726 strong

1212 strong

all other peaks are irrelevant

Table 1 IR absorbances of Compound A

The structures of Compounds A and B can be deducedwith great accuracy from the spectral data in the passage.When engaging in structure elucidation, some information ismore useful than others. As a general rule, NMR data areapplied last, as they have the most information.

3 0. A compound with one degree of unsaturation and twooxygens CANNOT be:

A. a cyclic ketone.

B. a cyclic ether.

C. a carboxylic acid.

D. an ester.

Copyright © by The Berkeley Review®

31. Which of the following 'HNMR shifts would beobserved 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 would beobserved for Compound B?

A. 1725 cm"1

B. 2200 cm"1

C. 1620 cm"1

D. 3550 cm"1

3 3. What is the molecular formula for the original alkyne?

A. C9H14

B. C9H16

C. C9Hi8

D. C9H20

34. Which of the following IR absorbance values isindicative of an alkene?

A. 3550cm"1B. 2200 cm"1

C. 1725 cm"1

D. 1620 cm"1

35. Which of the following structures corresponds toCompound A?

A. n B. O

H3C CH3 H3CH2C H

D. Q

CH3


36. Which of the following structures corresponds toCompound B?

A.

cr "ecC. D.

CHa



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 labels 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

8 = 2.5 ppm (broad) 1H triplet 3Hz

8 = 3.1 ppm 1H singlet NA

8 = 4.3 ppm 2H doublet 3Hz

Table 1 'HNMR signals of Compound T

Shorthand for describing nuclear magnetic resonancespectra describes the chemical shift value (8), the number ofhydrogen atoms (relative area under each signal), and thecoupling 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 = quartet,and t = triplet.

For the unknown compound, Compound T, three bandsat 8 50 (t), 73.8 (d), and 83.0 (s) appear in the carbon-13magnetic resonance spectrum. The splitting in the 13CNMRis due to the hydrogens directly bonded to the carbon. Theimportant absorbances in the infrared spectrum of CompoundT are found at 3350cm"' (broad), 2988cm"1,2116 cm"1, and1033 cm"1. The spectral data can beapplied todetermine thesymmetry of a compound as well as the functional groups ofthe compound. The IR peak at 2116cm'1 is indicative of atriple bond, indicating the presence of either an alkyne or anitrile in Compound T. There is no spectral evidence tosuggest that a nitrogen atom is present in the compound, sothe most logical assumption is that there is a carbon-carbontriple bond present in the structure.

From the spectral data, the lab technician concludes thatthe compound contains no carbonyl functionality. However,the compound contains a functionality that is involved inhydrogen-bonding, explaining the relatively slow evaporationof Compound T. Using chemical tests and a polarimeter, thelab technician determines that there is no mirror symmetry inthe molecule. This implies that the number of signals in the13CNMR spectrum is also the number of carbons inCompound T.

37. The shift at 1033 cm"' in the IR is caused by theC-Obond of:

A. an ether.

B. an ester.

C. a carboxylic acid.

D. an alcohol.


38. The absorbance at 2116 cm-' in the IRimplies that thecompound is an:

A. alkane.

B. alkene.

C. alkyne.

D. alcohol.

39. The 13CNMR doublet at 73.8 ppm can beattributed towhich type ofcarbon?

A. The terminal C of a terminal alkyne

B. The internal C of a terminal alkyne

C. C of an aldehyde

D. C of an ketone

40. The number of carbons in an asymmetric molecule thatshows seven peaks in the '̂ CNMR is:A. 3.

B. 4.

C. 7.

D. 14.

41. Hydrogens that are coupled to one another have thesame:

A. shift value.

B. peak integration.

C. areaunder their peaks.

D. same J-value.

4 2. Compound T is which of the following compounds?

A. B.

H-C=C— CH2

OH

c.

HO-C=C-CH

CH-

H-C=C-CH2 M

C

IIO

D.

H-CEC-CH,

CH-


43. The 13CNMR peak at 50 ppm can be attributed to the:A. CofaC=0.

B. C of a C-O.

C. HnexttoC=0.

D. H next to C-O.



From the leaves of a tree that grows wild along the coastof Ecuador, three esters were extracted using ether. The threeesters were separated by fractional distillation and purified bycolumn chromatography. Once isolated and purified, a'HNMR spectrum was obtained for each compound. Thethree spectra are shown in Figure 1 below. All three spectrawere obtained using the same NMR spectrometer in CDCI3solvent from the same bottle. A dead give-away for an esteris the shift for the hydrogens attached to the carbon bound tothe oxygen of the alkoxy group. These hydrogens are foundbetween 3.5 ppm and 4.0 ppm. Compound I has a molecularformula of CgHg02, Compound II has a molecular formulaof C9H10O2, and Compound in has a molecular formula ofC4H802.

5H

16 ppm

5H

6 ppm

3H

13 ppm

3H

A4 ppm

jr• ppm

2H

R2 ppm

Compound I

2 ppm

Compound H

3H

A

Compound HI

3H

A1 ppm

Figure 1 'HNMR spectra ofCompounds I, n, and HI

4 4. Which of the following structures is CompoundI?

A. o B« o

Or** Cr* OCH<

C. D.

olH ay CH«

4 5. It important that each sample use the same solvent inorder 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 for extraction of protic compounds.


46. According to IUPAC nomenclature convention, what isthe name of Compound HI?

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. 3500cm"1

B. 2980 cm"'

C. 1685 cm"'

D. 1300 cm-'

48. Which of the following structures represents Compoundn?

A. o B« O

CH2CH3 jj'̂ i^^OCHgCHs^ O

C. o D.

^^^ CH2CH3 ^^ O

OCH?

49. For the spectrum of Compound IQ, the three 'HNMRsignals are which of the following, respectively?

A. Triplet, doublet, triplet

B. Triplet, quartet, triplet

C. Singlet, doublet, triplet

D. Singlet, quartet, triplet

50. Which of the following would NOT work as an NMRsolvent for an ester?

A. CDCI3(chloroform)

B. CgDg (benzene)

C. D3CD2COD (ethanol)

D. D2CO (formaldehyde)


Passage VIII (Question 51 - 57)

Nuclear magnetic resonance spectroscopy is a powerfultool for distinguishing isomers. By comparing shift values(measured in ppm), coupling constants (J-values), andintegration of peaks (the area under the curve), it is possibleto deduce a structure with great accuracy. The couplingconstants help us deduce which hydrogens are neighbors ofone another. For instance, a carbon containing twoequivalent hydrogens influences the signal for hydrogens onthe neighboring carbon to form a triplet. The integration isthe area under the curve that is directly proportional to thenumber of hydrogens within that signal. Table 1 is a generallisting 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 3.8

-CH=CH2 5.5 -CH=CH2 5.3

-ArH 7-8 -CH2CH=CH2 2.3

-CH=0 9.7 -C02H 10-12

Table 1 'HNMR shift values for common hydrogen types

When 'HNMR information is coupled with IR shiftvalues, it is possible to narrow the structure down quickly toone possibility. Important IRabsorbances are 3500 cm"' (O-H), 1700 cm"' (C=0), and 1600 cm"1 (C=C). There areother absorbances, but from these three values, it is possibleto estimate many other values. As the strength of a bondincreases, it exhibits a higher absorbance value in the IR.The heavier the atoms in the bond, the higher the absorbancein the IR. This information makes it possible to predict shiftvalues for other bonds. For instance, a C-C bond must be

less than 1600 cm"1 in absorbance in the IR, because acarbon-carbon single bond is weaker than a carbon-carbondouble bond. However, the importance of NMRspectroscopy is greater than that of infrared spectroscopy.

51. 2-Methyl-3-pentanone contains:

A. 3 non-equivalent hydrogens in a 9 : 2 : 1 ratio.

B. 4 non-equivalent hydrogens in a 4 : 3 : 3 : 2 ratio.

C. 4 non-equivalent hydrogens in a 6 : 3 : 2 : 1 ratio.

D. 5 non-equivalent hydrogens in a 3 : 3 : 3 : 2 : 1ratio.

52. What in the proton NMR is a dead give-away for anisolated isopropyl group?

A. A 6H doublet and a 1H septet

B. A 6H singlet and a 1H sextet

C. A 6H septet and a 1H doublet

D. A 6H sextet and a 1H singlet


5 3. Pentanal is BEST distinguished from 2-pentanone by apeak:

A. above 1700 cm"1 in the IR.

B. between 2.0 to 2.3ppm in the !HNMR.C. near 3.5 ppm in the !HNMR.D. near 9.7 ppm in the 'HNMR.

54. 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 HCl and observe whether the broad peakgrows.

D. H2O with HCl 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 ppm

C. 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 theformula C4H8O, an IR absorbance at 1721 cm"', andthree NMR peaks at 9.7 ppm (1H), 2.2 ppm (1H), and1.0 ppm (6H)?

A. Tetrahydrofuran

B. 2-methylpropanal

C. Butanal

D. Butanone

157

57. A compound with one oxygen in its formula, onedegree of unsaturation, and no IR absorbance between1600 cm"1 and 1750 cm"1 must be:

A. a ketone.

B. an aldehyde.

C. an alkene.

D. cyclic.


Passage IX (Question 58 - 64)

A 1HNMR was run for the cis and trans isomers of themolecule shown in Figure 1 below. The spectra obtained forthe two geometrical isomers are also shown in Figure 1. Thecis compound can be distinguished from the trans compoundusing the coupling constants for the vinylic hydrogens. Thetrans 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 thecarbon adjacent to benzene can be found downfield from theother vinylic hydrogen due to the presence of oxygen."Downfield" implies higher shift values.

H3CV /

CH=CH—OCH3

Spectrum I

4H

1 H 1 H

JUL

3 H

11t 1 1 r

PPm 7 6 5 4

Spectrum II

4H

I H 1 H

JUL

3H

1 1 1 r

PPm 7 6 5 4

3H

JL

3H

7.25 ppm (4 H)5.37 ppm (1H)4.88 ppm (1 H)3.76 ppm (3 H)2.25 ppm (3 H)

1

7.25 ppm (4 H)5.37 ppm (1 H)4.88 ppm (1 H)3.76 ppm (3 H)2.25 ppm (3 H)

1

Figure 1 ^NMR spectra of the cisand trans isomers

The spectra as drawn are different enough to distinguishthe trans and cis compounds from one another. The shiftvalues does not help to distinguish the cis from the transcompound, 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.

59. The methyl group on benzene appears in the 'HNMRat:

A. 7.25 ppm.

B. 5.37 ppm.

C. 3.76 ppm.

D. 2.25 ppm.


60. Which of the following compounds has the samenumber of signals in their carbon-13 NMR as there arecarbons in the compound?

A. An ortho substituted benzene with two identical

substituents

B. A meta substituted benzene with two identical

substituents

C. A meta substituted benzene with two different

substituents

D. A para substituted benzene with two differentsubstituents

61. Where do the vinylic hydrogens appear in 1HNMR?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

62. Which of the following absorbances would be found inthe IR spectrum for the compound in Figure 1?

A. 3500cm"1 -

B. 2220 cm"1

C. 1720 cm"1

D. 1640 cm"1

6 3. Had the compound had meta substitution rather thanpara substitution, the ratio of the benzene hydrogenswould be which of the following?

A. 1 : 1

B. 1 :3

C. 1:2:1

D. 1:1:1:1

6 4. Had the compound had an ethyl group on benzene ratherthan a methyl group, which of the following would beobserved in the proton NMR?

A. Doublet (2H) and triplet (3H)

B. Doublet (3H) and triplet (2H)

C. Triplet (2H) and quartet (3H)D. Triplet (3H) and quartet (2H)


Passage X (Question 65-71)

An unknown compound, labeled CompoundB, has onlytwo singlets in its proton magnetic resonancespectrum. Oneshows a shift of 8 1.42 ppm and the other shows a shift of 51.96 ppm, with relative intensities (from the integration ofthe 'HNMR spectrum) of 3 : 1. The decoupled carbon-13nuclearmagnetic resonance spectrum for Compound B showsfour signals (with one of greater intensity than the otherthree) with shifts at 8 22.3 ppm, 28.1 ppm, 79.9 ppm, and170.2 ppm. The signal at 170.2 ppm is attributed to acarbonyl carbon. For carbons of equal hybridization, thehigher the shift value in the carbon-13 NMR, the greater theelectronegativity of the atoms bonded directly to that carbon.Figure 1 shows both the 'HNMR and 13CNMR spectra ofCompound B.

'HNMR 3H Compound B

1H

1ppm 2.0

1 1

1.5 1.0

1

0.5 0

13CNMR Compound B28.1 ppm

22.3 ppm

170.2 ppm

79.9 ppm

>«W'1».»WH*»*W»W* n/+M4t*t*mt*4 0*MW»Mirn*<tW»MM«niW> t\X i *M*»»

ppm"T-150 100

T50

Figure 1 'HNMR and 13CNMR spectra of Compound B

The carbon-13 NMR information leads to the conclusion

that there are four nonequivalent carbons in the compound.The important bands in the infrared spectrum of Compound Bare found at 1738 cm"', 1256 cm"', and 1173 cm"'. Theband at 1738 cm"1 is attributed tothe stretching ofa carbonylbond. Elemental analysis of Compound B shows that itcontains two oxygen atoms. The spectral data, inconjunction with mass percent values from elementalanalysis, can be combined to determine the structure ofCompound B. Overlapping spectral data, such as an IRabsorbance at 1738 cm"1 and a 13CNMR signal at 170.2ppm, can be used to verify aspects of the structure.


65. The shift at 1738 cm"1 in the IR can be attributed to the0=0 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 : 1

B. 6:2

C. 9:3

D. 12:4

67. The 13CNMR peakat 79.9 ppm can beattributed to:A. an H bonded to C adjacent to C=0.

B. an H bonded to C adjacent to C-O.

C. the C in the C=0 bond.

D. the C in the C-O bond.

6 8. Which of the following compounds would NOT show a13CNMR peak above 100 ppm?A. Methylbenzoate

B. 3-methyl-2-hexanone

C. 3-methyl-2-pentanol

D. 3-methylpentanal

69. The !HNMR peak at 1.96 ppm can be attributed to ahydrogen bonded to a carbon:

A. adjacent to a C=0 bond.

B. adjacent to a C-O bond.

C. of a C-O bond.

D. ofaC=Obond.


70. In 1HNMR, a singlet is explained as the evidence ofhydrogens of the signal being coupled to:

A. equivalent hydrogens on all adjacent carbons.B. non-equivalent hydrogenson all adjacentcarbons.C. only one hydrogen on an adjacent carbon.

D. no hydrogens, because there are no hydrogens onany of the adjacent carbons.

71. Compound B is which of the following?

A. OB

O

H3CCH3

H3CO

CH3

H3C CH3 H3C CH3

C. D.

H3C CH3 O H3C CH3 O



Carbon-13 NMR can be used to determine the number of

unique carbons in a compound. Carbon-13 NMR is similarto hydrogen-1 NMR in that it generates separate peaks foreach unique isotope, in this case 13C. Many 13CNMRspectra are recorded without coupling (known as decoupledspectra), so all peaks appear as singlets. Table 1 lists theapproximate shift value for selected types of carbons. It is abrief guide to determining the types of carbon represented byeach ,3CNMR peak. The range for theppm of eachsignal isapproximate and on occasion, a peak may fall outside of therange.

Carbon ppm Carbon ppm

R2C=0 205 - 220 -C-OH 50-70

RHC=0 185-200 -C-Cl 40-45

R2C=CH2 120 - 140 -C-NH2 35-45

R2C=CH2 110-120 -C-Br 25-35

RCeCH 72-85 H3C-C=0 20-35

RCsCH 65-70 R2CH2 10-30

Table 1 13CNMR shift values

It should be noted that the peaks for carbons with nohydrogen directly attached are less intense than other carbonpeaks. Carbonyl carbons and quaternary carbons thereforegenerate shorter peaks than those for other carbons in thespectrum. For peaks that representmore than one carbon, theintensity increases, but not in a way that is easily integrated.Integration isgenerally not carried outon 13CNMR spectra.

Aresearcher used 13CNMR todistinguish two structuralisomers formed when toluene (methylbenzene) was acylatedusing acetyl chloride by comparing the l3CNMR spectra ofthe two compounds. The molecular formula for both isomersis C9H10O. Spectra for both compoundswere recordedusingthe same NMR instrument for the same period of time in thesame concentration. The only difference between the twosamples involved the isomers themselves. The spectral datafor the two isomers are listed below:

Isomer I: 27, 34, 110, 117, 123, 130, 206

Isomer H: 31, 35, 108, 115, 119, 123, 126, 131, 209

The difference in the number of ,3CNMR signals listedfor each isomer is caused by the lack of symmetry betweenthe two isomers. Isomer I has nine carbons but shows onlyseven peaks in the 13CNMR, so it must have some type ofsymmetry that equates carbons within the structure. Bothisomers are benzene rings with a methyl substituent and anacetylgroup attached to carbonson the benzenering.


72. Which of the following reactions CANNOT bemonitored by a change in the shift values in the13CNMR spectra?

A. Oxidationof a secondary alcohol to a ketone

B. Nitration of ethylbenzene to para-nitroethyIbenzeneC. Deprotonation of a carboxylic acidD. Reduction of an alkene to an alkane with hydrogen

gas and nickel catalyst

73. Which of the following structures represents thestructure of Compound I?

A* O^ ^CH3 B- Q- .CH3

C* O^ ^CH3 D* O^ ^H

CH2CH3

7 4. Which of the following functionalities are NOT presentin either of the two isomers?

A. Aldehyde

B. Alkene

C. Ketone

D. Methyl

7 5. How can it be explained that the least intense peak inthe spectrum for Compound II is found at 209 ppm?

A. Carbons with .s/?2-hybridization show low intensity13CNMR peaks.

B. Carbons with ^-hybridization show low intensity13CNMR peaks.

C. Chiral carbons show low intensity 13CNMRpeaks.

D. Carbons with no hydrogen atoms attached showlow intensity 13CNMR peaks.


76. How many signals would be seen in the carbon-13NMRof para-methoxy benzaldehyde?

A. Four

B. Five

C. Six

D. Eight

7 7. What is true about the units of unsaturation for the two

isomers?

A. Isomer I, because it is more symmetric than IsomerII, has more units of unsaturation than Isomer n.

B. Isomer I, because it has less unique carbons thanIsomer II, has fewer units of unsaturation thanIsomer n.

C. Isomer I and Isomer II each have three units of

unsaturation.

D. Isomer I and Isomer II each have five units of

unsaturation.

78. For the compound 2,2-dimethylbutane, which carbonshows the peak of lowest intensity?

A. Carbon 1

B. Carbon 2

C. Carbon 3

D. Carbon 4



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 is

also known as the coupling constant and is the distancebetween adjacent peaks within a proton NMR signal.

The 'HNMR spectrum for two isomers of C4H8O2 arecollected under identical conditions using the same NMRmachine. Figure 1 shows the spectrum for Isomer I, whileFigure 2 shows the spectrum for Isomer n.

Figure 1 1HNMR spectrum of Isomer I

3H Spectrum nC4H802

3H

2 H

•_J1 1

-1

1

3 ppm•1 1

2 ppm 1 ppm1

0 ppm

Figure2 'HNMR spectrum of Isomer n

Both of the isomers are esters. The exact connectivity ofthe esters can be deduced from the shift values of the singletand quartet. 'HNMR signals around 3.5 to4.0 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.


79. Which of the following is the common name for thecompoundrepresented by Spectrum I?

A. Ethyl acetate

B. Acetone

C. Methyl acetate

D. Isopropyl formate

80. A triplet in the proton NMR is associated with thehydrogens:

A. ofaCH2group.

B. on a carbon next to a CH2 group.

C. of a CH3 group.

D. of a carbon next to a CH3 group.

81. What type of compound is represented by Spectrum II?

A. An ethyl ester

B. A methyl ester

C. An ethyl ketone

D. A methyl ketone

82. A 'HNMR signal in the range between 6 and 8 ppmindicates the presence of which of the following?

A. a hydroxyl proton.

B. an aldehyde hydrogen.

C. a benzene hydrogen.

D. a carboxylic acid proton.

83. Which of the following features in the proton NMRspectrum CANNOT be used to distinguish an ester formof a carboxylic acid when the two compounds have thesame molecular formula?

A. A sharp peak between 2.0 and 2.5 ppm

B. A sharp peak between 3.5 and 4.0 ppmC. A sharp peak between 10.0 and 12.0 ppmD. The observation that no peak integrates to a relative

ratio of 1.

8 4. How many unique proton NMR signals are expected for4-heptanone?

A. Two

B. Three

C. Four

D. Five


Passage XIII (Question 85 - 90)

A chemist sets out to determine the structural identity forthreestructural isomers, using data from NMR spectroscopy,ultraviolet-visible spectroscopy, and mass spectroscopy toidentify each compound. The following spectral data areobserved for three separate compounds, all with the formulaC5H10O.

Compound I:

13CNMR:

HNMR:

211 ppm (1), 31 ppm (1), 22 ppm (1), 17ppm (1), 11 ppm (1)

2.32 ppm (triplet, 2H), 2.08 ppm (singlet,3H), 1.12 ppm (multiplet, 2H), 0.96 ppm(triplet, 3H)

UV-Visible: 268 nm and 189 nm

Compound II:

13CNMR:

'HNMR:

UV-Visible:

Compound III:

13CNMR:

'HNMR:

UV-Visible:

72 ppm (1), 24 ppm (2), 13 ppm (2)

3.71 ppm (multiplet, 1H), 1.88 ppm (broad,1H), 1.33 ppm (multiplet, 4H), 1.14 ppm(triplet, 4H)

No intense peaks above 180 nm

68 ppm (2), 21 ppm (2), 10 ppm (1)

3.58 ppm (triplet, 4H), 1.28 ppm(multiplet, 4H), 0.92 ppm (multiplet, 2H)No intense peaks above 180 nm

All NMR spectra are obtained using deuteratedchloroform as solvent. The mass spectrometer show anintense peak at 86 amu for all three isomers, confirming theirmolecular formula. Elemental analysis shows that carbonand hydrogen are not the only atoms present.

8 5. Compound II is what type of compound?

A. An alcohol

B. An ether

C. A ketone

D. An alkene

86. Carbon-13 NMR is useful for determining all of thefollowing 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.


87. What can be concluded from the data obtained usingUV-visible spectroscopy?

A. Compound I is a conjugated diene.

B. CompoundsII and in are carbonyl compounds.C. Compound I is a ketone.

D. There is no rc-bond in Compounds I and n.

8 8. Which of the following is NOT a valid conclusion fromthe spectral data for Compound HI?

A. The absence of a broad peak in the 'HNMR meansthe compound cannot be an alcohol.

B. The absence of a peak above 175 ppm in the13CNMR means the compound cannot be acarbonyl.

C. The absence of an absorbance above 180 nm

confirms that the structure must contain a ring forits unit of unsaturation.

D. The large number ofpeaks in the '3CNMR meansthe 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

9 0. The integration of a signal in a proton NMR is usefulfor determining the:

A. localmagnetic 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 the

compound.


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. Thethree useful components of thespectral datafor extracting structural information are the shift value (5,measured in ppm), the coupling andcoupling constants (peakshape), and the integral (thearea under thecurve of a signal).From the shift value, information about the electronegativityof 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 compound. Figure 1 and Figure 2 show the'HNMR spectra for two simple organic structures,Compound A and Compound B, along with the molecularformula of the compounds they represent.

Spectrum AC7H8

5H

3H

J 1 i

1 1 18 ppm 6 ppm 4 ppm 1

1. ppm 0 ppm

Figure 1 'HNMR spectrum of Compound A

Spectrum BC3H602 [

/ j

•

_

A 1u\ 1

10 ppm 5 ppmi

0 ppm

Figure 2 'HNMR spectrum of Compound B

Structures can often be deduced using only some of the'HNMR information, such as the coupling information andthe integration. Often, the coupling information is the mostimportant of all the data. The coupling constants can giveinformation about the connectivity of the structure, as well ashints about the three-dimensional orientation of atoms within

the molecule.


91. Which of the following is the IUPAC name for thecompound represented by SpectrumA?

A. Phenylmethane

B. Toluene

C. Methylbenzene

D. Orthoxylene

164

92. In Spectrum B, what is the ratio of the areas under thethree peaks?

A. 3:2:3

B. 1:1:4

C. 2:3:5

D. 1:2:3

93. Whatpeaks are expectedfor 2-bromopropane?

A. A 6H sextet and a 1H singlet

B. A 1H sextet and a 6H singlet

C. A 6H septet and a 1H doublet

D. A 1H septet and a 6H doublet

9 4. Which of the following compounds would NOT have aquartet in its proton NMR spectrum?

A. o B. o

A A(H3C)2HC CH2CH3 H3CH2C(H3C)2C CH3

C. o D. 0

A X(H3C)2HCH2C CH3 H3CH2CH2C CH2CH3

9 5. Which of the following compounds would NOT have adoublet in its proton NMR spectrum?

A. 2-methyl-1-pentanol

B. 3-methyl-1-pentanol

C. 4-methyl-1-pentanol

D. All isomers of methyl-1-pentanol have doublets intheir proton NMR spectrum.


96. 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 iodoform testwould likely have which of the following signals in itsproton NMR?

A.

B.

C.

D.

Singlet

Doublet

Triplet

Septet

97. The broadness of the 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.



9 8. The following two molecules are best described as:

'''its' /\

A. structural isomers.

B. geometrical 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.

100. How many structural isomers of CsH]2 are possible?

A. 3

B. 4

C. 5

D. 6

1. B 2. C 3. D 4. C 5. C 6. D

7. B 8. C 9. A 10. A 11. A 12. D

13. D 14. D 15. B 16. C 17. D 18. B

19. A 20. B 21. D 22. D 23. D 24. C

25. A 26. D 27. A 28. D 29. C 30. A

31. B 32. A 33. A 34. D 35. B 36. C

37. D 38. C 39. A 40. C 41. D 42. A

43. B 44. B 45. C 46. B 47. A 48. B

49. D 50. C 51. C 52. A 53. D 54. B

55. A 56. B 57. D 58. C 59. D 60. C

61. B 62. D 63. D 64. D 65. C 66. C

67. D 68. C 69. A 70. D 71. D 72. C

73. C 74. A 75. D 76. C 77. D 78. B

79. A 80. B 81. B 82. C 83. A 84. B

85. A 86. D 87. C 88. D 89. B 90. D

91. C 92. D 93. D 94. C 95. D 96. A

97. D 98. B 99. C 100. A

ENOUGH CHEMISTRY... FOR NOW!

Structure Elucidation Passage Answers

PassageI (Questions 1-7) Chair Conformation

1. Choice B is correct. The most stable form of the cyclohexane ring is the chair conformation. The most stableposition on the chair form is referred to as equatorial. Combine these two facts and the result is choice B.

2. Choice C is correct. Because the three deuterium atoms are cis with respect to one another, they cannot all beaxial nor all be equatorial. The most stable orientation (most stable chair confirmation) has as many deuteriumatoms with axial orientation as possible. However, because the deuterium atoms are all mutually cis to oneanother, the structure must have at least one deuterium with equatorial orientation. The best choice (andconsequently your choice) is C, which is drawn below.

3.

2 axial deuteriums and

1 equatorial deuterium h

D

13

I-I

D

11

1 axialdeuterium and

2 equatorial deuteriums

Choice D is correct. As the reaction is written, the value ofKeq.^4 must be greater than 1, because the product ismore stable than the reactant. The reaction is favorable in the forward direction as written. This eliminateschoice A, a value less than 1.

The question now focuses on whether the conformational change with Keq.^4 is more favorable than theconformational change with Keq.^2, because the value of Keq-1,2 is 4-31. In the case of the 1,2-disubstitutedcompound, there are both diequatorial versus diaxial interactions as well as gauche versus anti interactions toconsider. The diequatorial orientation is better than the diaxial orientation, because with diaxial there areeclipsed interactions with the axial hydrogens. However, the anti orientation of the methyl groups is betterthan the gauche orientation.

Overall, diequatorial preference over diaxial is a more important factor than a preference for gauche over anti,so the value ofKgq.^2 is greater than 1. The 1,4-disubstituted compound hasno gauche-versus-anti interactionsbetween the methyl groups to consider, because thecarbons arefar apart. Thus, theconformational preference ispurely an effect of the diequatorial orientation being preferred over the diaxial orientation. This makes thevalue ofKeq-1,4 greater than the value ofKeq.^2, making choice D the best answer. Drawn below isone picture'sworth (approximately equal to 1000 words worth) of explanation.

Methyl groups are apart fromone another when diaxial.

CH,

Methyl groups are close to oneanother when diequatorial.

CH,

H3C

Methyl groups collide withaxial Hs when diaxial.

CH

Methyl groups do not collide withone another when diequatorial.

CHo

3 Anti > Gauche

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 1,4-equilibrium does.

Diaxial < DiequatorialCH

Diaxial < Diequatorial

No anti vs. gauche factor

1,4-diaxial orientation results in steric hindrance

from diaxial interactions with hydrogens, while1,4-diequatorial has no eclipsing steric hindrance.Equilibrium favors the products more than with 1,2.

4. Choice C is correct. In order forsubstituents to be gauche or anti to one another, theymust be bonded to carbonsthat are connected to one another. In the case of 1-4-dimethylcyclohexane, the methyl groups are not bonded toadjacentcarbons, so the two methyl groups cannot be gauche or anti to one another. This makes choice C correct.There arehydrogens on every carbon, soH can be gauche tomethyl. For theH to be anti to a methyl, the methylgroup must assume axial orientation. The only possible chair conformation of cis-l,4-dimethylcyclohexane hasone methyl group axial and the other in an equatorial orientation. This makes choices A, B, and D valid.

Copyright © by The Berkeley Review® 166 Section II Detailed Explanations

5.

6.

Choice Cis correct. Both of the chair conformations possible for cis-l,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 level of the product is equal to the energy level of the reactant. This makes choice Ccorrect. Do the correct thing and pick C. As apoint of interest, the value of Keq can never be less than orequal to0 (products and reactants alwayshave somepositive quantity), so choice A is an absurd answer.

H,C

H,C

Bothstructures have one methyl equatorial and the other methyl axial

Choice D is correct. With cis-l,3-dimethylcyclohexane, the cis-l,3-diaxial interactions (steric repulsion)between the two methyl groups makes the diaxial orientation less stable than the trans-l,2-diaxial orientation.This decrease in stability in the conformation drawn on the reactant side results in a greater value for Keq aswritten. This means that Keq-i,3 > Keq-1,2, which is choice D.

Methyl groups are apart fromone another when diaxial.

Methyl groups are close to oneanother when diequatorial.

Methyl groups collide withone another when diaxial.

Methyl groups do not collide withone another when diequatorial.

CHCH,

H3C

1,2-diaxial orientation results in anti orientation, while1,2-dieqatorial orientation results in gauche orientation.The equilibrium still favors product (the right), but not asmuch as the 1,3-diaxial-to-diequatorial equilibrium does.

CH CH, CHo

CH,

1,3-diaxial orientation results in increased sterichindrance. This has an effect on the equilibrium byshifting it heavily to the right. 1,3-diequatorialhas no eclipsing steric hindrance.

7. 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 axial for one and equatorial for the other. The structure is drawn below with the hydrogens noted.Choice B is the best answer.

Axial

EquatorialAxial

Equatorial -•H

Passage II (Questions 8 -14)

OH

Note that the two hydrogen atoms andthe hydroxyl group are all cis and up.

OH

Rotational Energy Diagrams

8. Choice C is correct. Because 240° is at the nadir (low point) of the graph in Figure 1, it correlates with thestaggeredstructure. This eliminateschoices A and B. Thecompound has R stereochemistry at carbon two whichmakes the correct answer choice C and eliminates choice D, which has S stereochemistry.

9. 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.


10. Choice A is correct. The 330° point on the graph is near (30° away from) the most stable conformation (whichhas anti orientation of the CH2OH group and the CH3 group of carbon 4). This eliminates choices C and D.Because of R stereochemistry, the correct choice is A. Stereochemistry is difficult to see in the Newmanprojection and can be seen more easily in other projections. Drawn below is a way to convert the Newmanprojection back to the stick-wedge drawing and a subsequent evaluation of stereochemistry.

11.

12.

13.

14.

P

CH2OH CH2OH CH2OH

Choice A is correct. Conditioning may cause you to respond automatically that the best orientation is the onewith the fewest repulsive interactions. This is often true, but it does not tell the entire story in this case. Themost stable orientation can also be the result of the strongest attractive interactions. Hydrogen-bonding betweenthe hydroxyl and amine groups occurs only from gauche orientation, where the two groups are close enough tobond. Hydrogen-bonding cannot occur between substituents with anti orientation. This makes choice A correct.

Choice D is correct. Hydrogen bonds have some acid-base character to them, so the most favorable protontransfer reaction is a good indicator of the strongest hydrogen bond. Because the amine is more basic than thehydroxyl group, the nitrogen is the lone pair donor. Likewise, the hydroxyl is more acidic than the amine, sothe hydroxyl is the hydrogen donor. This makes choice D correct.

Choice D is correct. Because of the planar nature of electron density in a Tt-bond, rotation about a double bondrequires that the jc-bortdbe broken. This is not observed under thermal conditions. To convert a cis 71-bond into atrans 7i-bond, UV light is needed. Pick D to tally big points. The drawing below shows that a 90° rotation aboutthe C—C breaks the 7t-bond. It requires substantial energy to break a rc-bond.

71-bond broken 71-bond

90° rotation

Choice D is correct. The onlysingle bond aboutwhich rotation isnot possible is a single bond between two atomsin a cyclic compound. Theonlycyclic compound of the answer choices is methylcyclopentane, choice D. The bestanswer is therefore choice D. Oneitem ofnotable interest is that both 71-bonds and rings lower the entropy of acompound by lowering its degrees of freedom (i.e., its ability to rotate freely).

Passage III (Questions 15 - 22) Deuterated Cyclohexane

15. Choice B is correct. The most stable form of cyclohexane is the chair conformation (as opposed to the boatconformation), with the smallest substituents (determined by bond length) in the axial position. The integralfrom the proton NMR shows that the ratio of peaks for :H is 1.12 : 1 in favor of the 8 = 1.25 ppm shift, the shiftdue to the equatorial hydrogen. This indicates that more hydrogens are located in the equatorial position (5 =1.25 ppm) than the axial orientation (8 = 1.51 ppm). Consequently, the preferred conformation has deuterium inthe axial position. For this reason, choose B. Drawn below are the two chair structures and their equilibrium.

8 = 1.51 ppm

8 = 1.25 ppm

6 axial H : 5 equatorial H (*HNMR integral 1.2 : 1) 5 axial H : 6 equatorial H (]HNMR integral 1 : 1.2)Actual ratio is 1.12 : 1, which shows H prefers the equatorial position, so the right structure is more stable.


16.

17.

18.

19.

20.

21.

Choice C is correct. Using D2 gas rather than H2 gas would have produced C6H5D7. The addition of the D2 isin 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 _ D

HUH

D H

3 axialH : 2 equatorial H 2 axialH : 3 equatorialH

H prefers theequatorial position, confirming that thestructure on therightismore stable. The Keq for this ring flip is >1. The ratio is 2axial: 3equatorial.

This question could have also been answered without knowing the exact chemistry. The passage infers thathydrogen prefers the equatorial position over deuterium. This means that in the most stable orientation, thereare more hydrogens with equatorial orientation than axial orientation. This eliminates choice D. Becausethere are only five hydrogens present (you start with five on deuterobenzene), the sum ofthe two numbers mustbe five. This eliminates choices A and B and makes choice C the best answer. Learn to answer questions asquickly asyoucan, whether youuseorganic chemistry knowledge orcommon sense.

Choice D is correct. The addition ofH2 gas to chlorobenzene results in the hydrogenation of the benzene ringand the formation of CHnCl (chlorocyclohexane). For chlorocyclohexane, there are two possible stableconformational isomers, one having a ring structure in a chair conformation with the chlorine equatorial(choice D) and one having a ring structure in achair conformation with the chlorine axial (choice C). Chlorineis larger than hydrogen, so itprefers to be in the equatorial orientation. Choice Dis more stable so choose it.

Choice Biscorrect. The hybridization of carbon remains sp3, whether it is bonded to deuterium orto 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°, whichare both nearest to 109.5° ofall thechoices. 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 assumeidentical orientation in terms of axial or equatorial. This means either choice A or choice D is thebest answer.Equatorial orientation is more stable than axial orientation, so choice Ais the best answer. The conformation isshown below:

CH,

Choice B is correct. Like benzene, C6H5D has three 71-bonds and one ring. The best answer ischoice B. Ifyourecall the formula for units of unsaturation, (2(#C) - (#H) + 2)/2, you can calculate the units of unsaturation,knowing thatD behaves thesame as H. This would also yield a total of4.

Choice Dis correct. Ahydroxyl group, OH, is larger than ahydrogen, so choice Ashould be eliminated. Theaxial orientation is more hindered in terms ofsterics, sochoice Bis invalid. Choice C is a good explanation forwhy the two hydroxyl groups would be found in adiequatorial orientation, but the question is looking for anexplanation for why diaxial, rather than diequatorial, is the preferred conformation. This eliminates choiceC. 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 theability to form hydrogen bonds. Thismakes choice D the best answer.


22. 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.

Passage IV (Questions 23 - 29) Infrared Spectroscopy

23. Choice Discorrect. An IR absorbance between 1700 cm"1 and 1740 cm"1 isthe result ofacarbonyl (C=0) group, asstated in the passage. Of the choices, only the ether doesn't contain a carbonyl group, so it is the ether thatdoes not have an IR absorbance between 1700 cm"1and 1740 cm"1. Pick D for best results.

Choice C is correct. A straight-chain monosaccharide has O-H groups, a C=0 functionality and C-H bonds.These groups have IR absorbances for bond stretching of3300 cm'1,1715 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 isfound around 2300 cm'1, sothat should lead you toanswer choice C.

Choice Ais correct. Abroad signal near 2850 cm"1 indicates the hydroxyl group ofa carboxyhc acid (O-H withhydrogen-bonding). A hydroxyl group exhibits hydrogen-bonding, so the signal is broad. Because the O—Hbond ofa carboxylic acid is weak, itsabsorbance is lower than thatofstandard hydroxyl groups. Oftheanswerchoices, only choice A has a carboxyhc acid functionality, letalone a hydroxyl group. Pick A for happiness.

26. Choice Dis correct. The region between 7.0 and 8.0 ppm inthe proton NMR can beattributed tohydrogens onanaromatic ring. This immediately eliminates choices A and B, which have no aromatic rings associated withthem. The difference between ethylphenol and methylphenol can be demonstrated by either the total numberofhydrogens (as shown in the integration) orby the coupling ofthe alkyl portion ofthe compounds. The ethylgroup exhibits a 2H quarter and a 3H triplet, while the methyl group exhibits a 3H singlet. The extra CH2group associated with theethylphenol compared to methylphenol is found in the2.0 to 2.5 ppm region, not thearomatic region. This eUminates choice C. To decide between the twocompounds in choice D, onecanlookatthe coupling of the hydrogens on the benzene ring, found in the range between 7.0 ppm and 8.0 ppm. Thesplitting for the hydrogens ona benzene ring with para-substitution is symmetric, while the splitting for thehydrogens on a benzene ring with meta-substitution is asymmetric. This is the distinguishing factor betweenthe para-substituted and meta-substituted phenolsin choice D. Thebest choice is answer D.

27. Choice A is correct. Because the broadness of hydroxyl peaks is associated with hydrogen-bonding, thedecreased hydrogen-bonding of a tertiary alcohol is reflected as a narrower absorbance in the IR spectrum ascompared to a primary alcohol. The alcohol peak is found well above3000 cm"1, so the best answer is choice A.

Choice D is correct. An IR absorbance at1715 cm"1 is indicative of a carbonyl group (the stretching of a C=0bond). Aldehydes and ketones contain a carbonyl group, so choices A, B, and Care eliminated. Only the ether(tetrahydrofuran) does not have a carbonyl functionality. Pick Dand bask in the glow ofcorrectness.

24.

25.

28.

29. Choice C is correct. The hydrolysis of an ester results in the formation ofa carboxylic acid and an alcohol.Both the ester and the carboxylic acid have carbonyl groups, so each has an IRabsorbance around 1700 cm"1.This means that both before and after hydrolysis, there is a signal around 1700 cm"1, eliminating choices AandB. An ester has no hydroxyl group, so initially there is no signal around 3300 cm"1. Following hydrolysis, bothan alcohol and carboxylic acid are formed, so a signal for the hydroxyl group appears. In particular, thehydroxyl group of an alcohol shows an absorbance around 3300 cm"1. This means that during the course ofthehydrolysis ofanester, an IR signal appears around 3300 cm"1, making the best answer choice C.

Passage V (Questions 30 - 36) Unknown Alkene Determination

30. Choice A is correct. Because a ring takes one degree of unsaturation and a carbonyl takes one degree ofunsaturation, a cyclic ketone has two degrees of unsaturation. The compound has only one degree ofunsaturation, soit cannot bea cyclic ketone. The correct answer ischoice A. Acyclic ether hasone ring andnon-bonds, so ithas one degree of unsaturation. This eliminates choice B. Acarboxyhc acid has norings and one n-bond, soithas one degree ofunsaturation. This eliminates choice C. An ester has norings and one 7t-bond, so ithas one degree of unsaturation. This eliminates choice D.


31. Choice B is correct. Ozonolysis cleaves tiie double bond of an alkene, converting it into two carbonyl compounds(either an aldehyde or a ketone). In this reaction, the products formed are aldehydes, as mentioned in thepassage. Aldehydes can be identified by their peak around 9.7 ppm in the ^HNMR. Choose Bfor best results.

32. Choice A is correct. Compound B, like Compound A, is an aldehyde. Like Compound A, it shows an IRabsorbance near 1725 cm"1 for thestretching mode of theC=0 bond. This makes choice A correct.

Choice A is correct. Compound A has one degree of unsaturation (due to the carbonyl), and Compound Bhas twodegrees of unsaturation (one due to the carbonyl and the otherdue to a ring). Thesecond degree of unsaturationhas to be from a ring, because there can be no double bond in the product from ozonolysis (had there been a it-bond, the ozone would have reacted with it.) This means that tiie original alkyne must have had a ring and atriple 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 C9H20 For every unit of unsaturation, two hydrogens are removed from theformula, so a hydrocarbon with three units of unsaturation must have the formula C9H14, choice A. You canverify this with the molecular mass, which is roughly 122 g/mole, as stated in the passage.

Choice D is correct. An alkene is distinguishable by an IR absorbance between 1620 cm"1 and 1660 cm"1 for thestretching of the C=C bond. You should know that acarbonyl C=0 bond absorbs at around 1700 cm"1, and beingthat a C=C bond is slightly weaker than a C=0 bond (they are both double bonds, but a carbonyl bond isslightly shorter), it takes slightly less energy to stretch the C=C bond than the C=0 bond. This means that aC=C bond has an absorbance slightly less than 1700 cm"1. Only choice Dis less than 1700 cm"1, so choice Disthe best answer. The test requires that you have some IR peaks in your memory; but for ones you don't recall,estimate them by comparison to the values you know. Choose Din this question for the feeling ofcorrectitude.

Choice Bis correct. The products from the ozonolysis ofthe alkene are both aldehydes, soCompound Amust bean aldehyde, which makes choice Bthe best answer. All of the choices have the correct units of unsaturation,so the best answer must be determined from the presence of the aldehyde proton.

Choice C is correct. As stated in the passage, Compound Bmust be an aldehyde, which eliminates choice A.The compound cannot contain a double bond (the product from ozonolysis cannot have a double bond betweencarbons), so choice Bis eliminated. The 1HNMR integration shows a peak ratio of 1:1:4:4, which indicatesthat there are no methyl groups in the compound. Amethyl group would have shown a relative ratio in theintegral of 3. This eliminates choice D. Choice Dcan also be eliminated, because itcontains 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 multiple-choice elimination process rather than structural deduction,because in a multiple-choice format, eliminating invalid structures is faster than elucidating the correct one.

33.

34.

35.

36.

Structure ElucidationPassage VI (Questions 37- 43)

37.

38.

39.

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 at3350 cm"1, the compound has H-bonding, so itmust be an alcohol. The correct answer is thereforechoice D. Choices Band C should be eliminated, because the passage states there is no carbonyl. Choice A canbe eliminated, because ethers exhibit no hydrogen-bonding.

Choice C is correct. An absorbance at 2116 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 value for an alkyne knowing thata C-C bond 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 bond strength increases (or bond length decreases), the IRshift valueincreases. Choices C is the bestanswer. The answer is given at the end of the third paragraph in the passage.

Choice A is correct. A doublet in the 13CNMR is due to a carbon with one hydrogen attached to it. Becauseaketone carbon and an internal alkyne carbon do not have hydrogens attached to them, they would be singletsrather than doublets. This eliminates both choice B and choice D. There is no peak in the IR spectrum,indicating that there is a C=0 present (no peak around 1700 cm"1), so choice Cis eliminated. The only choiceleft is choice A, the best choice.

.©Copyright © by The Berkeley Review 171 Section II Detailed Explanations

40. Choice C is correct. If a compound hasno symmetry, then there is a 13CNMR peak for every carbon in thecompound. Tiie number ofcarbons in a compound exhibiting seven 13CNMR peaks is therefore seven. Pick Cforbest results.

41.

42.

43.

Choice D is correct. When hydrogens are coupled to one another, they have the same coupling constants (J-values). This makes choice D the best answer. Two different hydrogen groups have different shift values, sochoice A is eliminated. Choices B and C are the same answer, so they should both be eliminated. This leavesonly choice D as the best answer.

Choice A is correct. From theNMR data, we know that Compound Thas three carbon atoms and fourhydrogenatoms, so the structure must containonly three carbons. Choices B, C, and D all have four carbons, so they areeliminated without involving the spectroscopic data. The only choice that fits the formula is choice A. Oncethis question is resolved, it can help to solve some of the otherquestions. This happens oftenon the MCAT.

Choice Bis correct. The 13CNMR peak at50 ppm is due to a carbon, not ahydrogen. This eliminates choices Cand D. Acarbonyl shows a13CNMR peak around 200 ppm, so choice Ais eliminated. Only choice Bremains, soyou should pick it. The absence of an IR peak around 1700 cm"1 should reaffirm that a carbonyl is notpresent inCompound T. This means that choice B is the best answer.


44.

46.

NMR and IR Spectroscopy

Choice Bis correct. The total number ofhydrogens in the compound, according to the NMR is eight, ofwhichfive are on the benzene ring. This eliminates choice C. The passage states that the compound is an ester, sochoice A, a ketone, iseliminated. The best answer is choice B, because the methyl shift is around 4.0 ppm. Thisis tough to know without an NMR chart. In the passage, it is mentioned that the alkoxy group has a shift inthe 1HNMR between 3.5 and 4.0 ppm. If you can't recall afact, search for it in the passage.

45. Choice C is correct. The three separate !HNMR spectra are to be compared to determine the compounds insolution. It is important that nothing varies between samples, so the same solventshould be used in each case.Any impurity peak would be common to all of the spectra and thus could be eliminated. This makes choice Cthe best answer. Acommon solvent has no effect on the pH or the behavior ofa protic species in the 1HNMR.Although !HNMR invokes an external magnetic field, the solvent has no bearing on the field.

Choice Bis correct. This question is typical ofspectra-to-structure-to-name questions. There arefour carbons inCompound III, so choices A and C are eliminated, because they have only three carbons, according to thenomenclature. Translating thenames ofchoices Band Dyields the following twostructures:

O

XH3CH2C OCH

Methyl propanoate

O

OCH2CH3

Ethyl ethanoate

Both Aand C would have identical peaks of a 2H quartet, 3H triplet, and a 3H singlet. The key feature is the3H singlet at approximately 3.5 ppm. This indicates that the methyl group is attached to the ester oxygen asin choice B. Youshould pick B to score.

47. Choice A is correct. The four answer choices are the result of the following bonds: O-H (3500 cm"1), sp3C-H(2980 cm"1), C=0 (1685 cm"1), and C-O (1300 cm"1). An ester contains an s/?3"C-H bond, a C=0 bond, and a C-Obond, but it does not contain an O-H bond. This means that there will be no O-H stretch (thus no IR peak atornear 3500 cm"1) associated with the IR spectra of an ester. Choose Aand be a stellar achiever at choosingcorrect answers.

48. Choice Biscorrect. Compound II has two oxygens in its formula, so choices Aand Care outimmediately. Thequestion here is reduced to placing the alkyl group either on the oxygen (as in choice B) or on the carbonylcarbon (as in choice D). Because the peak for the two hydrogens is around 4.0 ppm, the ethyl group must beattached to the oxygen, making choice Bthe best answer. You should note that when there is a 2H quartet anda 3H triplet in the spectra, there is an isolated ethyl group in the molecule.


49. 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 dueto the presence of hydrogens on the neighboring carbon.

50. Choice C is correct. An NMR solvent should be inert and show no peaks in the spectra. All of the compounds aredeuterated, (contains 2H rather than *H) so they show no peaks in the 1HNMR spectrum. The most reactivesolvent listed is the alcohol. 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 estershould 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.

PassageVIII (Questions 51- 57) NMR Data Table

51.

52.

54.

Choice C is correct. 2-methyl-3-pentanone has four nonequivalent hydrogens, according to the symmetryof themolecule. The structure of 2-methyl-3-pentanone is drawn below:

a

H3C

blHC

a S \H3C C

CH2. d

CH

0

6 a hydrogens1 b hydrogen2 c hydrogens3 d hydrogens

This results in a ratio of 6:3:2:1 for the four nonequivalent hydrogens, which makeschoice 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 ofCH 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.

53. 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 Aand Bshouldbe eliminated, because both pentanal and 2-pentanone have a peak in the IR just above 1700 cm"1 for the C=0bond and a peak inthe proton NMR between 2.0 and 2.3 ppm for the alpha hydrogens. Neither compound has aproton NMR peak between 3.5 ppm and 4.0 ppm, because that shift value is attributed to hydrogens on acarbonbonded to oxygen. Only choice D, a peak near 9.7 ppm, is for a peak that is unique to the aldehyde, so thepresence of the peak confirms that the compound is pentanal, while the absence of the peak supports that thecompound is 2-pentanone.

Choice Bis correct. An alcohol hydrogen ismildly acidic. This means that when a base isadded to analcohol,the protic Hon oxygen can be removed. If the alkoxide formed is placed in deuterium-labeled water, then thealkoxide can remove deuterium from water to form a deuterium-labeled alcohol. Deuteriums do not appear inthe!HNMR, so the peakfor the H on oxygen disappears, making choice B the bestanswer.

D- O,

\H

R- O.

D

©RO

©+ DO

55. 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 isasinglet between 2.0 and 2.5 ppm. Pick A.


56.

57.

Choice Bis correct. A1HNMR peak at 9.7 ppm indicates that the compound is analdehyde. This eliminateschoice A (an ether) and choice D (a ketone). The ratio of 1:1:6 for the area of the signals indicates that thereare two equivalent methylgroups (thataccounts forsix equivalent hydrogens) and two unique hydrogenson thecompound. Only choice B has two equivalent methyl groups, so choice B is the best answer. There are fourunique types ofhydrogens onbutanal, so butanal would show four signals inits THNMR, notjust three signals.

Choice Dis correct. Having no IR absorbance between 1600 cm"1 and 1750 cm"1 indicates that the compound hasneither a C=C bond nor a C=0 bond in its structure, both of which have IRsignals between 1600 cm"* and 1750cm"1. This eliminates all of the choices except choice D. The compound must be cyclic to account for the onedegree of unsaturation.


58.

59.

60.

Alkene-Coupling Experiment

Choice C is correct. In Spectrum I, the distance between the peaks in the alkene region (coupling constant) isgreater than it is in Spectrum II. A larger coupling between vinylic hydrogens is attributed to the transcompound. This is choice C.

Choice D is correct. The methyl group has three hydrogens, so the signal for the methyl group on benzenecannot be the peak at either 7.25 ppm and 5.37 ppm, because neither of those peaks contains three hydrogens.This eliminates choices A and B. The alkoxy methyl group on the oxygen is found farther downfield than themethyl on benzene. Thus from the values on the spectra, the methyl on benzene (3H singlet) comes at 2.25 ppm.,which is choice D.

Choice C is correct. In order for the disubstituted benzene compound tohave the same number ofpeaks in thecarbon NMR asit has carbons, the compound must not have symmetry. Symmetry results inequivalent carbons,resulting in fewer 13CNMR signals. The only asymmetric compound among the answer choices is the metasubstituted benzene with two different substituents. Choice C is correct. The choices are drawn below:

6 Carbons

313CNMR peaks6 Carbons

413CNMR peaks

C.

c,

c.

X

I-ca

"c" Y

6 Carbons

613CNMR peaks

D. >

.('

C.

C

6 Carbons

413CNMR peaks

61. Choice B is correct. The 1H peaks (the vinylic hydrogens) in the spectrum fall between 4.88 and 5.37 ppm,implying that the hydrogens on the alkene are found in this range. Hydrogens on an alkene are referred to asvinylic hydrogens. This means that vinylic hydrogens are found inthe range of 4.5 ppm to 7.0 ppm. This makeschoice B the best choice.

62.

63.

64.

Choice D is correct. The compound contains a C=C doublebond, whichhas an IRabsorbancebetween 1620 cm"1and 1660 cm"1. The best choice is 1640 cm"1, choice D. An IR absorbance of 3500 cm"1 isdue toa O-H bond, anIRabsorbanceof 2220 cm"1 is due to a OC bond, and an IRabsorbance of1720 cm"1 is due to a C=0 bond.

Choice D is correct. With meta-substitution, all of the benzylic hydrogens would be different, because thebenzene compound would be asymmetric. There would thus be four nonequivalent hydrogens in a 1:1:1:1 ratio.Picking D is a beautiful thing on this question.

Choice D is correct. An ethyl group on benzene iscomposed ofa CH3 group next to two equivalent hydrogens(making it a 3H triplet) and a CH2 group next to three equivalent hydrogens (making it a 2H quartet). Thisdescribes choice D.


Passage X (Questions 65 - 71) Unknown Compound Identification

65.

66.

67.

68.

69.

70.

71.

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 groupsattached to that carbon. If the other group was carbon-based, there would be more signals in the 'HNMRspectrum 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 asinglet). 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=0 bond is found at 170.2 ppm., which eliminates choice C. The bestanswer is choice D. Carbons bonded to an electronegative atom that are sp3-hybridized typically show asignal 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 170 and 220 ppm in the 13CNMR. Choice Aisanester, choice Bis 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 13CNMR 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. Tiie 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 ashydrogens ona carbon adjacent to acarbonyl (C=0) bond. This ischoice A; your choice for a question like this.

Choice D is correct. Asinglet in a XHNMR 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 isstated in the passage. Choice Bis 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 methylgroup is adjacent to the carbonyl. The correct structure is choice D.

Passage XI (Questions 72 - 78) Carbon-13 NMR

72. Choice C is correct. The oxidation of a secondary alcohol to a ketone results in a carbon that goes from a13CNMR signal of roughly 75 ppm to a signal of roughly 200 ppm. This conversion can bemonitored easily by13CNMR, so choice A is valid. Nitration of ethylbenzene results in a carbon that changes from a 13CNMRsignal of roughly 115 ppm to a signal of roughly 140 ppm, because the nitro group changes the immediatechemical environment. This conversion canbe monitored by 13CNMR, so choice Bis valid. Deprotonation ofacarboxylic acid does notchange the immediate environment (adjacent atoms) ofany carbon sono drastic changein change in any 13CNMR signal is observed. This conversion cannot be monitored by 13CNMR, so choice Cisinvalid, 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 by13CNMR, so choice D is valid.

73. Choice C is correct. According to the reaction in the passage, the compound is an aromatic ketone. The peak at206 ppm, according to the data in Table 1, confirms that Isomer I is a ketone. Choice D (the aldehyde) is thuseliminated. Because there are only seven peaks for the 13CNMR, the compound must have symmetry, sochoices A and Bare eliminated. The best answer (one that contains only seven unique carbons) is choice C.


74. Choice A is correct. An aldehyde, according to Table 1,is found between 185 and 200 ppm, and no peak is foundin this range for either structural isomer. Choice A may be the correct choice. It stated in the passage that onoccasion, the shift may not fall exactly in the range given. An alkene, according to Table 1, is found between 110and 140 ppm, and there are peaks found in this range for both structural isomers. Choice B can thus beeliminated. A ketone, according to Table 1,is found between 205 and 220 ppm, and peaks are found in this rangeforbothcompounds. Choice C is thus eliminated. A methyl group, according to Table 1, is found between 10 and35 ppm, and there are peaks found in this range for both structural isomers. Choice D is thus eliminated. Thebest answer is choice A.

75.

76.

77.

78.

Choice D is correct. According to the passage, theleast intense peakis caused by a carbon with no hydrogensattached. This makes choice D the best answer. The carbons with no hydrogens attached in a disubstitutedbenzene derivative with a carbonyl are two of the benzene carbons (the benzene carbons with a substituentattached) and the ketone carbon. It has sp2-hybridization, not sp3-hybridization, and it is achiral. Thiseliminates choices Band C. While it has sp2-hybridization, that is not the cause of its low intensity. ChoiceD is a better answer than choice A. This is a question that rewarded the test taker who sifted through theinformation in the passage. This will happen on occasion, because even the science sections are reading exams.

Choice C is correct. Para-methoxybenzaldehyde has a total of eight carbons in its structure, but the moleculecontains a mirror plane that reflects two pairs of equivalent carbons. Using symmetry, this means that thereare only six unique carbons, so the best answer is six signals,choiceC. The structure is shown below:

O^

,H

C°

CF

I fOCH

Choice Dis correct. Because the two compounds are isomers, they have the same molecular formula. Havingthe same molecular formula results in having the same units of unsaturation. This eliminates choices A and B.Each compound has a benzene ring and a carbonyl group. The benzene ring has three 7t-bonds and the one ring.There are four units of unsaturation due to the benzene ring alone, so choice Cis eliminated. When the carbonylrc-bond is accounted for, there are five units of unsaturation in the molecule, so the best answer is choice D.

Choice Bis correct. It is stated in the second paragraph of the passage that quaternary and carbonyl carbonsgenerate peaks oflow intensity. There isno carbonyl in the compound, but carbon 2 of2,2-dimethylbutane hasfourothercarbons attached, making it a quaternary carbon. The bestanswer is choice B, carbon 2.


79.

80.

Distinguishing IsomersUsing*HNMR

Choice A is correct. Spectrum I contains a quartet (2H), a triplet (3H) (this combination is a dead give-awayfor an isolated ethyl group), and a singlet (3H) (a dead give-away for an isolated methyl group). The bigquestion here is whether the ethyl group or methyl group is attached directly to the oxygen of the ester group.The degrees of unsaturation (1) and number ofoxygens (2) tell you the compound mustbe either an ester or acarboxylic acid. The lack ofa broad peak between 10 -12 ppm eliminates the possibility of the compound beinga carboxylic acid, so the compound must be an ester. Because the quartet is so far downfield (at a higher shiftvalue), the ethyl group is attached to the oxygen. This makes the best choice an ethyl group on the methylester, whose common name is ethyl acetate. Choose A for best results.

Choice Bis correct. Atriplet is the result ofcoupling to tiie neighboring hydrogens (there aretwo hydrogens onthe adjacent carbon in the case ofa triplet). The integral (quantity ofhydrogens for the signal) hasno effect onthe shape of the signal, meaning that the peak shape does not tell you any information about the hydrogens ofthe signal, only about the neighboring hydrogens. This eliminates choices A and C. A triplet is therefore theresult ofneighboring a CH2 group. Choose Bfor a grade A, genuine, altogether correct, bestanswer.


81. 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 singlet (3H) (a dead give-away for an isolated methyl group). Because the3H singlet is found around 4.0 ppm, the methyl group is attached to the oxygen. This makes the best choice amethyl group on an ethyl ester. Choose Bfor the happiness ofanother correct answer. The drawing below listshow the name and structure are determined for the compound.

Singlet in the 3.5 to 4.0 ppm rangeO

/O CH2CH3

j

methyl propanoate

Nomenclature rules state that the alkyl group on oxygen is named first, followed by the ester chain. Thismakes this compound methyl propanoate.

82. Choice C is correct. This is one of those trivial facts that you should know. A peak in the neighborhood of 7ppm is indicative of aromatic hydrogens, which are found in benzene compounds. Pick C, to score big!

83. Choice Ais correct. Acarboxylic acid has one proton that forms abroad peak between 10 ppm and 12 ppm inthe2HNMR. The hydrogen inquestion 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 Dareeliminated. An ester has an 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 are present in both an ester and acarboxylic acid. This means that a peak between 2.0 ppm and 2.5 ppmcannot be used todistinguish anester from a carboxylic acid. The correct answer ischoice A.

84. Choice Bis correct. 4-Heptanone is a seven-carbon ketone with the carbonyl directly in the middle. Thestructure is symmetric, so there are many equivalent carbons and hydrogens. There are four unique carbons, ofwhich three contain hydrogens. This results in three signals in the !HNMR for 4-heptanone. The best answeris choice B.

Passage XIII (Questions 85- 90) Structure Elucidation Using *HNMR and 13CNMR

85. Choice Aiscorrect. For Compound II, the absence of a13CNMR signal between 180 ppm and 230 ppm supportsthe idea that it has no C=0 group. This eliminates choice C. No peak above 5ppm in the ]HNMR spectrumconfirms that there is no double bond. This eliminates choice D. This means that the unit ofunsaturation in thecompound must be the result of aring. The presence of abroad peak in the ]HNMR spectrum supports the ideathat the compound is an alcohol, eliminating choice Band making choice Athe best answer. The CNMRspectrum shows that there is great symmetry in the structure. The choices are either cyclopentanol or 2,3-dimethylcyclopropanol. The integral of the proton NMR says that there are mostly CH2 groups present,which favors cyclopentanol over 2,3-dimethylcyclopropanol. Cyclopentanol is drawn below:

Broad

(1H)

H

-HO H-*-

\/

CHn

H2C c:

Triplet (4H)


Multiplet(1H) 13

CNMR:

There is a mirror planecutting through themolecule, so there are three unique carbons,resulting in three different signals.

Multiplet(4H)

177 Section II Detailed Explanations

86. Choice Dis correct. Each unique carbon within amolecule exhibits a unique signal in the 13CNMR spectrum, sochoice A is valid. This eliminates choice A. A carbonyl carbon has a signal around 180 ppm in a 13CNMRspectrum, which is different from other signals in the 13CNMR spectrum. The presence ofa carbonyl group canbe identified using 13CNMR spectroscopy, so choice B is valid and thus ehminated. 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 double bond, cisversus trans, does notexpress itself in 13CNMR spectroscopy. Typically, geometry is determined by looking atthe coupling constants ofthe vinylic hydrogens in^^HNMR spectroscopy. Choice Dcannot bedetermined using13CNMR spectroscopy, sochoice Dis the best answer.

87. Choice C is correct. According to the data in the passage, only Compound I has a UV absorbance above 175nm.This means that only Compound I has a 7t-bond. There is only one unit of unsaturation, so Compound I can have,at most, one Jt-bond. This means that Compound I cannot be a conjugated diene, which eliminates choice A. If itwere a conjugated diene, there would be a UVabsorbance above200nm for the n to n* transition. Compounds IIand III cannot be carbonyl compounds, according to their UV data. If there was a carbonyl group on themolecule, there would be a TC-to-Tt* transition around 190 nm and an n-to-7t* transition around 260 nm. Thiseliminates choice B. Because Compound I has two UV absorbances and only one 7t-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 in show no UV absorbance above 175nm,there is no 7t-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.

88. Choice D is correct. For Compound III, the absence ofa broad peak in its1HNMR spectrum confirms that thereis no alcohol in the compound. This makes choice A a valid statement, and thus the incorrect choice. Theabsence ofapeak around 180+ ppm inits 13CNMR spectrum confirms there isnoC=0present. 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 7t-bond in 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 13CNMRshows very few signals (only three), which implies that there is great symmetry in the structure. It must be asymmetric cyclic ether. This makes choice Dan invalid statement, making it the best answer.

89. Choice B is correct. Key features from each spectrum must be extracted. From the molecular formula, we knowthereis one unit of unsaturation and oneoxygen. This means that the compound must contain either a ring, aC=C bond, or a C=0 bond. All of the answer choices fit these criteria, so we must use the spectroscopic data.The 13CNMR 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 13CNMR due to its symmetry, which eliminates choice C. The fact thatno twocarbons are alike also eliminates choice D,which has twoequivalent methyl groups. The1HNMR datashows a 3Hsinglet at 2.08 ppm,and no peak between 9and10ppm. This means that thecompound is a methylketone and not an aldehyde, which eliminates choice A and choice C. The ratioof the hydrogen signals in the1HNMR (3:2:2:3) supports choice B. The structure isshown below:

o

IIcr*-

UV-vis: 189nmand268nm13,

H3CA

CH2A \ / \

CH

'CNMR: 211 ppm

CH

2HNMR: Triplet MultipletTriplet(3H) (2H) (2H)

Singlet(3H)

13CNMR: All five carbons are different, so there are five different signals.

There can be no branching, because the protonNMR shows that the integral values are 3, 2, 2, 3. This impliesthat the structure is linear.


90. Choice D is correct. Integration is used to determine the relative 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.

Proton NMR of an UnknownPassage XIV (Questions 91 - 97)

91. 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 methylgroup. Thisnow becomesa nomenclature question, rather than a spectroscopy question. The correctname for a methyl groupattached to benzene is methylbenzene. The common namefor methylbenzene is toluene. Choose C for optimalresults.

92. Choice D is correct. The formula contains six hydrogens in all, so the sum of the ratio values must equal 6. Tliefirst peak is shortest, the middle peak is the second tallest, and the last peak is the tallest. This means thatthe values must be ascending. The only combination ofascending values adding to 6 is 1:2:3. Choice D is tliebest answer.

93.

94.

95.

96.

Choice D is correct. The compound 2-bromopropane has two unique types ofhydrogens, so ithas two peaks inits!HNMR 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 ithas a signal with an integration of 1. Tliepeak 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 doublet of integration 6. The one hydrogen has sixhydrogen neighbors, so there is a septet ofintegration 1. Choice D is the best answer.

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 Cisthe only structure that shows no quartet initsproton NMR spectrum.

A. O R O

x(H3C)2HC CH^IH^ H3<:H2C(H3C)2C CH3

c. 0

X(H3C)2HCH2C CH3

D. o

X fi-H3CH2CH2C CH2CH3

Choice D is correct. A doublet is the result of hydrogens on a carbon that neighbors a carbon with only onehydrogen 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 (atertiary 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. Apositive iodoform test, as stated in the question, is caused by a compound with threealpha 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 isa singlet. Pick Afor 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.

.©Copyright © by The Berkeley Review 179 Section II Detailed Explanations

97. Choice D is correct. The hydrogen responsible for the broadness is the carboxylic acid proton. It is bonded to anoxygen, making choice A an incorrect answer. The coupling to eight or more other hydrogens would result in amultiplet made of many sharp peaks, not a broadened signal, which eliminates choice B. Choice C can beeliminated, because hydrogens on carbon do not form hydrogen bonds. Tliehydrogen forming the hydrogen bondin this molecule is bonded to oxygen. This makes choice D the best answer.

Questions 98 -100 Not Based on a Descriptive Passage

98. Choice B is correct. Because the compound on the left has cis orientation about the internal double bond, whilethe compound on the right has trans orientation about the internal double bond, the two compounds must begeometrical isomers of one another. Pick choice B, and you won'tbe sorry. In caseyou were considering choiceC, the two compounds have the same absolute configuration, so they cannot be optical isomers.

99. 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 chirality.This makes the two structures optical isomers, choice C. If you flip thestructure on the right to align with thestructure on the left, it can be seen that the chiral center has changed. A change in a chiral center results in anoptical isomer.

100.

\ /

> ?>*' *S**

Choice A is correct. The best way to do this problem is the systematic counting ofcarbon backbones, startingwith the longest carbon chain possible (five carbons). The tally for each possible carbon backbone is drawnbelow. There are only three possible structures: pentane, 2-methylbutane, and 2,2-dimethylpropane. Twostructures that are in fact structural isomers must have different IUPAC names. Pick A to be a correct answerpicker person.

C

c- c- e c- c c- e c- c c- c- c

c

"Chemistry, it's a beautiful thing!"


Section III

Stereochemistryby Todd Bennett

Molecule fits template!Reenter if •>•>•>•

Molecule does not fit template!ScenterifB>T>»>Q

Configurational Isomers

a) Stereochemistryb) Asymmetryc) Chiralityd) Determining Absolute

ConfigurationChiral Centers

ii. Prioritizing Substituentsiii. R- ans S- Designations

e) Shortcuts for R or S determination

f) Optical rotationi. Equation for specific rotation

g) Types of configurational isomersi. Enantiomers

ii. Diastereomers

h) Meso compoundsi) Stereoisomerization

Stereochemistry in Reactions

a) Stereoisomer formationi. Stereoselectivityii. Racemic Mixtures

iii. Enantiomeric Excess

Nucleophilic Substitution

a) Terminologyi. nucleophileii. Electrophileiii. Leaving Group

b) Sn2 Reactionc) Sfll Reactiond) Distinguishing Mechanismse) Reaction Kineticsf) Physical Propertiesg) Separating Stereoisomers

BERKELEYJJr-E-V-KE'W®


Stereochemistry Section Goals

•*Be able to identify stereocenters and chiral compounds.A stereogenic center (oftenreferred to as a chiralcenter)is most commonlymade up of a centralatom (usuallycarbon) with four unique substitutentsattached. The stereocenter is identified aseitherR (Latin, rectus) or S (Latin, sinister) to define its orientation in space. Questions will requirethat you identify the number of chiral centers and often label them according to convention. Thefrequently cited example ofanon-typical situation involves allene, which hassp-hybridized carbonsthatcan bechiral. Anexample ofallene enantiomers isdrawn below:

H H

CI'"/#.. / « H»"'.... /H^ \ C,^ —°—\

CI CI

Be able to recognize and classify stereoisomers.You must understand the differences between enantiomers and diastereomers. You should be abletoidentify meso compounds from their optical inactivity andstructure. You should know specialcases involving sugars suchasanomers andepimers. Most importantly, whengiventwostructures,be able to identify their relationship if they are stereoisomers of one another.

Be familiar with common biological examples of chiral molecules.Itshould beassecond nature toyou that sugars occur naturally intheDform, which isdefined byhaving thepenultimate carbon withRstereochemistry. The typical exception is seen withbloodtypes where one of the sugars in the antigenicdeterminant is L-fucose. It should be as second naturetoyou thatamino acids occur naturally mthe Lform, which isdefined byhaving thealpha carbonwith Sstereochemistry (except for cysteine). Atypical exception isseen with transcriptidase enzymeswhere the active amino add is D-alanine.

Be able to use optical rotation data to identify an unknown compound.Just like the boiling point and the melting point, the optical rotation isaphysical property that canbeused to identify a molecule. The optical rotation isa measurement oftherotation ora planedpolarized lightby a solution of the optically active compound. A common application is theidentification an unknown sugar.

©#H Be able to distinguish nucleophilic substitution mechanisms.jjjf There are two mechanisms for nucleophilic substitution that you must know. The first is the SnIi and the second isthe Sn2. They are defined bythe number ofreactants inthe rate determining step

ofthe mechanism. You must beable topredict the reaction from the initial conditions, recognizethe reaction from the intermediate ortransition state, andidentify the reaction from theproducts.The differences include solvent, strength ofleaving group, steric hindrance, ability to stabilize acarbocation intermediate, and stereochemistry.

Be able to recognize typical nucleophiles and leaving groups.You must recognize what makesa goodleaving group,and what effect thishas on the reaction. Thestrength ofa leaving group isdependent onthe solvent andcan bepredicted from theacidity oftheconjugate acidof theleavmggroup. Equally, thestrength ofa nucleophile canbe predicted fromthebasicity ofthenucleophile. Again, solvation andsteric hindrance playa role in thestrength ofa nucleophile.

Organic Chemistry Stereochemistry

StereochemistryStereochemistry 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 all 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 sterically 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 thatcategorization, configurational isomers can also be categorized as eitherenantiomers or diastereomers. Enantiomers are nonsuperimposable mirrorimages while diastereomers are nonsuperimposable 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 Sivjl-mechanism and theSN2-mechanism. In a nucleophilic substitution reaction that proceedsby an SnI-mechanism, the leaving group leaves to form a carbocation intermediatebeforethe nucleophile attacks. An Sisfl-reaction has a unimolecular rate-determiningstep. In a nucleophilic substitution reactionthat proceeds by an SN2-mechanism,the nucleophile attacks the electrophile from the opposite side of the leavinggroup and forces the leaving group offof the electrophile. An Sjvi2-reaction hasonlyone step, a bimolecular step. We will compare and contrast the conditionsand features of an Sul-reaction with that of an SN2-reaction to establish a set ofcriteria you can use when deciding which mechanism (SnI or Sn2) 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.

Introduction



*fc€

No plane of symmetry, sothe compound is chiral.

Figure 3-1


StereochemistryStereochemistry 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 IsomersConfigurational 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. Opticalisomers, due to their asymmetry, rotate plane-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 notsuperimposable. 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.

CI *C1

HoC CH,

Plane of symmetry, sothe compound is achiral.

ChiralityChiral is the term assignedto a molecule with no plane of symmetry, thereforeachiral molecule has an asymmetric structure. Simply put, chirality is the "leftand right handedness" of a molecule. From our perspective (keeping at the levelof this test), a chiral molecule has at least one stereogenic center present. Astereogenic center is an atom within the chiral compound that has asymmetryabout it. For ourneeds, chiral (asymmetric) carbons aresp3-hybridized carbonswith four unique substituents attached. Within 2-chloropentane there is onestereocenter (chiral 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.

HgCr^Cr^C

Hn»«...•a

ciV CH,

Object

planemirror

H3C

CH2CH2CH3

'C'HH'^"•»l

ImageCI

Each structure has sp -hybridization with four different groups attached.

Figure 3-2


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 (whichare mirror images of one another). The two structures,mirror images that haveidentical bonds, are stereoisomers that may exhibitdifferent chemical propertiesdespite identical physical properties as one another. Thebiological ramificationsofchiralityare important. For instance, humans digest only D-sugars (Drefers 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 followingeveryday analogiesmay help:a) A screw with right-handed threads does not fit into a left-handed nut.

b) Your right hand does not fit into a left glove.

c) A key with a groove on only one side and its mirror image do not open thesame lock.

Example 3.1 shows some examples of pairs of butane molecules that have fourunique substituents on carbon two. The goal of the question is to develop skillsfor quickly recognizing when two structures represent conformational isomers(discussed in section II) versus when they represent configurational isomers.

Example 3.1Which of the following structural pairs represents the same molecule and not apair of configurational isomers?

A- CHo CH,

H3CH2C\^H & H3CH2C'''>^/II H

B- OH CH3

^S^«H & HO^>l!"HCH2CH3 CH2CH3

H3C

c Br CI

H3C^CH2CH3& BrA^/,CH2CH3CI 'CH,

D- CH2CH3 CH2C1

C1H2C^X'/H &H^C^N^HCI CI

Solution

You must decide whether the compounds are either enantiomers or the identicalcompound with different spatial orientation. Rotate the molecules in your mindto see if the atoms overlay. If you do this successfully, then you will see that theyare identical only in choice C. However, if this is hard for you to visualize, try aset of models. A shortcut you may recognize is that when two of the substituentsare interchanged, the chirality of that stereocenter changes. In choices A, B, andD, two substituents have interchanged, making them enantiomeric pairs.

Copyright©by The Berkeley Review 185 Exclusive MCAT Preparation

Organic Chemistry Stereochemistry Configurational Isomers

We will start with a traditional approach to stereochemistry problems and thenslowly work our way into shortcuts and visualization tricks. The first pair ofmolecules in Example3.1are 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 two of the substituentsinterchange their locations (H and I in this case), because there is only one chiralcenter, the two molecules are mirror images of one another. Mirror images thatare non-superimposable are defined as enantiomers. The substituent exchangeshortcut should make it easier to recognize enantiomers.

The molecules in choice C in Example 3.1 are identical. If the left structure isrotated 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 astructure is rotated by 120° about a bond, the three other substituents interchangetheir locations on the molecule. The conclusion from this is that when three

substituents are different from one structure to another, those two structuresrepresent different conformational isomers (orientations) of the same compound.

CI

| W rotation,GH3. V Br^ \^//CH2CH3V BrA;'CH2

CH3

Figure 3-3

Rotating molecules in your mind becomes easier with practice, although if theskill is never fully developed, you can still answer stereochemistry questions byfollowing a few simple rules. If two groups are interchanged and the rest of themoleculeremains 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 inversionpoint. 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 samedistance, no matter which direction you proceed (positive or negative direction)on the line. Within asymmetric molecules, each stereogenic center is assigned aletter, R or S, to describe its stereochemistry.

Determining Absolute ConfigurationThe identification and naming of a chiral center is based on nomenclatureconvention. There is a set of guidelines, the Cahn-Ingold-Prelog rules, to followfor determining R and S for a stereocenter (chiral carbon). The Cahn-Ingold-Prelog rules to determine the stereochemical orientation (R or S) are as follows:

1) First, you must prioritize (from heaviest to lightest) the substituents that areattached to the carbon of the stereocenter according to the atomic mass of theatom 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 withpriority number four points behind the plane of the molecule.

3) Third, you must draw a circular arc from substituent 1 through 2 and on to 3.If the arc is clockwise, it is referred to as R. If the arc is counterclockwise, it isreferred to as S.



R is from the word rectus, 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 molecule and a generic S molecule 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.

Reposition byrotating by 90* about

Cchiral—Ci bond

Reposition byrotating by 90° about

-chiral' bond

Figure 3-4

Clockwise: R

Counterclockwise: S

Knowing the terminology is key; it is recognition, not recall, that is emphasizedon a multiple-choice exam. Here are some modified 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 ismeant by "unique substituents" is not four different atoms, but four uniquegroups 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 ithas a chlorine (priority 1), a propyl group (priority 2), a methyl group (priority3), and a hydrogen (priority 4) bonded to it. These four substituents are differentfrom one another, therefore they are four unique substituents.

R-center

A carbon center that when you look down the bond from the chiral carbon to thefourth priority substituent (usually a C—H bond) in a way that you can't see thefourth priority substituent, the remaining substituents form a clockwise arcwhen moving from priority one to priority two and on to priority three accordingto the priority rules. This can be thought of as a right-handed molecule whenplacing your thumb in the direction you're looking and curling your fingers tomatch your right hand to the structure.

S-center:

A carbon center that when you look down the bond from the chiral carbon to thefourth priority substituent (usually a C—H bond) in a way that you can't see thefourth priority substituent, the remaining substituents form a counterclockwisearc when moving from priority one to priority two and on to priority threeaccording to the priority rules. This can be thought of as a left-handed moleculewhen placing your thumb in the direction you're looking and curling yourfingers to match your left hand to the structure.




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 the highest priority. If two atoms are equal (as isoften the case with 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.

^"BrH

14 4

I>Br>OH 0>C>D>H Br>C = C>H

CII

o

c=c=c=c

CH'CHo

CH3 CH3 CH3 CH3

H-V* hcA«D H3CH2^"Br H^\4 1 1 H 6 2 H II fCH2CH3

CH2CH3> CH3 C=0 > C=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 questions 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 themolecule, 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 thecenter. In the interest of saving time, this works well for use on the 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 inthe conventional manner.

HoC

H in back of plane .*.Take what you observe as is.

Clockwise Arc = R

CH,

iHOCH2CH3H in plane far away from back group

Take what you observe as is.Counterclockwise Arc = S

Figure 3-6


H3CH2<

H in front of plane .'.Reverse what you observe.Clockwise Arc reverses to S

H3CH

HO CH3H in plane close to back group

Reverse what you observe.Clockwise Arc reverses to S

The Berkeley Review


Example 3.2The following molecule has what type of chiral orientation?

HOW^C CH2CH3H

A. R

B. S

C. The molecule has no chiral center.

D. The compound is meso.

Solution

Thecompound has one chiral center,so it cannot be meso (tobe meso requires aneven number of chiral centers). The compound is chiral, because carbon two hasfour different substituents attached it. The molecule is therefore either R or S.The priorities are OH > CH2CH3 > CH3 > H. Correct alignment 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 results.

Priority #3 H3C.

Priority #1 HON^JH

Priority #2CH2CH3

Priority #4

Priority #4 in frontCounterclockwise = R

Example 3.3What is the chirality of the triol below according to the Cahn-Ingold-Prelog rules?

O CH2OH

OH OH CH3

A. 2R,3S,4S,6RB. 2R,3R,4RC. 2R,3R,4SD. 2R,3S,4S

Solution

Carbon six is not a chiral center, because there are two methyl groups present.This eliminates choice A. The chiral centers are 2R, 3S, 4S (choice D), as shown.

O 2R CH2OH

Vi k kOH OH CH3

Clockwise Arc and #4

in back .*. take as R

CH2OH

3S= =OH OH CH3

Counterclockwise Arc and

#4 in back .*. take as S

OH OH CH3Clockwise Arc and #4

in front .\ reverse to S




Example 3.4Which of the following compounds have R orientation?

O

CI

,/>/HCH3H3CH2C

Compound I

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

H,N+.

H CH2OHCompound II

HOH2C

H OH

Compound HI

Solution

Hydrogen 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 > CH2CH3 > CH3 > H. Compound I has anR chiral center. Thepriorities in Compound II are: NH3+ > CO2" > 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.

Compound I(2-chlorobutane)

'a


#4 in front .\ reverse to R

Compound II(Serine)

O

Clockwise Arc and #4

in front .*. reverse to S

Compound HI(D-Glyceraldehyde)

HOH2<

H OH


#4 in front .*. reverse to R

Example 3.5What is the stereochemicalorientation of the following molecule?

H3C*7CA.

B.

C.

D.

2R,3R2R,3S2S,3R2S,3S

Br

vCHo

£*rl\

CI

Solution

For the first chiral 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 counterclockwise. 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. However, because H is in a reverse position, thechirality is S. The best answer is thus 2R, 3S which makes choice B correct.



Figure3-7shows a summary of the tricks presented in Figure 3-6and applied inExamples 3.2 through 3.5.

If Priority #4 is in the plane far fromgroup in back, then take arc as is.

If Priority #4 is in front,reverse the arc to opposite

If Priority #4 is in the plane neargroup in back, then reverse arc.

If Priority #4 is in back,then take arc as is.

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 interchanged, the chirality isinverted. Figure 3-8 shows the application of this method.

HOHoG? 3 CH

Priority #4 not in back .-.Interchange #4 and group in back

H3CPriority #4 not in back .*.

Interchange #4 and group in back

H,C3

2 CHoOH

2 0 filHOH2C VgJJ

Priority #4 not in back .\Interchange #4 and group in back

=>HOH2<

HO

Priority #4 now in back of plane /.New structure has a clockwise arc, so it is R

The original compound must be S

•=>HO H2OH

Priority #4 now in back of plane .'.New structure has a clockwise arc, so it is R

The original compound must be S

r=>H,C3

2

HOH2(Priority #4 now in back of plane .*.

New structure has a clockwise arc, so it is RThe original compound must be S

OH

Figure 3-8

These are two-dimensional tricks that may be done on paper. There are othertricks 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 frompriority 1 to 2 to 3. Axiother 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.




Optical RotationOptical rotation isaphysical measurement ofthe rotation ofplane polarized lightby a solution with a chiral molecule. Asolution containing a pure compound ofknown concentration (dissolved into solvent) in a standardized cuvette rotatesplane-polarized light the same amount each time. Consequently, the specificrotation (optical rotation under specific conditions) is a physical measurement(like melting point andboiling point) thatmaybeusedasa diagnostic testfortheidentity ofa compound. This iscommon withsugars which havemultiple chiralcenters. The direction of the rotation is signified by either (+) or (-) orientationfollowed by the degrees of rotation. The (+) denotation describes clockwiserotation of light while the (-) denotation describes counterclockwise rotation oflightby themolecule. If the R-enantiomer of a compound rotates the light in apositive direction by X°, then theS-enantiomer of thecompound rotates light byX° in thenegative direction. The measurement is takenwitha polarimeter whichis made from a sample tube sandwiched between two polarizing plates. Theplates are rotated in a manner to allow for the greatest amount of light to passthrough. 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 samplepolarimeteris shown in Figure3-9below.

Solution of Chiral Compound

1 polarizer

Figure 3-9

rotated light exits

nH2 polarizer

Chiral molecules may be assigned a "+" or "-" preceding their name to indicatethe direction that the compound will rotate plane polarized light. For instance,(+)-2-butanol rotates light in a clockwise direction while its mirror image, (-)-2-butanol, rotateslight in a counterclockwise direction. There are compounds withR-stereochemistry that rotate plane polarized light in a clockwise fashion andother compounds with R-stereochemistry that rotate plane polarized light in acounterclockwise direction. Consequently, R stereochemistry does notnecessarily correspond to either (+) or (-) rotation of plane polarized light. Inbiochemistry, the designation of D and L is based on threose, where (+) wasoriginally assigned to D-threose and (-) was assigned to L-threose. However,because of the vast multitude of sugars, there is no correlation between D and Ldesignation and (+)or (-) rotation of plane polarized light.



Having chiral centers does not always result in the rotation of planepolarizedlight. Mesocompounds have opposing chiral centers that cancelone another out,resulting in no net rotation of plane polarized light. Meso compounds aretherefore optically inactive, meaning they have a specific rotationof zero.

Polarimeters measure the optical rotationof a solution. Thespecific rotationof acompound, [a]o, is calculated from the opticalrotationusing Equation 3.1.

r ,t observed rotation indegrees . .

(length of sample in dm)(concentration ofsample in 2 )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 specificconditions.If the solution is too concentrated, then the rotation of light 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 chiraUty of molecules and their stereogeniccenters. A molecule 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, all of the chiral 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 itwould 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 molecule). 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.


Organic ChenUStry Stereochemistry Configurational Isomers

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 3-11shows two pairs of diastereomers.

H3CV yCH2CH3 h3Cv ^CH2CH3

vH H Br^ OH/

YEnantiomeric pair because both of the chiral centers are different

H3C ^ HO ^1

yHO CH2CH3 h3C CH2CH3/

YEnantiomeric pair because both of the chiral centers are different

Figure 3-10

H3C .CH2CH3 h3C .ch2ch3

l^y^CH. &H^^CH,\H H HO YLJ

Diastereomers, because only one of the chiral centers (the left one) is different

H3C H HO ^CH2CH3\ ^OCH2CH3 \*^

\tp ch2ch3 h ch2chJ

YDiastereomers, because only one of the chiral centers (the right one) is different

Figure 3-11

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 (s^-carbons with four unique substituents). From thatpoint, compare the comparable chiral centers in the two structures. If thestructure is oriented in a similar fashion, but two substituents are in differentpositions, then the chiral center is different between the two compounds. If thestructure is oriented in a similar fashion, but three substituents are in differentpositions, then the chiral center is the same between the two compounds.Finally, it is a matter of deciding if all, some, or none of the chiral centers on thetwo molecules are different and then deterrnining their relationship.


: yCH2CH3 H3C CH2


Example 3.6The following pairofcompounds isbestdescribed aswhich ofthefollowing?

CI C(CH3)3 H3C C(CH3)3

H3C^/ V"'OH H**/ Y""hH H CI OH

A. Diastereomers

B. Enantiomers

C. Identical achiral compoundsD. Identical chiral compounds

Solution

When the orientation of the molecule remains constant and three substituents

change 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 chiraUty. 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,making choice A the best answer.

Example 3.7How can the relationship of the following pair of molecules be described?

CH3 OH

A. Diastereomers

B. Enantiomers

C. Identical achiral compoundsD. Identical chiral compounds

Solution

In this example, the mirror plane between the two moleculescan be seen easily asthey are drawn. So without rotating or counting chiral centers, the twocompounds 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 H HQ, H

H3C^/ \ 7 V"CH3H OCH3 H OCH3

A. A pair of anomersB. A pair of constitutional isomersC. A pair of diastereomersD. A pair of enantiomers



Solution

The two structures are not aligned in an equivalent fashion, so one of the twostructures must be rotated into a structure equivalent to the other structure.

HH>

r^ \ Itwo60arotatioiqs>H

rV'"OCH3CH,OCH,

HO. HH3C, H

7 \,//OCH3H CH,

H3C^*

V'"CH,OCHa

&

H

After rotating the structure, it is easier to see that the left chiral center has twosubstituents that have interchanged between the two structures, thus it haschanged chiraUty. The same is true for the right chiral center after rotation, thusit too has changed chiraUty. This means that both chiral centers have changed, sothe compounds are enantiomers. There is no need to determine the chiraUty ofthe stereocenters (R or S) within molecules to determine whether they areenantiomers or diastereomers. Deciding whether two molecules are enantiomersor diastereomers is as easy as asking whether all of the stereocenters or just someof the stereocentershave changed their orientation between the two compounds.

(All centers differ = enantiomers; Some centers differ = diastereomers)

Example 3.9The foUowingmolecules are best described as:

CH,

A.

B.

C.

D.

H3C-

Br-

H-

CH3

H

CI

CH3

diastereomers.

enantiomers.

identical achiral compounds.identical chiral compounds.

&

H3C-

Br-

Cl-

CH,

CH,

CH3

H

H

Solution

The structures are drawn as Fischer projections, which represent the top view ofthe all eclipsed form of the molecule. In a Fischer projection, the side groups arecoming out at you in the three dimensional perspective. The two isomers havetwo chiral centers. Only the chiral center with the chlorine is different whencomparing the two compounds. The chiral carbon with bromine has notchanged, because the substituents have not moved. Only one out of the twochiral centers differs, so the compounds are diastereomers. This makes choice Athe best answer.



Example 3.10Thephysical propertiesof the following stereoisomers relate in what way?

CH2OH CH2OH

A.

B.

C.

D.

H3C-

Br-

CH,

OH

H &

HO-

H"

Same boiling point; different melting pointsSame density; different boiling pointsDifferent boiling points; different melting pointsDifferent densities; same boiling point

CH,

CH,

Br

Solution

The 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 differentbetween the two structures, because the H, CH3 and the Br have allinterconverted. When three substituents interconvert, the chiral center is notchanged. This means that one out of two chiral centers differ, so the twocompounds are diastereomers. Diastereomers have different physical propertiesincluding melting point, boiling point, and density. Pick C for best results.

Example 3.11The foUowing pair of isomers is best described by which of the following terms?

H3C/„„

\_/ H3C^\/""OH

A. Anomers

B. Diastereomers

C. Enantiomers

D. Identical meso compounds

Solution

The two compounds are mirror images of one another, so they are enantiomers,choice C. To see the mirror plane relationship, one of the compounds must berotated 180°. The drawing below shows the right structure being rotated.

H3C/''/.. ,v\\OH

H,C^'\/'""q

H3C/,//,.><'\,*\OH HO/,//f.>^V^\CH3

mirror images




Meso

Meso compounds are individual structures which contain a mirror plane sUcingthrough the middle of the compound and an even number of chiral centerssymmetrically displaced about the mirror plane. The net optical rotation of ameso compound is 0°. It is zero, because the opposing chiral centers on each halfof the molecule cancel one another out, leaving no net rotation of plane polarizedlight. Meso compounds are referred to as optically inactive. Remember thephrase, "Me so inactive", a high-energy rap lyric that describes your physicalstate while studying for the MCAT. Figure 3-12 shows a meso compound (it hasbeen rotated into a side view to see the mirror plane more easUy).

Ho^rOH HO. .OH

top view side view

Figure 3-12

A meso compound may be identified by an inversion center in the middle of themolecule. Figure 3-13shows 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.

Mirror plane.*

H3CV ViInversion point

H,C \ H

i^/\ i^/ Tv"/iH CH3 H V H

Figure 3-13

Example 3.12The reflectionof a meso compound can be classified as which of the foUowing?

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 socks

that disappear from laundry loads in our world gather.

Solution

A meso compound when viewed in a mirror reflects the identical compound.This makes choice A the best choice, although choice D is a close second. Anexample of a meso compound and its reflection is drawn below.

HO OH HO OH

mirror plane



Stereoisomerism

As mentioned before, stereoisomers are compounds that have identical bondsbut their atoms differ in spatial orientation. When a molecule contains more thanone chiral center, the maximum number of stereoisomersincreasesexponentiaUywith 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, 5R-trimethyloctane is justone of the eight possible stereoisomers. Table 3-1 shows the possiblestereoisomers for compounds with a variable number of chiral centers.

Chiral CentersMaximum

StereoisomersStereoisomers

1 2 RorS

2 4 RR, RS, SR, or SS

3 8 RRR, SRR, RSR, RRS, SSR, SRS, RSS, or SSS

Table 3-1

We are most concerned with stereoisomers when a molecule contains more than

one chiral center. If there is only one chiral center, there cannot be diastereomers.Stereoisomerism is important in biological sciences, because only a very fewbiological compounds have just one chiral center (some amino acids beingamongst of these few) with some proteins having in excess of 200chiral centers.

Example 3.13How many stereoisomers are possible for the following structure?

OH OH

H

O CH, CH,

A. 4

B. 8

C. 12

D. 16

Solution

The molecule is drawn in a way to make you mistakenly see four chiral centers ifyou don't pay close attention, but the number of chiral centers is only three.Carbon five is not a chiral center, because there are two methyl substituentsattached to it (one that is drawn as a methyl substituent and the other that isdrawn as carbon six of the longest chain). By having two methyl groupsattached, it does not have four different substituents attached, thus it is not achiral carbon. This means that only carbons two, three, and four are chiral. Themaximum number of stereoisomers is derived using the equation 2n, which isthis case is 23. Because there is an oddnumber ofchiral centers, the compoundcannot be meso, so there are eight stereoisomers. The correct answer is choice B.




Example 3.14The relationship of the foUowing pair of compounds is best described as:

A. Anomers

B. Diastereomers

C. Enantiomers

D. Structural isomers

Solution

The 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 recaU 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 foUowing compounds is/are opticaUy inactive?

I. 2R, 3S-dibromopentaneII. 2S,3R-dichlorobutane

III. lR,2R-dnodocyclopentane

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

Solution

To be opticaUy inactive, the compound must either be achiral or meso. All of thecompounds listed have stereocenters, so achiral is not a possibiUty. The questionis whether or not each structure is meso. To be meso, the compound must besymmetric and have an even number of chiral centers equaUy displaced aboutthe internal mirror plane (i.e., R on one side and S on the other). Compound I iseliminated,becauseit is not symmetricabout the a plane (the mirror plane wouldhave to sUce through carbon three to break the five carbon species into halves).To be symmetric, carbons two and four would have to have the samesubstituents, which they do not. Compound I is chiral, so it is opticaUy activeand therefore eliminated. Compound III is eliminated, because it does not haveopposite chiral centers (they are both R). To be meso, carbons one and twowould have to have opposite absolute configuration, which they do not.Compound HI is chiral, so it is opticaUy active and therefore eliminated. Theonly choice left is Compound II, which is symmetric, because it has a mirrorplane that slices through the bond between carbon two and carbon three. Thebest answer is choice B.


OrganiC ChemiStry Stereochemistry Stereochemistry in Reactions

.. i $«*&•*£ ,*J/m,vv3$C

Stereoisomer Formation

The utility of stereochemistry lies in the selectivity of chiral reactants for oneanother. When reactions involve chiral reactants, they are often selective for onestereoisomer over another. This is a staple of enzymatic selectivity. However,when a reaction involves reactants without any chiraUty, the formation ofstereoisomers is random and foUows basic probabiUty. Most reactions in organicchemistry produce stereoisomers. The type of stereoisomers formed depends onthe chiraUty of the starting reagents. When a symmetricnucleophnecan attack aplanar species from either side, there usually are two enantiomers formed inequal proportion. When a symmetric nucleophile is hindered from attacking aplanar species from one side more than the other (due to a chiral center in theelectrophile that creates greater steric hindrance ononesidethantheother), 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.

? ..e H3CO\r: :o\?CH3/Cl> + : OCH3 • .O & ^CL

H3C \s^CH2CH3^-S.. H3(= ^Ch2CH3 H3C CH2CH3Planar carbonyl

H3CCi «PH HO+ -P0^

H3C CH2CH3 H3C CH2CH3

Enantiomers

Figure 3-14

HOCH,

Example 3.16The addition ofalkyl magnesium bromide (RMgBr) toa carbonyl inether adds anew alkyl substituent to the carbonyl carbon, resulting in conversion of thecarbonyl into analcohol. The addition ofH3CMgBr toR-2-methylcyclohexanonein diethyl ether yields which products?

A. One meso compoundB. Two diastereomers

C. Two enantiomers

D. Twoepimers


Organic Chemistry Stereochemistry Stereochemistry in Reactions

Solution

In 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 chiraUty. The chiral centerpresent in the reactant does not change during the course of a reaction, so theproducts cannot be enantiomers. This eliminates 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 theyare diastereomers. ChoiceB is correct. Due to steric hindrance from the ethylgroup on carbon two, the product mixture of the two diastereomers is not 50/50.

H3C OH HO CH3

+ H3CMgBr ' ^ ' ^

When two enantiomers are formed, they are formed in equal quantity, and theproduct mixture is said to be racemic. When the two enantiomers are equaUypresent, there is no net rotation of plane polarized light. When twodiastereomers areformed, theyare formed unequaUy, so the product mixturehasa major and a minor product. When the two diastereomers are present inunequal amounts, there can be a net rotation of plane polarized light.Enantiomers canbe formed in an unequalratio if a chiralcatalystis present. Thisleads to the concept of enantiomeric excess, used to analyze product distributionsfrom reactions with a chiral catalyst (most often an enzyme).

Example 3.17Which of thefoUowing reactions produces no opticaUy active compounds?A. 2-Butanone treated withNaBH4 in etherfoUowed by acidic workup.B. (Z)-2-butene with KMn04 in base (syn addition oftwohydroxyl groups).C. Reduction ofHN=C(CH3)CH2CH3 using LiAlH4 in thf solvent.D. S-2-Bromobutane treated with NaCN in ether solvent.

Solution

For a compound to be opticaUy 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,KMn04 adds a pair of hydroxyl groups to the alkene carbons to from a synvicinal diol. Because the alkene is symmetric to begin with, symmetricadditionresults in a symmetric product. The product is a meso diol, so choice B is thebest answer. In choice C, L1AIH4 adds a hydrogen to the imine carbon fromeitherside, resulting in a racemic mixture ofamines foUowing workup. Choice Cis eliminated. In choice D,a goodnucleophile attacksan alkylhaUde, resultingininversion of stereochemistry. One chiral species is formed, so choice D is out.

Choice B \ / +KMn04 *~ \ / = \ /Symmetric 4 V <£ V. , Symmetric , „

Symmetric about plane addition HO OH HO OHSymmetric product: meso vicinal diol

Copyright ©byTheBerkeley Review 202 The Berkeley Review

Organic ChemiStry Stereochemistry Stereochemistry in Reactions

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 Excess

Enantiomers rotate light in opposite directions of one another, but with equalmagnitude. When both enantiomers are present in equal quantities in solution (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 excessfromthe observed specific rotation. The enantiomeric excess is the difference inpercentage between the more abundant enantiomer and the less abundantenantiomer.

. measured specific rotation imo/ ,,,.%ee (enantiomeric excess) = - x 1UU /o (3.2)

specific rotation of the pure enantiomer

Example 3.18What enantiomeric distribution would account for a specific rotation of +13.6" ifthe pureenantiomers havespecific rotations of+27.2° and-27.2° respectively?A. (+)-enantiomer = 25% and (-)-enantiomer = 75%B. (+)-enantiomer = 33%and (-)-enantiomer = 67%C. (+)-enantiomer = 67%and (-)-enantiomer = 75%D. (+)-enantiomer = 75%and (-)-enantiomer = 25%

Solution

Because the net rotation is positive, the (+)-enantiomer must be in higherconcentration than the (-)-enantiomer. This eliminates choices A and B. Todetermine the exactquantity, Equation 3.2canbe applied.

%ee = + 13.6 x 100% =U 100% = 50% in favor of the (+)-enantiomer27.2 2

Theexact ratio is found using the followingrelationship:%(+)-enantiomer +%(-)-enantiomer =100% and (+)-enantiomer - (-)-enantiomer =50%

.-. (+)-enantiomer = 75% and (-)-enantiomer = 25%, choice D


Organic Chemistry Stereochemistry Nucleophilic Substitution

WttCleoDlliHteSlra • WMNucleophiUc SubstitutionOne appUcation of stereochemistry is in nucleophilic substitution reactions.NucleophiUc substitution involves the attack of an electropositive carbon by anucleophile (Lewis base) to dislodge an atom or functional group (referred to asthe leaving group). This is a recurring reaction in organic chemistry and itinvolves the substitution of one functional group for another. NucleophiUcsubstitution can proceed by more than one reaction pathway. It can proceed bythe two-step SnI mechanism, or it can proceed by the one-step Sn2 mechanism.NucleophiUc substitution reactions are based on the fundamental chemistryconcept that negative charge seeks positive charge. The electron pair of thenucleophUe hunts for an electron deficient carbon to attach to. It wiU beimportant to understand the steps of the reaction for both mechanisms, asreactants proceed to products. Figure3-15 shows an example of a nucleophiUcsubstitution reaction.

H3CH2C ^^••©/*—*^_\ ( ^« ••©

nucleophile „"• leaving group A

electrophile

Figure 3-15

We will discuss the mechanism of this reaction shortly,but for now, there aresome fundamental definitions with which to be famUiar. Listed below are themost important definitions. Each definition is foUowed by some generalcomments about the relative reactivity of the species and how to discern itsreactive strength.

Nucleophile

The species donating an electron pair in a nucleophilic substitution reaction(Lewis base). As its name implies, it loves (philes) a positive charge (nucleo).Nucleophiles must have an avaUable pair of electrons to share. Nucleophilestrength is closely approximated by its base strength, although steric factors(nucleophUe size) affect nucleophiUcity. SmaU nucleophiles are generaUy betternucleophiles. This is to say that steric hindrance plays a larger role innucleophiUc substitution reactions than proton transferreactions. The strengthofthe nucleophile does notperfectly correlate with base strength, but it is closeenough to say that it parallels. A short list of nucleophiUc strength in watersolvent is as foUows:

SH" >CN- >I- >OR- >OH" >Br >NH3 >C6H50" >CH3C02" 2> CI" >F" >ROH >H20It should be noted that if the baseis too strong, an elimination reactioncan occur(as is the case with OR" and OH"). The solvent also has an effect in thatnucleophUes that canhydrogen bond are hindered in protic solvents, becausethey are solvated. The solvation by water binds the electron pair of thenucleophUe andreduces itsnucleophiUcity. 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, nucleophUicity moreclosely paraUels basicity. The big difference in nucleophilicity is that sizeof theanionisnot as important as it is withbasicity.


OrganiC ChemiStry Stereochemistry Nucleophilic Substitution

ElectrophileThe species accepting an electron pair in a nucleophilic substitution reaction(Lewis acid). The electrophileholds the reactive carbon and the leaving group.The weaker the bond between the leaving group and the carbon, the better theelectrophile. Electrophile strength can be approximated by the stabiUty of theleaving group once it is off of the electrophile. Electrophilic carbons typicallyhave a partiaUy positive charge.

Leaving groupThe functional group that dissociates from the electrophile in a nucleophiUcsubstitution reaction. The more stable the leaving group, the weaker it is as abase. This means that the strength of a leaving group can be predicted by thestrength (pKa) of its conjugate acid. The theory is that the more stable theleaving group, the less basic the leaving group, and thus the more acidic theconjugate acid of the leaving group. The strength of a leaving group increases asthe pKa of its conjugate acid decreases. This is most true in water, but can also beseen in organic solvents. Leaving group strength increases as the strength of thebond between carbon and the leaving group decreases. This is why iodine is abetter leaving group than fluorine.

Racemic mixture

A product mixture that has an even distribution of enantiomers, 50% of eachenantiomer, in the product mixture. A racemicmixture is the observed productwhen the mechanism involves an intermediate where the reactive site is an sp2-hybridized carbon(like a carbonyl or carbocation) and the molecule is symmetric(has no other chiral centers). There is no such thing as a racemic mixture ofdiastereomers, because diastereomers have at least two chiral centers associatedwith them, and a chiralcenterpresent in the reaction hinders attackofoneside oftheelectrophile relative to theother, causing thedistribution tonotbefifty-fifty.

All nucleophilic reactions involve a nucleophile, an electrophile, and'a leavinggroup. Some, but not all, nucleophilic reactions generate a racemic mixture.Whether a racemic mixture is generated or not depends on thereaction pathway.The fundamental question in a nucleophilic substitution reaction is, "does thenucleophUe come infirst, ordoes the leaving group leave first?" This is the basicdifference between the SnI and the Sn2 mechanisms. The nucleophile attacksfirst in an Sn2 reaction mechanism, whUe in the leaving group leaves first in anSnI reaction mechanism. We willlook at these twoscenarios in more detail.

Example 3.19Afavorable nucleophiUc substitution reaction has all ofthe following EXCEPT:A. a good leaving group.B. a reactant with a weak bond to the leaving group.C. a strong Lewis base as the nucleophUe.D. a weak Lewis base as the nucleophUe.



Nuc:

Solution

A favorable nucleophiUc substitution reaction is one that forms a stronger bondthan the one broken. A good nucleophUe is one that forms a stronger bond withcarbon than the bond between carbon and the leaving group. A good leavinggroup is one that forms a weak bond with carbon, thus minimal energy is neededto break the bond between it and carbon. When the leaving group is strong, thereaction is said to be favorable, so choice A is a vaUd statement. Choice A iseliminated. A weak bond to the leaving group makes it a good leaving group, sochoice B is also a vaUd statement. This eliminates choice B. The nucleophUeshould be a strong Lewis base, so choice C is a vaUd statement and choice D is aninvaUd statement. This eliminates choice C and makes choice D the best answer.You might note that to determine the favorabUity of a nucleophiUc substitutionreaction,you need to considerboth the nucleophile and leaving group.

SN2

In an Sn2 reaction, thenucleophUe attacks prior to the leaving group leaving. Inessence, thenucleophUe comes in from the backside and pushesthe leaving groupoff of the electrophile. An important factor to consider is the transition state thatforms during the reaction. Transition states cannot be viewed directly (theirUfetimes aretoo short), butevidence in theproduct (inversion ofconfiguration ata chiral carbon) infers they exist. Backside attack by a nucleophUe causes thisinversion at thechiral carbon. Certain features in the reactants (nucleophUe andelectrophUe) and the product (if it is chiral) indicate that thereaction proceededbyanSn2 mechanism. These are idiosyncrasies ofthe reaction, andthey dictatethe pathway the reaction chooses. Each property favors one of the twomechanisms. They can be used to distinguish an Sn2 reaction from an SnIreaction. Figure3-16 shows a genericmechanism for an Sn2 reaction.

Trigonal bipyramidalTransition state

t

L.G. Nuc

R

iH H

Figure 3-16

L.G.

NUC_"^H LG'H

The reaction takes place inone step, so the rate ofanSn2 reaction depends onboth the concentration of the nucleophile and the concentration of theelectrophUe. The nucleophUe initiates the reaction by attacking the electrophUeand forcing the bond between the carbon and the leaving group to stretch andweaken. At the same time that the nucleophUe approaches the electrophUiccarbon, the electron density of the nucleophUe repels the substituents on theelectrophUic carbon and thus they form the trigonal bipyramidal transition state.As the leaving group begins toleave, the substituents onthe electrophUic carbonbegin to fold in the direction of the less hindered side of the molecule (lesshindered because the leaving group has left). The hybridization finishes as sp3.Table 3-2 lists some key features associated with anSn2 reaction, according toobservation order. What is meant by observation order is that the first features(features of the reactants) are observed before the reaction begins, the secondfeatures (features of the transition state) are observed during the reaction, andthe last features (features of theproducts)are observed after the reactionends.



Reactant Features Course of Reaction Features Product Features

The preference for an Sn2mechanism is 1° > 2° > 3° in

terms of electrophiles

An Sn2 mechanism forms a five-ligand transition state duringthe middle of the reaction

A single enantiomeric productis formed

(No racemic mixture)

An Sn2 mechanism is favoredwith a good nucleophUe

The 5-ligand transition state isthe highest energy state and itexists for just a split second

Sn2 reactions exhibit secondorder kinetics

(rate = k[Nuc][Elect])

An Sn2 mechanism is favoredin polar, aprotic solvents suchas ethers and ketones

Steric forces destabilize the

transition state by forcing bondangles to values less than 109.5°

Sn2 reactions are one-stepreactions, so they have fastrates of formation

Table 3-2

The reaction in Figure 3-17 proceeds by an Sn2 mechanism, because theelectrophile is primary and it has a good nucleophile. With a primaryelectrophile, the reaction must proceedby an Sn2 mechanism. Theether solventis polar and aprotic, which further favors the Sn2 reactionpathway.

NC:® +Good

NucleophUe D

ElectrophUe

EtoO

Figure 3-17

CH3

8>*//D

Inversion Product

+ I

Stable

Anion

Example 3.20Allof the foUowing are associated with an Sn2 reaction EXCEPT:A. backside attack of the electrophUe by the nucleophUe.B. inversion of stereochemistry.C. nucleophUe concentration affecting the reaction rate.D. rearrangement ofalkylgroups from reactant toproduct.

Solution

This question focuses on the fundamentals of an Sn2 reaction. For an Sn2reaction to proceed, the nucleophile must attack the electrophile from theopposite side as the leaving group ina colUnear fashion relative to the bond tothe leaving group. This isreferred to as backside attack, so choice AisvaUd, andthus eliminated. Backside attack results in inversion of stereochemistry if theelectrophilic carbon is a chiral carbon. This makes choice B valid, whicheliminates it. Because thereis justonestep in an Sn2 reaction, the ratedependsonaU ofthereactants, including thenucleophUe. This makes choice CvaUd, andthus eliminates choice C. Rearrangement can occur when there is a carbocationpresent, because carbocations lack a bond. Carbocations are associated with SnIreactions,not Sn2 reactions, so choice D is invaUdand thus the correct answer tothis question.



SNli

In an SnI reaction, the leaving group leaves before the nucleophUe attacks. Inessence, the nucleophile waits until the leaving group has left, aUowingit moreroom to attack. The SnI reaction rate does not depend on nucleophileconcentration. Once the leaving group has dissociated, a planar cationicintermediate forms. Evidence for the intermediate comes from kinetics data as

weU as stereochemical 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 s/^-intermediate) are observed with SnI reactions. ThenucleophUe 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 anSnI reaction.

R

109.5

;^R"

R120V1

Nuc attacks

from the left

Nuc attacks

R11 £, from the right ^95Trigonal planar R

Carbocation Intermediate R"

Figure 3-18

R109.5"

»—UR>>v R" L.

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 fold in the directionof the less hindered sideof the molecule, aUowing the bond angles to increasefrom 109.5° to120° asthe carbon re-hybridizes from sp3 tosp2. This results inaslight increase in stability, accounting for the intermediate being at a lowerenergy level than the first transition state in the energydiagram. In addition tore-hybridization, the planar cation is solvated, which also increases the stabUityof the intermediate. ThenucleophUe can attack the carbocation intermediateanddisplace the solvent from either side of the carbocation intermediate. Thisdisplacement ofsolvent and the re-hybridization from sp2 back to sp3 causes adecrease instabUity from the intermediate to the second transition state. FinaUy,as the new bond is formed, the energyleveldecreases until it reaches the leveloftheproducts. Bond formation isanexothermic process. The hybridization ofthecentral carbon finishes at sp3. Table 3-3 lists features associated with the SnIreaction. Like in Table 3-2, the features are listedaccording to observation order.

Reactant Features Course of Reaction Features Product Features

The preference for an SnImechanism is 3° > 2* > 1°

in terms of electrophUes

Steric hindrance pushes theleaving group off of theelectrophUe

A racemic mixture forms

when the electrophilehas chiraUty

An SnI mechanism isfavored in a protic solventsuch as alcohol

The intermediate is a planar,three-ligand carbocation withsp2-hybridization

SnI reactions exhibitfirst order kinetics

(rate = k[Elect])

An SnI reaction is seenwith a poor nucleophUe

An intermediate is observed in

addition to transition statesSnI reactions are slowtwo-step reactions

Table 3-3



The SnI 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 (R3C+), if a tertiary carbocation ispossible. For alkyl carbocations, the relative stabUity is 3° > 2° > V, the samepreference that is observed with free radicals. The features in an SnI reaction areopposite of those in the Sn2 reaction. These features of a reaction can be used topredict whether a reaction wiU proceed by an SnI or Sn2 reaction mechanism.The reaction in Figure 3-19 proceeds by an SnI mechanism, because theelectrophUe is tertiary and it has a good leaving group.

H3CH2CH2C

H3CH2CH2C

H3CN7H3CH2C @

H3N: + .%J 1 • & + IH3C^/Average H3CX / CH CH CH Stable

NucleophUe H3CH2C ^223 AnioR3" ElectrophUe

Figure 3-19

w-LCH3

CH2CH3

Racemic Mixture of Products

Thereactionis favorable, because the leavinggroup is a good leavinggroup andthenucleophUe formsa strongerbond with carbonthan the leavinggroup.

Example 3.21Theaddition of ammonia to R-3-iodo-3-methylhexane at low temperaturewouldyield:

A. one product withRconfiguration exclusively (retention ofstereochemistry).B. one product withSconfiguration exclusively (inversion ofstereochemistry).C. two products in an enantiomer mixture.D. two products in a diastereomeric mixture.

Solution

First, we must determine whether the reaction proceeds by an SnI or Sn2mechanism. The electrophile (R-3-iodo-3-methylhexane) is tertiary, so thereaction proceeds by an SnI mechanism. The chiral center is lost with theformation of the carbocation intermediate, because the intermediate is planarwith symmetric sides. This results in a racemic product mixture of twoenantiomers. Choice C is a swell answer for this question.



In Figure 3-19, the electrophUe is tertiary. In cases where the reactive carbon issecondary, an SnI reaction canbe compUcated by rearrangement. Thisis shownin Figure 3-20.

H3C

\

H*y(H3C)2HC

H3?

H3C

I hydride shift_c+ ^

A"CH3

2° carbocation

H3CH2C H3CH2C

I H3N: \C+ 3 » J +NH3

H3C

achiral productH3C CH3

3° carbocation

Figure 3-20

H3C

H^C+ H/\

CH33° carbocation

%

H3C

Rearrangement is rapid, because it is an intramolecular process. In 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 haUdeleaving group is notbasic enough to deprotonate the ammonium cation formed from the substitutionreaction, so the product remains as a cation.

If the electrophUe has a chiralcenterat a site other than the electrophUic carbon,an SnI reaction wUl form both a major and minor product. The major productresults from the transition state with least steric hindrance.

CH2CH3 CH2CH3

CH, -CH3 H3N: .H+

Attack from the backside is

more favorable than frontside

attack due to steric hindrance.

CH2CH3

MajorProduct

CH2CH3

A.NH

Minor

Product

2

CH3

Figure 3-21

In the examplein Figure3-21, 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 (frontside attack in this example).



Distinguishing an Sn2 reaction from an SnI reactionThe first thing to look for when determining the mechanism by which anucleophilic substitution reaction will proceed, is the substitution of theelectrophUe. Tertiary and aUyUc (adjacent toa rc-bond) electrophUes wiU proceedbyanSnI mechanism whUe methyland primary electrophUes wUl proceed by anSn2 mechanism. This is the first factor to view. If the electrophUe is secondary,then the reaction can proceed by either mechanism. After considering thesubstitution of the electrophUe, the next feature to consider is the nucleophiUcstrength. The stronger the nucleophUe, the more likely the reaction wiU proceedby an Sn2 mechanism. The better the leaving group, the more Ukely the reactionwill proceed by an SnI mechanism. Lastly, you should consider the solvent. Ifthe solvent is protic (capable of forming hydrogen bonds), the reaction wiU havea tendency to proceed by an SnI mechanism. If the solvent is aprotic (notcapable of forming hydrogen bonds), the reaction will have a tendency toproceed by an Sn2 mechanism. These factors can be appUed when looking at thereactants.

If rate data are given, then the mechanism can be inferred without ambiguity.The rate law associated with an SnI mechanism is shown in Equation 3.3, whilethe rate law associated with an Sn2 mechanism is shown in Equation 3.4.

SnI Rate = k [Electrophile]

Sn2 Rate = k [NucleophUe][Electrophile]

(3.3)

(3.4)

If the rate of a reaction changes as the nucleophUe concentration is varied, thereaction is proceedingby an Sn2 mechanism. Conversely, if the rate of a reactiondoes not change as the nucleophile concentration is varied, the reaction isproceeding by an SnI mechanism. Because thesolvent canaffect thestrength ofa nucleophile, solvent and nucleophile are often considered together. The ratesofboth reactions vary with a change in electrophile concentration.

The energy diagrams for the two mechanisms also differ. There is nointermediate associated with an Sn2 reaction, only a transition state. There is anintermediate and two transition states associated with an SnI reaction. Figure 3-22 shows the energy diagrams for the one-step Sn2 reaction (on the left) and thetwo-step SnI reaction.

Transition state

Reactant

bou

W

Intermediate

Reactant


u

cw

Product Product

Reaction co-ordinate Reaction co-ordinate

Figure 3-22



The energy level increases at the start of each energy diagram, because a bond isbeing broken. In the caseof the SnI reaction, the intermediate is of lower energythan the transition state, because the carbocation can rehybridize to the lesscrowded s^-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 solvation to allow the nucleophile toattack. It is important to be able to recognize these diagrams and apply theinformation they contain to conceptual questions.

Example 3.22The addition of sodium methoxide to S-2-bromohexane at low temperaturewould yield:

A. one product with Rconfiguration 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.

Solution

First, we must determine whether the reaction proceeds by an SnI or Sn2mechanism. The electrophUe (S-2-bromohexane) is secondary so the reaction canproceed by either an SnI or Sn2 mechanism. The nucleophile is a strongnucleophUe, so we can assume the reaction wiU proceed via an Sn2 mechanism.This 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 E2product formed. An E2 reaction resultsin the formation of an alkene.

Example 3.23The following reaction shows what relationship between nucleophileconcentration and reaction rate?

H3CCH2S" + H3CCH2Br • H3CCH2SCH2CH3 + Br"A. The reaction rate increases in a linear fashion with increasing nucleophUe

concentration.

B. The reaction rate increases in an exponential fashion with increasingnucleophUe concentration.

C. The reaction ratedoes notchange withincreasing nucleophUe concentration.D. The reaction rate decreases in a linear fashion with increasing nucleophUe

concentration.

Solution

The reactionhas a primary electrophUe and a good nucleophUe, which favors anSn2 mechanism. The rate equation associatedwith a reaction proceeding by anSn2 mechanismis rate = k [ElectrophUe][NucleophUe]. The equation shows thatthe reaction rate is directly proportional to the nucleophile concentration. Therate increases in a linear fashion with increasing nucleophUe concentration, asstated in choice A. The best answer is choice A. Choice B should be eliminated,because the rate of a nucleophiUc substitution reaction does not depend on theconcentration of any species in an exponential fashion. Choice D should also beeliminated, because theratewUl not decrease withadditional nucleophUe. It wiUeither increasein a linear fashionor not change.


OrganiC ChemiStiy Stereochemistry Nucleophilic Substitution

Example 3.24Atransition state with no intermediate is associated with which ofthe followingreactions?

A. H3CCH2O- + H3CCH2Br • H3CCH2OCH2CH3 + Br

B. H3CCH2OH + (H3C)3CBr *- H3CCH2OC(CH3)3 + HBr

C. (H3O3CSH + (H3C)2CHOH2+ »- (H3C)3CS+HCH(CH3)2 + H20

D. NH3 + (H3CH2Q3CBr • (H3CCH2)3NH3+ + Br

Solution

As shown in Figure 3-21, no intermediate is associated with an Sn2 mechanism,so we must find the reaction most likely to proceed by an Sn2 mechanism. Thereaction most likely to proceed by an Sn2 mechanism should have a goodnucleophUe and ideally a primary electrophile. A low temperature is importanthere so that there wUlbe little to no E2 product formed. Choice C is the reactionof a secondary electrophUe with a poor nucleophUe, so it wiU Ukely proceed byan SnI reaction mechanism. This eliminates choice C. Choices B and D involvetertiary electrophiles, therefore they definitely wiU proceed by an SnI reactionmechanism. This eliminates both choice Band choice D. Choice A has a primaryelectrophUe and a good nucleophUe, which makes it the most likely to proceedby an Sn2 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 stiUthe best answer.

Reaction Kinetics

The rate of a reaction depends on several factors. The rate depends on theavaUable energy for the molecules to colUde, orient, and break the necessarybonds. The rate depends on the likelihood of the molecules colliding. For anSn2 reaction, the rate depends on the availabUity of nucleophUe, whUe it doesnot depend on the nucleophile concentration in an SnI reaction. Consider theSn2 reaction shown in Figure 3-23with an ethoxide concentration, [CH3CH2O-],of0.01 M a 2-bromopropane concentration, [CH3CHBrCH3], of 0.01 M and a krxof 2.53 x 10'2 L-moH-s-1 at 298 K.

H,C ^* ^ .CHl3H3CH2co:<T^-Br: CH3CH2QH> :p^ +h3ch2co—^H3C^ CH3

Figure 3-23

The concentrations are low, so the reaction is very slow. Plugging the values intoEquation 3.4 yields a rateof2.53 x10-6 Mpersecond. The reaction rate may beincreased by increasing the reactant concentrations, increasing the temperature,or by adding a catalyst. A catalyst stabiUzes the transition state complex andlowers Eact. 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 to generate theproducts. Figure 3-24represents the species of the Sn2 reaction in Figure 3-23atdifferent stages in chronological order over the duration of the reaction.



CH,

H3CH2CO..5- Io-—i—

" AHCH3

Transition State

,8-Brl

Reactants draw close to start

bond formation and form the

Early Complex (Ceariy).

Transition State starts to splitas the bond breaks to form the

Late Complex (Clate).

-CH3..8-Brl boH3CH2CO

H3C..8- \

H3C

H3CH2CO.. §_ / . .8-O—"fe--— Br:• • v'// • •

CH,

-early

..0H3C

-late

•CH3 ^

h3ch2co—^,, :bt:H3CH2CO..© "J-\

H3C3 J

H

CH,V.

Reaction co-ordinate

Reactants

Figure 3-24

V.Y—

Products

J

Physical Properties of StereoisomersEnantiomershave identicalphysicalproperties (boilingpoint, 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 ao Boiling Point Density Index of Refraction

(+) S +13.5° 99.4°C 0.808 1.398

(-) R -13.5° 99.4°C 0.808 1.398

(±) R/S 0° 101.2°C 0.840 1.442

Table 3-4

From the data in Table 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 aracemic soUd mixture has a higher melting point and greater density than eitherenantiomer.

Copyright ©by The BerkeleyReview 214 The Berkeley Review

OrganiC Chemistry Stereochemistry Nucleophilic Substitution

Separating StereoisomersOne of the most challenging tasks a synthetic organic chemist faces is theseparation of stereoisomers. If a reaction generates a new chiral center in theproduct, then it wiU be compUcated 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. ThechiraUty of the enzyme helps to select for the desired product. In organicchemistry, there are compounds known as chiral auxiliaries, which introducechiraUty to, or exaggerate existing chiraUty within, a reactant molecule. ChiralauxiUaries serve in a simUar fashion to an enzyme. Whenaimingfor 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 beappUed.

ChiraUy selective separation techniques come in two types. The first involvesemploying an enzyme (or chiraUy selective molecule) to react specificaUy 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 employed to return the compound back to its original form.An example of such an enzyme is porcine renal acylase, which selectively acylatesthe N-terminal of L-amino acids. By acylating the L-amino acid, it is no longer azwitterion at neutral pH, whUe the D-amino acid is a zwitterion. Because onecarries a net charge, it is easUyseparated from the other.

The second chirally selective separation technique involves invoking chiraUty inan existingseparation technique. For instance,a columnchromatographygel 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 selectively bind an antigen. Chiralcolumns in organic chemistry are not as specific as enzyme columns and theyhinder solutes, but do not actuaUy bind them. In theory, chiraUty 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 chiraUty, determine the relationship betweenstereoisomers, understand the nuances of nucleophiUc substitution, distinguishbetween the SnI and Sn2 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 chiraUty (Clockwise =R; Counterclockwise = S). If priority number four is one substituent away fromthe substituent in back, then determine the arc and assign the opposite chiraUtyof the arc.

Enantiomer: Enantiomers are stereoisomers which are nonsuperimposablemirror images (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 have

identical 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.

S}sjl Reactions: Preferred when an electrophUe is tertiary, when the solvent ispolar and protic, and when the electrophUe has a good leaving group. SnIreactions form a planar carbocation intermediate that can undergorearrangement to form a more stable carbocation. SnI reactions result in aracemic mixture when the reactive carbon is the only chiral center in the reactant.The rate of an SnI reaction depends on only the electrophile and not thenucleophUe (rate = k[ElectrophUe]).

Sfij2 Reactions: Preferred when an electrophUe is primary, when the solvent ispolar and aprotic, and when the nucleophile is good. Sn2 reactions form atransition state complex as the nucleophUe forces the leaving group off from theelectrophile. Sn2 reactions result in inversion when the reactive carbon is achiral center. The rate of an Sn2 reaction depends on both the electrophile andnucleophUe (rate = k[Nucleophile][ElectrophUe]).

Stereoisomer Mixtures: A fifty-fifty mixture of enantiomers is said to be a racemicmixture. The diastereomers in a product mixture formed from a chemicalreaction are referred to as major and minor, because they do not occur in a fifty-fifty ratio. Separating one diastereomer from another is easier than separatingone enantiomer from another. To separate enantiomers, chirality must beincorporated into the separation technique.



Key Points for Stereochemistry (Section 3)

Chirality and Asymmetry

1. Chiral Molecules

a) Have asymmetry within their structure due to atoms that are unevenlysubstituted

i. Stereogenic carbons are sp3-hybridized carbons with four uniquesubstituents

ii. Chiral molecules are predominant in many organic reactions

b) Stereogenic carbonsare assigned an absolute configuration of either R orS to describe their chiraUtyi. 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 #1, #2, and #3 in S -stereogenic centers,

ui. 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 deterrnining 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 thechiraUty. If priority #4 is drawn in the plane far away from thegroup going back, then the arc as is determines the chiraUty.

ii. Whenever two groups are switched, the chirality reverses


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 (7t-bondor 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 differ

iv. Diastereomeric optical isomers are better thought of as stereoisomerswhere some, but not all, of the chiral centers differ


Organic Chemistry Stereochemistry Section Summary

Stereochemistry in Reactions

1. Stereoisomers are formed when a nucleophUe attacks an asymmetricmolecule in multiple waysa) Racemic mixtures form whenthereis no preexisting chiraUtyb) Diastereomers areformed in a major/minordistribution whenoneof the

reactants has a chiral center at a non-reactive site

i. Mixtures are resolvedby using chiral reagents or lab techniques thatinvoke chiraUty.

U. Enzymatic reactions use a chiral catalyst to cause the reaction todrasticaUy favor the formation of one stereoisomer over all otherpossible stereoisomer products

in. 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 whUediastereomers have different physical properties.


1. Proceeds by either an SNl-mechanism or SN2-mechanisma) In an SNl-mechanism, the electrophUe is highly substituted, the solvent

is protic, and a carbocation intermediate is formed because the leavinggroup leaves in the first step. There is potentiaUy rearrangement and theproduct mixture is often racemic. The reaction rate only depends on theelectrophUe.

b) In an SN2-mechanism, the electrophUe is minimaUy substituted, thesolvent is polar and aprotic, and a transition state is formed because thenucleophUe 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 nucleophUe and the electrophUe.

"Better living through chemistry


StereochemistryPassages

13 Passages

100 Questions

Suggested schedule:

I: After reading this section and attending lecture: Passages I, II, VII & XGrade passages immediately after completion and log your mistakes.

II: Following Task I: Passages IV, V, IX, & XIII (28 questions in 36 minutes)Time yourself accurately, grade your answers, and review mistakes.

m: Review: Passages III, VI, VIII, XI, XII, & Questions 92 - 100Focus on reviewing the concepts. Do not worry about timing.

-\-

R-E-V-I-E-W®


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 S^i Synthesis

VII. nucleophilic Substitution

VIII. nucleophilic Substitution of Chlorocyclohexane

IX. Leaving Group Strength

X. Reaction Rates of nucleophilic Substitution

XI. Elimination versus Substitution Experiment

XII. Enantiomeric Excess

XIII. Stereochemistry in Synthesis


Structure, Bonding, and Reactivity Scoring Scale


84- 100 13 - 15

66 - 83 10 - 12

47 - 65 7 -9

34 -46 4-6

1 - 33 1 -3

(1 -7)

(8 - 14)

(15 - 21)

(22 - 28)

(29 - 35)

(36 - 42)

(43 - 49)

(50 - 56)

(57 - 63)

(64 - 70)

(71 - 77)

(78 - 84)

(85 - 91)

(92 - 100)


Isoleucine is one of eight essential amino acids. Theterm essential is applied to amino acids that humans cannotproduce, and therefore must take in through diet. Isoleucineexists as a zwitterion in aqueous solution. The natural formof isoleucine has the same chirality as other naturallyoccurring amino acids at carbon number two. Isoleucine isfound naturally in the L form. Naturally occurring aminoacids have an S chiral centeron carbon two (except cysteine).Isoleucine has a second chiral center in addition to the one atcarbon two. Seventeen of the amino acids we code for have

exactly one chiral center, with glycine (which has no chiralcenter) and threonine (whichhas two chiralcenters)being theother exceptions. The second chiral center of L-isoleucinehas fixed chirality, so a diastereomer of isoleucine varyingatthe side chain is not a biological substitute for L-isoleucine.Figure 1 shows the zwitterion form of L-isoleucine.

H. CH3

H3N+N *JF

jT CH2CH3/ **

0=f H

0-

L-Isoleucine

Figure 1 Zwitterion form of L-Isoleucine

The only other amino acid coded for by human beingsthat has a chiral side chain attached is threonine. The

threonine side chain contains an alcohol functionality. Likeisoleucine, the side chain chiral center must be specific forthe amino acid to be biologically incorporated. Threonine isalso an essential amino acid. The diastereomer of threonine

that varies at the side chain is known as "allo-threonine."

1. What is the stereoconfiguration for isoleucine?

A. 2R, 3R

B. 2R, 3S

C. 2S, 3R

D. 2S, 3S

2. If the side chain chiral center were changed, the newstructure would be which of the following?

A. An enantiomer of isoleucine.

B. A diastereomer of isoleucine.

C. An epimer of isoleucine.D. Identical to isoleucine.

Most naturally occurring amino acids have whichstereochemical orientation?

A.

B.

C.

D.

Copyright © byTheBerkeley Review® 221

4. The amino acid glycine has an H as its side chain. Whatwould you predict for the optical rotation for naturallyoccurring glycine?

A. (+)32°B. (-)32°

C. Oe

D. Moreinformation is needed. (Neverchoose this!)

5. If L-isoleucine were found to have an optical rotationof-62° from plane-polarized light studies, what wouldyoupredict for the optical rotation of its enantiomer?

A. -62e

B. +62°

C. -31°

D. +31°

6. Which of the following is the side chain of threonine?

A. -CH(CH3)2

B. -CH(0H)CH3

C. -CH2CH2OH

D. -CH20H

7. Only twenty-five percent of synthesizedisoleucinecan beused biologically. This is best explained by which ofthe following explanations?

A. Only 25% exists as a zwitterion in the body.

B. 75% 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 four 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 1 below, has been determined to reduceconstipation in the Southern European Red-Eared JumpingLizard during mating season. The effects are lesssignificantduring courting periods and does not occurat 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 activityof each stereoisomer and its subsequent reactivity.

HO H

CH2CH3

H H H OH

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 havebeen formed as to the effect of Compound P on constipationin other organisms. Being such an important chemical to thebowel process of reptiles and amphibians everywhere, a greatdeal of research 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 convertedinto Br groups through a reaction which proceeds by anSn2 mechanism. The dibromo product formed fromCompound P would show which of the following chiralassignments?

A. 3S, 4S

B. 3R, 4S

C. 3S, 4R

D. 3R, 4R

9. Assuming Compound P is made by treating an alkenewith KMn04 in basic water, what is the geometricalorientation of the alkene precursor to Compound P?(KMn04 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.


10. What is the stereochemistry of Compound P?

A. 3S, 4S

B. 3R, 4S

C. 3S, 4R

D. 3R, 4R

11. The following structure relates in what way toCompound P, the Southern European Red-EaredJumpingLizard's constipationmedication?

H OH

CH2CH3

A. It is an enantiomer of Compound P

B. It is a diastereomer of Compound PC. It is identical to Compound PD. It is a meso structural isomer of Compound P

12. What is the stereochemical orientation of the followingcompound?

H3C O

A. IS, 2S

B. IR, 2S

C. IS, 2R

D. IR, 2R

13. Enzymatic active sites are all of the followingEXCEPT:

A. chiral specific.B. size specific.C. functional group specific.D. isotope specific.

14. The enantiomer of Compound P has:

A. the same boiling point as Compound PB. a higher melting point than Compound P.C. A lower density than Compound P.D. The same specific rotation as Compound P.



Isomers are compounds with identical formulas but adifferent arrangement of atoms in space. Thereare two basictypes of isomers: structural isomers (also known asconstitutional isomers) and stereoisomers (whichcan furtherbe categorized as either configurational isomers orconformational isomers). Structural isomers are defined ashaving different connectivity of atoms, also referred to asdifferent bonds. Ethanol and dimethyl ether are anexample ofstructural isomers.

Stereoisomers are isomers which have the exact samebonds,but differ by the position of their atoms (substituents)within space. Stereoisomers can be classified as eitheroptical isomers or geometrical isomers. Optical isomershavedifferentspatial arrangementdue to asymmetry about anatom with in the molecule. A good example of opticalisomers are the R and S enantiomers of a given molecule.Geometrical isomers have a different spatial arrangement ofatoms with respect to a molecular plane. Cis and transbutene are a good example of geometrical isomers.

A common, and often challenging, task in an organicchemistry laboratory is to separate and distinguish theisomers formed from a chemical reaction. Of all isomeric

mixtures, it is easiest to separate and distinguish structuralisomers, which have varying physical and chemicalproperties. Geometrical isomers also have different physicalproperties, and can be separated readily. They can often bedistinguished from one another by their melting or boilingpoints. Of isomers, optical isomers are the most difficult toseparate. To separate optical isomers, a pure chiral materialcan be used in a chemical reaction, which is then followed byprecipitation or extraction. Once separated, the specificoptical rotation value may be used to identify the enantiomerthat was isolated.

Other traditional laboratory techniques can be modified toemploy chirality to help separate optical isomers from oneanother. As a general rule, the distinguishing featuresbetween isomers can be used to separate them in a laboratoryprocedure, although some are considerably easier than others.For identifying a purified isomer, spectroscopy and physicalproperties are most often employed.

15. The pure R enantiomer of some compound has aspecific rotation of +32.2°. A sample you make in labhas a specific rotation of +16.1°. This is best explainedby which of the following mixtures?

A. 100% R enantiomer present.B. 75% R and 25% S enantiomers present.C. 50% R and 50% S enantiomers present.D. 25% R and 75% S enantiomers present.

Copyright © by TheBerkeley Review® 223

16. Which of the following explanations doesNOTaccountfor an observed optical rotation greater than that of apositive literature value for thepurespecies?

A. Bothenantiomers arepresent in an unequal amount.B. The solution is too concentrated.

C. Thecell for the polarimeter is too long.D. The wavelength of light being used is greater than

thatof the standardsodiumlight used.

17. How many degrees of unsaturation are present in acompound with formula C11H21N02?

A. 4

B. 3

C. 2

D. 1

18. How many structural isomers are possible for theformula CgHi4?

A. 8

B. 7

C. 6

D. 5

19. The chiral centers for the following molecule are:(note: OH is located on carbon 2)

Br**;H3C

A. 2R, 3R

B. 2R, 3S

C. 2S, 3S

D. 2S, 3R

20. Which of the following CANNOT form opticalisomers?

A. A four carbon gem diolB. A four carbon vicinal diol

C. A four carbon secondary amineD. A four carbon secondary alcohol


21. How many stereoisomers arepossible for thefollowingstructure?

A. 32

B. 64

C. 128

D. 256



Stereocenters are important features in chiral organiccompounds. Stereocenters are responsible for physicalproperties, chemical reactivity, and biological function. Anexample of the correlation between physical properties andchirality is shown with the stereoisomers of tartaric acid.Drawn in Figure 1 is the generic structure of tartaric acid, anda data table of the physicalpropertiesof the stereoisomers:

HO O

H-

H-

(OH)

(OH)

HO^^OFigure 1 Tartaric Acid

Table 1 shows the physical properties of the threestereoisomers of tartaric acid and the physical properties ofthe enantiomeric mixture.

Form m.p. ocd Density H20 sol. @ 20°C

(+) 168-170 + 12° 1.7598 139 g/lOOmL

(-) 168-170 -12° 1.7598 139 g/lOOmL

meso 146-148 0° 1.5996 125 g/lOOmL

(±) 205-207 0° 1.7880 21g/100mL

Table 1

The differences in physical properties can be attributed tolattice formation in the solid phase. The compound can packmost easily when it is symmetric. This can be seen in thedensity and the melting point of each stereoisomer.Examples of biological activity are numerous. A commonexample involves the digestion of D-sugars. Our body canmetabolize D-glucose yet it cannot metabolize L-glucose (theenantiomer of D-glucose). A recent example involves thecompound L-Dopa. L-Dopa is used as an anti-Parkinson drugwhile D-Dopa has no effect and can in fact be toxic in largeenough doses. L-Dopa is drawn in Figure 2 below.

Figure 2 L-Dopa

22. Which of the following physical properties would havethe same value for morphine and a diastereomer ofmorphine?

A. Melting point.B. Density.C. Molecular mass.

D. Specific rotation.


23. How can a compound with an optical rotation of+233.0° be discerned from a compound with an opticalrotation of-127.0°?

A. The intensity of the light is greater with thepositive optical rotation.

B. The sample with +233.0° optical rotation 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 thedensity of thecompound.

D. It is not possible to distinguish the twocompoundsfrom one another.

24. 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 only onestereocenter.

H OH

HO- H

H OH

H OH

CH2OH

D-Glucose

A. It cannot be +52.6°, -52.6°, or 0°.

B. It must be either +52.6°, -52.6°, or 0°

C. Itis0°.

D. It is-52.6°.

2 5. How many stereoisomers are possible for penicillin V?

H

IN^ ^.sH2COC6H5

"TO /—N^

A. 1

B. 2

C. 4

D. 8

Copyright © by TheBerkeley Review®

CQ2H

225

26. What can beconcluded 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 packthesame.

B. The meso compound packs least tightly while the(+) enantiomer and (-) enantiomerpack the same.

C. The meso compound packs most tightly of all ofthe stereoisomers. The (+) enantiomer packsmoretightly than the (-) enantiomer.

D. The meso compound packs most tightly of all ofthe stereoisomers. The (+) enantiomer packs lesstightly than the (-) enantiomer.

27. How many stereocenters are present in the moleculecamphor which shows an optical rotation of +44.3°?

A. 0

B. 1

C. 2

D. 3

•&CH3

CH3

28. 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 requiresmore 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 uponitselfat 100°Cyields two enantiomers of limonene in a fifty-fifty ratio. Thediene of one isoprene molecule reactswith the less hindereddouble bond of another isoprene molecule to form theproduct. Because isoprene is a planar molecule, there is anequal chance for the reaction to occuron 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 similar

physical 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 lemon

flavoring or the lime flavoring, the chosen enantiomer mustbe separated from the product mixture by one of severalpossible laboratory techniques that involve chirality.

29. Which of the following techniques does NOT work toisolate one enantiomer from the presence of anenantiomeric mixture?

A. Adding the mixture to a chiral gel in a columnchromatography.

B. Distillation of the product mixture.

C. Crystallization of the mixture 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.


31. The lemon flavored isomer has what stereochemistryassociated 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

3 2. Which of the following physical properties is MOSTlikely different for the two enantiomers?

A. Boiling pointB. DensityC. Alcohol solubilityD. Optical Rotation

33. When hydroborane (BH3) reacts with limonene, it addsboron to the less hindered carbon of both alkenes and

hydrogen to the more hindered carbon of both alkenes.What is the major product when hydroborane reacts withR-limonene?

A. B.

H—B H—B

C. D.

H—B H—B

3 4. What can be concluded about the olfactory receptors?

A. They are symmetric (achiral) because they candistinguish between enantiomers.

B. They are asymmetric (chiral) because they candistinguish between enantiomers.

C. They are symmetric (achiral) because they cannotdistinguish between enantiomers.

D. They are asymmetric (chiral) because they cannotdistinguish between enantiomers.


35. If the following molecule were the reactantrather thanisoprene, how would the results differ?

vVA. The productmixturewouldno longerbe racemic.B. The product mixture would still be racemic.C. The products would be achiral.D. The products would have four stereocenters each.



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(lS,2R)-2-methylbromocyclopentane in the hopes ofcarryingout Reaction 1, which is shown below.

H3C Br OCH3 H3C OCH3

0CH3OH

majorproduct

Reaction 1

&

minor

product

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 beminimized at a lower temperature. The optical rotation forthe crude product mixture was found to be 0°, which iscontrary to what the chemist had expected. The chemist hadexpectedthat the crude product mixture would exhibit opticalactivity, although the exact value would be different than thatof the reactant.

The chemist reevaluated the proposed reaction, Reaction1, and decided that the temperature and the pH should bechanged. Under the reaction conditions used, the proposedreaction proceeds by a mechanism that is susceptible torearrangement. The chemist also failed to consider otherreactions that compete with nucleophilic substitution atelevated temperatures, such as elimination. Under differentconditions, the chemist found that optical activity could beretained.

36. The chemist attempted to carry out what type ofreaction?

A. SnI

B. Sn2

C. Ei

D. E2

3 7. Which of the following reaction mechanisms wouldexplain a 0° optical rotation?

A. An SnI reaction with inversionof configurationB. An Sn2 reaction with rearrangement

C. An Sn2 reaction with inversionof configurationD. An Ej reaction with rearrangement


38. The two products from the proposedreactionare relatedin what manner?

A. They are enantiomers of one another.

B. They are diastereomersof one another.

C. They are identical to one another.

D. They are different orientations of the same mesocompound.

39. To increase the amount of substitution reaction

observed, the chemist would likely change thetemperature 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

40. The small amount of substitution product isolated wasfound to have both the OCH3 group and the CH3 groupboth 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 placefollowed by an elimination reaction.

C. A hydride shift occurred.

D. A methyl shift occurred.

41. PBr3, when added to an alcohol, converts an OH groupinto a Br substituent with inversion of configuration atthe carbon. The mechanism is an Sn2 substitution ofthe Br for the OH. What alcohol can be used in this

reaction to produce the starting reagent in the proposedsynthesis in Reaction 1?

A. (lR,2S)-2-Methylcyclopentanol

B. (lR,2R)-2-Methylcyclopentanol

C. (1S,2R)-2-MethylcyclopentanolD. (lS,2S)-2-Methylcyclopentanol

Copyright © by The BerkeleyReview® 228

42. The following distribution of products can best beexplained by which of the explanations?

H3C CH3

CH3OH

H3C CH1 t*och3

Omajor

H3C OCH3

1 R%\\CH3

Ominor

A. The intermediate undergoes a hydride shift.

B. The methyl group on the carbon adjacent to thecarbocation influences the attack of methanol.

C. The methyl group on the carbocation influences theattack of methanol.

D. The intermediate undergoes a methyl shift.



Nucleophilic substitution is the process by whichfunctional groups on an alkane can be exchanged. It iscommonly viewedas a reaction which can proceed by one oftwo pathways. The first of the pathways is named SnI forits rate dependence on one reactant. The second pathway isreferred to as Sn2 for its rate dependence on two reactants.The data in the tables were gathered for reactions in which thenucleophile was varied in twodifferent experiments involvingtwo different electrophiles (one primary and the othertertiary). The reactions were monitored using UVspectroscopy where the magnitudeof the rate of disappearancefor the peak corresponding to the electrophile varies directlywith the reaction rate.

Nucleophile Reaction rate w/ CH3CH2CH2CI

H3CNH2 4.7 x 10"2M/sec

NH3 4.2 x 10"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

Nucleophile Reaction rate w/ (HaC^CCl

H3CNH2 3.3 x 10"4 M/sec

NH3 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 1 and Table 2 representthe initial rate of each reaction. All other variables that can

affect the reaction rate besides the nucleophile, such astemperature and concentration, were held constant betweentrials.

43. From the data presented in the passage, which of thefollowing is the best nucleophile?

A. H3CNH2

B. NH3

C. H3CSH

D. H3COH

44. 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.


45. Reaction of (2R,3S) 2-bromo-3-methylpentane withammonia yieldswhich of the following products?

A. (2S,3S)2-amino-3-methylpentaneB. (2R,3S) 2-amino-3-methylpentaneC. (2S.3R) 2-amino-3-methylpentaneD. 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 achiral center) is 0° is which of the following?

A. SN1

B. Sn2

C. SnEI

D. SnE2

4 7. Which of the following electrophiles is the best choiceto react with NaOCH3 to yield the following ether?

H

H3C-A;'''CH2CH3

"° ^CH3

A. (R)-2-chlorobutane

B. (S)-2-chlorobutane

C. (R)-2-aminobutane

D. (S)-2-aminobutane

4 8. Which of the following graphs BEST represents the CI"concentration as a function of time for the reaction of

ammonia with 1-chloropropane?

A. B.

A A

Time Time


49. Monitoring the following reaction by optical rotationwould yieldwhich of the following graphs?

BrH3CH2C

+ H3CSH

H3CH2CH2C CH3



It is possible to exchange one functional group on asubstituted alkane for another by performing a nucleophilicsubstitution reaction. There are two versions of nucleophilicsubstitution, known as Sp/I and Sfj2. The number in eachreaction describes the rate dependence. The rate of an SnIreaction depends only on the electrophile concentration andnot on the nucleophile concentration. The rate of an Sn2reaction depends on boththeconcentration of nucleophile andthe concentration of electrophile. The difference betweenthetwo mechanisms boils down to the sequence of the steps. InSnI reactions, a leaving group first leaves followed bynucleophilic attack of the carbocation intermediate. In Sn2reactions, the nucleophile attacks the electrophilic carbon,forcing the leaving group off of the molecule. In amechanistic study,a secondary alkylchloridewas treatedwithtwo different nucleophiles to get the same ether product, andrate data were collected for each.

oNaOEt

HOEt, 0° Cy°*Reaction 1

Cy»Mr(J^ OEt

Reaction 2

Tables 1 and 2 show the initial rate data for three separatetrials for each of the two reactions.

Reaction 1

Rate (M/s) [CH3CH2ONa] [C6HnCl]

1.32 x 10-2 0.05 M 0.05 M

2.63 x 10"2 0.10 M 0.05 M

5.25 x 10-2 0.10 M 0.10 M

Table 1 Initial Rates for R eaction 1

Reaction 2

Rate (M/s) [CH3CH2OH] [C6HnCl]

1.93 x 10"3 0.20 M 0.05 M

1.95 x 10'3 0.40 M 0.05 M

3.83 x IQ"3 0.40 M 0.10 M

Table 2 Initial Rates for Reaction 2

Based on the data presented, the nature of the mechanism(whether it follows SnI or Sn2 kinetics) can be determined.The key to the analysis is to observe the change in rate as thenucleophile concentration changes. As a rule, Sn2 reactionsare faster than SnI reactions.


50. Reaction 1 and Reaction 2 are best described as whattype of reactions?

A. Reaction 1 is an SnI; Reaction 2 is an Sn2B. Reaction 1 is an Sn2; Reaction 2 is an SnIC. Reaction 1 is an El; Reaction 2 is an Sn2D. Reaction 1 is an SnI; Reaction 2 is an E2

51. 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.

52. A product mixture from a nucleophilic substitutionreaction on an enantiomerically pure compound thatyields a product distribution of 87% R and 13% S canbest be explained by which of the following?

A. The reaction goes purely by an Sn2 mechanism.

B. The reaction goes mostly by an Sn2 mechanismwith some SnI mechanism transpiring.

C. The reaction goes mostly by an SnI mechanismwith some Sn2 mechanism transpiring.

D. The reaction goes purely by an SnI mechanism.

53. 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 SnI and Sn2 rates would increase.

B. The SnI rate would increase, while the Sn2 ratewould decrease.

C. The SnI rate would decrease, while the Sn2 ratewould increase.

D. Both the SnI and Sn2 rates would decrease.

5 4. Which of the following is the BEST nucleophile?

A. H3CO-

B. H3COH

C. CI"

D. HCl


55. Which of the following reactions is most likely toproceed by an Sn2 mechanism?

A. H3CO-+ (H3C)3CBr->

B. H3COH + (H3C)3CBr -»

C. H3CO- + (H3C)2CHCHBrCH3 -»D. H3COH + (H3C)2CHCHBrCH3 -»

5 6. Which of the following energy diagrams corresponds toan exothermic Sn2 reaction?

A.

60Hc

W

b

C.

S

W<D9>

Rxn Coordinate

Rxn Coordinate

B.

e

IPU

D.

c

tin

Rxn Coordinate

Rxn Coordinate



In nucleophilic substitution reactions, the reactivity ofan electrophile dictates the reactionrate. The reactivity of theelectrophile is correlated to the strength of the leaving group.The electrophile with the better leaving group is the morereactive electrophile, and thus reacts faster and undergoes themore favorable nucleophilic substitution reaction. A goodleaving group is stable once it has left the electrophile, so astable leaving group does not readily donate its electron pair.This implies that a good leaving group is a weak base. Theweaker the compound is as a base, the stronger its conjugateacid is. As acid strength increases, the pKa of the aciddecreases. This ultimately means that the lower the pKa ofthe conjugate acid of the leaving group, the more reactive theelectrophile.

Based on the pKa values, it is possible to predict therelative reactivity of various electrophiles. The favorabilityof a nucleophilic substitution reaction can be approximatedby comparing the pKa value of the conjugate acid of thenucleophile with the pKa value of the conjugate acid of theleaving group. Equation 1 can be employed to approximate areactivity constant, C, for the reaction:

C = l()[PKa(H-Nucleophile) - pKa(H-Leaving group)]

Equation 1

A reactioncan be classified anywhere from very favorableto very unfavorable. The C value can be used as follows toapproximate the favorability of a reaction:

IfC> 103, then the reaction isvery favorableIf 103 > C> 1, then the reaction isslightly favorableIf 1>C> 10"3, then the reaction is slightly unfavorableIf 10"3 >C, then the reaction isvery unfavorableTable 1 lists pKa values for the conjugate acids of some

common leaving groups.

Acid pKa Acid pKaHI -10.5 HCN 9.1

HBr -8.5 C6H5OH 10.0

HCl -7.0 H3CCH2SH 10.5

HF 3.3 H20 15.7

Table 1

The pKa data given in Table 1 can be used to predict thefavorability of a nucleophilic substitution reaction. Theaccuracy is within experimental error for substitution studies.Equation 1 does not hold well in protic solvents due tovariations in nucleophile strength from hydrogen bonding.

57. Which of the following compounds is the BESTelectrophile?

A. (CH3)3CI

B. (CH3)3CBr

C. (CH3)3CC1

D. (CH3)3CF


58. Which of the following compounds is the MOSTreactive when treated with cyanide nucleophile?

A. CH3CH2F

B. CH3CH2OC6H5

C. CH3CH2SCH3

D. CH3CH2Br

5 9. Which of the following is the MOST stable leavinggroup?

A. HCN

B. CN-

C. H3CCH2S-

D. H3CCH2SH

60. The best explanation of why NaSCH3 is a betternucleophile than NaSCH(CH3)2 is which of thefollowing?

A. Inductive effect of methyl is weaker than isopropylB. Resonance affects only three carbon fragmentsC. Hybridization of carbon varies with substitutionD. Steric hindrance is less with the methyl group

61. The reaction of sodium cyanide with 2-iodobutane is:

A. very favorable.B. slightly favorable.C. slightly unfavorable.D. very unfavorable.

62. The reaction of methylthiol (CH3SH) with R-2-butanolis which of the following?

A. Very favorable.B. Slightly favorable.C. Slightly unfavorable.D. Very unfavorable.

63. How can it be explained that fluoride, F~, is a betternucleophile than iodide, I", in ether solvent but a worsenucleophile in alcohol solvent?

A. In a protic solvent such as alcohol, F" is hinderedby hydrogen bonding, and cannot migrate as well.

B. In an aprotic solvent such as ether, P is hinderedby hydrogen bonding, and cannot migrate as well.

C. In a protic solvent such as alcohol, P exhibits nohydrogen bonding, so it cannot migrate as well.

D. In an aprotic solvent such as ether, P exhibits nohydrogen bonding, so it cannot migrate as well.



The strength of a leaving group can be determined byexamining the reaction rate of either the Sn 1 or Sn2 reaction.The rates for both reaction mechanisms show a linear

dependence on the concentration of the electrophile. It can beinferred from kinetic data that an electrophile with a betterleaving group undergoes nucleophilic substitution at a fasterrate than an electrophile with a worse leaving group. Aresearcher designed a study that varies the leaving group on anelectrophile while keeping all other factors constant. Factorsthat influence the rate include temperature, nucleophilestrength, solvent, and concentration. In a valid study, all ofthese factors should remain constant between trials.

A difficulty that can arise with this experiment iscompetition between substitution reactions and eliminationreactions. To alleviate the problem of the competingelimination reaction, a one-carbon electrophile is chosen.When the electrophile has only one carbon, the substitutionreaction must proceed via an Sn2 mechanism. Reaction 1shown below is a generic reaction representing each of thesixteen trial runs.

Nuc + CH3—X -> Nuc—CH3 + X

Reaction 1

There were four nucleophiles and four leaving groupsused in the experiment to account for sixteen combinations.Table 1 is a matrix showing the four nucleophiles and fourleaving groups used in the generic reaction. The value listedin each box in the table is the log of the reaction rate.

\x-*Nucix

OSO3CH3 I CI OCH3

H3N -3.82 -3.37 -4.11 NoRx

CN" -1.39 -1.17 -1.92 NoRx

H3CS- -2.16 -1.92 -2.42 NoRx

H3COH -5.21 -4.80 -5.70 NoRx

Table 1

The pKa values for the conjugate acids of the leavinggroups can be used to estimatereactivity. The relative pKasare pKa(H3COCH3) > pKa(HCl) >pKa(HOS03CH3) >pKa(HI)-

The rate for the reaction is measured in molar per second,therefore the less negativethe log value of the initial reactionrate, the faster the reaction. Because the rate of an Sn2reaction depends on both the nucleophile and the electrophile,thisexperiment can be used to determine the strengthof bothnucleophiles and leaving groups.

64. The relative strength of the nucleophiles is bestdescribed by which of the following relationships?

A. CN' > CH3S- > NH3 > HOCH3

B. CN" > CH3S-> HOCH3 > NH3

C. CH3S" > CN" > NH3 > HOCH3

D. CH3S- > CN" > HOCH3 > NH3

Copyright © byThe Berkeley Review®

65. What difficulty arises if Reaction 1 is carried out usinga secondary propyl electrophile instead of the methylelectrophile?

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 is

reduced.

6 6. Generally, nucleophilicity and basicity run parallel toone another. What can be said about the correlation

between the leaving group's basicity and the strength ofthe leaving group?

A. The less basic the leaving group, the better it is asa leaving group.

B. The less basic the leaving group, the worse it is asa leaving group.

C. The more basic the leaving group, the better it is asa leaving group.

D. There is no observable correlation between basicityand leaving group strength.

67. Which of the following does NOT directly affect 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 the

electrophile.

6 8. Whichof the following could NOT be determined froma similar experiment carried out with an SnI reactioninstead of the Sn2 reaction?

A. The strength of the leaving group.B. The strengthof the nucleophile.C. The effect of temperature.D. The effect of varying solvent.

69. The best explanation for the lack of any observedreaction when NH3 wasadded 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.

70. A nucleophile can also be classified as which of thefollowing?

A. A Lewis acid.

B. A Lewis base.

C. An oxidizing agent.D. A reducing agent.



It is theorized that under identical conditions, whilevarying 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 involve

higher temperatures (elimination is endothermic and oftenhas a greater activation energy than the competingsubstitution reaction). Elimination is carried out in thepresence of either a strong bulky base (E2) or a strong acid(Ei) in solution. This implies that weak bases at lowtemperature react as nucleophiles rather than as bases.Reaction 1,drawn below, was designedto verify this theory.

+ Nuc

Br H

I3C^ CH3H3C

Reaction 1

Product

The idea was to monitor the reaction by the opticalrotation of plane polarized light. Only the Sn2 reactionshows retention of some optical activity although the exactvalue of the specific rotation ([cc]D) is not predictable. Table1 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°.

Trial Temperature Nucleophile [a]D products

Trial I 10°C NH3 -36°

Trial II 60°C NH3 -28°

Trial m 10°C NaOH -38°

Trial IV 60°C NaOH 0°

Trial V 10°C NaNH2 0°

Trial VI 60°C NaNH2 0°

Table 1

The two competing reactions when a good nucleophileis present in solution, in the absence of an acid, are the E2and Sn2 reactions. The internal alkene is thepredominantproduct when elimination is observed. It is found thatdeuterium is less acidic than a proton due to the deuteriumisotope effect rooted in the shorter bond length associatedwith the deuterium-carbon bond. An E\ reaction cancompete if the leaving group is a good leaving group and itis situated on a tertiary carbon.

71. Based on the data listed in Table 1, Trial I ispredominantly what type of reaction?

A. SnI

B. Sn2

C. Ej

D. E2

Copyright © by The BerkeleyReview® 234

7 2. When a secondary alkyl bromide is treated with strongbase at 60°C, what type of reaction occurs?

A. SN1

B. Sn2

C. Ei

D. E2

7 3. The chirality of the reactant can BEST be described aswhich of the following?

A. 2R, 3R

B. 2R, 3S

C. 2S, 3R

D. 2S, 3S

74. In Trial n, the -28° optical rotation for the productmixture can BEST be explained by which of thefollowing explanations?

A. The reaction goes purely by an Sn2 mechanism.B. The reaction goes purely by an E2 mechanism.C. The reaction is an SnI reaction with some

competing elimination reaction side productpresent.

D. The reaction is an Sn2 reaction with somecompeting elimination reaction side productpresent.

75. Which of the following statements is true regarding theinterchanging of stereoisomers for the reactant?

A. When the enantiomer of the reactant is used, thesame geometrical isomers are formed, so either thereactant or its enantiomer may be used.

B. When the enantiomer of the reactant is used,different geometrical isomers are formed, so anenantiomer cannot be substituted for the reactant.

C. When a diastereomer of the reactant is used, thesame geometrical isomers are formed, so either thereactant or its diastereomer may be used.

D. When a diastereomer of the reactant is used,different geometrical isomers are formed, so adiastereomer cannot be substituted for the reactant.


76. An elimination product for this reaction would havewhat optical rotation?

A. +24°

B. 0°

C. -24°

D. -42°

7 7. To support the theory that an E2 reaction mechanism istaking place, it would be best to use chiral centers onwhich carbons of a deuterated 2-bromobutane?

A. Carbon 2 only

B. Carbon 3 only

C. Both carbon 2 and carbon 3

D. Carbons 1,2 and 3

Copyright © byThe Berkeley Review® 235


Not all stereocenters are chemically reactive. When areaction is carried out on a molecule with an unreactive

stereocenter present, there exists the possibility thatdiastereomers will be formed in unequal quantities, due to theasymmetry in the molecule. This influence is referred to asstereochemical control. Reaction 1, drawn below,demonstrates this principle.

CH2CH3 CH2CH3 CH2CH3

£cal.BH3(et20)

2. H2O2CH3

OH'(aq)A

[<x]D = +48

Reaction 1

CH3B

[ab = +18°

The two products, Compound A and Compound B, arenonsuperimposable and are not mirror images. Inhydroboration, the hydroxyl group prefers to add to the lesshindered carbon of the Ji-bond, so the reaction is referred to asanti-Markovnikov. The hydroborane prefers to add to the lesshindered face of the molecule, which means that the twoproducts are present in unequal amounts. Their percentagescan be found using Equation 1 below. The percentagesdetermined using this equation can be referenced against thequantitative valuesobtained using GC analysis.

otobs = xa«a + (1 - xa)(Xb

Equation 1

oc0bs is the observed optical rotation for the mixture andxa is the mole fraction of component a in the mixture.

The same phenomena can be observed any time anucleophile is attacking an $p2-hybridized carbon of anasymmetric molecule. This means that unequal amounts ofdiastereomers may be observed with SnI reactions,electrophilic addition reactions, and carbonyl additionreactions. Reaction 2, shown below, is an SnI reactioninvolving the formation of two diastereomers.

h3c. .1 H3C NH3 H3N. ^CH3

VQD*QD[a]D = +62°

Reaction 2

[a]D = +30°

78. For Reaction 2, what is the observed specific rotationfor the product mixture?

A. Greater than 62°

B. Between 46° and 62°

C. Between 30° and 46°

D. Less than 30°


7 9. What would be the specific rotation for Reaction 1 ifCompound A is 80% of the diastereomeric mixture?

A. 48°

B. 42°

C. 33*

D. 18°

8 0. The two products in Reaction 2 are best described as:

A. enantiomers.

B. epimers.C. diastereomers.

D. identical.

81. 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?

CH2CH3

83. What are the orientations of the three chiral centers inthe reactantin Reaction 2, startingwith the chiralcenteron which the iodine is attached and moving clockwisearound the cyclopentane?

A. S,R,S

B. R,R,S

C. S,S,S

D. R,S,S


84. How many stereogenic centers (chiral carbons) arepresent in the alkene reactant in Reaction 1?

A. 0

B. 1

C. 2

D. 3


Passage XIII (Questions 85-91)

A chemist sets out to perform a multistep synthesis.The first step, Reaction 1, is a standard Diels Alder reaction.

O O

H3C - ^ «3*O

Compound 1 Compound2 Compound3

Reaction 1

Compound 3 represents a mixture of enantiomers. Themixture undergoes a chirally specific laboratory technique toisolate Compound 3a, shown below, from Compound 3b.

O

HoC£AO

Compound 3a

In the second step, Reaction 2, Compound 3a is treatedwithmeta-chloro peroxybenzoicacid, mcpba, in ether to formCompound 4.

O

Compound 4

Compound 4 then undergoes Reaction 3 to formCompound 5.

O O

H,CNH,H3CHN-^V4L3^1NH2

K?Wo o

Compound 4 Compound 5

Reaction 3

In Reaction 4, Compound 5 is hydrolyzed using water at90°C to form Compound 6.

O

H3CHN>^^\^^OHOH™XX(

OCompound 6

Copyright © byThe Berkeley Review® 237

85. All of the following reagents, when added to Compound3a, result in a product with more asymmetric carbonsthan compound 3a EXCEPT:

A. Br2(l)/CCl4(l)

B. H2(g)/Pd(s)

C. cold KMn04(aq) at pH = 10

D. NH3(1)

8 6. Which of the following is Compound 3b?

A. o B.

87. Whatlaboratory technique wouldbe MOSTeffective inobtaining a pureenantiomer froma racemic mixture?

A. Adding the mixture to a chromatography columnfilled witha gel withbothenantiomers bound to it.

B. Adding the mixture to a chromatography columnfilled with a gel with just one of the enantiomersbound to it.

C. Distilling the mixture using a vertical columnfilled with beads that contained both enantiomersbound to their surface.

D. Using a chirally pure carrier gas in a gaschromatography experiment.

88. Whatis themajor product, mostabundant stereoisomer,formed in Reaction 2?

H3CO o


89. All of the steps in the overall synthesis shown in thepassage generate an optically active product mixtureEXCEPT:

A. Reaction 1

B. Reaction 2

C. Reaction 3

D. Reaction 4

9 0. What is a likely side product of Reaction 3, if excessamine is used?

NCH3

OH

NHCH>

NHCHjNHCHj

91. The final productmixture following Reaction4 can bestbe described as:

A. an enantiomeric mixture with ocd= 0°.B. a diastereomeric mixture with ocd= 0°.

C. a diastereomeric mixture with ocd * 0°.D. a meso mixture with ocd * 0°.



92. The following molecule has which of the followingstereochemical orientations?

A. 2R, 3R

B. 2R, 3S

C. 2S, 3R

D. 2S, 3S

OCH3

93. The following molecule has which of the followingstereochemical orientations?

A. 2R, 3R

B. 2R, 3S

C. 2S, 3R

D. 2S, 3S

H-

H-

CH3

OH

CH3

OCH3

9 4. The following pair of molecules can best be described aswhich of the following?

0 H OH

HO H HO CH2CH3

A. Diastereomers

B. Enantiomers

C. EpimersD. Anomers

H OH

CH2CH3

H OH HO CH3


95. How many stereoisomers are possible for the molecule1,2,3-trifluoropentane?

A. 2

B. 4

C. 6

D. 8

96. Addition of KMn04(aq) at pH = 10 generates a vicinaldiol with syn stereoselectivity. What does the additionof KMn04(aq) at pH = 10 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

98. 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 25% exists as a zwitterion in the body.

B. 75% of synthetic isoleucine does not have thecorrect side chain.

C. In synthesizing isoleucine, the two chiral centersresult in four stereoisomers being formed. Onlyone of the four is biologically correct.

D. In synthesizing isoleucine, the two chiral centersresult in eight stereoisomers being formed. Onlyone of the eight are biologically correct.


100. Which of the following compounds does NOT showany optical rotation of planar light?

A. 2R, 3R dibromobutane

B. 2R, 3S dibromobutane

C. 2R, 3R dibromohexane

D. 2R, 3S dibromohexane

1. D 2. B 3. B 4. C 5. B 6. B

7. C 8. C 9. B 10. B 11. B 12. A

13. D 14. A 15. B 16. A 17. C 18. D

19. B 20. A 21. B 22. C 23. B 24. A

25. D 26. B 27. C 28. D 29. B 30. B

31. A 32. D 33. A 34. B 35. A 36. A

37. D 38. B 39. B 40. C 41. B 42. B

43. A 44. A 45. A 46. A 47. A 48. A

49. A 50. B 51. A 52. B 53. A 54. A

55. C 56. B 57. A 58. D 59. D 60. D

61. A 62. D 63. A 64. A 65. C 66. A

67. D 68. B 69. D 70. B 71. B 72. D

73. B 74. D 75. A 76. B 77. C 78. B

79. B 80. C 81. D 82. A 83. C 84. B

85. D 86. A 87. B 88. B 89. A 90. C

91. C 92. A 93. D 94. A 95. B 96. A

97. C 98. D 99. C 100. B

LIFE BEYOND CHEM STARTS NOW

Stereochemistry & Nucleophilic Subst'n Passage Answers

Passage I (Questions 1-7)

4.

5.

6.

7.

Isoleucine and Threonine

Choice D is correct. Using the Cahn-Ingold-Prelog rules for substituent priority and drawing the appropriatearcs, the chirality for the two stereogenic carbons of isoleucine is determined as follows:

Priority #4 in back(an as isposition)

Countercloskwise Arc

.-. Chirality isS

+H3N

Isoleucine

CH2CH3

Priority #4 in back(an as isposition)

Countercloskwise Arc

.-. Chirality isS

Both of the chiral carbons have S chirality, which makes the compound 2S,3S. It would be swell of you tochoose D. The chiral center on carbon two ofan amino acid must beS according to the rules discussed in thepassage, so choices A and Bcould have been eliminated early. Regardless, the chiral center on the side chainof the amino acid must be solved for. Determining Rand Sis actually rather simple when you get the hang ofit. The key is to find a method that works foryouand honeit in by repeated use.

Choice B is correct. If the side chain (carbon three of isoleucine) were to change the orientation of its chiralcenter while carbon 2 retained the orientation of its chiral center, then only one of the two chiral centers woulddiffer between the two compounds. The two stereoisomers (isoleucine and the other compound) would beclassified as diastereomers. The correct choice is B.

Choice Bis correct. This question can be solved from straight memorization. Naturally occurring amino acidsare "L" as in life and natural. It is also stated in the passage that naturally occurring amino acids have Sstereoconfiguration and that S stereochemistry is associated with L-amino acids. The best answer is choice B.

Choice C is correct. With H as the side chain, carbon two of glycine (the alpha carbon) has two hydrogensattached, thus there is no chiral center present on glycine. Neither of the two carbons in glycine have fourdifferent substituents attached. The absence of a chiral center in glycine results in an optical rotation of 0°.Choose C and be happy.

Choice B is correct. Enantiomers are nonsuperimposable mirror images, thus all of the chiral centers aredifferent between the two structures. If all of the chiral centers are reversed, then the specific rotation shouldbe completely reversed, which would lead to a value of +62° rather than -62°. It wouldbe terrific if you were tochoose B. Enantiomers always have the same absolute value for the specific rotation, only the sign (directionof rotation) differs.

Choice B is correct. As stated in the passage, the side chain of threonine is an alcohol (eliminating choice A)and it is chiral (eliminating choices A, C, and D), all choices except answer B are eliminated. Only choice Bcontains a carbon that is asymmetric (chiral center).

Choice C is correct. Isoleucine contains two chiral centers, one for the alpha carbon and one in the side chain.Plugging into the stereoisomer equation 2n where n is the number of chiral centers, there are four possiblestereoisomers for the isoleucine structure. Because isoleucinecontains two chiral centers that must have specificorientation, only one of the four stereoisomers will have the correct chirality to be biologically usable. This isstated in the passage in two fragments. The best choice is C.

Passage II (Questions 8 -14) Unknown Stereochemically Active Compound

Choice C is correct. PBr3 converts the OH group of an alcohol into a Brgroup through an Sf\j2 reaction. Becausethe reaction is by way of an Sj\j2 mechanism, the chiral centers inverts. If you recall, chiral centers invert inS[\j2 reactions, but not S^jl reactions. This means the product shows stereochemistry of 3S,4R. Choose C for thesensation of correctness and satisfaction.

Copyright © by The BerkeleyReview® 240 STEREOCHEMISTRY EXPLANATIONS

Choice Bis 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. PickBfor best results.

CH2CH3

IWBota5fl^HOH H H CH2CH3

Hydxroxyl groups are syn

CH2CH3

So, alkene must have been cis

10. Choice B is correct. Using the rules of priorities, the following is determined:

Priority #4 in backClockwise = R

CH2CH3

Priority #4 in frontClockwise = S

This makes the compound 3R, 4S. Choose Bfor optimalcorrectness 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 thatchiralcenterhas not changed. Whenonlyone out of two (some, not all) chiralcenterschange, it is not a mirrorimage, nor identical (superimposable), making the twocompounds diastereomers. Pick answer B.

11.

OH

CH2CH3

OH H HO CH2CH3

Mystery CompoundCompound P

12. Choice A is correct. Usingthe rules of priorities, the following is determined.H3C

H in front, so reversechirality from R toS

~ ~Hin back, so chirality

is S as shown

This makes the compound IS, 2S. Choose A.

13. Choice D is correct. The active siteofan 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 ina 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 not isotope specific, making choice D the best answer.

Copyright ©by The Berkeley Review® 241 STEREOCHEMISTRY EXPLANATIONS

14. 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 in a sea of many choices (well maybe not many, but four.)

Passage III (Questions 15 - 21) Classification of Isomers

15. 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 enantiomer). 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 the S stereoisomer would result in an overall negative opticalrotation, making choice D invalid. If the mixture were in fact 100% of the R enantiomer, then 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 problemcould have been solved mathematically as follows:

%R (+32.2) + %S (-32.2) = +16.1

%R (32.2) - %S (32.2) = +16.1

32.2 (%R - %S) = 16.1

%R - %S = 0.5 = 50%; and %R + %S = 100%

Thus 2 (%R - %S) = %R + %S

2 %R - 2 %S = %R + %S

%R = 3 %S, so %R = 75% and %S = 25%

16. Choice A is correct. Because one enantiomer always has theopposite optical rotation of the otherenantiomer(one isnegative and one is positive with the same magnitude), if both enantiomers are present in solution, theabsolute value of the rotation must decrease from the absolute value of the pure enantiomer. The observedoptical rotation is an average of the enantiomers in solution. Therefore, choice A cannot account for an observedspecific 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 Band Cboth lead to an increase in the observed rotation. The standard rotation is based on monochromatic light from asodium lamp. If monochromatic light from another source is used, it will interact differently with thecompound, resulting in a different optical rotation. Whether the rotation is greater or lower is uncertain, butbecause it is possible that the rotation is greater, choice D is eliminated.

17. Choice Cis correct. The formula for units of unsaturation is Units of unsat = 2(#c) +1(#N) +2- \ (#H) For a2

compound with a formula of CnH2iN02, the calculation is: Units of unsat = 2(n) +*CO*2"1 (21) =4 = 2.2 2Pick C. Remember that oxygen is not included in the formula for degrees of unsaturation. If you didn't remember

the formula, then count the bonding electrons for the compound (44 for the eleven carbons, 21 for the hydrogens,3for the nitrogen, and 4for the two oxygens). There are a total of 72 bonding electrons, so the compound has 36bonds total. There are 35 atoms in the molecule, therefore only 34 bonds (minimum) are necessary to connect theatoms and form themolecule. There are36 bonds when only 34 areneeded, thus there are twoextra bonds. Thislong-winded path still leads to answer choice C.

Choice Dis correct. The only way to do this problem is to count isomers systematically. Start with the longestchain (6) and list all the possible isomers. Then look at chain lengths ofone less carbon (5) and list allof thosepossible isomers. This continues until the possible backbones are depleted. For this question, the carbonbackbone is shown for all of the possible isomers. There are five isomers total, so pick D.

18.

longest chain= 6 carbons

longest chain = 5 carbonsr

longest chain = 4 carbonsA

' * f >* |c—c— C—C—C—C c—c—c—c—c c—c—c—c—c c—c—c—c c—c—c—c

^

Copyright © by The Berkeley Review( 242 STEREOCHEMISTRY EXPLANATIONS

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. Tlie rationale is shown below:

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 from

the 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.

21. Choice B is correct. In polycyclic systems, if there is no plane of symmetry, then all tertiary and quaternarycarbons are stereogenic centers. In addition, anysecondary carbons with a functional group are also stereogeniccenters. The compound has two quaternary carbons, three tertiary carbons, and onesecondary 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 stereoisomers is 26, which is 64. Pick B,and feel the warmth of correctness.

Passage IV (Questions 22 - 28) Stereoisomers and Optical Activity

22. 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 ofdiastereomers 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.

23. Choice Bis correct. When using a polarimeter, an observed optical rotation of +233.0° and -127.0° would yieldthe 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-127.0°, then the lower concentration would show a rotation less than -127.0° (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.

24. Choice Ais 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 Dglucose), 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 ofstereoisomers (assuming that there is nomeso structure), can be determinedby raising 2 to the power of the number of chiral carbons (stereocenters). There are three chiral carbons(stereocenters) associated with penicillin V, thus there will be eight (23) stereoisomers for the structure ofpenicillin V. The 2n formula represents the maximum number ofstereoisomers. For every meso structure, youmust subtract one from the total. Choice D is correct.

®Copyright © by The Berkeley Review 243 STEREOCHEMISTRY EXPLANATIONS

26.

27.

28.

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-find-the-density question. The mesocompound 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 description. The correct choice is answer C. Drawnbelow is camphor with the two chiral carbons (stereocenters) labeled:

Top View Side View CH,H3C

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 greater in the pure compound. 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 iseliminated. Physical properties, such as melting point,result from intermolecular forces, not covalent bonds. Enantiomers are no more likely to form covalent bondswith one another than diastereomers. Choice Bis an untrue statement that does not explain the observedmelting points. Choice Bis eliminated. Covalent bonds are not affected when a compound melts, sochoice Cshould be eliminated. The best explanation is choice D, because when the molecules pack more tightly, theyexert stronger forces on each other. Because the enantiomeric mixture (R with S) has a higher melting pointthan 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.

Passage V (Questions 29 - 35) Enantiomers from Diels-Alder Reaction

29.

30.

31.

Choice Bis 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 gel in column chromatography. Thetwo enantiomers exhibit different migration rates down the column, because the two adhere to the column to adifferent extent. Choice A is a valid method. Distillation will not separate the two enantiomers, so choice Bis the correct answer choice. The mixture can be selectively crystallized with apure Rorpure Scompound. Thisis often carried out with tartaric acid. Enzymes are chiral, so chiral compounds pass through an enzyme filterat different rates. This makes choice C and D valid.

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 C is eliminated.The concentration does not affect the alignment ofthe molecules in the transition state, only the frequency withwhich the reactants collide to form the transition state. This eliminates choice D. The presence of a chiralcenter in the solvent does not affect the chirality of the transition state unless the solvent is involved in thetransition state. Choice A cannot be eliminated yet, but it is not a likely choice. The only change that willdefinitely affect the distribution of enantiomers is the addition of a chiral catalyst which affects thetransition state. This is the whole idea behind the activity and specificity of enzymes (chiral catalysts) inbiological reactions. The best answer is therefore choice B.

Choice A is correct. Both enantiomers have only one stereocenter, therefore choices C and D are eliminated.The lemon flavored extract is the structure on the left, which has its lone stereogenic center in Rconfiguration.This eliminates choice B and makes choice A the correct answer.

Copyright © by The Berkeley Review* 244 STEREOCHEMISTRY EXPLANATIONS

32.

33.

34.

35.

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 ischiral 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 theboron and the hydrogen add syn to oneanother, the methyl substituent on the ringmust be trans to the bridging boron. This eliminates choice Band makes choice A the best answer.

R stereocenter

First addition,of BH3(Et2Q) '

H,B

Second addition

of BH3(Et20)

Trans addition to methyl

H—B

Choice A

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.

Passage VI (Questions 36 - 42) Proposed SnI Synthesis

36. Choice A is correct. The proposed product is a mixture ofdiastereomers formed from a nucleophilic substitutionreaction. Because two diastereomers are formed, theproposed reaction musthavebeen predicted to proceed byan SnI mechanism. The best answer is choice A.

37. 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.

38. 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 isbyway ofan Ei mechanism, because nostrongbase is present. To reduce the amount ofEi product, the amount ofacid 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. Thebest 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 SnI 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, sothe reaction proceeds via a hydride shift before the nucleophile attacks. The best answer is choice C.

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 ofstereochemistry, the alcohol must have opposite chirality at carbon 1 as the bromoalkane in Reaction 1. Thepassage states that the reactant is (lS,2R)-2-methylbromocyclopentane, so the alcohol precursor must havechirality of IR 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

CH,

\ y Hin front on C-1,\ 1 so reverse it: R

CH CH,

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 sp*--hybridized, so it is planar. The only sterichindrance comes from the adjacent methyl group (which is above the ring). The adjacent methyl, by beingabove the ring, influences the nucleophile to attack from below the ring. The best answer is choice B.

Passage VII (Questions 43 - 49) Nucleophilic Substitution

43.

44.

45.

46.

47.

48.

Choice A is correct. The first reaction in Table 1 proceeds by way of an Sn2 mechanism, because theelectrophile is a primary alkyl halide. The second reaction in Table 1 proceeds by way of an SnI mechanism,because the electrophile is a tertiary alkyl halide. Only the rate of an Sn2 reaction depends on thenucleophile, therefore to determine the best nucleophile, the data from the reaction with the primary alkylhalide (first set of data in Table 1) should be used. The best nucleophile is the compound that has the fastestreaction rate, which according to Table 1, is methylamine (H3CNH2). Choose A if you're a table believer.

Choice A is correct. From the data in Table 1, the faster reactions are observed with 1-chloropropane an otherelectrophiles, so it is a safe and valid assumption that the reaction with the primary alkyl halide proceedsmost rapidly. Choose A.

Choice A is correct. The reaction can proceed by either an SnI or Sn2 mechanism with a secondary alkylhalide as the electrophile. If the reaction were to proceed by an Sn2 mechanism, then the product would bethe 2S, 3S stereoisomer. Only the stereocenter from which the bromine substituent left underwent a change inchirality (inversion), thus only that stereocenter will show a change in its orientation. If the reaction were toproceed by an SnI mechanism, then the intermediate would undergo rearrangement, and thus the ammoniawould attack the third carbon leaving an achiral product. Given the answer choices, choices B and C can'tform, and there is no achiral choice, therefore the reaction proceeded by an Sn2 mechanism. This means thatinversion of the second carbon will transpire to yield 2S, 3S. Choose A.

Choice A is correct. To have optical activity and lose it during thecourse of the reaction eliminates choice B,because the Sn2 reaction proceeds with inversion (thus an optically active product is formed). By definition,the product mixture asdescribed in the question is racemic. Aracemic product mixture is associated with theSnI reaction. Pick A.

Choice A is correct. The ether product has S stereochemistry as drawn and was formed by a substitutionreaction using NaOCH3 as the nucleophile. Sodium methoxide (NaOCH3) is a strong base, and thus it is also agood nucleophile. Because the nucleophile is good, the reaction must have proceeded by an Sn2 mechanism.Because the final product has S stereochemistry, the starting material (electrophile) must have had Rstereochemistry to form the S product from inversion. This eliminates choices B and D. For the substitutionreaction to proceed, the electrophile must have had a good leaving group. Ammonia is not a leaving group,therefore the chlorine leaving group is the better choice. This makes the correct answer choice A.

Choice A is correct. The reaction proceeds at the fastest rate during the first segment of the reaction becauseinitially the concentration of 1-chloropropane (a reactant in the rate determining step) is greatest and bothreactants (nucleophile and electrophile) are depleted over time. Chloride anion is the leaving group, thus itsconcentration will increase over time. All of the graphs show increasing concentration. Over time, theconcentration of 1-chloropropane gradually decreases, thus the reaction rate decreases gradually; this resultsin a slower production ofCI" anion. Graph A best depicts this gradual decrease in reaction rate. Choose Aandbe a wiinder student.


49. Choice A is correct. Because the electrophile is a tertiary alkyl halide, the reaction is an SN1 reaction, whichproceeds with racemization. The optical rotation of the product mixture following an Sj^l 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 SnI reaction is first order, 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 which isnotpossible withanSnI reaction. Choose Aandmake your support group proud.

Passage VIII (Questions 50 - 56) Nucleophilic Substitution of Chlorocyclohexane

50. 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 Sn2 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 SnI reaction. The rate of an SnI 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.

51. 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 SnI reaction. An SnIreaction 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 SnI reaction, so choice B is valid and thus eliminated. All nucleophilic substitution reactions,whether it is an SnI or Sn2 mechanism, 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 SnI reaction. This makes choice D valid, and thus eliminates it. ChoiceA is in fact the top dog of choices.

52. Choice B is correct. If the reaction were to proceed purely by an Sn2 mechanism, the product would be 100% R,because the reactant is enantiomerically pure and the Sn2 reaction results in complete inversion. If the reactionwere to proceed purely by an SnI mechanism, the product mixture would be 50% R and 50% S, because thereaction goes through a planar carbocation intermediate resulting in a racemic mixture. Tlie mixture is 87% Rand 13% S, which is closer to the products of an Sn2 mechanism than the products of an SnI mechanism. It isnot a pure reaction so the best answer is choice B.

53. 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 theSnI and the Sn2 mechanisms, because they both depend on the electrophile. This makes choice A correct.

54. Choice A is correct. The strongest nucleophile is most willing to donate its lone pair to carbon. The answerchoices include two conjugate pairs. Theconjugate base is the better nucleophile of the pair, so choices Band Dare eliminated. HCl 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.

55. Choice C is correct. AnSn2 reaction favors a primaryelectrophile over secondary or tertiary electrophile, anda good nucleophile is required. Choices A and Bcan both be eliminated, because the electrophiles are tertiary(and tertiary electrophiles proceed via the SnI mechanism). Choice C is better than choice D, becausemethoxide is a better nucleophile than methanol. Pick C for optimal results.

56. Choice Bis correct. An Sn2 reaction proceeds by way of a one-step mechanism, whicheliminates 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 SnI reaction, choice C is an endothermic SnI reaction, and choice D is an endothermic Sn2 reaction.

Copyright © by The Berkeley Review® 247 STEREOCHEMISTRY EXPLANATIONS

Passage IX (Questions 57 - 63) Leaving Group Strength

57.

58.

59.

60.

61.

62.

63.

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, because 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 inthis question is the bromide leaving group. The ranking of the conjugate acids for the leaving groups are: HBr >HF > HOQH5 > 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 aresult 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 group is the weakest base. Of the three choices left, CH3CH2SH has the strongest conjugate acid(CH3CH2SH2+ is a stronger acid than CH3CH2SH and HCN), thus CH3CH2SH is a weaker base thanCH3CH2S" and CN". CH3CH2SH is the weakest baseof 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 thequestion, 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 n-system. Choice Bis eliminated. The hybridization is sp3 in both cases, so the difference in nucleophilicitycannot be attributed to hybridization. This eliminates choice C. Steric hindrance predicts that the smallernucleophile has less interference in the transition state, thus it is a better nucleophile. In this case, sterichindrance plays a larger role than the inductive effect in the reactivity of the nucleophile. The correct answeris choice D.

Choice A is correct. Sodium cyanide is a good nucleophile and iodide is a great leaving group, therefore thisreaction should be very favorable. The difference between their pKa values is 19.6, which implies that theKeq for this reaction is near 1019-6 =4x1019. This defines a reaction that goes completely to product, whichsupports the evaluation that the reaction is very favorable. Do thecorrect 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 unfavorableby intuition. The difference between their pKa values is -5.2, which implies that the Keq for this reaction isnear 10"5-2 =6x10"6. This defines a reaction that stays predominantly as reactant, which 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 anaprotic solvent, there is no hydrogen bonding, sochoice Bis eliminated. Hydrogen bonding affects fluoride andnot iodide, so the fact that fluoride is a worse nucleophile than iodide in alcohol implies that hydrogenbonding reduces the nucleophilicity of fluoride. This can be attributed to hindrance to migration caused byhydrogen 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.

Reaction Rates of Nucleophilic Substitution

Choice A is correct. The strength of a nucleophile can be measured by its reaction rate in a second-ordernucleophilic substitution reaction (an Sn2 reaction). Tlie nucleophiles are listed in the first column of Table 1.Any other columncan be used to determine the relative strength of the nucleophiles, because all other factors inthe reaction are constant. The less negative the value in the table, the faster the reaction, and therefore thebetter the nucleophile. The CN- has the lowest value in all three columns, so the bestnucleophile is CN-. Thiseliminates choices C and D. The question now is to determine whether the ammonia (NH3) or methanol(HOCH3) is the better nucleophile. Tlie stronger nucleophile is ammonia, because in each column, the lessnegative value is associated with ammonia. The best answer is choice A, CN' > CH3S" > NH3 > HOCH3.


65. Choice Cis 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 double 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 across the reaction chart, so choices A andB can be eliminated.

66. Choice A is correct. The best leaving group is the functional group that takes electrons from carbonand 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 a leaving group is generallycorrelated (in a linear fashion) to the strength of the conjugate acid of the leaving group. As an acid becomesstronger, the conjugate base becomes weaker. This implies that it is valid to compare the strength of theleaving 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.

67. 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 forminghydrogen 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 aprotic and proticsolvents. The fluoride is the strongest nucleophile of the halides in aprotic solvents while it 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 nucleophile 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. Tlie leaving group is independent of the nucleophile in nucleophilic substitution reactions. Thismakes choice D the best answer.

68. Choice B is correct. Because the rate of an SnI reaction depends only on the leaving group breaking free fromthe electrophile in the rate-determining step, the nucleophile is irrelevant to the reaction rate for an S^jlreaction. 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 SnIreaction. Choose B for yet another chance to flash a happy "I just got another one right" smile.

69. 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). On the other hand, the data in Table 1 shows that ammonia(NH3) 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.

70. 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.

Passage XI (Questions 71 - 77) Elimination versus Substitution Expenment

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 Sn2 mechanism. Elimination and SnIreactions 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 SnI 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. Choice D is correct. At high temperature, the predominant reaction is elimination. This eliininates choices Cand D. Because a strong base (NaOH) was used, the mechanism must have been an E2 rather than an Eimechanism. The best answer is choice D. This can be verified by looking at Trial IV in Table 1, which showidentical reaction conditions as the reaction in the question. The loss of optical rotation in the product implythat 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.

73. Choice B is correct. There are two chiral centers in the reactant, located at carbons two and three. This can beinferred from the answer selections. The number four priority is hydrogen on both chiral centers andconveniently it is in the backin both cases. Thefirstchiralcenter(carbon two)has prioritiesbromine > 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 secondchiral center is thus S. The solution is drawn below:

carbon 2 = R

The correct answer is 2R, 3S, which makes choice B correct.

carbon 3 = S

74. Choice D is correct. If the reaction proceeded purely by an Sn2 mechanism, then the optical rotation would bethe same as was observed for trial I (-36°), a purely Sn2 reaction. This eliminates choice A. An eliminationreaction would yield an optically inactive product so the optical rotation observed for an elimination reactionwould be zero. This eliminates choice B. The optical activity observed implies that an Sn2 reaction must havebeen occurring to some degree. The reduction in optical activity must be attributed to the presence of someimpurities (from some side reaction). The best answer is choice D.

75. Choice A is correct. Because the enantiomeris a nurror imageof the reactant,it forms a transition state when itis eliminating that is a mirror image of the reactant's transition state. This means that the products are alsomirror images, but without stereogenic centers, they can be rotated to match as identical compounds. Thissymmetry presents itself in the product as an identical geometric isomer. Because a diastereomer varies at onlyone chiralcenter, it is not a mirror image of the reactant when it is eliminating. Thisasymmetry presents itselfin the product as different geometric isomers. This may not make sense in words so the drawing below showsthe products.

Br

H3C CH3Reactant

Diastereomer

H,C^

Enantiomer

rotation to correct

alignment

rotation to correct

alignment

rotation to correct

alignment H3C^

Elimnation with

a strong base

Elimnation with

a strong base

Elimnation with

a strong base

H D

xoC CHqH3CCis Methyls

H CH,

H,C D

Trans methyls

H,C

H D

Cis Methyls

CH,

The enantiomer forms the same products while a diastereomer forms geometric isomers,making the best answerchoice A.

Copyright © by TheBerkeley Review® 250 STEREOCHEMISTRY EXPLANATIONS

76.

77.

Choice Bis 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 Cis correct. An E2 mechanism involves the removal of a proton alpha to the leaving group which thenshifts the electrons into a 71-bond forcing the leaving group off. The mechanism is concerted, implying that bothevents occur simultaneously. This results in very predictable geometry with respect to the product alkene. Tosupport the operation ofthe E2 mechanism, both carbons that become involved in the 7U-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 carbon three. Therefore, the best answer is choice C.

Passage XII (Questions 78 - 84) Enantiomeric Excess

78. Choice B is correct. Because the electrophile is a tertiary alkyl halide, Reaction 2 proceeds by way of an SnImechanism. The two products have specific rotations of +62° and +30° respectively as written. A fifty/fiftymixture of the two products would yield an observed specific rotation of +46° (the average of the values for thetwo diastereomers). Because of steric hindrance from the bonds to the six membered ring, the nucleophile(NH3) prefers to attack from the backside of the molecule. This makes the major product Compound A, thefirst product in Reaction 1. The major product has a specific rotation of +62°, thus the specific rotation of themixture is closer to +62° than +30°. The specific rotation is between 46° and 62°, which makes choice B tlie bestanswer.

79. Choice B is correct. Tlie 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 Equation 1 does. Because there is moreof the component with an a = +48° than the component with an a = +18°, the averaged value should be closer to+48° than +18°. However, because it is not purely one component, tlie 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 valuecan be determined mathematically as follows:

aobs = 80% (+48°) + 20% (+18°) = 38.4 + 3.6 = 42°

80. Choice C is correct. The two products formed in Reaction2 both have identical bonds to one another and threechiral centers each. In comparing the two structures, only one of the three chiral centers differs, making thetwo structures diastereomers. If some, but not all, of the chiral centers differ between two stereoisomers, theyare not superimposable nor are they mirror images. This, by definition, makes them diastereomers. The bestanswer is choice C. The 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), they pack differently into theirrespective solid lattices and thus they exhibit different melting points. This eliminates choice B. Because oftheir varying asymmetry, they bind a gas chromatography column differently (due to a difference in sterichindrance) and thus they show different retention times on the gas chromatographer (GC). It can be inferredfrom the last sentence in paragraph two of the passage that the two products have different retention times onthe gas chromatographer (GC), since the concentration values can be determined (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 distinguish the two diastereomers by infraredspectroscopy. Diastereomers have identical bondsas one another. The best answer is choice D even if you 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 conformations must have ethyl and methyl up withthe hydroxyl down. This eliminates choices B and C, which have the exact opposite geometry (ethyl andmethyl are down and the hydroxyl group is up). The most stable conformer has the least steric repulsion,which ideally 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 up/down/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:

Substituent number four is in front,so the arc must be reversed from

clockwise to counterclockwise.

This makes the chiral centerS.

Substituent number four is in back,so the arc is correct as drawn. The

counterclockwise arc makes the

chiral center S.

The correct answer is S, S, S, which makes choice C the correct answer.

H3C J3 % ^

Substituent number four is in back,so the arc is correct as drawn. The

counterclockwise arc makes the

chiral center S.

84. Choice B is correct. In the alkene reactant in Reaction 1, only one carbon has four different substituentsattached, therefore only one carbon is chiral. The best answer is choice B. The chirality is specified in thereactant for the ring carbon with the ethyl substituent attached.

Passage XIII (Questions 85 - 91) Stereochemistry in Synthesis

85. Choice D is correct. This question is asking to determine which reactions generate new chiral centers. Whentreating an alkene with bromine, two bromides are added, one to each of the double bond carbons. The result isthat the two double bond carbons become new chiral centers. This eliminates choice A. When treating analkene with hydrogen gas in the presence of a catalytic metal such as palladium, two hydrogen atoms areadded, one to each of the double bond carbons. The result is that the double bond carbon with the methyl groupbecomes a new chiral center. This eliminates choice B. When treating an alkene with permanganate in basicwater, two hydroxyl groups are added, one to each of the double bond carbons. The result is that the two doublebond carbons become new chiral centers. Thiseliminates choice C. When Compound 3a is treated with a goodnucleophile like ammonia, the acid anhydride is cleaved. The result is that one carbonyl group becomes anamide and the other becomes a carboxylic acid. Neither of these groups are crural, so no new chiral centers areformed. The best answer is choice D.

86. Choice A is correct. It is stated in the passage that Compound 3 represents a mixture of enantiomers, soCompound 3b is the enantiomer of Compound 3a. Enantiomers are mirror images of one another, therefore all oftheir chiralcentersdiffer. Thereare two chiralcenters in Compound3, so Compound 3b must have the oppositechirality of Compound 3a at these two chiral centers. This is true in choice A, so choice A is the best answer.

87. Choice B is correct. To separate a compound from its enantiomer, the technique must select for one of the twoenantiomers. This requires that there is chirality (asymmetry) in the technique. If the mixture travelsthrough a column with one enantiomer attached, the other enantiomer is likely to adhere to the column as ittravels. This means that one enantiomer will migrate quickly while the other travels slowly. This makeschoice B a strong answer. If the column has both enantiomers bound, then both of the free enantiomers insolution will be hindered by the column, slowing equally. This does not result in separation. Choice A iseliminated. The same logic can be used to eliminate choice C . If the distilling column has beads with bothenantiomers, then both of the free enantiomers in solution will exhibit the same affinity for the beads, andwill not separate as well. Choice C is eliminated. The compounds do not interact with the carrier gas in gaschromatography to any notable extent. This means that choice D will do nothing to help separate theenantiomers. Choice D is eliminated.

88. Choice B is correct. Reaction 2 converts Compound 3a into Compound 4, using a peroxyacid. The oxirane canform from either side of the 7t-bond. However, because of the steric hindrance above the ring with the alkene(caused by the six-membered ring), it is preferential to form the epoxide on the back side. Both bonds to oxygenin the epoxide must be on the same side, so choices C and D are eliminated. Because of steric hindrance, choiceB is a better structure than choice A. The best answer is choice B.

Copyright © by TheBerkeley Review® 252 STEREOCHEMISTRY EXPLANATIONS

89. 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 isseparated, 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 material is optically pure at the start, the product of reaction 4 is also optically pure. The bestanswer is choice A.

90. Choice C is correct. Methyl amine, like all amines, is a good nucleophile. The 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 carbonyl, cleaving the anhydride andgenerating a carboxylic acid and amide. Choices A and B can be eliminated immediately, because the three-membered ring is highly unstable and will not reformonce broken. The OH group is a bad leaving group, so it isnot likely that an amine will substitute for the hydroxyl group, once the anhydride is cleaved, so choice D iseliminated. The result is that choice C is the best answer.

91. Choice C is correct. In Reaction 2, a mixture of stereoisomers is formed that is never resolved. As a result, thereactant in Reaction 4, Compound 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 chiral 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.

Not Based on a Descriptive Passage(Questions 92 -100)

92.

93.

Choice A is correct. If you are well versed in using 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 determine 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 2R, 3R

stereochemistry. Choose A; be a chem star.

CH,

OCH, H3CO

OH

CH,

CH,

H in back; R stereocenter

H

OH

H in back; R stereocenter

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 in a three dimensional perspective. Hence, whatever arcyoudetermine fromthe Fischer projection must subsequently be reversed to get the chirality of the stereocenter. In this example,both chiral centers generate clockwise circles in a two-dimensional perspective. But after reversing theclockwise circles to counterclockwise because of the hydrogen atoms are in front, both centers have S chirality.Choose D and be a scholar.

CHo

H -trH

Clockwise arc with H in front; S stereocenter

OH

CH3Clockwise arc with H in front; S stereocenter

OCH3


94. Choice Ais correct. Onthesecond carbon, the OH andHgroups have interchanged, so thatchiral center differsbetween the two stereoisomers. On the third carbon, the OH and H groups have not interchanged, so thatchiral center is identical in the two stereoisomers. On the fourth carbon, the ethyl and methyl groups haveinterchanged, so that chiral center differs between the two stereoisomers. When two out of three (i.e., some,but not all) of the chiral centers differ, the two stereoisomers are neither superimposable, nor are they mirrorimages ofone another, which defines diastereomers. Pick Aand be jovial.

95. Choice B is correct. In the molecule, carbons two and three are chiral centers. However, carbon one is not achiral center, because it contains two equivalent hydrogen atoms. The maximum number of stereoisomers isequal to 2n where n is the number of chiral centers in the molecule, which in this case is 2. This yields a total offour (22) stereoisomers. Be stellarand choose B.

96. Choice Ais correct. Syn addition ofequivalent toa trans double bond in an alkene results in theformation oftwoenantiomers (specifically the R,R and S,S enantiomers) as shown below. Choose A forbestresults. Amesocompound can beobtained from synaddition toa cis alkene, where the alkyl groups areequal on thealkene.

H3C H

H CH,

Trans reactant (asymmetric)

KMn04(aq)

pH = 10

Syn addition(symmetric)

HO OH

H3C*y\**HH

H

CH3

Asymmetric products

&

H CH,

HO OH

97. Choice C is correct. If a compound has stereocenters but is not optically active, this implies that the compoundmust be meso. To be meso,a compoundmust have a mirrorplane in the molecule about which the chiralcentersare evenly displaced. This mirror plane slices the molecule into two equivalent halves. 2R,3S-dibromobutaneand 2R,4S-dibromopentane each have equivalentmirror halves (thus mirror symmetry) so they are both mesocompounds. Cis-l,3-dichlorocyclohexane has equivalent halves as well, thus it is a meso compound. Only2R,4S-dibromohexane does not have two equivalent halves. The correct answer is thus choice C.

98. Choice D is correct. 4-chlorocyclohexanol contains no stereocenters (chiral carbons), therefore it cannot rotateplane polarized light. The result is that 4-chlorocyclohexanol is not optically active. Choose wisely andchoose D. The other three compounds have two chiral carbons and therefore are all optically active. It ispossible for the compound (4-chlorocyclohexanol) to exhibit isomerism in the form of geometrical isomers (cisand trans), but geometrical isomers do not rotate plane polarized light.

99. Choice C is correct. Isoleucine contains two chiral centers, one for the alpha carbon and one in the side chain.Plugging into the stereoisomer equation 2n, where n represents the number of chiral centers, there are fourpossible stereoisomers for the isoleucine structure. Because isoleucine contains two chiral centers that musthave specific orientation, only one of the four stereoisomers has the correct chirality to be biologically useful.Choose C to be a successful point collector.

100. Choice B is correct. For a compound with chiral centers to be optically inactive, it must be meso. and thuscontain an internal mirror plane of symmetry. The molecule must have R, S chirality to be meso. Thiseliminates choices A and C. The mirror plane must be through the C2-C3 bond according to the chiral centers inthe answer choices. This mirror plane is possible only with butane, so the correct answer must be choice B.

"Orient your life with chemistry!"


Section IV

Hydrocarbon

Reactionsby Todd Bennett

Me(

4.

Me

'3 •" "1 ^*3Geraniol

Limonene

OH

0Me

2

1

Me

2

1

J»Me

IT

Me

Alkanes

a) Structures and Physical Propertiesi. Aliphatic and Cyclic Alkanes

b) Alkane Reactivityi. Free Radical Halogenationii. Mechanism

iii. Energeticsiv. Selectivity

Hydrocarbon Reactions

a) Elimination Reactions

i. E2 Mechanismii, Ei Mechanismiii. Carbocation Stabilityiv. Electrophilic Additionv. Regioselectiviryvi. Stereoselectivityvii. Stereoisomer Formation

b) 1,2-Additiori versus 1^Addition

c) Pericyclic Reactionsi. Diels-Alder Reaction

ii. Cope Rearrangementiii. Claisen Rearrangement

Terpenes

a) Classification (Carbon Count)

b) Formation and Synthesis

c) Connectivityd) Spectroscopic Properties

BerkeleyUr.e-v-i^e-w®


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•>

Hydrocarbons and ReactionsSection Goals

Recognize the nomenclature associated with the alkanes and dienes.The Greek prefixes and suffixes associated withthehydrocarbons mustbe common knowledge.Know the terminology so thatwhennames arepresented in passages, youcandraw the structureorrecognize thestructure inananswer choice. Be able to recognize thepositioning oftherc-bondsin conjugated n-systems.

Know the general mechanism for free radical addition to an alkene.Free radical haloeenation ofanalkane follows a simple three-step process. Afree radical halogenis generated by the initiation reaction, requiring heat or light. Tnefree radical halogen starts thepropagation reactions which undergo a continuous chain reaction untilit is terminated. Terminationreactions 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 Ei and E2 mechanism in detail. Know which reactant, solvent, and catalystcombination will result in which mechanism. As a rule, Ei occurs in acid and E2 occurs in strongbase. Rearrangementcan offercomplications in an Ej reaction. E2reactionsrequire the acidicprotonand 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 it-bonds in adiene. You must recognize the difference between the 1,2-addition and 1,4-additions of haloacidsand water acrossa conjugateddiene. Recognize the stereochemical resultsassociatedwith electrophilicaddition reactions of alkenes.

Know the general mechanism for electrocyclic reactions of polyenes.Themost commonelectrocyclic in organic Aemistry is the Diels-Alder reaction. You must recognizeit and be able to predict relative reactivityof different reactions based on electronicfactors andsteric hindrance. You must also recognize that the six-electron transition state in the Etfels-Alderreaction is also found in sigmatropicrearrangementslike the Cope and Claisen rearrangements.Youshould have a basic understanding of 7C-systems.

Recognize the isoprene subunits in natural terpenes.Terpenes arenatural oils derived from a five-carbon species. Isoprene is2-methyl-l,3-butadiene,although the biologically active molecule isa derivative. Terpenes can beextracted and/or distilledfrom natural products. You should be able to identify the isoprene units in biologically commonterpenes and terpenoids.

OrgaillC Chemistiy Hydrocarbons and Reactions Introduction

Hydrocarbons and ReactionsThe crux of organic chemistry centers on reactivity. In organic chemistry thereare many classes of reactions. They are organized according to mechanisticfeatures. Common reaction types include acid-base reactions (section 1),nucleophilic substitution (section 3), free radical reactions (typically halogenationofanalkane or polymerization ofanalkene), elimination, electrophilic addition,electrophilic aromatic substitution, electrocyclic reactions, oxidation-reductionreactions, and carbonyl substitution reactions (section 5). There areother typesof reactions, name reactions, which we shall address by functional group.However, in this section, we shall address the reactions of alkanes and thereactions of Jt-bonds, as classified by their mechanisms.

Free radical reactions follow three basic mechanistic steps: initiation,propagation,and termination. Whether we considerfree radical halogenation ofan alkaneor freeradicalpolymerization of an alkene, the firststep of the processis initiation. Initiation entails the homolytic cleavage of a weak sigma bond inonly a small number of the molecules present in the reaction mixture. Inhalogenation reactions, the halogen-to-halogen bond is cleaved to initiate thereaction. In free radical polymerization, the oxygen-to-oxygen bond of aperoxide initiator is cleaved to generate a low concentration of free radicalcompounds. The product of a free radical halogenation reaction is an alkylhalide. An alkyl halide can undergo nucleophilic substitution reactions (as wesaw in the stereochemistry section) and elimination reactions. Eliminationreactions result in the formation of a 7t-bond, which opens the door to a plethoraof new reactions. Starting with an alkane, most any 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 7t-bond of an alkeneacts as a nucleophile by attacking an electrophile, this starts an electrophilicaddition reaction. The 7t-bond is not limited to being part of an alkene, as it canalso be part of a diene, triene, or any other conjugated system of multiple n-bonds. When there is more than one rc-bond in the system, we must consider thepossibility of the addition reaction occurring at different reactive sites. Thenucleophilic Tt-bond can also be found on benzene, although the aromaticity ofthe rc-system drastically reduces the nucleophilicity of a rc-bond. However, if theelectrophile is strong enough, the rc-bond of benzene can attack it and eventuallyexchange the electrophile for a hydrogen, retaining the aromaticity of the system.This is known as an electrophilic aromatic substitution. The last reaction we shallconsider for rc-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 compounds. Tobest prepare for the MCAT, your goal should 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 section interms of the nucleophilicity of a rc-bond,no matter what compound contains theJt-bond, then you have a solid grasp of the topic at the MCAT 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.


OrgailiC Chemistry Hydrocarbons and Reactions Alkanes

Alkanes

Alkane Structure

Alkanes contain only carbon and hydrogen atoms, and all of the carbons haves/^-hybridization. All of the hydrogens in alkanes use s-orbitals to form bonds.Alkanes are held together exclusively by a-bonds. In any alkane, there are onlytwo types ofbonds present: C—C bonds (which are aSp3.Sp3) and C—H bonds(which are asp3.s). Both types ofbonds present inalkanes are shown in Figure 4-1 as molecular orbitals.

°~sp3-sp3 Gs-sp3

Figure 4-1

Alkanes can be classified as either aliphatic (straight chain form) or cyclic(containing a ring in its structure). Aliphatic alkanes have a molecular formula ofCnH2n + 2 while cycloalkanes with one ring have a molecular formula of CnH2n-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.


CnH2n +2 for linear alkanes (no rings)

Methane CH4 Hexane C6H14

Ethane C2H6 Heptane C7H16

Propane C3H8 Octane C8Hi8

Butane C4H10 Nonane C9H20

Pentane C5H12 Decane C10H22

CnH2n for cycloalkanes (with one ring)

Cyclopropane C3H6 Cycloheptane C7H14

Cyclobutane C4H8 Cyclooctane QH16

Cyclopentane C5H10 Cyclononane C9H18

Cyclohexane C6Hi2 Cyclodecane C10H20

Table 4-1


Organic ChemiStry Hydrocarbons and Reactions

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. Theycan also be defined as lipophilic, whichimplies that they are highly soluble in oils such as the lipid membraneof a cell.Aswith all compounds, their physicalproperties vary with massand 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 physical properties can beobserved.

Isomer NameBoilingPoint

MeltingPoint

Density(g/mL)

Mass

(g/mole)

CH4 Methane -162°C -183°C 16.043

H3CCH3 Ethane -89°C -183°C 30.070

H3CCH2CH3 Propane -42°C -187°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)4CH3 n-Hexane 69°C -95.3°C 0.659 86.178

(H3C)2CH(CH2)2CH3 2-Methylpentane 60°C -154°C 0.654 86.178

H3CCH2CH(CH3)CH2CH3 3-Methylpentane 63°C -118°C 0.676 86.178

(H3Q3CCH2CH3 2,2-Dimethylbutane 50°C -98°C 0.649 86.178

(H3C)2CHCH(CH3)2 2,3-Dimethylbutane 58°C -129°C 0.668 86.178

H3C(CH2)5CH3 n-Heptane 98°C -90.5°C 0.684 100.205

H3C(CH2)6CH3 n-Octane 126°C -57°C 0.703 114.232

H3C(CH2)7CH3 n-Nonane 151°C -54°C 0.718 128.259

H3C(CH2)8CH3 n-Decane 174°C -30°C 0.7362 142.286

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. 268.51 grams/moleD. 280.54 grams/mole

Solution

First, 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 total. 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.

Alkanes


OrgaiUC Chemistry Hydrocarbons 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 radicalhalogenation (more specifically bromination, using Br2, and chlorination, usingCI2) and combustion. Reaction 4.1 is the free radical chlorination of methane.

CH4(g) +Cl2(g) -j^* CH3Cl(g) + CH2C12(1) + CHC13(1) + CC14(1) + C^g)80% 10% minor minor minor

Reaction 4.1

Free Radical Halogenation of AlkanesA free radical halogenationreactionstarts with the addition of activation energyto cleavea 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. Accordingto the octet rule, most atoms wishto have eight valence electrons. In a free radical, there are only seven, so freeradicals are highly 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 canbe angled 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 morestable than primary free radicals. Figure 4-2 shows the relative stabUity of thealkyl free radicals.

R—SSttf/'R > H—£s^ii//R > R-—SSUL'/H > H-

3° Free radical 2°Free radical 1°Free radical Methyl free radical

Figure 4-2

Once a free radical halogenation reaction has been initiated by the addition oflight (activation energy), a free radical halogen atom then attacks an alkane andabstracts a hydrogen from the alkane to leave behind an alkyl free radical. Thehalogen free radical abstracts the first hydrogen it encounters, but because alkylfree radicals can react with other alkanes, over time the distribution favors theformation of the more stable tertiary free radicals. The conversion from aprimary free radical into a tertiary free radical is shown in Figure 4-3below.

H2C \^ u*-" *--j^3^ ^3^- f^C

H\\*y CH2CH3 + Y$ CH2CH3 —• vu* CH2CH3 + J CH2CH3H3C H3C H3C H3CPrimary Free Radical Tertiary 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.


OrganiC Chemistry Hydrocarbons and Reactions

Free Radical Mechanism

The mechanism for a free radical halogenation is a chain reaction processinvolving an initiation reaction, followed by propagation reactions, andultimately a termination reaction. An initiation reaction breaks a covalent bondin a homolytic fashion to form two free radicals, so an initiation reaction goesfrom no free radicals to two free radicals. Homolytic cleavage is different fromheterolytic cleavage in that each atom gets a singleelectron (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 halogen-halogen bond that is broken. Propagation steps involve the abstraction of anatom 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 stepsinvolve 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 asReactions 4.2,4.3, and 4.4.

Initiation: x&Reaction 4.2

Propagation: R-^H + '-X • R- + H—XReaction 4.3a

R^ + X^X • R—X + -XReaction 4.3b

^^Termination: R*. + '.R *- R—R

Reaction 4.4a

r(*£\x • R-XReaction 4.4b

xT^.x • X-XReaction 4.4c

Overall Reaction: R—H + X—X • R—X + H—X

Reaction 4.5

The sum of the propagation steps for a free radical halogenation reaction givesthe overall reaction, Reaction 4.5. The initiation step is brief, for just a splitsecond at the start of a free radical reaction. However, the propagation stepscontinue until the reaction is quenched or the free radicals are completelyconsumed in termination reactions. There are always multiple termination stepspossible. 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.

Alkanes


Organic Chemistry Hydrocarbons and Reactions Alkanes

Example 4.2Which of the following reactions represents an initiation step?

A. H3CCH2CH2- + Br- -» H3CCH2CH2BrB. H3CCH2CH2- + H3CCH2CH3 -» H3CCH2CH3 + (H3C)2CH.C. (H3C)2CH. + Br2 -> H3CCHBrCH3 + Br-D. HOBr -> HO- + Br-

Solution

In an initiation reaction, free radicals 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 eliminates 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. Howcan this best be explained?

A. An ethane molecule attacked an ethyl chloride in a nucleophilic substitutionreaction 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 form

methyl free radicals, which rapidly combine to form long-chain alkanes.

Solution

An ethane molecule has neither a lone pair of electrons nor an available pair ofbonding electrons to share (likea rc-bond), so it is definitely not going to act as anucleophile. This eliminates choice A. An ethyl free radical can definitelyremove a hydrogen from an ethane molecule. However, that does not result inthe formation of butane, it simply regenerates the same molecules.

H3CCH2- + H3CCH3 -> H3CCH3 + H3CCH2-

Choice Bis eliminated. When two ethyl free radicals combine, they form a newsigma bond between the two free radical carbons. Combining the two two-carbon fragments results in the formation of a four-carbon fragment. The resultis that butane is formed from the termination reaction of two ethyl free radicals.The best answer is choice C. Carbon-carbonbonds are not easily cleaved, so theactivation energy added in the initiation step is not high enough to cleave acarbon-carbon bond. Even if it could cleave the carbon-carbon bond to form

methyl free radicals, there is little likelihood that four CH3 groups wouldcombine to form C4Hjq- The loss of two hydrogen atoms would not occur.


OrgaillC Chemistiy Hydrocarbons and Reactions

Free Radical Halogenation EnergeticsWith eachreaction, we consider how muchproductis formed (thermodynamics)and how fast the reaction proceeds (kinetics). AHand AG values determine howmuch is formed while the activation energy (Eact) dictates the speed. For thechlorination ofmethane, Reaction 4.1, the following enthalpy values apply:

CI—CI -> CI- + -CI AHinitiation = 58kcal

H3C—H + -CI -» H3G + H—CI AHi = 1 kcal

H3C- + CI—CI -> H3C—CI + -CI AH2 = -26kcal

Figure 4-4 shows the energy diagram associated with propagation steps inReaction 4.1.

u<VG

Wi—i

«3XSc

Sio

cr + ch4

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. However, this does not mean that bromination proceedsfaster than chlorination. The reaction rate is determined from the activationenergy 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, generating enough energy to explode.

AH values (Kcal/mole)

Halogenation Reaction X = F X = C1 X = Br X = I

Initiation: X2 -> 2X- 38 58 46 36

X- + CH4 -» HX + H3G -32 +1 +16 +33

H3G + X2 -> H3CX + X- -70 -26 -24 -20

CH4 + X2 -> H3CX + HX -102 -25 -8 +13

Table 4-3

Alkanes


Organic Chemistiy Hydrocarbons 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 : 3.8 :1 at room temperature. Attemperatures around 100°C, chlorination selects for tertiary over secondary overprimary by a ratio of roughly 4 : 2.5 : 1. As the temperature increases, thereaction proceeds faster, and is therefore less selective. For the brominationreaction, the relative reactivity of 3° : 2° : 1° carbons is 1600 : 62 : 1 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 distribution forcomparable chlorination and bromination reactions.

CI,

C4H10

Br,

C4H10

CI

Br

hv ^ ^ ' ^ "CI71.7% Secondary 28.3%Primary

hv ' ^ ' ^ Br97.6%Secondary 2.4% Primary

Figure 4-5

As a general rule, slower reactions are more selective than faster reactions,becausereactants have more time to selectthe 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 radical isformed, it has time to abstracta hydrogen from another alkane and form a new,possiblymore stablefree radical. If the new free radical is tertiary, it is likely toonly react with a dihalogen molecule. Consider Reaction 4.6 and the data inTable4-4corresponding to the distribution of mono-halogenatedproducts.

+ X2(g)hv

Reaction 4.6

Trial Halogen Temp (K) Product A Product B Product C Product D

I Br2 298 0.3% 89.4% 10.1% 0.2%

n Cl2 298 29.1% 24.3% 32.0% 14.6%

m Br2 373 0.4% 88.3% 11.1% 0.2%

IV Cl2 373 33.3% 22.2% 27.8% 16.7%

V 12 373 No Rxn No Rxn No Rxn No Rxn

Table 4-4


OrganiC Chemistry Hydrocarbons and Reactions Alkanes

The data in Table 4-4 confirms that bromination is more selective than

chlorination 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 really knows if Boltzmann said it, or if it's even a law. The keythingis that as energy is added, lessstablecompoundsare formed.

Example 4.4Why inTrial IIofTable 4-4 is Product A formed to a greater extent thanproductB?

A. Product A results from the more stable free radical, thus it is selected for.B. Product Brearranges 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 Bresults 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 choiceC. It is often possible to answer a questionwithout full 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 toProduct 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 aremore hydrogens available to form Product AthanProduct 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.

Solution

Because 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 chlorine-chlorine bond and the brornine-brominebond. BecauseCI2and Br2 are cleaved, it is safe to assume that I2 is even easierto cleave. This eliminates 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 completelyoppositereasons. Fluorine forms strong bonds to carbonand hydrogen, and thefluorine-fluorine bond is weak. The products are so much more stable than thereactants that the free radical fluorination reaction is explosive.

Example 4.6Why are there minimal di-halogenated products formed in the free-radicalchlorination of an alkane?

A. The addition of the halogen makes the alkyl halide less acidic than thealkane, so it is less reactive tosubsequent halogenation reactions.

B. The addition of the halogen makes the alkyl halide more acidic than thealkane, so it ismore reactive tosubsequent 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 halogen is added to the alkane, the weakest bond is a carbon-halogen bondand not the carbon-hydrogen bond. As a result, it is easiertoremove the halogen rather than the hydrogen from the mono-substitutedalkyl halide.

Solution

Halogens are electron-withdrawing, 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-halogenation preferable. Once analkane is halogenated, the weakest bond is the carbon-halogen bond, not acarbon-hydrogen bond. If a second halogen free radical reacts with an alkylhalide (rather than analkane), it preferentially removes thehalogen (breaking theweakestbond), forming a non-halogenated alkyl free radical. This is because thereverse halogenation reaction (in propagation) is more favorable than theremoval of a hydrogen and subsequent additional halogenation reaction. ChoiceD is the best answer.



Example 4.7To synthesize a primary alkyl halide from an alkane in highest yield, whatshould be done?

A. Bromination at 25°C

B. Chlorination at 25°C

C. Bromination at 100°C

D. Chlorination at 100°C

Solution

A primary alkyl halide is the least favorable product, so the best free radicalhalogenation reaction is the one with lowestselectivity. 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 is2-bromo-2-methylbutane following the free-radical bromination of2-methylbutane at 75°C?

A. 88.1%

B. 88.7%

C. 89.1%

D. 89.7%

SolutionTo answer this question, youneed to readTable 4-3. In trials I andIII, product Bis2-bromo-2-methylbutane. Therefore, we need toestimate how much Product Bis formed at 75°C. At 25°C, thereis 89.4% productBformed, while at 100°C thereis88.3% product Bformed. This means that the amount ofproduct Bformed at75°C should be between 88.3% and 89.4%, which eliminates choices A and D.The amount formed at 75°C should be closer to 88.3% than 89.4%, so choice B isthe best answer.

Example 4.9How many mono-chlorinated structural isomer products are possible when 2,5-dimethylhexane undergoes free radicalchlorination?

A. 3

B. 5

C. 6

D. 8

SolutionThis question isasking for how many structural isomers there are for chloro-2,5-dimethylhexane. Because of themirror plane through the carbon 3-to-carbon 4,bond, there are three unique carbons on 2,5-dimethylhexane. This means thatthere arejust three carbons thatcanbe chlorinated, so there areonly three mono-chlorinated structural isomers. If stereoisomers were included, the value wouldincrease to four, given that chlorination of the secondary carbon yields astereogenic center. The best answer is choiceA.


OrCjaniC C^lieiniSvry Hydrocarbons and Reactions Hydrocarbon Reactions

Hydrocarbon ReactionsElimination Reactions

The reaction that forms an alkene from a substituted alkane is elimination. It is

named from the fact that a functional group and a hydrogen on adjacent carbonsare eliminated in order to form a 7u-bond. The reaction requires elevatedtemperatures to help overcome the activation energy and to push the reaction inthe forward direction. Like the nucleophilic substitution reactions, there are tworeaction mechanisms, appropriately named E} and E2. As with nucleophilicsubstitution, the two versions are named also for their reaction orders (kineticrate dependence). E] is similar to Sjsjl and E2 is similar to Sn2, except that theproduct is an alkene. In an Ei elimination, the leaving group first leaves andthere is a carbocation formed. The empty p-orbital of thecarbocation eventuallybecomes one of the two p-orbitals in the new 7U-bond. In an E2 elimination, a baseremoves an alpha hydrogen to force the leaving group off of the neighboringcarbon. Elimination converts a functionalized alkyl group into an alkene.

E2 Reaction (Carried out Under Basic Conditions at High Temperature)The E2 reaction is concerted (one-step) like the Sn2 reaction, with one majorexception being that the E2 reaction occurs at higher temperatures (temperaturesabove ambient temperature) than theS^2 reaction. An E2 reaction also requiresthat the base be bulky. Because of the steric hindrance associated with a bulkybase, it is less apt to act as a nucleophile and thereby minimize the competingSjsj2 reaction. An important feature of the E2 reaction is that the substituentsbeing eliminated must beanti tooneanother (have a dihedral angle of 180°). Tliemechanism for an E2 reaction is shown in Figure4-6.

H

H3C

H,C

:or"

H

R.O

1

H

H

H3C

t - /H

^_ H-iO/Mr..^ rtttttlH

H^X"

H^7\A/\\H

H

X

l-I

Figure 4-6

In an E2 reaction, the compound may have to rotate to proper conformationbefore the reaction canproceed to form thealkene. This is tosay that theleavinggroup must be oriented anti to the hydrogen being eliminated before the baseattacks theproton tostart thereaction. Whenever thereare two alpha hydrogensthatcanbeeliminated from the starting reagent, there are two possible products.The consequence is that whenthe startingmaterial has a specific stereochemistryat one of the reactive carbons, a corresponding specific geometrical isomerproduct (either cis or trans) forms. Figure4-7shows an example where there aretwo alpha protons to choose between. Oneof the two protonshas been replacedby deuterium, so that the structure can be monitored and the product can beused to support the idea that the elimination reaction occurred from the antiorientation. Figure 4-7shows the rotation to anti and formation of both products.


UrC}aniC v^nemiStry Hydrocarbons and Reactions Hydrocarbon Reactions

120° rotationLiNRn

Br LiNRn

D H

H3C CH,

H3C H

•HH CH,

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 would include the cis and trans alkenesfeaturing the deuterium, and the cis and trans alkeneswith no deuterium. Figure4-8 shows the two alkenes not formed in the reaction in Figure 4-7.

H

H3C

H

~(CH,

H3q

H

D

CH,

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, only the products with an internal rc-bond aredrawn. The internal rc-bond is more favorable then the terminal rc-bond by about1.5 kcals, making 1-butene a minor sideproductin thereaction inFigure 4-7. TherelativestabUity of substituted alkenes is shown in Figure4-9.

R Rm R H

H > XR' R"

Tetrasubstituted

R' R"

Trisubstituted

R H R H

H R'

Disubstituted

H H

Monosubstituted

Figure 4-9

Terms such as anti-periplanar orientation shouldsound familiar and immediatelymake you think of an E2 reaction. Any time you see a strong bulky base, ahydrocarbon with a leaving group,and high temperature, youshouldthinkofanE2 reaction.


OrganiC CJoemiStry Hydrocarbons and Reactions Hydrocarbon Reactions

Ei Reaction (Carried out Under Acidic Conditions at High Temperature)The Ei reaction also yields an alkene, but it goes through a different mechanismthan the E2reaction. The mechanism for the Ei reaction is quite similar to that ofthe SnI reaction. In both the Ei and Sjvjl reactions, there is (1) a carbocationintermediateformed after the leavinggroup leaves in the rate determining stepof the reaction and (2) the possibility of rearrangement with the carbocationintermediate. A schematic for a typicalEi reactionis shown in Figure 4-10.

H3C

H

HÔH

\CH,

+ H20

Figure 4-10

To recognize Ei, as opposed to E2, it is easiest to look at the reaction conditions.Ei 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 smallquantity. The reaction in Figure 4-11 shows the formation of both the 2-alkenemajorproduct and the1-alkene side product.

ho: ^ch2ch3CH,

H3C\îHoC H

H3C///,,^®

H-CH2CH3

@o0 tH3C^lli

<CH2CH3

yi-îihuj®.h3ci

H20

CH2CH3

CH,

,rtuxCH3,CH2CH3 mill CH2CH3CH,

CH,

Figure 4-11


OrCjaniC l^nemiStry Hydrocarbons and Reactions Hydrocarbon Reactions

Example 4.10What is the major product when 2-methylcyclopentanol is treated withconcentrated sulfuric acid at 50°C?

CH, CH,

CH,

Solution

Atanelevated temperature in thepresence ofa strong acid, the reaction proceedsbyanEi mechanism, so theproduct is a highly substituted alkene. Choice A iseliminated, because it is a terminal alkene. Choice Bis 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 tertiarycarbocation. Thebest answer is choice C,becausethe rc-bond ishighly substituted andit includes the tertiary carbon, implying thatit was formed via a tertiary intermediate.

Rearrangement is possible with carbocation intermediates, so Ei reactions aresusceptible to forming rearranged products. Hydride and alkyl shifts are rapid,intramolecular processes that occur before the solvent can remove theneighboring proton to forman 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. Forcarbocations conjugated to a 7t-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 asecondary carbocation (R2CH+) is formed, it can rearrange to a tertiarycarbocation (R2R'C+), ifa tertiary carbocation ispossible. Figure 4-12 shows therelative stability of alkyl carbocations.

„r_®?....^CH3 _JiUuu\CH3 HC_sR^lH >h—^c<fH3 8^ S^CH3 6 H cTH

2° Carbocation 1°Carbocation Methyl Carbocation3° Carbocation

Figure 4-12

The conclusion that can be drawn from the relative stability of alkyl carbocationsis that methyl groups are electron donating toelectron poor carbons, which canbe extrapolated tosay that alkyl groups are electron donating. Figure 4-13 showsthree rearrangements where a less stable carbocation is converted into a more

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OrQaiHC ChemiStry Hydrocarbons and Reactions Hydrocarbon Reactions

stable carbocation via a hydride shift. The third example forms an allyliccarbocation, which is more stable than alkyl carbocations, because of resonance.

H H

I IH\© /CS(L'H H\ ©/^'H,w — „/-\ HH h h: ,. H1°

CH, CH3

.c—c© vrn • c—c. ^n3

H^ \ 3 H* *„H 2° H «o H

H H H H\ / \ /C=C H ^ C=C H

/ \ ©/ *~ / \ © /H C—C H JC—Q.

H_f—^ H H VjHri H

1° 2°Allylic

Figure 4-13

Example 4.11Which ofthefollowing carbocations isapt to undergo rearrangement?A. (H3C)3C+B. (H3C)2CH+C. H3CH2C+D. (H3C)2CHCH2+

Solution

Choice Ais already a tertiary carbocation, so it hasno reason to rearrange, leastofall toa primary carbocation. Choice Aiseliminated. Choice Bis a secondarycarbocation, but can only rearrange to form a primary carbocation, so it iseliminated. Choice Cisa primary carbocation that only hasprimary carbons, soit is eliminated. In choice D, a hydride shift can covert a primary carbocationinto a tertiary carbocation, making choiceD the best answer.


OrganiC OnemiStry Hydrocarbons and Reactions Hydrocarbon 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 7t-bond acting as nucleophile by attacking an electrophile toform a substituted alkane product. In the first step of electrophilic additionreactions, the 7i-bond of the alkene donates its electron density to an electrophile.The first step of a generic electrophilic addition reaction is shown in Figure 4-14.

\£~\_\_A - • .

\ Step IE+ represents anyelectrophile (lone pairacceptor)

Figure 4-14

After the electrons from the 7t-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. Thesecondstep of a genericelectrophilicaddition reactionis shown in Figure4-15.

H—Nul^ ©H—Nu

Step IIH—Nuc represents any nucleophile (lonepair donor)

Figure 4-15

It is important torecognize thatwhenthere isa carbocation intermediate, there ispotential for rearrangement. If the carbocation is unstable, then prior to theattack by a nucleophile, thecarbocation can rearrange by way ofa hydride shiftor alkyl shift to form a morestable carbocation. Rearrangement is not showninthis example, but it canoccur between steps one andtwo ofthe mechanism. Thenucleophile should bea weak base, otherwise itcan deprotonate the intermediatetocarry outthe reverse reaction (elimination) andregenerate analkene. To avoidthis, electrophilic addition reactions arecarried outunderacidic conditions. Thefinal step is theneutralization of thecationic product, which is carried out by asolvent basic enough to deprotonate the cationic species. The final step of ageneric electrophilic addition reaction, a workup step, isshown in Figure 4-16.

Solvl

Ht—Ni? x NuStep III (workup)

Solv represents any polar/protic solvent

Figure 4-16

A proticsolventis capable of forming hydrogenbondsand transferring protons.


Organic Chemistry Hydrocarbons and Reactions Hydrocarbon Reactions

It is a good idea to understand basic mechanism of electrophilic addition todienes and to recognize the class of reaction. Know how to draw 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 are completelyunderstood. Later in this section we will look at a variations on this same

mechanism 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 equally substituted. Because the two carbons are not equallysubstituted, one is less sterically hindered than the other. As a mechanistic rulein electrophilic addition reactions, the first substituent attacks the less hinderedcarbon of the alkene, leaving the second substituent to add to the other carbon.Generally, the less hindered carbon of the intermediate is the alkene carbon thatwasn't attacked by the first substituent. A Markovnikov 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 anti-Markovnikov addition product 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 additionrefers to a reaction where the two new substituents add to the same side (face) ofthe alkene reactant or intermediate. 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 stereochemistryof any newly formed chiral centers.

Stereoisomer Formation

You 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 likely 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 asymmetry influences the stereoselectivity ofthe reaction. There is a greater chance to attack the alkene from the less hinderedface 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 vJtiemiStry Hydrocarbons and Reactions Hydrocarbon Reactions

1,2-Addition to a Conjugated Diene vs. 1,4-Addition to a Conjugated DieneElectrophilic addition reactions are not limited to alkenes with isolated 7C-bonds,as these reactions also take place on conjugated dienes. With conjugated dienes,there are multiple potential products. 1,2-addition refers to an addition reactionthat adds substituents to the first and second atoms (usually carbon) in theconjugated 7t-network. 1,4-addition refers to an addition reaction that addssubstituents to the first and fourth atoms in the 7i-network. As a general rule, 1,4-addition is favored at higher temperatures, because the more stable product isformed (thermodynamic control). At lower temperatures, the more stableintermediate dictates the product, so 1,2-addition is favored. This is known askinetic control. There is always the option to add either 1,2 or 1,4 when thesystem has conjugation. Figure 4-17 shows the product distribution for thehydration of 1,3-butadiene at two different temperatures.

H

^OHH2C=CH H2S04(aq)> H3C—C^ H3C- CH

HC=CH2 °C HC=CH2 HC—CH2OH62% 38%

1,2-Additon Product 1,4-Additon Product

H^OH

H2C=CH H2SOf(aq)> H3C~< ^"^HC=CH2 5°C HC=CH2 HC—CH2OH

16% 84%

1,2-Additon Product 1,4-Additon Product

Figure 4-17

In the example in Figure 4-17, there are two possible carbocation intermediates.The mechanism and both intermediates are shown in Figure 4-18. To start thereaction, the proton adds to the terminal carbon (least hindered carbon) leavingthe carbocation on carbon two (a secondary carbocation). The carbocation isallylic meaning it can resonate through the rc-network. It is in essence apropylene cation that can have cationic density at either end of the 7C-network.What this means is that the carbocation can resonate to the terminal carbon(carbon number four). This forms a primary carbocation, which is notas 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 thatat higher temperatures there aremore primary carbocations than at lower temperatures. This is why we see theproduct distribution favoring substitution of the alcohol at carbon two at lowertemperatures while we athigher temperatures we observe the substitution ofthealcohol at the fourth carbon. Figure 4-18 shows the energy diagram andmechanism associated with this reaction, which includes two possible pathways.


Organic Chemistry Hydrocarbons and Reactions Hydrocarbon Reactions

©H

JB>C= CH

Kinetic Pathway Thermodynamic Pathway(favorableat lower temperatures) (favorable at higher temperatures)

ft2° carbocation is most stable

H

H,C—C

ft1° carbocation is least stable

H,C—CH

^ /HC C©

\

H

\HC=CH2

0d^\HC—CH2

H

0>

+ H2Q

H©£*OH2

H,C—C

\HC=CH2

-H+

H

H3C C

\HC=CH2

Monosubstituted alkene

is least stable

1" carbocation

+ H20

H,C—CH

^ ©HC—CH2OH2

-H+

H,C— CH

HC—CH2OH

EHsubstituted alkene

is most stable

Lower energy intermediateleads to the kinetic product.

Higher energy intermediate leadsto the thermodynamic product.

Reaction Coordinate

Figure 4-18

The energy diagramshows the two possible pathways for the reaction and theirrelative energetics. The dashed pathway represents 1,2-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 givenreactiontemperature.


OrganiC CJiemiStry Hydrocarbons and Reactions Hydrocarbon 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.Pericyclicreactions include cycloaddition reactions, which most notably include theDiels-Alder reaction, sigmatropic rearrangement, and electrocyclic reactions. We willaddress only cycloadditions and sigmatropic rearrangement. The significantdifference between a cycloaddition reaction and sigmatropic rearrangementinvolves the number of molecules. In cycloaddition, two separate compoundscome together, resulting in a single new compound. In sigmatropicrearrangement 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 of the atoms and not thespin of the electrons within the molecular orbitals. The first reaction we shallconsider is the Diels-Alder reaction.

Diels-Alder Reaction

The Diels-Alder reaction, an electrocyclic addition reaction, involves the additionof a conjugated diene (4 Jt-electrons) to an alkene (2 Jt-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.

0QBenzene Resonance

Diels-Alder Reaction

Figure 4-19

Sixit-electrons in a cyclic rc-network make benzene aromatic, so we refer to thetransition state of a Diels-Alder reaction as aromatic (containing 6 7t-electrons in aring). Diels-Alderreactions 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 drawn in Figure 4-20.

diene dieneophileO

cyclohexene derivative

CH,

C=>CH,

CH2CH3

First: draw the

cyclohexene ring

Figure 4-20


CH2CH3

Second: draw the rest of

the molecule connected

to the numbered carbons

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OrganiC ChemiStry Hydrocarbons 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.

O

iTiT^oo

Endo Product

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.

EWG EWG EDG©

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 4-23shows a Diels-Alder reaction where regioselectivity is an issue.

X

O

XH

H,CO^VS- S+^ CH,

Figure 4-23


H3COMajor

The Berkeley Review

OrganiC CnemiStry Hydrocarbons and Reactions Hydrocarbon Reactions

Example 4.12Which of the following Diels-Alder reactions is fastest?

^0*^ CH3

O

CH,

CH3 H3C CH3

C.

H3C

D.

H,CO^

CH,

H3C CH

CH,

O

H3CO^^Sv//"^CH3

H3C^^^CH3CH3

O O

HoAAabH3C CH3

Solution

The rate of a Diels-Alder reaction is increased by the presence of an electrondonating group on the diene and an electron 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 molecule alignsentering the transition state, one of the methyl groups will be in the way. InchoiceB, there is a methoxy group on the diene, while in choice A is just a methylgroup. 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 involving two pi-bondsand one sigma-bond. Figure 4-24 shows a simple Cope rearrangement carriedout on a 1,5-diene. In more complex examples, stereochemistry may be an issue,because stereocenters can be both formed and lost as hybridization changes.

Figure 4-24


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 allyl 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 4-25shows a Claisen rearrangement.

Figure 4-25

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 4.13Which 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 allyl etherwhen the vinylic ether carbon is unsubstituted.

Solution

Choices C and D are eliminated immediately, because Cope rearrangementresults in the conversion of one 1,5-diene into another 1,5-diene, not an aldehyde.The question is reduced to determining which carbon in the reactant forms thecarbonyl group following Claisen rearrangement. The reaction is shown below.

H H

vinylic etherL ^AaldehydeU"o^ o^Nallylic ether ^

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 ahydrogen, and no carbons, in order to form an aldehyde and not a ketone. Thebest answer is choice B.


Organic Chemistry Hydrocarbons and Reactions

Classification

Terpenes and terpenoids, biological molecules derived from terpenes, arenatural hydrocarbons found in plants and animals that are made from 5-carbonisoprene (2-methyl-l,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 vitro synthesis of terpenes and terpenoids is called natural productsynthesis. Some naturally occurring monoterpenes are shown in Figure 4-26.

Terpenes

y^^^^Ky^ X^^V.^A^Myrcene

(Oil of Bay)

CH,

OH

CH, O

Citronellal

I CH, O

CitronellolGeraniol

(Oil of Germanium)

Limonene

(or Limin)

O

oc-Terpinene(Oil of Coriander)

^ j^XXHy-Terpinene

(Oil of Coriander)

Citral

(Oil of Lemongrass)

H3C^^CH3

Carvone

(Oil of Spearmint)

OH

Menthol

(Oil of Peppermint)

Figure 4-26

CH,

CH3a-Pinene

(Oil of Turpentine)

Studies in biogenesis show that the large terpenes are synthesized usingisopentenyl 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.

H3C oCamphor

Isoprene

O OII

O-P-O-P-OHI I

O" O"Isopentyl pyrophosphate

O OII II

O-P-O-P-OHI I

O" O"Dimethylallyl pyrophosphate

Figure 4-27

These molecules add to one another in a way where the it-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 n-bond. The reactioninvolveshead-to-tail addition. Whencyclizing, the bond thatis formed to complete the ring is rarely connectedhead-to-tail. Figure4-28 showsthe reaction of isopentenyl pyrophosphate and dimethylallylpyrophosphate.


OrgailiC Chemistry Hydrocarbons and Reactions Terpenes

Isopentyl pyrophosphate

O OII II

HO-P-O-P-OI I

O" O"

Geranyl pyrophosphate

j-j Dimethylallyl pyrophosphateHO

OII

-P-1

O"

O

OII

-P-1

O"

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-15terpene). 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.

O OII II

O-P-O-P-OHI I

O" O"Isopentyl pyrophosphate or

(3-Methyl-3-butenyl pyrophosphate)

O OII II

O-P-O-P-OHI I

O" O"Dimethylallyl pyrophosphate

(3-Methyl-2-butenyl pyrophosphate)Isopentyl

pyrophosphate

O OII II

O-P-O-P-OHI I

O"Monoterpenes (C10) Geranyl pyrophosphate O(C10-pyrophosphate)

Isopentylpyrophosphate

Sesquiterpenes (C15) -<-

Diterpenes (C20)

Farnesyl pyrophosphate(C15-pyrophosphate)

Isopentylpyrophosphate

C20-pyrophosphate

*Tetraterpenes (C40)

Figure 4-29


O OII II

O-P-O-P-OHI I

O" O"

Squalene (C^)

ILanesterol (C^)

tChloesterol (C^)

The Berkeley Review

Organic Chemistry Hydrocarbons and Reactions

As Figure 4-29shows, the triterpene squalene can undergo further reactivityto generate cholesterol, which does not have a number of carbons 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 startingfrom isopentenyl pyrophosphate is shown in Figure 4-30.

O O

O—P— O— P—OH

Io^ a

Isopentenyl pyrophosphate

Cholesterol

Figure 4-30

As a general rule, smaller terpenes are found primarily in plants, while somelarger terpenes, such as lanesterol (a C-30precursor to steroid hormones) and 6-carotene (a C-40 source of vitamin A), are found in plants and animals. Forinstance, the monoterpene pinene is found only in plants while Vitamin Ai, aditerpene, is found in both plants and animals. Figure 4-31 shows some selectedlarger terpenes.

Elemene

(Oil of Coral)

C-Selinene

(Oilof Celery)

Farnesol

(Oil of Lemon)

X^-'^xA^^^

Vitamin A,

CH,

H3C CH3 CH3 CH3 OH

Figure 4-31

Terpenes


Organic Chemistry Hydrocarbons and Reactions Terpenes

Zingerberine

One of the skills that you must develop 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.

Patchouli 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 oils 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. Terpenescan alsobe extracted from pulp.When isolating terpenes, it is a mixture of geometrical isomers that is collected.Thedifferent geometrical isomers ofa terpeneare given the prefixesa-, C-, y-andso on. The different geometrical isomers have similar physical properties, butbecause ofdifferences in conjugation, they exhibitdifferences in the absorptionofphotons in the ultraviolet (UV) and visiblerange of the EMspectrum.

Terpenes are UV active, because of their rc-bonds. An isolated alkene has aUVabsorbance around 180 nm. A conjugateddiene has a UVabsorbancearound225 nm, which is significantly more intense than the absorbance of an isolatedalkene. As the conjugation of a rc-network increases, the wavelength ofmaximum absorbance, Xmax, and the intensity of absorbance, £, increase. Someterpenes contain oxygen, which is added in a way that does not alter the carbonskeleton. Carbonyls exhibit absorbances of greater wavelength than alkenes ofthesameconjugation. Forinstance, carvone (shown in Figure 4-26) is evidentbya carbonyl absorption at 1744 cm"1 in its IR spectrum and a strong UV (£ >10,000) absorption at Xmax = 242 nm. Terpenes are often isolated in educationallaboratory experiments. Because of theirbiological significance and the fact theyare isolatedin labexperiments, theyare highly representedon the MCAT. Ifyouhave a fundamental understanding of terpenes, then you should be fine.


Organic ChemiStry Hydrocarbons and Reactions Section Summary

Key Points for Hydrocarbons and Reactions (Section 4)

Alkanes

1. Hydrocarbon compounds with only carbons,hydrogens, and sigma bonds

a) Only contain C—C and C—H single bonds

i. Can be aliphatic (straight chain) or cyclicii. Low water solubility, low boiling point, and low melting pointiii. Relatively inert compounds that are used as solvents

b) Undergo free radical halogenation reactions with chlorine and brominei. Involves initiation, propagation, and terrnination in that orderii. Bromination is more selective than chlorination

iii. The relative stability for alkyl free radicals is: 3* > 2° > 1* > methyl

Hydrocarbon Reactions (Reactions involving 7C-bonds)

1. Elimination

a) Loss of an H and a leaving group to form a 7t-bondi. Requires high temperatureii. Competes with nucleophilic substitution reactionsiii. Goes by way of one of two mechanisms: Ei or E2

b) Ei reactions are similar to Sjsjl reactions

i. Requires a strong acid (Bronsted-Lowry or Lewis)ii. Forms carbocation intermediate so rearrangement is possibleiii. Forms most substituted and least stericallyhindered alkene

c) E2 reactions are similar to Sn2 reactions

i. 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. Electrophilic Addition

a) An electrophile can be added to a rc-bond followed by nucleophihc attacki. The 7t-bond is a weak nucleophileii. The reaction is driven by the strength of the electrophile

b) Electrophilic addition reactions exhibitregioselectivityi. Sterichindrance and carbocation stability influence the site of attackii. When the electrophile attacks the less substituted carbon, the

reaction is said to be a Markovnikov addition

iii. When the electrophile attacks the more substituted carbon, thereaction is said to be an anti-Markovnikov addition

c) Electrophilic addition reactions exhibit stereoselectivityi. Sterichindrance is the most influentialfactorin stereoselectivityii. When substituents add one at a time (the first is added before the

second on attacks) the product exhibitsanti addition stereochemistryiii. When substituents add simultaneously (both add at the same time)

the product exhibitssyn addition stereochemistry

Copyright©by The Berkeley Review 285 Exclusive MCAT Preparation

OrganiC Chemistry Hydrocarbons and Reactions Section Summary

3. 1,2-Addition versus 1,4-Addition

a) Electrophilic addition reaction can proceed in multiple ways withconjugated rc-systemsi. Typically, 1,2-addition is favored at lower temperatures (kinetic

control)ii. Typically, 1,4-addition is favored at higher temperatures

(thermodynamic control)

4. Diels-Alder Reaction

a) Reaction of a diene and a dienophile (alkene)

i. An electrocyclic reactioncarried out with either light or heatii. 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 (Copeand ClaisenRearrangements)

a) Both sigma-bonds and pi-bonds are broken and formed via a cyclictransition state

i. Cope rearrangement converts a Y»8-unsaturated alkeneinto anotherY,8-unsaturated alkene via a realignment of molecular orbitals

ii. Claisen rearrangement convertsa vinylicallyl ether into another 7,6-unsaturated carbonylvia a realignment of molecular orbitals

iii. Sigmatropic rearrangement requires heat

Terpenes

1. Natural productsderivedfromthe connecting of five-carbon unitsa) Derived via biosynthesis involving either isopentyl pyrophosphate or

dimethylallyl pyrophosphate

b) Can be cleaved into isoprene subunitsi. Terpenes are named for their carbon count:10C = monoterpene, etc.ii. Isolated by steamdistillation or extraction as natural oilsiii. Presence of Jt-bonds results in UV absorbances. As conjugation

increases, intensity and Xmax both increase.

^Wf^^^^^^M^i^^^M^}'WWM^

"3.14? Phooey, n is not a number, it's a nucleophile"


Alkanesand

HydrocarbonReactions

Passages

14 Passages

1 00 Questions

Suggested schedule:

I: After reading this section and attending lecture: Passages I, III, VII, X, & XIIIGrade passages immediately after completion and log your mistakes.

Following Task I: Passages IV, VI, VIII, & XII (27 questions in 35 minutes)Time yourself accurately, grade your answers, and review mistakes.

Review: Passages II, V, IX, XI, XIVI, & Questions 93 - 100Focus on reviewing the concepts. Do not worry about timing.

II

III:

"The

BERKELEYJLIr-e-v-i^e-w®


Alkanes and Hydrocarbon Reactions

I. Free Radical Halogenation Selectivity

II. Free Radical Reactions

III. Elimination and Stereochemistry

IV. Elimination Study

V. Phermones

VI. Green Synthesis

VII. Conjugated 7t-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


(1 -7)

(8- 14)

(15 - 21)

(22 - 28)

(29 - 35)

(36-41)

(42 - 48)

(49 - 54)

(55 - 60)

(61 - 67)

(68 - 73)

(74 - 80)

(81 - 86)

(87 - 92)

(93 - 100)

Alkanes and Hydrocarbon Reactions Scoring Scale


84 - 100 13 - 15

66 - 83 10 - 12

47 - 65 7 -9

34 -46 4 -6

1 - 33 1 -3


Alkyl halides can be formed from a halogen reacting withan alkane. Alkanes can be treated with either chlorine gas andUV radiation to form chloroalkanes, or with bromine liquidand UV radiation to form bromoalkanes. The bromination of

an alkane is a slower and more selective reaction than the

chlorinationof 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 free radicalsfollows the trend 3° carbon > 2° carbon > 1° carbon by afactor of 4 : 2.5 : 1 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 alone. 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 : 1, the product distribution instead is five 2-chlorobutanes to three 1-chlorobutanes. This implies that thepercentage of secondary products is 62.5% (rather than 40%whichis 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 : 1 ratio. Six hydrogens are secondary and six areprimary. The abundance of the primary hydrogens is 6hydrogensx 1 reactivity for 6. The abundance for carbon twoand carbon four is 4 hydrogens x 2.5 reactivity for 10. Theabundancefor carbon three is 2 hydrogens x 2.5 reactivity for5. This means that the chlorination product ratio is six 1-chloropentaneto ten 2-chloropentane to five 3-chloropentane.

1. How many degrees of unsaturation are there in thecompound C5H9CI?

A. 0

B. 1

C. 2

D. 3

2. How many structural isomers of C5H1iCl are there?

A. 5

B. 6

C. 7

D. 8


The chlorination of methylcyclopentane would yieldhow many different structural isomers?

A. 2

B. 3

C. 4

D. 5

What is the most abundant product in the brominationof 2-methylbutane?

A. l-bromo-2-methylbutane

B. 2-bromo-2-methylbutane

C. 2-bromo-3-methylbutane

D. l-bromo-3-methylbutane

Following the free radical monochlorination of pentane,what is the ratio of 2-chloropentane to 1-chloropentane?

A. 2:3

B. 5:3

C. 2.5: 1

D. 3: 1

In the monochlorination of n-hexane, how can a ratio of1.07 2-chlorohexane to one 3-chlorohexane be

explained?

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 of n-hexane can better stabilize a

free radical due to resonance.

D. The third carbon of n-hexane can better stabilize a

free radical due to the inductive effect.

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 (C3H7O

B. Cyclopropyl free radical (C3H5O

C. Butyl free radical (C4H9O

D. Cyclobutyl free radical (0^7*)



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 form asigma bond. The reaction involves two transition states, inwhich the second is of higher energy than the first. Table 1shows the bond energies of the halogens and the reactionenthalpies for the various halogenation reactions:

Compound B.D.E. AHrx

F2 154 ^mole

-483 "mole

Cl2 239 ^mole

-114 wmole

Br2 193 Umole

-33 Wmole

12 149 wmole

+27 wmole

Table 1

The enthalpy of a chemical reaction can be found byusing Equation 1.

AHrxn = X Energybonds broken - Energybonds formed

Equation 1

The average bond dissociation energy for a sigma bondbetween an .^-hybridized carbon and a hydrogen is413 kJper mole. A hydrogen free radical cannot be formed in thisreaction mechanism.

8. The MOST stable type of carbon free radical formed inthe monobromination of (R)-3-methylhexane is bestdescribed as:

A. primary.

B. secondary.

C. tertiary.

D. quaternary.


9. Which of the following energy diagrams corresponds toreaction of I2 with an alkane?

A.

cW

C.

uc

W

Reaction Coordinate

Reaction Coordinate

B.

n

D.

A

c

Reaction Coordinate

Reaction Coordinate

10. The first propagation step in a free radical reaction iswhich of the following?

A. X2 + R- —> RX + X-

B. RH + X- —> HX + R-

C. X2 —> 2X-

D. X- + R- —> RX

11. 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. X2 + R- —> RX + X-

C. RH + X- —> HX + R-

D. X- + R- —> RX


13. Which of the following conclusions can be inferredfrom the observation that usually only one halide reactswith the alkane and minimal multiple halogenatedproducts are isolated from the product mixture?

I. Halogens do not help to stabilize free radicalintermediates.

II. Halogens, once on an alkane, increase the C-Hbond strength.

m. Halogens, once on an alkane, decrease the C-Hbond strength.

A. Ilonly

B. monly

C. I and II only

D. I and III only

14. Given that a C-F bond energy is 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-^-mole

B. -116-^1-mole

C. +116 —kJ—mole

D. +483-kL_mole



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 reactions

alkenes 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 E\ 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 cis

geometrical 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 of a total synthesis process:

Br „H

H3c CH2CH

t-butOK

t-butOH1

CH2CH3H3C I

•HH CH3

Reaction 1

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 aligned for E2elimination, nor is the structure drawn in its most stableconformation. Counting the conformer shown in Reaction 1,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 elimination

reaction that generates the following data?

Trial (H3C)3CBr t-ButOK Rate

I 0.25 M 0.25M 4.61 x 10"3 M/s

n 0.50 M 0.25M 9.29 x 10-3 M/s

m 0.40 M 0.50M 1.47 x 10-2 M/s

A. The reaction proceedsbecause the data indicate

step is unimolecular.B. The reaction proceeds

because the data indicate

step is bimolecular.

C. The reaction proceedsbecause the data indicate

step is unimolecular.D. The reaction proceeds

because the data indicate

step is bimolecular.

by an Ei-mechanism,that the rate-determining

by an Ei-mechanism,that the rate-determining

by an E2-mechanism,that the rate-determining

by an E2-mechanism,that the rate-determining


16. Which of the staggered complexes is MOST stable forthe reactant?

A. The conformer drawn in the example.B. The conformer from which the E2 elimination

takes place.

C. The conformer with carbons 1 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 beexplained by which of the following statements?

A. E2 eliminates the carbon 2 chiral center.

B. E2 eliminates the carbon 3 chiral center.

C. E2 eliminates both the carbon 2 chiral center andthe carbon 3 chiral center.

D. The product is a pair of enantiomers.

18. Which of the following structures (from the perspectiveof the eye) is the Newman projection for the reactant?

Br.

A.

H^H3C

Br

H30^k H

C.

J&CH2CH3

Br Br

H3C^^4v .H h3C

<^CH3£>

CH2CH3

B.Br

H- J^ CH3

f CH3 H3C/V4-^HCH2CH3

D.

H3C | H h3C" -f- ^HCH2CH3 CH2CH3

19. The ENANTIOMER of the reactant has which of the

following stereochemical orientations?

A. 2R, 3R

B. 2R, 3S

C. 2S, 3R

D. 2S, 3S


20. What is the role of concentrated sulfuric acid in an E\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 helpsremove the proton allowing the leaving group toleave.

D. Sulfuric acid stabilizes the carbocation intermediate

by protonating the cationic carbon.

21. How many units of unsaturation are present in theproduct of chlorocyclohexane and strong base and heat?

A. 0

B. 1

C. 2

D. 3



Elimination results in the formation of a new Jt-bond

following the loss of two groups from the molecule. Thereaction can proceed by one of two possible mechanisms: E\orE2.

In Ei reactions, the leaving group first leaves to form acarbocation intermediate. The carbocation has the potentialto undergo rearrangement. The solvent serves as a base anddeprotonates a hydrogen off of the carbon adjacent to thecationic carbon. The result is the formation of a new rc-bond.

In E2 reactions, the leaving group leaves simultaneouslyas the proton is removed by a strong base. The proton andleaving 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 1 shows threeelimination reactions carried out concurrently.

Reaction I:

CI

Reaction II:

CI

Reaction III:

OH

t-butOK

t-butOH, 50*C

t-butOK

t-butOH, 50°C

cone. H2SQ4,

60"C

Figure 1. 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.

22. 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 HI only


23. Which of the reactions results in an optically activeproduct mixture?

A. Reaction I only

B. Reactions I and II only

C. Reactions II and in only

D. Reaction HI only

24. When the following reaction is carried out, why doesthe optical activity disappear?

OH CH2CH3

cone. H2SP4^

CH3 60°C

Major Product(ocD = 0°)

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.

n. The amount of alkene product is maximized atlower temperatures.

HI. Reaction H has a competing an Sn2 reaction.

A. I onlyB. nonlyC. I and HonlyD. I and monly

26. Why do Reactions I and II yield different majorproducts?

A. Ei reactions can undergo rearrangement if thehydrogen is on the correct side of the plane.

B. Ei reactions are influenced by steric hindrance.

C. E2 reactions require that the leaving group andproton are anti to one another.

D. E2 reactions require that the leaving group andproton are syn to one another.


27. How does the hybridization change for the carbonscommon to both rings in Reaction HI?

A. sp2 to sp3B. sp2 tospC. sp3 tosp2D. sp3 tosp

28. If diethyl amine, (H3CCH2)2NH, is used in Reaction Iinstead of t-butOK, what is the major product?

B.

C.

Et2N= H



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 (CnH^) or tridecane(Ci3H28) 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-Methylheptadecane)

O

xO CH3

Oriental Fruit Moth sex attractant

[(E)-8-Dodecen-l-yl acetate]

H17C8 C13H27

WH H

House Fly sex attractant(Muscalure [(9Z)-Tricosene])

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-l-yl acetate as a sexattractant in roughly the same concentration that the OrientalFruit Moth uses (Z)-8-dodecen-l-yl acetate.

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 following compound is:

O

XCH.

A. a conformational isomer.

B. a geometrical isomer.

C. an optical isomer.

D. a structural isomer.

31. Which of the following statements accurately relates thefour structures shown in Figure 1?

I.

n.

m

Muscalure has a shorter wavelength of maximumabsorbance in UV-visible spectroscopy bombykol.

The Tiger Moth sex phermone has the more unitsof unsaturation than undecane.

The Oriental Fruit Moth sex phermone can beclassified as a terpene.

A. I only

B. Hlonly

C. I and II only

D. I and monly

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 solubility

D. The ability to rotate plane-polarized light

33. Which of the phermones in Figure 1 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 defense phermone is shownbelow.

What is NOT true of the structure?

A. It can be synthesized from isoprene units.

B. It has more jp2-hybridized carbons than sp3-hybridized carbons.

C. It can potentially undergo 1,4-addition.

D. It is highly flexible.


3 5. Which spectroscopic observation does NOT correlatewith the corresponding compound?

A. The Oriental Fruit Moth sex phermone: an IRabsorbance at 1741 cm"1

B. Bombykol: a UV-visible absorbance at 227 nmC. Muscalure: two signals around 5.00 ppm in its

!HNMRD. The Tiger Moth sex phermone: 14 signals in its

13CNMR 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 calledsustainable chemistry, aims to developchemical reactions and processes that are environmentallysafe. The goal is to reduce waste generation rather thanemploying 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 fluid.

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 atom-economy, the tracking of how many atoms used in thereaction end up in the product, by not using solvents orprotecting groups. Figure 1 lists three atom-economicalreactions used in green synthesis.

Reaction I:

X*

Reaction II:

C02CH3

IIIonic liquid "Nf 71^

C02CH3C02CH3

98% yield

O

xH CH2CH2CH3

H2C=CHCH3 + CO + H2cat. Rh o

xH CH(CH3)2

Reaction HI:

O O

Pd,H2

Supercritical C02

CH3 H3C CH3H3C

> 99% yield

Figure 1. Three Atom-economical Green Syntheses

Reaction 1 is a Diels-Alder reaction, Reaction 2 is ahydroformylation reaction, and Reaction 3 is a hydrogenationreaction. All of the reactions in Figure 1 start with alkenes,a common starting material in the production of plastics andpolymers. Green chemistry is ideal for polymerizationreactions, which by design aims to minimize the materialneed to carry out the propagation reaction. Green synthesistechniques can be applied to any reaction.


3 6. In ReactionI, the alkyne is best 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 H?

A.O

c

/\H2C CHCH3

C.

H2C C

I IH2C O

//CH2

B.O

/\H2C ^2

CH2

D. CH3

HC:

H2C O

38. Which of the reactions in Figure 1 involves theformation of new stereocenters?

A. Reaction III only

B. Reactions I and H only

C. Reactions I and HI only

D. Reactions n and m only

3 9. Which of the following changes does NOT fit with thephilosophy of green chemistry?

A. Using supercritical fluid as a solvent.

B. Using protecting groups and not removing themuntil the very last step of the reaction.

C. Using solid-state catalysts built into the labequipment.

D. Running a constant stream of oxygen gas throughthe reaction vessel for oxidation reactions.

40. Reaction III can be described by all of the followingterms EXCEPT:

A. reduction.

B. hydrogenation.

C. stereoselective.

D. regioselective.


41. What is the major product of the reaction below?

O

A.

C.

O

O

O

0

o

B.

D.

Copyright © by TheBerkeley Review®

O

0

o

o

p

o

297


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 Ej mechanism to form aconjugated diene is shown in Reaction 1.

OH

cone. H2S04

Compound I Compound H

Reaction 1

Reaction 1 is monitored using UV spectroscopy. Overthe course of the reaction an intense UV absorbance at 179

nm diminishes as a new peak at 222 nm appears.

When the product of Reaction 1 is treated with acidicwater, two products of detectable quantity are formed. Figure1 shows the distribution of the two hydration products at35*C, labeled Compound 3a and Compound 3b.

OH

Compound HI39.2%

Compound TV60.8%

Figure 1

Thepercentage of the secondary alcohol formed increasesas the temperature of the hydration reaction increases. This isattributed to a shift from kinetic control to thermodynamiccontrol.

It is found that if the allylic alcohol in Reaction 1 isreplaced by a new compound containing both an alcoholgroup and a carbon-carbon rc-bond, with the exception of avinylic alcohol, a conjugated diene is formed upon treatmentwith concentratedstrong acid at elevated temperatures.

4 2. Which of the following statements accurately reflectReaction 1?

I. Rearrangement is possibleduring the reaction.

U. A vinylic carbocation is formed as an intermediatein the reaction.

in. The first step of the reaction is the protonation ofthe hydroxyl oxygen.

A. I onlyB. II onlyC. I and U onlyD. I and monly


43. Which spectroscopy technique is MOST effective indistinguishing Compound in from CompoundIV?

A. !HNMR

B. Infrared

C. Ultraviolet

D. Visible

4 4. Compound 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.

45. Which of the following compounds is an allylicalcohol?

CrOH

cCr °'Crm46. 1,2- and 1,4-addition is possible in all of the following

compounds EXCEPT:

A. 2,4-Hexadiene

B. 2-methylcyclopentadiene

C. 1,4-cycloheptadiene

D. 1,3-cyclohexadiene

47. Whichof the following species is NOT an intermediateformed duringthe hydration of Compound II?A. 0 B.

©

DOC. © D.

OO CDCopyright © by TheBerkeley Review® 298

4 8. What is the major product of the reaction below?

H2Q/H2SQ4

a 75°CCH3

"aOH

CH3



The Diels-Alder reaction is quickly becoming the premierway to synthesize polycyclic compounds. New carbon-carbon bonds are formed when a conjugated diene reacts witha compound containing a rc-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 1 lists the

data for the eight trials.

B A

C D

Reaction 1

Trial A B C D XRelative

rate

1 H H H H H 1.0

2 CH3 H H H H 5.3

3 CH3 H H CH3 H 8.1

4 CH3 CH3 H H H 0.0050

5 H H H H YCH"0

82

6 CH3 H H H Y"0

411

7 CH3 H H CH3

O

673

8 CH3 CH3 H H

O

0.34

Table 1


The product of a Diels-Alder reaction is a cyclohexenederivative. The Diels-Alder reaction is classified as an

electrocyclic reaction and it is believed to proceed by aconcerted mechanism that goes through a so called aromatictransition state, where the six Tt-electrons in the reactant

interact 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. Thisinformation is in agreement with the rate shift associatedwith the addition of electron-withdrawing groups to thedienophile.

49. The product shown in Reaction 1 is one of twoenantiomers that is formed. Which of the followingpairs of molecules represents the product mixtureformed in Trial 6?

CH3 O

c- CH3 O

tX"-S-

CH3

CH3

&

&

&

&

CH3 O

CH3 O


50. Howwould the relative ratechangeif ethylgroups 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.

51. Addition of which of the following species to Trial 7would increase the rate?

A. A1C13

B. CCI4

C. KH

D. UAIH4

52. 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?

D.CH3 p<

jYn*53. How can the significantly lower reaction rate in Trial 4

than 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 Trial 4.


54. Which of the following compounds is the WORSTdienophile?

B. COCH2CH3

COCH2CH3

CH2CH3



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.

X O X O X

O

Compound A Compound B

Figure 1 Asymmetric Diels-Alder reaction

If the reactants are asymmetric, the product distributionof Product A to Product B is never 50-to-50. When X is

electron 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 1lists 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 94% 6%

OCH3 CH3 88% 12%

CH3 NHCH3 87% 13%

CH3 CH3 63% 37%

COCH3 NHCH3 18% 82%

COCH3 CH3 31% 69%

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=0), making the alkene reactantsymmetric, what would be predicted for the distributionbetween Product A and Product B?

A. > 50% Product A; < 50% Product B

B. < 50% Product A; > 50% Product B

C. 50% Product A; 50% Product B

D. ProductA and Product B are the same compound


5 6. Which of the following conclusions can be drawn fromthe data in Table 1

B.

D.

OCH3 is more electron donating than CH3 becauseOCH3 in the X position yields more product A.

OCH3 is more electron donating than CH3 becauseOCH3 in the X position yields more product B.

CH3 is more electron donating than OCH3 becauseCH3 in the X position yields more product A.

CH3 is more electron donating than OCH3 becauseCH3 in the X position yields more product B.

5 7. Predict the major product for the following reaction:

H3CH2CO O

0

CH2C1

A. B.

H3CH2CO O H3CH2CO O

O

R

CH2C1

C. D.

H3CH2CO H3CH2CO

AxHaCl ^L.CH2C1

Co Cv

5 8. What is the major product for the following reaction?

150°C

"OO


59. Two structural isomers are formed from Diels-Alder

reactions 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 following reaction?

H3CO

LA. 2

B. 4

C. 8

D. 16



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 Coperearrangement and the Claisen rearrangement. 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

[3,3]

N*^A new 1,5-diene

o^

A 1,5-diene

Claisen rearrangement

An allyl vinylic ether A y,6-unsaturatedcarbonyl compound

Figure 1 Cope and Claisen rearrangements

When the reactant in a Claisen rearrangement includes abenzenering, the ketone formed from the allyl phenylicetherundergoes 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.

otOt*

OH

CH3 H3C,^JLX,CH3Step m

"xLCH3 ^ CH3

Figure 2 Synthesis using Claisen and Cope rearrangements


61. How can it be supported that the Cope rearrangement isconcerted rather than a multistep process involvingsubstitution?

A. All of the stereocenters are retained

B. All of the stereocenters are inverted

C. 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.

63. 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 form sigma-bonds

64. Which of the following orbital arrangements representsthe transition state of a Cope rearrangement?

A. B.

6 5. The reactant in Figure 2 is best described as a:

A . allyl benzylic ether.

B. allyl phenylic ether.

C . vinyl benzylic ether.D. vinyl phenylic ether.


6 6. What spectroscopic evidence supports the formation of aproduct in a Claisen rearrangement?

A. Appearance of a signal at 9.5 ppm in the 'HNMRB. Appearance of a broadabsorbance around3400cm'1

in infrared spectroscopy

C. Disappearance of an absorbance around 1700 cm"1in infrared spectroscopy

D. Disappearance of two signals between 5 and 6 ppmin the "HNMR

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 Clasienrearrangement; 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-methy1-1,3-butadiene) reacts at carbonsone and two or one and four with other isoprene molecules toform 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.

H3C CH3

Vitamin Aj

Citronellol

Figure 1. Common Terpenes

Both plants and animals synthesize terpenes. Pinene, amonoterpene, is found in plants and Vitamin A\, a diterpene,is found in both plants and animals. Smaller terpenes arefound primarily in plants, while some larger terpenes, such aslanesterol (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 form other terpenes and terpenoids.Figure 2 shows a basic schematic for the biosynthesis ofcholesterol starting from isopentenyl pyrophosphate.


Cholesterol

Figure 2. Biosynthesis of Cholesterol

6 8. Which of the following compounds is NOT a terpene?

A. Limonine (C10H16)

B. Geraniol (CioHigO)

C. Patchouli alcohol (C15H26O)

D. Stearol (C18H38O)

69. What irregularity in a sample of a sesquiterpene (15carbon terpene) would indicate that the compound wassynthesizedin 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?

O O

O— P— O— P— OH

A. Carbon 1

B. Carbon 2

C. Carbon 3

D. Carbon 4

O" O'

71. Isoprene units are believed to be formed from threeacetyl coenzyme A molecules. What is a likely sideproduct from the reaction?

O

AH3C SCoAAcetyl Coenzyme A

A. Carbon dioxide gas

B. Ethanol

C. Acetic acid

D. Isopropanol

72. The biosynthesis of which of the following moleculeslikely involved a Diels Alder reaction with isopreneunits?

A. Caryophyllene

B. Citronellol

C. ct-Pinene

D. Vitamin Ai

73. Which of the labeled bonds in y-terpinene was formedin the biological synthesis from isoprene units?

-Bondd

A. Bond a

B. Bondb

C. Bondc

D. Bondd

Bond a T Bondc

Bondb



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 fifteen 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 variousfoods and the active molecule in many drugs. Much oforganic chemistry research involves the development of invitro synthesis of terpenes and terpenoids. Some naturallyoccurring monoterpenes are shown in Figure 1.

Carvone

Myrcene

Figure 1. Four Common Monoterpenes

Limin

H3C 0

Camphor

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 IRspectrum and astrong UV (e > 10,000) absorption at Xmax = 242 nm.

74. When limin is converted into carvone, what type ofreaction has to transpire?

A. Oxidation of carbon

B. Reduction of carbon

C. Hydrolysis of a rc-bond

D. Nucleophilic substitution

75. Which compound in Figure 1 is LEAST likely toundergo ozonolysis when treated with O3?

A. Camphor

B. Carvone

C. Limin

D. Myrcene


76. How many singlets does camphor show in its protonNMR spectrum?

A. Two

B. Three

C. Six

D. Nine

7 7. If myrcene reacts with another isoprene unit, what kindof terpene is formed?

A. Diterpene

B. Monoterpene

C. Sesquiterpene

D. Sesterterpene

78. Camphor is likely to show which of the followingphysical and chemical properties?

I. High water solubility

H. A boiling point above 298K

HI. No specific rotation of plane polarized light

A. I only

B. Honly

C. I and Honly

D. I and monly

79. Does limin display a strong (log e > 4) UV absorption?

A. Yes, because the 7i-bonds are spaced far apart.

B. Yes, because of the six-membered ring.

C. No, because there is no carbonyl group.

D. No, because the rc-bonds are not conjugated.

80. Which compound is the direct product of a Diels Aldercondensation of two isoprene units?

A. Camphor

B. Carvone

C. Limin

D. Myrcene


Passage XIII (Questions 81- 86)

Many processed food products often contain partiallyhydrogenated vegetable oil as one of their ingredients. Partialhydrogenation serves to reduce some of the 7i-bonds found innatural oils. Naturally occurring fatty acids, such asvegetable 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 melting point, andoften converts a naturally occurring liquid into a solid. Thefatty acids can be found as either the carboxylic acid or as partof a fatty acid triglyceride. Figure 1 shows the enzymaticallycontrolled conversion of a fatty acid triglyceride into glyceroland three fatty acids.

O

O

AR2+H2QLip^O

O'

— O'

u-0

r-OH

O

UHO' "Ri

O

O

— OH+ HO

•—OHHO

Figure 1 Enzymatic Hydrolysis of a Triglyceride

The fatty acid is isolated when a fatty 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 functional group, naturally occurring fatty acids have aneven number of carbons. This is attributed to the fact that

fatty acid biosynthesis occurs two carbons at a time, viaacetyl coenzyme A. Natural fats can be distinguished fromsynthetic fats by their carbon chain length. Table 1 listssome common fatty acids that are naturally found in animals:

Acid Formula 7t

Arachidic CH3(CH2)i8C02H 0

Arachidonic CH3(CH2)4(CH=CHCH2)4(CH2)2C02H 4

Behenic CH3(CH2)20CO2H 0

Laurie CH3(CH2)ioC02H 0

Lignocaric CH3(CH2)22C02H 0

Linoleic CH3(CH2)4(CH=CHCH2)2(CH2)6C02H 2

Linolenic CH3CH2(CH=CHCH2)3(CH2)6C02H 3

Myristic CH3(CH2)i2C02H 0

Oleic CH3(CH2)7CH=CH(CH2)7C02H 1

Palmitic CH3(CH2)i4C02H 0

Palmitoleic CH3(CH2)5CH=CH(CH2)7C02H 1

Stearic CH3(CH2)i6C02H 0

Table 1 Common Fatty Acids


Vegetable oils generally have more unsaturation thananimal fats. For instance, corn oil is 63% linoleic acid and

26% oleic acid, with the rest being made of other saturatedfatty acids. Safflower oil is 75% linoleic acid, 14% oleicacid, and 4% 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 ashormones (intercellular messengers). Fatty acids are alsoused for fuel through fatty acid metabolism.

81. What is the structure for the MOST abundant fatty acidfound in corn oil?

A.O

HO VN^VH/S

c o

HO V\^\/V\

82. The 7t-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:

A. with lower molecular mass and a lower meltingpoint than the reactant.

B. with higher molecular mass but a lower meltingpoint than the reactant.

C. with a lower molecular mass but a higher meltingpoint than the reactant.

D. with a higher molecular mass and a higher meltingpoint than the reactant.


8 3. Addition of D2 with Pd catalyst reduces 7c-bonds byadding deuterium to each 7i-bond carbon. Treatment ofoleic acid with D2 and palladium yields a compoundwith how many chiral centers?

A. Zero

B. One

C. Two

D. Four

84. 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 KMn04 under basic

conditions?

A. Carbons 8, 9, 11, 12, 14, and 15

B. Carbons 7, 9, 11, 13, 15, and 17

C. Carbons 9, 10, 12, and 13

D. Carbons 9, 10, 12, 13, 15, and 16

85. Bromine liquid is used as a quantitative test reagent todetermine the amount of rc-bonds per molecule of acompound. Which of the following acids consumes theMOST Br2 per molecule?

A. Arachidic

B. Arachidonic

C. Linoleic

D. Linolenic

86. Complete hydrogenation of palmitoleic acid yieldswhich of the following acids?

A. MyristicB. Palmitic

C. Stearic

D. Arachidic


Passage XIV (Questions 87 - 92)

Terpenes and terpenoids are natural compounds found inplants and animals that are built from 5-carbon reactants.Figure 1 shows the sesquiterpenoid(±)-occidentalol.

Figure 1 (±)-Occidentalol

A multistep synthesis leading to occidentalol beginswith the conversion of Compound 1 into Compound 6,which can further react to form occidentalol. Figure 2 showsthe synthesis of Compound 6.

CO-Me

Compound 1

Xi0150eC, -CQ2

•

Step 1

Step 2

S03Py, THF, 0°C;-*

LAH, THF

Step 3

Compound 4

Step 4 1 WHCl.HOAc

EtOv.. X

_ ^ P SMeEtO 11

Ou •

H HMPA, DME, 62°C

Compound 5 Step 5 Compound 6

Figure 2 Synthesis of Compound 6

H

COjMe

Compound 2

1. HO^°HTsOH, PhH, A

y1 2. LAH,EtOEt

Compound 3

SMe

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 aketal 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 reform the ketone.Compound 5 undergoes a variation on the Wittig reaction toform an alkene in Compound 6.


8 7. How many chiral centers are present in occidentalol?

A. 1

B. 2

C. 3

D. 4

88. Terpenes containing fifteen carbons are best describedas:

A. monoterpenes.

B. diterpenes.

C. triterpenes.

D. sesquiterpenes.

8 9. Which compound has the longest maximum wavelengthof absorbance, A^ax, in ultraviolet spectroscopy?

A. Compound 1

B. Compound 2

C. Compound 3

D. Compound 4

9 0. What intermediate compound forms in Step 1 before thedecarboxylation takes place?

A. Acyclohexadiene

B. A cyclohexene

C. An a,B-unsaturated ketone

D. A lactam

91. To synthesize occidentalol, Compound 6 is firstconverted into a methyl ketone, which is then treatedwith MeLi in Et20 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.

9 2. If Compound 4 were treated with strong acid, at whichcarbon in the jc-network is it most likely to gain H+?

A. Carbon a

B. Carbon b

C. Carbon c

D. Carbon d


Questions 93 through 100 are NOT based on adescriptivepassage.

93. Leukotriene A4, LTA4, is derived from arachadonicacid. Its structure is shown below:

C02H

What is NOT true of LTA4?

A. LTA4 has six units of unsaturation.

B. LTA4 has six rc-electrons in a conjugated system.

C. LTA4 is capable of undergoing 1,2-, 1,4-, 1,6-, or1,8-addition when treated with an electrophile andnucleophile.

D. LTA4 has more .sp2-hybridized carbons than it has^-hybridizedcarbons.

94. What is the major organic product of the reactionbelow?

H3C H

X^H

HjCCHCH3

Br2

hv

B.A. BrH2C N\H 'H3C x\Br\j2aH \jjSaH

r^CCHCH3

\^BrHjCCHCH3

HjCCHCH3

C. D.H3C H3C H

H}CCBrCH3

95. Which of the following reactions is a propagationreaction?

A. H3C- + H3CCH2CH3 -> 2 H3C- + H3CCH2-

B. H3C- + H3CCH2- -» H3CCH2CH3

C. H3CH2C- + H3CCH3 -> 2 H3CCH2- + V2 H2D. 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.


97. A conjugated diene is necessary in which of thefollowing reactions?

A. Claisen rearrangementB. Clemmensen reduction

C. Cope rearrangementD. Diels-Alder cycloaddition

98. 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.

99. All of the following are physical properties of a terpeneEXCEPT:

A. High lipid solubilityB. High boiling pointC. Low volatility

D. High specific rotation

100. Which of the following molecules have dipoles NOTequal to zero?

I. E-butene

II. C2H4

HI. Z-butene

A. I only

B. II only

C. monly

D. I and HI only

1. B 2. D 3. C 4. B 5. B 6. B

7. B 8. C 9. D 10. B 11. B 12. A

13. C 14. A 15. D 16. C 17. C 18. D

19. C 20. A 21. C 22. D 23. B 24. C

25. D 26. C 27. C 28. B 29. A 30. D

31. A 32. D 33. A 34. D 35. D 36. A

37. A 38. A 39. B 40. D 41. C 42. D

43. A 44. D 45. A 46. C 47. B 48. A

49. D 50. A 51. A 52. D 53. A 54. D

55. D 56. A 57. A 58. A 59. D 60. C

61. D 62. D 63. A 64. B 65. B 66. A

67. C 68. D 69. C 70. D 71. A 72. C

73. B 74. A 75. A 76. B 77. C 78. B

79. D 80. C 81. B 82. D 83. C 84. D

85. B 86. B 87. C 88. D 89. A 90. B

91. B 92. A 93. D 94. D 95. D 96. A

97. D 98. C 99. D 10C>. C

ALL DONE, NO MORE!

Alkanes and Hydrocarbon Reactions Passage Answers

Passage I (Questions 1-7) Free Radical Halogenation Selectivity

Choice B is correct. Halides are like hydrogens when considering degrees of unsaturation, because they, likehydrogen, make only one bond with carbon. Plugging into the foUowingformula for units of unsaturation yieldsan answer of 1. Pick B for best results.

Units ofUnsturation = 2(#C) +2- (#H) • (*C1) „ 2(5) +2-(9) - (1) = xq +2-9-! . J2J0 . Z. 12 2 2 2 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 one of the following structures, but drawn differently, violate the units of unsaturation, ornot be structuraUy possible.

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 straight 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 four bonds, therefore there is only one monochloro isomer of 2,2-dimethylpropane.This means that there are eight structural isomers total, so pick D for greatest success.

C—C—C—C—C C—C—C—C—C C—C—C—C—CI I ICI CI CI

1-chloropentane 2-chloropentane 3-chloropentane

C—C—C—C C—C—C—C C—C—C—C C—C—C—CI I /\ II I IC1 C C CI C CI c ci

l-chloro-2-methylbutane 2-chloro-2-methylbutane 2-chloro-3-methylbutane l-chloro-3-methylbutane

C—CI

IC— C l-chloro-2,2-dimethylpropane 8 total

Choice C is correct. By symmetry, there are four unique carbons on methylcyclopentane, therefore chlorinationcan occur at a total of four different sites (four different carbons). This yields a total of four structural isomers.The correct answer is choice C.

CH2C1 H3C CI CH,

&CH

Copyright © by The Berkeley Review® 310 HYDROCARBONS EXPLANATIONS

4. 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 Ha lead to 1-chloropentane and hydrogens symbolized by Hb lead to 2-chloropentane. They are multipliedby their respective reactivity factor.

1-chloropentane

CI

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, so the product ratio should be 1 : 1. Because the ratio is 1.07 : 1 and not 1 : 1, choice A iseliminated. There is no resonance in the molecules (because there are no lone pairs and no n-bonds), so choice Cis eliminated. The inductive effect would favor the formation of 3-chloropentane, so choice D is eliminated.Steric hindrance is the bestexplanation why the carbon-2 is selected over carbon-3, so choice B is best.

Choice Bis 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 ringhasmore ringstrain than a four membered ring, so tlie bestanswer is choice B.

PassageII (Questions 8 -14) Free Radical Reactions

9.

10.

11.

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 offour bonds and a free electron oncarbon would exceed the octet rulefor carbon. The best answer is therefore choice C.

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!

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.

Choice Bis correct. According to die 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 isanexception 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.


12. 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.

13. Choice 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 electron withdrawing by the inductive effect. Halogenstherefore do not help 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 reaction

will 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. This 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 halide, itwould abstract the halide, not a hydrogen. This is the more logical explanation for the absence of multiplesubstitution products with free radical halogenation.

14. 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 fooled into thinking every question on the MCATwill be difficult.

Passage III (Questions 15 - 21) Elimination and Stereochemistry

15. Choice D is correct. When 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 electrophile concentration has increased (by sixtypercent), and the rate has more thandoubled. If the reaction depended just on the electrophile, the rate wouldbe less than 9.2 x10-3 M/s. Because the rate in Trial III is more than the rate in Trial I, the rate-determiningstep depends on both reagents, making it bimolecular. E2-reactions are bimolecular, so choice D is correct.

16. 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 ofcarbon 1and the bulkiest group oncarbon three is the ethyl group comprised ofcarbons 4 and 5, sochoice Ciscorrect. The Newman projections forthe three staggered conformations are drawn below:

17.

18.

CH2CH3

Conformer from

the example

CH2CH3

\ N*CH3J>CH2CH3Hw^r~\

H3C H

Conformer from which

elimination takes place

CH2CH3

CH:

XH2CH3<£* H

Conformer with

Cj and C4 anti

H3CH2C

CH,

CH,

Choice C is correct. Carbons in the reactant have sp3-hybridization while two carbons in the alkene producthave s/72-hybridization, and thus those carbons cannot 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 chiral centers, it can have no optical activity. The best answer for the question ischoice C.

Choice D is correct. From the perspective ofthe eye as shown, the bromine atom isbonded 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 wouldstickout to theright. The best choice is thus answer choice D.


19. 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 of both of the reactant's chiral centers isshown below. The chiral centers of the enantiomer are 2S and 3R. The best answer is choice C.

CH,

H3Q CH2CH3

Carbon 2 = R

Compound is (2R,3S), so theenantiomer must be (2S,3R) fCH2CH3

Carbon 3 = S

20. 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 E^ reaction rather than the E2 reaction. The acid 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 eliminated immediately, because acarbocation cannot be protonated. The best answer is choice A.

21. Choice C is correct. Chlorocyclohexane when treated with a strong base undergoes an elimination reaction byway of an E2 mechanism to yield cyclohexene. There are two units of unsaturation in cyclohexene, one unit ofunsaturation for the 7i-bond and one unit of unsaturation for the ring. Tlie correct answer is choice C.

PassageIV (Questions 22 - 28) Elimination Study

22. 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, because of the presence of the strong, bulkybase. The base must be strong enough to removea proton from carbon and it must be bulky enough to minimizecompetition with an SN2-reaction. Reaction III proceeds by an Ei-mechanism, because of the presenceof strongacid, which protonates the hydroxyl group. This makes a better leaving group, and ultimately facilitates theformation of a carbocation. In fact, the location of the double bond in the final product can only be explained bya hydride shift. The hydride shift results in the conversion from a secondary carbocation into a tertiarycarbocation. Only Reaction III exhibits rearrangement, so the best answer is choice D.

23. Choice B is correct. This question involves counting thechiral centers in each product in Figure 1 to see whichmolecules are chiral, and then analyzing to see if chirality is involved in the reaction. Tlie major 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 active. 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 final product mixture exhibits nooptical activity. The best answer is choice B.

24. Choice C is correct. The reaction involves the use of concentrated strong acid at high temperature, so thereaction isapt to proceed via anEi-mechanism. An Ei-mechanism entails the hydroxyl group being protonatedand then leaving, producing a secondary carbocation. Ahydride shift results in the conversion from a secondarycarbocation into a tertiary carbocation, which is the intermediate from which deprotonation to form thealkene product occurs. All 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.

25. 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 substitution product and minimize thecompeting elimination product, temperatures are reduced. This makes Statement II invalid, which eliminateschoices Band C. Reaction II is an E2-reaction, so it has a competing SN2-reaction occurring in the same flask.This makes Statement III valid, and makes choice D the best answer.

Copyright © byThe Berkeley Review® 313 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 hydrogen being lost be oriented anti tothe leaving group, in this case chlorine. In Reaction I, there are two anti hydrogens, and the one that is chosen,is the one that leads to the more substituted alkene product. In Reaction II, there is only one anti hydrogen, soonly one product may be formed. The hydrogen on the 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 double bond carbons in Reaction III, so theyboth finish with sp^-hybridization. Only choice C finishes with s/?^-hybridization, so it is the best answer.

28. 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 electrophile is a secondary alkyl chloride, which could go byeither an Sjsjl or Sjsj2 mechanism. Because the nucleophile is a good nucleophile, the best choice is an Sjsj2reaction. An Sjm2 reaction results in the inversion of the reactive site, which puts the amine group behind theplane of the molecule and makes choice B the best answer.

Passage V (Questions 29 - 35) Phermones

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 carbons are secondary. There areno tertiary carbons, so choice C is eliminated. Only one single bond separates the two double bonds, so they are infact conjugated. This eliminates choice D. All of the carbons have at least two identical substituents or theyhave sp2-hybridization, so there are no stereogenic centers. This makes choice Athe best answer.

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 has cis geometry, while the phermone in Figure 1 hastrans geometry. This can be misleading and tempt you to pick choice B. But, the connectivity is not the same, sothe structures are structural isomers. The 7t-bond is between carbons 8 and 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 eliminateschoice A. There is no stereogenic carbon, so the two structures cannot be optical isomers.This eliminates choice C. Tlie two structures are structural isomers, making choice D the best answer.

31. Choice A is correct. UV-visible spectroscopy is used to detect rc-bonds. Bombykol has conjugated 7t-bonds, whilemuscalure just one 7t-bond. Conjugation reduces the transition energy,so the wavelength of maximum absorbanceincreases with conjugation. This means that bombykol has a greater Xmax than muscalure, making Statement I avalid statement. Choice B is eliminated. The Tiger Moth sex hormone is an aliphatic alkane, so it has no unitsof unsaturation. Undecane is an 11-carbon aliphatic 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 phermone 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 miscibility. Choice A is a valid statement, and thereby 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-polarized light. This makes choice D an invalid statement, so choice D is the best answer.

33. Choice A is correct. All of the compounds in Figure 1 have straight chains (no rings), so each has at least twoprimary carbons. Only 2-methylheptadecane has branching, so it has an extra primary carbon. No otherstructure in Figure 1 has any branching, so 2-methylheptadecane, the Tiger Moth sex phermone, has the mostprimary carbons. Choice A is the best answer.

Copyright © by The BerkeleyReview® 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 bonds, so it has eight s/?2-hybridized carbons. With fifteen carbonstotal, more than half of the carbons are s/^-hybridized. Choice Bis 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 7i-bonds in the Green Peach Aphiddefense phermone, it is not very flexible. Choice D is an invalid statement, which makes it the best answer.

35. Choice D is correct. The Oriental Fruit Moth sex phermone has an ester group. A carbonyl has an IR absorbanceabove 1700 cm"1, so an ester accounts for the IR absorbance at 1741 cm"1. Choice A is a valid correlation ofstructure to spectroscopic observation, so it is eliminated. Bombykol has two it-bonds in a conjugated network.The presence of Tt-bonds in a structure results in an absorbance in the ultraviolet-visible range of the EMspectrum, so an absorbance of 227 nm seems viable for a conjugated diene. Choice B is a valid correlation 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, resulting in two signals with values around 5.00 ppm in 1HNMR. Choice C is a valid correlation ofstructure to spectroscopic observation, so it is eliminated. The Tiger Moth sex phermone has eighteen carbonsand no plane of symmetry. Thereare two equivalent methyl groups, so there are seventeenunique carbons. The13CNMR would show 17 signals if it were of high enough resolution. The reality is that many of the signalswould overlap, so itwould likely show less. It will not show fourteen signals in its 13CNMR spectrum, so choiceD is an invalid correlation of structure to spectroscopic observation, making it the best answer.

Passage VI (Questions 36 - 41) Green Synthesis

36. 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 7t-bond. The alkyne meets this requirement, so it is a dienophile. The best answeris choice A. There isnonucleophilic substitution going on, sochoices Band Careeliminated. There isnochangein oxidation state, so the compound is not an oxidant or reductant, eliminating choice D.

37. Choice A is correct. Each of theanswer choices is lacking two hydrogens from the formula of the final product,C4H8O. This means thathydrogenation converts the intermediate to the final product. The final product is analdehyde and hydrogenation cannot convert a cyclic ether into a carbonyl, so choices Cand Dcan 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 hydrogenation mustbreak the strained ring by adding a hydrogen to the carbonyl carbon and a hydrogen to the neighboring atom. Inchoice A, the hydrogen atoms are correctly displaced on the intermediate (CH2-CH-CH3) to form bothaldehyde products (CH2-CH2-CH3 and CH3-CH-CH3) when a single hydrogen is added to the alkyl chain.This is not the case in choice B (CH2-CH2-CH2), so it is eliminated.

38. Choice Ais correct. On the product of Reaction I, all of the functional groups are bonded to sp2-hybridizedcarbons, so there are no stereocenters. This eliminates choices Band C. Based on the remaininganswer choices,Reaction III must have formed a new stereocenter. The new stereocenter is located on the ring carbon with themethyl substituent. Hydrogenation can occur from above or below the ring, so the methyl is above the plane infifty percent of the product mixture and below the plane in fifty percent of the mixture. Reaction III forms aracemic mixture, hi 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 container ending up in the product.Using a supercritical fluid as a solvent makes for easy recovery and reuse of the solvent. Reaction III usessupercritical C02, so choice A is valid and thus eliminated. Using protecting groups adds extra atoms to thesolution that are not destined to bepart of the product, so it violates the principle of atom-economy. Protectinggroups are difficult to recycle without spending a great deal ofsolvent, so they do not fit the green chemistryphilosophy. Choice B is anexception. If catalysts are part of the lab equipment, such as catalytic beads, thenthey are easily recovered and reused. This makes choice C in philosophical agreement with the principles ofgreen chemistry. This eliminates choice C. It isstated in the passage that direct oxidation using oxygen fits inthe philosophy ofgreen chemistry, so choice D is eliminated. Choose Band be on top ofyour game.

Copyright ©by The Berkeley Review® 315 HYDROCARBONS EXPLANATIONS

40.

41.

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 B. The gain of bonds to hydrogen, a less 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 hydrogen atoms in a syn fashion,which means that the reaction is stereoselective. This eliminates choice C. Both carbons gain a hydrogenatom, so there is no regioselectivity. This makes choice D the best answer. Choose D for the sake of correctness.

Choice C is correct. The reaction is a Diels-Alder reaction, similar to Reaction I, exceptan alkene is servingasthe 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 diene and an alkene, theproduct is a cyclohexene ring, so choices Band D are eliminated. To form choice B, the dienophile would haveneeded to be an alkyne, like Reaction I. The 7t-bond in the product is located between the two internal carbonsof the original conjugated diene, so they should be found on the left side of the central ring. This makeschoiceC the best answer. Choose C for a brighter smile when scoring like you did.

O

c^c°o


42.

43.

Conjugated 7t-Networks

Choice D is correct. Reaction 1 is an elimination reaction by way of an Ei mechanism. In an Ei reaction, acarbocation is formed, so rearrangement is possible. Whether it is observed depends on the compound, but it ispossible. Statement I is a valid statement. The intermediate is an allylic carbocation, where the cationiccarbon is bonded to oneof the carbons in the double bond, not a vinylic carbocation, where thecation carbon isone of the carbons in the double bond. Statement II is invalid, which eliminates choices Band 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. This generates a good leaving group, which thenleaves in the second step. Because the hydroxyl group is protonated to start the reaction, Statement III isvalid. This makes choice D the best answer.

Choice Ais correct. Compound III and Compound IV are structural isomers of one another. They each have ahydroxyl group and an alkene functionality, so infrared spectroscopy yields the same key absorbances. Thismakes infrared spectroscopy ineffective at distinguishing the two allylic alcohols, so choice Bis eliminated.Ultraviolet spectroscopy is great for determining the amount of conjugation in a system. However, bothcompounds have the same number of 7t-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 theymention that the peak at 179 nm disappears. That peak is associated with Compound I, which happens to beoneof the two enantiomers represented by Compound III. Choice D is eliminated, because 179 nm is not in thevisible range of the EM spectrum. The best method is ^NMR, which can distinguish structural isomers bytheir equivalent hydrogens. Choice A is the best answer.

44. 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 arenot possible (there must be atleast two chiral centers for diastereomers to be possible). This eliminates choice C. Because the hydroxylgroup can be above the plane or below the plane, there ismore than one structure possible for Compound IV.This eliminates choice B and makes choice D the best answer.

OH

Compound IV

> -_. / and

OH OH

Enantiomers

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45. 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, hi 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 JC-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.

46. Choice C is correct. The option for either 1,2-addition or 1,4-addition occurs when the reactant has conjugatedrt-bonds. Choices A and D should be eliminated immediately, because when the two numbers describing the k-bonds differ by 2, then the 7t-bonds are conjugated. Cyclopentadiene has only five carbons, so one Ti-bondmust bebetween carbons one and two. The second 7t-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 7t-bondsnot conjugated, so choice C is the best answer.

Choice A Choice B Choice C ^\ Choice D

CH,

47. Choice B is correct. Tlie hydration of Compound II starts with the addition of a proton to the conjugated k-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 canundergo resonance to generate thestructure 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 7i-system.

Choice C Choice A

Choice D

48. Choice A is correct. At 35°C, the hydration of Compound II using sulfuric acid and water yields about 40%secondary alcohol and 60% 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 also solve this question.

Passage VIII (Questions 49 - 54) Diels-Alder Reaction Rate Study

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 oneof 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, sochoice Bis eliminated. In choice D, the groups are cis, soit is the best answer.

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50. Choice A is correct. In Trial 4, with methyl groups in positions A and B of the conjugated diene, the rate isroughly 0.005 that of 1,3-butadiene. The reason for this reduced rate is the steric hindrance associated withthe methyl group pointing to the middle of the transitionstate. No matter how the molecule contorts, one ofthe two methyl groups is always 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.

51. 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, AICI3, is a Lewis acid, because the aluminum lacks a complete octet.Choice B, CC14, 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 readily donateselectrons. Because KH acts like a base and not a Lewis acid, choice C is eliminated. Choice D, L1AIH4, is astrong reducing agent where each atom has a satisfied octet. Because LiAlHj acts like a reducing agent and nota Lewis acid, choice D is eliminated. The best answer is choice A.

52.

53.

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 thedienophile has an electron-withdrawing substituent, so let 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 dienophile shouldcarry a negative charge (because it accepted electrons). Thisis observed in eachanswer choice except choice A, so choice A is eliminated. On the bright side, based on theanswer choices, we know that our assumption about the diene beingthe nucleophile and dienophile being theelectrophile is valid. The drawing below shows the formation of the first intermediate after nucleophilicattack.

CH3 O0

The second intermediate drawn matches choice D, so choice D is the best answer. Choices B and C could havebeeneliminated by the incorrect location of the positive charge.

Choice A is correct. The dienophile is the same in Trial 2 and Trial 4, so the difference in reactivity must beattributed to theconjugated diene. This eliminates choice D. Methyl groups, when bonded to a structure, donotactas Lewis acids, because their octets arecomplete. This eliminates choice C. The methyl groups are mildlyelectron-donating, notelectron-withdrawing, sochoice Bis incorrect. The most significant factor in thereactionrate is that the extra methyl group on the diene in Trial 4 causes steric hindrance in the transition state. Nomatter howthediene contorts, one of the two methyl groups interferes with the incoming dienophile. The bestanswer is choice A.

CH, CH,CH,

Significant steric hindrance

54. 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 enhanced by the carbonyl groupsconjugated to thealkene, so choice A is eliminated. Choice B, albeit an alkyne and not an alkene, has electron-withdrawingester groups conjugatedwith the 7t-bond, so it is enhancedas a dienophile. Choice Bis eliminated. Choice C isenhanced by the carbonyl groups conjugated to the alkene, 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.


Passage IX (Questions 55 - 60) Diels-Alder Reaction

55. 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 indistinguishable. BothProduct A and Product B are the same compound, if the reactant is symmetric, so choice D is the best answer.

56. 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 the major product isproduct A, not product B.

57. Choice A is correct. By analogy, OCH2CH3 (ethoxy) is an electron donating group like OCH3 (methoxy). Thepresence of the electron-donating group makes Product A the more favorable product. Product A from thegeneric reaction of tlie passage is choice A. Be careful not to choose B without paying attention to the locationof the double bond. The double bond in choice Bis on the side opposite from where it should be.

58. 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 of 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:

59. 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.

60. 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 Bin 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 such as this, the 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.

Passage X (Questions 61 - 67) Claisen and Cope Rearrangements

61. 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. Thestereochemistry can be lost at centers thatgo from sp^-hybridization to s/?2-hybridization and it can be gained at centers that go from sp2-hybridizationto sp3-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 the71-bond from thecarbonyl to the benzene ring, so it is the resultoftautomerization, not reduction. The best answer is choice D.


63. 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. Choice B should have been

eliminated, 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 eliminated.

64. Choice B is correct. The Cope rearrangement involves a 1,5-diene, so there are six carbons within the molecularorbital of the transition state. Choice A is eliminated because it has only four carbons. Choice C is eliminatedbecause the orbitals show no 7t-overlap 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 on the left andtlie terminal orbitals are aligned correctly to form a pi-bond.

65. Choice B is correct. The oxygen is directly bonded to the benzene ring in the reactant, so it is phenylic and notbenzylic. This eliminates choices A and C. The oxygen is also bonded to the carbon alpha to the alkene. Thismakes the carbon allylic, so choice B is the best answer.

66. 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 show a signal around 9.5ppm in the ^HNMR, so the formation ofan aldehyde would in fact correspond with the appearance of a signalaround9.5 ppm. Choice A is the best answer. Infrared absorbances around 1700 cm"1 indicate the presence ofacarbonyl group and broadinfrared absorbances around 3400 cm"1 indicate thepresence of an alcohol group. Nohydroxyl group appears in either the reactant or product, so choice B is eliminated. The reaction would besupportedby the appearance ofan absorbance around 1700 cm"1 in infrared spectroscopy, not a disappearance, sochoice C is eliminated. Signals between 5 and 6 ppmin the -^HNMR correspond to vinylic hydrogens bonded toalkene carbons, which are present before and after the reaction, so choice D is eliminated.

67. Choice C is correct. The first reaction in the synthesis in Figure 2, Step I, involves the oxygen. The Claisenrearrangement involves oxygen as the ether is converted into a ketone. This eliminates choices 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 during the Claisen rearrangement or whether they remainconstant at five. In all compounds in Figure 2, there are four rc-bonds and one ring, so there are always five unitsof unsaturation. This makes choice C the best answer. You could also concluded that the units of unsaturation

do not change by looking at the Claisen rearrangement in Figure 1, where there are two 7t-bonds in both thereactant and product.

Passage XI (Questions 68- 73) Isoprene Units

68. 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 number of carbons that is divisible by five. Stearol has eighteen carbons, so it cannot bea terpene. The correct choice is D.

69. 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. Tlie dead give-away would be an impurity with sixteen carbons. Terpeneshave multiples of five for their carbon values. Because sixteen carbons is not possible, choice C is the bestchoice. A twenty-carbon impurity is a terpene, thus it is naturally occurring.

70. Choice D is correct. The carbon that is most susceptible to nucleophilic attack is the carbon with a leavinggroup attached. Carbon four, with the pyrophosphate leaving group, is the most electrophilic. Alkene carbonsdo act as electrophiles on occasion, but in this compound, carbon four is more electrophilic than an alkenecarbon. The best answer is choice D.

71. 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 and isopropanol contains threecarbons. Choices B, C, and D are all eliminated.


72.

73.

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 oc-pinene and Vitamin Aihave a cyclohexene moiety, so we must look closer. Diels-Alder reactions involve a diene and dienophile, sowe can look at the compounds in a retrosynthetic fashion. In Vitamin Ai, the retro Diels-Alder reaction doesnot generate terpene fragments, so choice D is eliminated. Choice C is thebestanswer by default.

Choice B is correct. Bond a can be eliminated immediately, because the fragments formed from the break arethree carbons andseven carbons. Bond ccan be also eliminated immediately, because the fragments formed fromthe break are nine carbons and one carbon. These are not multiples offive, 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.

H3C

H3CBond a T 4 Bondc

Bondb

Bond b must have been formed

to connect the isoprene units.

'Bondd

CH,or

Bond d

CH,

Bond a T Bond c

Bond b

None of the labeled bonds were

formed to connect the isoprene units.

Passage XII (Questions 74 - 80) Terpenes

74.

75.

76.

Choice A is correct. Carvone differs from limin by a carbonyl group. To go from limin to carvone, a carbon mustlose two bonds to hydrogen and gain a double bond to oxygen. This is oxidation, 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 reactantmust containan alkene functional group. Allof 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 whichcoupling can take place. In camphor, all of the methyl groups are bonded to quaternary carbons, so they all fitthis description. Because the cyclic 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 isthateach 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 withhydrogens, so there are no other singlets thanthe onesfrom the methylgroups. This generates three protonNMR singlets, so the best answer is choice B.

No Hs on neighbor, cannot berotated to be equivalent withother bridge methyl group.(3H singlet)

QuarternaryCarbons have

no Hs attached

HoC

No Hs on neighbor, cannot berotated to be equivalent with

'CH3 other bridge methyl group.H (3H singlet)

No Hs on neighbor, isolatedmethyl group. (3H singlet)

77. Choice C is correct. Myrcene contains tencarbon atoms, so the addition ofanother isoprene unitwould result hi 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 Berkeley Review 321 HYDROCARBONS EXPLANATIONS

78. Choice B is correct. Camphor has a carbonyl group (water soluble) and a large alkyl ring system (not watersoluble). It ishard to decide based on the structure. Ithappens that the compound is water soluble, which youmay know first hand from using camphor-containing cleaning agents for skin. The question iswhether ornot itis highly water soluble. Because there is some ambiguity, let's say for now that it likely not highly watersoluble, and consider statement I to be invalid. Camphor is a liquid at room temperature, as you might haveseen ifyou synthesized it ina lab experiment. Being a liquid at standard temperature, itsboiling point is above298 K. Statement II is valid. Camphor has two chiral carbons, so it rotates plane-polarized light. This makesstatement III invalid. Choice B is the best answer, but not with one hundred percent certainty.

79. Choice D is correct. It is stated in the passage that carvone has a strong UV absorbance (e > 10,000). Carvonehasconjugation, which causes its intense UV absorbance. Onthe other hand, limin hasnoconjugation, so itsUVabsorbance is not as intense as that of carvone. This means that the UV absorbance for limin has an e less than10,000 (therefore, loge < log 104 =4). The best answer ischoice D. This question required some backgroundinformation on UV spectroscopy. The minimum you should know is that rc-bonds are UV active, and withconjugation, the intensity of the absorbance increases and the energy of the absorbance decreases.

80. 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,eliminatingchoices A and D, but carvonehas a carbonyl group and isoprene contains only Cs and Hs. Thebestanswer is limin, choice C.

Passage XIII (Questions 81 - 86) Fatty Acids and Oils

81. Choice B is correct. It is stated in the passage that corn oil is 63% linoleic acid. Looking at table 1 shows thatlinoleic acid is made up of eighteen carbons and has two rc-bonds. Choices C and D are eliminated, because theyonly have sixteen carbons. Choice A is eliminated, because it has three 7t-bonds. The best answer is choice B.You can try to match the exact location of each 7t-bond from the formula in Table 1 to the drawing in the answerchoices, but doing so is not time efficient.

82. Choice D is correct. Linoleic acid contains two rc-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 the acid (byfour), and the loss of unsaturation results in more molecular flexibility, which results in a higher melting point.Unsaturated fats, with less flexibility and therefore less ability to engage in intermolecular interactions, havelower melting points than saturated fats of comparable mass. This is common organic chemistry knowledgethat you should have addressed when comparingvegetable and animal fats. The correct answer is choice D.

83. Choice C is correct. Oleic acid is an eighteen-carbon acid with one rc-bond between the eighth and ninthcarbons. The rc-bond in oleic acidhas cisorientation. Treating oleic acid with deuterium (D2) and a catalyticmetal like palladium adds two deuterium atoms across the 7t-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 two chiral centers. The best answer is choiceC.

84. Choice D is correct. Potassium permanganate reacts with alkenes to form diolsby adding two hydroxylgroupsis a syn addition fashion to the carbons of the 7i-bond. Linolenic acid has three 7t-bonds, located betweencarbons 9 and 10,carbons 12and 13, and carbons 15and 16when 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 9,10,12,13,15, and 16,as listed in choice D.

85. Choice B is correct. Themost bromine per molecule is consumed by the fatty acidwith the greatest numberofn-bondspresent. Forevery7t-bond present, onemolecule ofbromine liquidwillbe consumed. Arachidonic acid hasfour 7U-bonds. Arachidic acid has no 7t-bonds present in itsstructure, linoleic acid has two7t-bonds presentin itsstructure, and linolenic acid has three rc-bonds present in its structure. Arachidonic acid is the most unsaturatedof the choices. The correct answer is thus choice B.

86. Choice B is correct. Palmitoleic acid has sixteen carbons and one 7t-bond. When palmitoleic acid is fullyhydrogenated, it forms the aliphatic acid of sixteen carbons (listed in Table 1 as palmitic acid). The bestanswer choice is B.

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Passage XIV (Questions 87 - 92) Occidentalol Synthesis

87.

88.

89.

90.

91.

92.

Choice C is correct. Figure 1 shows occidentalol with three specified chiral centers, so choices A and Bcan beeliminated. All of the carbons in the ring structure either have s/?2-hybridization or they are methylene(CH2) groups. The methyl group and tertiary alcohol carbon do not have stereocenters, so there are no otherchiral 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 reasonable conclusion that sesquiterpenes have fifteen carbons. Choice D is the bestanswer. Monoterpenes have ten carbons, diterpenes have twenty carbons, and triterpenes have thirty carbons.

Choice A is correct. The maximum wavelength of absorbance, Xmax, increases as the conjugation of the n-network increases. All four choices are conjugated dienes, but Compound 1also has a carbonyl in the conjugatednetworks. As such, the compound with the longest ^^ is Compound 1,choice A.

Choice B is correct. Step 1 involves a Diels-Alder reaction followed by decarboxylation. The intermediatecompound is the Diels-Alder product. There isnonitrogen present in either reactant, so thecompound cannot bea lactam (cyclic amide). This eliminates choice D. The product of a Diels-Alder reaction involving a dieneand an alkene (dienophile) is cyclohexene, so choice B 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

CyclohexeneLactone (not lactam)

CO Me CO Me Carbonyl is NOT conjugated to alkene

Compound 1 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 methyl group than Compound 6, so the role of methyl lithium,MeLi, must be to add a methyl group to Compound 6. This eliminates choices C and D. Methyl lithium has ananionic carbon, so it acts as a nucleophile rather than an electrophile. This eliminates choice A and makeschoice B the best answer. The methyl anion attacks the carbonyl carbon in the same fashion as an alkylmagnesium bromide anion attacks a carbonyl in a Grignard reaction.

Choice A is correct. A conjugated diene can be protonated at either terminal carbon of the 7t-network, becausethe carbocation that results is resonance stabilized. This eliminates choices B and C. Carbon a is a secondarycarbon while carbon d is a tertiary 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.

Not Based on a Descriptive PassageQuestions 93-100

93. Choice D is correct. Leukotriene A4, LTA4, has four alkene rc-bonds, one carbonyl Tt-bond, and an epoxide ring.This results in six units of unsaturation. Choice A is a valid statement, which eliminates it. LTA4 has threealkene rc-bonds in conjugation, resulting hi six 7t-electrons hi a conjugatedsystem. Choice Bis a valid statement,which eliminates it. Because of the extensive conjugation, there are several sites at which a nucleophile andelectrophile may add. For instance, if the epoxide oxygen were protonated, a nucleophile could attack the ringor the left carbon of any 7t-bond to add across the system. This means that 1,2-addition, 1,4-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 earned the coveted "huh?", meaning you can't eliminate it, because you're justnot sure. On a multiple choice exam, this is not a problem. Youjust need to look at choice D and use your testinglogic. LTA4 has four alkene 7t-bonds and one carbonyl rc-bond, so there arenine sp2-hybridized carbons. Thereare twenty carbons total in LTA4, soeleven ofthem are s/^-hybridized. There are more s/?3-hybridized carbonsthan sp2-hybridized carbons, so choice D is an invalid statement, which makes it the best answer.


94. Choice D is correct. Bromine, in the presence of light, adds to themostsubstituted 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. The productshown, as well as its enantiomer, are both formed.

95. Choice D is correct. Free radical propagation reactions keep the free radical reaction going, so to be apropagation reaction, there mustbe the same number offree radicals on each sideof 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 sideand nofree radicalson the product side, so it is a termination step and not a propagationstep. Choice Bis 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 propagationstep. 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.

96. Choice A is correct. Bergamontene contains fifteen carbons, so it is likely made from three 5-carbon isopreneunits. We can't be sure without analyzing the structure 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 7t-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 cyclohexane 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 (2(15) + 2 -24)/2 = 4, sochoice C is eliminated. There are three chiral centers on bergamontene, so 8 (2^) is the maximum number ofstereoisomers, not 16. This eliminates choice D and secures choice A as the best answer.

97. Choice D is correct. Both the Claisen rearrangement and the Cope rearrangement require dienes, but they neednot be conjugated. The two rc-bonds must be separated by three sigma bonds, so choicesA 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 conjugateddiene and a dienophile, so it musthave a conjugated diene. This makes choice D the best answer.

98. Choice C is correct. Foran E2 reaction, thebase mustbestrong enough toremove a protonfrom carbon andbulkyenough to not undergo substitution. Thismakes choice A a valid statementand thereby eliminatesit. For an E2reaction, the leaving group and proton being lost from carbon must be positioned 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 Bis a valid statement and thereby eliminated. For an Ei reaction, aleaving group first leaves, resulting in a carbocation. With carbocations, rearrangement can be observed, so it iswith Ei 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 drivebothEi and E2 reactions, so choice D is a valid statement. It is eliminated, leaving choice C as our choice.

99. Choice D is correct. Terpenes are hydrocarbons of10,15,20, etc... carbons, so theyare somewhatmassive 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/mole, etc., so they have somewhat high boiling points. High is a relative term, so we can't be certain ineliminating choice B. However, choices Band C are the same concept, so they mutually exclude one anotherfrom consideration. It is important that you use all of your test taking skills. The specific rotation of acompound is dictated by its chiral centers, which a terpene may or may not have. Given that there is nogeneral rule about the chirality of terpenes, we cannotconclude that they have high specific rotations. ChoiceD is the best answer.

100. Choice C is correct. Tohave a dipole not equal to zerois to have a dipole. Tohave a dipole is to be polar. Ciscompoundsare alwayspolar so Compound IIIis polar. Ethylene is perfectly symmetric, so choice Compound IIis nonpolar. Thequestioncomes down to: "IsCompound I polar?" Compound I is not polar, becausethe methylgroups on the alkene cancel one another and sum to a resultant vector of 0. Choose C for best results.


Organic Chemistry Sections I - IV Section Answers

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ChemistrySection

Raw

Score

EstimatedScaled Score

328 The Berkeley ReviewSpecializing in MCAT Preparation

'ifeDERKELEYIJR'E'V'i'e'W"

PERIODIC TABLE OP THE ELEMENTS

1

H

2

He

1.0 4.0

3 4 5 6 7 8 9 10

Li Be B C N O F Ne

6.9 9.0 10.8 12.0 14.0 16.0 19.0 20.2

11 12 13 14 15 16 17 18

Na Mg AI Si P S ci Ar

23.0 24.3 27.0 28.1 31.0 32.1 35.5 39.9

19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

39.1 40.1 45.0 47.9 50.9 52.0 54.9 55.8 58.9 58.7 63.5 65.4 69.7 72.6 74.9 79.0 79.9 83.8

37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54

Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe

85.5 87.6 88.9 91.2 92.9 95.9 (98) 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3

55 56 57. 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86

Cs Ba La+ Hf Ta W Re Os Ir Pt Au Hg TI Pb Bi Po At Rn

132.9 137.3 138.9 178.5 180.9 183.9 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 (209) (210) (222)

87 88 89 KAc§

104 105 106 107 108 109 110 HI 112

Fr Ra Rf Db Sr Bh Hs Mt Uun Uuu Uub

(223) 226.0 227.0 (261) (262) (263) (262) (265) (266) (269) (272) (277)

58 59 60 61 62 63 64 65 66 67 68 69 70 71

f Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tin Yb Lu

140.1 140.9 144.2 (145) 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.0 175.0

90 91 92 93 94 95 96 97 98 99 100 101 102 103

5 Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr

232.0 (231) 238.0 (237) (244) (243) (247) (247) (251) (252) (257) (258) (259) (260)


OrganicChemistry

the berkeley review mcat organic chemistry part 1

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