tbr ochem2 opt

Organicffiemist

Ya IISections V-VilI

Section VCarbonyls and Alcohols

Section VI

Section VIINitrogen Compounds

Section VIil ' '

Organic ChemistryLaboratory Techniques

Kr E. V. I. E.IilI @

Specializing in, MCAT Preparation

Section VCarbonyls

and

Alcoholsby Todd Bennett

oil

-c\"R CH,I

oll n

OH -Cr + Cl3

-L,/\-- R - -ott

+ HCI3yeliow oil

Oxygen Containing Compounds ,

a) Alcohol Propertiesb) Alcohol Reactivity ,

c) Alcohol Spectroscopy ,

d)AldehydeandKetrrnePro,p.ertie.se) Aldehyde and Ketone K€activity,f) Aldehyde and Ketone spectroscopyg) Ketals and Acetals :

i. Protecting Oroups:h) Carboxylic Acids and Derivatives

i. Carboxylic Acidsii. esters ,r,

iii. Lactonesiv. Acid Anhydridesv. Acid flalides ,

vi. Amides

Carbon5rl Reactivity l

a) Attack at Carbonyi Carbonb) Deprotonation of cr-Protonsc) Oxidation,Reduction ReactionC

Name Reactions ll. .,

a) QrignardReaction " I,'b) Aldol Condensationc) Claisen Condensationd) Transesterificatione) Wittig Reaction ' ' '

f) Pinacol Rearrangementg lodoform Reaction ,r,

,

h) n/olff-Kishner Reaction 1 .

$ynthetic r.o'gic ,

a) Reactions of Acetoacetic ,Esterb) Reactions of Malonic Esterc) Decarboxylationd) Protecting Croups

"

Carbonyl Biocheqistry '

a) Biological Oxidation-ReduCtionb) Biochemical Reagents

oilJI OH't-\

.,/ - \ -l-R CH3

oilll lr(-

-\- ,r'- \o -

./R CHr R-

ooll 2 more times ll

R-t\ctr, + R-c\cI.

O

ll oc-.. +ICHrI

oII

-C\R CIr

oI

* -c\

ooll, rrl ,

-\

(-. -,: __ -/-\ OOHROIodoform Test

REKI{ELEYI)R.E V.l.-E-w'

Speci altztng in MCAT Preparation

Carbonyls & AlcoholsSection Goals

Kecoqnize the carbonyl functional qrouDs and frpes of comoounds.oB You must be able to recognize functional grouFS sr-i;h as ar,lles, anhvdrides, acid halides, andespeciallr'ketonesandaldehydes. Youmustkn('\\', 1-.:'-.:.:..'.lrdsaremostreactivetowardssubsbitution (which is based cin the leaving groun :lrc: i:: .'.- * r-*.ra:t electrophilic (which is basedon the electron withclrawing or -d-ona:i:.: :.-::..:-. ,': the functional group).

Be able to identify infrared peaks for carbont-l compounds.Carbonrl compounds will have a peak in the inirar=i =:=:::-

.:: r: ::.e area of 1700+ cm-1. This willm.rsr Lleh be useful when compaiing two carbonr'. . --:': . -_ -: . : --i:rtiJr ing an unknown carbonvl.r\Itl'--r'LLlld. \oumaywishtoknowr6ughlr'rr'hercci:=:i - -i:i'.-r::.rJketjnesfallinthelRrang'e.

a?

carbon ds.-i:r- -:-. Lrcluded in this group should- : :- -; :- - Claisen reactioil. The Aldol

Ti.:"::t-onalphato,thecarbonylc_an!edeprotrr:.:::-t - - .-:: i::-,:.-,nbondedtoit. ThepK6of a-:::"j:riketoneistherangelT!2. Oncedepr..:::.::..-i::.r:i-,::.:=:ormed. Theenolat'ehisan;: *].-t::lum of its own with the enol structurE. I:.. - - r :,-: - . -: :, r:iorr€ into an enol is referred:: :s :f -i:omerization.

r 1 tg nlu(rr:: :--.*', :-:'.- a role in select biochemical

Ee a,ble to the aciditv of al carbons.

Ee able to identifv ketals, hemiketals, acetals and hemiacetals.

L'nd erstand the difference between thermo"dr-na m ic and kinetic enolates.*?

Iinos the mechanisms for acidic and basic calbonrl reactions.sw . ' = : - -- - --.:l for transesteriJication is preser.: --- i- - -.-'.grize the steps. Also recogniz._-: n*:r ,_- _" --:la:iic chemistry so it important:r-.: :' : .*::=.-.:r- to carry "out the procesS.

Recoqnire €ommon oxidizing and reducilq aqents.@p

fi.nolr common reactions by both name and re,ageots.

,: he loss of bonds to hydrogen.= -i -r.- the loss of bonds to oxvsen---'.ude LiAlHa and NaBFIa.

up: -' - - . ::.)\\ by name. Ttis a good idea- - -'- : - -: :s.rcfion conditions. Hishliehtsj . ---

= '.'. itting reaction, the io"dof6rm

:,1-, 1

Organic Chemistry Carbonyls and Alcohols Introduction

The carbon-oxygen bond is a major part of organic and biological chemistry. Asignificant part of organic chemistry on the MCAT involves compounds thatcontain carbon-oxygen bonds. In the case of carbonyl compounds, the carbon-oxygen n-bond is easily broken to form new bonds to the carbonyl carbon andsubsequently form a new compound. The carbon-oxygen o'-bond found inalcohols and sugars can undergo several reactions, but it is generally not asreactive as the carbon-oxygen n-bond. Our goal is to organize the vast multitudeof reactions involving carbonyl compounds and alcohols. Figure 5-1 showsseveral types of carbonyl compounds and carbonyl derivatives with which youshould be familiar.

Types of Carbonyl Compounds

oil

,zc:.H

Aldehyde

oil

./\ ,/RNI

HImide

o

1---L_/

Lactam

Enolate Resonance Forms

Figure 5-1

Ketal

oR'Ester

oil

NHz

Amide

/ NHPhNlt

,ZC\RRPhenylhydrzine derivative

.+

R'O OR'

\/-,.\

Hemiketal

oolr il

,zc:. .zc\RORAcid anhydride

oil

RR

R'O OR'

\/,zc:.RH

Acetal

R'O OH\/

Ketone

R'O OH\/

,ZC\RR

oil

,zc:.ROHCarboxylic Acid

oil

Acid halide

oil

ollC\

R'R

H

o/\c*r,

oil

.zc\ -R CHz

oI

,ZC\

,z c:.RH

Hemiacetal

-zc\RX ,zc:.

.zc:.

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Organic Chemistry Carbonyls and Alcohols Oxygen Containing Compounds

ioxy fi |cu tfii ing I Gbffid,uffi$],

'', r,i

!,:.i -,,i, r, i iLr;'

."'

oxygen containing compounds, because of the highly electronegative nature ofoxygen/ are very reactive. Much of organic chemistry revolves around alcoholsand carbonyls, so it is imperative to get a fundamental understanding of theirproperties, reactivity, and spectroscopic evidence that supports their existence.

Alcohol PropertiesBecause of their ability to form hydrogen bonds, alcohols typically have highboiling points and are generally miscible in water. Alcohols mike good solveritsas they are often liquids at room temperature and they have u lurg" rangebetween their melting and boiling points. Alcohols are hydrophilic, poiarmolecttles that become less hydrophilic (more lipophilic) as tireir iarbon chainiength increases. The smaller alcohols (three carbons or less) are highly watersoluble, but as the size of the alkyl group increases, their wateisoiubititydecreases. As with all compounds, their physical properties vary with mass andbranching, as well as the position of the hydroxyl gtonp. As themolecular massincreases, the boiling point increases, but the effect on the melting point is lessclear. As the branchrng increases, the boiling point decreases. Tatle s-1 showsthe physical properties of several alcohols, from which the effects of mass,branching, and positioning of the hydroxyl group on the physical properties canbe ascertained.

Isomer IUPAC Name(Common Name)

BoilingPoint

MeltingPoint

Density(g/rnL)

WaterSolubility

(g/100 mL)CH3OH Methanol 64.6'C -98"C 0.797 HighH3CCH2OH Ethanol 78.4'C -115"C 0.789 HighH3CCH2CH2OH 1-Propanol (n-Propanol) 97.2"C -727'C 0.804 High(H3C)2CHoH 2-Propanol (l-Propanoi) 82.3"C -90"c 0.786 HighH3C(CH2)3OH 1-Butanol (n-Butanol) 117,3C -90"c 0.810 8.2

H3CCH(oH)CH2CH3 2-Butanol (sec-Butanol) 99.6"C -115'C 0.806 12.8

(H3C)2CHCH2OH 2-Methyl- 1 -propanol (r-Butanol) 707.7'C -122'C 0.802 11.3

(H3C)3CoH 2-MethyI-2-prop anol ( / -Butanol) 82.0'C 24'C 0.789 HighH3C(CH2)aOH 1-Pentanol (n-Pentanol) r37.6"C -79'C 0.814 2.7

H3CCH(OH)CH2CH2CH3 2-Pentanol 119.3"C 0.809 5.0

(H3CCH2)2CHOH 3-Pentanol 115.9'C 0.815 5.6

H3C(CH2)aCH3 l-Hexanol (n-Hexanol) 157.5"C 0.814 0.8

C6H11OH Cyclohexanol 161.5"C 0.956 2.7

H3C(CH2)6CH2OH n-Octanol 194.7'C 0.817 0.3

Table 5-1

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Example 5.1\Alhat type of alcohol is the following molecule?

A. Primary alcoholB. Secondary alcoholC. Tertiary alcoholD. Phenol

SolutionThe compound has the alcohol functional group attachedbonded to two other carbons. This is defined as a secondaryanswer is choice B.

to a carbon that isalcohol, so the best

Alcohol ReactivityAlcohols are nucleophilic reagents in organic chemistry. Th"y are not goodnucleophiles in their protonated (neutral) state, but they can be deprotonated andconverted into their anion (alkoxide) form under basic conditions. Becausealkoxides (the deprotonated form of the alcohol) are strong bases, they are notthe ideal nucleophile, but they are generally better than alcohols. Alcoholchemistry also involves oxidation into a carbonyl as we shall see later in thissection. Alcohols are commonly formed from the reduction of carbonyls, whichwe shall also postpone for the moment. The common reactions to form andconsume alcohols that do not involve carbonyl compounds center aroundnucleophilic substitution. Figure 5-2 shows nucleophilic substitution reactionsthat convert alkyl halides into alcohols. Figure 5-3 shows nucleophilicsubstitution reactions that convert alcohols into alkyl halides.

Alkyl halides to alcohols

R

OI{\

H

+X

R

',.'F*H

R'

o,.Fu,H

R"

ot,,F*R'

1. RCO?-

---+2. OH-(aq)

H"O

-z --acetone

R'

HOJ.,,,.\.,,R

H

+ Br- + RCO2H

R''\

ou,r o + HX

R'

Figure 5-2

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Organic Chemistry Carbonyls and Alcohols Orygen Containing Compounds

Alcohols to alkyl halides

R

,ruroH

R'

ouroH

R"

ou,|o*,R'

oXR

cll"\cl - 'r*F.,H

PBr"---L

R'

Ut frrO

Inversion

H

R"

RuFu,R'

Retention

RacemizationHBr__-+

Figure 5,3

Spectroscopic Evidence for AlcoholsAlcohols can be detected using either infrared or NMR spectroscopy. In IRspec_tra, hydroxyl groups present a distinct absorbance between 3200 and 3500cm-l that is medium in intensity and broad due to hydrogen bonding. In,lHNvn spectra, hydroxyl gronpr present a signal betweln l ind 5 ppm i'hat isbroad due to hydrogen bonding, although the broadness varies with ihe solvent.They have no definite 6-value (it varies with concentration and solvent). Thepeak slowly disappears with the addition of D2o to the NMR tube. The oHgroup does not couple-well, so we rarely consider splitting patterns for alcohols.Figure 5-4 shows the lHNMR spectrum for 2-propanol in carbon tetrachloridesolvent. Figure 5-5 shows the IR spectrum for 2-propanol obtained neat on saltplates.

HOH

XHsC CHs

Copyright @ by The Berkeley Review

Figure 5-4

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1392 cm'l

3422 cm-lHOH

"r.X.*r,

l4J) cm ' & l-165 cm-l

2894 cml

2984 cm'l

Figure 5-5

Aldehyde and Ketone PropertiesBecause aldehydes and ketones do not form hydrogen bonds, they typically haveboiling points only slightly higher than alkanes of equal mass. Because of thepolarity of the carbonyl bond, they are slightly miscible in water. Aldehydes andketones are aprotic, polar molecules that become less hydrophilic as their carbonchain length increases. The smaller aldehydes and ketones (three carbons or less)are generally water soluble but as the size of the alkyl group increases, theirwater solubility decreases. Table 5-2 shows the physical properties of severalaldehydes and ketones, from which the effects of mass, branching, andpositioning of the carbonyl group on the physical properties can be ascertained.

Table 5-2

[somer IUPAC Name(CommonName)

BoilingPoint

MeltingPoint

WaterSolubility

(g/100 mL)HCHO Methanal (Formaldehyde) -27'C -92"C HighH3CCHO Ethanal (Acetaldehyde) 21"C -127'C InfiniteH3CCH2CHO Propanal (Propionaldehyde) 49'C -81'C 76.3

H3C(CH2)2CHo Butanal (n-Butyraldehyde) 76"C -99'C 6.8

HgC(CHz)gCHO Pentanal 103"C -92'C J.J

H3C(CH2)aCHO Hexanal 728'C -56'C 2.1

C6H5CHO Benzaldehyde 778"C -26"C 0.3

H3CCOCH3 Propanone (Acetone) 56'C -94'C InfiniteH3CCOCH2CH3 Butanone (Ethyl methyl ketone) 80"c -86"C 25.6

H3CCO(CHz)zC}{s 2-Pentanone 102"C -78"C 5.7

(H3CCH2)2CO 3-Pentanone 101'C -47'C 4.9

H3CCo(CHz)eCHe 2-Hexanone T28"C -55"C r.6

H3CCH2CO(CH2)2CH3 3-F{exanone I24'C 1..3

H3CCOCH2CH(CH3)2 4-Methyl-2-Pentanone T79'C -85"C r.9

C5H19O Cyclohexanone 156"C .C2.2

C6H5COCH3 Acetophenone 202'C 27'C Insoluble

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Organic Chemistry Carbonyls and Alcohols Orygen Containing Compounds IExample 5.2What is the IUPAC name for the following compound?

OH

I

./\,/\'--Hilo

A. 1-A1do-4-pentanolB. 4-HydroxypentanalC. 5-Oxo-2-pentanolD. 2-Hydroxypentaldehyde

SolutionThe longest chain is five carbons and the highest priority functional group is thealdehyde. The functional group with the most oxidized carbon receives thehighest priority according to IUpAC convention. For naming aldehydes, the ',e,'is dropped from the alkane chain of the same length and an ial', suffix is added.This makes the compound pentanal, which makei choice B correct. The oH is:9""'*Aldehyde and Ketone ReactivityAldehydes consist of a carbonvl with a hydrogen bonded to the carbonyl carbonalong with either an alkr-l group or in the-case of formaldehyde, a secondhydrogen. Ketones consrst of a carbonyl group with two alkyl substituentsattached. The cher,-rrstn- occurs primarily at the electrophilic carbonyl center.Aldehydes and ketones are reactive with most nucleophiles, but not by atraditional nucleoph-L:

'u'nstirr-rtion mechanism. once a nucleophile attacks acarbonyl carbon, il io::--. a fcur-ligand intermediate with u ,-r"gutirru charge onoxygen known as a:::.,;.,'..-;.--;.. :'::ermediate. This intermediatu iuitt Uu shown inthe mechanism oi :na:r'! :a:r.::., I reactions in this section. The chemistry ofaldehydes is similar r- '-r= -i=rr.--.::r-of ketones except that an aldehyde can beoxidized into a ca:'r-.,-.'--: ::-.:-,.,-hil" k"tones cannot be oxidized easily.oxidation in carbcr.-" r ::.e::---::-. :a:, be r.iewed as either the gain of bonds iooxygen or the loss -.: : ::: :- ... j:--:er. \ve shall thoroughly idd."ss carbonylreactions throughc-: --i -: ::-i .l

Eilejea

.q..

M.

D.

Smffi

AiltiTrr;:

-iiinn

u[1iJ]11

*d.n i1

$lPi"u

lnixf

Cffi

Berkeley Review The Berkeley Review fil,f


o

AHsC cH2cH3

2ppm

Figure 5-5

o

rAcH2cH2cH3

1117 cm'1

cm'1 1457 cm-r 1388 cm I

and

1ZZA "^-t

1362 cm'1

Figure 5-7

Example 5.3Pentanal can be distinguished from 3-pentanone by all of the following EXCEpT:

A. a signal at9 - 1,0 ppm in the IHNMR.B. fir'e signals in the lHNMR rather than two signals.C. an IR absorbance at L826 cm-1.D. an ultraviolet absorbance at 230 nm instead of 240 nm.

Solution-rn aldehyde hydrogen is found between 9 ppm and 10 ppm in the 1HNMR, so::-oice A is a valid way to distinguish an aldehyde from a ketone. choice A is--minated. Pentanal has five unique hydrogens while 3-pentanone has two--nique hydrogens, so the two compounds can be distinguished by their:espective number of signals in the 1HNMR. Choice B is etiminated. Aldehydes=C ketones have different n-bonds, so they have different carbonyl absorbances-:- 'he ultraviolet absorbance region, This eliminates choice n. Ataenyaes andAetones have different IR absorbances, but they are observed around 1TO0 cm-l ,:.;'t at 1826 .m-1. This makes choice C an invalid technique, which makes choiceC 'ie best answer.

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Organic Chemistry Carbonyls and Atcohols Orygen Containing Compounds

Ketals, Hemiketals, Acetals, and HemiacetalsKetals and hemiketals are derivatives of ketones while acetals and hemiacetalsare derivatives of aldehydes. Ketals occur when a ketone roses the carbonylgrgup and gains two alkoxy functional groups (R-o). The oxidation state ofcarbon does not change, because the carb-on ,till hus two bonds to oxygen, butnow-it is,two sigma-bonds to two different oxygen atoms rather thanJsigma-bond and pi-bond to the same oxygen. A hemiketal occurs when the ketone hasits carbonyl group converted into ahydroxyl group and gains one arkoxy group.Acetals are similar to ketals, except it is ihe" ataenyae"tnat loses its carbonylgroup to gain the two f]<o1y groups. Hemiacetals are similar to hemiketais,except it again is an aidehyde, rather than a ketone, that converts its carbonylqtoYp into a hydroxyl group while gaining an arkoxy group. Figure 5,g showsthe formation of the four compounds.

o

ARR'

o

ARR'

Ketone

xs R"OH/OH-=-=.

R"O OH

XRR'Hemiketal

R"O OR"

Ketone

o

ARHAldehyde

Ketal

'o oH

XH

Hemiacetal

xs R"oH/H' \ /RR'

R

xs R'OH/OH

R

o R'O OR'

ll ,*, R'oH/Ff -\ /-"nAs

-E- nAn

Aldehyde Acetal

Figure 5-g

Acetals and ketals are useful as protecting groups in organic synthesis. Acetalsand ketals can be formed and removea onty ,t d", aciiic .orrditio.rr, where ashemiacetals and hemiketals are formed oniy under basic conditions but removedunder any conditions.

1

n

ul

e

s

s,[

d

m

fi,d

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Figure 5-9 shows a general mechanism for ketal formation under acidicconditions. It is the same mechanism for the formation of an acetal from analdehyde, except that an aldehyde is the reactant, rather than a ketone.

+ 2 R'OHH*

HzO +

Or"r.^"r"'lf/t'll

deprotonatetl

n'- oH"\*.

lbreak I

tH\6..

.l

R'- OH

'll,-*"

,""41,

"o7o :o/

.:+.n'- o+r

11,**-

H\.+ C)' R'

deprotonate-iiA

protonate

Figure 5-9

T'he steps of the mechanism are labeled to emphasize the predictable nature ofacid-catalyzed mechanisms. When you draw a mechanism for an acid catalyzedreaction, the intermediates must carry positive charges and no molecule shoulder-er carry a negative charge. With the exception of rearrangement steps inselected cases, acid catalyzed mechanisms follow this same pattern of: 1)p'rotonate (making the leaving group a better leaving group), 2) break (theleaving group leaves), 3) make (the nucleophile attacks the carbocation), and 4)Ceprotonate (returning the molecule to a neutral state). Base catalyzedrrechanisms follow the exact opposite pattern of: 1) deprotonate (to make as'Tong nucleophile),2) make/break (where the nucleophile attacks and dislodges5e leaving group), and 3) protonate (returning the molecule to a neutral state).

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These mechanisms should be kept as simplistic as possible. The mechanism forthe formation of a hemiacetal from an aldehyde and alcohol in the presence ofstrong base is shown in Figure 5-10.

.o.A\

make/breakH

-

..1r1 f 'Base.- t\\,, !!

a"or.-ranua"H

R,onH<) /vrotonate

o

AcP:,s1:i

:-.-t-E--

::i

\rs.E

:.:- -:-E_

{;

!lflxl;urH$

dflni

ilhn

Figure 5-10

when you draw the mechanism for a base catalyzed reaction, be sure that theintermediates carry negative charges ancl no molecule ever carries a positivecharge. Ketals and acetais serve as protecting groups for carbonyl groups insynthesis involving ketones and aldehydes. Hemiacetals and hemiketals are notuseful as protecting groups, but they are important in sugar chemistry.

Example 5.4

Addition of ethanol at a pH of 4 to propanal yields which organic product?

A. B. C. D.

"'oy-o" tY_"'' 'oyou' "oyo'EtHEIOHEt-HEiH

SolutionAddition of an alcohol to an aldehyde under acidic conditions (pH = 4 is acidic)yields an acetal. The only acetal in the choices is choice A. Choice B has toomany ethoxy groups, choice c is a hemiacetal, and choice D is a geminal diol.

Example 5.5

Addition of sodium methoxide in methanol to acetone vields:

A. isopropyl alcohol.B. acetaldehyde.C. a ketal.D. a hemiketal.

SolutionAddition of an alcohol to a ketone under basic conditions yields a hemiketal. Themethoxide anion attacks the carbonyi carbon of acetone to generate thetetrahedral anion intermediate, which then deprotonates the methanol to formthe hemiketal and regenerate the methoxide anion. The best answer is choice D.

-:-:,lL

:',r,,

Er

;.rlr:rt

_4

5olrl

rrilnt

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Acetals and Ketals as Protecting GroupsProtecting groups are used in synthesis to prevent a reagent from reacting at asite where it is undesirable to have a reaction. We shall only discuss protectinggroups for aldehydes and ketones at this time. Both of these carbonylcompounds employ the same reaction to add the protecting group, formingeither an acetal or ketal. As we have seen, aldehydes and ketones, in thepresence of alcohols and acid, form acetals (from aldehydes) and ketals (fromketones). Because acetals and ketals are less reactive than aldehydes andketones, they are an ideal protecting group. The alcohol that is typically used toform the protected carbonyl compound is a vicinal diol (a l,2-diol), such asethylene glycol (HOCH2CH2OH). Figure 5-11 shows the protecting ofcvclohexanone using ethylene glycol.

n5+U Hzo

o

Cvclohexanone Protected as a Ketal

Figure 5-11

The mechanism for forming the protected ketone is the same generic acid-catalyzed mechanism shown in Figure 5-9, only instead of using two moleculesof alcohol, a vicinal diol is used, so the second "mske" step involves the secondhvdroxyl group of the vicinal diol rather than a new alcohol.

Example 5.6The addition of ethylene glycol (HOCH2CH2OH)orovides a protecting group for ketones. \Alhich of theform of 2-pentanone?

.{./ \

u'l-f

xr^--

in the presence of acidfollowing is the protected

C.

oo\><,,.

o'nooY

SolutionThe original ketone (2-pentanone) has two bonds from carbon 2 to oxygen, so the:roduct must also have two bonds from carbon 2 to oxygen. This eliminates all:f the answer choices except choice B. The product of a diol and a ketone in:i:r,hr-drous, acidic conditions is a cyclic ketal.

There is not much to using an acetal or ketal protecting group. The last thing to:rnsider is when to use a protecting group in synthesis. As a guideline, any time--j-.at vou have a molecule with more than one reactive site, you must protect thesr:es at which you wish to have no reaction. The exception to this rule is when--:e site vou wish to react at is significantly more reactive than any other sites on:::e mo1ecule.

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Carboxylic Acids and Their DerivativesCarboxylic acid derivatives are different from aldehydes and ketones in that thecarbonyl carbon has a functional group that possibly can act as a leaving group.Much of the chemistry of carboxylic acids and acid derivatives centers aroundchanging the group on the carbonyl carbon. We shall address each functionalgroup starting with the carboxylic acid and look at their chemical reactions.

Carboxylic AcidsCarboxylic acids are weak acids with a pKu between 2 and 5. They are readilyconverted into esters, anhydrides, or acid halides. They carry out similar orgurri.reactions as esters, but are less reactive than esters. Carboxylic acids cin beformed by saponification (treating an ester with strong base in water), by treatinga methyl ketone (RCocH3) with 12 and strong bise, by oxidizing primaiyalcohols and aldehydes in water, or by hydroryzing a nitriie o, un uniide ,rirrgstrong acid at high temperatures. These reactionr ut" shown in Figure s-12.

saponification: 3 xs H.O/Olr ?r _______::*

OAO + HOR,R,Aoo,--Ester Carboxylate

Ester hydrolysis: O

oAooEster

Iodoform reaction: O

o4.",

xs HrOlFf

-12/oH

o

A + HoR,ROH

Carboxylic acid

o

A + I,CHRO-Carboxylate Iodoform

oil + NHn

ROHCarboxylic acid

r:

Oxidation:

Oxidation:

Methyl ketone

oll xrranonZorr

_

-/\ :-

RHAldehyde

HH\/ Krc,gr/.L _

-/\ --R- OHPrimary alcohol

oA

Amide hydrolysis: O

AR

xs H2OlFf

-

NHR'Secondarv amide

Nitrile hydrolysis:

It- c: N .t H'zolFI.

-A

ROHCarboxylic acid

olt

-A-ROHCarboxylic acid

oilil + R'NH3}

ROHCarboxylic acid

s",S{

,tru

frfi

pW

ffiffidlW

cn

milf'pdLmtfrtilJlli


Figure 5-12

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Carboxylic acids can be reduced into primary alcohols or converted into othercompounds such as acid halides, acid anhydrides, or esters. Figure 5-13 showsfour reactions of carboxylic acids with which you are expected to be familiar.

HH

XROHPrimary alcohol

ooAARORAcid anhydride

Figure 5-13

with ethanol and acid yields which of the following

o

oA.'Acid halide

"fo

AR OR'

Ester

Example 5.7Treatment of benzoic acidcompounds?

A'oB' o o D.c. o

d*oa"*o,oao'oEt

SolutionAddition of an alcohol to a carboxylic acid under acidic conditions with enoughheat to overcome the activation barrier yields a new ester by way of atransesterification reaction. The final product is an ester with an ethoxy group inplace of the hydroxyl group of the carboxylic acid, along with a water molecuieside product. Choice B is the best answer.

EstersEsters are carbonyl compounds with an alkyl group and an alkoxy groupattached to the carbonyl carbon. Esters have a leaving group (the alkoxy group),so they undergo more reactions than ketones or aldehydes. Their reactivitycorrelates with the pKu of the conjugate acid of the leaving group. Leavinggroups that are more stable (are less basic and their conjugate acids have a lowerpKu value) are better leaving groups. Esters can easily exchange their alkoxygroup in the presence of acid and an alcohoi in what is referred to as a

t r ans e s t er ifi ca t i on rcaction.

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Organic Chemistry Carbonyls and Alcohols Oxygen Containing Compounds jExample 5.8What is the major organic product for the following reaction?

.d

I

s.nr

:l

.te

:;r

o

A+H3CH2C OCH2CH3

H3coH/Ff 1+

B.A. C. D.

OCH3 ;X::"' :X;o

Ao

,,Ao.ru EtO

SolutionAt high temperature under acidic conditions, an alcohol can undergo atransesterification reaction when mixed with an ester. The final product ii thenew ester with the methoxy group attached instead of the ethoxy grorp. Ethanol

jt" ut""t" rta" t

In Example 5.8, a C-o bond and an o-H bond were both broken in the reactantsand- formed in the products. This implies that the enthalpy (AH) for the reactionis close to 0. Going from an ester and non-cyciic alcohoi to an ester and non-cyclic alcohol generates a change in entropy (AS) of roughly 0, because thereaction starts and finishes with roughly the same degreei of freedom. ThisilPlies that the change in free energy for transesterification is around 0 (AG = AH- TAS). This means that the equilibrium constant for a transesterification reactionis_ approximately 1. The reaction can be driven to products by focusing on LeChatelier's Principle. It will proceed in high yield if ihe producis are removed ora reactant is constantly added. This also has biological significance in thattransesterification can be used for shuttles in the cell membrane. For theformation of a lactone (cyclic ester), there is a ring formed and an alcoholmolecule produced from a linear system. Depending on the reaction, there canbe either a gain or a loss in entropy, so lactonelormati,on is less predictable.

LactonesT,actones are cyclic esters, as shown in Figure 5-1. Lactones undergo the samechemistry as esters, only entropy is a factor in the reaction,s flvorability.Illgtones can be synthesized by treating cyclic ketones with peroxyacijs(RCO3H) in what is known as the Baeyer-Villiger reaction. Lactone chemistry iseasier to understand when you recognize that the compound is an ester. Figure5-14 shows the formation of a lactone from an intramolecular transesterificulionreaction of a hydroxyester.

a

-{t&Cri

5!

iilMi;

ffi

.u'f

rfud

ury

H*__+OHA

+ ROH

:-JleAudru

dfrM

,ffimtu

mflW

LiS[M

l{lLr!][Copyright @ by The Berkeley Review

Figure 5-14

The Berkeley Review


Acid AnhydridesAcid anhydrides are formed by a condensation reaction (dehydration) of twocarboxylic acids at elevated temperatures (as shown in Figure 5-13). Therefore,when you add water to an acid anhydride, it breaks into two carboxylic acids. Itis easiest to predict the product if you focus on the inorganic side product (H2Oin this case) more than the organic product. If you circle and remove an H fromone reactant and an OH from the second reactant and then connect the twoleftover fragments, then you can derive the organic product rather effortlessly.This is shown in Figure 5-15 as a short-cut method for deducing the product ofdehydration of two carboxylic acids.

ROR'Connect the atoms

Figure 5-15

Example 5.9What is the major product from the following reaction?

o

A

.;.,4o,'L.";,AoA.*,,Atoms are connected Anhydrides are good

o

^*Ao

Ao

,,.11.... * AAROR'

Viva la anhydride

oo

Find the water

OH HsC---

OH

A. DiketoneB. Acid anhydrideC. EsterD. Aldehyde

SolutionAddition of heat to carboxylic acids yields an acid anhydride by driving offwater. The final products are an acid anhydride and a water molecule. Choice Bis the best answer There are three possible acid anhydrides that can form. Oneof the three possible acid anhydrides that may form is drawn below. The othertwo acid anhydrides that can form result from the condensation of each acidupon itself. Probability says that the product below is the most likely.

o

Aooo oo

HrrCo

Water is

Acid HalidesAcid halides are similar to esters, but with a halide (Cl, Br and I) in place of thealkoxy group. They are the most reactive of all the carbonyl compounds, becausethe halide is a great leaving group. They undergo the same substitution reactionsas other carbonyl compounds that have a leaving group, but they react faster.For some reactions, acid halides can be too reactive. They are a usefulintermediate product in many synthetic pathways, such as the conversion of acarboxylic acid into an amide for instance, where the carboxylic acid is firstconverted to the more reactive acid halide which then goes on to form the amide.

found

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Organic Chemistry Carbonyls and Alcohols Orygen Containing Compounds Or

Example 5.10what are the two products formed when propanoyr chroride, H3CCH2CoCI, istreated with methyl amine, H3CNH2?A. N-methyl propanamide and hydrochloric acidB. 1-amino-2-butanone and hydrochloric acidC. Propanal and chloromethylamineD. Carbon dioxide and N,N-ethylmethylamine

SolutionA' amine is a good nucleophile capable of attacking the carbonyl carbon anddisplacing the chloride anion to form an amide, choicl A. Amides are biologicalstructures that you are required to know according to the MCAT student Manuril.This reaction can be used as a precursor to synthJsizing an amine. Amides canbe reduced to an amine using LiAlH4 in ether followediy neutralization with aweak acid. There are problems with direct synthesis of primary amines usingammonia and an alkyl halide due to multiple additions. The point here is thaiamides can be products themselves, or intermediates in amine synthesis.

AmidesAmides are carbonyls with-an amine group bonded to the carbonyl carbon.Amides form the backbone of proteins, ut-ra tn"y are found in most of the bases ofnucleic acids (i.e., DNA and RNA). An amide bond that links amino acidstogether is referred to as a peptide bond. Amides can be reduced into aminesusing a strong reducing agent such as lithium aluminum hydride, LiAlHa.

Example 5.11\A/hich of the following compounds does Nor contarn an amide bond?A. GuanineB. UracilC. IsoleucineD. Cytochrome

SolutionMost amino acids do not contain an amide bond, although polymers of aminoacid (polypeptide chains) i: J, is in proteins that amiio icids form peptide(amide) bonds. The peptide bonds of proteins are broken down in acidicaqueous environments to regenerate the individual amino acids, Guanine anduracil are bases in RNA, and they contain amide bonds (as drawn below). It isthe amide functional group that forms the hydrogen bonding in base pairing.Cytochrome is an enzyme (protein), so it contains amide troids in its peptidelinkages. The best answer is choice C, isoleucine, an amino acid.

Cr

Ca:

ll"p-- "1->uL

:eairaei!dl -

I€Aithri

-{rhkr.

A,>L

'2+-q

E',:E

a-L'l

-,f Jt::i

j-*

-t--ld

i::E:-

o/\ .HfiT-*Ao

HUracil

(pyrimidine base)Guanine

(purine base)

oH

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Organic Chemistry Carbonyls and Alcohols Carbonyl Reactivity

rcffiH$ffiiiiiiffiH ftiffiffiCarbonyl reactions can be categorized in one of three ways. The fist type ofreaction involves a nucleophile attacking the electrophilic carbon. The secondtype of reaction involves the deprotonation of an alpha proton and thesubsequent nucleophilicity of the anion that is generated. The last type ofreaction falls into the realm of oxidation-reduction chemistry, although mostmechanisms for carbonyl redox reactions involve a nucleophile attacking thecarbonyl carbon. If you keep things simple in terms of the three types ofreactions, you should be able to summarize all of carbonyl chemistry and workthrough any questions they may present.

Attack at Carbonyl CarbonBecause carbonyl compounds contain a C=O bond, they are good electrophiles.We shall consider ketones and aldehydes first, but other carbonyl compoundsalso act as electrophiles. The difference in reactivity between a ketone and acarboxylic acid derivative, such as an ester, centers around the presence of aleaving group on the carbonyl carbon. The C=O bond is polar with a partialpositive charge on the carbon atom and a partial negative charge on the oxygenatom. It is the partial positive charge on the carbon that makes a carbonyl such awonderful electrophile. Figure 5-16 shows a generic carbonyl reaction, where acarbonyl compound is attacked by a nucleophile to form a tetrahedral intermediste.

Nuc:----------+

oil

Nuc O-

\/

Carbonyl Tetrahedral Intermediate

Figure 5-16

\{e have already seen aldehydes and ketones serving as electrophiles in theformation of acetals, hemiacetals, ketals and hemiketals. As far as electrophilicchemistry of carbonyls is concerned, there is no major variation betweenreactions. In other words, it is best to view carbonyl reactions as all basically thesame with slight variations of the nucleophile. The tetrahedral intermediate inFigure 5-16 represents every generic intermediate in carbonyl addition reactions.

For carbonyl compounds that have leaving groups, the reactivity of the carbonylcompound is based on the strength of the leaving group. Stronger leavinggroups make for a more reactive (more electrophilic) carbonyl compound. Thestrength of a leaving group can be inferred from the pKu of its conjugate acid.Leaving groups are considered to be good when they form a stable compoundupon leaving. So, as the leaving group gets stronger, it gets more stable, whichmakes it less basic, and thus makes its conjugate acid stronger. This means thatgood ieaving groups generally have conjugate acids with low pKu values. Figure5-17 shows a carbonyl reactivity chart (relative reactivity of substitutedcarbonyls) for an acid halide, an acid anhydride, an ester, and an amide. Theconjugate acid of each leaving group is shown in the same diagram.

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organic chemistry carbonyrs and Alcohors Carbonyl Reactivity

o

H

PKu = -1.0 to -7 very goodvery strong acid leaving group

H-- oRi -[oR OR'

The better X is at leaving from H, the better it will be at leaving from C,so the best leaving group has a conjugate acid with the lowest pKu.

ii -_ ii fr flR.,,t\x) ,ezt-o o-tr- oa.to\_,

n

jIt^d

f,(D

sTGdiTnn

nuT&dtelh

oil

oil

o

o

PKu=3to5average acid strength

good leaving group

PKu= 14to 17

weak acidseml-poor

leaving group

oo-ll il)r

x o-'- o/'\R

pKu = 33 to 35 very poorvery weak acid leaving group

*r_4rro, --CCR NHR'

o.il-zL\OR' R- NHR'

Relative Reactivity:

-zC:.R o/c\

Figure 5-17

The relative reactivity implies that an acid halide can easily be converted into ananhydride, ester, or amide. An acid anhydride can easily be converted into anester or amide, but it is difficult to convert the anhydride into an acid halide.This technique is a good predictor of the reactivity of all carboxylic acidderivatives. The generic reaction and its tetrahedral intermediate are shown inFigure 5-18.

olt

--zY:.

-R T L.G.

L:Nuc

op L.G.( \ //7'i t =::R Nuc

o\il) o

il

/'cR

o+ L.C.\o

Nuc

Figure 5-18

Carbonyl substitution reactions proceed via a tetrahedral intermediate as shownin Figure 5-18. If the leaving group is not a good one, then the reaction cannot gofurther than the tetrahedral intermediate and will ultimately shift back to tf,ecarbonyl reactant. This results in an equilibrium between the original carbonyland the tetrahedral intermediate. This is observed with ketoner u"na uta"nya"twhen they are present in a solvent that has nucleophilic capability.

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Example 5.12\Mhich of the following is the MOST electrophilic carbonyl compound?

A. AmideB. EsterC. AldehydeD. Acid anhydride

SolutionThe reactivity of a carbonyl is dictated by the leaving group on the carbon of thecarbonyl. As the acidity of the conjugate acid of the ieaving group increases, sodoes the reactivity of that particular carbonyl. In the choices above, the conjugateacids of the leaving groups are an amine, an alcohol, H2, and a carboxylic acidrespectively. The most acidic is the carboxylic acid, so the anhydride is the mostreactive carbonyl.

Deprotonation of cr-ProtonsThe hydrogen on the alpha carbon (the carbon adjacent to the carbonyl carbon) isacidic (its pKu is approximately 19), so it can be removed using a strong base.Enolates are formed when a hydrogen on the alpha carbon is deprotonated. Theenolate can regain a proton at either the carbon or the oxygen. If it is protonatedat the oxygen, an enol is formed. There is an equilibrium between the ketone and

=nol. The conversion from a ketone into an enol is known as tautomerizotion,iecause a ketone and its enol are tautomers, structural isomers that vary in the:osition of a n-bond and a hydrogen. The tautornerization of acetone is shown inF:zure 5-19.

Ketone

.u.il

Carbanion

.u.\ll)^

-./t\

._ Q.z--,_f) CH: uru

ilA: uuru

Enolate Enol r r

, o;o,-r.sf|r" :ot n

tlurc,zc- -cHr._= ,c2c\

.eCHe CHs

Figure 5-19

-- : - carbanion that forms is a good nucleophile. When an alkyl halide is added:: :.!e solution, the carbanion can attack the alkyl halide in a nucleophilic; -:siitution reaction to form a new carbon-carbon bond. This results in a longer. =::i-,e. The halide can be any halide, but the reaction works best with an alkyl,'--re compound. Alkyl bromides and chlorides yield more O-alkylation side

:: :.i:cts than alkyl iodides. The generic reaction is shown in Figure 5-20.1

)e.1

S

Ketone Carbanion

.u. .u.ilil,/ -\lJ:F\ CH: Hzg. CH:

l) -\

HA:base \ o4Figure 5-20

Longer Ketone

:o:il

.zC\H"C CH"'l J

R

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Organic Chemistry Carbonyls and Alcohols Carbonyl Reactivity lWhgn the carbonyl is asymmetric, the possibility of two different enolates arises(each formed by deprotonating a different arpha hydrogen). If one of the twoalpha carbons is more substituted, then it is both -or" It"ri.ully hindered andmore stable. This creates a situation where the reaction can be dictated bytemperature and base size. At low temperature with a bulky base, the lesshindered (less substituted) alpha carbon gels deprotonated to form the so calledkinetic enolate (lower energy transition staie). This is referred to as kinetic control.At a higher temperature with a smallbase, the more hindered (more substituted)a-lpha carbon gets deprotonated to form the so called theimodynnmic enolate(leading to the more stable product). This is referred to as thermodynamic controlWe shall apply this concept to aldol condensation reactions of asvmmetricketones later on in this section.

Example 5.13what is the final product after acetone is treated first with NaH, followed byiodoethane, and subsequently followed by workup?A. 2-Methyl-2-butanolB. 3-Methyl-2-butanoneC. 2-PentanoneD. 2-Butanone

SolutionThis reaction is simiiar to the generic reaction in Figure 5-20. The carbon chainle1sfh is_ increased by two carbons (from three to five) when the electrophile isethyl iodide. This eliminates choice D. The product is a ketone, so choice A iseliminated. The final product is 2-pentanone, as shown below, which makeschoice C the best answer.

,ll

!o! ioiilil^

--- ^./ t\ -

--- ozL\HzGf -CH.. H,C- - CH,l)1'rHAtbase

"\+HsC

oxidation and reduction are recurring in organic chemistry, so working from alogic-based foundation is key. If the oxophilic carbon (carbtn containin[ a bondt9 gxygen) has hydrogens, it can be oxidized. Primary alcohols are oxidized intoaldehydes, which can be further oxidized into caiboxylic acids. secondaryalcohols are oxidized into ketones. Tertiary alcohols cinnot be oxidized (thealcohol carbon has no hydrogen to lose). Reduction is defined as the opposit! ofoxidation, so the reverse of each reaction just mentioned represents reduction.To make the processes more clear, we shall define oxidation and reduction interms of bonds to oxygen and bonds to hydroge'. More than just oxygencontaining compounds do redox chemistry. when two cysteine residues form acrosslink, they undergo dehydrogenation (1oss of hydrogen), an oxidativeprocess. when the n-bond in a fatty acid is hydrogenated to form an aliphaticchain, it has undergone a reductive process.

. (_).

ll-->(- ,/-\H,C CH:'l

H3CH2C

2-Pentanone

B,A

ru!lllm

{l$r.l

.Frul

rtu.d@,ri

frnm

SMrm

(fill!$l

ffm

Jlimm

$Mf,

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Oxidation is defined as an increase in oxidation state, which is caused by eitherlosing a bond to a less electronegative atom (in most cases hydrogen) or gaining abond to a more electronegative atom (in most cases oxygen). Reduction isdefined as a decrease in oxidation state, which is caused by either gaining a bondio a less electronegative atom (in most cases hydrogen) or losing a bond to anore electronegative atom (in most cases oxygen).

)etermining oxidation states using the method learned in general chemistry is a

=tatter of assigning oxygen a -2 (because it is more electronegative than the atoms:c rvhich it bonded, and it makes two bonds) and a +1 to hydrogen (because it is,ess electronegative than the atoms to which it bonded, and it makes one bond).ll:re oxidation state of any remaining atoms is found by difference. In organic::.emistry, oxidation states for specific atoms can easily be found by considering=iectron sharing in each bond. If the bond is between two atoms of unequal=-ectronegativity, then the more electronegative atom is assigned a -1 and the less

=-ectronegative atom is assigned a +1. The oxidation state of an atom is foundrv summing the numbers from all of the bonds and any formal charge it mayhar-e. Figure 5-21 shows this method as it applies to the oxidation state of carbonI n 2-propanol and acetone.

OxidationGain of bonds to O

Loss of bonds to HIncrease in oxidation state

CHs

ReductionLoss of bonds to O

Gain of bonds to HDecrease in oxidation state

Hl+

lol >lost a bond to H

gained a bond to O

-o CHs

Figure 5-21

::-:. oi the four bonds to carbon is analyzed for its relative electronegativity: ::'.pared to the atoms to which it is bonded. Bonds to hydrogen give a negative: - :arbon and a positive to hydrogen, because carbon is more electronegative- ::. hvdrogen. Bonds to oxygen give a positive to carbon and a negative to,-"'j€fl, because carbon is less electronegative than oxygen. Both carbons in a

:,:::r-carbon bond get zero, because there is no difference in electronegativity.-: r .econdary alcohol, the oxophilic carbon has an oxidation state of 0 while in a,-:r--r.€, the oxophilic carbon has an oxidation state of +2. This means that the:,:::::r rvas oxidized by two electrons, which is predictable, because it has lost a: -: ' to hvdrogen (oxidizing it by one electron) and has gained a bond to oxygen, '':-zrng it by another electron). Figure 5-22 shows that oxygen and hydrogen: :.-t change oxidation state when going from 2-pentanol to acetone.

H A-,\11

Hrclo ocfu cH,CHs

Figure 5-22

, , ::n has a -2 oxidation state, as is expected. Hydrogen has a +1 oxidationr:. as expected. Oxidation states should be made this simple.

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Organic Chemistry Carbonyls and Alcohots Carbonyl Reactivity (

Oxidizing and Reducing Agentsoxidizing agents are rich in oxygen (given that they deliver oxygen to thereactant) and reducing agents are rich in hydrogen (given that itrey deliverhydrogen to the reactant). This perspective is appfcable in both organicchemistry and biochemistry. It should be noted that NAD+ acts like Cro3-withpyridine, NADH acts like NaBH4, and FADH2 acts like H2/pd. Ethanol can beoxidized into ethanal using either Cro3 in the presence of HCI and pyridine(known as PCC) or by NAD+. in the reduction half-reaction, NAD+ is redlced toNADH, because for every oxidation half-reaction, there is a reduction half-reaction. The coupled half-reactions are shown in Figure 5-23.

NAD* NADH +H

Figure 5-23

oxidizing agents cause oxidation (and in doing so, they get reduced). oxidizingagents are rich in oxygen, poor in hydrogen, and have an atom in a highoxidation state with high electron affinity. Reducing agents cause reduction (aidin doing so, they get oxidized). Reducing agents are poor in oxygen, rich inhydrogen, and have an atom in a low oxidation state with low ionizai=ion energy.Some common oxidizing and reducing agents are listed below.oxidizing Agents (Rich in o; Poor in H) Reducing Agents (poor in o; Rich in H)

.l :rl-;-_E

: l':;ler'ff"tS

lmruils

i,;.m

Typical oxidation and reduction reactions involve the conversion betweenalcohols and carbonyl compounds. Figure s-24 is a summary of commonoxidation-reduction reactions in organic chemistry (1' alcohol + aldehyde -+acid and 2' alcohol -+ ketone -- 3" alcohols undergo no oxidation reactionsj.

Primary alcohols have two hydrogens to lose, so they are oxidizedtwice (first into an aldehyde and then into a carboxylic acid.) Thecarboxylic acid is reduced back into a primary alcohol using LiAlHn.Hloontou o,*'ot'5,lt"nt- o ll ,_,

oxidizingagent- o ll o'Lr <-r. ullof,ot arboxytic acid

reducing agent

KMnO4 CrO3 03RCO3H ROOR ClzNAD+ NADP+ FAD

Secondary alcohols have one hydrogen to lose,

H

n-f ouR'

2" aicoholreducing agent

Figure 5-24

LiAlHa NaBH4 H2NNH2Hz/Pd HCI/Zn HOCH2CH2SHNADH NADPH FADH2

so they are oxidized once.

o

oll o'ketone

oxidizing agent

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OxidationFrom an organic chemistry perspective, oxidation is the process of gainingoxygen (and/or losing hydrogen) at the carbon which has an electronegativeatom attached. The first rule is to count the number of carbon-oxygen bonds tothe carbon of interest in the reactant. As the number of carbon-oxygen bondsincreases, the oxidation level (and oxidation state) of the compound increases. Interms of oxidation levels, every carbon-oxygen bond counts as one. If you lose acarbon-hydrogen bond and replace it with a carbon-oxygen bond, the oxidationlevel has increased by one while the oxidation state of the carbon has increasedby two. The oxidation of a primary alcohol into an aldehyde and then into acarboxylic acid is shown in Figure 5-25.

''PCC''

CrOeCl-

----9+A

\-.r* -tPyridinium

HOH]li(-,/^-\H.C H

oiidation State'0+1-1-1=-1

CrO"/HCl

-+Pyridine

o*r ll *t

(-,/^-\HsC H

Oxidation State:0+1+1-1=+1

K"CrnOo

H2SOa(aq)

KMnOa

---+KOH(aq)

KeCrrOz

HrSOa(at)

o*r ll *r

,/atr\HsC OH

Oxidation State:0+1+1+1=+3

Figure 5-25

You should be able to recognize reagents, the reaction type, and correctly predictthe product for redox reactions. As a point of interest, Cro4 with Hct inpvridine is known as pyridinium chlorochromate, or PCC for short. Mostorganic compounds lose carbon-hydrogen bonds and gain carbon-oxygen bondsrlhen oxidized, so the reagent (oxidizing agent) must be rich in oxygen. The two:nost common oxidizing agents in organic chemistry are Cro42- and Mno4-.Some common oxidation reactions of alcohols are shown in Figure 5-26.

o\

"\rr'"- Ott

ttJH

Primary alcohol

o\

-\rurc- oH

,t1H

Primary alcohol

t\

oYt-otR

Secondary alcohol

Tertiary alcohol

Figure 5-26

oila\

tot

RHAidehyde

oll(-

,/-\ROHCarboxylic acid

otl(

,/-\RRKetone

No Reaction(No Hs to lose)

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These reactions should be known well enough that you can recognize them in agroup of many reactions. For other oxidation reactions, you should simplyrecognize the oxidizing agent and know that the product is more oxidized thanthe reactant. Figure 5-27 shows some oxidation reactions of compounds otherthan alcohols. The common theme is that one of the reactants is rich in oxygen.

RH\:J/\

HRDisubstituted alkene

o.-----+(H"C)2S

HAldehyde

HIO* has excessiveoxygen; diol reactantis being oxidized

HVicinal diol Aldehvde

C6H5CO3H has excessiveoxygen; ketone reactantis being oxidized

Ketone Lactone (Ester)

Figure 5-27

The three oxidation reactions shown in Figure 5-27 are ozonoiysis of an alkene,oxidative cleavage of a vicinal diol using periodic acid, and the Baeyer-Viiligeroxidation of a ketone respectively. These reactions are rare enough that theydon't need to be memortzed,but you should recognize that each is an oxidation.More important than memorizing organic chemistry reactions is the ability topredict results and explain observations based on the data given about thesereactions in the passages associated with them. For instance, excessive oxygen inthe reactant should provide a ciue that the organic molecule is being oxidized.

Example 5.14

R

HIO4............--------'-H3COH/H2O

c6HscoqH

-------l-

R

H

What is the major organic product for the following reaction?

OH OHI I KMnOa

,.\-) oHlfi*

tulrildh

8fill

.{'-

n,cmi

S.iilmilm

i(Mm

mMm

t-

ffimimM@u

ili.n"LI]L-

@.

$ild!!fihs ifum-il

frum'Jl

-llllluOWII

B'o oH\) ruoHAA""SolutionThe secondary alcohol is oxidized into a ketone, which eliminates choice D. Theprimary alcohol is oxidized all the u'ay to a carboxyiic acid, which makes choiceC the best answer.

\ 03 has excessive oxygen;

F O alkene reactant is being/ oxidized

(

:s''

iilli

,4.

&c.n

5lt,r

t;r.['Jd

Al&

l'"r

{LIfr"

cm

Sut

friEE

o



Example 5.15\Ahich of the following compounds is NOT an oxidizing agent?

A. Peroxybenzoic acid (C6H5CO3H)B. Ozone (O3)C. Diisobutylaluminum hydride [((H3C)2CHCHz)zAlH]D. Chromic acid (H2CrOa)

Solutionorganic oxidizing agents generally contain oxygen atoms, like choices A, B, andD' Diisobutylaluminum hydride (DIBAH) contains no oxygens, and in factcontains hydrogens. DIBAH is a selective reducing agent that is weaker thanLithium aluminum hydride (LiAtH4). The best answer is choice C.

Example 5.16oxidation of an aldehyde into a carboxylic acid requires the addition of:

A. an oxidizing agent.B. a reducing agent.C. a strong base.D. a strong acid.

SolutionOxidation requires the addition of an oxidizing agent to carry out the oxidation.Oxidation can be carried out in either acidic or basic conditions, so choices C andD are eliminated. The best answer for this question is choice A.

Example 5.17i\'hen a primary alcohol reacts with potassium permanganate, the primaryalcohol is acting as which of the following in the reaction?{, An oxidizing agentB. A reducing agentC. A strong baseD. A strong acid

SolutionBecause the alcohol is being oxidized, it is losing electrons. This means that it is:ausing the reduction of Mno4- into Mno2. Reducing agents get oxidized, so the:rimary alcohol is the reducing agent. The best answer is choice B.

Erample 5.18Cridation of a secondary alcohol leads to:

{. an aldehyde.B. a carboxylic acid.C. a ketone.D. a tertiary alcohol.

Solution-{s shown in Figure 5-26, a secondary alcohol is--.rrther, because it has only one hydrogen to lose:or this question is choice C.

oxidized into a ketone and noto oxidation. The best answer

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Organic Chemistry Carbonyls and Alcohols Carbonyl Reactivity o

Example 5.19\Alhich of the following compounds CANNOT be oxidized by chromic acid?

A. 2-Methyl-3-pentanolB. ButeraldehydeC. 3-PentanolD. Phenol

SolutionPrimary and secondary alcohols can be oxidized while tertiary alcohols andphenols cannot be oxidized. Choices A and C are secondary alcohols, so they canbe oxidized into ketones. Aldehydes are oxidized into carboxylic acids, so choiceB is eliminated. This leaves choice D, phenol, as the best answer. The carbonbonded to oxygen in phenol has no hydrogen atoms, so it cannot be oxidized.

Example 5.20To oxidize a primary alcohol into an aldehyde without further oxidation into acarboxylic acid, what reagent should you add in conjunction with chromic acid?

A. Pyridine and hydrochloric acidB. Toluene and hydrochloric acidC. Pyridine and sulfuric acidD. Toluene and sulfuric acid

SolutionA primary alcohol is oxidized into an aldehyde and no further when pyridiniumchlorochromate is added to the solution. This is known as PCC or Collin'sreagent. The absence of water from solution prevents the aldehyde from furtheroxidation into a carboxylic acid. The best answer for this question is choice A.

ReductionReduction is the process of losing oxygen (and/or gaining hydrogen) at thecarbon which has an electronegative atom attached. As with oxidation, the firststep is to count the number of carbon-oxygen bonds in the reactant to the carbonof interest. As the number of carbon-oxygen bonds decreases, the oxidation leveldecreases (as does the oxidation state). If you lose a carbon-oxygen bond andreplace it with a carbon-hydrogen bond, the oxidation level decreases by one andthe oxidation state of carbon decreases by two. From a ketone or aldehyde to analcohol, the oxidation state decreases by two and the level decreases by one. Thereduction of an ester into a primary alcohoi is shown in Figure 5-28.

OHOH*r ll*r 1. LiArH,(tho I i],/^\ ,-\rHrci*) - ,z^c\ + HoCH'CH'

HsC vf'| ocH2cH3 + '' HSC v-r HOxidationState:0+ L + L -+1 = +3 OxidationState:0+1- 1- 1=-1

Figure 5-28

The two most common reducing agents in carbonyl chemistry are LiAlH4 andNaBH4. Aluminum is less electronegative than boron, so LiAlH4 is more apt todonate an H- to the carbonyl carbon, and thus is more reactive than NaBH4. Assuch, NaBH4 reduces only ketones and aldehydes while LiAlHa reduces allcarbonyl compounds. Figures 5-29a and 5-29b show these carbonyl reductions.

Copyright @ by The Berkeley Review 2A The Berkeley Review


1. LiA1H4(thfl----+2. NHaCI(aq)

1. LiAIH4(thf)

---+

2. NHaCI(aq)

1. NaBHa(et2O)

2. NHaCI(aq)

1. NaBHa(et2O)

2. NHaCI(aq)

FADH2oI

--,>>

S_ CH"CH"OH+l LL

s- cH2cH2oH

oil

o/t\o*uCarboxylic acid

oil

o/t\oo'Ester

oil

o/t\r,Aldehyde

oil

o/t\ o,Ketone

Hs-

olt

1

S

t

rdto\srl1

:w

o/t\r,e

rt

n,t

ddn.e

Reduced Linkage

HH

o/t\*,Aldehyde

oil(.

-/._\li R'

Ketone

lr';ulfide linkage

F.H\zH2:

Pd

.. cH2co2H

-':<aturated fatty acid

Figure 5-29a

HCI/Zn(Hg)____=>or

1. HS(CH2)2SH

2. Raney Ni Alkane

HHH2NNH2 \ r--.-->KOHlaq) ^,ZC: _. + N, + HrO

R- R'

Alkane

2 equivalentsHSCH2CH2OH

HOH\r-zc:. + H'o

RHPrimary alcohol

HOH\s-zc:. + HoR'

R'HPrimary alcohol

HOH\s(-

n/-\nPrimary alcohol

HOH\s-zc:.RR'

Secondary alcohol

H

\

\Jou7 Y""H CH2CO2H

Saturated fatty acid

Figure 5-29b

= aldehyde-to-alkane reactions are Clemmensen reduction and Raney nickel:;;lion, respectively. They reduce either an aldehyde or ketone to an alkane.

it-t


OfganiC ChemiStry carbonyrs and Arcohors Carbonyl Reactivity

Example 5.21what is the major organic product after pentanal is treated with sodiumborohydride (NaBHa)?

A. A primary alcoholB. A secondary alcoholC. A hemiacetalD. A carboxylic acid

SolutionAldehydes can be oxidized into a carboxylic acid or reduced into a primaryalcohol. sodium borohydride is a reducing agent, so an alcohol is formed. Thiseliminates choices C and D. The carbonyl in an aldehyde is on the C-1 carbon, sothe hydroxyl group forms on C-1 carbon, resulting in the formation of a primaryalcohol. Choice A is the best answer.

Example 5.22Acetone, when reacted with lithium aluminum hydride and quenched withweakly acidic water, yields which of the foliowing products?

A. PropaneB. 1-PropanolC. 2-PropanolD. Propene

SolutionAcetone is a ketone, thus when it is reduced, it forms a secondary alcohol. Theonly secondary alcohol listed is choice C, so choice C is the best answer.

Example 5.23To carry out the following synthetic transformation, what reagent should beadded?

cH2oHH --+

A. LiAlHaB. NaBH4C. H2N=NH2D. FADH2

SolutionIn this reaction, an aldehyde has been reduced into a primary alcohol on acompound where an ester functional group is unaffected. FADH2 caries outhydrogenation of a n-bond, so choice D is eliminated. Hydrazine, H2NNH2reduces aldehydes and ketones to alkanes, so choice c is eliminated. BothLiAlHa and NaBH4 reduce aldehydes to primary alcohols, but LiAlH4 wouldalso reduce the ester. This eliminates choice A and makes choice B the bestanswer. NaBH4 is selective for aldehydes and ketones.

jE

Ats

cD

Si

:l:

Et

q.

Ss{Stfu,l

,Jtffift

"@nilI

oo

Copyright @ by The Berkeley Review The Berkeley Review30 'ffiW


Example 5.24\A/hat is the oxidation state of aluminum in LiAlHa?A. +3B. +1c. -3D. -5

SolutionThe term hydride implies that hydrogen has a negative charge associated with it.Lithium is +1, so the Al must have an oxidation state of +3 to have the charge onthe compound be zero. The best answer is choice A.

Example 5.25\'Vhat is the major organic product for the following reaction?

A-.'Jl-"HffiOH OH B. oH c'o o D'oH o

oo

uo" uoHSolutionSodium borohydride, NaBH4, reduces aldehydes and ketones only. Both the<etone and the aldehyde functional groups are reduced to alcohols (secondaryand primary respectively). This yields a 1,3-diol, so the best answer is choice A.

I

t2

I

It

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Organic Chemistry Carbonyls and Alcohols Name Reactions:

NaulG,,,Keactions : ,, ,

Name ReactionsThere are name reactions such as the Grignard reaction, aldol condensation,Claisen condensation, transesterification, the wittig reaction, pinacolrearrangement, the iodoform reaction, and the Wolff-Kishner reaction that youshould know. There are other name reactions in organic chemistry, but these arethe name reactions you must know according to the MCAT student Msnual. wewill address each reaction according to their frequency on previous MCATexams. Where appropriate, we will emphasize the reaction mechanism or itsregioselectivity and stereoselectivity. of most importance to the MCAT, we shalladdress any biological applications of the reactions or analogous reactions.

Grignard ReactionGrignard reactions involve the addition of a carbanion (carbanion metal halide)to an electrophilic carbon center in a carbonyl to form a new carbon-carbon bond.when alkyl magnesium bromide, the Grignard nucleophile, attacks the carbonylcarbon, a tetrahedral intermediate forms. Almost all Grignard reactions g"r,"tui"alcohols. No matter what the reactant, Grignard reactions generate a hydroxylgroup in the product where the carbonyl group originated. Figure 5 -30 showsthree such Grignard reactions, the third of which is a double addition to an ester(an acid anhydride or acid halide yield the same alcohol product as an ester).

Grignard Reaction

No L.G. .'. one alkyl group adds: one C + two C (probably chiral)

&ffilcbwrfri

ffiwmismirssr

qfrW@

hufim

uw

cl,l,G.n.

ffi&M

mHO

Aldehyde 2" Alcohol

No L.G. .'. one alkyl group adds: two C -> three C (possibly chiral)

lO BrMs'-O R'

-+".-i'NIgBr --+ oXnwork-up-

oXr_,

-S-.'MgBr --->

BrMg+ -O R' Ho R'"-Xo4 oXo3" Alcohol

-/ ketone

-X:RMgBr

R'

#ft@,qp

Ketone

Has a L.G. .',

HO R BrMg*

X work-up

RR'3" Alcohol

two alkyl groups add: one C -+ three C (never chiral)

o____+ ARR'

Ester(or anhydrideor acid halide)

Copyright @ by The Berkeley Review 32

Figure 5-30

The Berkeley Review

Organic Chemistry Carbonyls and Alcohols Name Reactions

I,

,1

u

e

r.S

t1

Crignard nucleophiles add once to aldehydes, which have one R-group, to formsecondary alcohols. Ketones have two R-groups, so they form tertiary alcohols.Carbonyls with a leaving group (esters, acid anhydrides, and acid halides) startrr-ith one R-group, but because they have a leaving group, they add a new R-group twice. Because the Grignard reagent adds twice to esters, acid halides andacid anhydrides, the products are achiral tertiary aicohols. Grignard reagents arestrong bases, so it is critical that no protic hydrogens are present. This is why thereaction uses anhydrous ether solvent. The second step in the Grignard reactioniequires a weak acid, as opposed to a strong acid, to avoid protonation of thel'droxyl group, which can lead to an E1 elimination reaction. The Grignardreaction can also take place with carbon dioxide to form a carboxylic acid.

Example 5.26;\-hen 2-butanone is added to 1-propylmagnesiumbromide in ether then worked'-:c rvith ammonium chloride in water, it yields a major organic product of:

4.. 2-propyl-2-l:utanol.B, 3-ethyl-2-pentanol.C. 3-methyl-3-hexanol.D. 3-bromo-2-butanone.

SolutionTr.e Grignard reagent attacks the carbonyl carbon to form a new carbon-carbon:-.nd. The reactant is a ketone, so the product is a tertiary alcohol, eliminating::.cice D. The IUPAC name in choice A is not possible, because a2-propyl group'".

-',r-rld actually be part of the longest chain. The reaction is drawn below:

BrMsO CH"CH.CH

* "XL L'T:''LH:C CH2CH3

-:.e longest chain in the product is six carbons, so the product is hexanol. The-lir {C name is 3-methyl-3-hexanol, so choice C is the best answer.

Erample 5.27:::.'"'1 magnesium bromide, when added to 3-pentanone, yields which of the

--..rving organic products following weak acid workup?

\. Triethyl etherF. 3-Ethyl-3-pentanonet-. 3-Ethyl-3-pentanolD. ,l-Heptanol

S rtrution*^-. reactant is a ketone, so the product is a tertiary alcohol. This eliminates: :,:es A and B. The reaction is drawn below:

HO + CH'CH"CH,\/ '5'o

;]C- z rH:C CH2CH?

/ O BrMsO CH.CH' HO 4 CH"CH"

tll + HTCH2CMgBT > "X L r

!L ,, ).r.._." ,'' -n.aU ----

rr,,arrra- -cHrcHr-HrcH2c' -cH2cH3

- = ,ongest chain in the product is five carbons, so the product is pentanol. The

- - \C name is 3-ethyl-3-pentanol, so choice C is the best answer.

+ H3CH2CH2CMgBT

rew - -:'" right @ by The Berkeley Review 33 Exclusive MCAT Preparation


Aldol CondensationAldol reactions start with deprotonation of an alpha hydrogen from a carbonyl(ketone or aldehyde). The pKu of most o-hydrogens that are adjacent to only onecarbonyl is between 17 and 19. This means that they are weakly acidic and canbe removed by using a strong base. The resulting anionic compound (enolate)nucleophilically attacks a neutral carbonyl compound (electrophile) to form a G-

hydroxy carbonyl compound. This species is rarely isolated, because under thereaction conditions, it undergoes elimination to yield an cr,G-unsaturatedcarbonyl. Regioselectivity is involved, because the final product can have eitherE or Z structural geometry. The anionic species can also be protonated on theoxygen, resulting in the formation of an enol, but that is in equilibrium with thecarbonyl. Conversion from a ketone (or aldehyde) into an enol is referred to as

tautomerization, which was shown in Figure 5-19. Figure 5-3L shows the aldolcondensation reaction of acetone.

Aldol Condensation

:o:Or<+l

IzA-HrC- - CH,

:o:

A9HzI,/t\HsC I CHs

:.o.:o

CHs

Figure 5-31

It is often easier to predict a reaction by focusing on the side product. In an aldolcondensation that goes all the way to an cx,,B-unsaturated ketone, water is the sideproduct. Finding the atoms in the reactants that make up the water makespredicting the product easy. Figure 5-32 shows how to align two acetonemolecules for the aldol condensation.

oo*.A.* "..4.r,Hur- >- r'

uI rr o

'nl I orient th; iater, connect the:{ JtH:C -CH: HrC

:o:

A:o:5A.

CH3 withOH-, HeC- -.QHz

ketone-to-anionratio- - (

is about 1000: 1 "3'\f

/ zo:\-7

HsC

\ ",.

-PHaC

HgC CHs

cr,B-Unsaturatedketone

:o:

ACH

,(

eliminationI

-

HeC 9HzI,'f\H:C I CH,

:9lrB-Hydroxyketone

Copyright O by The Berkeley Review 34

Figure 5-32

The Berkeley Review

ns Organic Chemistry Carbonyls and Alcohols Name Reactions

ry1

neaIte)

B-

heedrer

heheAS

to1

The aldol condensation offers the potential for thermodynamic control versuskinetic control. when the ketone is asymmetric, then the choice of which alphahydrogen is deprotonated is dictated by reaction conditions. The reactionconditions that are most significant are steric hindrance of the transition state andthermodynamic stability of the intermediates and product. When the reaction)'ields a major product that was selected to minimize steric hindrance in thetransition state, this product is said to be the kinetic product. This term refers tothe lower activation energy required to go through a less sterically hinderedtransition state. When the reaction yields a major product that was selected tomaximize stability of the intermediates or product, this product is said to be thellrcrmodynamic product. This term refers to the greater amount of free energyreleased when forming the more stable compound. Figure 5-33 shows the kineticand thermodynamic products for the aldol condensation of butanone.

Kinetic Controlless steric hindrancelower transition state

(lower Eactivation)

prefered at low tempwith strong, bulky base

t.t.4 .-tnu

Thermodvnamic Controlmore substituted intermediate

greater energy released(more negative AG .*r. )prefered at high tempwith strong, small base

/\Hti

Krnetic Control

t-butOK___€t-butOH, -78'

Figure 5-33

o

/Hn Htn

\H,t.,

{3

1o1

ide(es

!ne

{"l

oH\A

P CH'CH]

H3CH2C CH:

*r"cA c .c.'''rffil

:(H:C CH2CH3

o Thermodynamic Control

ot-gAcH2cH3

Jt+H3CH2C CH.:

major

o

t-.AcH,c

It H3

HsC CH2CH3

minor

o

.r.cAc ,CHz

JtH3CH2C CH:minor

o

KH.-*50"c

.r"cAc 'cHt,AHsC CH2CH3

major

Erample 5.28lhe following reaction is best described as:

1. NaFlthfl

----+2. workup

\. a Claisen condensation via the kinetic enolate.B. a Ciaisen condensation via the thermodynamic enolate.C. an aldol condensation via the kinetic enolate.D. an aldol condensation via the thermodynamic enolate.

o')u.

ew -rpyright @ by The Berkeley Review 35 Exclusive MCAT Preparation


SolutionThe reaction involves the condensation of an asymmetric ketone upon itself,which eliminates choices A and B, because the Claisen condensation requires thereaction of esters with esters, not aldehydes or ketones. Because the morehindered alpha carbon is involved in the addition, the reaction must have beencarried out by way of the thermodynamic enolate (more substituted enolate).This makes choice D the best answer.

\Arhen a reaction has a biological example, its likelihood of appearing on theMCAT is increased. The biological application of the aldol reaction is the fourthstep of glycolysis, which is a retro-aldol reaction. In the retro-aldol step, a six-carbon B-hydroxyketone breaks into two three-carbon fragments. The two three-carbon species, after neutralization, are dihydroxyacetone phosphate (DHAP)and glyceraldehyde-3-phosphate (G-3-P). It is not necessary to know themechanism, but you should know that the enzyme responsible for this step isaldolase. Figure 5-34 shows the fourth step of glycolysis.

Step IV GlYcolYsis cH"cm)

ft'@ t;il-' + 1",o,,LIr\-II- Urrv rr DHAP

ff"cH2@

-n*I_T-OHcH2@

H

H

The G-ketoester exists in equilibriumwith the deprotonated species untilthe solution is neutralized.

Figure 5-35

Trurf,mfi

dqt!fism

rfo

!mah

Mmffi@do

0-m

ffihffi

HO

H

HOH

B-hydroxyketone cH2@G-3-P

Figure 5-34

Claisen Condensation ReactionThe Claisen condensation reaction is essentially the aldol reaction with an ester.Esters have a leaving group, so the product is not an c[,8-unsaturated carbonyl, asis seen with aldol condensation reactions. Figure 5-35 shows the Claisencondensation reaction. Note that without acidic workup, the carbon between thetwo carbonyl species would be deprotonated by the ethoxide in solution.

Claisen Condensation

EtO-Na*

Aldolase-

-

+ -OEt

oH

H

:o:

A:o:

AEtO CHsEtOH

The base is chosen to matchthe leaving group so that atranesterification reactionwon't generate a new ester.

EtO 9Hz

I

"r.Ag.B-Ketoester

EtO\o

CH"L

,cto,/ll

[:oz\-tz

H

Copyright @ by The Berkeley Review 36 The Berkeley Review


3

:1

S

:t

e

The B-ketoester product has a pair of hydrogens that are now conjugated to twocarbonyl groups. As such, their acidity is enhanced, so the sodium ethoxidedeprotonates the B-ketoester product preferentially over the ester reactant. Oncereutralized, a B-ketoester is formed. A B-ketoester when treated with strong acidin water can hydrolyze to form a B-ketoacid, which when heated readilyr-rLdergoes decarboxylation into a ketone. This ultimately leads to a ketone witha longer alkyl chain than the starting ester.

Fatty acid synthesis is a biological example of a Claisen condensation reaction,-.r-here the ester is actually a thioester. Synthesis of a fatty acid involves Claisencondensation of acetyl-CoA onto another thioester. The chain gl'ows two carbonsat a time as shown in Figure 5-36.

Fatty Acid Synthesis (Claisen Condensation with a Thiosester)

:o: :o: :o:

c"ASA.";-T.orA?'r, T"orAcH2 coorA.r,,fhe enzyme acts as a .. ^ (

--ase by ieprotonating H3c t scoA Hrc-f{scoa HrcA g_

-*re thioester /-]A: :9.:A c-.*o,n,o"r,".\-:7

:o:

A hydrogenationCoAS 9Hz FADH2

| 1

r-H,cf H

H

:o:

JICoAS CH

trtA

Figure 5-36

dehvdration(elimination )

-H

:o:

A1CoAS 9HzI

",ctSi.uH

Transesterificationlransesterification involves the exchange of one alkoxy group for another on anrster. Because the equilibrium constant is roughly one, the reaction is driven byLe ChAtelier's Principle. Either a product is removed or a reactant is added. An:cid-catalyzed transesterification reaction is shown in Figure 5-37.

oll + HoR" -- HoR' +

R/\ oR, cat. H'

o

J(R OR''

Figure 5-37

Transesterification can be carried out in either acid or base. When acid isemployed, the mechanism is a typical acid-catalyzed process. In base, thealkoxide nucleophile causes the reactivity. If the base is hydroxide, a carboxylateand alcohol. are formed as the two products. When a fatty acid triglyceride is,ieated with strong base, glycerol and fatty acid carboxylates are formed from ahvdrolysis reaction. Figure 5-38 shows the acid-catalyzed mechanism for:ransesterification.

reductionNADH

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Organic Chemistry Carbonyls and Alcohols Name Keactions

Fi+ A

+ R,,oH

R OR'

o-.^",.1i

Transesterification mechanism

;H--io!

R'oH . ll

RAoR.,'ll a"oroto,",ut"fi

n" - ijsH\o.o:

il)nAon,lbreak I

I

HVo.

I

ofo*o"-Ht

-"uJ Ill nti- n"

H

,:6tnAR OR''ll-rk"i

,:n/n!nAon"

../i..R'- O-@

protonate \/'./\. R OR"

1 |

0.""0

")

n'ci: .'iin,'

.)t'-.deprotorlate

OH

Figure 5-38

The biological applications of transesterification involves membrane transportand the conversion of fatty acids into triglycerides. Fatty acids are found ilcellmembranes, where they are part of a phospholipid (comprised of glyceror, twofatty acids, and a phosphate group). when glycerol binds three fatty acids,triglycerides are formed. Fatty acids that have been esterified with glycerol arebroken down in basic water in what is known as saponification, the basedcatalyzed cleavage of the ester bonds of the triglyceride, as shown in Figure 5-39.

Saponificationo

o4*o4-*

oH--

oAo

OH

OH +3oo

OH

Copyright @ by The Berkeley Review 5B

Figure 5-39

The Berkeley Review

Organic Chemistry Carbonyls and Alcohots Name Reactionsns

Wittig reactionAnother useful carbonyl reaction for synthesis is the Wittig reaction. The Wittigreaction essentially undoes an ozonolysis reaction. It is yet another way to makea carbon-carbon bond. The Wittig reaction converts a carbonyl into an alkene,rvith stereoselectivity. You can get either the E or Z product depending whatreagents you choose. The double bond you form can be added to by electrophilicaddition reactions, with high stereoselectivity. What this means is that there arenow synthetic routes to get high yietds for stereoselective synthesis. we aregoing to see it in its most simplistic form however. It goes through a four-membered intermediate, similar to the hydroborane addition reaction with analkene. Figure 5-40 shows the synthesis of a phosphonium ylide (the reactivespecies in a Wittig reaction).

oGP:+ R2cHBr--+

ouq a*ro, + Li-but ->

oGo

oGo

Figure 5-40

ooP- CR^

rrtoll

'\'ols,lre,ed

\9.

/Rp:O + H"C=={.\R

azP

ll+o

a *''ls:: .--.>CH.

il'CRz

oGo

R

R

/P:CH" .. g{ +z\

oGo

it is the final product here that is the reactive species in a Wittig reaction. Therers a resonance form which has a double bond between the carbon and:hosphorus. You should recall that the phosphorus is able to make more than:our bonds due to the availability of 3d-orbitals. The overlap between a 3p-,rrbital (from P) and a 2p-orbital (from C) is not as strong as the overlap between:''r'o 2p-orbitals however, so the n-bond between phosphorus and carbon is not=at strong. Figure 5-41 shows the Wittig reaction and various intermediates inre mechanism in an aprotic environment.

Wittig reaction

3P:= CH2

\o: cR2

Figure 5-41

b. an aprotic environment, the transition state is a four-centered transition state..re phosphonium ylide @3P=CH2 can be synthesized from @gP: and H3CBI;nder basic conditions, as shown in Figure 5-40. The wittig reaction prefers theE-alkene product whenever there is asymmetry in the reactants.

ew -opyright O by The Berkeley Review 39 Exclusive MCAT Preparation

Organic Chemistry Crtoryts ad lllcohols ttime Eeactions

Pinacol RearrangementThe pinacol rearrangement is a topic in the MCAT Student Manual, so we shallcover it. It is a classic example of rearrangement with some applications insynthesis. The pinacol rearrangement converts a vicinal diol into a ketone. Thereaction proceeds with an alkyl migration, which converts a tertiary carbocationinto a resonance stabilized carbocation. Figure 5-42 shows the pinacolrearrangement of pinacol into pinacolone.

Pinacol rearrangement

o!ilr

ilmileCmpFffir&Fu@ffimmd

H\ ..

HO: '\< OU

\/tuto"f \"'t*

H:lC CHr

f I p,otonat"ff

\ if+tu.*'f \"'.*

HeC CFI3

:O CH"

\ /

J

f- \,,.*H"C CFL

o"o.o,o.u,"\ r

(c).H-A.' CHsv\/

i-\"tcbrsH:C I CH:

/t"rt"7.. V**

HO:\@

,,*"?3HsC

H

..,.rrl CF{, t"uttu$"\att, ---

oil

--C-....oR CHz

HO\i)

I-I -O* r

CFL

/J@'r- r

/- '("'a*r,\

HsC CF{3

Figure 5-42

Iodoform ReactionThe iodoform reaction falls in a general class of reactions known as the haloformreactions, which convert a methyl ketone into a carboxylic acid and a haloform,HCX3. The iodoform reaction is used as a chemical test for a methyl ketone,because iodoform is yellow in color and insoluble in water. Methyl ketones reactwith iodine, 12, in the presence of base to generate a carboxylic acid and a

deprotonated iodoform molecule. The carboxylic acid is deprotonated to yield acarboxylate and iodoform. The iodoform reaction is shown in Figure 5-43.

Iodoform reaction

oil

o /c\.rr,

olt* --C\R CH^

l"Iotl] '-*""1:l

-zc-. 4R CI:

oOH*

o

ll + HCr.

R '/-\ oo J

o

: ll -Cr-o&-_ -zL\ROH


Figure 5-43

The Berkeley Review40


nallsinIheionrcol

)rmrm/trI€r

ract)^Jd

da

-N

"r', ii1 f f-Kishner Reduction.::e are three common methods for reducing a ketone or aldehyde into an.-.:ne. There is the Clemmensen reduction, which is carried out under acidic: :.:ritions using HCl and Zn(Hg). There is the Wolff-Kishner reduction, which is,.:::ed out under basic conditions using hydrazine (H2NNH2) and basic, -:(up. And last there is the treatment of the carbonyl with a 1,2 dithiol, -SCH2CH2SH) followed by treatment with Raney nickel (fine nickel in ther:.sence of aluminum oxide). The good thing about having three reactions is,, :.: each is carried out at different pH values. This allows for many different.,:::onvl compounds to be reduced independent of any pH sensitive functionall:- jps elsewhere on the molecule. For the MCAT, it is important to know the:::.hanism for the Wolff-Kishner reaction. The Clemmensen reaction- -:hanism is not thoroughly understood, so for the Clemmensen you really only=:i to know that most transition metals have a positive oxidation potential and

.:. 'leerefore strong reducing agents. When added in their zero oxidation state,

-.=ials donate electrons to the organic molecule, causing the molecule to get:=-ruced. Figure 5-44 shows the three reactions carried out on cyclopentanone.

Wiilff-Kishner reductionHzN:.

N

HH

-->

+Nz

o

dHHF3

Raney nickel reduction

o f-fA r-^g" X RanevNi-

L_,/ cat HCr I

ti # tj + HSCH'GH'sH

f2

Clemmensen reduction

+ HrO + ZnCI2

Figure 5-44

.:u must apply the right reaction under the right conditions, which depends on.--e presence of other functional groups on the molecule. The Wolff-Kishner:=action is carried out in two steps. The first step of the Wolff-Kishner reaction

-...'olves the substitution of hydrazine for the carbonyl oxygen, resulting in the. , -nation of a hydrazone. Treating the hydrazone with strong base in water:=nerates nitrogen gas. The first step of the reaction is reversible with the:,rlition of water and acid. That is to say that you can regenerate the carbonyl by:--:ding water and a catalytic amount of acid to the hydrazine deril'ative.-:ibonyls can also react with secondary imines, but because there is only one::cton to lose from the amine, the reaction generates an enamile rather than an-::'.ine. We shall consider such reactions in the nitrogen compound section.- rzure 5-45 shows the mechanism for Wolff-Kishner reduction.

j

HH

-op;rright @ by The Berkeley Review 4t Exclusive MCAT Preparation


Step 1:

:o:lt

.zc:. +

RRStep 2:

H.r-. ..OA. -:.gtt

il)*\'.il ^-

.zC:.R

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R

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H

t^

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sril-..Nlt

..o:OH

/H

Iffifilq&MfiJFGcm@

l

't/

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il

Iilt

J..t

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./:.\ .t'R- I -R .-O-H\"/

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.29:.RlnH

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*)'n{r*t""iS:4

,,,,1

.1*s,,


Organic Chemistry Carbonyls and Alcohols Synthetic Logic)ns

3

StftthGtiru:::::D$giCSynthetic LogicSvnthesis is a recurring topic throughout organic chemistry. The method for:esting synthesis in most lecture courses is not feasible on the MCAT, because the\ICAT is a multiple choice exam. However, synthesis appears on the MCAT.Svnthetic pathways are shown in a step-wise road-map fashion similar to how a

:athway appears in biochemistry. The passage will present the steps, which mayrr may not include the reagents. Questions can focus on stereochemical:onsequences, reaction types, and the logic behind a step in the overall synthesis.lhe strategy behind a synthesis is based on the idea that it is easier to solve a

raze backwards than forwards. The steps to analyze a synthetic pathway are:

1) Identify the new bonds (and functional groups) formed by counting theatoms and noting the changes in bonds from reactant to product.

l) Break the product into fragments that match the reactant skeleton.

3) Reconnect the fragments with the proper chemical reagents.- et's consider the conversion of 1-propanol into 2-pentanol. Figures 5-46 through-:-f8 show the stepwise analysis of the synthesis and retrosynthetic analysis.

1) A five carbon secondary alcohol is formed from a three carbon primaryalcohol, so two carbons must be added to the reactant.

OH

?"':r?frr 9o., ----+ -^--I- ;"':rfi:,

Figure 5-46

Using retrosynthetic logic, the product is broken into two fragments.The fragment with the oxygen must have come from the carbonyl.

OH

,.-J,! E+ "\>retron

OHIand

! r̂etron

Svnthesisrequires:

carbonyl electrophile

Figure 5-47

--: reaction calls for a three carbon nucleophile adding to a two carbon carbonyl

= :ctrophile). The more oxidized fragment (oxygen containing fragment, or:' ::-t fragment in this case) comes from the carbonyl. The oxygen is bonded to

= first carbon, so it requires an aldehyde. The rest of the problem requires.' .tresizing the Grignard reagent from 1-propanol and choosing the solvents.

,- This Grignard reaction requires propyl magnesium bromide and ethanal.

3.oOH

a-.\r,nucleophile

\-^""ffiyrgu.Figure 5-48

rew 43- -:vright @ by The Berkeley Review Exclusive MCAT Preparation

Organic Chemistry Carbonyls and Alcohols Synthetic Logic 3f,cMa

8r0{-rrtl

rutpmffIr

ciltaddtof

HOH

XRR'Alcohol

Protection and deprotection of an alcohol

(H3C)3Siq H(H3C)3SiCl \ / NaF or H"O*

ffi -r .. ffiprotecting group R R' protecting group

Protected alcohol(sityl ether)

E|

hfrk,rhn''''lt+rytusffit

.;

,,t"&

oo

Protecting GroupsProtecting groups are used when a reactant has multipre reactive sites, but youonly wish to react at one of them. you must determine which functional groupscan react under the reaction conditions. In cases where more than orru ,It"

"unreact, the functional group you wish to not have react is converted into a lessreactive functional group. Aldehydes are converted into acetals, ketones areconverted into ketals, and alcohols are converted into silyl ethers. you must beable-to remove the protecting group at the conclusion of the reaction withoutinvoking extreme conditions which may cause the molecule to react. Figure 5_49shows the conversion of an alcohol into a silyl ether and then back into analcohol' The ketal protecting group has already been shown in Figure 5-11.

HOH

XRR'Alcohol

Figure 5-49

Reactions of Acetoacetic EsterAcetoacetic ester has a carbon that is alpha to two carbonyl groups. As aconsequence, its protons are more acidic thin typical alpha proto*. rr-,u pKu ofthe protons alpha to both carbonyl groups in acetoacetic estei is roughly ri. as aconsequence, it can be deprotonated in mildly basic conditions, rresulting in acarbanion. The carbanion is a strong nucleophiie, capable of displacing a telvinggroup to form a new carbon-carbon bond. Fig.rte 5-t0 shows the addition of analkyl group to acetoacetic ester.

1. NaOEt (HoEt)

2. R'-X

Figure 5-50

The base of choice matches the alkoxy group of the ester, so that anytransesterification reaction that takes place does not result in any change. ThLsolvent is the conjugate acid of the akbxy group, an alcohol, beciuse it is beingformed by the reaction anyway. The reactint ii a B-keto ester, which is formedfrom a Claisen condensation reaction of two esters. The product can be treatedwith acidic water and heat to hydrolyze the ester into a B-ketoacid, which canundergo decarboxylation into a ketone if the temperature is high enough.


rc Organic Chemistry Carbonyls and Alcohols Synthetic Logic

lups

tnSS

rebe

utJO

1n

Reactions of Malonic Ester\4aionic ester, like acetoacetic ester, has a carbon that is alpha to two carbonylgloups. As a consequence, the same reactions carried out with acetoacetic estercan also be carried out with malonic ester. When malonic ester undergoes thereaction shown in Figure 5-50, the same addition takes place, although theproduct is a diester rather than a ketoester. The significant difference betweenacetoacetic acid and malonic ester is that a diester has two functional groups thatcan undergo substitution, resulting in cyclization. Figure 5-51 shows theaddition of an alkyl group to malonic ester followed by the addition of a diamineto form a six-membered ring.

1. NaOEt (HoE|2. R-X

oEr

Figure 5-51

iarbiturates form when urea is used in place of the diamine in the second step.

Decarboxylationrs mentioned with the acetoacetic ester synthetic pathway, a B-ketoacid can-:ndergo decarboxylation when heated. This occurs readily, because thertramolecular hydrogen bonding of a B-ketoacid aligns the molecule for a shift-i. the n-electrons. Figure 5-52 shows decarboxylation, the six-membered ring-:eated by hydrogen bonding, and the decarboxylation products formed from-:.at conformation.

Decarboxylation

oooo

a

ofia

a

1g

]n

---+A

CO2 +o

AHsC R1V

1e

1g

:d:dtrl

hrtramolecular:{-bonding in B-ketoacid

Carbon dioxideand an enol form

Figure 5-52

oll_-\

H:C R

Enol to ketone(tautomerization)

"X$-.-r[:^.

:w -r.pyright @ by The Berkeley Review 45 Exclusive MCAT Preparation

Organic Chemistry Carbonyls and Atcohols Carbonyl Biochemistry

G lffio.GBiological Oxidation-ReductionThere is a great deal of oxidation-reduction chemistry in biology. using anorganic chemistry perspective can simprify many biological reictions. L ugeneral rule, you should know that anabolism is-a redultive process and theresult is the build up of a molecule. Examples of anabolism lnclude fatty acidsynthesis and gluconeogenesis. You should know that catabolism is an oxidativeProcess and the result is the breakdown of a molecule. Examples of catabolisminclude glycolysis and B-oxidation. Typically, catabolism rele-ases energy whileanabolism requires, and thereby stores, energy. you should know thii in us,oxidations tend to take place in the mitochonaiiat matrix while reductions tendto take place in the cytoplasm, with glycolysis being a notable exception.

Example 5.29which of the following cofactors or coenzymes is most likely required in thefollowing step of glycolysis?

o\ " ove-pe^2-tt

H-J-oH GAPDIL u*ou|

- 'Ir I "'HzC- 9- P6r2- HzC- 9- pgr2-

A. BiotinB. AMPC. ZnD. NAD+

SolutionIn this reaction, carbon 1 gains a bond to oxygen, so it has been oxidized. Thismeans that it must be coupled with a reduition half-reaction. The enzymeGAPDH does not undergo reduction, so the cofactor,/coenzyme must undergoreduction. of the choices, only NAD+ (a species poor in H) is an oxidizing age"ntthat undergoes reduction. Choice D is the best u.,r*"r.

Example 5.30In humans, where does fatty acid synthesis take place?A. CytoplasmB. Rough endoplasmic reticulumC. MitochondrialmatrixD. Mitochondrial inner membrane

SolutionFatty acid synthesis involves molecular build up, so it is a reductive process.Reductions occur mostly in the cytoplasm, so choice A is the best answer.

There are several examples of oxidation-reduction chemistry in biology, so weshall only focus on a few here and a few more with carbohydrate che"mistry insection 6. We shall consider oxidative decarboxylation, which shows itself inaerobic and anaerobic metabolism. Figure 5-53 shows the decarboxylation ofpyruvate, a three-carbon species, into acetyl coenzyme A, where ihe acetylmoiety contains two carbons.

IM

ffii

m

&cMq


;ry Organic Chemistry Carbonyls and Alcohols Carbonyl Biochemistry

o-

aniahe,id.\.e

;mile-1S,

nd

Q:g:Q

CHg NAD+ NADH + H+Pyruvate Carbon dioxide

NAD+(Nicotinamide adenine dinucleotide)

o\ _,,, S-CoA+C

I

CH:

Acetyl CoA

NADH(Nicotinamide adenine dinucleotide hydride)

he

Figure 5-53

Biochemical ReagentsFor biological redox chemistry, there are two common reversible reactions. Thefirst reaction is the inter-conversion between NAD+ and NADH, and the secondleaction is the inter-conversion between FAD and FADH2. The two moleculesand their reduction reactions are shown over the next few pages. Figure 5-54shows NAD+ (the oxidized form), which acts as an oxidizing agent in:iochemical reactions, and NADH (the reduced form), which acts as a reducingagent in biochemical reactions.

NHz oil-o- P- o- cH2

HOH HO OH

o o

/l4lN

N

z.)

9-o

z.)ooa.o

11S

ne

3ont

Adenine N

N

iq

teininofr'l

Figure 5-54

\-{D+ serves as an oxidizing agent by picking up a hydride anion on its pyridine:-rg to yield NADH. The species being oxidized gives off H+ as a side product.,ris biological redox reaction plays a major role in many cycles including

"-lr colysis. NADH carries out the reduction of aldehydes and ketones, much like

\aBH4 does in organic chemistry reductions. A second biological reducing:qent is flavin adenine dinucleotide dihydride (FADH2), which carries outr',-drogenation of n-bonds, much like H2 and a catalyst such as palladium.::gure 5-55 shows the oxidized form, FAD, and the reduced form, FADH2.

H

:w -opyright @ by The Berkeley Review 47 Exclusive MCAT Preparation

Organic Chemistry Carbonyls and Alcohols Carbonyl Biochemistry I

:..g$.I

CHr

"l o"

"I,r"'-f o"

CH.t'oI-O- P: OI

oI

-N"'

AoHsC

HsC

za

oo-o

HOs$tlCH. HIH-f OH

,,-l- o,,IHTOHCH.t'oI-O- P: OI

oI-o- P- o- cHilLol\

oL_

Flavin

NHz

-A-\AN

/O --fI

N

)N

2

L1--

H

HO OH HO OHFADH2 (Reduced form)FAD (Flavin adenine dinucleotide)

Figure 5-55

FAD serves as an oxidizing agent by picking up a hydrogen on the nitrogen ofthe central pyrimidine ring and a second hydrogen on the imine nitrog"t of th"third ring, to yield FADH2. one hydrogen is gained as a hydride, while the otheris gained as a proton, so there is no change in the charge of the compound.

Acetyl coenzyme A is also very common in biochemistry, so we shall considerthe acylation process of coenzyme A. The acetyl group is transferred from areactant to a thiol when acetyl CoA forms. The structure of acetyl coenzyme A isshown in Figure 5-56. The biochemical role of coenzyme A is in transfer of anacetyl group to a biological species.

O OH CH?

Acetyl Coenzyme A

-Frgure 5-5b

'*iil

Si

,*,a

Nucleotide

Copyright O by The Berkeley Review 4a The Berkeley Review

Organic Chemistry Carbonyls and Alcohols Section Summary

tlf

IiI

Key Points for Carbonyls and Alcohols (Section 5)

Oxygen Containing Compounds

1. Alcohols (Compounds with hydroxyl groups)

a) They have boiling points higher than alkanes of roughly equal mass

i. Boiling points of alcohols are: primary > secondary > tertiaryii. Water solubility is high for short chain alcohols

b) The hydroxyl group can be converted into a better leaving groupi. SOC12 and PBr3 convert an aicohol into an alkyl halideii. They are mesylated or tosylated to form a good leaving group

c) They have broad IR absorbances between 3300 and 3600 cm-1

l. Aldehydes and Ketones (Carbonyl compounds with H or R-groups attached)

a) They have boiling points higher than alkanes of roughiy equal mass

i. Boiling points decrease with branchingii. Water solubility is high for carbonyls up to four carbons

b) They readily form acetals and ketals with alcohol in acid and formhemiacetals and hemiketals with alcohols in base

c) Aldehydes have strong IR absorbances around 7725 cm-I and. ketoneshave strong IR absorbances around 1710 cm-1

-r. Carboxylic Acids (Compounds with a COOH group at an end of the chain)

a) They have melting and boiling points significantly higher than alkanes ofroughly equal mass because they form strong H-bonds.

i. They have pKu values in the 2-5 rangeii. They represent the end point of oxidation for most molecules

b) They form more reactive derivatives by converting the hydroxyl groupinto a better leaving groupi. SOC12 converts a carboxylic acid into an acyl chloride, heat converts

two carboxylic acids into an acid anhydride, and an alcohol in acidconverts a carboxylic acid into an ester

ii. Carboxylic acids can be made by complete oxidation of a primaryalcohol or aldehyde or by hydrolysis of acid derivatives such as

esters, acid anhydrides, acid halides, amides, and nitriles

= Carboxylic Acid Derivatives (Compounds with a carbonyl group and a

functional group other than OH at an end of the carbon chain)

a) Esters (OH group of a carboxylic acid is replaced by an OR group)

b) Lactones (Cyclic esters)

c) Acid Anhydrides (forms from the dehydration of two carboxylic acids)

d) Acid Halides (OH group of a carboxylic acid is replaced by a halide)

e) Amide (OH group of a carboxylic acid is replaced by an amine)

Carbonyl Reactivity

a) They are electrophiles that can be attacked at the carbonyl carbon by a

nucleophile, forming a tetrahedral intermediateb) Strong bases deprotonate them at the cr-carbon to form a nucleophilic C

c) Theyundergooxidation-reductionchemistry

i. Oxidation: gain of bonds to O and,i or loss of bonds to Hii. Reduction: loss of bonds to O and/or gain of bonds to Hiii. Oxidizing agents are rich in O and reducing agents are rich in H

Copyright @ by The Berkeley Review 49 Exclusive MCAT Preparation

Organic Chemistry Carbonyls and Alcohols Section Sumrrary

Name Reactions

1. For the reactions rn the MCAT student Manual, know the basic reactiona) Grignard Reaction (addition of an alkyl magnesium halide to a carbonyl)

i. Adds one R-group to aldehydes and ketones, forming alcoholsii. Adds twice to carbonyls with a leaving group, forming a 3" alcohol

b) Aldol Condensation (deprotonation of the c-carbon of an aldehyde or aketone and the subsequent addition of the anion to the carbonyi carbonof either a second aldehyde or ketone)i. Forms an c[,8-unsaturated ketone or an cr,B-unsaturated aldehydeii. Exhibits kinetic versus thermodynamic preference with ketonesiii. Has a biological application in the fourth step of glycolysis

c) Claisen Condensation (deprotonation of the s,-carbon of an ester and theaddition of the anion to the carbonyl carbon of a second ester)i. Forms a B-ketoesterii. Has a biological application in fatty acid synthesis

d) Transesterification (conversion of one ester into another by exchangingthe alkoxy grol;lp, in either acidic or basic conditions)

e) wittig Reaction (conversion of a ketone into an alkene via @3p=CR2)

0 Pinacol Rearrangement (conversion of a vicinal diol into a ketone in acid)g) Iodoform Reaction (conversion of a methyl ketone into a carboxylic acid)h) Wolff-Kishner Reduction (conversion of a ketone into an alkane in base)

Synthesis

1. synthesis Thought: 1) changes, 2) retrosynthesis, and 3) choosing reagentsa) Retrosynthesis (product-to-reactant analysis of changes in a compound)b) Protecting groups are used on reactants with multiple functional groupsc) -oH group is protected as -orMS and c=o group is protected as C(oR)2

2. Reactions of Acetoacetic Ester (Atpha-proton has a pKu around 1L)a) Great nucleophile when deprotonated, so it readily attacks C=Ob) \rvhen treated with acid water and heat, it hydrolyzes and decarboxylates

3. Reactions of Malonic Ester (Alpha-proton has a pKu around 11)a) Readily adds an R-group at the methylene carbonb) Can cyclize with a reactant with two nucleophilic sites (i.e., barbiturates)

4. Decarboxylation (observed with B-keto acids or 1,3 diacids)a) Proceeds through a six-membered ring held together by an H-bondb) Forms an enol which equilibrates with the carbonyl compound

Carbonyl Biochemistry

1. Oxidation and Reduction in Biochemistrya) Catabolism (oxidative breakdown of bioiogical molecules)

i. Occurs mostly in the mitochondrial matrix (i.e., B-oxidation)b) Anabolism (reductive build up of biological molecules)

i. Occurs mostly in the cytoplasm (i.e., gluconeogenesis)

2. BiochemicalReagents

a) NAD+/NADH are a redox pair nhere NADH is the reduced formb) FAD/FADH 2 ate a redox pair rvhere FADH2 is the reduced formc) Acetyl Coenzyme A (Picks up acetvl group to form thioester)

(

I


v

Carbonylsand

AlcoholsPassages

14 Passages

I OO Questions

Suggested schedule:I: After reading this section and attending lecture: Passages I, lV, V, X, & XI

Qrade passages immediately after completion and log your mistakes.

II: Following Task I: Passages II, VI, IX, & XIII (26 questions in 34 minutes)Time yourself accurately, grade your answers, and review mistakes.

III: Review: Passages III, VII, VIII, XII, XIV & Questions 95 - IOOFocus on reviewing the concepts. Do not worry about timing.

c=ffi6 ffiH ffiftdHdl#"$tfi 5$ffi

Passr

-{or_qaLl

the al

iradidalcoh,:

I.

II.

III.

IV.

V.

VI.

VII.

VIII.

IX.

X.

XI.

XII.

XIII.

XIV.

Alcohols, Acidity, and Nucleophilicity

Alcohol Reactions

Acetals and Ketals

Carbonyl Reactivity Study

Thermodynamic versus Kinetic Control

Alcohol Oxidation

Unknown OxygenContaining Compounds

Wolff-Kishner versus Clemmensen Reduction

Crignard Reaction

Aldol Condensation and Alpha Hydrogen

Claisen Condensation Reaction

Transesterification Reaction

Malonic Ester Synthesis

Carboxylic Acids

Questions not Based on a Descriptive passage

Carbonyls and Alcohols Scoring Scale

Raw Score MCAT Score

85 - IOO t5-1565-82 LO-12

46 64 7 -9JJ-4b 4-6|-32 t-5

(1 - 6)

(7 - 12)

(15 - 18)

(te - 25)

(26 - 52)

(33 - 3e)

(4o - 46)

(47 - s3)

(54 - 60)

(6t - 67)

(68 - 73)

(74 - Bo)

(Br - 87)

(BB - e4)

(e5 - lOO)

\m

ttl

Di

Passage I (Questions 1 - 6)

Alcohols are a common protic solvent used in manyrrganic reactions. Alcohols have organic properties (due to,he alkyl chain), and they exhibit hydrogen bonding (not:aditionally thought of as an organic property). Low-weight.lcohols are the only organic solvents that can be used to,:mulate aqueous conditions while dissolving organic: rmpounds. They are not an ideal solvent, because they can:.act as a nucleophile, as an acid, or as a base. Alcohols are

-rosen in many reactions involving esters as a reactant,'; rere the ester leaving group is the conjugate base of the.-:ohol. This is done to eliminate the chance for a

:rn sesterification reaction.

In some reactions, alcohols serve as both the reactant and::.e solvent. For instance, when an alkoxide is used as a base

'. it is in elimination reactions), it is convenient to use the::njugate acid of the alkoxide as the solvent. This is done,::Jause the conjugate acid (an alcohol) is formed when the...oride is protonated. One drawback to alcohol solvents is::ir relatively high boiling point, due to hydrogen bonding.,1"1en the solvent has a high boiling point, it is hard to

- stil1 the solvent away to isolate the product mixture. Of:anic compounds with equal carbons (and therefore roughly

.:ual molecular mass), alcohols and carboxylic acids::rerally have the highest boiling and melting points.

- . The pKu of methanol (H3COH) is 15.5, while the pKuof phenol (HSCOOH) is 10.0. Which of the followingstatements are true?

I. Phenol is less acidic than methanol due to theelectron donation from the benzene ring throughresonance.

II. Phenol is more acidic than methanol due toelectron withdraw by the benzene ring throughresonance.

m. Methanol is more acidic than phenol, because ofelectron withdraw through the inductive effect.

IV. Methanol is less acidic than phenol, because ofelectron donation through the inductive effect.

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

: . Which of the following bases will deprotonatemethanol completely?

A. H3CCO2Na

B. NH3

C. NaOHD. H3CCH2Li

,-rpyright @ by The Berkeley Review@ 53 GO ON TO THE NEXT PAGE

3. Which of the alkoxides is the BEST nucleophile whenundergoing a substitution reaction with methylbromide?

A. H3CCH2ONa

B. HjCCCI2ONa

C. HsCeONa

D. para-ClH4C6ONa

4 . Which of the following phenols is the strongest acid?

A. para-ClC6HaOH

B. para-O2NC6HaOH

C. para-H3COC6HaOH

D. C6H5OH

5 . The pKu of H3COH is 15.5, and the pKu of CI3COHis 11.2. Which of the following values is the BESTapproximation fbr the pKu of F3COH?

A . t9.4B. 13.8

c. 11.8

D. 10.4

6 . In which of the following reactions would ethanol be

the solvent of choice?

A. HrC\H:N - ,.frt +

H

B.

(HjC)2CHMgBr +

C.

HjCH2CH2C

o

A

4H:C CH2CH3

+ CrO3/FI2SOa

->

o

A

D.ll ll r. KOEI

H.cgocH2cH3ffi

Passage ll (Questions7 - 12)

Alcohols are useful reagents in a vast number of organicreactions, most often used as a nucleophile. Their ,"u"tiuityas a nucleophile correlates directly to steric hindrance. Forinstance, because of steric hindrance, a primary alcohol is abetter nucleophile than a secondary alcohol, which is in turna better nucleophile than a tertiary alcohol.

Alcohols do not readily react as electrophiles, becausethe hydroxyl group is a poor leaving group. To increase theelectrophilicity of an alcohol, the hydroxyl group is treatedwith a Br6nsted-Lowry acid or a Lewis acid to form a betterleaving group. Once a hydroxyl group is protonated to forma water molecule, it becomes a stronger leaving group thatcan be displaced by weak nucleophiles. A hydroxyl groupcan also be made into a better leaving group by converting itinto a tosylate group according to the reaction in Figure 1.

RoH + .r-fi-G CH3 _-> RoSo2c6HacHi

oFigure 1 Tosylation of an alcohol

After a hydroxyl group is converted into a better leavinggroup, the compound readily undergoes dehydration reactionsat elevated temperatures, which yield an alkene. Adehydration reaction is commonly known as an eliminationreaction. Elimination reactions compete with substitutionreactions in most cases where an alcohol nucleophile ispresent. The elimination reaction is most favorable withtertiary alcohols under both basic and acidic conditions,because the alkene product is highly substituted in that case.

The nucleophilicity of alcohols is enhanced by treatmentwith a strong base, which converts it into an alkoxide anionafter it is deprotonated. A nucleophilic substitution reactionwith an alkoxrde acting as the nucleophile also hascomplications with the competing elimination reaction,which proceeds by an E2 mechanism.

7 " Which of the fbllowing alcohols would MOST readilyundergo an E1 elimination reaction?

o

aHsC OH

D.

8 " Which of the following alcohols will MOST readilyundergo transesterification with ethyl acetate and H+?A. H3CCH2CH2CH2CH2OHB. H3CCH2CH(CH3)CH2OH

C. H3CCH2CH(OH)CH2CH3

D. (H3C)2COHCH2CH3

Copyright @ by The Berkeley Review@

C.

OH

9 . Which statement is NOT true about alcohols?

A . Alcohols cannot be oxidized.B. Alcohols are among the f'ew organic compounds

capable offorming hydrogen bonds.C . Primary alcohols are better nucleophiles than

secondary and tertiary alcohols, because theyexhibit less steric hindrance.

D . Alcohols can be identified by their broad peak ininfrared spectroscopy between 3250 and 360b cm-1.

10. Which of these alcohols can be oxidized to a ketone?

A. H3CCH2CH2OH

B. (H3C)2CHCH2OH

C. H3CCH2CHOHCH3

D. (H3C)2COHCH2CH3

I 1. What is the IUPAC name for the following compound?

OH CI

CH2CH3

A. 4-chloro-3-ethyl-2-pentanol

B. 2-chloro-3-ethyl-4-pentanol

C . 4-chloro-3-ethy1-sec pentanolD. 2-chloro-3-ethyl-sec-pentanol

1 2. Which of the following srrucrures represents the productafter isopropanol reacts with tosvlchloride?

nr

@mfr

iliml

iM,&hhmwnmm

l1milnr

,&sE

&/t[@

lh@[

WaTuhlrfufq

d

tqtd

A.

f,C.

H3c_il

CH:

CH:

CH./'O- CH

\CH:

olt

54

D.

GO ON TO THE NEXT PAGE

akcm

)unds

thanthey

1n-l

e?

oduct

11+

-L-

Passage lll (Questions 13 - 18)

Acetals and ketals are formed by adding excess alcohol to

'..her an aldehyde (which forms an acetal) or a ketone (which,,r:ns a ketal) in the presence of an acid. Hemiacetals and

-:niketals are formed by adding an alcohol to either an

:-:ehyde or a ketone in the presence of a base. Acetals andr j:ais do not react at high pH and cannot be formed under..ic conditions, which makes acetals and ketals ideal

:::tecting groups for the aldehyde and ketone functionality in-:;-'tions carried out under basic conditions. Figure 1 shows: :eneric reaction forming a ketal protecting group.

o

Ar3tOh€ vicinal diol ketal water

Figure I Formation of a ketal from a ketone

und? The ketone can be regenerated by adding water to the,:..i. as shown by the reverse reaction in Figure 1. The

'--ilibrium is shifted back to the ketone by adding water., -: ketal must be broken under acidic conditions. Figure 2, '"-. s the acid catalyzed conversion of a ketal into a ketone.

-tx"-n

."><".:"

n.'&)o'

OH

IHI S

--Xi,-"

OH4o \^r

==::- .)<i'-OH

/

'-."k:' '

H

OHJt,,'-x: o*'H":oA.: uo&oH

,Ho*o

.A* - *A*+ HjO+

I :zure 2 Acid catalyzed conversion of a ketal into a ketone

The ketal protecting group is commonly employed in';r:: ,ions such as the Grignard reaction, where the Grignard'::iint could attack the carbonyl carbon. Because the- - ::rard reaction is carried out under basic conditions, the...t remains intact during the course of the Grignard

.::::ion, until it is removed at the end.

, :', nght @ by The Berkeley Review@I,GE 55 GO ON TO THE NEXT PAGE

13. What is the product for the following reaction?

(X + oH.(aq) ----+

A. Acetone

B. Formaldehyde

C . |,2-Ethylene diol

D. No reaction transpires, so no product is formed


H*-Hzo

B.

x": . :.o

15. What is the product for the following reaction run underacidic conditions?

Hov.'VoH * H.c

>"."D >".::O

^ x").::x)

o

A.r,-*

'::>o

x"DA.

x":.

'x"l

16. Which of the followingvicinal diol?

A . Acid and an ester

B. Base and an ester

C . Acid and a ketone

D . Base and a ketone

would be added to protect a

17. The following molecule has what functionality?

("DA. Ketal

B. Acetal

C. Hemiketal

D. Hemiacetal

18. Mixing methanol, sodium methoxide, and acetoneresuits in the formation of which of the followingcompounds?

A. Ketal

B. Acetal

C. HemiketalD. Hemiacetal

Copyright @ by The Berkeley Review@ 56 GO ON TO THE NEXT PAGE

Passage lV (Questions 19 - 25)

The reactivity of a carbonyl compound is correlated in alinear fashion to the acidity of the conjugate acid of theleaving group. Generally speaking, the more acidic HX is,the more reactive RCOX is. This relationship allows us topredict the reactivity of a carbonyl compound using pKuvalues. Figure 1 shows a generic carbonyl substitutionreaction where the X is a variable leaving group.

o

A

o

A

It:

o

/..NaOCHT___+HOCH3

XR+X-

OCH3

Figure 1 Generic carbonyl substitution reaction

The leaving group can be a variety of groups fromhalides to carboxylates to alkoxy groups. The relativereactivity can be quantified by comparing reaction rates ofvarious carbonyl compounds. When comparing the relativereaction rates of compounds, a standard is selected. For thisstudy, a researcher chose the reference reaction in Figure 2.

NaOCHT

HOCH3OCH2CH3 HsC ocH3

Figure 2 Reference reaction for carbonyl substitution study

A reactivity number can be calculated for a given leavinggroup by comparing the log of the reaction rate with the logof the reaction rate for the reaction with a reference leavinggroup, such as the ethoxy leaving group. The reactivitynumber is derived by means of the following equation:

B=log kx

kreference

Equation 1

where B is the reactivity number, k* is the rate constantfor the reaction, and k."1"."n." is the rate constant for thereference reaction with an ethoxy leaving group. Table lshows the pKa values along with reactivity numbers for a

series of X's (leaving groups):

X HX PKa B

I HI -11.2 2.8

C1 HCI -1.O 2.5

H3CCO2- H3CCO2H 4.8 t.lC6H5S- C6H5SH 1.8 l l

CN- HCN 9.1 0.3

H3CO- H3COH 15.5 0.2

Table 1

R!t- I

{

J

III

ffiqmcl-Lo-

o

"raAm

ItIop

X-

lin a

'f theX is,us to

PKaution

fromlativeles oflativen this:2,

CH:

study

avingre 1og

aving;tivity

nstantor the

rble 1

for a

19. Formic acid has a pKa of 3.78. What b value isexpected for methylethanoate ?

a. 2.1

B. 2.0

c. 1.8

D. 0.3

f 0. In the reference reaction shown in Figure 2, the otherorganic product is:

A. -OCH2CH3

B. HOCH2CH3

C. H2O

D. H3COCH2CH3

I I . What b value is expected when X- is bromide?

A. 1.9

B. 2.4

c. 2."7

D. 2.9

For the last example in Table 1, the leaving group isH3CO-. This is also the nucleophile. How could youcarry out this study?

-{. By using 24Na in rhe NaoCH3B. By using a carbonyl with 18O labelingC. By using an R group of CD3 on the carbonyl

compound

D. By using NaOCD3/HOCD3

:J. AnegativeB-value:

-{. occurs when the observed reaction has a greaterequilibrium constant than the reference reaction.

B. occurs when the observed reaction is slower thanthe reference reaction.

C. occurs when the observed reaction is faster than thereference reaction.

D. is not possible.

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2 4. Which of the following types of carbonyl compounds isthe LEAST reactive?

A. Ester

B. Anhydride

C. Amide

D. Acid halide

25. Which of the following relationships accuratelydescribes the relative reactivity of the carbonylcompounds listed?

A. Anhydride > Acetoyl iodide > Ester.

B. Acetoyl iodide > Anhydride > Ester.

C. Ester > Anhydride > Acetoyl iodide.D. Acetoyl iodide > Ester > Anhydride.

Passage V (Questions 26 - 32)

Hydrogens on alpha carbons adjacent to a single carbonylfunctional group have pKu values between li and 19.Hydrogens on an alpha carbon adjacent to two carbonylfunctional group have pKu values between 11 and 13. Thismeans that alpha hydrogens can be deprotonated with theaddition of a strong base.

Upon treating a ketone with one equivalent of a strongbase, a small amount of enolate is formed. The enolateexists in the resonance shown in Figure 1 below:

:o:il

-zc\..oR CHz

Figure I Resonance forms of deprotonated ketone

The resonance structure on the left in Figure 1 (the onecarrying a negative charge on the alpha carbon) is a strongnucleophile, capable of undergoing an 5512 reaction with analkyl halide, resulting in the addition of rhe alkyl group tothe alpha carbon.

A researcher carried out a study on the effects oftemperature on the reaction of an asymmetric ketone. Theresearcher chose 2-pentanone and methylated it according tothe reaction shown in Figure 2 below:

oll ,. LiN(cH3)2

2-cHrl*A-BH.rC CH2CH2CH j

Figure 2 Methylation of 2-pentanone

The reaction was carried out at five different temperaturesand the percentages of the two products, A and B, wererecorded for each trial. Table I below shows the percentagesofthe products and the corresponding reaction temperature;

Temperature Product A Product B-78'C 93.6Vo 6.4Va

-33'C 68.1% 3I.9Vo

0"c 4l.3Vo 52.17a

+25"C 21.9Vo 72.lVo

+50'c 8.47a 91.6V0

Table 1

At lower temperatures, the base deprotonates the lesshindered alpha proton, because the transition state is of lowerenergy. As the temperature of the system is increased, thereaction tends to form the more stable intermediate. Anequilibrium exists between the two intermediates. As thetemperature is increased, the percentage of the thermodynamicintermediate increases. This shift in equilibrium explains thechange in product distribution with temperature. The morestable intermediate leads to the more substituted enolate.

:ii:o€lR/- \a",

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2 6. Which of the following CANNOT be synrhesized from2-pentanone using the same reaction sequence as shownin Figure 2 with any alkyl halide?

A. 4-Heptanone

B. 3-Ethyl-2-pentanone

C. 2-Heptanone

D. 3-Ethyl-2-hexanone

27 . What is the percentage of 3-hexanone formed atusing the same reactants shown in Figure 2?

A . 80Vo 3-Hexanone

B . 65Vo 3-Hexanone

C. 35Vo 3-Hexanone

D, 20Vo 3-Hexanone

2 8. Enolates react in Michael addition reactions as

1. LiN(CH3)2

sCI- I

1

III

.lt- nffi

m

ffi

M

dt

IcD

mhil-l,0.L

o

"..4.n2.o

, =A*.-_..>3. Workup

o

H.cAcr, o- (A_Which of the following products can be formed usingthe kinetic enolate of 2-butanone in a three-stepMichael addition reaction with H2C=CHC(O)CH:?

A. 6-Oxo-2-octanone

B. 3-Methyl-6-oxo-2-heptanone

C. 5-Oxo-2-heptanone

D. 6-Oxo-2-heptanone

29. The kinetic intermediate is preferable when the base isbulky. If the experiment were repeated usingLiN(CH2CH3)2 instead of LiN(CHj)2, what would be

observed?

A . The percentage of Product A would be greater at alltemperatures.

B. The percentage of Product A would be less at alltemperatures.

C. The percentage of Product A would be greatertemperatures below -4"C, and it would be lesstemperatures above -4'C.

D. The percentage of Product A would be lesstemperatures below -4"C, and it would be greatertemperatures above -4"C.

o

rtR

romown

r5"c

)ws:

using:-step)

r at all

, at all

ater at

less at

less at)ater at

rase is

usinguld be

,i U. Product A is the result of the methyl iodide reactingwith which intermediate?

A. The transition state intermediate.

B. The kinetic intermediate.

C . The more stable intermediate.

D. The thermodynamic intermediate.

: n. When treating a ketone with a small base, thethermodynamic intermediate is preferred. What is themajor organic product when 2-methylcyclohexanone isirst treated with potassium hydride at 40'C followed bytreatment with ethyl iodide?

.\ . 2-Ethyl-2-methylcyclohexanone

B. 2-Ethyl-6-methylcyclohexanone

C. 3-Ethyl-2-methylcyclohexanone

D . 2-Ethyl-3-methylcyclohexanone

\\hich of the following reactants offers no competitionr,-tween the thermodynamic and kinetic enolate?

.\ . 2-methyl-3-heptanone

B. 2,2-dimethyl cyclopentanone

C. 2-butanone

D . 2,3-dimethyl cyclopentanone

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Passage Vl (Questions 33 - 39)

Alcohols can be oxidized into a carbonyl (aldehyde orketone) and/or a carboxylic acid upon the addition of anoxidizing agent. The reaction of alcohols varies with thesubstitution of the alcohol. For instance, tertiary alcohols donot oxidize. Figure 1 shows a guideline of reactivity for thealcohols.

o

Ao

A

RCH2oH [ox] >l'Alcohol

R

R2cHoH [ox] >2" Alcohol

R

[ox] I---------> ,\HROH

R3coH [ox] > No Reaction3'Alcohol

Figure 1 Synopsis of oxidation fbr various alcohols

The [Oxl listed in the reactions above represents an

oxidizing agent. Because an oxidizing agent gets reduced, itmust start in a high oxidation state. Oxidizing agents can be

recognized by their abundance of oxygen. In the table beloware some common oxidizing agents and their yields in theoxidation of various alcohols:

Oxidizing Agent Alcohol Yield

KMnO4/OH- 2-pentanol 81.]Vo

PCC l-hexanol 13.9V0

CrOj/H2SOa 2-butanol 65.2Va

Na2Cr2O7/H2S04 2-pentanol 87.4Vo

KMnOa/OH- 1 -butanol 92.37o

K2Cr2O7|H2SO4 l -butanol 52.3Vo

Cu/CuO/H2S04 1 -propanol 45.6Va

H2CrO4 Cyclohexanol 97.2Vo

In the absence of water, primary alcohols can be

converted into aldehydes. The lower yield in the reactionsinvolving sulfuric acid is attributed to the formation of an

alkene side product. The choice of oxidizing reagent can be

made based on the sensitivity of the alcohol compound toeither acid or base. While the alcohol itself may not be pH-sensitive, other functionalities on the molecule may be

sensitive to the pH of the solution.

3 3. What is the change in oxidation state experienced by thereactive carbon in the oxidation of a secondary alcohol?

A. From O to +2B. From -2 to 0C. From}to-2D. From +2 to 0

3 4. What is the major organic product and its approximateyield following the treatment of 2-hexanol with aqueouspermanganate solution at a pH of 10?

A. 92.3Vo 2-hexanone

B. 9l.lVo2-hexanone

C. 73.9Vo 2-hexanone

D. 92.3Vo pentanal

3 5. Which of the following alcohols will NOT undergooxidation upon treatment with chromic oxide in sulfuricacid?

A. 2-methyl-3-hexanol

B. 2-ethylcyclopentanol

C . 3,3-dimethyl-1-pentanol

D. 2-methyl-2-butanol

3 6. Addition of which of the following alcohols to orangeNa2Cr2O7 will turn the solution green?

A. 2-methylphenol

B. 1-methylcyclohexanol

C. 3-ethyl-3-heptanol

D. 2-rnethyl-1-octanol

3 7. Which of the following oxidation reactions is NOT'possible?

A. 1-butanol to butanone

B. 2-methyl-1-pentanol to a carboxylic acid

C. l-pentanol to an aldehyde

D. Cyclohexanol to a ketone

3 8. How does the oxidation state of chromium change inthe treatment of chromic oxide in sulfuric acid withethanol and 2-methyl-2-butanol?

A . The oxidation state of chromium increases with theaddition of both ethanol and 2-methyl-2-butanol.

B . The oxidation state of chromium decreases with theaddition of both ethanol and 2-methyl-2-butanol.

C . The oxidation state of chromium increases with theaddition of ethanol and remains constant with theaddition of 2-methyl-2-butanol.

D . The oxidation state of chromium decreases with theaddition of ethanol and remains constant with theaddition of 2-methyl-2-butano1.

Copyright @ by The Berkeley Review@ 60 GO ON TO THE NEXT PA

3 9. In the following reaction, where R - CH3, what is toxidation and reducing agents?

oHoHoo*U*+ riQ -* *A-[* +ri(oH

A.B.C.D.

Oxidizing agent: C5H1 2O2; reducing agent: TiO2.Oxidizing agent: C5H16O2; reducing agent: TiO2.Reducing agent: C5H12O2; oxidizing agent: TiReducing agent: C5H16O2; oxidizing agent:

t is the

i(oH)z

TiO2.

TiO2.

TiO2.

TiO2.

Passage Vll (Questions 40 - 46)

A researcher wants to distinguish and identify three-rknown isomers (Compound A, Compound B, andlrmpound C), all of which have the molecular formula,-<H12O. To help distinguish the structures from one.:l,rther, the researcher performs some standard reactions used- qualitative analysis tests for organic compounds. To-,-mpounds A, B, and C, she adds Jones's reagent,-rO3/H2SO4). Only compound A turns green upon the.:lition of Jones's reagent. Compounds B and C remainr.nse in color for an extended period of time. The researcher

- :lcludes that only Compound A is an oxidizable compound.*-, three compounds maintain a brown color when treated.,: Br2 liquid in the presence of CCl4 solvent, indicating

- r' the brown bromine liquid did not react with any of the--:.nowns. The bromine liquid tests for the presence of any-*: -rrjugated alkene n-bonds.

Only Compound B reacts with 12 in basic medium to.:r- a yellow oil at the bottom of the flask (the iodoform

';. used to test for the Z-keto functionality). Only- .:pound A, after being treated with strong base, yields a:-:;:pitate when added to AgNO3 in water solvent. This is, : . ,r, n as Tollen's test, which indicates the presence of a'r,:llv oxidizable carbonyl functional group. Compound A

-s a precipitate with H3CCI in the presence of silver- :::e. The precipitate is silver chloride, formed when' , ide anion is displaced as a leaving group during a, - : rphilic suhsritution reaction.

The product of the reaction of Compound A with Jones's'::irtlr. was isolated, collected, and tested with blue litmus"r.;r The oxidized derivative of Compound A turned the-: lllmus paper red immediately. The information from

:,: .;nous tests gave clues as to the type of compounds that.-- -nknown was likely to be. The researcher followingl::: .istS was able to conclusively identify the functionality

..:h unknown. Table 1 summarizes the results of each,, . -,r Compounds A, B, and C.

Table 1

.in alcohol, when oxidized, forms a compound that

.rrns blue litmus paper red. In terms of substitution,:-; alcohol is best described as a:

A. primary alcohol.B. secondary alcohol.C. tertiary alcohol.D . primary alcohol or secondary alcohol.

Test -+ Jones's I2lKOH(aq) Tollen s RX/Ag+:.:ound A Positive Negative Positive Positive

.lund B Negative Positive Negative Negative

-:rund C Negative Negative Negative Negative

PAGE ': ,:,,rht O by The Berkeley Review@ 6r GO ON TO THE NEXT PAGE

4 1. Which of the following structures is NOT possible forthe formula C6H120?

A. Cyclic alcohol

B. Linear ketone

C. Linearaldehyde

D. Cyclic ketone

4 2. Which of the compounds summarized in Table I isNOT correctly identified?

A . Compound A is a primary or secondary alcoholB. Compound B is a methyl ketone

C. Compound C is an aldehyde.

D . None of the compounds can be a cyclic ether.

43. Which of the following compounds CANNOT beoxidized?

A. Primary alcohols

B. Aldehydes

C . Esters

D. Hemiacetals

4 4 . All of the following are true for a compound that can beused to test for an aldehyde EXCEpT:

A. it undergoes a color change or phase change whenit is reduced.

B. it is a compound that is rich in hydrogen atomsC. it is a transition metal in a high oxidation srate.

D . it is in lower concentration than the aldehvde.

4 5. Which of the following is NOT an oxidizing agent?

A. KMnO4

B. H3CCO3H

C. Mg(oH)2D. HIOa

4 6. Which of the following reagents can reduce an ester to aprimary alcohol?

A. H2lPd

B. HCllZnC. LiAlHaD. BH3

Passage Vlll (Questions 47 - 53)

Wcilff-Kishner reduction converts either a ketone or analdehyde into an alkane. The two-step process involves theformation of a hydrazone (the hydrazine derivative) which isreduced under basic aqueous conditions. Figure 1 shows theWolff-Kishner reduction of a generic ketone.

oll . H2N-NH2

*A**-*tll oH-

*A* Hzo

*- n"'

__-____+.4.

HH

*x*+Nz

Figure I Wrjlff-Kishner reduction of a ketone

Reduction of a carbonyl into alkane can also be carriedout under acidic conditions using Clemmensen reduction.Zinc metal provides the electrons for the reduction, as shownin Figure 2.

oHH

*4. #-*

*X*+ H2o + ZznCt2

Figure 2 Clemmensen reduction of a ketone

47 . The reactant in a Wrilff-Kishner reduction reaction couldalso react with all of the following compoundsEXCEPT:

A. LiAlHaB. H3CMgBr(et2o)

C. O31H2O2

D. HCllZn

4 8. Over the course of a Clemmensen reduction of a ketone,how does the oxidation state of the carbonyl carbonchange?

A . It increases from -2 to +2.

B " It decreases fiom +2 to -2.

C . It decreases from 0 to -2.

D. It decreases from +2 to 0.

4 9 " What is true following the Wolff-Kishner reduction of aketone?

A . One new chiral center is formed.

B . The hybridization of carbon goes from rp2 to sp3.

C . Both nitrogen and carbon are reduced.

D . Three net water molecules are formed as a product.

Copyright O by The Berkeley Review@ 62 GO ON TO THE NEXT PAG

mn$

5 0. What accounts for the greater acidity associated with a

proton on H2N-N=CR2 than a proton on ammonia?

A. The hydrazone has greater steric hindrance thanammonia, which increases the acidity.

B. The hydrazine molecule becomes aromatic uponlosing a proton.

C . Ammonia has less of an inductive effect thanhydrazone.

D. The lone pair formed upon deprotonation from thehydrazone is stabilized by resonance.

5 1. What product is expected when the following moleculeis treated with hydrazine and basic water?

o

A. 3-EthylcyclohexanolB. 3-Hydroxyethylcyclohexane

C. Ethylcyclohexane

D. 3-Ethylcyclohexanone

52. Under what conditions is Wrilff-Kishner reductionprefered over Clemmensen reduction?

A . When the reactant has an acid sensitive functionalgroup.

C.

When the reactant has an amine sensitivefunctional group.

When the reactant has a base sensitive functiondgroup.

D . When the reactant has a water sensitive functiondgroup.

HO

lPass

{a

11.e,:r

lmE ;:I:tLTe :ri

:;:r:e.t;l

!m0)r

:tffe -r*,1

TllrirLrna'rm:r,

Im;fu r,o,lnl

gmm unr

r(@lmhuMwrl

m

iffilmm

(lro{

rltmDnm

,dh{

B.

W.

Wld

5 3. Whatphenyl

A.

H:C

C.

H:C

is the product of 2-methylcyclopentanone a

hydrazine, which reacts like hydrazine?

Ph B. Ph\\

,zN- H -zN- oHNN-

*

HH

{

*

*a nn

*

m.'

mn

mH:C

D.

H:C

t

Passage lX (Questions 54 - 60)

Carbonyl compounds are reactive with nucleophiles.Their electrophilicity can be attributed to the polar nature of:e carbonyl bond which houses a partial positive charge on-:e carbon in the carbonyl. Carbonyl reactivity can be::'rrelated to the pKo of the conjugate acid of the best leavingroup on the carbonyl carbon. This is to say that the better:,; leaving group the more reactive the carbonyl compound.l.:le 1 shows the pKu values of conjugate acids of various..r'ing groups on carbonyl compounds:

PKa Acid'74 HCI

4.2 H5C6CO2H

4.8 H3CCO2H

9.1 HCN

10.5 H3CH2CSH

t5.1 H:CHzCOH

33.1 NH:

Table 1

h addition to the leaving group in a carbonyl reaction,iir : -u;leophile is also considered. A Grignard reagent is a" :,:,: nucleophile that is selective in its reactivity withr,,::r1'ls. Three Grignard reactions are shown in Figure l.

Reaction I (Grignard reaction with an aldehyde)oR'OHll l. n'rvrgxer,g A.l

nAs l. NFlcltaqr nAn

Reaction II (Grignard reaction with a ketone)OR'OHll r. n'r,rgx.r,g o...1

nAn 2. NFlaClraqr nAn

Reaction III (Grignard reaction with an ester)

Requires two equivalents of R'MgBr

l. R'Mgxer,g U...fn

X 2. NFIaCkaqr

nA n,;|unrme I Grignard reactions with three different carbonyls

::::nard reagents react most readily with acid halides and,itttll l i" .-:', drides (to which two equivalents of the Grignard canillrrr,,Li l3n-snard reagents do not react with amides. Despiteliiil': r:3ngrh of the Grignard reagent as a nucleophile, it

.ri:lr..: ri:,: jlsplace an amine leaving group. These observationsLuru"j i''l :.'rord with the acidity of the conjugate acid of therLllr:ii. r.: ,5oups. This information can be used to predict therrltrirl:--:t I product formation and the favorability of a carbonylIt(llil]ill]l'

-i _ -_

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54. Which of the following compounds is the LEASTreactive carbonyl compound?

A. An ester.

B. A thioester.

C . An anhydride.

D . An acid chloride.

55. To make 2-pentanol, it would be best to use which ofthe following reaction sequences?

A. H3CMgBr + pentanal.

B. H3CMgBr + 2-pentanone.

C. H3CMgBr + butanal.

D. H3CMgBr + butanone.

5 6. Which of the following Grignard reactions would formstereoisomers as the product?

A. Butanone + ethyl magnesium bromide.B. 3-Pentanone + methyl magnesium bromide.C. Ethyl acetate + 2 equiv. ethyl magnesium bromide.D. Propanal + methyl magnesium bromide.

57. When treated with RMgBr, the following compoundwould yield which of the following products?

A. A primary alcohol.B. A secondary alcohol.

C. A tertiary alcohol.D. A ketone.

5 8. A Grignard reagent adds how many times to an ester?

A. Once.

B. Twice.

C . Three times.D . It depends on the temperature of the mixture.

H.9 o

o/

5 9. Which of the following conditions is NOT true about a

Grignard reaction?

A . Ether solvent is required for the Grignard reaction.

B. The Grignard reaction involves the reduction of thecarbonyl carbon.

C . The Grignard reaction can be employed to formeither primary, secondary, or tertiary alcohols.

D. For the highest yield, the solution should be acidicwhen the Grignard reagent is added.

60. If propanoic acid has a pKu value of 4.9, what can be

concluded about the reactivity of propanoic acidanhydride?

A . It is more reactive than benzoyl chloride.B. It is less reactive than N-methylbutamide.C. It is more reactive than methyl acetate.

D. It is less reactive than acetic acid anhydride.

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Passage X (Questions 61 - 67)

The Aldol condensation reaction is a main staple insynthetic organic chemistry. The reaction involves thedeprotonation of a hydrogen from the alpha carbon (thecarbon adjacent to the carbonyl group) and the subsequentreactions of the carbanion as a nucleophile. The carbanionform is in resonance with the enolate form. In an Aldolcondensation, the electrophile to which the carbanion isadding is a neutral carbonyl compound. The preliminaryproduct (the product of Step I in Figure 1) is referred to as a

./S hydroxy carbonyl. This B-hydroxy carbonyl can thenundergo elimination (shown as Step II in Figure 1) to form a

conjugated alkene-carbonyl compound. The conjugatedalkene carbonyl compound is referred to as an a,t3-unsaturatedcarbonyl. The Aldol reaction is shown in Figure 1 below.

Step I: Formation of B-hydroxyketone

62. $,

{B.C.D.

ffi,4 -,iqr:u

:: [r:

:

!{vI

oilR- -\-c- -R

/\HH

*-."\o HCrraor

I

-

R:.\ou a

/\ H

HH

A.B.C.D.

o

LpAretzol *-.r\*IR-r"\oH

/\ HHH

Step II: Formation ofu,B-unstaurated ketoneo

*-.An/\

HHFigure I Generic Aldol reaction

The pK6 of an alpha hydrogen falls in the range of 11

19 depending on the number of carbonyls to which itadjacent. The hydrogens of acetone have a pKu of roughly 1

while the central alpha hydrogens of4-oxo-2-pentanone ha

a pK2 of roughly 11. The pK2 information can be used

predict the most reactive site on a molecule. It can also ! tr..used to predict the degree ofreactivity for a compound.

61. Which of the labeled hydrogens on the molecule belcis the MOST acidic?

? ? oc+ lii Ig.cHucl t,.a-.V

CHHI

d----+Il3Q

Tllhu um

umr' mdfiu

fl- md

l- rfir

G-@lD-"Ifil

a

b

d

*-.4

62. What is the IUPAC name for the compoundundergo an Aldol condensation reaction withform the following compound?

ooH

that can

itself tople ines then (the

equentbanionAldol

rion is

minarytoasaLn thenform a

iugatedturatedlow.

)

(R

: of 11

rich it

CH:

H:C

A. Butanaldehyde

B. 2-Pentanone

C. 2-MethylbutanalD. 2-Methylpropanal

CHI CH:

Treatment of an aldehyde with excess alcohol inpresence of an acid would lead to which oflollowing products?

A. An acetal

B. A hemiacetal

C. An alkene

D. An ester

t 4. \\/hat is the major organic product isolated from thetbllowing reaction?

H

tr *1 thethe

H

rughly 1

rone ha

e used

n alsord.

ule be

C.

H:C

D.

H:CCH:

The chemical inter-conversion from a ketone to an enol:s referred to by which of the following terms?

-\. Resonance

B. Conjugate pairingC . Hydroxyl exchange

D. Tautomerization

I PAG nght @ by The Berkeley Review@ 65 GO ON TO THE NEXT PAGE

6 6. Which of the following bases is NOT strong enough todeprotonate the hydrogen on the alpha-carbon of a

ketone?

A. NaNH2

B. NaH

C. Na2CO3

D. NaOCH2CH3

6 7. Which of the following compounds CANNOT undergoan Aldol condensation reaction?

A. A ketone

B. An aldehyde

C. AcarboxylicacidD . All of the above can undergo an aldol reaction

Passage Xl (Questions 68 - 73)

The Claisen condensation involves the dimerization ofan ester. The first step is deprotonation of an ester to forman enolate, which is followed by the subsequent addition ofthe enolate to the carbonyl carbon of a neutral ester. Thetetrahedral intermediate that is formed loses an alkoxideleaving group to yield a B-ketoester product.

The base chosen to deprotonate the alpha hydrogen of theester should match the alkoxide leaving group so as to avoidcomplications with transesterification. This alkoxide leavinggroup is an anion that is also a strong enough base todeprotonate the alpha hydrogen of the B-ketoester product.To isolate the B-ketoester product, the solution is neutralizedwith a weak acid. Figure 1 shows the equilibrium mixturesin a Claisen reaction.

H3CH2C

HjCH2C

fr ot,ct ?)A o.n r." i-" 3c HC+oc Hzc'H rI

o e zH'cH'cfRocH2cHjA.'A o{*^,

I

CH1- OCH2CH-1B-Ketoestcr

\ O.A

H3CH2C

o

Ac- ocH2cH3I

cHrHOCH2CH3

Figure 1 Claisen reaction scheme

Without an alpha hydrogen, the last step (equilibriumstep) of the scheme above is not possible. If the B-ketoesterthat forms has no alpha hydrogen, then the yield is reducedbecause the alkoxide anion acts as a nucleophile (rather thanas a base) and causes the retro-Claisen reaction. To increasethe yield (from as low as zero), less alkoxide base is added,and the reaction is run at a lower temperature.

6 8. Which proton has the lowest pKu value of thehydrogens on the following B-ketoester?

HjCH2COCH2CHj

dH CH'CH.b c-

A. Hydrogen a.

B. Hydrogen b.

C. Hydrogen c.

D. Hydrogen d.


HHa

oo

6 9. Which of the following ester reactants would give theLOWEST yield in a Claisen reaction?

A. H3CCO2CH2CH3

B. H3CCH2CO2CH2CHj

C. (H3C)2CHCO2CH2CH3

D. (H3C)2CHCH2CO2CH2CH3

7 0. A Claisen condensation reaction is NOT possible withwhich of the following esters?

A. HCO2CH2CH3

B. C6H5CH2CO2CH2CH3

C. (C6H5)2CHCO2CH2CH3

D. C6H5CH2CO2CH3

71. Which of the tollowing is a product of a Claisencondensation reaction ofone ester upon itself?

A . H3CCH2COCH2CO2CH2CH3

B. H3CCOCH(CH3)CO2CH2CH3

C . (C6H5)CH2COCH2C O2CH2CH3

D. (C6H5)CH2COCH(C6H5)CO2CH2CH3

7 2. What is the major organic product for the followireaction?

NaOEtoEt

--*-->HOEt

B.o

&'D.o

U"

GO ON TO THE NEXT PA

73. Counting stereoisomers, how many products arepossible for the following reaction?

ooll ll NaOEt

EroA * \-Anr, ffi

2

J

4

6

A.B.C.D.

rvrng

'AG - ;riright @ by The Berkeley Review@ 61

Passage Xll (Questions 74 - B0)

In transesterification, an alkoxy group of an ester isexchanged by the addition of an alcohol in the presence ofeither acid or base. The equilibrium constant for the reactionis approximately one, so the reaction must be driven by theaddition of excess alcohol. A researcher carries out a

transesterification reaction by adding methyl propanoate toethanol in the presence of HCI at 25"C for thirty minutes(Reaction I). The product is isolated by neutralization and

then purified using fractional distillation. The identicalreaction is then repeated under reflux conditions for thirtyminutes (Reaction II). The products from Reaction II are

then neutralized and isolated by neutralization and thenpurified by fractional distillation.

oll H*-

-'\- + H3CCH2OH ;IiHTCH2C OCH3 30;,Reaction I

o H+-ll + HjCCH2OH ;ffi-A-H3CH2C- -OCH1 30 min

Reaction IIFigure 1 ancl Figure 2 show the 1HNMR spectra of the

major components from Reactions I and II respectively.

Major component isolated from Reaction I

Figure 1 Reaction I's maior component lUNVR spectra

Figure 2 Reaction II's major component lUNVR spectra

GO ON TO THE NEXT PAGE

Major component isolated from Reaction II

The ratio of the integrals of the peaks are 3 :2 :3 forthe spectrum in Figure I and2l.2:3 :3 for the spectrum inFigure 2. The 1HNMR spectra were collected indeuterochloroform solvent.

7 4. What is the IUPAC name and formula for thecompound in the 1HNMR spectrum in Figure l?A. Methyl propanoate (CH3CH2COZCH)

B. Methyl ethanoate (CH3CO2CH3)

C. Ethyl ethanoate (CH3CO2CHZCH1)

D. Ethyl methanoate (HCO2CH2CH3)

75. What can be concluded about Reaction II from theltnn4R spectrum in Figure 2?

A. Methyl propanoate, CH3CH2COZCHZ, did notreact with ethanol.

B. The ethyl and methyl groups exchanged, resultingin the formation of ethyl acetate,CH3CO2CHzCHz.

C. Propyl acetate, CH3CO2CH2CH2CH3, formedfrom the reulrrangement of a carbocation.

D. Ethyl propanoate, CH3CH2CO2CH2CH3, wasgenerated in a greater than 5OVo yield.

7 6. An ester is more reactive as an electrophile than whichof the following compounds?

A . Acid anhydrides

B. Amides

C. Acid halides

D. An ester is the most reactive carbonyl compound.

77. Had ethyl amine (CH3CH2NH2)ethanol is Reaction II, the finalbeen which of the following?

A. N-ethylpropyl amineB. l-aminopentane

C. N-ethylpropanamide

D. N-Hydroxypropanamine

been used instead ofproduct would have

helping

helping

helping

helping

78. The presence of acid in the reaction mixturewhich of the following purposes?

To protonate the ester electrophiledecrease reactivity.To protonate the ester electrophileincrease reactivity.To protonate the alcohol nucleophiledecrease reactivity.To protonate the alcohol nucleophileincrease reactivity.

to

to

to

A.

B.

C.

D.


to

68 GO ON TO THE NEXT PAG

7 9 . The presence of water in the reaction mixtureresult in which of the following impuritiesproduct mixture?

A. Carbonic acid (H2CO3)

B. Formic acid (HCO2H)

C. Acetic acid (CH3CO2H)

D. Propanoic acid (CH3CH2CO2H)

wouldin the

8 0. Why must the NMR be carried out in deuteratedsolvent?

A. To prevent reaction from occurring in the NMRtube.

B. To induce magnetization into solution.C. To prevent hydrogens from the solvent from being

detected by the NMR spectrophotometer.D. To decrease the pH of the solution.

aassage Xlll (Questions 81 - 87)

1. NaOEr(HoEt)

2. RX(et2O)

Figure 1 Synthetic

lrlalonic ester is a common starting reagent in the'i' r -:,:sis of many bio-organic compounds. Figure 1 shows.)Mr :::l'erent synthetic routes starting from the malonic ester,illr :ading to a carboxylic acid and the other leading to a

,uur. *r:. cvclic amide). The first route leads to a carboxylicrirlu,:r i -: specific composition. The carboxylic acid can be anrtilrj jf-:3Ciate product that is subsequently converted into onen ,r":,::al possible carbonyl compounds, such as an ester or

irrnr ,:,: The second synthesis route leads to compounds in therltttur.i -:l-ic acid derivative group. These are useful in the,r': [:asls of barbiturates.

I-; first part of both synthetic pathways involves therlirrLrlr'':-,rnation of the hydrogen alpha that is adjacent to both:uriiLu:"T ri:'*l groups. The proton that is on a carbon that is alphaI t,'i carbonyls has a pKa value of roughly 11, so an

riilll ]1ii x-13 is strong enough to remove that proton. Thisrr""i-rL:. a nucleophile that can react with alkyt halides,iiillilrfl r!:si other electrophiles. In these two examples, sodium,lrrlLlr: t,:e is chosen to prevent competition with the

ulriliiitLtlu lE:s::ifi cation reaction.

,\ hat is the structure of urea?

,H3cH2oAr(ocH2cH3 A-J(I

R

70 to 85Vo yield

ooo

CH3CH2O

o

4",R

80 to 90% yield

o

AHN NH

roUo"

IocH2cH3

\\_*

Urea

o

uses of malonic ester

H:C CH: Hrc CH:

8 3. Which of the following molecules will NOT undergo a

decarboxylation when treated with heat?

^,,.j\.L",

c'oD'o\oH

",.\o To "\:r"

8 2. When (CH3)2C(NH2)2 is used in lieu of urea, what isthe final product?

A' g B' Hrc. .cHr

"NAN' "*X.nt

R

HrC CH"-XHN NH

D.C.

{"o

",*Aot{f_

B.o

"r*A*", B. oo

NHz

NHz

D.o

NH

o

o

*AI*Y

o

65 to 75Vo yield

ri: -:::nt O by The Berkeley Review@ 69 GO ON TO THE NEXT PAGE

84. Why is sodium ethoxide chosen in the synthesis?

A . Sodium ethoxide is a strong base that will notresult in transesterification of the ethyl ester.

B. Sodium ethoxide is a weak base that will not resultin transesterification of the ethyl ester.

C. Sodium ethoxide is stereo-specific.

D. Sodium ethoxide is regio-specific.

What signals would be observed in the proton NMR ofmalonic ester?

A. A 2H doublet, a 2H doubler, and a 3H tripletB. A lH singlet, a2Htriplet, and a 3H doubletC. A 2H singlet, a 2H rriplet, and a 3H quarrerD. A lH singlet a2Hquartet, and a 3H triplet

Which alkyl halide must be chosen to tbrm thefollowing molecule when carrying out a synthesis frommalonic ester?

H:C CH2CH3

A. 2-bromopropane.

B. 2-bromobutane.

C. 2-bromopentane.

D. Two alkyl halides must be used; methyl bromideand ethyl bromide.

87. What is the final organic product after malonic ester isrreared wirh: 1) NaOEt, 2) CH3CH2CH2I, and 3) urea?

A.o

AHN NH

CH:

o

AHN NH

H:C cHs H:C

85.

86.

o

A

B.o

AHN NH

CH2CH2CHj

D.o

AHN NH


cH2cH2cH3


Passage XIV (Questions 88 - 94)

Carboxylic acids can be produced by one ofseveral ways.Carboxylic acids are useful in synthesis as an intermediate innumerous reaction pathways. Carboxylic acids are weakacids that when added to water will partially dissociate intotheir carboxylate conjugate base and hydronium ion.Carboxylic acids play a distinct role in biological chemistry.because of their acid-base properties and their syntheticflexibility. Figure I shows six of rhe ways in whichcarboxylic acid can be synthesized:

Grignard Reduction:RMgBr + CO2 --* RCO2MgBT

H+

--+ RCO2H

Oxidation of an Aldehyde:

*An. cro3 HzSo+_

oAo"Hydrolysis of an Anhydride:

.A".JLor+,*Ao,Hydrolysis of an

oEster:

*Ao*,* Hzo

Ozonolysis of an

ot lH t.q o o

"><_ ,,1o,5- _-\on*oA

Hydrolysis of a Nitrile: O

H+/ARCN + H2o:i

*Ao, * NH+*

Figure I Six reactions used to synthesize a carboxylic

The large number of starting reagents offers varietl,synthesis based on the availability of starting compouThe acid products can react further to form either e

(through transesterification under acidic conditions) or ami(by way of acid chloride formarion using SOCI2). An a

is formed by treatment of either an acid halide oranhydride with an amine.

8 8. To synthesize butanoic acid, which of the followisynthetic pathways will NOT work?

A. Oxidation of 1-butanol.B . Hydrolysis of ethylbutanoare.C . Ozonolysis of 4-octene, followed by oxidari

rvorkup with hydrogen peroxide.D. Treatment of 1-bromobutane with NaCN, follou

by strong aqueous acid treatment.

oH+/L ltrr /d!

*Aon* R'o

AIkene:

ways.

Liate inr we8k.te inton ion.mistry,nthetic'hich a

rylic ac

varietympoun

8 9 . How does propanoic acid exist after it has been added towater given that it has a pKu value of 5.0?

A. Fully deprotonated.

B. 50Vo deprotonated.

C. 25Vo deprotonated.

D. Less thanTTo deprotonated.

9 0. Which of the following reactions does NOT result inthe formation of a compound that can turn blue litmuspaper red?

A. Treatment of 1-octanol with CrO3/H2SO4.

B. Treatment of 3,4-dimethyl-3-hexene with ozonefollowed by reductive workup with dimethylsulfide.

C . Treatment of maleic anhydride with acidic water.

D . Treatment of ethylbenzoate with acidic water.

4 1 . What is the product for the following reaction?

1. PBr3(et2O)H3cH2cv' cH3 2. Ms(etro)

I 3.cQ(e)oH 4. H3o+/

C.

oA. B' , O

oH AA'HoD.o

her

orAnle or

followi

oxidati

{, fol

Which of the following reactions is NOT reversible?

A . Hydrolysis of an ester.

B. Grignard reagent plus carbon dioxide.

C. Hydrolysis ofan acid anhydride.

D . Hydrolysis of a lactone.

u l. Which of the following is NOT associated withcarboxylic acid?

A. An aqueous pH less than 7.0.

B. A broad IR peak between 2500 and 3000 cm-l.

C . The loss of one peak in the 1HNMR when D2Oadded to the NMR sample tube.

D. A pKu value between 9.0 and 10.5.

t PAGlrri'right @ by The Berkeley Review@ GO ON TO THE NEXT PAGE71

9 4. What sequence of reagents will convert an aldehyde intoan amide?

A. l. NH3 2. SOC12

B. 1. CrO3/pyridineiHCl 2. SOC12 3.NH3C. 1. CrO3/H2SOa 2. NH3 3. SOC12

D. 1. CrO3iH2SOa 2. SOCI2 3. NH3

Questions 95 - 100 are NOT basedon a descriptive passage

95. Which of the following structures represents theintermediate for a transesterification reaction under basicconditions?

A. B.RO-\,s

*O/"ao*,,

*/"\o

9 6. Esters are NOT used as a reactant in which reaction?A. Claisen condensationB. Friedel-CraftalkylarionC . Grignard reactionD . Transesterification

9 7. What is the major product of the reaction below?

oil

R'O-C- OR"

I

R

D.C.oil

+C

I

R

->oHA

B.A.

c(" c("

Copyright @ by The Berkeley Review@ YOU ARE SO VERY FINISHE

98. Which type of compound can be reduced by NaBHa?

A. Phenol

B. Tertiary alcoholC. Ketone

D. Ether

Which of the following describes the hybridization ofthe carbon number three in 3-hexanol?

A. rp

B. tp2

c. rp3

D. The carbon is not hvbridized.

10 0. The product of 2-butanol with chromium (VI) oxide insulfuric acid would have which of the followinglHNuRpeaks?

A. Singlet (3H), doublet (2H) and triplet (3H).B. Singlet (3H), triplet (2H) and doublet (3H).C . Singlet (3H), quarret (2H) and triplet (3H).D. Triplet (3H), doublet (2H) and triplet (3H).

"Oh, what a feeling; CHEM-IST-RY!l!',

99.

l.D 2.D6. D 1. D

11. A 12. B16. C 11. B21. C 22. D26. C 27. C31. A 32. B36. D 31. A41. D 42. C46. C 47. C51. A 52. A56. D 57. B61. B 62. D66. C 61. C71. D 72. A16. B 17. C81. B 82. B86. B 87. B91. D 92. B96. B 97. A

3.A 4.8 s.D8.A 9.A 10.C

13. D 14. D 15, c18. C 19. C 20. A23. B 24. C 25. B28. A 29. A 30. B33. A 34. B 35. D38. D 39. C 40. A43. C 44. B 45. C48. B 49. B 50. D53. A 54. A s5. c58. B s9. D 60. C63. A 64. A 65. D68. B 69. C 70. A13. D 74. A 75. D78. B 19. D 80. C83. C 84. A 85. D88. D 89. D 90. B93. D 94. D 95. A98. C 99. C 100. c

.?+

ion of

'ride inlorving

Choice D is correct. According to the answer choices, either statement I or II must be true and either statementIII or IV must be true. The pKo values are given in the question, so the question centers around determining therelative acid strength of phenol and methanol. Phenol has a lower pKo, so it is a stronger acid than methanol.Vethanol is therefore a weaker acid than phenol. This makes statements II and IV the true statements, whichmakes choice D the best answer. Resonance is a stronger effect than the inductive effect, so statement II is moresignificant than statement IV. Pick D, and let correctness set you free.

Choice D is correct. For the base to be strong enough to deprotonate methanol, its conjugate acid must be a weakeracid (have a higher pKu) than methanol. Choices A and B can immediately be eliminated, because they areroth weak bases, and therefore are not strong enough to deprotonate methanol, a very weak acid. The conjugateacid of NaOH is water, which has a pKo of 1,5.7 (roughly equal to methanol), so it will deprotonate about halfrf the methanol molecules. The conjugate acid of H3CCH2Li is H3CCH3, which is one of the weakest acidsinor'\,n, so H3CCH2Li is definitely strong enough to deprotonate methanol. Pick D, and correct you will be. This:-'-restion can be reworded to read "which of the following bases is the strongest?", because the one base strong:rough to deprotonate methanol must be the strongest base of the choices.

Choice A is correct. Because the electrophile has only one carbon, elimination is not possible, resulting in a

:-.action where only the substitution product is observed. The best nucleophile is also the strongest base, becauser e strongest base donates eiectrons most readily. Of the four answer choices, C and D can be thrown out, because--::\- are phenoxides (the conjugate base of a phenol) while choices A and B are the conjugate bases of alcohois.- :,enols are more acidic than standard alcohols, so phenoxides are less basic than standard alkoxides. Between-: rice A and B, the chlorine atoms on choice B (H3CCCI2ONa) are electron withdrawing and thus decrease the, - :r.pound's basicity and nucleophilicity. Pick A to make your day.

tlhoice B is correct. Electron withdrawing groups increase the acidity of a compound by stabilizing the negatir-e- ,'.:ge on its conjugate base. As the conjugate base becomes more stable, the acid becomes more acidic. The

:::-:,g€st acids of the phenols listed in the question (and it could be any type of compound, not just phenols, as

...: as all four choices are the same class of compound), is the one with the strongest electron withdrarving::.-rP. The strongest electron withdrawing group is the nitro group, which withdraws by resonance, making.- ,:e B the most acidic. Pick choice B and a correct choice is what you'll see.

-F.oice D is correct. Because fluorine is more electronegative than chlorine, it is more electron withdrar,r'ilg:--,.:. :hlorine. This means that fluorine atoms on the backbone of the molecule increase the acidity even more so-,:. :he chiorine atoms. An increase in acidity results in a lowering of the pKu. The only value for pKu 1ou-er

'," :: ' I.2 is choice D,1.0.4. You'll no doubt feel glee, if you just picked choice D.

-:.:ice D is correct. The reaction in choice A is an Sp2-reaction, because the electrophile is primary and.--::'"-ria is a good nucleophile. Sp2 reactions require an aprotic solvent, so ethanol is not a good choice. Chorce

-' .ilnhated. The second reaction is a Grignard reaction, which cannot be carried out in a protic solvent. h-r a

,: solr'ent, the alkyl magnesium bromide species can deprotonate the protic hydrogen, thereby destror.ins.::glard reagent. Grignard reactions must be carried out in an aprotic solvent, so choice B cannot use ethano.:.lr-ent. Choice B is eliminated. In choice C, an aldehyde is being oxidized into a carboxylic acid. -\.:'.i must be inert if it is to be useful. Because primary alcohols can be oxidized by ]ones reagent

- : H2SOa), ethanol is reactive under the reaction conditions. This makes ethanol a bad choice for the::.:, so choice C is eliminated. The last reaction involves the deprotonation of an alpha hydrogen on::etic ester. Because the ester can undergo both deprotonation and attack at the carbonyl carbon, the base

-. chosen matches the leaving group of the ester, in this case ethoxide. The ideal solvent is the conjugate-: :he alkoxide, so that it cannot generate another possible nucleophile for transesterification. Given that, -:e is the base being used in choice D, ethanol is the ideal solvent. Choice D is your kind of choice.

--:r:e D is correct. The E1 elimination reaction is a two-step process that involves the formation ot a

. -: ::-:ion intermediate. The most stable carbocation is a tertiary carbocation. The only tertiary alcohols of:-- -i .'. er choices is choice D, which makes choice D the best answer.

DlC

CABBD\CDCC

D

DCDB.\C

SHED:: , The Berkelcv Review@ /c CARBONYLS & ALCOHOLS EXPLANATIONS

8. Choice A is correct' The best nucleophile will undergo transesterification most readily. The nucleophile thatmost readily undergoes transesterification must have-the greatest partial negative charge and the least sterichindrance' Considering they are all five carbon alcohols'ana ,o,rgnty "qrrJt

in basicity, their differences innucleophilicity revolve around steric hindrance rather than partial negative charge. The least sterichindrance is associated with the primary alcohol. This eliminates choice C (a secondary alcohol) and choice D(a tertiary alcohol). Choice A iJa straight chain alcohol with less steric hindrance than choice B (a primaryalcohol with a branched R group), so choice A is the best answer.

Choice A is correct' Both primary and secondary alcohols can be oxidized to carbonyi compounds, thereforestatement A is a false statement. Choice A is not true in regards to alcohols. Alcohols do ror* nydrogen bonds,and outside of alcohols, few other organic compounds forri hydrogen bonds (amines, amides, acids are some ofthe few)' Statement B is a valid statement. Piimary alcohols mak"e the best nucleophiles, because they exhibitthe least steric hindrance in the transition state, therefore statement C is a valid statement. The infraied peakfor an alcohol is broadened due to hydrogen bonding, and hydroxyl bonds show in th" t;001;*-i'L"g", irr*"t"*statement D is a valid statement. Pick A to start your correct answer filled day.

Choice C is correct' A primary alcohol will oxidize into either an aldehyde or a carboxylic acid, depending onthe conditions, but it cannot be oxidized into a ketone. This eliminates choices A and B. A tertiary alcohol doesnot undergo oxidation, which eliminates choice D. The oxophilic carbon (carbon bonded to the hyhroxyl oxygenatom) in a secondary alcohol has two bonds to carbons, so when it is oxidized into a carbonyl, it still has twobonds to carbon' A carbonyi compound where the carbonyl carbon has two bonds to other carbons is a ketone.Choice C is pretty unbeatable when it comes to correctness on this question.

Choice A is correct' The alcohol group takes priority according to IUPAC nomenclature rules, so the compound L.automatically something-2-pentanol. This makes only choiJe A possible. It is not necessary to determine theexact name in light of the answer choices. Pick A to be correct. In the interest of learning, you should note thalthe substituents are listed in alphabetical order according to nomenclature rules.

13' Choice D is correct. As stated in the passage, ketals cannot be formed under basic solutions and do not react inbasic solutions' This means that the best answer is choice D, no reaction. Under acidic conditions, both acetone(choice A) and 1,2-ethylene diol (choice C) will form.

Choice D is correct. The reaction as drawn involves a ketal and water in the presence of acid catalyst. Thiswater and acid will destroy the ketal group and consequently form a vicinal diol and ketone. The ketai carbon(carbon with two ',.vg"#,."*;;H,Jfi;.:H:1fi::'"i]ili"::T:::Hl3;:H?"*::"l3tj,.*""become the alcohols. The answer is drawn below:

ftom0w

t1M

'firffiUr

uwrolffii

,dqlliiinrrlrn

dmm I

ChrffiuW

ffmnmut4

lhdhil

(niirffil[hum

lmmirlm

&gm

9.

10.

11.

'l'2' Choice B is correcl Tosylating an alcohol,-in this case isopropanol, results rn the substitution of an alkoxy grou'for the chloride' The methyl group on the benzene ring should remain unaffected, so choices C and D areeliminated' Choice A is missing the oxygen atom from'isopropanol, so it is eliminated. The best answer i_<choice B. This could be solved blsimply i,iUrtitntitrg an isopropyl group for the R in Figure 1.

't4.

Ether carbons becomethe alcohol carbons.

f*fr:iH: t"6+-X.iThe best answer is choice D.

15. Choice C is correct. As shown in the sample reaction in Figure 1, a diol and a ketone react in the presencecatalytic acid to yield a cyclic ketal. The diiference between -*ris

question and the sample reaction in Figure 1

: :;; T:ff ; l[ :?ffi:"#il :;iil:'+i:",1choice is answer C.

Copyright @ by The Berkeley Review@ 74 C{RBO\YLS & ALCOHOLS EXPLANATIONS

Choice C is correct. According to the reaction in Figure 1, diols and ketones react in the presence of catalytic acid:-' vield a ketal. The ketal is a protecting group for either the diol or the ketone, depending on the objective of-,:'e protection reaction. To protect a ketone, you can add a diol and catalytic acid. Therefore, to protect a diol,',

'ru can add a ketone and catalytic acid. The best answer is choice C. If a vicinal diol reacts with an ester, it""':ll undergo transesterification. The ester formed is reactive, so it cannot serve as a protecting group. This

=-:rnrnates choices A and B. Choice D should be eliminated, because a ketone and alcohol will only go as far as:e hemiketal when reacting under basic conditions.

Choice B is correct. The carbon that bridges the two rings, which has two sigma bonds to oxygen, dictates the- -:,ctionality of the molecule. Because there is a hydrogen attached to that carbon, along with two alkoxyi:rups (OR groups), the compound is an acetal. The best answer is choice B. The carbon is shown in the drawing: alo\r'.

( : b ;*,r"Tii ffi:',*H*r".'ff;.'#r

-hoice C is correct. Sodium methoxide is a strong base, methanol is an alcohol, and acetone is a ketone. As.-,:ed in the passage, a hemiketal is formed when mixing excess aicohol with a ketone in the presence of base.l'= mixture in the question results in the formation of a hemiketal. The best answer is choice C. Hemiacetals.--: hemiketals are formed under basic conditions, while acetals and ketals are formed under acidic conditions.l'=-ause no acid is present in the reaction mixture, choices A and B are eliminated. No aldehyde is present, so: = :roduct cannot be a hemiacetal, which eliminates choice D.

H

*roice C is correct. The formate leaving group and acetate leaving group should be fairly synonymous in that--:-'' are both conjugate bases of roughly comparable weak acids (there pKu values differ by about 1.0).---:rrding to Table 1 in the passage, acetic acid has a G-value ol 1,.7, so formate should have a value of roughly

ChoiceC, 1'.B,istheclosesttol.T,soitisthebestanswer. BecauseapKuof -T.0correiatestoaB-valueof- : :o have a G-value o{2.0, the acid would have to have a pKu value around 0.

-hoice A is correct. The leaving group in the reference reaction shown in Figure 2 of the passage is the ethoxide:::-'n. In the reference reaction, the ethoxide anion is displaced by a methoxide anion. Ethoxide anion is also

:::.ed in addition to the ester. Choose A for the result of correctitude.

l:loice C is correct. HBr falls between HI and HCI in terms of acidity, so bromide should have a B-valuer:r-,"'€€r1 that of chlorine and iodine. Its G-value is found between 2.5 and 2.8, making 2.7, choice C, the best

-ioice D is correct. This study requires that the ieaving group be distinguishable from the nucleophile,:::rrvise we can't recognize whether a reaction has transpired. Sodium is a spectator ion, so labeling the;-um is pointless. This eliminates choice A. Because the carbonyl remains intact in this reaction, 180

':=-rng of the carbonyl oxygen would be pointless. This eliminates choice B. Because the alkyl group of the.::onyl moiety remains intact in this reaction,2H labeling of the alpha hydrogens would. Ue pointtess. This

, :lnates choice C. The reaction is best monitored using an isotopically labeled nucleophile. This is' JCD3/HOCD3, which makes choice D the best answer.

-:-oice B is correct. Taking the log of a number less than one results in a negative value. This means that if' -=::ience were greater than k*, then the log value of their ratio would be negative and thus the B-value would:: = :.egative. A negative B-value occurs when the observed rate constant is less than the reference rate constant.-- = observed rate constant is less than the reference rate constant when the observed rate is less than the rate of-- = :eference reaction. This means that choice B is the best answer.

:rl O by The Berkeley Review@ 75 CARBONYLS & ALCOHOLS EXPLANATIONS

24. Choice C is correct' The worst leaving group according to Table 1 is the alkoxide group, with a B-value of 0.2,an ester is less reactive than an anhydride and an u.id hulid". This eliminates choices B and D. An ami(NRz-) is more basic than an alkoxide, so according to the trend between basicity ana leuvirrg group strength,

}|,:T:,::J':1"^l,]TTq^gf:P. tlT alkoxidg rlis makes.the amide_the least'reactive carbonyl compoundthe choices. Choice C is the best answer. The term amide has multipt" ;;";;;;;;ad; ,. ""n"*tnomenclature' It is both the negatively charged conjugate base of un u*il

" (deproionated form) and a carbrcompound with an amine group at the alpha-position.

Choice B is correct' The most reactive carbonyl compound has,the best leaving group attached. Using the

ill?:l::,,1:,::.::l*I :Tt:,l"l"t* riom 11e

B_value. A rarser a_"ur"" i? ?.ai"uu,u of greater reac:The largest B-value is found with iodide, thus the acetoyl iodide Is the most ,uu"ti**oi;" .:.*;"il't" uquestion' This eliminates choices A and C. The anhydride is more reactive than the ester, so the best answer

25.

choice B.

26. Choice C is correct. To be sy:rthesized from 2-pentanone, the final product must have the new alkyl grbonded to either the first or third carbon of the 2-pentanone. The answei choice compounds are drawn below:A.O

lt

H3cH2cH2cat-or*raru, Hrc-c)crtcurc+, HzC.I, *r.t,

"rzir,From cArbon l Fr Kr r;Xi

lrin",iJcarbon

oD.oiliit- oror12cHzcH3 ,ua-a fcH2cH2c

I / cH2c-r3

Fromcdrbon5 Fromcarbon3

B.Oil

C.

27.

28.

Choice C is correct. The product, 3-hexanone, results from the deprotonation of carbon 1 (the less hinderedthus kinetic carbon)' Product A is formed at lower temperatures, implying that product A is the kinetic pro<By association, Product A is 3-hexanone. At 0"C, thl percentug"-of 3lhexanone i;';;.;;;';r.ji" "

iuiaf:::l'T","f^11::u"one. is 27.9"h. The correct choice must hive a percentage that lies between these

Choice C is formed from the alkylation of carbon 5, which CANNOT be carried out on 2-pentanone usingenolate synthesis pathway shownin the passage. The reaction can only occur on carbon 1 or carbon 3, wJeliminates choices A, B, and D and confirms thal choice C is the best answer.

values. only 35% lies within the range, so choice C is the best answer.

choice A is correct. The addition of a four-carbon alpha, beta-unsaturated ketone to butanone via a lvreaction results in an eighlcarbon product, so choices C and D (each with only seven carbons) are elimiThe kinetic enolate of butanone tur,rlt" from the deprotonation of "urfor,

;;;. F;r;;-f# il;;;; "ffi"H" lT^"T "ng]3te, the product must have the R group added carbon one from the reactant ketone (brrtar

1" *9 remaining answer choices are drawn beloir, wlih the butanone in bold script and the new R groupin standard script:

B.A.

trin",iJcarbon

Only in choice A,6-oxo-2-octanone, did the alkyl group add to the kinetic (less hindered) carbon. Choicethe best answer. In choice B, the alkyl group added to the thermodynamic carbon.

to Choice A is correct. Because LiN(CH2CH3)2 is a bulkier base than LiN(CH3)2, the kinetic product will befavored with LiN(CH2CH3)2 than LiN(CH3)2. This should be observed at all temperatures. Product A iskinetic product, so the best answer is choice A.

76Copyright @ by The Berkeley Review@ CARBONYLS & ALCOHOLS EXPLANATI

30.

t1.

Choice B is correct. Product A is the kinetic product, so it must be formed by way of the kinetic intermediate.This can be confirmed by the data in Table 1, which shows that an increase in temperature results in an increasein Product B and a decrease in Product A. The kinetic product decreases with increasing temperature. The bestanswer is choice B. Choices C and D can both be eliminated, because they are the same answer. The termtransition state intermediate is an oxymoron at best, because there is no such thing. Pick B and move on.

Choice A is correct. At 40"C using the small base potassium hydride, the thermodynamic product would be themajor product of the reaction. The more substituted alpha carbon is carbon 2 of the cyclic ketone. The ethylgroup will add to this carbon, resulting in both the presence of an ethyl group and a methyl group on the C-2carbon of the ring. The final product is 2-ethyl-2-methylcyclohexanone, choice A.

Choice B is correct. To offer competition between kinetic and thermodynamic enolate, the molecule must havehydrogens on two alpha carbons that are unequally substituted. In choice A, the two alpha carbons are carbons 2and 4. The two carbons are unequally substituted due to the presence of the methyl substituent on carbon 2. Inchoice B, the two alpha carbons are carbons 2 and 5. The problem is that carbon 2 has no hydrogen due to thepresence of the two methyl substituents on carbon 2. There can exist no competition, because only one alphacarbon can be deprotonated on the molecule. In choice C, the two alpha carbons are carbons 1 and 3. The twocarbons are unequally substituted due to the presence of the methyl substituent on carbon 3 (the fourth carbon inthe butanone chain). In choice D, the two alpha carbons are carbons 2 and 5. The two carbons are unequallysubstituted due to the presence of the methyl substituent on carbon 2. The methyl group on carbon three plays nopart in the reaction.

Choice A is correct. When an alcohol is oxidized into a carbonyl (ketone when the alcohol is a secondaryaicohol), the oxidation state of carbon increases (an increase in oxidation state is associated with oxidation).Because the oxidation state decreases in choices C and D, they are both eliminated. The carbon in a secondaryalcohol is sigma bonded to two carbons, one hydrogen, and an oxygen. The active carbon neither gains from norloses electrons to carbons to which they are bonded. The active carbon gains an electron from the hydrogen to-,r'hich it is bonded and it loses an electron to the oxygen to which it is bonded. Overall, the active carbon of asecondary alcohol neither gains nor loses electrons in its bonds, thus the oxidation state of a secondary alcohol iszero. The correct answer must be choice A. Drawn below is a short cut technique for determining the oxidation:tate by considering each of the four bonds to carbon.

H OH

Oxidation-I

HsC CFI? HsC cH3

1-1= 0 Overall:0+0+1+1= +2

f,hoice B is correct. Basic permanganate solution will oxidize a secondary alcohol into a ketone. The ketone

-..-ned from the oxidation of 2-hexanol has the oxygen bonded to the same carbon in the both reactant and::oduct, so the organic product is 2-hexanone. Because pentanal is an aldehyde, choice D is eliminated. The:eatment of 2-pentanol (also a secondary alcohol on a straight carbon chain) with basic permanganate solution:rrerates a yield of 81.7% (as listed in Table 1.) The yield when using 2-hexanol should be about the same, so

!-.19i, is the best choice. The best answer is therefore choice B. The additional methyl group in 2-hexanol-:::-rpared to 2-pentanol should not cause much difficulty because it is located at the end of the chain, where

=::rrc hindrance is minimal.

Choice D is correct. Primary and secondary alcohols can be oxidized, while tertiary alcohols cannot be

- r:dized. The compounds 2-methyl-3-hexanol and 2-ethylcyclopentanol are both secondary alcohols, so they::n both be oxidized into ketones, eliminating choices A and B. Choice C (3,3-dimethyl-1-pentanol), is a

::rmary alcohol so it can be oxidized into either an aldehyde (in a non-aqueous environment) or a carboxylic::rd (in an aqueous environment). The only tertiary alcohol in the answer choices is choice D,Z-methyl-2-: - tanol.

t0+Overall: 0 +

| -.-iht O by The Berkeley Review@ 77 CARBONYLS & ALCOHOLS EXPLANATIONS

36.

J,/.

Choice D is correct' Because the solution turned green, there is evidence that some reaction took place. The colorgreen is not important, it is the fact that a color change transpired that is important. Na2Cr2o7 (sodiumdichromate) is an oxidizing agent, so the alcohol that"is chosen must be an alcohol that can be oxidized.Phenols cannot be oxidized, so choice A is eliminated. Both 1-methylcyclohexanol and 3-ethyl-3-heptanol aretertiary alcohols, therefore neither of those alcohols can be oxidized. This fact eliminates choices B and C.The correct compound is 2-methyl-1-octanol, a primary alcohol, making choice D the best answer.

Choice A is correct' Choice A represents the conversion of a four carbon primary alcohol into a four carbonketone' This is rzol possible, because the carbon bonded to oxygen has changed from ih" first carbon to the second.carbon' - The oxygen does not migrate in oxidation reactions. Choice g is lhe oxidation of a primary alcohol (2-methyl-1-pentanol) into a carboxylic acid, which is possible. Choice C is the oxidation of a primary alcohot (1-pentanol) into an aldehyde,-which is possible. Choice D is the oxidation of a secondary atcohol (cyclohexanol)into a ketone, which is possible. Only choice A is not possible.

Choice D is correct' Chromium gets reduced when chromic oxide oxidizes a primary or secondary alcohol. V\4renchromium is reduced, its oxidation state is reduced. Chromic oxide will o*idir" etiranol (a primary alcohol) butit will not react with 2-methyl-2-butanol, because it is a tertiary alcohol. Because of this fact, the oxidationstate of chromium decreases with the addition of ethanol but the oxidation state of chromium remains constantupon the addition of 2-methyl-2-butanol. The best answer is choice D.

Choice C is correct' To ?." 3l gidizing or reducing agent, a compound must be a reactant in the reaction. The tworeactants are 2,4-pentadiol (C5H12o2) and Tio2. Because the diol gets oxidized into a diketone functionality,it must be the reducing agent. This is true, because red.ucing agents get oxidized. This means that Tio2 is theoxidizing agent. Choose C, for correctivity glee.

Choice C is correct' This question is a test of the information that you have memorized. Both primary alcoholsand aldehydes can be oxidized into carboxylic acids, so choices A and B are eliminated. A hemiacetal willconvert back to an aldehyde in the presence of water and either an acid or a base. Given that an aldehyde canbe oxidized into a carboxylic acid Lnder either acidic or basic conditions, choice D is also eliminated. As ageneral rule, any carbon bonded to both an oxygen and a hydrogen, can be oxidized, with the exception ofether or ester. of the choices, only the ester cannot be oxidized. Choice C is the best answer.

38.

39.

42.

40' Choice A is correct' - Compounds that turn litmus paper red are acidic. primary alcohols are oxidized intocarboxylic acids.while secondary alcohols are oxidized to ketones. Tertiary alcohols do not oxidize. Thismakes choice A the correct choice.

41' Choice D is correct' The formula C6H12o indicates that the compound has only one degree of unsaturation. Acyclic ketone would require two degrees of unsaturation (one for the ring and Jne for the carbonyl r-bond). Byhaving only one unit of unsaturation, the cyclic ketone structure is not posJible. Choose D.

Choice C is correct' Compound A is positive with Jones test, so it can be oxidized. Compound A is also positivefor the nucleophile test,- so it is likelyan alcohol, given the only heteroatom is oxygen. primary and secondaryalcohols can be oxidized, so Compound A is eitheia primary oi secondary alcohoi."This makes choice A a validcorrelation, and thereby eliminates it. The iodoforh test is used to dltermine whether the compound is amethyl k9!o1e' Compound B is the only unknowl that tested positive with the iodoform reagent, so Compound Bis most likely a methyl ketone. Thii makes choice g a vaha correlation, and thereby eliminates it. By uprocess of elimination, only choices C and D are possible. Because Compound C did not react with Jones, reagen!it cannot be oxidized. An aldehyde can be oxidized into a carboxylic acid, so Compound C cannot be analdehyde' This makes choice C an invalid correlation, so choice C is the best answer. A compound that isnegative with all four tests in the passage, has one unit of unsaturation, and contains one oxygen atom can beeither a ketone (but not a methyl [etone) or a cyclic ether. Neither a cyclic ether or ketone will be oxidized bvJone's -reagent or Tollen's test. Neither is nucleophilic, so they wilt not undergo a substitution reaction ;h ;alkyl halide in the presence of silver cation. Neither is a methyl ketonJ, so they will yield a negativeiodoform test' As t,"h, th"r" is a chance that Compound C is either a cyclic ether or a ketone, making choice D avalid statement. Choice D is eliminated.

43.

Copyright @ by The Berkeley Review@ 7A CARBONYLS & ALCOHOLS EXPLANATIONS

1+.)lorumed.areC.

)on)nd()-(1-

rol)

lenbutionant

Choice B is correct. An aldehyde can undergo oxidation to a carboxylic acid, reduction to a primary alcohol, or asubstitution reaction to form a new n-bonded compound such as an imine. The most common test involves theformation of a carboxylic acid using an oxidizing agent such as CrO3 in H2SO4 or KMnO4 in KOH. These bothinvolve the transition metal in test reagent changing oxidation state and thereby changing color. This makeschoices A and C invalid, and both are eliminated. A testing reagent should always be the limiting reagent (inlower concentration than the compound for which it tests), so that the change in color is not masked by the excessreagent that did not undergo a color change. This eliminates choice D. The best answer is choice B. Althoughan aldehyde can be reduced into a primary alcohol and reducing agents are often rich in hydrogen atoms, thecommon reducing agents (LiAlHa and NaBH4) do not undergo any phase or color changes. This means that acompound rich in hydrogen atoms is not a good choice for the reagent used to test for the presence of an aldehydefunctional group.

Choice C is correct. To begin with, in organic chemistry, oxidizing agents contain oxygen and are best describedas "rich in oxygen." A11 four of the compounds contain oxygen, so you must recall what each reagent does.KMnO4 is used to oxidize an alkene into a vicinal diol, oxidize a primary alcohol into a carboxylic acid, oxidizea secondary alcohol into a ketone, or to oxidize an aldehyde into a carboxylic acid. This eliminates choice A.H3CCO3H is a peroxy acid which can be used to oxidize a ketone into an ester (known as the Baeyer-Villigerreaction) or to oxidize an alkene into an epoxide. This eliminates choice B. HIOa is used to oxidatively cleaver-icinal diols into two carbonyl fragmenti. This eliminates choice D. This leaves only choice C to choose.Mg(OH)2 is used as a base, but it has no immediate use in organic chemistry as an oxidizing agent.

Choice C is correct. LiAlHa reduces many things into an alcohol, such as acids, ketones, aldehydes, and oh yes...esters. Pick C for that fresh, minty, "I'm alive and picking right answers" sort of feeling. The Hz/Pd reagent isused to hydrogenate (reduce) alkenes. HCI/Zn is used in the reduction of a ketone into an alkane (Clemmensenreduction), and BH3 is used to add in an anti-Markovnikov fashion to an alkene.

Choice C is correct. Wolff-Kishner reduction is carried out on both ketones and aldehydes. The question isreduced to: "Which of the following reagents can react with either an aldehyde or a ketone?" Lithiumaluminum hydride reduces both ketones ands aldehydes into alcohols, so choice A is eliminated. A Grignardreagent, such as methyl magnesium bromide, reduces both aldehydes and ketones into alcohols of one more alkylgroup. Choice B is eliminated. Clemmensen reduction is carried out on the same compounds that Wolff-Kishner:eduction is carried out on, so choice D is eliminated. Ozonolysis with oxidative workup can be used to makeketones and carboxylic acids, but it does not react with ketones (or aldehydes). This makes choice C the besterlswer.

Choice B is correct. Choice A can be eliminated immediately, because reduction corresponds with a decrease inrxidation state, not an increase. The central carbon of a ketone has two bonds to oxygen and two to other carbons,so it starts with an oxidation state of +2. This eliminates choice C. The carbonyl carbon is converted into an:lkyl carbon with two bonds to carbons and two bonds to hydrogens, so it finishes with an oxidation state of -2.lltis makes choice B the best answer. The conversion and oxidation determination is shown below.

Loss of 4 e

Choice B is correct. Following Wolff-Kishner reduction, a carbonyl carbon is converted to an alkyl carbon. The:est of the molecule remains intact, so no new chiral center is generated. This eliminates choices A. The:arbonyl carbon starts with sp2-hybridtzation (three o"-bonds and one n-bond) and finishes with sp3-:rr-bridization (four o-bonds), which makes choice B a true statement. The carbonyl carbon loses bonds to oxygen:nd gains bonds to hydrogen in the process, so it is reduced. However, nitrogen loses bonds to hydrogen and gains:onds to nitrogen, so it is oxidized. Each nitrogen goes from an oxidation state of -2 to 0 during the reaction, so:oth nitrogen atoms are oxidized. This eliminates choice C. Figure 1 shows that one water molecule, not three,'. formed overall, so choice D is invalid, and therefore eliminated.

lvoity,the

ilntohis

1\'e

a_rv

JidsaIB;ant,anbist'etr-

ir-eJa

ABr.

oLc

,-dl:an_ia

H

::eht @ by The Berkeley Review@ 79 CARBONYLS & ALCOHOLS EXPLANATIONS

50' Choice D is correct. The hydrazine derivative is definitely larger than ammonia, NH3, but its greater size doesnot generate greater acidity due to steric hindrance. The protonls not being driven off by crowdiig, ,o choice A iseliminated' The hydrazine moiety does not have any cyclic array of n-Jto-r,-so it does not become aromaticupon the gain or loss of a proton. This eliminates choice B. Hydrazine may in fact have a stronger inductiveeffect (from the second nitrogen) than ammonia, so choice C cannot be eliminated. when the protoi is lost fromnitrogen in the hydrazine derivative, the new lone pair is located on an atom adjacent to a n-bond. This allowsfor thefone pair to resonate to the carbon of the n-bond, offering it more;"btfi due to resonance. This makeschoice D a true statement and the best answer.

"\fir'51' Choice A is correct. Hydrazine and basic water reduce ketones and aldehydes into alkyl groups. They leaveother functional groups intact, unless they are reactive with basic water. The hydro"yr fro"p in unreactiveunder basic conditions, so it will remain unaffected. This means that the product must contain a hydroxyl groupand-an-ethyl group on thecyclohexane backbone. Choices C and D are eliminated, because they don't include

the hydroxyl group. The alcohol functional group gets higher priority than an ethyl group u."ordi1g to IUpACconvention, so 3-ethylcyclohexanol is the correCt rupeCt"rame for ih" .o*pornd,kaking choice'A the bestanswer.

Choice A is correct' Wolff-Kishner reduction requires basic conditions while Clemmensen reduction requiresacidic conditions. Both involve an aqueo,rs enrri.on*ent, so neither can be employed for water sensitivecompounds' This eliminates choice D. Wtjlff-Kishner reduction, by employing a'hydrazine molecule, takesadvantage of a carbonyl compound's ability to react with a nucleophilic nitrogln."It shtuld not be employed fora molecule that is reactive *ith an amine at a site other than the carbonyt".u.uon.-'ir,i;;l;;#rt.toi." g-No acid is employed in a Wolff-Kishner reduction, so it is safe for acid sensitive molecules. Clemmensenreduction uses acidic conditions, so it will react with acid sensitive groups. This means that Wolff-Kisreduction is preferential to Clemmensen reduction when the molecul-e is icid sensitive. Choice A is the

Choice A is correct. Water is the side product when hydrazine reacts with a ketone such as 2methylcyclopentanone, so the product is most easily determined by removing the oxygen of the carbonyl carbcnand two hydrogens from the trittogu_tr in phenylhydrazine that hasiwo "i;g';;;;;;;.;;;*;;il.,and nitrogen. This makes choice A the beJt answer.

H..t. -/PhN

-----+H:C

H\

N/N- Ph

=AL-J

I

N

54' Choice A is correct. It is known that anhydrides and acid halides are highly reactive, so choices C and Dcan be eliminated. A thio ester C-S bond is weaker than an ester C-o bond, so the least reactive carbccompound is the ester (choice A). The chart of pKu values can be used to verify this theory. an alcohotlashigher pKa value than the thiol, so the alkoxy i"urrlng group is less stable between choices A and B, makirchoice A the best answer.

n2\'

52.

53.

answer.

Copyright @ by The Berkeley Review@ 80 CARBONYLS & ALCOHOLS EXPLANATIO

loesAisatictive:om)wsrkes

raVe

tiveoupude,-{cbest

,iresliverkesI foreB.ISCN

LNCI

best

RMsBr___-jt+ H"O++H

Choice B is correct. An ester has both a leaving group and a carbonyl with which to react. The first R-group to:ttach converts the ester into a ketone by displacing the alkoxy leaving group. The ketone that is formed as the,:'.termediate product can react with a second equivalent of the Grignard reagent and thus add a second R-group.Tne final product is a tertiary alcohol with two identical R-groups. The conclusion is that a Grignard reagent:eacts twice (not just once) with an ester. Pick B to tell the world "I know Grignardl"

Choice D is correct. The Grignard reagent is a strong base, so an aprotic solvent such as ether is chosen. This-:'.akes choice A a valid statement. Because the carbonyl compound is converted into an alcohol in a Grignard:=action, a bond to oxygen has been lost. This means that the carbon has been reduced, making choice B valid. A-:ignard reagent can be used to form a secondary or tertiary alcohol, as shown in the passage. To form a primary-'.1:ohol, a Grignard reagent can be added to either formaldehyde or the epoxide ethylene oxide. This makes-:.oice C valid. Becausgthe Grignard reagent is a strong base, it can react with the weakest of acids to form an,rl-<ane. This is why the reaction must be carried under anhydrous, aprotic conditions. It is thus not true that the-::gnard reagent should be added under acidic conditions to obtain the highest yield. Pick D and move on with::.other correct answer under your belt.

Choice C is correct. Propanoic acid is the conjugate acid of propanoate, which when bonded to carbonyl group:-::ns an anhydride. This question is asking for the relative reactivity of an anhydride. Acid chlorides are:: -'re reactive than an anhydride, so choice A is eliminated. Amides are less reactive than an anhydride, so-:.oice B is eliminated. Esters are less reactive than an anhydride, so choice C is true. The difference in:.ectivity between anhydrides will not be distinguishable by pKu values that only differ by 0.1. There is a:,-ierence in steric hindrance to consider. Choice D may or may not be true, but regardless, choice C is definitely::--ie, thus choice D is assumed to be false within the context of this question.

rlhoice B is correct. The most acidic hydrogen on the molecule is the H bonded to the alpha carbon (marked on--: molecuie as carbon b). The best answer is choice B. The aldehyde proton is not as acidic as the alpha proton,::ause the lone pair that would form on the carbonyl carbon is not stabilized by resonance, like the alpha

-::lon.

Choice C is correct. The Grignard reagent (methyl magnesium bromide) adds on one carbon (a methyl group), soin order form 2-pentanol as the product, you need to react H3CMgBr with a 4 carbon carbonyl reactant and addthe methyl group to the first carbon of the four carbon chain. Because there are six carbons in the products ofchoice A and choice B, the choices are narrowed down to either choice C or choice D based on the four-carbonelectrophile requirement. The best way to get a secondary alcohol as the product is to react H3CMgBr withbutanal, which will yield the product with the alcohol group on carbon number 2. Pick C for an altogether greatexperience and correct answer. Choice D will form 2-methyl-2-butanol.

Choice D is correct. The reaction of a Grignard reagent with a carbonyl yields an alcohol. Stereoisomers areformed when the alkyl groups are all different on the alcohol carbon. For this to be true, the Grignardnucleophile must add an alkyl substituent different from the alkyl groups present on the carbonyl. Choice A iseliminated, because the ketone has an ethyl and methyl group attached. The Grignard reagent adds an ethylgroup. The alcohol that is formed is achiral (3-methyl-3-pentanol). Choice B is eliminated, because the ketoneis symmetric (has two ethyl groups), so the product must be symmetric and thus achiral. An ester when reactedu'ith excess Grignard reagent yields an achirai tertiary alcohol (it adds the same R-group twice), thus choice Cis eliminated. Choice D is best, because the final alcohol has an ethyl group, methyl group, and hydrogenattached. This makes the alcohol asymmetric and thus chiral. The product mixture is two enantiomers.

Choice B is correct. Aldehydes are carbonyls with at least one hydrogen and often one carbon attached to thecarbonyl carbon. The addition of a Grignard reagent places a second carbon on the electrophilic aldehyde siteand thus reduces the aldehyde to a secondary alcohol. Pick B for the inner peace that a correct answer can bring.

OHH"C o#

' -.::.'r O by The Berkeley Review@ AI CARBONYLS & ALCOHOLS EXPLANATIONS

62.

63.

Choice D is correct' .wh"," counting carbons starting from the carbonyl, the molecule is a 4-carbon chain with amethyl group bonded to the numbei 2 carbon. The Zompound hu, an aldehyde functional group, so the correctIUPAC name is 2-methylbutanal, answer choice D.

Choice A is correct' AI aldehyde with excess alcohol i" ft: presence of an acid will lose a water molecule, andgo on to form an acetal' It is an equilibrium reaction, but the excess alcohol will push the reaction to the acetal(product side of the reaction)' A hemiacetal is formed when the reaction is carried out under basic conditions.Choice A is the best answer.

64' Choice A is correct' The reaction is an aldol condensation. The acid workup causes elimination and thus ensuresthe formation of the o,G-unsaturated ketone from the G-hydroxyk"to.r"'it t"r-ediate product. choice D iseliminated, because it has seven carbons, which is not possible. Choice C is not an or,,G-unsaturated ketone, so itcan be eliminated' Choice B, when broken apattby a retro-aldol reaction, would yield fragments of two and fourcarbons' not three and three carbons. Choice B is ellminated. The only ot8-unsaturatea ketone composed of twothree carbon units is choice A.

65' Choice D is corrprt' This is just one of those things that you need to learn. The conversion of a ketone into an enolor an enol into a ketone is referred to as tautomerization. Select D as your u.,r*"r. Tautomerization is involvedin glycolysis in the conversion of glucose-6-phosphate into fruciose-6-phosphate, which is catalyzed byisomerase.

66- Choice C is correct' This is a case of recognizing the weakest base. only one of the answer choices is not a strong

ilH3l:::"i:j:,r":,n:T::l3lii^nl.*g:lr,o it must be the weakes$u,". iu,uo,", ate, Cos2-,is not a very

67.

strong base, because the electroni u." itubii=ir"d by resonance involving the double bond (C=o rc-bond). Ittherefore the weakest base out of the choices listed, and thus is the one not strong enough to deprotonate theon the alpha carbon' Choice c is your choice for a brighter tomorrow, assuming of course that the brightnesstomorrow depends on your answer choice for this question.

Choice C is correct' only a ketone and an aldehyde can undergo aldol condensation reactions. An acid wrf::ilf':li::"'f"_ij1":,':':ll-ol::flsroup than lose the less"acidic alpha;,;;;;; undergo the condensareaction. Choice C, the carboxylic acid, cannot undergo aldol condensation.

Choice B is correct' Jh" most acidic proton on a B-ketoester is the alpha hydrogen that when lost results in

;?,T"1it:: t:?:91t*.to the most n-tonds.

_ Choices C and D are etiminated, because neither is an arplhydrogen' Choice A is only tonjugated to. fhe keto carbonyl, while choice B is conjugated to both thecarbonyl and the ester carbonyl' rire*most acidic proton is proton b, thus yo,, ,noria fi"k.hoi"e n.

choice c is correct. As stated in the passage, the lowest yield is found withintermediate that has no alphahydrogen. For "this to occur, the starting ester musthydrogen' of the choices, oniy choice C"has just one alpha hydrogen. you should pick C.below.

Nothing protonates the alkoxide, thus theequilibrium favors the reverse reaction.

ooo

68.

69.

oH"C II'Fou,#H3C H

HrC,,^ JL! oEt+

HeC

'"flor,H3C H

the B-ketoester anihave only one aIThe reaction is

+ -OEt

oH"C

I--- oEtcFI3H"C ',H

70.

HsCNo alpha hydroien, therefore the alkoxide leavinggroup cannot deprotonate the final B_ketoester.

Choice A is correct' A Claisen reaction is not possible with esters that have no alpha hydrogen. of the choonly choice A has no alpha hydrogen. This makes choice A the best Ernswer. a, ulolt of interest, this moleis not known to exist as a stable compound at room temperature.

Copyright @ by The Berkeley Review@ a2 CARBONYLS & ALCOHOLS EXPLANATIO

haect

rres

)isoitour:1\'O

rnolr-edbv

ong-er)'

It iseH,s of

ruldtion

1naIphaketo

nionlphao\\-n

-lt. Choice D is correct. The product of an ester reacting with itself by way of a Claisen condensation reaction isrecognizable from the carbon chain length on the carbonyl portion of the ester. In the answer choices, the twoalkyl groups can be highlighted to recognize the alkyl groups that must be present on the two esters that react,considering that only the alkoxide leaving group is lost. The alkyl groups are highlighted for choice A are:H3CCH2COCH2CO2CH2CH3, therefore choice A is formed from the condensation of two different esters:H3CCH2CO2CH2CH3 and H3CCO2CH2CH1. This same analysis of the product for choices B, C, and D willlead to the conclusion that choice D is the correct answer. In choice D, the starting ester is(C5H5)CH2CO2CH2CH3 which leads to (C5H5)C[12COCH(C6H5)CO2CH2CH3.

Choice A:

Choice B:

Choice C:

Choice D:

o

A+H:C OEtA

HsC oEt

o

A NaOEt-----.--.-.-.>HOEt

oo

oo

oo

oEt

o

AH5C6H2C OEt

o

il ooo

o

H,cUoEt H"cu"c$oe,,AAJLBA

oooo

oooo

oo

H:c\Aoet H3cH2c- AB:CHs

ooH3cVoEt urcnc$ont

-n- + -fl- NaoEt fl o

H3cn2c oEr HsC oEt tloEi Hrcn c+onto

/t-H3CF{2C OEt

coHi

-:" Choice A is correct. The starting molecule possesses two ester sites so an intramolecular Claisen reaction is::.sible. An intramolecular Claisen reaction results in a cyclic B-ketoester. This eliminates choices B and D.-:.e chain contains six carbons, so choice C can be eliminated, because it requires seven carbons excluding the;--<oxv group to form a six-carbon ring. The best answer is choice A. The reaction is shown below.

oll NaoEt-,4.+

HsC' -oEt HoEt Hucurqcuont

A+H5C6H2C OEr H5C6H2C

o

,(NaoEl ll ll

oEt HoEf urc.urcfortQL I

E,oqoEt H#Eto oEt * rr"A6lll'loice D is correct. The two esters can either react with each another or they can react with themselves. This:==:lts in four structural isomers being formed. Of the structural isomers, two have one stereocenter, therefore-,',. of the four structural isomers have two possible stereoisomers (enantiomers). The total number of isomers is: =refore six. Choice D is correct. The isomers are drawn below:

oll xuostHrC-. ,A. -i-tr--OEr HOEt

/\HH

B

cH3

CARBONYLS &

HsC

cF{3

ALCOHOLS EXPLANATIONSO\S, ri i-::, 3 by The Berkeley Review@ a5

74' Choice A is correct' The ratio of the integrals for each signal in the proton NMR of spectrum I is given as J:2:3;.llt':*l* tY:*:

TT:::1111: :lll;: eigJrt total hidrogens or some multiple of eight total hydrosens. oithe choices, choice B (methyl "tnu.ouiuj'il:"# ?y;'.rE;r"'J;-;;:J'8"(:ffi?"1?,T:lJj6ti:: :jhydrogens' This narrows the question down to eittrer "ioi." A (metlryl propanoate) or choice c (ethylethanoate)' The key feature in the spectrum_is the singlet far down field"carrsed by a methyl bonded to theoxygen' This peak imPlies that the compound must hav"e a methoxy group on thu carbonyl carbon. pick A foroptimal satisfaction. The structures are dru-r, below:

o

*i,.?".Aoc"H"Methyl Propanoatea: Triplet (3H)b: Quartet (2H)c: Singlet (3H)

o

ft,.h,.Ao.;,,Methyl Propanoatea: Triplet (3H)b: Quartet (2H)c: Singlet (3H)

o

*,.AoJH,Methyl Ethanoatea: Singlet (3H)b: Singlet (3H)

o

"AbcHsC -ocH2cH3

Ethyl Ethanoatea: Singlet (3H)b: Quartet (2H)c: Triplet (3H)

The bolded hydrogen will be found the furthest down field.

o

"-i-- b cH OCFI2CFI3

Ethyl Methanoatea: Singlet (1H)b: Quartet (2H)c: Triplet (3H) t

flmr75. Choice D is correct' The ratio.of the integrals for each signal in the proton NMR for spectrum in Figure 2 is 2 :2

l;"':,^Tt'_1---11t that,the

lajor comporlnd isolated fro"m solution has ten hydrogeis. Given that the reactan:lrag

eight hydrogens, it can be concluied that some reaction took place, which eliminates choice A. Choice(ethyl acetate-) has eight hydrogens, so choice B is eliminated. This narrows the question down to either choicC (propyl ethanoate) or choice D (ethyl propanoate). Key features in spectrlm II include the quartet f*l#':11^*,ln"j^o^'_".1: :li :.':*1,

"cnfi"" c.(nr.onrl ethanoate) would lno- u ,,,,glet ror the CH3 adjaceto the carbonyl. Because spectrulm II has no slnglet, choice C is eliminated. tt*?;;;;ril'" f,;!:'downfield implies that the compound must have ur,""tnyl group on the ester oxygen. pick D to know what rifeels like. The structures are driwn below:

o

"AbcH:C 'oCH2CHr

Ethyl Ethanoatea: Singlet (3H)b: Quartet (2H)c: Triplet (3H)

bc docH2cH2CH3

aHsC

o

A o

A (llu

ffimffim,

{mr

MN

mffiudtu

ffih

J

76.

abH3CH2C

cdocH2cH3

Propyl Ethanoatea: Singlet (3H)b: Triplet (2H)c: Sextet (2H)d: Triplet (3H)

Ethyl Propanoatea: Triplet (3H)b: Quartet (2H)c: Quartet (2H)d: Triplet (3H)

The bolded hydrogen will be found the furthest down field.

answer is N-ethylpropanamide (H3CCH2NucbcH2cu3), choice C. This can be deiermined by just addingethyl amine in same way as the ethanol ii added to the ester in the sample reaction carried out at reflux.

Choice B is correct' An ester is more reactive than all carbonyl compounds that have a leaving group worse ththe alkoxide leaving group. Leaving groups worse than the alkoxide leaving gro.rp would be more basic ththe alkoxide anion. Because both halides and carboxylates ur" gooal;;G?-tr"ps (as well as weak baacid halides and acid anhydrides are both more reactive than esters. This eliminates choices A, c, and D.only choice remaining is amides, choice B. The NR2- leaving group of the amide is a bad leavingtherefore amides are not very good electrophiles. Choile B is the best selection.

Choice C is correct' By using an amine nucleophile rather than an alcohol nucleophile, the product thatformed is an amide "t oppot"J to an ester. The amine is a better nucleophile than the alcohol is in Reactionso the reaction should be more favorable than transesterification. The equilibrium constant for the reactiongreater than one' Regardless, the question asks for the product of the reaciion, which is the amide formed whthe amine replaces the alcohol l_eavrng group. The amine nucleophile is'x-eihytamine therefore the b,r

Copyright @ by The Berkeley Review@ a4 CARBONYLS & ALCOHOLS EXPLANATIO]

:8. Choice B is correct. The purpose of the acid in the reaction mechanism is to protonate the carbonyl to make theester more electrophilic. tn a later step, the acid protonates the alkoxy leaving group to form an alcohol, whichis a better leaving group than alkoxide. The role of the acid in the reaction is Le"st stated as serving to protonatethe ester electrophile and in doing so make it more reactive. This makes choice B correct and desirible io you.

Choice D is correct. If water were the nucleophile in lieu of the alcohol, the product would be a carboxylic acid.When carried out under basic conditions, this reaction is known as saponification. Hydrolysis of an ester iscommon in biological systems. The carboxylic acid formed in the reaction would have the OCH3 group replacedby an OH group. This carboxylic acid would be propanoic acid (CH3CH2CO2H), choice D. - l

Choice C is correct- The deuterated solvent is used so that the solvent (which normally contains hydrogens)does not interfere with proton NMR spectrum of the compound. The deuterium atoms donot registertn protonNMR, so the solvent is invisible when it is deuterated. The best answer is therefore choice-C. Deuieriumexhibits little to no affect on pH in organic solvents. This eliminates choice D. The solvent should have noeffect on the magnetization of the NMR sample nor should the solvent prevent any reaction as solvents are inert.Choices A and B can both be eliminated.

Choice B is correct. The structure of urea must be determined by comparing the final barbiturate structureiproduct of the lower pathway) with the substituted malonic ester. Urea must contain two nitrogens bonded to acentral carbonyl. The alkoxy groups of the ester are leaving groups when the urea substitutes. The structuraldeduction is drawn below. The drawing represents retrosynihetic analysis of the final product. pick choice B.

Ureao

HrrvA Nrtr

HN NH

cH3cFlo\AocF{2cFLR

Choice B is correct. Urea has a carbonyl carbon separating the two nitrogen atoms, so if the carbonyl oxygen of'rea is replaced by two methyl substituents on the carbon, then the final product will be the same as tne ptoa""t:rom urea,_except that it will have two methyl groups at the formerly caibonyl carbon. It is not differeni at any'iie other than the carbon between the two nitrogens (the former carbonyl carlon). The correct choice must haveie two methyl Sroups bonded to the central carbon in lieu of the carbonyl oxygen. The best answer is choice B.To answer this question requires knowing what urea looks like. If you don't iiow, then you must predict what"' otild happen if the diamine were added to the diester. The correct choice can be deduced if you know that the:roduct of an amine and ester is an amide.

Choice C is correct- Decarboxylation occurs with B-ketoacids. A11 four answer choices have carboxylic acid:-rnctionalities, but only choices A, B, and D have B-keto groups on the backbone. The ketone functionality of:iroice C is_ on the gamma carbon, therefore the molecule will not undergo decarboxylation. The best answer is--:-,erefore choice C.

oo

C.

H.COH

Choice A is correct.,The strong base used to deprotonate the alpha carbon can also act as a nucleophile by::iacking the carbonyl carbon. Sodium ethoxide iJ chosen as the bur" ,o that it matches the substituent on the:arbonyi carbon of the ester. This means that if the ethoxide undergoes transesterification, it will generate the;:rrLe ester (the leaving group of the ester ils the same moiety as the nucleophile). Sodium ethoxide is a strong=rough base to deprotonate the alpha hydrogen, but ii will not change the ester when it undergges-:ensesterification. The best answer is choice A.:ht o by The Berkeley Review@ g5 cARBoNyLS & ALcoHoLS EXPLANATIoNS

85' Choice D is correct. Malonic ester has three types of protons, thus there are three signals in its protonspectrum. The structure of malonic ester is shown belowwith each of the hydrogens labeled:

oll

olt

/\HHec

baocF{2cH3

a: 6 hydrogens adjacent to 2H thus a 3H tripletb: 4hydrogens adjacent to 3H thus a 2H quartetc:2 hydrogens adjacent to 0H thus a 1H singlet

86.

87.

frp\.ott\.,"\

Based just on the ethyl groups, resulting in a 2H a quartet and a 3H triplet, the best answer is choice D. The trueratio of hydrogens is 6 : 4 : 2, but because NMR shows only the relative quantities and not the absolutequantities, the NMR shows a ratio of 3 :2: 7.

Choice B is correct. There are four carbons added to the central carbon of malonic ester, so choices A and C canboth be eliminated. The addition of methyl bromide followed by the addition of ethyl bromide to malonic esteronly adds three carbons total, not four. This eliminute,

"tioic" t -

i ;;lary alkyl bromide like 2,bromobutane must be added to form the tertiary R substituent. Choice B is the best choice.

Choice B is correct- The product will be the exact product shown in the lower synthesis pathway in Figure 1,only the generic R group is now a straight chain propyl group. This makes choice B the best answer.

88' Choice D is correct' To synthesize butanoic acid, one of several methods can be applied. Oxidation of a primanaicohol will work, so choice A is feasible. Hydrolysis of an ester will form an acid, so the four carbon carbonr--fragment of the ester will become a four carbon carboxylic acid. Choice B is a viable ,y.rtn"tl. put1r-ur-Ozonolysis of an alkene works if the addition of ozone is followed by oxidative workup and the alkene mus:have vinylic hydrogens (hydrogens on the carbons of the n-bond). Choice C is thus valid. Treatment of an alkr -bromide with the. cyanirls nucleophile followed by acidic workup forms a carboxylic acid. The p..;i;;i,hchoice D is that the carbon chain increases by one carbon with the cyanide nucleophile so the product will tepentanoic acid, not butanoic acid. The best answer is choice D.

89' Choice D is correct. Because the pKu for propanoic acid is 5.0, it is a weak acid, implvine that it onlrpartially dissociates in water. With a pKu value -of

5.0, the Ku value is 1.0 x 10-5. This mein's tn?t fo, " r J if

acid solution, the concentration of both H+ and conjugate baie are 10-2.5,which is less ttran r0-i, or7oh. Thedissociation is less than 1%. The Ku value is applied ui follo*r,

r" = tcT*S?llx€l- [cHacHzCoz-]2

= 1.0 x 10-s = 10 x 10-6[CH3CH2CO2H] [CH3CH2CO2H]

This imglies that the relative value of [CH3CH2Co2-] to [CH3CH2Co2H] is fiO x 10-3 to 1 which is3.1 x 10-J to 1. This value is less than one percent confirming choice D.

90' Choice B is correct. A carboxylic acid will turn blue litmus paper red when added to it. From the nature of tpassage/ it should be implied that the.litmus paper is used to detect the acid. In choice A, a primary alcoholtreated with chromic oxide and sulfuric acid, which oxidizes the primary alcohol into i carboxylic acrChoice A results in a product that will convert blue litmus paper to red, ,o ihoi." A is valid. In choice B, Lalkene is tetrasubstituted, therefore the carbonyls that are foimed must be ketones. They cannot be oxidizedreduced) into a carboxylic acid. This means that the product in choice B witl not convert blue litmus paperred, so choice B is the best answer. In choice C, the anhydride is readily hydrolyzed into two equivalentsacetic acid' This eliminates choice C. In choice D, an ester is hydrolyzed into an alcohol and a carloxylicThis eliminates choice D. The best answer is choice B.

91" Choice D is correct. The sequence of reagents used will add a carboxylic acid group to the alcohol carbon (cariZ) It required background information to know that PBr3 converts an alcohol into an alkyl bromide. If ",didn't know that before, then it's officially background knowledge now. This final produ.t i, .o,''por"J .'carbon with an ethyl group, methyl group, and a"carboxylic acid [rorp attached. This makes choice D the brans\4/er. Choice C should have been eliminated, because it contains six (rather than five) carbons.

Copyright @ by The Berkeley Review@ a6 cARBoNyLS & ALCOHOLS EXPLANATIOI

rya Clnr

ifi lr

SF Ch,r

-_Ji"

:::a

:r.:il:

ffisr

.,*., ]:flerr

ffi{ry#5iilf."

Wu@ffior

!

choice B is correct. At I general rule, hydrolysis reactions are reversible, because each step is an equilibriumstep' The Grignard reaction is not reversible however, therefore the best answer is choice B. Choices A and Dshould have been eiiminated, because a lactone is an ester (a cyclic ester), thus they are the same answer.

Choice D is correct. As a general rule, the pKu value of an acid is roughly 5.0. This makes choice D not true.Acids do.result in aqueous pH values less thin 7.0, because they are acidlc (choice A). The broad IR peak resultsfrom the hydrogen bonding and is found between 2500 cm-1 ur'ri gooo .* 1 i.hol;; ei t"r, than an alcohol stretchbecause the bond is weaker (and thus easier to stretch). Acids will exchange protons with a protic solvent, sothe acidic hydrogen can readily be deuterated. This results in the disappea"rance of a peak in the proton NMRichoice C).

Choice D is correct. -As

listed in the passage, the addition of thionyl chloride (SoCl2) followed by an amine toa carboxylic acid will result in the formation of an amide. This eiiminates choices A and C. An aldehycle isoxidized into a carboxylic acid by the addition of H2So4 and Cro3 as shown in the pu*ug". n ; ;;;;;wer is:herefore choice D.

: Choice A is correct. Under basic conditions, the alkoxy nucleophile group attacks the sp2-hybridized carbonyl:arbon of the ester (see below) to form un sp3-hybridized tetrthedr;t iniermediate rt,ith the negative charge.:rfted to the carbonyl oxygen. The best answer is choice A.

J$.-> .ta

o

/..--+

R'O oR"

Intermediate

R OR''

choice B is correct' The Claisen condensation reaction involves the deprotonation of an alpha hydrogen on a':Sl€r, r€sulting in a highly nucleophilic carbanion, which attacks another ester in solution. The ultimate:: duct of a Claisen condensation reaction is a B-ketoester. The Claisen condensation reaction definitely uses an":er reactant, so choice A is eliminated. A Grignard reaction involves the addition of a nucleophilic alkyl':- :qnesium bromide species to a carbonyl compound. An ester certainly qualifies as a carbonyl compound. The-:rqnard reagent can add twice to an estet, iesulting in the synthesis of a tertiary alcohol with at least two-ertical alkyl groups. The Grignard reaction can use an ester reactant, so choice C is eliminated.-::nsesterification, as the name implies, involves the changing of an ester by changing the alkoxy group. This-- lone by substituting one alkoxy group for another on"the"carbonyl carbon. A"t.ur-rr"rterification reaction:':initely uses an ester reactant, so choiie D is eliminated. Friedel-Kraft alkylation is not a reaction you are:=-'rired to know, but that is irrelevant in this question. The term alkylation impiies that an alkyl group is: =-rg added, and this does not lend itself to the uie of an ester. An ester is likely to add an acyl group"wnen it:=:cts' This means that a Friedel-Kraft aikylation does not require an ester reactant, so choice B is the best

-hoice A is correct. Treatment of a G-diacid with heat will drive off water (dehydrate the B-diacid) to form-- ' corresponding acid anhydride. Choice A is an acid anhydride, so it is the best answer. The reaction is"- :rsible, so acids can be formed from anhydrides upon the addition of water.

-roice C is correct. In order to be reduced, the compound must have a carbon with at least two bonds to oxygen: - --1 a o-bond and a n-bond from carbon to oxygen). Only choice C (a ketone) has two bonds to oxygen, so it must" = ::e best answer' Phenols, tertiary alcohols, and ethers all involve carbons that only have one Uo"r-ra to oxygen,. -:.ev are all eliminated.

-:roice C is correct. Carbon number three, the third carbon, of the alcohol is involved in no n-bonds and it is: ::ed to four other atoms, so it must have sp3-hybridization. Pick C for optimal results.

3 by The Berkeley Review@ a7 CARBONYLS & ALCOHOLS EXPLANATIONS

100. Choice C is correct. Secondary alcohols, when oxidized, form ketones. The product following the oxidation of2-butanol using a chromium VI species is 2-butanone. The product is drawn below along with its proton NMRpeaks. Pick choice C.

oll "

a: 3 hydrogens next to a carbon with no hydrogens: Singlet (3H)

u - C- b -- CFL b: 2 hydrogens next to a carbon with 3 hydrogens: Quartet (2H)HzC' CH{ "

c: 3 hydrogens next to a carbon with 2 hydrogens: Triplet (3H)

aaCopyright @ by The Berkeley Review@ CARBONYLS & ALCOHOLS EXPLANATI

Monosaccharides

SeCtiOn VI "' :"11i:'":J:[r,::[*".ii. Pentoses and Hexosesiii. Common D-Sugars

Carbohydrates '' i"il:;""fferismii. Anomers

byTodd Bennett iii. tlaworth Projectionsiv, Mutarotation

D-Glucose nVO v. Isomer ization

"zf o" Oligo' and polysaccharidesHO{-H a) Linkages (Acetal Connectivity)u3f oH b) Disaccharides,, sl c) Trisaccharides-f "" d) polysaccharides

cH2oH i. Linkages

o ii. Cellulose and StarchesrH HOH?C iii. Branching- \_ -o1

= "flffio" chemical Keactions and restsi t - bn L a) Nitric Acid Sugar Oxidation

JOH{-O- (G-D-glucopyranosyl) -B-D- glucopyranoside

b) Osazone Testc) Benedict's Testd) Tollen's Teste) Periodate Oxidationf) Kiliani-Fisher SynthesisS) Ruff Degradationh) fMohl Degradation

Biological Appticationsa) Blood Typesb) Glycolysis

B-D-glucopyranose

OHOH

B

J 2

Speci altztng in MCAT Preparation

CarbohydratesSection Goals

oB Be able to recognize €ommon monosaccharides.

av

There are certain monosaccharides that recur in both organic chemistry and biochemistry that youshould recognize and be able to draw. These include glucose, ribose, fruciose, mannose, andgalactbse.it may be easiest to recall how the other sugars relate to glucose. For instance, mannose is the C-2epimer of glucose. If you know the structure of glucose, then you know the structure of mannose.

Be able to inter-convert between Fischer and tlaworth Prqiections.Some passages and questions will assume that you can translate between structures. For instance,the Fischer projection may be given in the passage, but the question may center around the structurein a Haworth projection. In translating structures, there are three separate points to observe. Thechirality of the anomeric carbon is dependent on the direction of the attack. The chirality of thepenultiinate carbon (carbon five in aldbhexoses) is constant according to the D or L status,'and thebackbone hydroxyl groups on the left in the Fischer projection are up in the Haworth projection.

Be able to solve the chiralitv of an unknown suqar ftom reaction data.From organic chemistry class, you may recall solving the identity of a sugar by evaluating the reactiondata presented. This makes for an ideal passage, althougbr the MCAT'writers raiely use thisinformation in a passage, just single questions. _You may recall evaluating whether a suiar wouldoxidize into an optically active or inactive diacid.

Be able to recognize the chirality of cyclic and linear sugars.It is rather simple to solve for the chirality of selected. carbons in both the Fischer and Haworthproiections, aftbr you have done this once. The rules are the same, so determine a quick way toidentify the R and'S configuration of each hydroxyl group of the sugar

Be able to identify common disaccharides.As with the monosaccharides, there are certain disaccharides that recur in organic chemistry andbiochemistry. The disaccharides you must recognize include sucrose, lactose, and maltose. Knowthe two monosaccharides that comprise the disatcharide and the glycosidic Iinkage binding them.

Be able to recognize the linkage associated with a given disaccharide.The linkage of a disaccharide is defined by the chirality of the anomeric carbon and the carbon ofthe sugar containing oxygen of the linkag"e. The disac-charide that presents the greatest difficultyis sucrbse, which is conipbsed of two anomeric carbons (one that is cr-1 and thebther that is G-2).

Be able to distinguish epimers and anomers.You must be able to identify the difference in chirality between diastereomers, which in the sugarquestions are commonly presented as either alpha or beta in regards to an anomer or epimels.

I{now the common reactions involving sugars and sugar derivatives.This booklet contains a large number of sugar reactions, perhaps more than you need. Key reactionsinclude the Kiliani-Fischer synthesis, Ruff degraclation, osazone formation, and nitric acid oxidation.The passage usually provides plenty of reactions, so don't memorize them, understand them.

Know the common polysaccharides and their biological significance.You must recognize the difference between glycogen and cellulose. The structural difference simplyinvolves the li[kage, but the difference in reattiv-ity is significant.

@v

@?

'v

@v

I:

:]I

lll

$.

*d&L

,0rl

ml

mn

fifid,l1

0Drun

,ed

yi\[i

Mm,ni

1ml&m

MdflW{fllmfl

Organic Chemistry Carbohydrates Introduction

Carbohydrates are organic compounds that contain carbon, oxygen, andhydrogen in a 1 : 1 : 2 ratio. The term is derived from carbon that has beenhydrated in a ratio of one carbon to one water, which explains the C : H : o ratioof 1 : 2 : 1. when we consider carbohydrates, we typicarly think of them as,qugars. For this section, we will focus exclusively on sugars and their organic;hemistry and biochemistry applications and exampres. while we tike a'lecidedly organic chemistry perspective on this material, you should pay extraattention to any subjects that bridge organic chemistry and biochemistry. Some.rf this material will overlap with topics in the metabolic components andmetabolic pathways sections of the biology books, but it will be piesented in arnanner that aims at chemistry concepts and short cuts.

"\-e shall consider rules and definitions for sugars first. There is generic

,--omenclature that describe sugars by their functionality and carbon count. In'"ddition to knowing generic nomenclature, you must be able to identify specific=ramples of aldohexoses, aldopentoses, and ketohexoses. Common sugais that' _ou

should recognize in all traditional structural representations (Fischer,:{arvorth, and chair) are glucose, ribose, fructose, galactose, and mannose. we"'ill present mnemonic devices to help with the recali of these structures.::ereochemistry is a significant part of sugar nomenclature, so it is necessary to:.row the stereochemistry associated with the prefixes of D and L.

l lonosaccharides of five carbons or more are typically found in cyclic structures::ther than straight chain structures. To form the cyclic structure, a hydroxyl:roup attacks a carbonyl carbon to form either a hemiacetal (if the sugar is in:. Jose) or a hemiketat (if the sugar is a ketose). Converting between linear and--rlic representations involves knowing how to draw thehydroxyl groups at=:ch carbon. we shall also consider the relationship of sugars. In the cyclic ftrm,

-'u must be able to determine what anomer is represented. In both the straight...ain and cyclic forms, epimers are possible. we shall emphasle, . :reochemistry terminology.

.' -'nsidering much of sugar chemistry involves multiple sugars, we will address:-, cosidic linkages and how to recognize the type of linkage. we will consider=---1nv examples from disaccharides, to trisaccharides, to oligosaccharides, and-:.a11y polysaccharides. Alpha and beta linkages impact the reactivity and'::rctural nature. To cleave iinkages, specific enzymes are required. oi most--,:oriety is the enzyme to cleave aipha linkages in glycogen, amylose, and,-:--r-lopectin, which aliows us to break down starches. we lack the enzyme to,.=:.r'e beta-linkages, so we are unable to breakdown cellulose." : n'ill also consider chemicai reactions of sugars. There are three types of- :mical reactions we shall address. one type is used to identify the chiiility at'::cific stereocenters, such as treatment witn HNo3 and osalone formation.-:,other type increases the length of the carbon chain of the sugar. The last type::e aks down the sugar, one carbon at a time. The last feature we will consideiis-. biologicai reactivity. We wiil look at glycolysis from a practical perspective,::rer than the detail seen in biochemistry. we will also ionsidertlood typ",

:1 the impact of stereochemistry on the recognition of the glycoproteins that=:ermine the blood type.

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Organic Chemistry Carbohydrates Monosaccharides gIvlonosaccharidesStraight Chain MonosaccharidesMonosaccharides have the basic formula Cr.,H2r.,Orl, giving them one oxygenattached to every carbon and one unit of unsaturation. The one unii ofunsaturation is present as either a carbonyl group or a ring. To form the cyclicstructure from the carbonyl structure, a hydroxyl group must attack the carbonylto form a hemiacetal or hemiketal, depending on whether the carbonyl form is inaldehyde or a ketone. Monosaccharides have generic nomenclature for theirstructures that depends on their carbon chain length and the type carbonylfunctional group. Aldehyde sugars are aldoses and ketone sugars ariketoses. Thelength of the sugar is also considered in the name. Four carbon sugars aretetroses, five carbon sugars are pentoses, and six carbon sugars are hexoses.Combining length and functional group yields sugar names such as:

Aldohexose: Aldo - hex - ose = aldehyde - 6 carbon - sugar.Ketotetrose: Keto - tetrose = ketone -4carbon-sugar.

within each chain length, there are a set number of stereoisomers, each with adifferent common name. The number of stereoisomers can be derived from 2n(where n = the number of chiral centers in the molecule). For example, analdotetroses has two chiral centers (carbon 2 and carbon 3), so there are fourpossible aldotetroses. The Fischer projection for each of the four aldotetroses areshown in Figure 6-1.

L-(+)-Erythrose

Figure 6-1

The last chiral center (on the penultimate carbon) in the D-isomers of the sugarshas R-stereochemistry, while the last chiral center in the L-isomers of the sugarshas s-stereochemistry. There is no correlation between D and L with (+) and (-)optical rotation. As seen when comparing L-threose to D-threose, D- and L-sugars with the same name are enantiomers of one another. That is why one isdesignated as (+) and the other as (-), indicating they have opposite directionrotations of plane polarized light.

sugars are routinely represented by Fischer projections, which are short hand forthe top view (top perspective) of a compound in its fully eclipsed conformation.The Fisher projection is derived from the straight chain form which is not themost stable form, but is the easiest form to draw. Any substituent on the right orleft side of the main backbone is projecting out at you. In bold wedge drawing,Fischer projections look like bow ties. When the fourth priority substituent iscoming out at you, you may recall that the chiral center is the opposite of whatyou see. This means that in Fisher projections, if the fourth priority substituent ison the side of the chain, then you must take the opposite rotation of the arc that isviewed for the chiral center in two dimensions to ascertain its chirality. Thetranslation into a Fisher projection from a three dimensional structure is shownin Figure 6-2.

f.;

$-I

r*.d[ r

:{r

,etllodl{m

&r,lffrUwu

,MngtrumrE

Mffmmw1mmillii{rnmmo

"Yry*--#-

'LtlT

@-,4!uilfli{

H

notnno*H

cFI2O

"Y"HoJ-HH-t- oH

cFI2OH

D-(-)-Threose

nIoHIHOtHcFI2OH

L-(+)-Threose

J""H-F oH

cFI2OH

D-(-)-Erythrose

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Organic Chemistry Carbohydrates Monosaccharides

Rotate the sugar tothe all eclipsed form

ViewingPerspective

cFI2OH

L-Ketotetrose

AOH

OH

cH2oH

"Y:HoJ-nHoIs"**

cH2oH

FilI"'".1_ H oHH H OH

Se"' '"J cH2oH

u H oHoHH

Figure 6-2

l",doses and Ketoses---'oses are sugars with an ardehyde functional group. Ketoses are sugars with a':lne functional group. Figure 6-3 shor.is two ardotetroses and two=:rtetroses.. A hydroxyl group in parenthesis indicates that the chirality is*-:-:,o!vfl. The only ketotetrose known is ervthulose.-*o

I-T(oH)- -f- oH

CFLOH

- --:.-,lotetrose

gFr2oHt"H-l- on

cH2oHD-Ketotetrose

cFI2OH

L-Aldotetrose

H

HO

Figure 6-3

:',: toses and Hexosesr i - -i€s are five carbon sugars, whether it be a ketose or aldose. Hexoses are six

r': -- r -r. sugars. Most sugars encountered in biology are pentoses and hexoses.I ;--:e 5-4 shows a pair of D-pentoses and a pair of 6_h"*oser.

nloIH-t- (oH)

H-l- foHrHtroHi

IHIOHcH2oH

D-Aldohexose

Figure 6-4

-''<o- (oH)

lroul-oH:FI2OH

"-iopentose

CH,)OHt;H--l- roHlu-F oH

cH2oHD-Ketopentose

CH"OH

t'",,-l-ro.,lu-l- roul

IH-t- oH

cH2oHD-Ketohexose

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Organic Chemistry Carbohydrates oMonosaccharides

Example 6.1How many aldohexoses are possil-.le?

4,48.8c. 16

D. 32

SolutionAn aldohexose is a six carbon aldehr-de sugar. Carbon one is an aldehyde, andcarbon six is a primary alcohol, so neither carbon 1 nor carbon 6 is chiral.Carbons 2, 3, 4, and 5 are all chiral, so there are 1.6 (24) possible combinations olchiral centers (i.e., 2R, 3R, 4R, 5R etc.), n'hich makes choice C the best answer.There are sixteen aldohexoses of n'hich 8 are D-stereoisomers (carbon 5 has Rstereochemistry) and 8 are L-stereoisomers (carbon 5 has S stereochemistry).

Example 6.2How many 2-ketohexoses are possible?

4,48.8c. 16

D. 32

SolutionA 2-ketohexose is a six carbon ketone sugar. Carbons 1 and 6 are primart'alcohols, and carbon 2 is a ketone, so neither carbon 1, carbon 2, nor carbon 6 ischiral. Carbons 3,4, and 5 are chiral, so there are 8 (23) possible combinations otchiral centers (i.e., 35,45,55, etc.) which makes choice B the best answer. Thereare eight 2-ketohexoses of which 4 are D-stereoisomers and 4 are L-stereoisomers.

Example 6.3How many D-aldopentoses are there?

4.28.4c.8D. 16

SolutionA D-aldopentose is a five carbon aldehyde sugar in which carbon 4 (thepenultimate carbon) has R-stereochemistry. Carbon 1 is an aldehyde and carbor5 is a primary alcohol, so neither carbon 1 nor carbon 5 is chiral. Carbons 2, 3

and 4 are chiral, so there are 8 (23) possible combinations of chiral centers (i.e-

2R, 33, 43; 2F., 3R, 45; 2R, 35, 4& 2& 3R, 4R; 25, 35, 43; 25, 3R, 45; 25, 35, 4R; ani25, 3R,4R). There are a total of eight aldopentoses of which four are D-isomersand 4 are L-isomers, which makes choice B the best answer. If you are wise, picirchoice B.

j1'

ltu

T]ll"""""""'U dffiurarCIiil

[.-sur,ry

tuWm[Wri0r.u*ch.

Mlliltmume

,n- -,sd0[t" swr( mmilD' [n6

$.n$tniliirnn

lMhrnf*rtt',oq,q

'tlilnme mumr,m

(0mumlms

i@Uiim,4IMTW,lilhs nr,mmn

trd

fdw

Copyright @ by The Berkeley Review 94 The Berkeley

aft ,u


t:7o t-FoH-f-ou u*outt--.|- o" "o-l- "no*n no*r,

cll2oH cll2oHL-(+)-Lyxose L-(+)-Arabinose

Figure 6-5 shows the eight aldopentoses along with their common names. Again,the D-isomer and L-isomer of Jparticurar sugar are enantiomers (mirror r-ig"rjof one another and haveopposiie optical rotations from one another (tne + uia -in parenthesis indicates the direction of the rotation of plane polarized light.)

tYoHo-+- H

"o-f *,rtotH

cFI2OH

L-(+)-Ribose

H\/oto-f- tro-l- "IH-l on

cFI2OH

D-(-)-Lyxose

"J",,-l- o'.to-l- tt

cFI2OH

L-(-)-Xylose

"J".r-l- o'H-F ou

cH2oHD-(-)-Arabinose

t-y'ot-l-.,t

,,o-l-.tn-f- o.t

cFt2oH

D-(+)-Xylose

"T:"u-]- o".r-l- o.t

cFI2OH

D-(-)-Ribose

In-;is;ofereL-

dt.rfr.R

(therbon)?(i.e.,and

merspick

Figure 6-5

* e.above sugars represent the complete group of aldopentoses (there are eight- -:al). As pointed out in Example 6.3, ther" are equal arnounts oi D-r.rgurc Jr-,d-sugars in each group of equivalent carbon sugars.

Erample 6.4

" rich of the following terms best describes the relationship between D-(-)-

:---r 3 ose and L-(+)-Ribose?

{, Anomersts. EpimersC. EnantiomersD. Diastereomers

Solution:ecause D-(-)-ribose and L-(+)-ribose vary at all of their chiral centers (comparere two structures in Figure 6-5), they are enantiomers. when all of the chiral:enters vary, the two structures are mirror images, which makes them=nantiomers. This makes choice C correct. Choice A is eliminated, because to belrlomers, the structures would have to be cyclic. If they were cyclic structures,--1e nomenclature would have included either nlphaorbeta at the start and the::ame would have been ribofuranose, which implies that it is a five-membered ring:lgar. choice B is eliminated, because the sugars vary at more than one carboiirey vary at carbons 2,3, and 4). Diastereomers refer to stereoisomers rn which

:-ct all of the chiral centers change (diastereomers are not mirror images). This=-:rninates choice D. Pick C for optimal results.

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Organic Chemistry Carbohydrates Monosaccharides (

Example 6.5VVhich of the following terms best describes L-(+)-Lyxose and L-(-)-Xylose?

A. AnomersB. C-2 EpimersC. C-3 EpimersD. Enantiomers

SolutionBecause L-(+)-Lyxose and L-(-)-Xylose vary at only the second carbon (comparethe two structures in Figure 6-5), they are referred to as epimers. If all of thechiral centers vary, the two structures are mirror images, which makes themenantiomers. Not all of the chiral centers drffer, so choice D is incorrect. ChoiceA is eliminated, because to be anomers, the structures would have to be cyclic,which they are not, as implied by the names. choice C is eliminated, because thesugars vary at carbon 2 and not carbon 3. This narrows it down to choice B. Thetwo structures vary at carbon 2, so they are C-2 epimers.

To complete our coverage of monosaccharides, let us consider aldohexoses.Figure 6-6 shows the eight D-aldohexoses. Do enjoy looking at them.

HO

H

HO

H

OH

OH

c.,

--1

::

:*

J[

t.C-,D.

o

H

H

o

OH

OH

H

Enat,i'urF,Lu

$ufouu

ffire#r

i:nmrmre

A lrugtm

,qlrlirtrrrnnrrrin

furuun:

-ffinnom[

lll_.u_4"Lll{.

mi*uttlriln

lhqwd

)t;.r

trffihwdu

OH

cH2oHD-(-)-Gulose

o

OH

H

OH

OH

20H

HO

HO

OH

OH

cH2oHD-(+)-Mannose

o

H

H ffi- f,-m[-c- [-m" f,

Figure 6-6

WllMh- d[

CH.OH

D-(-)-Idose

H-roHo4rruJ- oH

u-l- ouuIoH

I

cH2oH

D-(+)-Altrose

Example 6.6D-mannose and L-gulose vary u-r chiraLity at which carbon(s)?

A. 2 onlyB. 3and4onlyC. 5 onlyD. 2and5only

t-ooIH--l- oH

HO+HHoIHulou

I

cH2oHD-(+)-Galactose

n*

"IHo+"+H+

I

CH

"I""u-f oH

H-l- oH

H-l- oH

cH2oHD-(+)-Allose D-(+)-Glucose

"YoHO+Huo-l- H

uoIHulon

I

cH2oHD-(+)-TaIose

Copyright O by The Berkeley Review 96 The Berkeley Review


SolutionBecause D-sugars differ from L-sugars at the penultimate carbon, the twoaldohexoses must differ in chirality at carbon 5. This eliminates choices A and B.lhe question comes down to whether D-mannose and L-gulose vary at carbon 2-'t not. According to Figure 6-6, D-mannose and D-gulose differ al carbon 2, so--gulose and D-mannose must have the same chirality at carbon 2, because D-r.rlose and L-gulose differ at all chiral centers. This makes choice C the bestl:lswer. The two strucfures are shown below.

D-(+)-Mannose L-(+)-Gulose

20H

:r.ample 6.7:,ch of the following sugars contains the fewest carbons?

t. D-Threose: )-Galactose- L-Fructosel' L-Idose

: : "ution..::tose and idose are listed in Figure 6-6, so both are aldohexoses. They each, : the same number of carbons, so choices B and D are eliminated. Fructose is'nrrhexose, which means it too has six carbons, Iike the aldohexoses. This--:-rates choice C. Threose is an aldotetrose. Because aldotetroses contain only.-: :arbons, choice A is the best choice.

: ,.::'.ple 6.8' -- rse is all of the following EXCEpT the:

'I --a epimer of D-altrose.I .--3 epimer of D-giucose.

---i epimer of D-galactose." :-i epimer of L-talose.

;tionrxlg to Figure 6-6, D-allose has all of its hydroxyl groups on the right in its

.--=: projection. D-Altrose has all of its hydroxyl groups on the right except- -rroxyl 2, so it is in fact the C-2 epimer of D-allose. This eliminates choice- -'-lucose has all of its hydroxyl groups on the right except for hydroxyl 3,- ' the c-3 epimer of D-allose. This eliminates choice B. D-Talose has ail of

r:oryl groups on the left except for hydroxyl 5, so L-talose must have all of' ::orvl groups on the right except for hydroxyl 5. L-Talose and D-allose are:' :r,in€rS, which eliminates choice D. Galactose has two hydroxyl groups on-:: and two on the right, so it differs from D-allose at two chiral centeri. As

-- rs not the C-4 epimer of D-allose, making choice C the best answer.:

;:Tr+

CH

Haouo-f- H

noluu-l- oH

nlor-rI

cH2oH

o

H

H

OH

H

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Organic Chemistry Carbohydrates Monosaccharides I

Common MonosaccharidesMemorizing the names and structures for all sugars is no doubt a little pointless,but knowing the common monosaccharides is not a bad idea. to**onmonosaccharides in biology include ribose, glucose, mannose, fructose andgalactose. To help recall the common sugars, we shall present a mnemonic for afew (glucose, mannose, and ribose) and then know how the others relate to them.To help recall glucose, we have a not so nice way to remember the position of thehydroxyl groups. If you flip the page off with the fingernail oi yout middlefinger pointing towards the page, your finger tips are positioned" just as thehydroxyl groups of glucose are positioned in the Fischer projection. This helpswith enantiomers too. D-glucose is found using your right hand while L-glucoseis found using your left hand. Given that your left and right hands areenantiomers (mirror images), D-glucose and L-glucose are also enantiomers.Mannose can be recognized using a similar finger trick. By making a gunfigurine using your middle finger and index finger as the barrel, your finger tipsagain mimic the positions of the hydroxyl groups in the Fischer projection formannose. You can say "man uses guns, so the gun hand position representsmannose." Ribose has all the hydroxyl groups on the right side of the Fischerprojection, so "ribose is all right." This provides three solid mnemonic devices forrecalling three common sugars. Other monosaccharides can be learned relativeto those three. The straight chain Fischer projection of some of the more famousD-monosaccharides (famous implies that they are found in biological systems)and a suggested memory aid for each one are shown in Figure 6-7.

"YH+

Ho+Ho+H+

n{ono-l-Huo-l-unA ori

IHIOHcH2oH

fY:"-l- o"*o"

H

HHO

H

H

o

OH

H

OH

OH

o

OHHH

OHcH2oH

D-GalactoseC-4 epimer of glucose

HOH

H

D-Fructoseketose of glucose

cH2oHD-Ribose

"Ribose is all rightl"

fl

d

u

o

cH?oHIo

*"{o.t*o"

cH2oH

H

HHH

cH2oHD-Glucose D-Mannose

"p*x* glucosel" "Man uses his gun!"

Figure 6-7

The terms below the name of each sugar refer to a way in which each structurecan be recalled. "F***'r of course stands for "flip off", right?

Stereoisomerism in SugarsThe majority of sugar chemistry centers around the stereochemistry of thehydroxyl groups. As such, we shall focus on stereochemistry of sugars morethan the chemistry of alcohols, carbonyls, hemiacetals, or acetals. In sugarchemistry, it is critical that you be able to determine how stereoisomers relate tcone another, whether they are enantiomers, epimers, anomers, or structura-isomers. The term epimers should not be mistaken with anomers, which refers tatwo sugars in their ring forms that vary at the hemiacetal (in the case of aldosesor hemiketal (in the case of ketoses) carbon in the ring. It is easiest to say tha:anomers vary in chirality at the most oxidized carbon while epimers vary i::chirality at any one of the less oxidized carbons.

Anomers vary in chirality at the most oxidized carbon while epimersvary in chirality at any one of the less oxidized carbons.

T

E

td

m

m

ffiffim

ffi@

,m

Jfn

tmh

@,'@iimfir'Itl

'dtrf,r,muG&

WWM,mmplmffi

Copyright O by The Berkeley Review 9a The Berkeley Kevier


:1

Ia

I.

.e

.e

\S

;e

:eq

LrI

fS

oritseror\-e

US

rsl

the:lore_rgar

te torrralrs torses)thatrvin

II

II

,'"*

EpimersEpimers are sugar diastereomers that vary at one stereocenter (chiral center) inthe carbon backbone. D-Threose and D-erythrose are C-2 epimers, because theyvary at carbon 2. D-Glucose and D-Galactose are C-4 epimeis, because they varyat carbon 4. In the straight chain form of a monosaccharide, the carbonyl carbonis most oxidized, but because ithas sp2-hybridization, it has no chirality. The lastcarbon has no chirality, because the carbon is primary with two hydrogens. Thismeans that two sugars cannot be epimers at the last carbon or the carbonyl;arbon in the straight chain form. The straight chain form is not the most stableform, but it is an intermediate when one anomer is converted into the otherlnomer.

AnomersThe most stable form of a sugar is the cyclic form, which is formed when a:r'droxyl group attacks the carbonyl carbon yielding a hemiacetal. A new chiral:enter is formed when the hydroxyl attacks the carbonyl. Because the attack can:;cur from either top or bottom, there are two diastereomers formed. The twoiiastereomers that result when the ring structure is formed are referred to asrromers and the new chiral carbon is the anomeric carbon. The two anomers are::ferred to as the alpha (axial for D-pyranoses) and thebeta (equatorial for D-:r'ranoses) anomers. The beta anomer is typically more stable than the alpha:romer. It is only in the cyclic form (furanose form for five-membered rings and::,'ranose form for six-membered rings) that anomers are possible. Figure 6-g::.ows the anomers of glucose, o-D-glucopyranose and B-D-glucopyranose.

HOH2C.HOH"C

Ho \\ o

Ho$oHOH

B-D-glucopyranoseAll substituents haveequatorial orientation

Figure 6-8

:,r membered rings are referred to as pyrans while five membered rings are:=:erred to as furans. The name B-D-glucopyranose comes from the anomeric-'.'Jroxyl group being cis to carbon 6 (B), D because the last chiral center (carboni has R-stereochemistry, gluco from the sugar glucose, and pyranose because it-' a six-membered ring sugar. All cyclic sugars are named in a similar fashion.'l-rnosaccharides come as either five-membered or six-membered rings, so you:: -rpld be familiar with the term furanose as well as pyranose. Ribose in its-'-" :Lic form is a furanose as you are no doubt familiar with from biology.

In cyclic sugar structures, the term alpha technically means that theanomeric oxygen is trans with the last carbon and the term betatechnically means that the anomeric oxygen is cis with the lastcarbon.

HOHO

OHcr-D-glucopyranose

All substituents have equatorialorientation, except the anomeric OH


Organic Chemistry Carbohydrates MonosaccharidesT

MutarotationThe two anomers of a sugar exist in equilibrium in aiao and can interchange via astraight chain intermediate. conversion from one anomer into the other isknown as mutarotation. Figwe 6-9 shows the key structures in mutarotation.

OH

HWftrd

mfum

wd

WOM

,im

Mn

ffim

Wtrrmrmi@Iu

sffiil

o

OH

-:--

OH

H +

-OH

OH

Ho--4HO

H

HO

3 'oH

G-D-glucopyranose(64%) cH2oH

Straight chain(< 0.1%)

Figure 6-9

o-D-glucopyranose(36%)

Example 6.9Mutarotation involves which of the following?

A. Sugardecomposition.B. Sugar dehydration.C. Sugar oxidation.D. Stereoisomerization.

SolutionAs shown in Figure 6-9, mutarotation involves the conversion between anomers.sugar decomposition occurs with degradation or digestion, which does notdescribe mutarotation, so choice A is eliminated. Sugar dehydration involves theloss of water by the sugar, which does not occur with mutarotation, so choice B iseliminated. Dehydration is accomplished chemically by a highly hygroscopicsubstance such as concentrated sulfuric acid. Sugar oxidation would require-anoxidizing agent, which is not present in mutarotation. Choice C is eliminated.The process of mutarotation involves the conversion from one stereoisomer intoanother, which is described by the tercn stereoisomerization Choice D is oneterrific answer for a question like this.

Example 6.10In aiao , D-glucose is MOST stable in its:

A. alpha anomer of the pyranose form.B. beta anomer of the pyranose form.C. straight chain form.D. furanose form.

SolutionThe pyranose (six-membered) ring form is more favorable than the straight chainand the furanose (five membered) ring form. This eliminates choices C and D.The beta anomer, with the anomeric hydroxyl group in equatorial orientation, ismore favorable than the alpha anomer. This eliminates choice A and makeschoice B the best answer.

6

HOH,C

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Organic Chemistry Carbohydrates MonosaccharidesLes

iaaris

)'OH

)mers.es not,-es the.ce B is

'scopicrire anmated..er intois one

-

ht chain, and D.tation, isd makes

-'Review

Haworth ProjectionsThe Haworth projection is a simplified representation of cyclic sugars. Theyhave no physical relevance, but are easy to draw (just like Fisher projections).They do not show the three dimensional spacing that chair conformations show,but they emphasize positions above and below the ring, which is the major focusrvhen considering sugars. Converting from the Fischer projection to the Haworthprojection is easy once you get the hang of it. To go from the straight chain to thering form, recall the phrase "downright uplefting," which applies to carbons 2, 3,

and 4 in an aldohexose, but not the penultimate carbon (which has its oxygen inthe ring) or the anomeric carbon. Any hydroxyl groups on the right side in theFischer projection end up below the ring in its cyclic structure. Any hydroxylgroups on the left in the Fischer projection end up above the ring in its cyclicstructure. Numbering the carbons makes it is easier to convert to the Haworthorojection, because it helps to keep track of carbons 2,3, and 4. The anomericcarbon is given the far right position in the Haworth projection. Figure 6-10

'hows the generation of the Haworth projection of D-galactopyranose.

OH

H

OH

OH

H

HO

OH

HHct-D-mannopyranose

on right.'. down HO

6tC FI2OH

rL left,..w

H "ffi"HOH

HI-{2O6'C

Figure 6-10

-:.e anomeric carbon is shown at the far right in furanose rings just as it is in:i'rarlose rings. Fructose and ribose are classic examples of naturally occurring:Jarlose rings. The carbonyl group of fructose is on carbon 2, so when oxygen of. ','droxyl 5 attacks the carbonyl, the ring is five atoms in size. Figure 6-11 shows:.e Haworth projections of cr-D-mannopyranose and B-D-fructofuranose.

6HOH2C OH

5 Hln

1

2'cH2oH

HOHOH

B-D-fructofuranose

Figure 6-11

ucHroH

rl- 6HO

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Example 6.11The following structure is the Haworth projection of what sugar?

A. B-D-glucopyranoseB. G-D-mannopyranoseC. a-D-glucofuranoseD. u-D-mannofuranose

SolutionChoices C and D are eliminated immediately, because the ring in the question issix-membered, not five-membered. Translating from the Haworth projection tothe Fischer projection might make it easier to determine the identity of the sugar.Hydroxyl 2 is up, hydroxyl 3 is up, and hydroxyl 4 is down in the Haworthprojection, so hydroxyl2 is left, hydroxyl3 is left, and hydroxyl4 is right in thestraight chain form. This corresponds to the Fischer projection of mannose. Thebest answer is choice B. It might be easier to decide whether the compound isglucose (choice A) than it is to deduce the sugar's identity. B-D-glucose has thehydroxyls alternating between up and down as you move from carbon to carbon.The structure in this question doesn't show this, so it can't be glucose. The onlychoice left is mannose, the C-2 epimer of glucose. You may viant to verify thatonly the second carbon is different from B-D-glucose. The structures of thevarious sugars are drawn below:

6CH2OHOH

l:Ifl

]T

pot:

HH

OH

H

OHHO

OH

HOHB-D-glucopyranose

HOHo-D-glucofuranose

HHB-D-mannopyranose

HHcx-D-mannofuranose

H

OH

OH

cH2oHD-mannose

H

cH2oHD-glucose

The important skill to develop here is the ability to quickly convert from the ringstructure to the straight chain structure, as these are the common ways in whichyou will see the sugars.

o

osOH HO

2

ucrtrott

H

Copyright @ by The Berkeley Review r02 The Berkeley Revieu


IsomerizationLr addition to mutarotation, monosaccharides can convert from an ardose into aketose via an enediol intermediat". ihil conversion is known as ketose_ardose::omerization In glycolysir, tt"

".rrv_" io_-nto

rructose 6?h?;p;;;;lJ;;"1,;:r,:::';i::,ru"il',r:i;T:"":*ll',r"i,:,;)rocess similar to tautomerization converts dihydrJxya"itor^," pr.,orphate intoslvceraldehyde 3-phosphate- Figur" o-rz ,rrr*s the step-by-step conversion ofan aldose into a ketose in both u"ia una Ju

".Ketose-Aldose Isomerization (base and acid catalyzed)

Ho- Hprotonate Y'o-on C-t '- OH

Ho#u-!r._ IH+OH

u-f- oHI

cH2OH

HIH-r_ oH

protonate F O*tttl Ho*H-__l- HtOHHtOH

cH2OH

"YtH*otr

uo-l- uH-l- oHur-l- ort

cH2oH

',"::H,|f".

H.o\-g

H#oHuo-FuH*or-rH-f- out

cH2OH

deprotonateH on C-2

+ry* HOHH

H

OHHOHOH

deprotonateOH on C,2

-=-

deprotonateOH on C-2

protonate

OH

HO

{o-H.H-

H-

o-H-H-

OH

OH

zoH

+CH,CF

t,Htf+CH,

H

I

H(

H

o-HOHOH

zoH

c-1

l-Glucose

cH2oH

Figure 6-13

o\-Huo*H

"o-l-u,.r-l o"H-l ou

cF{2oHD-Mannose

HO^_ H

deprotonate ||- o"*{'no-l-nH#ou

IHtOHcH2OH

oH

Figure 6_12

* e boided hydrogen atoms are the active hydrogens that are being either gained: -ost in the reaction mechanism. The requirement for the conversion is the'::ration of either the enol (under acidic conditions) where tn" .u.uo'yl oxygen'' :rotonated and the alpha hydrogen is deprotonated, or tne rormation of the':"rate (under basic conditrons; #nere the arpha hydrogen is deprotonated.1en isomerization is catalyzei a,y un "-,'yrne, it often proceeds through an'" :nine intermediate. As Figure ol-n shows, conversion tetween C-2 epimers;--icose into mannose) arso pr"oceeds tnto"g^ an enedior intermediate.

i

r\- uIt-t+oHI* -'+H +H*| +.-+-oH=-I

-' -t- oHcFI2OH

Isomerization of glucose into mannose (acid catalyzed)H'oy. H Ho\-H H-o\- H

Ht oH l- on uo-f- HHo+H -H* _uo-Fn *H.__- a,o#n _Ho.r-: H--oH

- H-+-oHFHt ou H-J- oH ,_f_ on

cFI2oH cFI2oH cFLoHEnediol intermediate

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Example 6.12In the conversion from glucose into mamose, one of the intermediates is:

A. ribose.B. a six carbon enediol.C. galactose.D. a five-carbon ketose.

SolutionUnder basic conditions, the conversion involves the deprotonation of an alphahydrogen, which forms an enolate intermediate. protonation of carbon 2regenerates the monosaccharide as one of two epimers, mannose or glucose.Under acidic conditions, the conversion involves the protonation of the cirbonyloxygen and the deprotonation of an alpha hydrogen, which forms an ened.iolintermediate. Protonation of carbon 2 regenerates the monosaccharide in one oftwo stereoisomer forms, which leads to either mannose or glucose. The bestanswer is therefore an enediol, choice B. If enolate and enediol were bothchoices, then the reaction conditions would have been required. Because enolateis not a choice, it is safe to assume that the conditions are acidic. Figure 6-13shows the conversion. No carbon is gained or lost during the conversion pto""ss,

_^r-

Example 6.L3what is the name of the process that converts 3-phosphoglyceraldehyde into 1-phospho-1,3-dihydroxyacetone?

A. AldehydosisB. EnediolosisC. Ketose-aldose isomerizationD. Transphosphorylation

SolutionThe numeral preceding the phosphate group changes from 3 to 1, which wouldlead one to believe that the phosphate group changed positions. However, thechange in numeral is a result of a change in the functional group prioritynumbers, and the phosphate group in fact never changes its position. As suctr,transphosphorylation is an incorrect answer choice designed to trip up peoplewho didn't draw the structures and assumed the numbering changea le"arrt" *ttgroup moved. Choice D is eliminated. An aldehyde is converted into a ketone,so enediolosis is not an apt description. This eliminates choice B. The overallconversion is from an aldose into a ketose, so the best term to describe theprocess is ketose-aldose isomerization. This eliminates choice A and makeschoice C the best answer.

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Organic Chemistry Carbohydrates Oligo- and Polysaccharides

H_o20CH

I

l

ri

e.:

Off$6s*dHfiffidu$;Hilfl,,,,ildffiab.6 desThis section addresses disaccharides, trisaccharides, oligosaccharides, andpolysaccharides. Disaccharides result from combining two monosaccharides,trisaccharides result from combining three monosaccharides, oligosaccharidesresult from the combination of two to eight monosaccharides, andpolysaccharides result from combining several monosaccharides.

Linkages (Acetal Connectivity)sugars combine when a hydroxyl group of one sugar attacks the hemiacetalcarbon of a second sugar, displacing the hydroxyl group from the hemiacetalcarbon to form an acetal. This forms a glycosidic linknge, resulting from adehydration reaction of the two hydroxyls involved. The terms used to describethe orientation of the anomeric hydroxyl group in a cyclic monosaccharide (alphaard beta) are also used to describe the orientation of the oxygen in the glycosidic-rnkage. When the anomeric oxygen is cis with respect to the last carbon (carbonrumber 6 in an aldohexapyranose), the linkage is said to be beta.

DisaccharidesThe names for disaccharides are derived from the two component sugars where--he left sugar (using hydroxyl 1 of an aldopyranose) is considered a substituent-'n the right sugar (using either hydroxyl 4 or 6). The left sugar gets an "osyl".uffix and the right sugar gets an "oside" suffix. For instance, sucrose consists of:lucopyranosyl and fructofuranoside. In sucrose, the linkage is 1,2, which is rare.-r most cases, the linkage is 7,4, as in the disaccharides maltose, cellobiose,ormed from the hydrolysis of cellulose), and lactose. These common:saccharides all contain at least one glucose residue in one of its anomeric forms.-nkages are named for the carbons involved as well as the actual structural.tereochemical orientation) of the anomeric carbon. A picture says so much:.ore than words (at last check the picture was valued at 1000 words), so for that:eason, Figure 6-74 is a picture of four common disaccharides.

HO

oOH HO

cH20HB-glycosidic

linkage

Sucrose

OH

a-D-glucopyranosyl)-G-D-fructofuranoside)(o-1-B-2 -linked disaccharide)

HOFI"Ct.-rG\--O. HOH,C

Ho\,a\,oS/oroH Ho\-za-\-.oHB-glycosidic OH

linkage

Cellobiose

-O-(G-D-glucopyranosyl)-B-D-glucopyranoside)(B-1, 4-linked disaccharide)

Figure 6-14

Lactose( -O-(C-D-galactopyranosyl)- c-D-glucopyranoside)

(G-1, 4-linked disaccharide)

HOFI,Cuo ]-\-'- 01 a-g] ycos id i c

Ho\'^{\ Iinkage'or, I uourc.

oHO

OHMaltose OH

(4-O-(o-D-glucopyranosyl)-a-D-glucopyranosicle)(o-1, 4-linked disaccharide)

HOFI2C

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TrisaccharidesTrisaccharides are an oligosaccharide comprised of three monosaccharides.Trisaccharides contain two glycosidic linkages. one of the most commontrisaccharides is raffinose, which entails a a-D-galactopyranose forming a linkagewith its anomeric hydroxyl group with hydroxyL6 of iucrose. Figure 6--15 shor.israffinose in the Haworth projection and three dimensionally more valid chairform.

OH

cH2oH

HOHCXRaffinose

Trisaccharide of a-D-galactose, u-D-glucose, and B-D-fructose(u-7,4-and o-1-B-2 linked trisaccharide)

HO

HO

HO

cH2oH

\

oHlo

Figure 6-15

PolysaccharidesPolysaccharides are exactly what the name implies: multiple sugars. we shalltreat polysaccharides as polymers made of four or more monosaccharides. Mostare linked from the anomeric carbon of the sugar on the left (carbon 1) to thefourth carbon on the sugar on the right (carbon 4), a'J.,4 glycosidic linkage. Longchains of sugars (starches like amylose and amylopectin) are common inbiological systems. we store our excess sugar in long polymers such as glycogen(poly u-D-glucopyranose), which has 1,4-alpha linkages between ialacentglucopyranoses and 1,6-branching about one out of every ten glucopyranoseresidues in the polymer. The most abundant of all polysaccharides is-cellulose(poty G-D-glucopyranose) which is used for structural purposes in plants. weare unable to digest cellulose, because we lack the enzyme needed to break B-1,4-glycosidic linkages. we produce alpha glucosidase (which cleaves only alpha-linkages of poly glucopyranoses) but not beta glucosidase. Figure 6-1-6 showscellulose, amylose, and glycogen.

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o

"t'oH HOH2L

Cellulose(B- 1,4-linked polysaccharide)

HOHHOHHOHAmylose

( ct- 1 ,4Jinked polysaccharide--no branching)

oHO

OH

o HocH. or^v!./

Glycogen( a- 1,4-l inked polysaccharide-- 1,6-branchin g)

Figure 6-16

There are no 1,6-linkages found in cellulose, only 1,4-linkages. This means that;ellulose is a linear polymer with no branching. Strands of cellulose are heldtogether by hydrogen bonding between the ring oxygen and the hydrogen onhvdroxyl 3, making the structure uniform on the microscopic level. Each glucose:esidue is approximately 5.1 angstroms from carbon 4 to carbon 1. Thellucopyranose monomers alternate orientation in cellulose to allow hydrogen:onding, as shown in Figure 6-16.

E-xample 6.143ranching in amylopectin is attributed to what type of linkage?

.\. 4,4B. 4,1

c. 7,6D. 6,4

Solution3ranching is attributed to 1,6-linkages, although in dextran the monomers are-rnked by 1,6-glycosidic linkages. This is memorization for the most part.\mylopectin and glycogen have branching, while cellulose and amylose do nothar.e branching. The best choice is C.

s;ph"

l" o "tOH

f{

H

1

t3

Il

1

Ite

e

e

t-t-

S

cH2oH

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Organic Chemistry Carbohydrates Chemical Reactions and Tests

ffiHffid ruffiofi$:l 6ii#EgffiNitric Acid Sugar OxidationNitric acid is rich in oxygen, making it an oxidizing agent. vvhen an aldohexoseis treated with nitric acid, both terminal carbons are oxidized into carboxylic acidgroups. If the product, an aldaric acid, is meso, then the compound is opticallyinactive. using retro-analysis, the chirality of the hydroxyl grolps in the originalmonosaccharide can be determined. Figure 6-12 shows the aldaric acidformation reactions of four aldohexoses.

Nitric acid sugar oxidation

OH

OH

H

OH

OH

OHGlucaric acid

(optically active)Mannaric acid

(optically active)

OH

("I

u

rfr

d

o

M

nilifiru

ffi

OH

H HNo?__+oH H2o, A

OH

H

HO

H

H

o

HO

H

oGalactaric acid

(optically inactive)

HNOl

-+H?o, a

HN03

-:>

H2o, a

HO

HO

H

HO

HO

H

H

o

H

OH

OH

OH

OH

OHcH2oH

Glucose

H

OH

H

H

OH

HNO3

-_-_+H2o, L

cH2oH

Mannose

H

OH

OH

OH

OH

Allose

O-r- -OH

OH

OH

OH

OH

OHAllaric acid

(optically inactive)

H

HO

OH

H

H

OH

HHO

HO

cH2oH

GalactoseoH cH2oH

Miiifrffiil@&d

h&r@tum

Figure 6-17

Example 6.15The addition of nitric acid to galactose results in:A. an optically active compound.B. an optically inactive compound.C. the loss of a carbon.D. one formaldehyde and five formic acids.

SolutionNitric acid oxidizes an aldose into a diacid (aldaric acid) product. Galactosebecomes optically inactive, because the diacid form is meso (it has a mirror planethrough the bond between carbon 3 and carbon 4). This can be seen in Figure 6-17. Choice B is the best answer.

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uo-#Hno-I- ru

" -l-- o"

"-F ortcFl2oH

D-Marrnose

Osazone TestThe osazone test has appeared frequently on the MCAT, despite its obscurity inthe minds of most organic chemists. The repeated use of the reaction in passagesis a testament to the test writer's goal to present topics that are not commonplacefor the sake of asking questions that at initial glance look unfamiliar, Theosazone test should be more aptly named t}.e epimer test. Vlhen treated withthree equivalents of phenyl hydrazine, an aldose or 2-ketose undergoes asubstitution reaction to form an imine, followed by the oxidation of an alcoholinto a carbonyl, and lastly a second substitution reaction forming a second imine$oup. Imine moieties form at carbons 1, and2, so any chirality at either carbon islost. As a consequence, C-2 epimers yield the same osazone. Figure 6-18 showsthat D-mannose, D-fructose, and D-glucose, all yield the same osazone.

Osazone (C'2 epimer) test

Figure 6-18

3e osazone test works in conjunction with other information to zero in on the-dentity of an unknown sugar. For instance, if an unknown D-aldohexose yields-.ne same osazone as D-galactose, but does not have the same optical rotation as)galactose, then it must be either the C-2 epimer of D-galactose (D-talose) or theietose of D-galactose (D-tagatose.)

tsenedict's Test (Fehling's solution)f,enedict's test oxidizes the sugar as copper dication gets reduced. In getting:educed, the copper solution goes from blue to red for a positive test. The aldose:ecomes an aldonic acid, which under basic conditions is deprotonated to form

'r aldonate. Figure 6-19 shows Benedict's test.

Benedict's test (Fehling's solution)

2Crt2*-+5 0H-

Figure 6-19

H+OHuo-J- H

n-l- oHH*ou

cI-I2OH

D-Glucose

o\-uHf on

Ho-+- H

H*ouu-F on

cFI2OH

Glucose

N-NFUI

=N-NIjZuo-l-uHJ-on

IH -1- oH

cFI2OH

D-Glucosephenylosazone

o\on4on

HoIHrilonn*ou

cFI2OH

Gluconate

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Tollen's TestThe Tollen's test involves the reduction of silver ion by an oxidizable sugar. Assilver ion is reduced to silver metal, if the solution is spun within the coitainer,thel silver precipitates on the inside walls of the container, resulting in a socalled silver miror. As silver is reduced, the sugar is oxidized. the ildehydefunctional group oxidizes quickly, resulting in a rapid reaction with an aldose.Primary and secondary alcohols can also be oxidized, so eventually a ketose willreact too. The Tollen's test is said to be positive if the silver mirror forms rapidly,indicating the presence of an aldose. Figure 6-20 shows the Tollen's test.

Tollens test

o\Yo-*unn.H+OH

2 Ag(NH?)r. HO*U---- I2OH- H+OH

n*oricFI2OH

Ammoniumgluconate

"Y"H+OHo-l- H

H-]-onH*ou

cFI2OH

Glucose

H

"Yo_ NFr4*

Ho-#t-l2 Ag(Nu.)r. no--{- u

-

|2 OH- H-l- oH

"*o.,cFI2OH

Ammonium mannonate

Figure 6-20

Example 6.16The Tollen's test tests for:

A. a reducing sugar.B. an oxidizing sugar.C. an acidic sugar.D. a ketohexose.

SolutionThe Tollen's test results in the reduction of silver, so the sugar must be a reducingagent. This makes the Tollen's test a test for a reducing sugar (hemiacetal oialdose). Choice A is the best answer.

Periodate OxidationA strong oxidizing agent such as periodate , ro4-, oxidatively cleaves the carbon-carbon bonds of a sugar. After complete oxidation, the terminal carbons increaseby one bond to oxygen and internal carbons increase by two bonds to oxygen,forming several one carbon compounds as products. The complete oxidation oferythrose, an aldotetrose, is shown in Figure 5-21.

Periodate oxidation

H--{- OHu-l- oH +

cFI2OH

Erythrose

IOa- -----+ 3 HCO2H + H2CO + 3IO3

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Figure 6-21

The Berkeley Review


Example 5.17The addition of excess periodic acid (HIOa) to fructose yields:A. one formaldehyde and five formic acids.B. one formaldehyde, four formic acids, and carbon dioxide.C. one formaldehyde and four formic acids.D. two formaldehydes, three formic acids, and carbon dioxide.

SolutionFructose is a 2-ketohexose. with excess periodic acid, carbon-carbon bonds areoxidatively cleaved. The terminal carbons (carbons 1 and 6) are oxidized up onelevel from primary alcohols to formaldehydes (from one C-o bond to two C-obonds). Because two formaldehyde molecules are formed, the best answer ischoice D. The carbonyl carbon (carbon 2) is oxidized up two levels from a ketoneio carbon dioxide (from two C-o bonds to four C-o bonds). The secondaryalcohol carbons (carbons 3,4, and 5) are oxidized up two levels from alcohoh ttformic acids (from one C-o bond to three C-o bondi), resulting in the formationrf three formic acid molecules. The result is two formaldehyd.es, one carboncioxide, and three formic acids, choice D. An reaction summary is shown below.

HO- CF{2OH =

HOI

-oHO=

H\-^

Ho=c=o

OH

no*uOHOH

n*onOHOH

H*onOH

H_ L_^-rvHO

H

- \_^-ruHO

H_ \,_^-FU

HOH

Ho- cF{2oH = !oH

Kiliani-Fischer Synthesis}e Kiliani-Fischer synthesis increases the carbon chain of an aldose by one. The

:::::::::::::::erv carbon is added as a cyano group to the carbonyl carbon, so the chain-:',creases by one carbon at the aldehyde end. After reduction to an imine and:idrolysis, the cyano group is converted into an atdehyde. The result is the:rrmation of two aldose epimers with an additional carbon. For instance, when.:'ing D-arabinose in Kiliani-Fischer synthesis, both D-mannose and D-glucoseC-2 epimers) are formed. Glucose is72% of the product mixture, because of the,:.rral influence of carbon 2 of the aldopentose (D-arabinose). Figure 6-22 shows--:..e Kiliani-Fischer synthesis of two D-aldohexoses from D-aralinose. An oH:roup in parenthesis represents the possibility of both chiral centers.

CFLOH

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Kiliani-Fischer synthesis

oY"uo-tsr+t-l.ot*,-f- o.,

cH2oHD-Arabinose

KCN__>H"O

NlilC

u*ronrHO+Ht-J-,rt.,-f- o.,

cH2oH

1..H2/PdBaSOn

z.urolt{

o\-H

to-l- "to-l-*.,,,-l-,r0,.,-f- o.,

cH2oHD-Mannose

(28%)

o\-ut-J-,rt

toJ-rtt-l- ot

'-l- o'cH2oH

D-Glucose(72%)

Figure 6-22

Example 6.18The Kiliani-Fischer synthesis on a D-aldopentose produces:A. a fifty-fifty mixture of enantiomeric D-aldohexoses.B. a major-minor mixture of diastereomeric D-aldohexoses.C. a major-minor mixture of diastereomeric D-ketohexoses.D. a fifty-fifty mixture of diastereomeric D-aldotetroses.

SolutionFischer-Kiliani synthesis adds a carbon to an aldose, so with a five-carbonreactant, a six-carbon product is formed. This eliminates choice D. The productis an aldose, because the carbonyl group must form at a terminal carton, sochoice C is eliminated. The products are diastereomers (C-2 epimers), so choiceA is eliminated. Because attack at the carbonyl carbon is lnfluenced by theasymmetry of the rest of the molecule, the mixture is not fifty-fifty. Choice B ist"*

Example 6.19The addition of cyanide to a sugar is involved in which of the followingprocesses?

A. Nitric acid oxidationB. Nguyen degradationC. Kiliani-FischersynthesisD. Ruff degradation

SolutionAs seen in the first step of the Kiliani-Fischer synthesis in Figure 6-22, the cyanideanion adds to the carbonyl carbon of an aldose to yield u cyutro extension of theoriginai aldose. The best answer is choice C. Because oi the influence of theother chiral centers, the cyano group does not add in afifty-fifty manner to thetop and bottom of the carbonyl. This results in a diastereomeric mixture ofproducts. Nitric acid oxidation involves nitric acid, so choice A should havebeen eliminated. Degradation of a sugar results in the loss of a carbon, not thegain, so the reagents should not contain carbon. This eliminates choices B and D,although choice B is not an actual sugar reaction, so you may have eliminated it:y'"*

:s*.ff,r.h

m,f,_.

!

Siuu

Logr

'U tl:

Tlne

TilrXilr

..4r

nc.F}

ftwiiu*-lllb I

r0mwLlt

rflqp

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Ruff DegradationThe word degradation implies that something is breaking down. In Ruffdegradation, an aldose loses its first carbon, resulting in an aldose of one lesscarbon. As we saw in the carbonyl section and in several biochemistry examples,iosing one carbon is readily accomplished by decarboxylation. In order todecarboxylate the terminal carbon, it must be oxidized first to a carboxylic acidand then to carbon dioxide. Bromine in water effectively converts the aldehydeinto a carboxylic acid (or in the ring form, a hemiacetal into a lactone). Thecarboxylic acid is oxidized into carbon dioxide by peroxide in the presence of aferric cation catalyst. CO2 leaves and decreases the chain length by one carbon.Figure 6-23 shows the Ruff degradation of D-glucose. It is rarely used withaldopentoses, because when the product is an aldotetrose, the yield is low, likelylo to the difficulty encountered with a smaller ring in the cyclic form.

Ruff degradation

HO+HHIor-iu -[- oH

cH2oHD-Arabinose

Y""+" ab-# ou Fe"'

*""

o

H

H

H

Br, ca(oH),--' - -lI2-' Ho

o\-tH-]-or.

uoJ-nu-l- onuf oH

cH2oHD-Glucose

Hzo Hzo

cH2oHD-Gluconate

Figure 6-23

Lxample 6.20Ruff degradation results in:,\. the gain of a carbon onto carbon 1 of the chain.B. the loss of carbon 1 from the chain.C. the gain of a carbon onto the last carbon of the chain.D. the loss of the last carbon from the chain.

5olution)egradation implies breakdown, so Ruff degradation is associated with a loss of:arbon, not a gain. This eliminates choices A and C. As shown in Figure 6-23, it-o

the first carbon that is lost, so choice B is the best answer.

Erample 6.21lhe addition of bromine in water to a sugar is involved in which of the following:rocesses?

{. Nitric acid oxidationB. Wohl degradationC. Kiliani-Fischer synthesisD. Ruff degradation

Solution-{s seen in Figure 6-23, the combination of bromine and water will oxidize aronosaccharide into an aldonic acid (monoacid) in the first step of the RuffJegradation. The best answer is choice D.

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H2NOH.-.>- Hzo

H-F OH

rIo*,"f o"

cH2oHD-Ribose

Wohl DegradationWohl degradation is essentially a Kiliani-Fischer synthesis in reverse. An aldoseis first converted into-its oxime using hydroxylamine to which acetic anhydrideand acetate are added to acylate the hydroxyi groups and dehydrate the oximeinto a cyano group. The cyano group is removed under basic conditions, whichalso hydrolyze the esters on the sugar. The result is an aldose of one less carbon.Figure 6-24 shows the Wohl degradation of D-ribose.

H--F OH

H-J- oH

Hf oH

cH2oH

Figure 6-24

Wohl degradation

a o ili o\'

*# :+oA"NuocH,-;f::N"ofc,f "t1ec -

cH2oHut oac D-Erythrose

CH2OAc

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Organic Chemistry Carbohydrates Biological Applications

Biid[##fi $ffi iiiiffi ffi plicariffiBlood TypesAs a general rule, natural sugars have D-chirality. This is because our enzymesrecognize D-sugars, and therefore select for D-sugars. One notable exception tothe "D only" rule includes the presence of L-fucose, a reduced form of L-galactose, in the glycoproteins of blood type markers. An oligosaccharidedefined as a polysaccharide with anywhere from two to eight monosaccharides)

Ls bound to the amino terminal of a protein on the surface of a red blood cell tofunction as the antigenic determinant. Figure 6-25 shows the terminal sugars of:he three common antigenic markers. You should be able to ascertain from theslructures why type O is the universal donor.

Type O (Glycoprotein)

ln

Types A and B (Glycoprotein)

HO CF{.OHcF{2oH

=OH(for Type

HO

HO

CFLOH"1

HO

Figure 6-25

--- ucose is the sugar with a CH3 group, rather than a CH2OH group, at the:::bon 6 position. It is bonded to the central galactose residue via an alpha-7,Z-

:*. :osidic linkage. Where type O differs from types A and B is that is does not-::.-e a fourth monosaccharide linked to hydroxyl 3 of the central galactose::,i:Cue. Types A and B differ from one another at the carbon 2 position of the' ':rth sugar. Type B has an ordinary galactose residue as the fourth sugar, and:--:s has a hydroxyl group on carbon 2, while type A has the carbon 2 hydroxyl::up replaced by an N-acyl group.

o:1 6\1CH. prote

o

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Organic Chemistry Carbohydrates Biologicat Apptications

ATP

Glyceraldehyde 3-phosphutu Aldolur"Triose phosphate isomerase

Step V:

CF{2oPo"2- AG = o'6 kcal/mole

Glyceraldehyde-3-phosphate Dihydroxyacetone phosphate

2H2O

4 ADP

2 NADH

Step II:AG = -0.6 kcal/mole

Phosphoglucoseisomerase_.--

Step III:AG = -5.3 kcal/mole

Phosphofructokinase

cttoPo32-

OH

OHHFructose-1,6-bisphosphate

4 ATP CFI3

Pyruvate

f'11'3t fu,ou

OHHFructose-6-phosphate

K:

ll

i(

r{

;[

fi

m

0

l0

M

tfi

fit

Glycolysis HighlightsGlycolysis is a ten-step metaboLic process that converts glucose into two pyruvatemolecules. During the process, two net equivalents of adenosine triphosphate(ATP) are formed from adenosine diphosphate (ADp) and inorganic phospnate.Glycolysis is oxidative in nature (given that glucose is being bioken do#n;, sotwo equivalents of the oxidizing agent NAD+ are required". However, becauseglycolysis takes place in the cytoplasm, the NAD+ levels are kept low, so it getsused up quickly. Figure 6-26 shows the overall glycolysis reaction.

o-o

2 ATP 2 ADP

qd,m

s,l@

rd

HOHGlucose

Step I:AG = -8.0 kcal/mole

Hexokinase

ADP

Figure 6-26

Glycolysis is presented in detail in the Biology II book. The goal in presentingaspects of glycolysis in the organic chemistry book is provide biologicai example!of some common organic reactions. The MCAT is a thinking test, so we willfocus on the big picture and the logic behind selected steps. Tire energy valuesfor each step are derived from in aiao cond,rtrons, not standard .or-rditior-rr.Glycolysis can be segmented into three stages, the first of which converts glucoseinto two equivalents of glyceraldehyde-3-phosphate (steps I - v), the seJond ofwhich__converts glycera.ldehyde-3-phosphate into a-phosphoglycerate (steps VIand vII), and the last of which converts 3-phosphoglyceraie ;ito pyrrr.rute ('stepsVru - X). Figure 6-27 shows Stage I, the first five steps, of glycolysis

t&TN

Mffithrffi

mfin

m!L@0

Arn

L@m

m@

ffi

OHHOHGlucose

HOHGlucose-6-phosphate

o\-tI

Ho-1_HcFI2OPO32-

""Y;Ct-t.OI'O.--

-

ILJ

t., Step IV:| .-

AG = -0.3 kcal/molecFI2OH

ft@t@

IWT

@-whr{lrlndj

iil ?

@"

q-,l0[

2 NAD+

cFI2OPO32-

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Figwe 6-27

The Berkeley Review


step I (First committed step in glycolysis): Conversion of glucose into glucose-r-phosphate releases a great deal of energy, so it is the first committei step inllvcolysis. A pathway's first committed step is the first essentially irreversible:eaction in the pathway. This is true in many mechanisms. once a highly:avorable step transpires, the rest of the steps in the mechanism rapidly follow.-n the case of step I, in addition to the energetics not allowing for the back:eaction, once glucose is phosphorylated, it carries a negative charge, so it cannottffuse through the plasma membrane. Although glucose is too ptlar to diffuse-:"rrough the membrane on its own, it crosses with the aid of proteins called:lucose transporters. Glucose transporters act like enzymes in th;t their activity-' reversible (if glucose levels get too high, then they can pump it out) and.:ecific (they recognize glucose, not glucose-6-phosphate). step I ii a regulatory':ep, meaning that the activity of the enzyme is controlled (allosterically)by the-.r-els of ATP and other metabolites. Regulation always occurs at irreversible.:eps (rather than reversible steps), so as a helpful hint, you could look for steps::upled with the conversion of ATP into ADp, because that generally makes the::ep exergonic (AG < 0). Steps I and III are points of regulation.

5teps IV (Reverse aldol reaction); Following isomerization of an aldose into a-*:tose and the addition of another phosphate group in steps II and III, the: rlecule is poised to undergo a reoerse aldol rcaction. tn step IV, a six-carbon:.:lecule is cleaved into two three-carbon molecules by a reveise Aldol reaction:-at breaks the bond between carbon 3 and carbon 4 of fructose-1,6-diphosphate

^lorm dihydroxyacetone phosphate (DHAP) and glyceraldehyde-3-phosphate

--3-P)' Fructose-1,6-bisphosphate is a B-hydroxyketone, which undergoes the::.'erse aldol reaction. In step v, DHAP is converted into G-3-p by wiy of an-'''nerization reaction. By having so many steps, some of the intermediate:rLrducts in glycolysis can be used in other pathways. one example is the:,*'ersion of G-3-P into glycerol and then into fatty acid triglycerides.

---' of step v, no oxidation has occurred. In stage II, the carbon-based:: rpounds are oxidized, releasing energy. The energy is "stored" in a phosphate:',rd. It is at this point that NAD+ is required. Figure 6-28 shows stuge ri ori ,-;olysis, in which G-3-P is converted into 1,3-bisphopshoglycerate.

Phosphoglyceratekinase --Z-T

ADP ATP

'-. ceraldehyde-3-phosphate ^ Step VI: l,3-Bisphosphoglycerate Step VII: 3-phosphoglycerateAG = -0.4 kcal/mole AC = 0.3 kcal/mole

Figure 6-28

i:ep VI (Oxidation of an aldehyde into a carboxylic acid derivative): Oxidative:r:sphorylation of G-3-P into 1,3-bisphosphoglycerate is the only oxidation step- :ir-colysis. It is the oniy step where the number of bonds from carbon to

, - ien increases. oxidation releases energy, which is "stored" by adding a----<phate group. oxidation occurs at carbon L as it gains a bond to o*yg"t-tr::n converted from an aldehyde into an acid carboxylate. The term pir j-cates that there is an inorganic phosphate gained by the reactant. At pH =- - inorganic phosphate exists mostly in its dianion form. Dephosphorylati,on of

:-bisphosphoglycerate into 3-phosphoglycerate in Step ViI, releases a" - - sphate, which serves to regenerate ATP that was invested in earlier steps.

o) oPo32-

H-r oH

cH2oP032-

o\zH-f ou

cH2oPo32-

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In the last stage of glycolysis, the compound loses a phosphate to generate moreATP and then undergoes isomerization and elimination to generate the finalproduct, pyruvate. These three steps are shown in figwe 6-29.

(

(

a

0

s

ffi

@

Iffi

M

ffi

M

3-Phosphoglycerate 2-Phosphoglycerate

"Y"H-roH

curoror2-

Step VIII:AG = 0.2 kcal/mole

Phosphoglyceromutase"\"uf oPor2-

cFI2OH

Step IX:AG = -0.8 kcal/mole

o-

o

o\ol"ll-

oPo3-

cttPhosphoenolpyruvate

Enolase_

YHzo

ADP + H*

Step X:

CHs

Pyruvate

ATp AG = -4.0 kcal/mole

Figure 5-29

Steps IX (Enolate formation): Conversion of 3-phosphoglycerate into 2-phosphoglycerate positions the phosphate for enol formation in Step IX. Becausethese two structural isomers have the same bonds, this reaction results in aminimal change in energy. Conversion of 2-phosphoglycerate into phosphoenolpyruvate involves dehydration. The double bond formed from elimination has aphosphorylated hydroxyl group attached, making the compound an enor.Conversion of phosphoenolpyruvate into pyruvate completes glycolysis.

You should not memorize glycolysis, as the amount of time invested to do sowould not have a good return in terms of increasing your MCAT performance.However, you should have a basic idea of what happens overall. It is moreimportant that you recognize the type of reaction than it is to know details of anyparticular step. A passage may present one or more steps of glycolysis and askyou questions from both an organic chemistry and biochemistry perspective.You should know that glucose is converted into pyruvate and that there is a netgeneration of 2 ATP and a release of energy. It is an oxidative process, so amolecule is broken down. You can appty the general principles discussed here toother pathways, so learn the concepts and insights.

1) Regulatory steps usually involve ATP.

2) Multipstep pathways produce intermediate products that can feedother pathways.

3) Oxidation breaks down molecules and requires an oxidizing agent (aspecies poor in FI, such as NAD+). Oxidation releases energy.

@il

mhmuq

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Organic Chemistry Carbohydrates Biological Apptications

Oxidation and Reduction of PyruvateOnce formed, pyruvate has a few potential fates. It can undergo either aerobic oranaerobic breakdown. Pyruvate can undergo fermentation, where carbon 2 isreduced from a ketone into an alcohol, resulting in the formation of lactate,l'hich regenerates NAD+. we cannot reduce lactate any further (only yeast andbacteria can convert pyruvate into ethanol), so it is excreted as waste. pyruvatecan also undergo oxidative decarboxylation. Converting a carboxylic acid groupinto co2 results in the gain of a bond to oxygen, so it is oxidation and m.tittherefore be coupled with reduction of NAD+ into NADH. To make thesereactions easier to grasp, you should focus on the oxidative or reductive nature of'Jre pathway. Figure 6-30 shows glycolysis, followed by three potential pathwaysJrat pyruvate can take.

Glycolysis

-__+

IRI

o\o

F" rRl>

CHs [o]>Pyruvate

O Lactate

OH

CHs

CO2 + CH3CH2OH

CO2 + Acetyl CoA

o

H

HOHO

OHGlucose

Figure 5-30

latabolism is the biological process of breaking down a sugar. Given that the:eneration of ATP is considered critical to energy storage, let us finish by briefly-:oking at an ATP molecule. It is made of a nucleoside (comprised of adenine:nd ribose) and a triphosphate group linked at hydroxyl 5. Adenosine:rphosphate is shown in Figure 6-31.

Nucleoside

-O

- P-; O- P- O-P - O-rt I Io-l o- o-I

4Adenine

oil

N

(

Hydrolysis of ATP cleaves hereand releases -7.3 kcal/mole

OH HORibose

Figure 6-31

-- nucleoside is a monosaccharide (usually ribose) bonded at carbon 1 to one of,--e bases in either DNA or RNA. Linkages between phosphate groups generate- 3 kcal/mole when they are hydrolyzed, meaning that when ATp gains H2O, it,-.aves to form ADP, P1 (inorganic phosphate), and H+ with a LG = -7.3, :al./mole. This is a valuable tool for storing energy in small increments.

Triphosphate

r19 Exclusive MCAT Preparation

Organic Chemistry Carbohydrates Section Summary

Key points for Carbohydrates (Section 6)Monosaccharides

1. Sugars are named for carbon count and carbonyl typea) Tetrose = 4 C, Pentose = 5 C, and Hexose = 6 Cb) Aldose = aldehyde sugar; Ketose = ketone sugarc) D-sugar has last chiral center = R; L-sugar has last chiral center = sd) common monosaccharides from biorogy shourd be memorized

i. Ribose: "Ribose is all rightl"ii. Glucose: "F*** glucose!", and flip it off.iii. Mannose: "Man goes with gun.,'iv. Galactose: C-4 epimer of glucosev. Fructose: ketose of glucose

2. Cyclic sugars are either furanoses or pyranosesa) Pyranoses are six-membered ring sugars

i. Beta anomer has last carbon cis to anomeric hydroxyl groupii. Anomeric carbon is the most oxidized carbon (has two bonds to o)iii. B-Glucopyranose has all substituents in the equatorial position

b) Furanoses are five-membered ringsi. Ribose and fructose are the two most common examples

Oligosaccharides and Polysaccharides

1. oligosaccharides contain 2 - 8 monosaccharides, polysaccharides contain ?a) Glycosidic linkages bind them (water is released when linkage forms)

i. Disaccharides include sucrose, lactose, and maltoseii. Polysaccharides include cellulose, amylose, and glycogeniii. Linkages are o or B, named for the component hydroxyl groupsiv. Enzymes that cleave glycosidic linkages are specific

b) Branching is typically done using hydroxyl 6 of a sugar in the polymeri. Cellulose (B-linked) and amylose (*-linked)have no bran&ingii. Amylopectin (cx-linked) has branching about 1 out of 30 monomersiii. Glycogen (u-linked) has branching about 1 out of 10 monomers

Chemical Reactions and Tests

1. Chemical reactions and tests determine chirality, add Cs, or remove Cs.a) Nitric acid (HNo3) oxidizes terminal Cs to acids: optical activity testb) Phenylhydrazine (OHNNH2)forms an osazone: C-2 epimer testc) Tollen's test (Ag+) oxidizes C-1 of aldoses: aldose testd) Periodate test (IOa-) oxidizes sugar by cleaving all C-C bondse) Kiliani-Fischer synthesis increases an aldose by one C; forms epimers0 Ruff and Wohl degradation decrease an aldose by one C

Biological Examples

1. Blood types have glycoproteins with different stereochemistrya) Type o contains 3 sugars (Glc-Gal-Fuc) and Types A and B contain 4b) L-Fucose is the third sugar, de'iating from the natural sugars are D rule

2. Glycolysis breaks down glucose into two pyruvate molecules and energya) Forms a net of 2 ATP and 2 pyruvates for further reactionb) FIas one oxidation step; mostly isomerization and phosphate exchange

Copyright O by The Berkeley Review t20 The Berkeley Review

Organic Chemistry Carbohydrates Section Summary

Key Points for Carbohydrates (Section 6)

Monosaccharides

1. Sugars are named for carbon count and carbonyl typea) Tetrose = 4 C, Pentose = 5 C, and Hexose = 6 Cb) Aldose = aldehyde sugari Ketose = ketone sugarc) D-sugar has last chiral center = R; L-sugar has last chirar center = Sd) Common monosaccharides from biology should be memorized

i. Ribose: "Ribose is all right!"ii. Glucose: "F*** glucose!", and flip it off.iii. Mannose: "Man goes with gun."iv. Galactose: C-4 epimer of glucosev. Fructose: ketose of glucose

2. Cyclic sugars are either furanoses or pyranosesa) Pyranoses are six-membered ring sugars

i. Beta anomer has last carbon cis to anomeric hydroxyl groupii. Anomeric carbon is the most oxidized carbon (has two bonds to o)iii. B-Glucopyranose has all substituents in the equatoriar position

b) Furanoses are five-membered ringsi. Ribose and fructose are the two most common examples

Oligosaccharides and Polysaccharides

1. oligosaccharides contain 2 - 8 monosaccharides, polysaccharides contain ?

a) Glycosidic linkages bind them (water is released when linkage forms)i. Disaccharides include sucrose, lactose, and maltoseii. Polysaccharides include cellulose, amylose, and glycogeniii. Linkages are o or B, named for the component hydroxyl groupsiv. Enzymes that cleave glycosidic linkages are specific

b) Branching is fpically done using hydroxyl 6 of a sugar in the polymeri. Cellulose (B-linked) and amylose (u-tinked) have no branchingii. Amylopectin (u-linked) has branching about 1 out of 30 monomersiii. Glycogen (o-linked) has branching about 1 out of 10 monomers

Chemical Reactions and Tests

1. Chemical reactions and tests determine chirality, add Cs, or remove Cs.a) Nitric acid (HNo3) oxidizes terminal Cs to acids: optical activity testb) Phenylhydrazine (OHNNH2) forms an osazone: C-2 epimer testc) Tollen's test (Agr) oxidizes C-1 of aldoses: aldose testd) Periodate test (IOa-) oxidizes sugar by cleaving all C-C bondse) Kiliani-Fischer synthesis increases an aldose by one C; forms epimers0 Ruff and Wohl degradation decrease an aldose by one C

Biological Examples

1. Blood types have glycoproteins with different stereochemistrya) Type O contains 3 sugars (Glc-Gal-Fuc) and Types A and B contain 4b) L-Fucose is the third sugar, deviating from the natural sugars are D rule

2. Glycolysis breaks down glucose into two pyruvate molecules and energya) Forms a net of 2 ATP and 2 pyruvates for further reactionb) Has one oxidation step; mostly isomerization and phosphate exchange

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CarbohydratesPassages

l5 Passages

I OO Questions

Suggested schedule:I: After reading this section and attending lecture: passages I, III, V, VI, & XQrade passages immediately after completion and tog"your mistakes.II: Following Task I: passages II, IV, vil, & XI (za questions in J6 minutes)Time yourself accurately, grade your answers, and review mistakes.III: Review: passages VIII, IX, XII, XIII, & euestions 92 - I OO

Focus on reviewing the concepts. Do not worry about timing.

,1fu

B E.V.f.E.Speci altztng in MCAT preparation

I. Qlucose and Qlucopyranose

II. Fischer and Haworth Projections

III. Sugar Conventions

IV. Monosaccharides versus Disaccharides

V. Amylose, Amylopectin, and Cellulose

VI. Unknown LAldopentose trlucidation

VII. Unknown D-Aldohexose Elucidation

VIII. Kiliani-Fischer Synthesis

IX. Osazone Derivative Test

X. Clycolysis Reactions

XI. Blood'lYpes

XII. Biochemistry of Sugars

XIII. Combustion of Sugars

Questions not Based on a Descriptive Passage

(r -7)

(B - t4)

(r5 - 22)

(23 - 2e)

(5O - 56)

(37 - 43)

(44 - so)

(sr - 57)

(58 - 64)

(65 - 7r)

(72 - 78)

(7e - 85)

(86 - er)

(e2 - lOO)

Carbohydrates Scoring Scale


85 - 100 15-1565-84 10 l247 64 7 -953-46 4-6l-32 l-3

oassage I (Questions 1 - 7)

The most stable form of glucose is a hemiacetal ring--!-ture. The ring is formed when the hydroxyl group on

".::on 5 attacks the aldehyde group to form a hemiacetal.*-': h1'droxyl can attack the carbonyl carbon (sp2-center) from

, :.3r the top or bottom of the n-bond. As a result of this. ..x. there are two possible diastereomers that can form.*::

aldehyde carbon forms the new stereocenter. These two- ,.rble diastereomers are referred to as the alpha andbeta' -:ers of the glucopyranose. The beta isomer (anomer) is

": red as having the anomeric hydroxyl (on carbon one) cis. . respect to carbon number six in the ring form. Because- ,:i: itatural sugars have D-configuration, it is perhaps easier

:.:nk of the beta anomer as the one with the hydroxyl-: *r up (equatorial) in the ring form. Both anomers of D_: -: -rvranose are drawn in Figure I below.

OH

; -D-glucopyranose

Figure 1 Anomeric forms of D-glucopyranose

Ire B-anomer is favored by a ratio of 64Vo to 36Vo overll ,r r-:nomer. Hydroxyl groups on adjacent carbons in the

: ,.r form hydrogen bonds from all orientations except. When the hydroxyl groups are equatorial they

':-,rltco less steric repulsion than when they have axial: - :Iion, but hydrogen bonding is slightly reduced., --:: the B-anomer is more abundant, the reduced steric' -,:: .ce of the equatorial orientation must be of greater

r r."1nce than the slight loss in hydrogen bonding..- _:;n bonding is possible from the gauche orientation" :::d with diequatorial hydroxyl groups.

',i-hich is the more stable anomer of D-glucose?

\. The alpha-anomer, because the hydroxyl group hasequatorial orientation.

I

The beta-anomer, because the hydroxyl group hasequatorial orientation.The alpha-anomer, because the hydroxyl group hasaxial orientation.The beta-anomer, because the hydroxyl group hasaxial orientation.

D.

I."Ir-l- o"

*o"cH2oH

H

HO

H

H

\HOHoC

Ho \\ qHo-\--\

oHla-D-glucopyrunor" oH

ts.

HOH,C

OH

;rt @ by The Berkeley Review@ t23 GO ON TO THE NEXT PAGE

2. D- Glucose and D-galactose are best described as whichof the following?

A. Anomers

B, C-2 epimers

C. C-4 epimers

D. Enantiomers

D-Glucose and L-glucose are:

A. anomers.

B. C-5 epimers.

C. diastereomers.

D. enantiomers.

D-Glucose has whicharrangements?

A. 2R, 3R, 4R, 5RB. 2R, 35, 4R, 5Rc. 2s, 3R, 4R, 5RD. 25,35,4R,5R

of the following stereochemical

5 . How many stereoisomers are possible for a linearaldohexose?

4.4B. 8

c. 16

D. 32

In the beta-anomer of D-glucopyranose, the first carbonhas the hydroxyl with what orientation?

A. CisB. Planar

C. AxialD. Equatorial

The oxygen present in the 1,4-linkage in thedisaccharide below is part of what functionality?

3.

4.

6.

7.

HOHIC

"n-1-\.-QHo\-1\-o

A. Ketal

B. Acetal

C. HemiketalD. Hemiacetal

OH

Passage ll (Questions 8 - 14)

The most stable form of most monosaccharides is thecyclic form, rather than the straight chain form. This isparticularly true of the aldopentoses and aldohexoses.Although it is common to represent aldohexoses in theirstraight chain form, for most sugars, less than one percentexist naturally in the straight chain form. The straight chainform of a monosaccharide is represented by the Fischerprojection, while the cyclic form is often represented by theHaworth projection. Cyclic structures come in five and six-membered rings, referred to as furanose anrJ, pyranoserespectively. The Fischer projection of D-mannose ancl theHaworth projection of B-D-mannopyranose are shown inFigure l.

nvono-l- H

no-J- nHI os

Is-l- oH

cH2oHD-mannose

Figure I Fischer and

HHB-D-mannopyranose

Haworth projections of mannose

Polysaccharides, such as cellulose, are formed from adehydration reaction of cyclic monosaccharides. One water islost fbr every glycosidic linkage that is formed. Theglycosidic linkage between the two monosaccharides in adisaccharide involves the anomeric hydroxyl of one sugar(referred to as the glycosyl group) and a hydroxyl group of asecond monosaccharide (referred to as the glycoside). Forinstance, 4-O-(B-D-galactopyranosyl)-a-D-glucopyranoside,lactose, results from a B-1,4-linkage between hydroxyl-4 ofu-D-glucopyranose and carbon- l of B-D-galactopyranose. Aglycosidic linkage is named for the two hyclroxyl groups thatundergo dehydration to fonn the linkage.

I . Which ol the fbllowing Haworth projections representsthe C-3 epimer of R-D-mannopyranose?

A. cH,oH B.

CHrOH

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10.

A polysaccharide consisting of only B-anomers wi1,4-glycosidic linkages is commonly known as:

A. glycogen.

B. cellulose.

C. sucrose.

D. lactose.

A11 of the following are true of aldopentoses EXCEPT

A. they are likely to form furanose rings.B. carbon 4 determines the designation as D or L.C . carbon 2 is the anomeric carbon in the cyclic fornD . they have four chiral centers in their cyclic form.

1 1. What is the molecularfrom two glucoses and

A . 414 grams/mole

B. 486 grams/mole

C. 504 grams/mole

D . 540 grams/mole

weight for the trisaccharidea mannose with 1,4-B-linkages

12. The glycosidic linkage in most polysaccharides is wh_

of the following?

A. 1,1

B. 4,4c. 1,4

D. 2,s

I 3. The cyclic form of an aldopentose is a:

A. furanose ring with an acetal functional group.B. furanose ring with an ketal functional group.C. pyranose ring with an acetal functional group.D. pyranose ring with an ketal functional group.

14. The following structure has what glycosidic bond?

A

B.

An O-glycosidic bond between the carbormethanol and the anomeric oxygen of cr-mannc,,An O-glycosidic bond between the carbonmethanol and the anomeric oxygen of B-mannor*.An O-giycosidic bond between the oxyge:methanol and the anomeric carbon of c[-manno*An O-glycosidic bond between the oxygermethanol and the anomeric carbon of B-mannose

m

m'

C.

D.

aassage lll (Questions 15 - 22)

Sugars are organic compounds made exclusively of-;ir:on, hydrogen, and oxygen. In a monosaccharide, the ratioi :arbon-to-hydrogen-to-oxygen is I :2 : l. As the name;-:ohtdrate implies, for every carbon atom, there is a water

l:-;cule in the formula. By convention, sugars are nameclirr: rhe number of carbons and given the suffix "ose', torT,:-:ate that they are a sugar. The highest priority functional;:"::o of the sugar is indicated by the prefix. For instance, an, :.,hexose is a six carbon sugar with an aldehyde functionalr'::1. In an aldohexose, the first carbon has an aldehyde'.1:r;rional group and the remaining five carbons have" ,::r,\yl groups. The stereocenters have variety, but the' r.r.-:ctivity is the same. Figure I shows the eight possible- *,'iohexoses.

o

OH

H

H

OH

OH

I - - rGalactose

o

OH

OH

H

OH

cH2oH

-* - i-Gulose

Figure 1 Eight D-aldohexoses

:ructose is also a six-carbon sugar, but it has a ketone,rr:: rnality on carbon 2, rather than an aldehyde functional

rLll"r rr: It is referred to as a ketohexose.

,, aoHoJ-gnolHn-l- oH

u-l- onI

cH2oH

D-(+)-Mannose

"YoHO--+- H

HIoHH-l- oH

n-l- onI

cH2oH

D-(+)-Altrose

HyoH+OHn-]-osn-l- oH

u-f oH

cH2oH

D-(+)-Allose

"YoH--+- OH

no-l- un{osHIon

I

cH2oH

D-(+)-Glucose

"Y"HO-+- H

nlosHoJ-nn-l- on

I

CH2OH

D-O-Idose

CHz

HyoHO-{- H

Ho-l- nso-l- nH-l.-on

I

cH2oH

D-(+)-Talose

,. litl,i:iEht @ by The Berkeley Review@ 125 GO ON TO THE NEXT PAGE

15. All of the following statements about a D-aldohexoseare true EXCEPT:

A . It contains five secondary alcohol groups.B. Its straight chain form has four chiral centers.C . Its most oxidized carbon is carbon l.D. Its fifth carbon has R-chirality.

16. Which of the following sugars can be classified as aketohexose?

A. D-Mannose

B. D-Fructose

C. D-Glucose

D. D-Galactose

17. What is the C-5 epimer of L-glucose?

A. D-Idose

B. L-Mannose

C. D-AlloseD. D-Glucose

18. The C-2 epimer of D-Gulose is also the:

A . C-3 epimer of D-GlucoseB. C-3 epimer of D-Idose

C . C-4 epimer of D-AlloseD . C-4 epimer of D-Altrose

1 9. The Tollen's test involves an oxidizable sugar reducingsilver cation to silver metal, which can then beidentified by its precipitate. Which of the followingsugars yields no silver precipitate?

A. Mannose

B. Sucrose

C. Glucose

D. Galactose

2 0. Sugars naturally exist in cyclic conformations in theirmost stable form. The ring is formed when an OHgroup attacks the carbonyl carbon. The most stableform of fructose would be which of the following?A. Hemiacetal

B. Acetal

C. HemiketalD. Ketal

I1. Which of the following aldohexoses is NOT an epimerof D-talose?

A. D-Galactose.

B. D-Idose.

C. D-Mannose.

D. D-Glucose.

22. Treatment of which of the following sugars withNaBH4 will yield an optically inactive product?

A. Glucose

B. Mannose

C. Ribose

D. Fructose



Monosaccharides typically have a formula of CnH2nOn.where n is an integer. Depending on the number of carbon-sin the monosaccharide, it may have a preference for a cyclicstructure (a hemiacetal or hemiketal) over the straight chainform. To cyclize, the hydroxyl group on the penultimatecarbon attacks the carbonyl carbon, which turns the carbon$carbon into the anomeric carbon. Generally, aldopentoses and

ketohexoses form furanose rings, while aldohexoses forrmpyranose rings, as shown in Figure l.

H-ts (oH)

n*toHrn* rontn*on

cH2oH

D-aldohexose

H-# roHl

n-l- rouln-F oH

cH2oH

D-aldopentose

CH,OH

t-on*ronrn-l- roHr

n-F on

cH2oH

D-ketohexose

HHD-aldohexapyranose

HHD-aldopentafuranose

(oH)

cH2oH

HHD-ketohexafuranose

Figure 1 Generic cyclic monosaccharides

\\tren t*'o cyclic monosaccahides combine, they loservater molecule as a side product. The resulting disacchariis descnbed b1, the common oxygen between the rings.glvcosidic linkage is described by its respective positioneach ring. In almost all disaccharides, one sugar uses

anomeric hvdroxyl group to form the glycosidic linkage.disaccharide is typically drawn with the sugar usinganomeric oxl,gen in the linkage on the left and thesusar on the right. When drawn in this standard method.sugar on the left is given an "osyl" suffix while the sugarthe ri_eht is -eiven an "ide" suffix.

,dt

D.rS.

t"

m@1u

CH2OH

13. The C-2 epimer of galactose is the same as which of thefollowing sugars?

A . The C-2 epimer of glucose.B . The C-2 epimer of mannose.C . The C-4 epimer of mannose.D . The C-4 epimer of glucose.

l{. The linkage in sucrose is best described as:

HOHA. ul-a2B. al-82C. t31-a2

D. B2-a2

OH H"ffd",""

The glycosidic linkage offunctional group?

maltose is what type of

HOHO

A. Hemiacetal

B. Hemiketal

C. Acetal

D. Ketal

The following monosaccharide relates in:o B-D-glucopyranose?

{ . It is an anomer of B-D-glucopyranose.

B . It is an conformer of B-D-glucopyranose.C . It is an enantiomer of B-D-glucopyranose.D . It is an epimer of B-D-glucopyranose.

rnn',ight @ by The Berkeley Review@

o

what manner

27 . A trisaccharide made from three unique aldohexoses has:

A. two acetal linkages.

B. three acetal linkages.

C. two hemiacetal linkages.

D. three hemiacetal linkages.

2 8. Exhaustive methylation of a disaccharide converts all ofthe hydroxyl groups into methoxy groups. Uponhydrolysis under acidic conditions, what is observed forthe fully methylated disaccharide?

A. Al1 of the OCH3 groups return to being OHgroups.

B. Only the glycoside loses OCH3 groups, nor rheglycosyl group.

C . The disaccharide racemizes at every chiral center.D. The two carbons involved in the linkage will have

hydroxyl groups and no methoxy groups.

29 , In the following disaccharide, rhe glycoside is:

H

A. D-galactose.

B. L-galactose.

C. D-mannose.

D. L-mannose.

OH

t21 GO ON TO THE NEXT PAGE


Human beings possess the enzyme for cleaving thealpha-1,4-glycosidic linkage of starches. We can break downpolysaccharides such as amylose, amylopectin, and glycogen.However, we are unable to digest cellulose, because itcontains B-D-glucopyranose residues connected by B-1-4-glycosidic linkages. Foods rich in cellulose pass through us

undigested. Both amylose and amylopectin are made from c-D-glucopyranose, drawn in Figure 1.

HOHO

OH

Figure 1 cr-D-Glucopyranose

Amylose is a polysaccharide made of a-D-glucopyranosewith exclusively cr-1,4-glycosidic linkages. Amylopectin islike amylose, except that about one out of every thirty of theglucopyranose monomers has an a-1,6-glycosidic linkage toa branched strand of o-glucopyranoses. Glycogen also hasbranching, but significantly more than amylopectin. Allthree polysaccharides are ultimately broken down into smallerfragments, such as the disaccharide maltose. We absorb onlymonosaccharides into our bloodstream, so starches must beenzymatically cleaved in the mouth and small intestine.Even disaccharides are broken down into monosaccharides tobe absorbed from the intestine into the blood.

3 0. Which of the following disaccharides is maltose?

A. HOH.TC

"o-1-\- Q HoHoco\\z-ot

OH HO-\-za-\- OH

B. HOHrCHo+\zo.uo\ t \

oH IHOH,Cot-\-o.Ho\-.a-\- ou

C. HOH,Cno's\zo. HoH2c

no\ \\ o*orOH HO\/-t\

D. HOH.C

"""M\b"lXL=o.

Ho\,.a-\oHl

OH

OH

OH


OH

128 GO ON TO THE NEXT

31. Why is amylase released in both saliva and bypancreas into the small intestine?

A . Salivary amylase is destroyed by the gastric fluids.

B. Pancreatic amylase is weaker than salivary amy

C . Salivary amylase is weaker than pancreatic amy

D. Pancreatic amylase cleaves amylose with B-linwhile salivary amylase cleaves o-linkages.

3 2. Which sugar below CANNOT be absorbed into blood?

A. Fructose

B. Galactose

C. Glucose

D. Lactose

3 3. The difference between amylose and amylopectin is

A . amylose has more branching.

B. amylose has B-1,6-branching, while amylohas a-1,6-branching.

C. amylose has no branching, while amylopectina- 1,6-branching.

D. amylopectin has no branching, while amylosea- 1 ,6-branching.

3 4 . What is NOT found in glycogen?

A . Monomers of a-D-glucopyranose.

B. o-1,3-linkages

C. Acetal functional groups

D. o-1,6-linkages

3 5. Which of the labeled hydroxyl groups are involvedlinkages in amylopectin?

A. I and II only

B. I and III only

C . I, II, and III only

D. I, III, and IV only

3 6. Amylose is eventually broken down into:

A. cellulose.

B. lactose.

C. pyruvate.

D. sucrose.

lassage Vl (Questions 37 - 43)

Compound A, an unknown L-aldopentose, is treated with-':,: acid to yield Compound B, a diacid referred [o as an." -.;ric acid, as shown in Reaction I below. Compound B'.'",,s no specific rotation ([up] of 0'). This is to say that. :iacid does not rotate plane polarized light when an

u -3truS solution of the diacid is analyzed using a polarimeter.

Compound A HNOr -

ComPound B( apl ol0")

Reaction ICompound A, when treated with a sequence of reagents

'r i. .s a slight modification of the Kiliani-Fischer synthesis, --:h increases the length of an aldopentose by one carbon),

' :. trvo diastereomers that vary in their chirality at only.: :rrbon. The two epimers are referred to as Compound C- : :ompound D. Carbon I of Compound A is labeled withI The I3CNMR peak that starts at 202 ppm in:round A but is fbund at 78 ppm in both Compound C

-: ,lompound D. This implies that the l3C label is not on., .-Jehyde carbon in Compounds C and D. Reaction 2 is'

' n below.

ItrmPound A

1. HCN

2. H2lPd(BaSO a)Compound C

+Compound D3. I{jO+ / A

Reaction 2

,ltrmpound C when treated with nitric acid yields a diacidr :-.I (Compound E) that shows a specific rotation ([op])

-:'. Compound D when treated with nitric acid yields a

.*.: product (Compound F) that shows a specific rotationol 0'. This is to say that compound E is optically

, :. while compound F is optically inactive. The:-,, .1nS ore shown below as Reaction 3 and Reaction 4:

i-ompound C HNO: -

ComPound E( ap.l of -26")

Reaction 3

-ompound D HNo: -

ComPound F([ upl oi 0".r

Reaction 4

-- sing the optical rotation data presented, it is possible to

r,. --3 the identity of the unknown L-aldopentose (Compoundihe chirality of each hydroxyl group can be determined

r : -: r time from the optical activity data.

- -{orv many stereoisomers are possible for an:ldopentose?

\. 2

B. 4

c. 8

D. 16

" :reht @ by The Berkeley Review@ 129 GO ON TO THE NEXT PAGE

38. Compounds C and D are BEST described as which ofthe following?

A. AnorrersB. C-2 epimers

C. C-3 epimers

D. E,nantiomers

3 9. Which of the following sugars is NOT possible for theidentity of Compound A?

A. Ribose

B. XyloseC. Glucose

D. Lyxose

4 0. Compound B has which of theITIASSES?

A . 132 grams/mole

B. 148 grams/mole

C . 164 grams/mole

D . 180 grams/mole

following molecular

41. The melting point of Compound B:

A . is greater than the melting point of Compound A.B . is equal to the melting point of Compound A.C . is less than the melting point of Compound A.D. is less than the boiling point of Compound A.

42. Aldohexoses in their most stable form would NOTshow which of the following IR peaks?

A. 1140 cm-1

B. 1715 cm-1

C. 2980 cm-lD. 3480 cm-l

4 3. The second carbon (C-2) for Compound A:

A . must have R stereochemistry.

B. must have S stereochemistry.

C . can have either R or S stereochemistry.

D. is achiral.


Compound X, an unknown D-aldohexose, undergoes asequence of chemical reactions. Compound X is first treatedwith nitric acid and the purified product (Compound y) isplaced into a polarimeter to analyze its optical activity.Compound Y shows an optical rotation, [sO = 26t].Compound X is then treated with the following sequence ofreagents: first bromine in water, second ferric carbonate inwater, and finally hydrogen peroxide. The product(Compound Z) has a molecular mass of 150 grams/mole.Compound Z is then treated with nitric acid to yield a newproduct (Compound W), which when placed into apolarimeter shows no optical activity.

Compound X is then treated with one equivalent ofperiodic acid, which breaks the bond between carbons 3 and4,oxidizing carbons 3 and 4 up one level to aldehydes. Thissix-carbon hemiacetal fragment (Compound e) is thenhydrolyzed out of the hemiacetal form and converted into twosmaller fragments, both three carbons long (Compound R andCompound S). From the optical rotation data, the researcherconcludes that carbons 3 and 4 have opposite chirality. Thereactions are summarized in Figure I below:

o\" ^o, -l-Ht(oH)

H_F fOgr L Br2(aq)

Ht (OH) 2. Fq(CO3)3raq)

Hf oH 3. Hzoz(aq)

cH2oH

Compound X

Compound Yoptically active

Compound Z(MW = 150 gimol)

l*o,V

Compound Woptically inactive

"5*" '\'""'."i:i":.ooo

HOHCompound R

HIOa

Compound Q

HOH &Compound 5 ComPound S

Figure 1 Reaction summary of Compound X experiment

The reaction involving the Br2 liquid in water is knownas Ruff degradation, which removes carbon I and oxidizescarbon 2 into an aldehyde, while leaving the sugar chainintact from carbon 3. The stereochemistry of the backbonedoes not change during this reaction. The structure ofCompound X can be deduced from this information.

44. Treatment of Compound X with sodium borohydride(NaBHa) would yield which of the following?

A. An optically active diacid.B. An optically inactive diacid.C . An optically active hexa-ol.D. An optically inactive hexa-ol.

Copyright @ by The Berkeley Review@ 130 GO ON TO THE NEXT P

45. The stereochemistry of carbon 3 in the Compound X iA. definitely R.

B. definitely S.

C . variable (either R or S).D. achiral (no chiral center).

46. Compound X, when treated with hydrogen cyfollowed by hydrogen gas over palladium metalbarium sulfate and then treated with acidic wateryield which of the following?

A. A D-aldopentose.

B. An L-aldopentose.

C . A D-aldoheptose.

D. An L-aldoheptose.

47. A reverse aldol reaction involving Compound Xan aldotetrose (formed by breaking the C-2:C-3The aldotetrose when treated with sodium borohyyields an optically active tetra-ol. This confirms whiof the following conclusions about the chiral centersCompound X?

A . Compound X has chirality of 3R and 4R.B. Compound X has chirality of 35 and 4R.C . Compound X has chirality of 4R and 5R.D . Compound X has chirality of 45 and 5R.

4 8. How many stereoisomers are possible for thechain form of a D-aldohexose?

A. 4

B. 8

c. 16

D. 32

49. Which of the following reagents, when addedaldohexose, would yield the same conclusionchirality as adding HNO3?

A. NaBHa

B. H2SOa

C . Br2(aq)

D. KMnO4(aq)

5 0. Treatment of an aldohexapyranose with one equof HIO4 breaks the sugar between:

A. carbons 1 and2.B. carbons 5 and 6.

C. carbons 2 and3.D. carbons 3 and 4.

Passage Vlll (Questions 51 - 57)

The length of a sugar can be increased by one carbon-,:ns the Kiliani-Fischer synthesis. Kiliani-Fischer, --:hesis employs a cyano nucleophile in the first step. The:, ino group attacks the carbonyl carbon in an addition-'.,-tion. The cyano group is then reduced into an imine' .r;h is subsequently hydrolyzed into an aldehyde. An: .nple using ribose is shown in Figure 1.

o

H

OH T

OH

OHCH2OH

- - +)-Ribose CH2OHD-(+)-Altrose

Figure I Kiliani-Fischer synthesis starting with ribose

\ researcher carried out the Kiliani-Fischer synthesis on,: different D-aldopentoses to determine the effects of steric

tr - -.ance on stereoselectivity. The four D-aldopentoses are,ir .,. n in Figure 2.

yo 'yo Hyo

To" HtoH HotH Ho-lHTOH HtOH HtOH HtOH

cH2oH CHrOH cH2oH CH2OHD-Xylose D-Lyxose- -Ribose D-Arabinose

Figure 2 Four D-aldopentoses

Table 1 lists the product distribution for each synthesis.

Table 1

To synthesize the following D-aldohexose, what.ldopentose should be used?

o

oH r. HCN/KcNOHffioH 3. Hro+/ A

HO

H

H

H

H;4,oH{onHIoHuI oH

u-f oH

cH2oHD-(+)-A11ose

rldopentose Major Product Major Product

D-Ribose 81% D-Altrose 197o D-Allose

l-Arabinose '72Vo D-Glucose 28% D-Mannose

)-Xylose llVoD-Idose 297o D-Gulose

)-Lyxose J6Vo D-Galactose 24Vo D-Talose

O OHOHH OH

cH:oHH H H OHH

-\. D-Arabinose

B. D-Lyxose

C. D-Ribose

D. D-Xylose

:lsht @ by The Berkeley Review@

5 2. Which of the following pair of sugars are C-2 epimers?

A. D-Ribose and D-Altrose

B. D-Glucose and D-Talose

C . D-Mannose and D-AlloseD . D-Idose and D-Gulose

5 3 . In which of the steps in Kiiiani-Fischer synthesis is thesugar reduced?

A. Steps I only

B. Steps II only

C. Steps I and II only

D. Steps I, II, and III only.

54. Which of the following is anKiliani-Fischer synthesis ?

A. Aldehyde

B. Amine

C . Ester

D. Imine

intermediate in the

55. The product mixture following Kiliani- Fischersynthesis is best described as:

A . two structural isomers in equal concentration.

B. two structural isomers in unequal concentration.

C. two diastereomers in unequal concentration.

D. two enantiomers in equal concentration.

5 6. Which structure represents B-D-talopyranose?

5 7 . Kiliani-Fischer synthesis converts an:

A. aldohexose into an aldopentose.

B. aldohexose into a ketohexose.

C. aldopentose into an aldofuranose.

D. aldopentose into an aldohexose.

B.A.

131 GO ON TO THE NEXT PAGE


An unknown sugar is extracted from the fruit collectedfrom the Racaniqui Tree native to Willoughby, Montana.The sugar is isolated by coiumn chromatography in 99.9Vopurity as shown by lgNUR. The moleculai*us was fbundto be approximately 342 +2 grams per mole. To have thatmolecular mass, the unknown sugar must be a twelve-carbondisaccharide. The sugar shows a specific rotation of +126.2. .

The disaccharide linkage is cleaved rather easily withmild acid, and two six-carbon monosaccharides, Compound Iand Compound II, are isolated. When treated with threeequivalents of phenylhydrazine, both monosaccharides formosazones. Compound I yields an osazone with a specificrotation of +42.6" while Compound II yields an osazone witha specific rotation of +31 .2' . Figure I shows the reaction ofglucose with three equivalents of phenylhydrazine.

o

3 OHN-NH2

H./, N-NHO

I

- N-MroHO-t- H

n{ onu-f oH

cH2oH

H

HO

H

H

OH

H

OH

OH

HOAc

cH2oH+ NHj + C6H5NH2

Figure 1 Osazone formation starting from glucose

The specific rotation of the osazone of D-glucose is+54.6'. Both Compound I and Compound II are naturallyoccurring sugars, so they are assumed to be D-sugars. Whenboth sugars are treated with nitric acid, each one is oxidizedinto an aldaric acid (a diacid with carboxylic acid groups atboth carbon one and carbon six). Both of the aldaric acidsformed are optically active. This implies that neither aldaricacid is meso and that both sugars are aldohexoses, rather thanketohexoses. A 2-ketohexose can also react with threeequivalents ofphenylhydrazine to form an osazone derivative.

The information from the derivatives naffows the identityof each aldohexose to one of four choices. To identify theoriginal sugar, the physical properties such as melting pointand specific rotation can be compared to the values of the fourpossible D-aldohexoses. To determine the linkage of thedisaccharide, the hydroxyl groups can be glycosidicallylabeled. After hydrolyzing the labeled disaccharide into itstwo monosaccharides, the sugar with hydroxyl four unlabeledis the glycoside (sugar on the right in standard notation).This allows for the disaccharide to be identified precisely.

58. How many stereocenters are present in the originaldisaccharide?

4.1B. 8

c. 10

D. 11


5 9. Which of the following pairs of structures crepresent Compound I and Compound II?o''

(oH--+ OH

uo-l-r+H-J- oH

H-f oucH2oH

"x""ao-|.uHoInH-f on

cH2oH

HO..F H

nolHuoIHH-f on

cH2oH

u' '' (o

HO-+- H

n-l- osuI oH

H-f oH

cH2oH

t

""y" ""y"HO-+H HC

uo-l- u H{ onH-l- ou HoInH-f ou H-f oH

CH2OH CH2OH

" "Y*u-]-on

Ho-l--nn-f on

cH2oH

o

H

OH

H

o'n (o

H+- OH

H-l- oH

r-f- otCH2OH

"'tYHO-+

'+HotCH

o

H

OH

H

zoH

B.H

H

HO

H

"

""Yx+

CH

o

H

OH

OH

zoH

ild

d

I(l

,m,iltr(D

HO

H

HO

H OH

CH"OH

60. Which of the following aldopentoses will yieldoptically active aldaric acid?

T",#n*o"

cH2oH

132 GO ON TO THE NEXT PA

6 1 . Which of the following sugars CANNOT be Compoundi or Compound II?

I. Mannose (the C-2 epimer of glucose)

II. Talose (the C-4 epimer of mannose)

m. Galactose (the C-4 epimer of glucose)

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

Which of the following choices shows a sugar pairedwith the incorrect reason for eliminating it as a possiblestructure for either Compound I or Compound II?A. oB. o

cH2oHWill yield an opticallyinactive aldaric acid

H

H

OH

OH

cH2oHWill yield the sameosazone as glucose

D.

CHrOH

fono-l- su-]-on

IH-]- OH

cH2oHWill yield the sameosazone as glucose

H

H

H

H

HO

HO

H

H

OH

OH

OH

OH

C.o

HO

H

HO

H

H

OH

H

OH

CHrOHWill yield an opticallyinactive aldaric acid

Which of the following choices contains two sugarsthat will form the same osazone?

A. D-ribose and D-xylose (C-3 epimers).B . D-idose and D-altrose (C-4 epimers).C . D-talose and L-talose.D. D-mannose (C-2 epimer of D-glucose) and D-

fructose (ketose of D-glucose).

\\rlrat happens to the three phenylhydrazine molecules?

{ . One adds to the sugar and two are oxidized.B . Two add to the sugar and one is oxidized.C . One adds to the sugar and two are reduced.D. Two add to the sugar and one is reduced.

n"4.

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ATP ADP

Phospofructokinase-ry

uJ- onHo-l-HnJ-oHn-l- on

CH.,OH

CHrOH

HonotnuIoHH-l- oH

"YoH-F OH

HoInHtonn-f- oH

CH2OPOj2

CH"OPO"2-

fouo*nH*osuf os

CH2OPOj2-


The early part of glycolysis involves the conversion ofD-glucose into glyceraldehyde-3-phosphate. Glycolysis iscarried out in ten enzymatically catalyzed steps. The firstfour steps of glycolysis, forming glyceraldehyde-3-phosphate,are shown below:

STEP I: Glucose to Glucose-6-phosphateAG"=-4.0kcal/mole

cH2oHrLHtoH FoHo-t-H tsomerase no-f-njT-n

-

a n-Tun

HtoH -louCH2OPO32- CH2OPO32-

STEP II: Glucose-6-phosphate to Fructose-6-phosphate

AG"=+0.4kcallmole

cH2oPo32-ATP ADP

STEP III: Fructose-6-phosphate to 1,6-BisfructophosphateAG'=-3.4kcallmole

CHTOPOT2-L:^l-' cHroPOr2- 'voHO--H Aldola.^ i -- -J

l-tt"-En+uH+os € -ntoHI CH2OH CH2OpO32-Hf OH

CH2OPO32-

STEP IV: 1,6-Bisfructophosphate to Dihydroxyacetone-phosphate and Glyceraldehyde-3-phosphate

AG'=*5.7kcallmole

Dihydroxyacetonephosphate undergoes isomerization toform glyceraldehyde,3-phosphate. Steps I-IV correspond tocommon reactions in organic chemistry, so the mechanismfor each step follows standard organic chemistry processes.

The addition of the phosphate group in Step I is similar toesterification of a carboxylic acid with an alcohol. The roleof the enzyme is to align the molecules so that the hydroxylgroup on carbon 6 reacts to gain the phosphate, rather thananother hydroxyl group. The conversion of an aldehyde intoa ketone in Step II goes through a process known astautomerization. Step III is similar to Step I. The breakingof the carbon-carbon bond in Step IV is an example of theretroaldol reaction. The formation of a B-hydroxyketone froma carbonyl is a common reaction in synthetic chemistry.

65. G-3-P, glyceraldehyde-3-phosphare, and DHAp,dihydroacetonephopshate are best described as;

A. configurational isomers.

B. diastereomers.

C. epimers

D . structural isomers.

6 6 . If D-galactose (the C-4 epimer of glucose) undergoes thesame four reactions shown in the passage, how will theproducts differ from the products of glucose?

A . G-3-P will have a different chiral center.B. DHAP will have a different chiral center.C. Both G-3-P and DHAP will have different chiral

centers.

D. The products will be identical to the productsformed by the glucose reaction.

67 . If carbon 5 of a single D-glucose is enriched with 13C,

where will the enriched carbon (label) be found in theproducts?

A . Carbon 1 of DHAP.B . Carbon 3 of DHAP.C . Carbon 2 of G-3-P.D . Carbon 3 of G-3-P.

6 8. What is the overall free energy change from Step I toStep IV of glycolysis?

A . + 2.1 kcals per mole.B. + 1.3 kcals per mole.C. - 1.3 kcals per mole.D . - 2.1 kcals per mole.

6 9. How many chiral centers are present in the two finalproducrs (c-3-P and DHAp)?

A. 0

B. Ic.2D. 3

Copyright @ by The Berkeley Review@ t34 GO ON TO THE NEXT P.

gH2oPg32'

7 0. Which of the following sugars is fructose-6-phosphate.'

CH"OH cH2oPo12

71 . Glycolysis converts glucose into:

A. two pyruvates and generates a net two ATps.B . two pyruvates and generates a net four ATps.C . three pyruvates and generates a net two ATps.D. three pyruvates and generates a net four ATps.

Pd

ptrr

pil-I

mhr

Ifi[B

lfmrr

hdim,

Ben

mfu

dMSmffiWruflmr

D.


Blood types in human beings are determined by the,,,:r saccharide derivative attached to the amino terminal of a: :tein in the wall of the red blood cell. The success of a

: . od transfusion depends on the compatibility of antibodies

--.. recognize the markers on red blood cells as equivalent or,,:isn. If the antibodies do not recognize the foreign red- 'rd cells as invaders, then the transfusion is possible.

- :,rd type O, the universal donor, contains a trisaccharide,

' -,,1e types A and B contain tetrasaccharides. Figure 1 shows.-. oolysaccharide associated with Type O blood, Figure 2' '.,, s the polysaccharide associated with Type A blood, and

: :-re 3 shows the polysaccharide associated with Type B

cH2oH

\ -o*o-\-/"Ho\$o NHIo NHI

J o{oH*+ o4 o, I f,o,.in'ry CH:

luoHO

I .ure 1 The polysaccharide associated with Type O blood

HOH,cyzo|.$jtq ct{]oH

,['PS1]-"e\- T/4 o NHIT,! -o).o "Y-"i_.1 CH2OH 7f u \ protein- H.rC CH-r

\HoHO

i zure 2 The polysaccharide associated with Type A blood

"f ."ro" CH2oH

"pPvs-"d\MoNHIT.,{ -n).o, "Y-"-q

_ I CH2OH .4- ,___t \ proteinH.rC L/f cHr

\"oHO

ri llure 3 The polysaccharide associated with Type B blood

In the three blood types, the first sugar attached to the':' :-n is N-acetylglucosamine. Attached to the first sugar

" . i.4-glycosidic linkage is galactose.

. nght @ by The Berkeley Review@ 135 GO ON TO THE NEXT PAGE

CH"OH

In all three blood types, the galactose forms a bridge viahydroxyl two to a fucose residue. In blood type A, thegalactose has a N-acetylglucosamine linked to hydroxyl threewhile in blood type B, the galactose has a second galactoseresidue linked to hydroxyl three. These polysaccharides arereferred to as the antigenic determinants of the red blood cells.These, along with the Rh factor, determine the feasibility ofa transfusion.

7 2. What linkage is present between N-acetylglucosamineand galactose in blood type O?

A. u-1,2-linkageB. B-1,2-linkage

C. u-1,4-linkageD. B-1,4-linkage

7 3 . How many acetal and ketal functionalities are present inthe type O blood antigenic determinant?

A. 0

B. I

c.2D. 3

7 4. If the CH3 group of fucose were replaced by CH2OH,how would the molecule differ from glucose?

A . It would be the C-2 epimer of D-glucose.B . It would be the C-2 epimer of L-glucose.C . It would be the C-4 epimer of D-glucose.D . It would be the C-4 epimer of L-glucose.

7 5. Which of the fbllowing molecules is B-D-2-acetamino-2-deoxyglucose?

A. B.CH"OH

t

Hod/orNH

/Q:Q \

cHr

D ' cH2oH

Ho$/orHo-\//$/o/

Q:Q \NHe

/oHQ:Q

OH

OH

CH:

c ' cH2oH

"n-^\-o.HO-\//1l6l/oH

NHz

o-c

7 6. The formation of L-fucose from an L-aldohexose wouldrequire what type of reaction?

A . Reduction specific for carbon 6.B . Oxidation specific for carbon 6.

C . Reduction specific fbr carbon 4.D . Oxidation specific for carbon 4.

7 7 . In the type B antigenic determinant, what has been addedto hydroxyl three of the central galactose?

A. a-galactofuranose

B. B-galactofuranose

C. o-galactopyranose

D. B-galactopyranose

7 8. How many chiral centersantigenic determinant?

A. 12

B. 16

c. 20

D. 24

are present in the type A

Copyright @ by The Berkeley Review@ 136 GO ON TO THE NEXT P

Passage Xll (Questions 79 - 85)

Glycolysis is an example of a biological processgenerates energy that is ultimately stored in the formadenosinetriphosphate, ATP. ATp is able to storebecause its phosphodiester bonds are very weak and resultthe release of energy when they under go hydrolysis.hydrolysis of ATP into ADp releases 7.3 kcal/moleenergy. Table 1 shows the energy of hydrolysis for sephophorylated compounds that are common in bi

Compound AG'' (kcal/mole)

Phosphoenolpyruvate -14.8

Acetyl phosphate -10.3

Pyrophosphate -1.9

Adenosine triphosphate (ATp) - t.J

Glucose 1-phosphate -5.0

Glucose 6-phosphate -3.2

Glycerol 3-phosphate aa

Table 1

The AG'' term represents the standard free energy cfor the reaction at a pH of 7.

The conversion from one structure into another stin Table I involves a phosphate group transfer.favorability of the transfer reaction can be ascertainedthe energies of hydrolysis for the two reactions. Thephosphate group transfer agent is phosphoenolpyruvate.loss of the phosphate group by phosphoenolpyruvatesubsequent conversion from pyruvate enol into pyruvshown in Figure 1.

};,",,+;*TFigure I Conversion of phosphoenolpyruvate into

pyruvate enol and pyruvate

The substantially large energy of hydration awith phosphoenol pyruvate is attributed in part tolavorable conversion from an enol into a ketone. Theis the more stable of the two tautomers, so the equiliconstant is greater than I for the conversion of an enolketone.

7 9 . What is the approximate change in free energy forconversion of pyruvate enol into pyruvate?

A . +16.4 kcal/mole

B. +6.7 kcal/mole

C. -6.1kcallmoleD. -16.4 kcal/mole

: I,l . Which of the following compounds can couple with theconversion of ADP into ATP in an overall favorablefashion?

A. Acetyl phosphate

B. Glucose l-phosphate

C . Glucose 6-phosphate

D. Glycerol 3-phosphate

! n. The high energy associated with the hydrolysis of a

compound with a phosphate group can be attributed to:

A. the strong phosphodiester bond, where the strengthcan be attributed to its aromaticity.

B. the strong phosphodiester bond, where the strengthcan be attributed to its strong resonance.

the weak phosphodiester bond, where the weaknesscan be attributed to electrostatic repulsion.

the weak phosphodiester bond, where the weaknesscan be attributed to the inductive effect.

\\rhich enzyme catalyzes the conversion ofphosphoenolpvruvate, PEP, into pyruvate?

C.

D.

{. PEP dehydrogenase

B. PEP reductase

C. Pyruvate kinase

D. PEP aldolase

'Ihen glyceraldehyde-3-phosphate, G-3-P, converts to-.-3-bisphosphoglycerate, what is NOT true?

OPO32-

OH

CH2OPOj?-

A . Carbon I is oxidized.B . NAD+ is required.

C . Carbon 2 undergoes no redox chemistry.

D . Biotin is required.

o\o o\oI AG = - 15 kcal/mole I

,foeo:2 F foH il H ATP ADP cHr

I

l- o" ----->CH2OPO32-

,i' ::_iht O by The Berkeley Review@ 137 GO ON TO THB NEXT PAGE

84. Hydrolysis of which of the following compoundsreleases approximately 5 kcal/moles of free energy?

QPo:2-

cH2oPoj2-

85. How is the structure for fructose-1,6-bisphosphate bestdescribed?

CH2OPOj2- CH2OPO32-

A. It has a furanose ring with alpha orientation.

B. It has a furanose ring with beta orientation.

C . It has a pyranose ring with alpha orientation.D . It has a pyranose ring with beta orientation.

D.

OH

CH2OPO32

CHoOH

Passage Xlll (Questions 86 - 91)

Sugars, when metabolized by animals, generate asubstantial amount of energy. Many naturally occurringmonosaccharides and disaccharides can be extracted from thefruits of plants. Among the most common naturallyoccurring sugars are fructose, sucrose, and glucose. Figure Ishows six common biological monosaccharides.

Figure 1 Six common monosaccharides

The more heat released per gram of a sugar, the moreeffective the sugar is as a fuel. Most thermodynamic data,such as the heat of combustion (AH) and free energy changes(AG) under standard conditions, are listed in terms ofkcals/mole. Table I shows the thermodynamic valuesassociated with the sugars in Figure 1.

noJ-HHo-l-nn-l--,r"u*oH

CH2OH

D-Mannose

CHrOH

tnHo-.|-HH-l- o"u*oH

CH2OH

D-Fructose

"r""noJ-Hno-l-n

"**cH2oH

D-Galactose

"YoH-{- OH

n-l- oHIH-r oH

cH2oHD-Ribose

'vo"-l-.,""oI"H-l- o"H*os

CH2OHD-Glucose

'voH-l- o"

no-l-nn-f- o"

CH2OHD-Xylose

Sugar AHcombustion(kcal/mole)

AGcombustion(kcaUmole)

Glucose -2538 -2827

Mannose -2512 -2801

Fructose -2482 -2711

Galactose -2410 -2759

Xylose -2102 -2341

Ribose -2016 -2315

Table 1

8 6. What is the stereo-configuration for D-xylose?A. 2R, 3R, 4RB. 2R, 35, 4RC. 25, 3R, 4RD. 25, 35. 4R

Copl right @ by The Berkeley Review@ 138 GO ON TO THE NEXT PA

8 7. How much CO2(g) is produced from the metabolism c*ff

10.0 grams of D-glucose, if all of the glucose is usedup in the reaction?

A. 10 x --l - x 44 grams COZG)180

B. 10x I xLx44gramsCO2(g)i80 I

C. l0 x 18q x 6 grams COz(g)44

D. l0 x 180 * L* -Lsrams CO?re)644

88. The BEST explanation for the differences incombustion for the aldohexoses is whichfollowing?

A . The bonds are different in the different sugars.B. The arrangement ofthe hydroxyl groups is di

between sugars, causing differences in hydrobonding.

C . Only some of the sugars have resonance stabilin,.D . Only some of the sugars are aromatic.

89. If the heat capacity of water is 4.18 J/g K, how muchwill the temperature of 1000 g of water insulatingbomb calorimeter rise if it absorbs all of thegenerated by burning 15.0 grams of ribose?

A. 4.91 'C

B. 41.4 "C

c. 49.7 'C

D. 201 "C

90. The amount of CO2(g) produced by combustion wbe greatest from 1.0 gram of which of the followisugars?

A. Glucose

B. Sucrose

C. Ribose

D . A11 monosaccharides produce the same amountCOZ(g) per gram.

91. The amount of energy produced by combustionbe greatest using 1.0 gram of which of the followisugars?

A. Glucose

B. Xylose

C. Ribose

D. All monosaccharides produce the same amountenergy pef gram.

heats

of tk

il

II

m

It,tr

dI(I

mr

m

JII(D

W]{rum

il"f-C.D-

Questions 92 - 100 are NOT basedon a descriptive passage

I2. Decomposition of glucose with excess HIO4 wouldyield which of following products?

A. 4 formic acids and 2 formaldehydes.B . 5 fbrmic acids and carbon dioxide.C . 4 formic acids, carbon dioxide and 1 formaldehyde.D. 5 formic acids and 1 formaldehyde.

1.3 . What is the mass percent of carbon in an aldopentose?

A. 33Vo C by mass

B. 31% C by mass

C . 40Vo C by mass

D . 44Vo C by mass

r-$. After treatment with nitric acid, a sugar is convertedinto a diacid (aldaric acid). Which of the followingsugars would yield an optically active diacid uponrreatment with nitric acid?

A. D-Mannose

B. D-Galactose

C. D-RiboseD. D-Allose

In order for a D-aldopentose toinactive upon treatment with nitric,,r'hat stereochemistry?

become opticallyacid, it must have

{. 2R, 35B. 25, 3RC. 25, 35

D . All of the choices above will lead to an opticallyinactive diacid

lrr r fubose differs from deoxyribose in all of the following',IAYS EXCEPT:

.{. Ribose has more stereoisomers.

B. Ribose has one more chiral carbon.

C . Ribose is the C-2 epimer of deoxyribose.

D . Ribose is more oxidized than deoxyribose.

Lrr - ,\-hen two monosaccharides form a disaccharide, what,.:nes NOT occur during the reaction?

r . The loss of a water moleculeB . The formation of an acetal from a hemiacetal

C . The formation of a glycosidic linkageD . The oxidation of the anomeric carbon

p,, - -1-'1 O by The Berkeley Review@ 139 NO MORE CARBOHYDRATES!

"Just studv itl"

9 8. The Fehling's test is positive when copper goes from anoxidation state of +2 to +1. What change in the sugarcorresponds to a change is in color for the solution inFehling's test?

A . An aldose becomes a carboxylate

B. An aldose becomes a hexa alcohol

C. An aldose becomes an acetal

D. A ketosebecomes acarboxvlate

9 9. What is the molecular formula of a trisaccharide formedfrom two aldohexoses and one aldopentose?

A. C17H36O15

B. C'7H34O15

C. C13H32O16

D. C13H36O1g

10 0. An aldopentose has how many units of unsaturation?

A. 0

B. 1

c.2D. 3

1.B 2.C6. D 1. B

11. c 12. C16. B 11. A21. D 22. C26. D 21. A31. A 32. D36. C 31. C41. A 42. B46. C 47. D51. D 52. D56. C s7. D61. D 62. C66. D 61. C71. A 72. D76. A 71. C81. C 82. C86. B 87. B91. A 92. D96. C 97. D

3. D 4.8 5. C8.D 9.8 10.C

13. A 14. C 15. A18. D 19. B 20. C23. C 24. B 25. C28. D 29. C 30. D33. C 34. B 35. D38. B 39. C 40. D43. B 44. C 45. A48. B 49. A 50. D53. C 54. D 55. C58. C 59. B 60. D63. D 64. D 65. D68. C 69. B 70. A13. D 14. D 15. B78. C 19. C 80. A83. D 84. A 85. A88. B 89. C 90. B93. C 94. A 95. A98. A 99. A 100. B

]IChoice B is correct. According to the rules of steric hindrance, the equatorial orientation is more stable than th.axial orientation for the anomeric hydroxyl group on carbon 1. As stated in the passage, the G-anomer of D-glucopyranose has its anomeric hydroxyl gtot p irthe equatorial position. Because ihere is 64"/obetaanomer an;only 36"h alpha anomer formed ,tpon .yciiration, the beia anomer must be the more stable of the two anomersFrom the drawing in Figure 1 and the information presented in the passage, this question should have been eas,,You really should choose B.

Choice C is correct. Anomers are diastereomers of the same sugar that vary in chirality at the anomeric carbor.Jo b9 the same sugar, the sugar must have the same name. ThiJ eliminates choice A. Enantiomers vary at all c:the chiral centers. In the case of enantiomeric sugars, they are named the same, except for the D or L prefi1,.This eliminates choice D. Glucose and Galactor" ut" C-4 epimers of one another. Epimers are suga:diastereomers that differ in chirality at only one of the carbons ln backbone of the straight

"t ui' form of ih.

sugar' To get this correct requires knowing some sugar facts. Welcome to MCAT preparation and have a nice da,,after you choose C. I -

Choice D is correct. By definition, the D and L forms of the same sugar are mirror images of one another, whirmeans (by definition) tlll lhe two compounds are enantiomers of onJanother. The n aia L forms of a sugar 'tirhave opposite values (differing only in sign) for their optical rotations of plane-polarized light. This'mean,that one-rotates plane-polarized light in a clockwise fishion while its enantiomer rotates plane-polarizeclight in the counterclockwise directi,on. In this question, as would be the case when comparing the D and Lisomers of any other sugar, D-glucose and L-glucose are enantiomers of one another. Choose D for best results.

Choice B is correct. Hopefully the structure of glucose is permanent in your memory (using the ,,right-hancmeJho$' for D-glucose), glucose in its straignt cniin Fischer projection is 2 right (R), 3 iefi (S),"a right in) ana :tlsnl (n) It is important that you remembel tnat hydroxyls on tire teft are s #a nyaroxyls on the iighi are R i:-the Fischer projection of standard aldohexoses. This makes choice B the best unr-"r.

Choice C is correct. An aldohexose has chiral centers present at carbons 2, z, 4, and 5 for a total of four chira-centers in the molecule. The maximum number of stereoisomers is found using the formula 2n where n is thenumber of chiral centers present in the molecule. In this question, plugging in fJur for the value of n yields 2n =24, which equals 16, choice C.

Choice D is correct. Looking at the structures in Figure 1, it can be seen that the anomeric carbon (carbon 1) ha=the hydroxyl group in the equatorial position in the G-anomer. Choose D for optimal satisfaction. Cis and tranare poor answers, because to be cis or trans, it must be relative to somethrng. In actuality, the B-orientation isdefined as cis with respect to carbon six in an aldohexapyranose, but thai t,aJnot implied ir, tni, question.

Choice B is correct. The 1,4-glycosidic linkage between any two monosaccharide units involves the hemiaceta,OH of carbon-1 of one sugar with a hydroxyl oJ carbon 4 of the other sugar. These combine to form an acetal whencarbon 1 is anomeric. Pick B for correctitude sensations. Had the anomeric carbon been carbon-2 (as in the case o:a cyclic ketose), the glycosidic linkage would be2,4 and the functionalitr.n'ould be a ketal.

ffi.

2.

4.

5.

6.

Acetals are formed from an aldehydeqroup, so the monosaccharide is an aldoseHOFIC U I

' \ HCFTCHot\c- o

-\--q,,o\)i\.*:i\Ab"l\ \lH\ "l'Anomeric C L. bo:ri.i :.':OR, OR', C, ANd H lmI

._

:

- i':' *Copl'right O by The Berkeley Review@ 140 C\RBOHYDRATES EXPLANATIONS:

Choice D is correct. When the straight chain form of a sugar is converted into its Haworth projection, thehydroxyl groups originally on the left side of the sugar are found above the ring and hydroxyl groups originallyon the right side of the sugar are found below the ring (recall the mnemonic "downright uplefting;;. Mat''oosehas the hydroxyls on carbons 2 and 3 on the left. The C-epimer of D-mannose therefore has a itraight chainwith only the hydroxyl on carbon 2 on the left. In the Haworth projection of the C-3 epimer of B,D,mannopyranose, the hydroxyl groups on carbons l and2 are above the ring and the hydroxyl groups on carbons 3and 4 are below the ring. This makes choice D the best answer. Choices A and B should have been eliminatedearly on, because they have s-orientation at carbon 1. As a point of interest, choice C is B-D-glucopyranose.

Choice B is correct. This question is best answered from straight memorization, although a good guess can bederived from the passage. Lactose is a disaccharide made from glucose and galactose, not a polysaccharide, sochoice D is eliminated. Sucrose is a disaccharide made from glucose and fructose and not a polysaccharide, sochoice C is eliminated. Glycogen has alpha linkages between saccharide monomers, while cellulose (thepolysaccharide we can't digest) has beta linkages between the saccharide monomers. We lack the enzyme tocleave the B-1,4-glycosidic linkage. Choice B is the best answer.

Choice C is correct. When you consider the most common aldopyranose, ribose, it is reasonable to assume that afuranose ring is the most stable cyclic form for aldopentoses in general. Choice A can be eliminated. Thepenultimate carbon determines the designation of D or L for a sugar. In a five carbon sugar, carbon 4 is thepenultimate catbon, so choice B can be eliminated. In the cyclic form, furanose ring, all of the carbons exceptcarbon 5 have chiral centers, so it is true that a cyclic aldopentose has four chiral centers. Choice D iseliminated. In an aldopentose, carbon t has two bonds to oxygen, so in its cyclic form, carbon 1 still has two bondsto oxygen, making it the anomeric carbon. Carbon 2 is not the anomeric carbon, so choice C is correct.

Choice C is correct. For any trisaccharide, the total weight is the weight of the three monosaccharides minusthe weight of the two waters lost in forming the linkages. When three 6-carbon monosaccharides are combined,there are two bridging linkages. Calculating this value gives 3(180) - 2(18) = 540 - 36 = 504 grams/mole.Choose C to show off those math skills of yours. The fact that the linkages were 1,4-Iinkages is i moot pointrvhen considering the mass of the compound. Either way you look at it, the polysaccharide loses one water perlinkage, regardless of the exact linkage.

Choice C is correct. This question tests straight memorization, so remember it correctly, then choose C. Shouldvou not recall the fact that polysaccharides form linkages from carbon one to four (when dealing withpyranoses), then the passage also gives you a subtle hint, by giving only one example of a linkage, and theexample is a 1,4-linkage.

Choice A is correct. An aldopentose is most likely to form a furanose ring when the hydroxyl on the penultimatecarbon (carbon 4) attacks the aldehyde carbon. This eliminates choices C and D. Because the original sugar is analdehyde, the structure is a hemiacetal and not a hemiketal. The best answer is choice A.

Choice C is correct. The glycosidic bond forms when a nucleophile displaces the anomeric hydroxyl group of thesugar. The nucleophile is methanol, which attacks using the lone pairs on its oxygen atom. This eliminateschoices A and B. This question now focuses on whether it is the c[-anomer or the G-anomer. The methoxy group isirans to carbon 6, so it is the u-anomer, making choice C the best answer.

: Choice A is correct. An aldohexose has one aldehyde functional group, one primary alcohol functional group,and four secondary alcohol functional groups. This makes choice A the best irlr-"t. An aldohexose has chiralcentets at carbons 2,3, 4, and 5, so choice B is eliminated. The aldehyde carbon has the most bonds to oxygen ofany carbon, so it is most oxidized. This eliminates choice C. For a D-sugar, the penultimate carbon has R-chirality, so for a D-aldohexose, carbon 5 has R-chirality. Choice D is eliminated.

: Choice B is correct. A ketohexose is a six-carbon sugar with a ketone functionality (likely on carbon 2) in itshnear structure. If it were in the cyclic form, the term becomes pyranose or furanose depending on the ring size.Since fructose is the only ketose listed in the choices given, it is best for you to choose B.

:rright @ by The Berkeley Review@ t4t CARBOHYDRATES EXPLANATIONS!

17.

18.

20.

27.

1.9.

Choice A is correct' W1:" the hydroxyl group on carbon 5 of an aldohexose is inverted, the sugar becomes theopposite type of sugar. This means that the C-5 epimer of L-glucose must be a D-sugar, eliminating choice B. lnorder.to form D-glucose from L-glucose, all of ihe chiral cinters must differ, because the two structures areenantiomers. Only one chiral center differs, so choice D is eliminated. The C-5 epimer of L-glucose hashydroxyl SrouPS on the right on carbons 3 and 5, and hydroxyl groups 2 and,4 on the left, which a"ccording trFigure 1 is D-idose. The best answer is choice A.

Choice D is correct. D-Gulose has hydroxyl groups on the right for all carbons except carbon 4. The C-2 epimer ofD-gulose has hydroxyl groups on left for carbons 2 and 4. The C-3 epimer of giucose has all of the hyarox.,-groups on the right, so choice A is eliminated. The C-3 epimer of D-idose has thJhydroxyl group on the left forcarbons 2,3, and 4, so choice B is eliminated. The C-4 epimer of D-allose has the nyato*yt Erorp on the left oncarbon 4 only, so choice C is eliminated. The C-4 epimer of D-altrose has hydroxyt groups on the left for carborx2 and 4, so choice D is the best answer.

Choice B is correct. Ketals can not be oxidized by Tollen's reagent, because they are stable under basic conditiomand they cannot be converted into aldehydes (which can be-oxidized). The key feature in a ketal preventinqoxidation is the absence of a C-H bond on the ketal carbon (which is the feature that prevents a ketone fror:being oxidized) ' The C-H bond is necessary for oxidation to occur. oxidation from the organic chemislrrperspective involves the loss of hydrogen as well as the gain of oxygen. For this question, chooJe B for greatestsatisfaction.

Choice C is correct' Fructose is a ketose, so its cyclic structure cannot be an acetal or hemiacetal. This eliminateschoices A and B' Ketones, when added to alcohols in basic medium, go on to form hemiketals. The hemiketaih'i1l be in equilibrium with ketone. The question does not state whether the medium is basic or acidic, so ).oinmust answer this from experience and memory. Monosaccharides for hemiketals (or hemiacetals if the structurestarts as an aldose). This means C is the best choice.

Choice D is correct. Epimers differ by one chiral center in their backbones. Talose differs from galactose mcarbon 2, differc from idose at carbon 3, and differs from mannose at carbon 4. This means talose is air epimer crlgalactose, idose, and mannose, eliminating choices A, B, and C. It differs from glucose at carbons 2 and 4.,meaning it is not an epimer of glucose. This makes choice D the best answer.

Choice C is correct. Treatment of an aldehyde with a reducing agent such as NaBH4 reduces the carbonyl gror:(C=O) into a primary alcohol. Following that same reactivit|, treatment of an aldohexose (or ketohexose) rr-:convert the sugar into a hexa-ol (a six carbon chain containing six hydroxyls, one on each carbon). Treatmentan aldopentose will convert it into a penta-ol. The results fiom an- analytical reasoning perspective are simil

t

T

)',

tYoo--f- Ho{uo-l- nu-l- ou

I

cFI2OH

D-(+)-Talose

H

H

H

Y:"HolHuo-l- H

u-l- oHI

cFI2OH

D-(+)-Galactose

Hvoo-l- ruAoHoIuHAon

I

cFI2OH

D-(-)-Idose

H

H

;Y"o**,H*oHu-l oH

cFI2OH

D-(+)-Mannose

H

H "Y""uoJ- H

u--l- onu-f oH

cFI2OH

D-(+)-Glucose

to treatment of the sugar with nitric acid, because the product now has ihe possibility oifehg meso (indicatbv a lack of optical activity). Fructose is eliminated, because the new hydroxyl would result in the formationrL/uru^yr vvuutu rcbulr ut LIte IOrIIlaIlon ia new chiral center, thus there would be the formation of two diastereomers. At least one of the t,,,,-

1i1-:*i"?*"rs has to be opticallyactive Glucose and mannose are eliminated, because they have no mir:;plane between carbons three and four. This makes ribose the best choice. Ribose will generate a penta-ol withmirror plane slicing through carbon three. Pick choice C for the tingly sensation o1 correctivity. GalactcYrqrLL'rLurb rrtruu6rt Lcuuurt Lrlree. rlcK cnolce L lor tne tlngly sensation of correctivity. Galh'ould also yield an optically inactive product although it is not given ai un ur,r*", selection.

Cffiu

roe

ffih

I'10

@

G

ru

L@

UD

@

Gcfr.d[,

Wi@

mryi'@

'i@mcm!

M

Copvright @ by The Berkeley Review@ t42 CARBOHYDRATES EXPLANATIO]

n*o n_?o

,,Ion "-l-o'ttn\J-ra (- A nL-r- n r- 1lL-+_lL-._

H-l- oH ePrmers Hot u ePlmers

H-T- oH H--l- oH

<-S1+-eplmers

CFLOH

D-Glucose

cFI2OH

D-Galactose

III

d

23. Choice C is correct. Galactose is the C-4 epimer of glucose. The C-2 epimer of galactose varies from glucose atboth carbon 2 and carbon 4, so choices A and D are eliminated. The C-2 epimer of mannose is glucose, so choice Bis also eliminated. The only remaining choice, and the correct answer, is choice C. Talose (which is the C-2epimer of galactose as shown in Figure 1) is the C-4 epimer of mannose. Pick choice C.

Jr"",rola,.,-J- o"

"+o"cFI2OH

D-Mannose

Choice B is correct. The glycosidic linkage involves carbon 1 of the glucose on the left and carbon 2 of the fructoseon the right. This eliminates choice D. The glucopyranose ring has c,-orientation, so choice C is eliminated. Inthe fructose structure, the anomeric oxygen (involved in the linkage) is cis with carbon 6, so it has B-orientation.This makes choice B the best answer.

Choice C is correct. Glycosidic linkages involve an anomeric carbon with two OR groups, so the functionalitycannot contain the prefix "hemi". This eliminates choices A and B. Because maltose is formed from glucose, analdose, the functional group is an acetal. This makes choice C the best answer. Aldoses go on to formhemiacetals as monosaccharides and acetals as polysaccharides. Ketoses go on to form hemiketals asmonosaccharides and ketals as polysaccharides.

Choice D is correct. In G-D-glucopyranose, all of the substituents on the pyranose ring have equatorialorientation. This is a piece of information you should have committed to memory. The structure in questiondiffers from B-D-glucopyranose at carbon 3, where the hydroxyl group has axial orientation. This makes thestructure a C-3 epimer of G-D-glucopyranose, making choice D the best answer. An anomer would vary inchirality at the anomeric carbon. This eliminates choice A. A conformer is the identical molecule rotated orcontorted. If only one substituent changes from equatorial to axial while keeping the ring in the sameorientation, then the structures are not conformers. This eliminates choice B. Enantiomers differ at every chiralcenter, so all of the centers would need to be axial for it to be an enantiomer. This eliminates choice C.

l" Choice A is correct. A trisaccharide has two glycosidic linkages. This eliminates choices B and D. When therisaccharide is made from three unique aldohexoses, then the linkages will be acetal linkages, rather thanketal linkages. The linkages involve carbons with two OR groups, so they are acetal linkages, rather thanr.emiacetal groups. This makes choice A the best answer.

Choice D is correct. Hydrolysis under acidic conditions hydrolyzes acetals and hemiacetals, but does not affectethers. When a disaccharide is exhaustively methylated, all of the hydroxyl groups become methoxy groups.-{11 of the methoxy groups are ethers except for the anomeric carbon of the glycoside (sugar on the right), which:oes from a hemiacetal to an acetal. This means that the anomeric carbon of the glycoside and the anomeric:arbon of the glycosyl group making the linkage are both acetals. When hydrolyzed, they will generate:l-droxyl groups. All of the other oxygen atoms will be part of methoxy groups. Not all of the OCH3 groups:elurn to being hydroxyl groups, so choice A is eliminated. The glycoside loses the methoxy group on its anomeric:arbon, but not any others. The glycosyl group does in fact not lose any methoxy groups. Choice B is not a goodrlswer, but it cannot be eliminated. The disaccharide only racemizes at the anomeric carbons, so choice C is.,rminated. Following hydrolysis, the two carbons involved in the glycosidic linkage gain hydroxyl groups,.,-hen water is added. The best answer is choice D.

":I:HO+HHo--Fuu*oH

cFI2OH

D-Talose

r,'-ght @ by The Berkeley Review@ 143 CARBOHYDRATES EXPLANATIONS!

29' Choice C is correct. The glycoside is typically the sugar on the right, recognized because it does not use it-.anomeric carbon for-the linkage. G-D-Glucopyranose his hydroxvl lroups tliat alternate up-to-down-to-up-tc-down in the Haworth project. The sugar ot-t ih" right follows the saie puit"rrl as B-D-glucopyranose, except fo..carbon 2, which has its hydroxyl group up insteaJof down. This means that the glycoside is the c-2 epimer o:G-D-glucopyranose, making choicJ C, D-rr-rir-r.ror", the best answer.

37" Choice C is correct. An aidopentose has three stereocenters, located at carbons 2,3, and.4. The maximum numbe:of stereoisomers possible is found by 2n where n is the number of stereocenters. The reason the word ,,maximurr,is chosen is that if one of the stereoisomers happens to be a meso compound, then the total number of possibl.stereoisomers will_decrease by one. This won;t be a problem with the sugars, because sugars are not mescBecause there are three stereocenters, there will be eighi possible stereoisomJrs. Choose C.

o---* -' -

30' Choice D is correct. Maltose is a disaccharide made from two rx-D-glucopyranose structures. The linkage must b.alpha, so choices A and C are eliminated. Choice B is eliminatea,"uecause the hydroxyl group on the glycosid-(sugar on the right) has beta orientation. The best answer is choice D.

31' Choice A is correct. Amylase breaks down the cr-giycosidic linkage. However, it is not stable under highir.acidic conditions, where it is reaclily hydrolyzed. Gistric. fluids uti nignty acidic, so amylase is destroyed rthe stomach' It must be released uguitl try thl pancreas. This makes cho"ice'A the best answer. Both pancreatr-and salivary amylase cleave u-glycosidic linkages, so choice D is absoluteiy wrong. The two forms of amylas;are identical enzymes, so they are equivalent in strength. This eliminates choices ts and C.

32' Choice D is correct. The blood can only absorb monosaccharides. Fructose, galactose, and glucose ar-monosaccharides, so choices A, B, and C are eliminated. Lactose is a disaccharide of"galactose and giucose, so r:cannot be absorbed into the blood. Choice D is the best answer. In lactose intolerance, we don,t have th..disaccharase lacta-se, therefore you cannot break lactose into galactose and glucose. Since we cannot absorb th-disaccharide, the iactose does not get absorbed and instead g6es to feed our"bacteria (which metabolize it to ga:and toxic metabolites). This results in pain and an increase in the number of molecules in the gut and lumer.which draws water in and results in diirrhea.

33' Choice C is correct. It is stated in the passage that amylose has no branching, so choices A, B, and D ar.eliminated' It is also stated in the puttug" thit amylopectin has o,-1,6-branchiig at about one out of ever..thirty glucose residues. This confirms thatihoice C is the best answer.

34' Choice B is correct. Glycogen is a starch made from exclusively a-D-glucopyranose with a significant amount c:1,6-branching' This eliminates choice A. The standard glycosidicii"r.ug; in the polyrucJharide is s-1,4, s:glycogen contains both s-1,4-linkages and s-1,6-linkages. This eliminates Jhoice D and makes choice B the bes:answer' Because glucose is an aldehyde,_all of the linkages involve anomeric carbons that are part of an acet.-functional group. This eliminates choice C.

35' Choice D is correct. Amylopectin is a polysaccharide of a-D-glucopyranose that is held together by a-1,=-glycosidic linkages with an occasional 1,6-glycosidic linkage ro. "b.ur-r.i,irrg.

This means that thi oxygen ator.:on carbons 7,4, and 6 are involved in glycosidi. lit-rkug"s. Hydroxyl I is oi carbon 1, so the correct choice mu:-contain I' This does not help, because every answer selection contains I. Hydroxyl II is on carbon 2, so the corre."choice must not contain II. This eliminates choices A and C. Hydroxyi rfu is on carbon 4, so the correct choic;must contain III. This does not help, because the remaining answer seiections (choices B and D) both contains IIjHydroxyl IV is on carbon 6, so the correct choice must contain IV. only choice D contains IV, so it is correct.

36' Choice C is correct. Cellulose is another polymer of giucose, similar to amylose, but with 13-1,4-glycosidr:hnkages' This etiminates choice A' Lactose il a disacchaiide made from glucose and galactose, so lactose cannc:be formed from the break down of amylose. Choice B is eliminated. Sucr6se is a disaccharide made from glucos.and fructose, so sucrose cannot be formed from the break down of amylose. Choice D is eliminated. Amylose -broken down into glucose, which is absorbed by the blood and transported. Some of the glucose undergoe,glycolysis (that which isn't stored), resulting in the formation of pyr.r.rit". Choice C is the best answer.

Copyright @ by The Berkelcy Review@ I44 CARBOHYDRATES EXPLANATIONS:

Choice B is correct. Because the cyano nucleophile can attack the carbonyl carbon from either side (not

necessarily equally from either side however), carbon one in the reactant monosaccharide (which becomes

carbon 2 in the product) generates a mixture of two chiral centers on carbon two of the product. No other chiralcenters will change, so only the chiral center at carbon two differs between the two product molecules. Thismakes the two compounds diastereomers (non-superimposable and not mirror images). More specific thandiastereomers when dealing with sugars is the term epimer, given to sugar diastereomers that vary at only one

carbon in the backbone. This makes choice B the best answer.

Choice C is correct. Compound A is said to be an L-aldopentose. You should know from your information base

that glucose is a six carbon sugar (aldohexose). This means that the unknown sugar cannot be glucose. The correct

answer is choice C. Ribose you should know is an aldopentose, so choice A definitely should have been

eliminated. Without knowing exactly what xylose and lyxose are, they cannot be eliminated' Xylose is the C-3

epimer of ribose and lyxose is the C-2 epimer of xylose.

Choice D is correct. Compound A is an aldopentose with a molecular mass of 150 grams per mole. When oxidized

at both terminal carbons, the mass increases. This eliminates choices A and B. The formula for the aldaric acid

formed is C5H3O7 so the molecular mass is equal to 60 + 8 + 772 = 180 grams per mole. Choice D is therefore the

correct answer. You could also have solved this knowing that when it is oxidized, it gains two oxygen atoms and

loses two hydrogen atoms, resulting in a net gain in mass of 30 grams/mole.

Choice A is correct. Compound B has increased hydrogen bonding compared to Compound A, because carboxylicacids form stronger hydrogen bonds than alcohols. The increase in hydrogen bonding will manifest itself as an

rncrease in melting point. The best answer is therefore choice A. Choices C and D should exclude one another,

because if choice C were true, then choice D would have to be true.

Choice B is correct. Aldohexoses in their most stable form are actually aldohexapyranoses, the six-memberedring form of a monosaccharide. The ring forms when the hydroxyl group on carbon 5 attacks the carbonyl carbon.

This means that there is no carbonyl absorbance observed in the IR spectrum of the most stable form of the sugar.

The carbonyl peak in the IR is just above 1700 cm-l, so no peak is obserr.ed there. Choice B is the best answer.

Choice B is correct. Because Compound B is optically inactive, it must be a meso compound. In order for the

compound to be meso, there must be a mirror plane through the molecule. The molecule contains five carbons so

the mirror plane must slice through carbon three reflecting carbon 2 onto carbon 4. Compound B is drawn withcarbon 2 having accurate stereochemistry as determined by the optical inactivity of the diacid.

HO On the left becauseof the mirror plane

On the left because H HOit is an L-suear \'\u=

\ "o-

(oH)

(oH)

H4

5 crLou

Compound AUnknown L-aldopentose

HNG-+ H

HO

HO

Compound B

optically inactive diacid

:ir:

:

I: .r{

if the hydroxyl group on carbon 2 is on the left in the Fischer projection of compound B, then it must be on the left

Ln the fiscner projection of compound A (the original sugar). Hydroxyl groups on the left in the Fischer

projection are assigned S stereochemistry. Pick B.

Choice C is correct. Treatment of Compound X, an aldohexose, with NaBH4 will reduce the aldehyde

iunctionality of carbon 1 to a primary alcohol. Because carbon 6 is also a primary alcohol, the two groups are

tclentical. Generating matching groups at the terminal ends of the sugar is similar to what occurs when a sugar is

oxidized with nitric acids (in which case both terminal carbons become carboxylic acid groups). Because the

diacid of Compound X is optically active (it has an optical rotation associated with it), the reduced form (a

hexa-alcohol) is also optically active. The best answer is choice C.

- right @ by The Berkeley Review@ 145 CARBOHYDRATES EXPLANATIONS!

t F*tn-l- roHl

.t-J- ro"ln-f-roruH--F(oult*o.,

I

cFI2OH

3ctt"oH

u,-[j,o'ut-L o" =

.1,cFI2OH

tetra-ol

;(OH) NaBFd+ +

45. Choice A is correct. Because Compound W (the oxidized aldaric acid derivative of Compound z, thealdopentose formed from the Ruff degradation oi Compound X) shows no optical activity, it must be a mesodiacid' This means that the second and fourth carbons oico*pourrd w have opposite stereochemistry.This means that the

-two hydroxyl groups in question are on the same side of the backbone in the Fischerprojection' Carbons 2 and 4 in Compou"a w are originally carbons 3 and 5 from Compound X (the originalaldohexose)' Because the original aldohexose is a D-iugar, the fifth carbon has R chirality. This makes thethird carbon R, because the hydroxyl group is on the sarie side as hydroxyl five in the Fischer projection. Alisugars drawn on the right in a Fischer projeition have R stereochemisiry. lict a.

Choice C is correct' Treatment of an aldohexose with the reagents for the Kiliani-Fischer synthesis generates asugar with one more carbon in the backbone (an aldoheptosei. Because the chain increases by one carbon, analdopentose is not possible, so choices A and B are eliminated. f.he addition of C=N- occurs at carbon 1 of theoriginal aldohexose. The chirality of the penultimate carbon remains the same, because the penultimate carbonis still the same carbon. This makes choice C the best answer.

Jr:"",rll-roHrH--|.- roHlu--.1-roHlu-F oH

cFl2OHThe penultimate carbon (next to the last carbon) has D chirality in the final sugar (aldoheptose).

17 ' choice D is correct. A reverse aldol reaction cleaves between the alpha and beta carbons of a beta hydror.rcarbonyl compound' This means that Compound X, an aldohexose, is cleaved into a two carbon fragment (froncarbons 7 and2) and a four carbon fragmeni(from carbons 3 through 6). The four carbon fragment as stated in thequestion is an aldotetrose.

After the aldotetrose is reduced with_sodium borohydride, the four carbons (3 through 6 from the origina;aldohexose) have hydroxyl groups. Because the tetra-ol of the aldotetrose is optically active, its two chiralcenters have like stereochemistry (If they were R and S, then the compound would t" -"ro and opticallrinactive)' This means lft the two hydroxyl groups are on opposite sides of the sugar backbone in the Fischmprojection' Carbon 3 of the aldotetrose (carbon 5 fiom the original aldohexose) must be R given that the origiruilaldohexose is a D-sugar. This means that carbon 4 from the"original D-aldohexose has S chirality (which pu:sit on the opposite side from the fifth hydroxyl in the Fischer projection). pick D for best results. " The draw',gbelow summarizes the reactions.

3ct-troH

Ho-f--uH-loH

.15ncF{2oH

optically active tetra-ol

48' Choice B is correct. The number of possible stereoisomers is 2n where n is the number of chiral centers. The ,",a.1:of n in the case of a straight chain uldoh"tor" is four. 24 = 16, so there are sixteen possible stereoisomers for raldohexose' of these 16 possible stereoisomers for an aldohexose straight chain,

"ight ur" D-sugars and e

are L-sugars. This make B the best answer. Pick B, and do what is best... L terms of answering that is.

46.

NlltC

-l- lour-]- ,ott,

trottr-F(oH)*o.t

I

cFI2OH

H

H

H

H

H

Jr:"",n-l- roHlH-l- roni

"+oH'cFI2OH

OH

CN- H2

-=+

Pd (BasQ)H*

--+Hzo

(oH)

OH

(oH

OH5

6crtrou

aldotetrose

Reverse aldol__-+)

H

H

Ts[

I

1[

rI

{

Copyright O by The Berkeley Review@ 146, CARBOHYDRATES EXPLANATIO]

49. Choice A is correct. The addition of HNO3 results in a compound with identical groups on the terminal carbons(with HNO3 they are both carboxylic acid groups). The addition of NaBHf to ihe sugar will reduce thealdehyde to a primary alcohol making both terminal carbons identical without affectinglthe other carbons.This is analogous to the results when using nitric acid. Nitric acid is cheaper than sodium btrohydride, but as ageneral rule, the MCAT has stayed away from questions about cost effeciiveness of chemical reictions. pick Aand show off your correct answer selection skills. Choice B (H2SO4) dehydrates a sugar to carbon, while choiceC (Br2(aq)) will oxidize only carbon 1 into a carboxylic acid, not both terminal carbons. Choice D (KMnOa(aq))will oxidize everything on the sugar, so choice D is a bad answer. It is bad to pick bad answers; don't be badl

Choice D is correct. The most stable form of an aldohexose is the pyranose ring structure. This ties up the OH ofcarbon 5' Because periodic acid (HIOa) cleaves 1,2-diols, the sites are limited io neighboring diols aiCl and C2,C2 and C3, and C3 and C4. The least sterically hindered is the C3 and C4 pair of alcohols. Pick D. Thisinformation could be extracted directly from the passage, which is common on the MCAT.

Choice D is correct. Using the Kiliani-Fischer method, a D-aldopentose is used to synthesize a D-aldohexose.The addition occurs at carbon 1 of the D-aldopentose, turning carbon L into carbon 2 in the D-aldohexose products.This means that the chirality of carbons 2,3, and 4 in the D-aldopentose match that of carbons 3, 4, and 5 in theD-aldohexose. To determine the D-aldopentose needed to synthesize the D-aldohexose in the question, usingretrosynthetic analysis, carbon 1 must be removed and carbon 2 must be turned into a aldehyde group.

Upon matching structures with the four D-aldopentoses shown in Figure 2, the best answer is D-xylose, choice D.

Choice D is correct. To be C-2 epimers, the sugars must be diastereomers that vary in their chirality at carbon 2.Ribose is an aldopentose and altrose is an aldohexose, so they are not even isomers, let alone epimers. Choice Ais eliminated. The only way to answer this question, without actually knowing what the sugars are, is toemploy the data in Table 1. The products of Kiliani-Fischer synthesis arc C-2 epimers. Glucose and talose arerot C-2 epimers and neither are mannose and allose. This eliminates choices B and C. Idose and gulose are:ormed as products when xylose undergoes Kiliani-Fischer synthesis, so idose and gulose are C-i epimers,naking choice D the best answer.

Choice C is correct. Reduction in organic chemistry is most simply viewed as either the gain of bonds to:.r-drogen or the loss of bonds to oxygen. In step 1, the carbonyl carbon loses one bond to oxygen (in forming a new:ond to carbon), so carbon 1 is reduced in Step I. This eliminates choice B. in Step II, reduction occurs, because the:\-ano group gains a hydrogen on carbon and a hydrogen on nitrogen. The symbol [H] refers to reduction, so Step II,s definitely reduction. This eliminates choice A. Step III involves the hydrolysis of an imine into an:idehyde. This is where the general definition needs to be a bit more specific. Both oxygen and nitrogen arerore electronegative than carbon, so when carbon goes from being bonded to nitrogen to being bonded to oxygen, it::as not been oxidized or reduced. This means that Step III is not oxidation or reduction, so choice C is the bestr:'swer. This is a case where oversimplification can hurt. Technically speaking then, oxidation occurs when the: --rmber of bonds to more electronegative atoms increases and / or the number of bonds to less electronegative atomsr:creases. So, for those of you who considered Step III to be oxidation, you still got lucky in thai you didn't.-*ure it to be reduction and choice C was still your choice. But for edification, know the more correct definition: oxidation and reduction for organic chemistry.

tlhoice D is correct. In the Kiliani-Fischer synthesis as shown in Figure 1, and aldehyde is first converted into a-..Jroxy nitrile compound. The nitrile is reduced into an amine. The amine is hydrolyzed into the aldehyde.': the choices, only choice D, an imine, is an intermediate in the synthesis. Choice D is the best answer.

i0.

H-l- oHuo*HH*ou

cFI2OH

D-Xylose

H*-OHuIoH

uo_{_uu-l- ou

cF{2oH

,:t O by The Berkeley Review@ 147 CARBOHYDRATES EXPLANATIONS!

55.

56.

Choice C is correct. The two products formed from Kiliani-Fischer synthesis, according Table 1, are C-2epimersof one another. Epimers are diastereomers, which eliminates choices A, B, and D. The two diastereomers areformed in unequal concentration, because the presence of a chiral center in the reactant makes one side of thealdehyde easier to attack than the other. Choice C is the best answer.

Choice C is correct- Talose differs in chirality from glucose at carbons 2 and 4. In the Haworth projection of G-D-glucopyranose, the hydroxyl groups are up (above the pyranose ring) on carbons 1 and 3, and dtwn in carbons land 4. This means that talose has the hydroxyl groupJ on carbons i and. 4, up, so all of the hydroxyl groups ar€up in l3-D-talopyranose. For talopyranose, the hydroxyl groups on carbons 2,'3, and.4 must bL up, io Jnoice c i:the best answer.

Choice D is correct- Kiliani-Fischer synthesis adds one carbon to the sugar structure. Choice A is eliminatedbecause a carbon is lost, not gained. Choice B is isomerization, where thi number of carbons does not changeThis eliminates choice B. In choice C, the nurnber of carbons in the product is not listed, so the choice can neithe:be confirmed nor refuted. In choice D, the number of carbons has increased by one. This makes choice D the bes:answer.

Choice C is correct. The unknown disaccharide in the passage is comprised of two cyclic monosaccharides formejfrom aldohexoses (aldopyranose structures). A cyclic aldohexose molecule has five stereocenters (one at evertcarbon except the terminal carbon, carbon 6). Given that each monosaccharide has five chiral carbons, ;disaccharide must have ten stereocenters total. pick choice C.

Choice B is correct. Choice A is eliminated, because the structure on the left is glucose. Neither sugar can b,;glucose, because the specific rotations of the osazones of both sugars do not mat"ch the specific rotation of tl-:osazone of glucose. Choice C is eliminated, because the structure on the left is rrlu.nor", the C-2 epimer u:glucose. Neither sugar can be mannose, because the specific rotations of the osazones of both sugars do not matc:the specific rotation of the osazone of glucose (which would also be the osazone of mannoie). Choice D --.eliminated, because the two structures ate C-2 epimers of one another. Because the two sugars form osazon*with different specific rotations, the two sugars cinnot be C-2 epimers of one another. In choice B, neither of tlo:sugars wiil lead to an osazone that matches either glucose or mannose, neither sugar leads to an optical-;inactive aldaric acid, and the two sugars are not epimers of one another.

Choice D is correct. When an aldopentose is oxidized with nitric acid to form an aldaric acid, the two carbons ;:observe for optical activity are carbons 2 and 4. if the hydroxyl groups on carbons 2 and.4 are on the same side idthe chain in the Fischer projection, then the compound is meio ind inus optically inactive. It is only in choice Dthat the two hydroxyls on carbons 2 and 4are on opposite sides. Choice D is the best answer.

61. Choice D is correct. The osazone of Compound I has specific rotation +42.6", the osazone of Compound il hnrspecific rotation +37.2' , and glucose yields an osazone with a specific rotation of +54.6' . The osazone of manncsewill also show an optical rotation of +54.6', because ^urlnor"l, the C-2 epimer of glucose. This means that feunknown sugars cannot be either glucose or mannose (selection I). It is possibi" for thJ unknown sugars to be eil",stalose or galactose according to the osazone test. Because the aldaric aiids of both Compound I anl Compoun; Iare optically active, they are not meso. Galactose leads to an optically inactive (meso) aldaric acid. I::smeans that the unknown sugars cannot be galactose (selection III). Compounds I and II cannot be mannose :u'galactose, making choice D the best answer. For those of you who chose B, because you forgot it u,a. n"CANNOT" question, try to limit your scream to a few seconds... "AAAAHHHHHHHHFiHHHHHHHHII:

62. Choice C is correct" Choice A wilt yield an optically inactive aldaric acid when treated with nitric ac.i.because it will have a mirror plane slicing the carbon-carbon bond between carbons three and four. NeifisrCompound I nor Compound II yields a meso aldaric acid, so choice A cannot be Compound I or CompounCChoice B is mannose, the C-2 epimer of glucose. Choice D is fructose, the ketose isomer of g1.r.or". Both choicesand D are eliminated, because they will both yield the same osazone as glucose and Comp6unds I and Ii genei;a different osazone than glucose. Choice C will not lead to an opticalty inactive aldaric acid, because it will :rbe meso after oxidation. The best option for you is choice C.

57.

58.

59.

60.

I

Iu

Copyright @ by The Berkeley Review@ t4a CARBOHYDRATES EXPLANATIO]

53. Choice D is correct. Identical osazones will form from sugars with matching chiral centers from carbon threedown. This means that C-3 and C-4 epimers will form different osazones, so choices A and B are eliminated.Choice C is a pair of enantiomers, which would generated enantiomeric osazones, so choice C eliminated.Fructose is the ketose formed when mannose isomerizes from an aldose into a ketose. Mannose and fructose areidentical from carbon 3 on, so they form identical osazones. The best choice is thus answer D.

";r:uo-]- rIn-]- oHH*oH

cFI2OH

D-marrnose

a'4' Choice D is correct. Because the final osazone product after treatment of the sugar with three equivalents ofphenyl hydrazine contains two phenyl hydrazine moieties, it is safe to assume that two of the threeequivalents are substituting onto the sugar. The third phenyl hydrazine gains two hydrogens to form ammonia(NHs) and aniline (C6H5NH2). The gain of hydrogen is reduction, thus one of the three phenylhydrazinemolecules is reduced. The best answer is thus choice D.

Choice D is correct. Glyceraldehyde-3-phosphate and dihydroxyacetonephosphate are linear sugar structureswith the same formula, so they are isomers of some sort. Dihydroxyacetonephosphate has no chiral center,while glyceraldehyde-3-phosphate has one chiral center, so the two structures cannot be configuration isomers(isomers that have the same bonds, but different spatial arrangement). Epimers and diastereomers areconfigurational isomers, so if the two structures are not configurational isomers, they cannot be diastereomers orepimers' This eliminates choices A, B, and C. An aldose and ketose of the same carbon length are structuralisomers. In this example, they each have one phosphate group, so choice D is the best answer.

FH2oPo32- C3gso6p2- ' {ok5!os9 with no ts O (same formula but g--L Ou al&se with onechiral centers I -. -- I chiral center

CFI26H different bonds) CFI2opo32-

Dihydroxyacetonephosphate Glyceraldehyde-3-phosphate

Choice D is correct. Carbon 4 of glucose^becomes the aldehyde carbon in glyceraldehyde-3-phosphate. It losesits original chirality when it takes on sp2-hybtidization (associated with the aldehyde carbon), thus it has nochirality in the end. Galactose (the C-4 epimer of glucose) should therefore yield the exact same products asglucose if the enzymes were able to recognize galactose. The remaining chiral centers are identical between D-glucose and D-galactose, so the chirality of the products is the same. The best answer is choice D.

Choice C is correct. The lower three carbons (carbons 4,5, and 6) of D-glucose form the glyceraldehyde-3-rhosphate molecule, so carbon 5 finishes as the middle carbon (C-2) of glyceraldehyde-3-phosphate. Choice Crs the best answer. Because of isomerization, the label is also found at C-2 of DHAP, but that is not a choice.

3uq

H@N-N

HH

HqN-N

HH

H F

N-NFIZ

= N-NFIZno-J- H

ulonn-F ou

cF{2oHOsazone product

CFLOHt;HoJ-HutoHu*oH

cFI2OHD-fructose

3eq

Step I g9

CFI"OPO.2-l'JFocFI2OH

+

H OHHO H

H OHH-*C-OH

I

cFi2oH

H

OHH.*C-OH

I

cI-I2OPO32-

OHil

!1

fii

H- *C. OHI

cltoPo32-

t49

H-*C-OHI

cFI2OPOq2-H-*C-OH

I

cFI2OPO32-

'1":-ght O by The Berkeley Review@ CARBOHYDRATES EXPLANATIONS!

Choice C is correct. This is a question of Hess's Law as applied in thermodynamics. The overall free energychange is the sum of the individual free energy changes foi Step I through Step IV. The math to determine thesum of the four steps is carried out as follows: -4.0 + (0.4) + (-3.4) + (5.27 = -7.4 * 2.1 = _1.3, which is choice C.

Choice B is correct. The two terminal carbons in dihydroxyacetone phosphate (DHAP) have two equivalenthydrogens and the middle carbon has sp2-hybridization, so none of the

"uibor,, are chiral. The first carbon in

glyceraldehyde-3-phosphate (G-3-P) has sp2-hybridization and the third carbon has two equivalenthydrogens. Only the second carbon of G-3-P has four different substituents attached to it. The total number oichiral centers between the two compounds is therefore just one. The best answer is choice B.

Choice A is correct. Fructose is a ketohexose with a ketone on carbon 2, so carbon 2 is the anomeric carbon in thefuranose ring structure. The phosphate group is located on the last hydroxyl group (carbon 6) of fructose.Fructose, s'-D-fructofuranose, fructose-6-phosphate, and s-D-fructofuranose-6-phoiphate are shown below.

CFI"OH

,[ouo-l- H

H-J-or-ru-l- oH

CI-I"OH

,[oHo-l-H,r-l- o.t

IH-r- oH

6 CF{2OH

Fructose Fructose-6-phosphate s-D-Fructofuranse-6-phospha:

Choice C should be eliminated, because carbon 3 has no hydroxyl group. Choices B and D are eliminatedbecause the phosphate is on the wrong carbon.

7'l', Choice A is correct. This question tests your knowledge of glycolysis. Glucose is a six-carbon sugar whilepyruvate is a three-carbon structute, so two pyruvates form from one glucose molecule. This eliminates choices Cand D. Early in glycolysis, two ATPs are invested, but two ATPs are formed per pyruvate, so the net result rsthat two ATPs are formed in glycotysis. This makes choice A the best answer.

72. Choice D is correct. The linkage can be determined directly by looking at the two sugars of interest out of thethree sugars in the type-O blood antigen. N-acetylglucosamine is the ]irst sugar link"ed to the protein (the \-acetyl should give this away), thus the galactose must be the second sugar fiom the protein ln the antigeru:determinant trisaccharide. The linkage is from carbon 1 of the galactose t6 carbon 4 of irl-acetylgiucosamiie boway of the B-anomer of galactose. This means that the linkage is a B-1,4- glycosidic linkage, so cloice D is bes;The structure is drawn below.

%.frSA,ovottOH \ proteinCFI,

73. Choice D is correct-. All three sugars of the type O blood antigenic determinant are formed from the cycliza.;of aldoses, therefore there are three acetal functionalities (one for each sugar) rather than ke:functionalities. All three sugars are linked to something, so they are all acetals rather than hemiacetals. Tlbest answer is therefore choice D.

58.

69.

70.

CFLOH 6cF{2oPo32- 9nzoH

OHH

c-DFructofuranse

4

HO

6 CFI2OPO32-OHH

IW

,Iffi

rux

&mfr

{u0!

ruffi

HO6

HO

Copyright @ by The Berkeley Review@ 150 CARBOHYDRATES EXPLANATIO

HO

74' Choice D is correct. Fucose, differs from a normal sugar in that carbon 6 has been reduced from a primary alcoholi1!o a methyl group' If the methyl group is repLced with a primary alcohol group, then it becomes analdohexapyranose. This question centers on the relationship of the aldohexapyranose derivative of fucose andglucose' B-D-Glucopyranose has all of its substituents in equatorial position. Ho-"rr"r, because carbon 5 of fucosein blood type o has S-chirality, it is an L-sugar rather than a D-iugar. B-L-Glucopyranose also has ali of itssubstituents in equatorial orientation. The fuiose derivative with thJ hydroxyl group on carbon 6 has all of itssubstituents in equatorial orientation except the hydroxyl groups or, .urton, i

"l^ra q. The group on carbon 1 is

irrelevant in this question. The best ut t*e} is choice p, ine c-+^epimer of glucose. This means that when carbon6 of galactose is reduced from a primary alcohol into an alkyl group, ficose is formed. L-Fucopyranose, L-galactopyranose (L-fucopyranose with a hydroxyl group added to-carbon 6), and. L-glucopyraro.u *L all drawnbelow.

OH OH OHla 10 1cr

OH OH HOFICOHHO

HOHO

o-L-Galactopyranose(a-L-Fucopyranose w/ 6-OH)

a-L-Glucopyranose

The second structure_ (L-fucopyranose with a hydroxyl group added to carbon 6) is similar to that of L-glucopyranose, with the difference coming at the ihiral ."trl"t o? carbon 4. It is thus concluded that the secondstructure (the sugar formed when the CH3 group of fucose is replaced by CH2OH) is the C-4 epimer of L-glucose.The correct answer is choice D.

Choice B is correct. The 2-deoxyglucose portion of the name hints that the second carbon has no oxygen bonded toit, so choices C and D can be eliminated. Only choices A and B are deoxygenated at carbon z of thi iyranose ring.Choice A is an alpha sugar while choice B is a beta sugar. Choice B is therefore the best answer.

Choice A is correct. A normal L-aldohexose has a hydroxyl group present on carbon 6, while fucose lacks ahydroxyl grouP on carbon 6. This means that fucose is more tuirr.ua tirun u normal aldohexose, so fucose resultsfrom the reduction (loss of oxygen) of carbon 6. The best answer is choice A.

Choice C is correct. The sugar is a six-membered ring (a pyranose), rather than a five-membered ring (afuranose), so choices A and B are eliminated. AII of the choic"r-ur" cyclic galactoses, so identifying the sugar isnot necessary. As drawn in Figure 3, the galactose that has been added to hydroxyl group of .u.bo1l of the thirdsugar/ is upside down compared to the standard chair projection. The hydroxyl l-inkage is still from the axialorientation (trans to carbon 6), so it is the alpha anomer bf galactose. The sugar'is an cr-"galactopyranose, so thebest answer is choice C.

Choice C is correct. The type A antigenic determinant has four sugar rings, each of which has chiral centers atcarbons 1 through 5. This means that there are five chiral centeti p", sugar ring, and four rings total. Fivecenters times four sugars results in a total of twenty chiral centers. The .otr"c1 unr*", is choice C.

-i Choice C is correct. According to Table 1, conversion of phosphoenolpyruvate into pyruvate releases 14.8kcal/mole. Most oJ the other phosphate compounds in Table 1 release between 5 and 1b'kcals per mole whenhydrolyzed' It can be assumed that hydrolysiJ of phosphoenolpyruvate into pyruvate enol releases between 5 to10 kcal/mole. This means that the conversion from pyruvate'enol into pyruvate releases between 5 and 10kcalslmole, given that the overall conversion from phosphoenolpyrurrut" lr-rto pyruvate releases about 15kcals,/mole. The release of energy makes AG a negatirr" rrulru, so-c-hoices A and B are eliminated. The bestanswer is choice C, given that it is the only choice that fits within the range.

i "' Choice A is correct. The conversion of ADP into ATP involves the gain of a phosphate, which according to Table1 requires 7.3 kcals/mole. Only acetyl phosphate releases *or* thur"r 7.3 kcals/mole when it is hydrJyzed,, sochoice A is the best answer. Choices B, C, and D all release less than 7.3 kcals/mole.

HO

, .:r'right O by The Berkeley Review@ 151 CARBOHYDRATES EXPLANATIONS!

81' choice C is correct. \Atrhen bonds are broken, energy must be added to the system. when bonds are formed, energl-is released from the system. An energy releasinf reaction such as hydrolysis of a phosphodiester thus resu]Ffrom the breaking of weaker bonds and formatioi of stronger bonds. This means tnat tfre phosphodiester boncmust be weak, eliminating choices A and B. The weakness can be attributed to repulsion from the negatir-tcharges located on the oxygen atoms in the phosphate groups. Repulsion causes the bond to elongate, wherelonger bonds are weaker bonds. Choice C is the best answir.

82' Choice C is correct. Hydrogens are not gained or lost w^hen phosphoenol pyruvate, pEp, loses its phosphaiegroup to ADP, so the enzyme is not PEP dehydrogenase. Choice A ii eliminaiea. nnospnoenol pyruvate, pEp,

=not oxidized or reduced in the conversion, so the enzyme is not PEP reductase. Choice B is eliminateiPhosphoenol pyruvate, PEP, does not make or break any cirbon-carbon bonds in the conversion, so the enzyme rsnot PEP aldolase. Choice D is eliminated. A phosphate group is transferred, so the enzyme is a kinase. The bes:answer is pyruvate kinase, making choice C the best answer.

83' Choice D is correct. When glyceraldehyde-3-phosphate, G-3-P, converts to 1,3-bisphosphoglycerate, th.number of bonds to oxygen on carbon 1 increases. This means that carbon t has been oxidLed, elim"inating choiceA' The oxidizing agent used to oxidize the aldehyde into a carboxylic acid is NAD+ (which is poor in H, ,property of an oxidizing agent), so choice B can also be eliminated. Carbon 2 starts and finishes with the exa.-:same bonds, so it is neither oxidized nor reduced. This eliminates choice C. By process of elimination, choice Dis the best answer.

. The passage_ provided no information to know this fact, so you either had to know fror*

memory that biotin is not required or you had to use your chemistry logic to eliminate incorrect answer choices.

84' Choice A is correct. According to Table 1, 5 kcals/mole is released when glucose-1-phosphate is hydrolyze;This means that you need to determine which structure represents glucose-"1-phosphite. bhoi.", B and D areliminated, because the phosphate group is on carbon o. tn B-o-glu"copyranoi", tr," hydroxyl groups alternarnfrom up-to-down-to-up-to-down when vilwing from carbon 1 throrfih

"urUo.,4. This is only oUr"irr"a ir-, choice A

so choice A is the best answer

85' Choice A is correct. The ring is a five-membered ring, so choices C and D are eliminated. The anomeric carbon $carbon 2, on the right half of the sugar as drawn. rhe hydroxyl group on carbon 2 is down, while carbon 6 is u:This means that the two groups are trans to one another, -utirig thl sugar an alpha sugar. rhe besiln;;;;;choice A.

85' Choice B is correct. A shortcut for evaluating the chirality of sugars in the Fischer projection is to realize tha:all hydroxyl groups on the right side of the baikbone are located on an R stereocenter while all hydroxyl grouF$on the left side of the backbone are located on an S stereocenter for standard sugars drawn according to conventicnwith the carbonyl carbon at or near the top. The procedure for obtaining"the stereochemistr/ of one of 1'cstereocenters is drawn below:

m

lu

Method 1 Method 2

nvoSolving the first I -\ I

.t iruti"li". yi.ii' Hti< oH

ssffsn-l- on

i*o'D,XytJse

(r@

0m

mffiMm

m

mffip

moffi,SU,

mMrms

m

S for hydroxylson the left"

OH

H

OH

cFI2OHD-Xylose

R for hydroxylson the right.

nYoH>i< OH

Ho-].-HI

g>I< 6gicFI2OH

D-Xylose

H

HO

H

For sugar is in the standard Fisherprojection, hydroxyls on the righthave R chirality and hvdroxyls onthe left have S chiraiity

Draw the Fisher projection with Counterclockwise normally indicate3-dimensional accuracy. The Hs an s stereocenter, but because tf,eare in front in Fisher projections. hydrogen is coming out of the plana

the stereocenter is reversed to R. mmfll

mThe compound has chirality of 2R, 3s, 4R, so choice B is the best answer.

Copyright @ by The Berkeley Review@ 152 CARBOHYDRATES EXPLANATIONS!

i-. Choice B is correct. From the balanced equation, there are six CO2(g) formed for every one glucose that reacts. Tosolve this question, you must convert the glucose from grams to moles, convert moles of glucose to moles of CO2(g),and then convert moles of CO2(g) into grams of CO2(g). This process requires dividing by the molecular mass ofglucose (180), multiplying by the ratio from the balanced equation for carbon dioxide and glucose (6:1), and thenmultiplying by the molecular mass of carbon dioxide (a ). The long-hand math goes as follows:

10.00gC6H12o6" lmole * 6COz ,44grams=10x 1 x6x44q.Co)- 180 grams 1 C6H12O6 1 mole 180 1 "The set up is what is requested in the question, so the best answer is choice B.

:i. Choice B is correct. The best explanation for the difference in heats of combustion between the aldohexoses(glucose, galactose, and mannose) is structural in nature. They all have the exact same bonds (thus they arestereoisomers) so according to the simplistic application of Hess's Law, they should yield the same amount ofheat. Because the heats are so different, there must be a difference in stability. The more stable the molecule,the less heat that will be given off. Because all of the aldohexoses have the same bonds, choice A iseiiminated. Sugars are neither aromatic, nor do they have any stabilizing resonance forms. This eliminateschoices C and D. The only choice left is answer B, stating that the sugars have different orientation is space andthus different hydrogen bonding. The sugars also exhibit differences in their steric interactions. The best answerof the choices is choice B.

' I Choice C is correct. The given in this question is 15 grams ribose which, when divided by the molecular weightof ribose (150 g/mole), gives 0.10 nroles of ribose. The AH.o-bustion for ribose is lisied in Table 7 as -2076kJlmole. This means that the 0.10 moles of ribose generates 207.6kl (207,600 J) of heat when it is burned

(oxidized). Plugging into the equation E = mCAT yields ::::ry^K, which is approximut"ty ?90'9001000 x 4.18 1000 x 4

(roughly 50), making choice C the best choice. The mathematical layout is drawn for you below:

15gC5H1so5x lmole x2076kL=207.6kJ&E=mCAT...AT= E - 207'600 =207.6 =200 =50"C

150 grams 1 mole mC 1000 x 4.18 4.I8 4

It is doubtful that you will see a calculation this lengthy on your MCAT, but you should still understand the setup and the theory behind the math.

-- Choice B is correct. The most COz(g) results from the combustion of the compound with the largest mass percentof carbon. This is because for every carbon within a sugar, one CO2(g) molecule is formed. Therefore, thl morecarbon by mass in the compound there is, the more COZ(S) that is formed upon complete oxidation. The masspercent of carbon in any monosaccharide is 40%. The mass percent of carbon in sucrose (C9H22O1) is greaterthan the mass percent of monosaccharide, because the difference in the ratio of C:H:O between sucrose and themonosaccharides is due to a loss in H and O from the water. This increases the relative abundance of carbon insucrose compared to a monosaccharide giving sucrose a greater mass percent of carbon than a monosaccharide.The mass percent of carbon in sucrose it' 1'44 = 144 = 72 and. 72 > 72 = 0.40 ... 72 > +o%. Becatrse the

1.M+22+776 342 177 171 180 777mass percent of carbon in sucrose is greater tltan 40"/o, more CO2(g) will be produced from the combustion of onegram of sucrose than from the combustion of one gram of any monosaccharide. Choose B and be huppy.

" , . Choice A is correct. For this, you must use the AH in terms of k] per gram, where the kJ/g value is found bydividing the kJ/moles by the molecular mass of the sugar. Monosaccharides have masses that are multiples of30 g/mole, so the numerators should be simple to work with. The following calculations show the energy pergram values for each of the sugars given in the answer choices. The correct answer is the sugar with the highestnumerical value.

Glucose.2538 =423180 30Xylose '2102 -420'4- 150 30

Ribose .2076 - 415.2

150 30Ribose gets eliminated immediately, because it is equal in mass to xylose but produces less energy. The possibleanswers are narrowed down to either glucose or xylose. Instead of solving the math, it is easier to reduce the twol'alues to something over a common denominator (in this case 30). Glucose results in the greatest numerator,making choice A correct.

:vright @ by The Berkeley Review@ 153 CARBOHYDRATES EXPLANATIONS!

92. Choice D is correct. Addition of an excess of periodic acid to a monosaccharide results in the oxidative cleavag"of ail of the carbon-carbon bonds in the sugar. Aldehydes are oxidized into formic acid , primary alcohols ar;oxidized to formaldehyde and the secondary alcohols associated with internal carbons are oxidized by losir:both carbon-carbon bonds to form formic acid. The product distribution for the decomposition of glucose is dran:below:

CFI2OH ------+D-glucose H

In the end, the full oxidation of glucose yields 5 formic acids and 1 formaldehyde. Choose D for correctness.

Choice C is correct. The molecular formula for xylose is C5H16O5. The molecular weight of xylose (or any oth":aldopentose or ketopentose) is 60 + 10 + 80 = 150 grams/mole. The mass of the five Iarbons'is 60 grams. Tl-means that the mass percent of carbon in xylose is 60/150 x 100% = 40"/o. As a point of trivial interJst, the ma..percent of carbon in any monosaccharide is always 4O'h. Choose C and feel happlly correct.

Choice A is correct. In order to be optically inactive, the compound must be either meso or achiral. In the case .:sugar derivatives, the diacid (aldaric acid) formed after oxidation of the terminal carbons must be meso to :,.optically inactive. D-mannose has stereochemistry of 25,35,4R,5R, which does not yield a meso aldaric ac-:when oxidized, so the aldaric acid derivative of mannose is optically active. D-galactose has stereochemis5of 2R,35,45,5R, which yields a meso aldaric acid which is opticallyinactive. D-"ribosehas stereochemistrr':r-2R, 3& 4R and is five carbons long (which you should know from memory), so the third carbon would no longer :*a chiral center in the aldaric acid form. The aldaric acid of D-ribose ir *ero and therefore optically inactii':D-allose has stereochemistry of 2R, 3R, 4R, 5R, which would form a meso aldaric acid *ni.n is opticai-rinactive. The aldaric acid of D-allose is meso and therefore optically inactive. The only sugar to ftrm ;:optically active aldaric acid derivative of the given choices is D-mannose. The best answer is choice A.

HO

o

-J( H

93.

94.

OH

OH

HOAladirc acid derivative

of D-Mannose(not meso .'. active)

H

HO

HO

H

HO

Aladircacid derivativeof D-Galactose

(meso .'. inactive)

o

OH

oFt--

OH

acid derivativeD-Ribose.'. inactive)

HO

H

--+t

H

HOAladirc

o

OH

OH

of(meso

HO

Aladirc acid derivativeof D-Allose

(meso .'. inactive)

:!_

..:

:_

-----+H OH+

HO H_+

H

H

Copvright @ by The Berkeley Review@ 154 CARBOHYDRATES EXPLANATIO\SI

95. Choice A is correct. When treated with nitric acid, the two terminal carbons (carbon 1 and carbon 5) of analdopentose get oxidized into carboxylic acid functionalities. This results in the formation of a 1,S-dicarboxylicacid (aldaric acid). After the oxidation reaction, only carbons 2 and 4 can be chiral centers, because the thirdcarbon now has two identical substituents attached to it (making carbon 3 achiral.) In a D-aidopentose, carbon 4has R stereochemistry. This would mean that carbon 2 must aiso have R stereociremistry in the original sugar,although that changes in the aldaric acid, because the priorities are different. Onty choice A has iiris feature.The stereochemistry of carbon 3 is irrelevant.

HO HO Ho /'oIH--Fos

\ll-> u-T-iourr-rT vrf]'

I

cr-LoH

96' Choice C is correct. Ribose differs from deoxyribose in that carbon 2 has no hydroxyl group in deoxyribose. Thismeans that carbon 2 does not have four unique substituents, so it is not a chiral

"Lnt"t. fhis matls choice B avalid statement, which eliminates it. By having one more chiral center, ribose has more possible stereoisomers,making choice A a valid statement, and thus it is eliminated. Ribose, by having a hydroxyl group rather than ahydrogen on carbon 2, is more oxidized than deoxyribose. This makes cho]ice D a valid statement, whicheliminates it. Ribose and deoxyribose do not have the same molecular formula, so they cannot be stereoisomers,let alone epimers. This makes choice C a false statement and therefore the best answer.

i-' Choice D is correct. When two cyclic monosaccharides form a disaccharide, a glycosidic linkage is formed.Linkage formation occurs when a hydroxyl on one sugar attacks the anomeric carboi bn the other stigar. A watermolecule is released as the leaving group, as the hemiacetal with the reactive anomeric carbon gels convertedinto an acetal' This makes choices A, B, and C all true. The anomeric carbon starts and finishes *ith t*o bondsto oxygen, one bond to carbon, and one bond to hydrogen. This means it is neither oxidized or reduced, so choice Dis not observed, making it the best answer.

HO HOOH+ HO HO o

HO

Choice A is correct. When copper goes from an oxidation state of +2 to +L, it has been reduced. The color of thesolution changes when copper changes oxidation state, so a color change corresponds to an oxidation-reductionreaction involving copPer ion and the sugar. Because copper is reduced, the sugar must be oxidized. If an aldoseforms a carboxylate (deprotonated form of the carboxylic acid), it has been Jxidized. This makes choice A acorrect answer. If a sugar forms a compound with only hydroxyl functional groups and no carbonyl group, it hasbeen reduced. This eliminates choice B. If an aldose becomes an acetal, the rurmber of bonds to"o*yg"n andhydrogen has not changed, so it has neither been oxidized or reduced. This eliminates choice C. Ketonei do notoxidize into carboxylic acids, so choice D is eliminated.

no {oIHa-(oH)

ronl sQt-H-l-- oH

I

cFI2OH

D-Aldopentose(opticallyactive)

H

--+I

H

HO

(oH)

(oH|

OH

H

--+{

H

HO

OH

(GH}

OH

Aldaric acid derivative hractivg so it must be mesoof D-Aldopentose OH-4 on right ... OH-2 on right(opticallyinactive) OH-3 is unknown b / c C-3 is achiral

2&3unkrown

anomeric C is bonded to:OR, OR', C, and H

H\Hemilcetal becauseanomeric C is bonded to:OR, OH, C, and H

HOFLC

Ho+c-q

,:..ight @ by The Berkeley Review@ 155 CARBOHYDRATES EXPLANATIONS!

99.

100. Choice B is correct.bonds and one ring.

Choice A is correct. A trisaccharide has two glycosidic linkages. For every glycosidic linkage, one watermolecule is lost. An aldohexose has a formula of C5H12O6 while a aldopentor6 irut a formula"C5H1gO5. Atrisaccharide from two aldohexoses and one aldopentose has seventeen carbons, so choices C-and D areeliminated. Choice B represents the sum of the formulae of all three monosaccharides, but with the loss of twowater molecules, the formula is C17H3go15, making choice A the best answer.

An aldopentose has one n-bond and no rings. It can form a cyclic structure, but that has nEither way, the compound has only one unit of unsaturation. The best answer is choice B.

Copyright @ by The Berkeley Review@ 156 CARBOHYDRATES EXPLANATI

Nitrogen Compounds (Non-Biological)al AmlnesSection Vtr

Nitrogen

| -+H+

?*nT,;

Hisiidinea(R) = o.u/ (Basic Amino Acid)

Compoundstty Todd Bennett

ir, +H."-

b)

c)

i. Basicity and Acidityii. Nuelcophilicityiii. FormaLion of Aminesiv. Reactions of Aminesv. Hofmann E,liminationImine Chemistryi. Formation of Iminesii. Imine-Enamine lsomerizationAmidesi. Stru ct u reii. Formation of Amidesiii. Reactions of AmidesAmino Acid Synthesisi. Hell-Volhard-Zetinskii Synthesisii. Strecker Synthesisiii. Reductive Amination of g-Keto Acids

d)

CHr

I

^*\:J

Nitrogen Compounds (Biologicat)a) Amino Acids

i. Strcutures and Classificationii. Isoelectric pointsiii. trasy Calculation of pI

b) Proteinsi. Structure Featuresii. Structural Levels

c) Biological protein processes

d) Biochemistry Lab Techniquesi. Gel Electrophoresisii. Affinity Chromatographyiii. Sequencing (?rimary Structure)iv. Cutting/Fragmentationv. Edman's Reagent

pK

Nitro gen- Containing Compound sSection Goals

oB Know the amino acids that affect the tertiary structure of proteins.

aB

You must know the structures for cysteine and proline. You should know that cysteine forms disulfidebridges that result in cross-linking"within protbins. Proline, because of its cyclic structure, will causestructural abnormalities likes bends, kinks, and turns. You should have air idea of what structuralfeatures are most affected.

Be able to recognize and classify amino acids according to their side chain.Amino acids are classified in many ways, including hydrophobic, hydrophilic, acidic, basic, polar,and aromatic. Each of the classificatiohs eives vou iriforniation about tfie reactivitv of the aminoacid side chain. With

s gives you information about the reactivity of the aminoc6ain ii most imoortant. becarrse the amino and carhoxvlacid side chain. With proteins, the side chain is most important, because the amino and carboxyl

terminals are involved'in the peptide tinkage.

@3 Be able to determine the isoelectric point for amino acids and proteins.

@?

The isoelectric pH is the pH at which the compound carries no net charge. For all amino acids exceptthose with a basic side chain, it can be determined by averaging the values for pKrl and pKu2. Fbrproteins and the basic amino acids (histidine, lysine, and arginine), you must dverage thb two pK6values that involve the zwitterion (neutral molecule). This is most easily found by first determiningthe charge of the protein (or amind acid) when it is iully protonated.

Ilave an understanding of lab technigues such as gel electroDhoresis.The basics of gel electrophoresis and isoelectric focusing involve placing the protein or ami.no acidbetween the charged plates of a capacitor, and allowing the compound to migiate through a viscousgel that offers reslstance. The natuie of the charge on the compound is deteriiined by the migrationfeatures.

Know the structural definitions and structural features of proteins.It is important to know what is meant by primary, secondary, tertiary, and quaternary structure,although some definitions overlap. You shbuld,recognize G-pleated sheeis and hbw they are arranged(parallel or anti-parallel). Have -a

basic idea of s-turns and u-helices.

Be able to determine the sequence of a peptide from chemical information.Sequencing a protein involves treating the protein with denaturing reagents, and then sequentiallydetermining the component amino acids or peptides that fragment from fhe protein when it-is treatedwith a sequencing enzyme (digestive enzyme). You don't necessarily need to memorize the reagents,but you must be able to sequence a protein when provided with the reagent (or enzyme) and itsfunction. You must know Edman's reagent.

Know the basicity of amines and the effects of alkyl substituents.Perhaps the most common weak base in organic chemistry is the amine. The pKu of a standardprotonated alkyl amine_is in.the range of 9 to 11. This makes the amine a weak base, and theammonium cation a weak acid.

"?t?

@? Know that amine compounds can exist in either the free base or acid salt form.There are several common amine compounds used for medicinal purposes as antibiotics, antipyretics,and analgesics. The MCAT likes to present organic molecules that 6ave a definite biological apflication.

Know how amines react to form amides.There are several instances in biochemistrv where an amine will react with a carbonvl comnoundto form an amide bond. The most obvious'case is the formation of proteins from u.ni1'to acids. Forin aitro reactions, you must be aware of the solvent, because of the acid/base properties of the amines.

'v

lftt:

,i,.1-li

:.t:li:OTU

t::ilrr:

f,e{f

'ifilr,lr

nriIili:r1",:

ifinl.iLLm

Organic Chemistry Nitrogen Compounds Introduction

Compounds containing nitrogen make up a surprising large percentage of theknown organic molecules. Many oxygen-containing

"o*polrttai such as alcohols

and carbonyls have equivalent compounds where the oxygen is replaced by anitrogen. For instance, alcohols and primary amines are similar in that sense.Because nitrogen has two hydrogen atoms instead of one, Iike an alcohol, itsreactivity is not identical to an alcohol, but it is nonetheless similar. Ketones andaldehydes share a strong structural similarity to imines, and their relativereactivity is close to one another. The significant difference in an imine is that itstautomer, an enamine, has different reactivity than an enol, because nitrogen andoxygen are different in electronegativity. we shall draw as many analogies tooxygen-containing compounds as we can.

We wiII start by considering amines of all types. Unlike the drastic difference inreactivity between an ether and an alcohol, amines do not change their reactivitysignificantly as they become more substituted. Steric hindrance and electrondensity on the nitrogen are impacted by substitution, but not to the degree thatiertiary amines are significantly more or less reactive than primary amLes. ofsignificance when considering amines is their basicity and their nucleophilicity.This makes up the bulk of reactions carried out by amines. Amine basicity andnucleophilicity is not as clear cut and easy to undeistand as one might hope. f.he:rnpact of steric hindrance, hydrogen bonding in solution, and eleitroniis is notas predictable as it is with most other compounds.

-{fter considering amines and imines, we will address some of the less commonrilrogen-containing compounds such as hydrazines, oximes, and azides. Thereale not a great deal of examples of their reactivity, and it should be expected that:: they appear on the MCAT, then the test writers will provide a subsLntial pool:: information with them. we shall consider amides last, as they will serve as a:ansition from non-biological examples to biological examples. Amide bonds,:eferred to as peptide bonds when shared between two amino acids, make up the::r'rndation of the primary structure of a protein. Amides are formed and:",-drolyzed readily under biological conditions, so they are an ideal building:,ock for much of biological chemistry. we will address in aitro synthesis o1;rino acids as our segue from non-biological examples into biologicil examples: i nritrogen-containing compounds.

"''e shall spend a significant amount of time covering biological examples of"-:rogen-containing compounds. The majority of the text will be devoted to the;:rno acids and proteins, but we shall also address briefly other compounds;' '.h as the common bases in DNA and RNA as well as some common clisses of:::epounds found in neurobiology. we will focus on the structure and function:r ;he compounds, leaving the biological aspects to the biology books. of-reatest interest will be isoelectric points and an easy way to determine them.

'*<tlv, we will address biochemistry laboratory techniques that pertain to:'::-'teins and amino acids, The MCAT has placed an emphasis on iaboratoryrr,:eriments in their previous exams, so we will address the logic and theoryl"-::nd electrophoresis,

_affinity chromatography, and sequenciirg. Althougirtrt:e are many other techniques in biochemistry, these three are chosen becau"serr ::reir frequency on the MCAT and their applicability to proteins and aminou-:-. A11 other biochemistry laboratory technlques covered by o.tt course can be'r,:rd in the biology books.

. ;c",right O by The Berkeley Review 159 Exclusive MCAT Preparation

Organic Chemistry Nitrogen Compounds Non-Biological

r- *{" "H

Ammonia

,r-ilv" o"H

Hydroxylamine

o-*v""H

1" Amine

:'-

"''

N;'r xH,H

Hydrazine

o -*v" "R'

2" Amine

o-*$"o"R'

3'Amine

Non-Biological Nitrogen CompoundsAminesAmines are compounds with a central nitrogen that has three sigma bonds toeither carbon or hydrogen. The degree of substitution of the amine is describedas primary, secondary, or tertiary, where a nitrogen with al1 three sigma bonds tocarbon is known as a tertiary amine. To complete the octet about the nitrogen,there is also a lone pair of electrons on the nitrogen. The lone pair is said to be inarrsp3-hybrid orbital. If the nitrogen has a double bond (both a pi and a sigmabond) to carbon, the compound is known as an imine. If a hydrogen on the aminenitrogen is replaced by a hydroxyl group, the compound is known as a

hydroxyamine. Nitrogen when bonded to nitrogen by a sigma bond is ahydrazine.The structures of some nitrogen-containing compounds are shown in Figure 7-1.

R -N\cn , Ho/*oao.Oxime

OA"H

SolutionCompound A has a double bond between carbon and nitrogen making it animine. Imines can be reduced to form amines, but they are not aminesthemselves. Choice A is thus eliminated. Choice D has the nitrogen conjugatedto a carbonyl group. This functionality is known as an amide. Choice D is thu-.eliminated. Choice C is a hydrazine, because of the nitrogen-nitrogen singlebond. Choice C is thus eliminated. The only choice left is choice B. Thecompound is a primary amine, because only one carbon is bonded to nitrogen.

Amines are named in a similar fashion as ethers. The alkyl groups attached tonitrogen are named first, followed by the word amine. They can also be namedas amino alkanes. Because of the muitiple name option, the likelihood of aminenomenclature questions is not too high. We'lI do one example just to cover allbases.

Imine

Figure 7-1

Example 7.1

\A4rich of the following

A.H"C H"\ /

L-1\/

H:C

compounds is an amine?'QCNHz

D.

NI

NHz

imim,[@

rfu

Wi@

fttmffir@

@

Copyright O by The Berkeley Review r60 The Berkeley Revier- @


Example 7.2\Alhat is the ever-so-wonderful IUPAC name for the following compound?

CHs

A. N-methylphenylamineB. N-benzylmethanamineC. Methyl anilineD. Methyl amino benzene

Solutionft".?Tp:yl9 h?r l benzene ring (phenyt group) and a methyl group bonded toan spJ-hybridized nitrogen, making the compound an amine. The ctmpound isthus named N-methylphenylamine accordlng to the wild and adventurousIUPAC naming gang. The best answer is choice A.

-{mines are both weak bases and good nucleophiles. They are found in manyneurobiology compounds, because many amines have stimulating effects on thlbrain. some of the more biologically famous amines are shown in Figure z-2.

HO

H

N

ocH3

o-R, '',\F"6"

(-)-Morphine

H"N

Nicotine Cocaine

G.':l-'"' -."Qr"-"'HO t*- at, ocH3

MescalineAmphetamine Epinephrine (adrenaline)

FigureT-2

The above compounds are found in various medical and household products,.ome of which you may recognize. Because of their medicinal and industrial;ses, synthetic pathways to each are of some interest. Natural product synthesis'' a significant part of organic chemistry. what is meant by

-tne term "natural

:roduct" is a compound that is synthesized in a plant or animal. These-ompounds are targets of a great deal of synthetic laboi. In some cases however,-t is far cheaper to extract the amines from plants (i.e., nicotine) using both acid-:ase extraction techniques, as well as organic solvent extraction. Foiinstance, it*' through extraction that caffeine is removed from the coffee bean. Extraction::n be far easier than synthesis. For this reason, it is more useful to consider the:atural synthetic pathway rather than an in uitro synthesis.

-opyright O by The Berkeley Review l6l Exclusive MCAT Preparation


Because the nitrogen of the amine has a non-bonding pair of electrons that it isfree to donate, nitrogen atoms can serve as a nucleophilic center for the molecule.The lone pair of electrons can be donated to a proton making an aminecompound a Brsnsted-Lowry base. The basicity of amines is well-documented.The protonated form of the amine is often isolated from aqueous solutions as theammonium salt. For instance, ethyl amine (CH3CH2NH2) can also be found asthe ammonium chloride salt (CH3CH2NH3+C1-) after the free base is treated withhydrochloric acid. The free-base form is defined as the deprotonated form of theamine. It is with the basicity of amines that we shall start their study.

Amine BasicityA common use for amines is as weak bases. Amines have pK6 values in the 4 to5 range, which means that the K6 associated with the protonation of an amine isbetween 10-4 and 105. The varying basicity of amines ian be attributed to factorssuch as the inductive effect, steric hindrance, hydrogen bonding, and solvation.The seemingly chaotic order of relative amine basicity demonstrates the presenceof both the inductive effect and steric factors. Steric factors would favor a smallerbase, because of its ability to bind a proton with little to no hindrance. Theinductive effect predicts that the more substituted amine is the most basic. Therelative basicity of the methyl amines is: (H3C)2NH (pK, = J.27) > H3CNH2 (pK6= 3.35) > (H3C)3N (pKu = 4.27) > NH3 (pK6 = 4.74).

A lower pK6 value corresponds to a stronger base, so according to the pK6 data,a secondary amine is the most basic of all amines in water (with the same alkylgroup on all of the amines). In organic chemistry, pK data is applied todetermine which way an acid-base equilibrium will lie (as we considered insection I, pages 37-40 of Organic Chemistry book I). Reaction 7.1 shows afavorable proton transfer reaction involving amines of different substitution.

H3CNH2 + NHn*

-

H3CNH3+ + NH"

Reaction 7.1

Because the methyl group is electron donating, methyl amine is more basic thanammonia. This means that the ammonium cation is more acidic than the methviammonium cation. The stronger base and the stronger acid lie on the reactantside, therefore the product side of the reaction is more stable than the reactantside of the reaction. This means that the reaction will proceed in the forwarddirection as written when starting from standard conditions. At equilibrium, theproducts are more abundant than the reactants, so the equilibrium constant (Kgqis greater than 1. The equilibrium constant for a proton-transfer reaction can bedetermined using EquationT.T (listed also as Equation 1.13 in section I).

Keq = lgPKa(product acid) - PKa(reactant acid) 0.1,1

If you don't recall the common acid-base equations, you should consult page 4C

of Organic Chemistry book I. In organic chemistry, we are concerned with thepredominant species in solution, so it is a good idea that you fully understanithe Henderson-Hasselbalch equation for buffers, shown in a modified form as

Equation 7.2.

=

I::

ts

ID

5r

:t:

:"-J-a'

:E"r,

(q

pH = pKu (protonated species) a 1o, [Pgprotonated species].-

[Protonated species](7.2t

H.

D.

5,r lt,--_i -*

:*'Acid-base chemistry has been thoroughly covered in section I and in genera-chemistry, so we shall assume you have reviewed it thoroughly.

Copyright @ by The Berkeley Review 16,2 The Berkeley Reviex


Example 7.3Civen that the pK6 of NH3 is 4.74, what is the approximate pK6 of Et3N?

A. -0.32B. 3.25c. 5.31D. 9.26

SolutionAlkyl groups are electron donating, so triethyl amine (Et3N) is more basic thanammonia. As a consequence, the pK6 for triethyl amine must be less than 4.24.This eliminates choices C and D. A negative pK6 corresponds to a strong baseand amines are weak bases, so choice A is too low. Choice B is the best answer.

Example 7.4Diethyl amine is more basic than ethyl amine in an aqueous solution. If the pK6for ethyl amine is 3.29, then the pKu for diethyl u**o.irrrn chloride is:

A. 3.05.B. 3.89.c. 10.11.D. 10.95.

SolutionThe question is asking for the pKu of a protonated amine, which lies between 9and 11. This eliminates choices A and B. Because diethyl amine is more basicthan ethyl amine, the pK6 for diethyl amine is lower than ihe pK6 for ethyl amine(which is 3.29). The sum of pKu for diethyr ammonium catLn and pK6 fordiethyl amine is 14. For the math to work out, the pKu for diethyl u*.norri.r*cation must be greater than 10.71. This makes choice D the best of a1l possibleanswers in this the best of all possible acid-base questions involving amines.

Example 7.5\Alhat is the relative base strength of the following three compounds?

G*"' ""iG*"' ','o$*',Compound I " Compound II Compound III

A. Compound II > Compound I > Compound IiIB. Compound I > Compound II > Compound IIIC. Compound III > Compound I > Compound IID. Compound III > Compound II > Compound I

SolutionAll three compounds are amines. The question comes down to the electrondonating and withdrawing nature of the other functional group. Becausecarbonyls are withdrawing by resonance, Compound II is less basic thancompound I. Because alkoxy groups are donating-by resonance, Compound IIIis more basic than Compound I. The relative order for basicity is thusCompound III > Compound I > Compound II, making choice C correct.



A.B.c.D.

Example 7.5\A/hat is the pH of 0.1 M aniline, which has a pK6 of 4.2?

2.68.877.413.0

SolutionAniline is a base, so the pH is greater than 7.0. This eliminates choice A. Anilineis not a strong base, so a 0.10 M solution must have a pH less than 13. Thiseliminates choice D. For a pure weak acid or pure weak base, it is best to use thesimplified equation learned in the general chemistry book. For a weak base, lt'euse: pOH =1/zpKA-r/21c9 [Base]. Plugging the values into the equation r,r'e

get: pOH =1/2(4.2) -1/z(1) =2.L +0.5 =2.6. If the pOH is 2.6, then the pH is11.4, making choice C the best answer.

NucleophilicityWeak bases, such as amines, make good nucleophiles and the nucleophilicity otamines parallels their basicity for the most part. Deviations from this trend aredue to steric hindrance. The majority of amine reactions in organic chemistn-involve the amine behaving as a nucleophile by attacking an electrophile anieither replacing a leaving group or breaking a n-bond. Amines do nucleophiJrcsubstitution reactions without competition from elimination reactions. Oneproblem with amines is that they are increasingly nucleophilic as more alkv-groups are added. This makes the synthesis of primary amines rather difficultbecause once formed, they are more reactive than ammonia. Typical aminereactivity questions involve comparing the relative basicities andnucleophilicities of a series of amines, looking at the effects of substituents boththrough resonance and the inductive effect on reactivity, and amine structurequestions.

ExampleT.TWhat is the major organic product when ammonia, NH3, is treated excess eth';-iodide, H3CCH2I, following workup?

A. H2C=CH2B. H3CCH2NH2C. (H3CCH2)2NHD. (H3CCH2)3N

SolutionBecause secondary amines are beiter nucleophiles than primary amines which i;turn are better nucleophiles than ammonia, the reaction proceeds until the alloihalide is exhausted. With excess alkyl iodide, the reaction is free to add as mantrethyl groups to the amine as possible. The best choice shows three alkyl groupsadded to the amine. This makes the correct answer choice answer D. Choice,\would not form without a strong base, so it should have been eliminate,Jimmediately. It is possible that (H3CCH2)aN+, a quaternary amine, would forrr*but it is not listed as an ans\{'er choice.

Copyright O by The Berkeley Review t64 The Berkeley Reviet


Example 7.8An amine can react with all of the following EXCEPT:

A. an anhydride.B. benzoic acid.C. an alkyl halide.D. an ether.

SolutionAn amine can react as a nucleophile with a highly reactive electrophile such as ananhydride. This eliminates choice A. An amine can react as a base, so whenadded to a carboxylic acid such as benzoic acid, it carries out a proton transferreaction. This eliminates choice B. An amine can react as a nucleophile with analkyl halide in a nucleophilic substitution reaction. This eliminates choice C.Ether is usually a solvent for many of these reactions, because an amine does notreact with an ether. The best answer is choice D. The three reactions for choicesA, B, and C are shown below:

-,,nt."jtto * *t, -----+ oio*,

* ,r*loo

Ao" * + NH4+

HsCa

NHs * *ru.r.Jl-..o

+ NH, ----> i + HX

R NHz

o

ARX

Example 7.9lVhat is the major organic product for the following reaction?

o

{"

NH B.

+ NH3

--->C. D.o o o

+,"1}" H,NUoH H,NUN',

Solution-{mmonia is a good nucleophile that can break open the strained four-membered:it g by attacking the carbonyl carbon. The final product is not going to have a:our-membered ring, so choices A and D are eliminated. Amines react withesters to yield amides, so choice C is eliminated and the best answer is choice B.

Copyright Oby The Berkeley Review 165 Exclusive MCAT Preparation


Example 7.10\Alhat is the hybridization of nitrogen in a tertiary amine?

A. so

B. sp'^

c. sprD. d2sP3

SolutionIn a tertiary amine, nitrogen makes three bonds and has a lone pair of electrons.This means that there are four electron pairs surrounding ttitrog"r,, so thehybridization must involve four atomic orbitals. The hybridi"itionof-nitrogen inany amine, whether it is primary, secondary, or tertiary, is sp3. This makeschoice C the best answer.

Example 7.11How can it be explained that Compound AWLg3g6 is more basic thanCompound KMF8889?

A. Aromaticity reduces the availability of the electron pair for donation inCompound AWL8386.

B. The inductive effect is greater in Compound KMFgggg.C. Steric hindrance is greater in Compound AWLg3g6.D. The nitrogen in Compound KMF8889 lacks hybridization.

SolutionThe lone pair on nitrogen in compound KMF8889 is tied up in the aromatic ringof the pyrrole. This makes it less available for sharing wlth a protic hydrogeriand thus makes the compound less basic. This makes choice A tire best ur,r-ur.

CH"t'N

OCompound KMF8889

CH.I

oCompound AWL8386

I

I

Copyright O by The Berkeley Review r66 The Berkeley Review


Formation of AminesPrimary amines may be synthesized in a variety of ways. The synthesis ofprimary amines is not as simple as it may initially appear. You cannot just addammonia to an alkyl halide, because the product (an alkyl amine) is a betternucleophile than ammonia and thus it can react a second and third time withanother alkyl halide. This results in the formation of multi-substituted amineproducts (2' and 3' amines) with random distribution in the product mixture.Figure 7-3 shows six synthetic methods for forming primary amines. Some ofthe reactions can also be used to form secondary and tertiary amines.

Gabriel phthalimide synthesis (for 1' amines only)

oo o

e(*Ho

KOH-------->Hzo

Hofmann

oil

.z C:.R NHz

Imine reduction (for 1" or 2' amines)

Azide reduction (for 1" amines only)

NaN" - + H"/MetalR-X ----l-> R -N- \\==N '' >

Nitrile reduction (for 1'amines only)

R-X NaC=N- R- C=N Hz / Metal-

-.+ RXl(--->

OH+ HrN*R

OH

R- NH2 + N2(s)

R- CH2- NH2

RNH2+CO2+KX+HX

o

rearrangement (for 1" amines only)

X2 / KOH(aq)

oil

,zc:.RR'

NH"+A

-/HNil

.zC:,RR'

t-*-t

I

.z 9:.RIR'H

NI

H

oil

,zC:. ./HRNI

H

Amide reduction (for 1', 2', or 3' amines)

1. LiAlHa

--j+-

2. NHaCI(aq) n

Figure 7-3

c9

1. NaBHr(thO+2. NHaCI(aq)

HH\r,zC:. ./H

o(".



Reduction of an azide or nitrile, the Gabriel phthalimide synthesis, and Hofmannreaarangement all produce a primary amine as their majoi product after workup.Imine reduction can form either a primary or secondaryu*ii.", depending on theimine, and amide reduction can form either a primary, secorrdiry, oriertiaryamine. Figure 7-4 shows the formation of secondaty and tertiary amines.

Imine reduction

N/R t-*-o

RNH2* ll r NaBHd(rho I

^* _,,,t_-,, tNtrffi o,-[-oR' R" z""Ti""

oll

.', C:.R' R''

Amide reductionHH

i. LiAtH. \ t"-- /_\./_.2. NHaCI(aq) R' N

I

H2" amine

i] H

1. LiAlH, \ I

-

z-\ -/"2. NHaCI(aq) R' NI

R'3" amine

\zvhich of the following compounds CANNoT be reduced to form an amine?A. An amideB. An imineC. A nitro compoundD. An alkyl hydrazine

SolutionAccording to Figure 7-3, when an amide is treated with lithium aluminumhydride, it is reduced into an amine. when an imine (R2G=NH) is treated withsodium borohydride, it is reduced into an amine. This eliminates choices A andB immediately. Figure 7-3 shows neither a nitro group nor hydrazine beingreduced into an amine, so memorization won't finish lhis question. A nitrocompound contains one nitrogen, while hydrazine contains two nitrogen atorrr<-Considering an amine only contains one nitrogen, it is more likely tf,at a nitrogroup is reduced into an amine (primary amine to be specific) than irydrazine. Anitro group can be reduced to form an amine by adding Clemmensen reagents(HCl(aq) and a metal catalyst), which eliminates choice C. An alkyl hydrizne(RHN-NH2) cannot be reduced to form an amine, because reduciion-will notcleave the nitrogen-nitrogen bond. This makes choice D the best answer. Thkalso emphasizes a point about your review. Despite our best efforts to track the

oil

R/c\N/ R

I

H

oil

R/t-N/ R

I

R'

tuWh

MMM

ndffim

dhtu0mtu@ffiaffiPtum

Copyright @ by The Berkeley Review r6a The Berkeley Review {rq


MCAT and incorporate information they release about the exam, there will besome topics we miss. If you plan on memorizing all the information in ourmaterials, you will be fine for the most part. But there will some subjects thatthey introduce that may surprise you. To prepare for this exam, you mustemphasize general concepts and logic more than memorization.

Reactions of AminesThe lone pair of electrons on nitrogen is readily shared, making an amine a goodnucleophile. The crux of amine reactivity involves the lone pair on nitrogenattacking an electrophilic carbon. As a general rule, either a ieaving gro.rp i,displaced or water is formed as a side product. This should help to predicftheproducts of amine reactions. We will consider the addition of amines to alkylhalides, diazonium formation and the Hofmann elimination first and imineformation and hydrazine reactions later. Figure 7-5 shows a few amine reactions.

Addition of an amine to an alkyl halideH3N: + R-X

--+ H3N+RX-, H2N+R2X-, HN+R3X-,This reaction results in multiple additions, because the moresubstituted an amine, the better it is as a nucleophile.

Reactions of diazonium saltsNaNO"

R-NH2 ff R-N*=N: Cl- + NaCl + 2H2o

N2 is about as good a leaving group as there is,so diazonium salts are excellent electrophiles.

NaNOr.\r-NH2 HC- Ar-N+f N: Cl-

Hofmann Elimination

*-tHr - ar{ NH2 3 cH3l- o-atr-ari.*,cll3)r

I' \Ag2o / A

\p11g: CH2 + (HrC)rN + AgI

Least substituted alkene

Figure 7-5

in the first reaction, addition to an alkyl halide, there is a leaving group. F{ence,Jre nitrogen displaces the leaving group by way of a nucleophilic substitution:eaction to form an amine. As noted in Figure 7-5, there is the good chance thatnultiple alkyl groups will add to the amine. In the second reaction, formation of: diazonium salt, there is no leaving group. Hence, we need to look for the water5at forms. one nitrogen has the hydrogen atoms and the other has the oxygen:toms, so upon losing the hydrogens and oxygens, nitrogen atoms in need of:onds remain, so this reaction must form nitrogen-nitrogen bonds. In the last:eaction, Hofmann elimination, there is a leaving group, so the nitrogen displaces:1"1e leaving group to form an amine. With excess methyl iodide, it exhaustivelyrethylates to form a quaternary amine. The quaternary amine carries a positive::1arge, making it an excellent leaving group. The long alkyl chain attached to-ie nitrogen can undergo E2 elimination, kicking out the leaving group and:rrming a least substituted alkene. We shall address the Hofmann elimination inrore detail, as it has appeared on previous versions of the MCAT.

CuX(X = Cl, Br, C =N:)

Ar-X + N2(S)

-opyright O by The Berkeley Review 169 Exclusive MCAT Preparation


The beta hydrogen and thequatemary amine are anti toone another for E2reaction.

(TlcH,), r

o{--4;" -+H -OH

\-/

Hofmann EliminationThe mechanism for the Hofmann elimination reaction is a straightforward E2mechanism. The Hofmann elimination reaction involves forming a cationiiquaternary amine (nitrogen with four alkyl groups attached). when thequatemary amine is anti to a B-hydrogen (hydrogen on the carbon adjacent to thecarbon bonded to the quaternary amine), it canbe etminated with strong base.Ag2o, when hydrated by water, yields two hydroxide anions (oH-) urid t-osilver cations (Ag+), so it is a strong base. The silver cation can bind the iodideanjon and precipitate out of solution while the hydroxide anion deprotonates theB-hydrogen resulting in elimination. The E2 reaction does not haverearrangement associated with it, therefore the product is the least substitutedalkene. This is one of-few ways to synthesize a terminal alkene. Figure 74shows the orientation of a generic molecule in a Hofmann eiimination reiction.

N(crusH lrr..

a...rrlH/t

prCpO,H-O\

H

Figure 7-5

It is important that the compound have B-hydrogens relative to the quaternaryamine, otherwise there can be no elimination product. Be particularl! u*ur" ofthis when dealing with cyclic amines. Figure 7-7 shows Ho?mann elimination ofa cyclic amine.

T *trt, /.tt, ftu tr.\ /.",f*) 2cH3I f*) ++ r,,.'*) tt'L ,r..-*\\-,J \-J a

?-) z-)

(H3C)3N + AgI + 4..'\Figure 7-7

Copyright @ by The Berkeley Review 170 The Berkeley Revier


l

I:rines'"-hen

amines are added to a carbonyl, a nucleophilic attack at the carbonyl,::bon takes place. If there is no leaving group on the carbonyl, then the reaction- irns an imine and water and is driven by its equilibrium. If water or imine is

:=:noved (usually by distilling away the product with the lower boiling point),-:-tn the reaction is pushed forward. This is one of several reactions that can be---:ognized by the water side product more easily than the organic product. The: - rnation of an imine, oxime, and hydrazone are shown in Figure 7-8, to present*--t ease of predicting the product after identifying the water that is formed.

1'AmineR'

I

-N-HH

Aldehyde

:l'droxyamine

OH

I

-N-.HH

Aldehyde

Hydrazine

H-N"l

-N-HH

N,R--r ARHImine

1" AmineR'

I

.z N:.HH

Hydroxyamine

OH

I

,u N:.HH

Hydrazine

H,N-l

,u N:.HH

-+cat. HCI

R

N.Ro

A

o

A

o

A

o

AKetone

o

AKetone

o

AKetone

AImine

H

--------->A

________->cat. HCI

-oHN

ARROxime

OH

HR

H

H2N\ N

ffioA'.Hydrazone

R

N/

AOxime

HzN-..N

^*ARHHydrazoneRH

Aldehyde

Figure 7-8

These reactions proceed considerably better with an aldehyde than a ketone.This is attributed to the greater electrophilicity of an aldehyde than a ketone.when a primary amine is used, the imine that is formed is referred to as a shiffbsse. Hydrazones are used in the identification of carbonyl compounds,particularly ketones and aldehydes. Other carbonyl compounds have derivativesas well, but the majority of derivatives are formed by substitution reactions withnitrogen compounds. The basic idea here is that by saturating an aldehyde orketone with a compound that contains at least two hydrogen atoms on thenitrogen (ammonia, primary amines, hydroxyamine, or hydrazine), a C=O isconverted into C=N. With excess water, the reverse reaction is observed.

Copyright @by The Berkeley Review t7t Exclusive MCAT Preparation

Organic Chemistry Nitrogen Compounds Non-Itiological

Example 7.L3\zVhat is the major organic product of the reaction shown below?

o

oAo'.H

H

Ph

H

A.o

-Jr.-RNHI

Ph

D.NPh

ANHPh

R

N/

AR'R'

.. /, R"Nll *

o4ro:o/\

HHImine

SolutionFind the water and all will be fine. The oxygen comes from the ketone, so the

two hydrogens must come from the nitrogen-containing species. Only one of the

two nitrogen atoms has two hydrogen atoms, so that is the nitrogen that replaces

the oxygen in the ketone. The result after removing the water and connecting the

atoms is choice D. Connect-the-dot skills you developed in kindergarten may

prove more useful on the MCAT than the volume of information you memorizedduring college. Probably not, but it felt good to say.

Imine'Enamine Tautomerization]ust as ketones and aldehydes undergo tautomerization in the presence of mildacid or mild base to form an enol, imines also undergo tautomerization to form

an enamine. Figure 7-9 shows the equilibrium between an imine and an

enamine.

H\ ., R"

HEnamine

Figure 7-9

An imine can form only if the nitrogen compound has two hydrogen atoms on

the nitrogen. When a secondary amine reacts with an aldehyde or ketone, the

product L ut uttu-ine, because an imine cannot form. The formation of an

enamine from a secondary amine and ketone is shown in Figure 7-10'

R-n -R" R-iu'R"

tl"'txo

Eliminationr "\*Ho/\

HHHCarbinolamine Enamine

Figure 7-10

.. .,R"Nr

&R<+H

:111n"

-\H

+H'T)r-

o

-l-r'oIIN+

HHAldehyde

HII

NT_->/tt\ AR' R''

2" Amine

,N-N. A*

B' NH c'

*Ao

Copyright @ by The Berkeley Review 172 The Berkeley Keview


AmidesAmides are carbonyl compounds with a nitrogen bonded to the carbonyl carbon.They are formed when an amine or ammonia is added to a carbonyl compoundwith a leaving group. Figure 7-11 shows the three types of amides.

oil

R-c- N- R

I

H2'Amide

Figure 7-11

ruPAC nomenclature rules call for naming the alkyl groups on the nitrogen witha prefix of "N" first, followed by the carbonyl chain. The degree of substitutionimpacts the physical properties. Acetamide, H3CCONH2,has a melting point of82"C and a boiling point of 227'C. N-Methylacetamide, H3CCONHCH3, has a

melting point of 28"C and a boiling point of 204'C. N,N-Dimethylacetamide,H3CCON(CH3)2, has a melting point of -20"C and a boiling point of 165'C.Amides form H-bonds between oxygen's lone pair and the hydrogen on nitrogen,which explains the drop in boiling and melting points as substitution increases.

Formation of AmidesAmides are typically formed from the addition of an amine to an acid anhydride,acid halide, or an ester, as covered in section V of Organic Chemistry book IL Aprimary amide can be formed by partial hydrolysis of a nitrile (R-C=N) and canalso be formed from an oxime by way of the Beckmann rearrangement. TheBeckmann reamangement results in the formation of an amide from an oximethrough the addition of either a strong acid or PCI3O/PCI5. An oxime is formedby the addition of hydroxyamine to either an aldehyde or a ketone, as shown inFigure 7-8. Beckmann rearrangement is shown tnFigureT-12.

oil

R-c- *.tI

H1" Amide

oil

R,/c- N,/ R

I

R"3" Amide

rr Beckmann Rearrangement

.'l:, t?-*' :o:

^:frJ'-I' oA*.o/t oAo R/ -l

;ph,o il

,(H--1l, . rr "-'dYi

..d..atii n_.g.'.n H

ttt/gX

->J-, - n\ !r;o'. R/ 'o,R .-(.0*.T-- ::)':*'o/

H

FigureT-12



Reactions of Amideswe have seen that amides can be reduced into amines using lithium aluminumhydride. In addition to reduction, an amide can be hydrotyied into a carboxylicacid and amine. Primary amides can undergo Hofmann r"urrur,g"-ent to form aprimary amine of one less carbon and carton dioxide. The rlaction does notwork with substituted amides; it works only with primary amides. Figure z-13shows the step-by-step mechanism for Hofmann rearrangement.

Hofmann Rearrangement

o-c:oaa

/+ R-N:

\

R NI

H

piGYc:N.o/J '

\

il .i,. lr>H

\t /H. C_N:

H-61-*n$; {t

to

iol ll[i^- -

'.ztT ^rIsocyanate H-9: I R

: O q:*;/ "S-"':.- |

H-Oi/ R

.>Co:

.\-

t-''i-"

Figure 7-13

In the Curtius Rearrangement an acid chloride is treated with sodium azideNaN3 to form an isocyanate. The conversion goes through an acyl azide thatwhen heated decomposes to_yield an isocyanate ana nitrogeln gas (a great leavinggroup). Just as seen in the Hofmann rearrangement, an iiocyanate ian ,rndergJdecarboxylation in basic conditions to geneiate a primary imine. Figure 7-14shows the key intermediates of the Curtius rearrangement.

Copyright O by The Berkeley Review 174 The Berkeley Revier


Curtius Rearrangemento

oA.,

Hell-Volhard-Zelinskii synthesis

o

OJON' W R

H

ltllN"-li J

-o x:it:N

o:c:o + o-*,h

uo'lltlO:C:N\

oX.' \ o\ ll --------oA*:it-N

?/oA*-ir:N

FigureT-'1,4

Amino Acid SynthesisAn amino acid, as the name implies, is a molecule with an amine group and acarboxylic acid group. In an aqueous environment, the carboxyl terminal isdeprotonated and the amino terminal is protonated, so it has at least two chargedsites despite having no net charge. However, in organic solvents they exist as anuncharged molecule. There are three synthetic pathways for making aminoacids: Strecker synthesis, Hell-Volhard-Zelinskii synthesis, and reductiveamination of cr-keto acids. The three syntheses are shown in Figure 7-15.

Strecker synthesis

oil"t'o,-,

NHz

H"N CO"H.XLRH

H"N H.XR CO2H

NH"----i>

o

AN.H

-oL-4,.+HRH

Reductive amination of g-keto acids

A *t'-R CO2H

N,H

AR CO2H

Figure 7-15

NaBH, NH,CI.* "-.Et2o Hzo


Organic Chemistry Nitrogen Compounds Biological

ffifinu$itHl' gen,,Comp6#HftAmino AcidsNatural amino acids (s-2-amino carboxylic acids) are a critical part of biology asthe building blocks of proteins. There are a total of twenty standard amino icidsthat humans code for. Although there are over 500 amino acids in nature,humans code for just twenty. In mRNA, amino acids are coded fotby a sequenceof 3 bases (codons) using adenine (A), cytosine (C), guanine (G), and u.acil 1u;.For instance, histidine is coded as cAC or CAU. with 64 (43) combinations ofcodon triplets, there are multiple codes for some amino acids. Figure 7-16 showsthe generic structure of an S-amino acids in both aqueous environment and lipidenvironment.

oil

H.N+- c' \c'l.-\o/2,

HRZwitterion form of amino acid

(aqueous environment)

oltt'*-.-t-on

I ,,,

HRUncharged form of amino acid

(lipid environment)

Figure 7-"16

Because amino acids are compounds which contain an amino group on thesecond carbon and a carboxylic acid, they have acid-base properties that you areexpected to know. Amino acids are essentially just polyprotic acids where the allof the protons are weakly acidic. The amino terminal when protonated forms acation with a pKu of roughly 9.7 + 9.9. The carboxylic acid terminal has a pKu ofroughly 2.2 + 0.4. In long chains, they form proteins, which you are also expectedto know. This section covers functional features of the amino acids, the acid-baseproperties of the amino acids, protein structure, the structural effects of hydrogenbonding, cysteine, and proline, protein sequencing, and some enzymaticchemistry. We shall start with terminology and classification. Two terms thatwill be used repeatedly to describe the amino acids are hydrophilic andhydrophobic (based on their side chains). within the hydrophilic group, someamino acids are acidic, some are basic, and some the others are polar.

Structures and ClassificationThere is a chiral center at the cr-carbon in all amino acids except glycine.Naturally occurring amino acids have s-stereochemistry at carbon 2, although inbiochemistry we refer to that as an L-amino acid. The exception to the Lcorresponding to S rule is cysteine, where L-cysteine is the natural form, but ithas R-stereochemistry at the o,-carbon. For the MCAT, you should recognize thetwenty amino acids for which we code, along with general pKu data for theamino terminal, the carboxyl terminal, and the side chain of protic amino acids.Amino acids are categorized according to behavior, and although it may varyfrom textbook to textbook, it is generally based on water solubility and degree ofprotic activity. we shall classify amino acids into one of five categories:hydrophobic, semi-hydrophobic, hydrophilic, acidic, and basic. There are aminoacids that seemingly should fit into muitiple categories, like aspartic acid, whichis hydrophilic and acidic, but we sha1l have each amino acid appear only once.

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Hydrophobic amino acidsHydrophobic amino acids are "water-fearing", which means that they are notwater soluble and they prefer to be situated in an organic (lipid) environmentrather than an aqueous environment. In aqueous environments, hydrophobicamino acids are found in the core of a folded protein. Figure 7-17 shows thehydrophobic amino acids glycine, alanine, valine, leucine, and isoleucine as theyexist in an aqueous environment at pH = 7.

Hydrophobic Amino Acidso

pKaz =el8 ll PKaz =9'87

HrN\C

olt

oil

IH

o--t?q

PKaz = 9j2

o-= 2.17

= Z.JJ

opKaz = e.74 llsrN\ .zc\co-

| | PKor=2:tH CH2CH(CFI?)2

Leucine(Leu, L)

PKaz =10'64

opKaz =e.74 llHrN\ .zc\co-

I r2 pK,' =z.zz

Valine(Val, V)

Phenylalanine(Phe, F)

oII

opKaz =e.27 llHsN\./t\

I ''.- rKuH CHs

Alanine(Ala, A)

i PKat =2'29

cH(cFi")2

H CH(CFI3)CH2CH3

lsoleucine(Ile, I)

PKaz = 9'24oil

:'*\./t\o-\ i\ PKar=1'e5

\,J rrnProline(Pro, P)

IH

Figure 7-18

2 PKat

I rz. PK*=z'ssH CH^l'

ocH2cH2scH3Methionine

(Met, M)

srN\ -zc\co-| '2. PKot

HHGlycine(Gly, G)

FigureT-17

S emi-hy drophobic amin o acidsThese amino acids are typically considered to be hydrophobic in most text books,but we segregate them from other hydrophobic amino acids, because they don'tpack as well as the alkyl side chains. These include the aromatic side chains(except for tyrosine) and the amino acids that are more water soluble thanexpected for an aliphatic side chain. Figure 7-18 shows the semi-hydrophobicamino acids proline, phenylalanine, tryptophan, and methionine as they exist inan aqueous environment at pH = 7.

Semi-Hydrophobic Amino Acids

PKaz =9'44

.zc\ HsN\ -zc\co-

H'N\./t\o


Organic Chemistry Nitrogen Compounds Biologicat

Hy drophilic amino acidsHydrophilic amino acids are "water-1oving", so they are water soluble and preferto be situated in an aqueous environment rather than an organic 11lpia}environment. In water, hydrophilic amino acids are found on the exterior of afolded protein. These are sometimes referred to as protic side chains. Figure 7-19shows the hydrophilic amino acids serine, threonine, cysteine, tyrosine,asparagine, and glutamine as they exist in an aqueous environment at pH = 7.

PKaz = 8'80

Hydrophilic Amino Acidso

pKaz = e.13 llH:N'\ -zc:.co-

OHTyrosine(Tyr, Y)o

pKaz = e.46 llH'N\./t\o-

I r2 PK,.=zloH CH2OH

Serine(Ser, S)

Example 7.14Which of the following amino acids is LEAST hydrophilic?A. SerineB. TyrosineC. CysteineD. Leucine

SolutionHydrophilic is defined as "water-loving," which implies that the compoundshould be water-soluble. Serine has an alcohol side group, so it is water solubledue to the hydrogen bonding of the side chain. Tyrosine has an alcohol sidegroup as well, so it too is water soluble due to the hydrogen bonding of the sidechain. This eliminates both choices A and B. Cysteine has a thiol (sH) sidegroup, so it is water soluble due to the polarity of its side chain. Leucine has analkyl side chain, so it is hydrophobic. The best answer is choice D.

nuN\ -zc\co-I tz- PK,- =z'oz

H 9HzI

o4Nu,Asparagine

(Asn, N)

opKaz = e.1o llsrN\ -zc\c-o

| '7 vK,r- =z'ov

H CH(CFr3)OH

Threonine(Thr, T)

I r'-- PK* =2'tzH CH"t-

9HzI

o4Nu,Glutamine(Gln, Q)

oPKas = 10 33 llHsN\ -zc\c- o-

t r--- PK*=taaH CH2SH

PKaz = 8'36Cysteine(Cys, C)

Figure 7-19

opKaz = e.12 llHrN\ -zc\co-

I r2 PKot=z'ztH CH.

Q, =1007

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Example 7.15Which of the following amino acids contains a benzene ring?

A. ValineB. TyrosineC. HistidineD. Isoleucine

SolutionPhenylalanine, tyrosine, and tryptophan are the only amino acids with sidechains that contain a benzene ring. of those three amino acids, only tyrosine isoffered as a choice, so choice B is the best answer. Histidine has an aromaticring, but it is not a benzene ring.

Example 7.16\\4'rich of the following amino acids is classified as hydrophilic?A. ProlineB. ThreonineC. ValineD. Phenylalanine

SolutionProline, valine, and phenylalanine are hydrophobic, because they contain alkylside groups. This eliminates choices A, C, and D. The only choice reft isthreonine, choice B, which has an alcohol group as at the end of its side chain.The hydroxyl group forms hydrogen bonds, making threonine hydrophilic.Choice B is the best answer.

.lcidic amino acids{cidic amino acids have side chains that lose a proton from their neutral state.They have three acidic sites, the N-terminal, the C-terminal, and the side chain.ln the case of Glu and Asp, the side chain is a CooH. Cysteine and tyrosine canalso lose a proton from their neutral side chain, but it happens at a high enoughpH that we generally think of them as polar more than as acidic. Figure 7-20shows the acidic amino acids aspartic acid and glutamic acid as they exist in anaqueous environment atpH =7.

O Acidic Amino Acids

pKae = 10.01 ll

"r*\.-.-o_| '7

vx", =t'ooH 9Hz

I

AK*=E'sao- o-Aspartic acid

(Asp, D)

oPKas = e.e5 llHrN\ ,zc:.co-

| 'tz-

PKot =zlzH CH"

l'CH"t-

AK^z=a'zzo- o-Glutamic acid

(Glu, E)

Figure 7-20

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pKaT = 9.20

Basic amino acidsBasic amino acids have side chains that gain a proton from their neutral state.The three examples, lysine, arginine, and histidine, have nitrogen-containing sidechains. Like acidic amino acids, basic amino acids have three pKu values. youmight wish to recall the mnemonic "his lies are basic," which means thathistidine, lysine, and arginine are basic. For histidine, the lower nitrogen in theimidizole ring is the basic site. Figure 7-21 shows the basic amino acids lysine,arginine, and histidine as they exist in an aqueous environmen t at pH = 7 .

Basic Amino Ac

PKaz = 8'99

ids

oll

oilH'N\c/t\o-

|'t7 PKot =zteH (CH,)ot-

pKaa = 10.80 +pg,

Lysine(Lys, K)

HeN\./.\o_

I 17 PKot=t'ezH (CHr).t-"

NH

0"";,';:21-,u*,,

Arginine(Arg, R)

HrN\./.\

opKaa = e.15

ll

f 'r- vKu

PKaz

H 9HzI

^N-H= 6.0s\ t

N:./Histidine(His, H)

o-= 1.81

FigureT-21

Amino acids can be categorized in many ways, not just by their side chain, as wehave done here. For instance, let us consider essential and non-essential aminoacids. There are eight to nine essential amino acids, depending on a person's age.Essential amino acids are the amino acids we cannot synthesize in our body, iotherefore we must take them in through our diet.

Example 7.17\A/hich of the following amino acids has a charged side chain at pH = 7?

A. ValineB. PhenylalanineC. ArginineD. Cysteine

SolutionAt a pH of 7, the side chain for valine is CH3, the side chain for phenylalanine isCH2C6H5, the side chain for arginine is CH2CH2CH2NHC(NH 2)2+, and the sidechain for cysteine is CH2SH. The only charged side group is found on arginine,which makes choice C the best ansl\'er.

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Example 7.18which of the following amino acids carries a net positive charge at pH = 8?

A. ThreonineB. ArginineC. PhenylalanineD. Histidine

SolutionAt a pH of 8, the amino terminal is protonated and carries a +1 charge and thecarboxyl terminal is deprotonated and carries a -1 charge. The compound existsas a zwitterion if the side chain is neutral. So, to have a net positive charge at pH= 8, the side chain must carry a positive charge. The only amino acid of ihechoices to have a positively charged side chain at pH = 8 is arginine, which has aside chain pKu of 13.2. This means that arginine has an overall positive chargeup to the pH where the amino terminal is mostly deprotonated. This makesarginine, choice B, the best answer. At pH = 8, the structures are as follows:

ThreonineProtonated O

pKuz > 8 llHsN\c-C-o-

l'r, ,""""i:tHCH

/\HO CH:

HistidineProtonated O

PhenylalanineProtonated O

ArginineProtonated

PKoz > 8oll

H"N*\ cr t-ar-

pKu:>S ll pK"rr8 llHsN\c-C-o- HsN\c.C-o-

l',- ,?""i:, 1", i:i':,H (CH")"

I -"Protonated NHPKo:>B IHzNy'L: *rr',

overall charge = 0 overall charge = +1 overall charge = 0 overall charge = 0

Example 7.19\A/hich of the following amino acids is involved in cross linking within a protein?A. CysteineB. ThreonineC. ProlineD. Tryptophan

SolutionTwo cysteine residues form a disulfide bridge (-CH2-s-s-CH2-) within a proteinor between two separate primary structures (protein strands). The formation ofthe crosslink involves oxidation of the thiol groups, one of those facts you shouldknow. The best answer is choice A.

There are some fundamental structural features of proteins that result fromproperties of their component amino acids. Cross-linking of cysteine residues isone example. Another is strucfural turns, forced alterations in the structure of aprotein, caused by proline. Proline is cyclic and thus unable to rotate freelyabout its sigma bonds. It is locked into a cyclic conformation, which locks theprotein backbone into a tum.

1", ,t-""i:t

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Example 7.20\Arhich of the following amino acids is Mosr soluble in water?A. AlanineB. SerineC. IsoleucineD. Methionine

SolutionThe compound that is most soluble in water of the choices is the hydrophiliccompound. Because the side chain of serine contains a hydroxyl grorrp, it.*form hydrogen bonds. This makes serine hydrophilic, whiie alanine, isoleucineand methionine are hydrophobic. Choice B is the best answer.

Example 7.21which of the following amino acids has a chiral center in the side chain?A. ProlineB. ThreonineC. TryptophanD. Methionine

Solutionof the twenty amino acids for which we code, only threonine and isoleucinehave chiral centers in the side chain. The side chain of threonine is naturally RThe diastereomer of naturally occurring threonine (that differs from threonine inthe chirality of the side chain from R to S) is referred to as allo-threonine. Thebest answer is choice B.

Isoelectric PointsIn aqueous solution at a pH of T, an amino acid exists as an ionic species. Theamino terminal is protonated and the carboxyl terminal is deprotonated. The netresult is zero overall charge (excluding the side chain in some amino acids), andthe structure is referred to as a zutitteiion. The zwitterion is neutral overall butcontains oppositely charged sites. The zwitterion form of an amino acid, becauseit is a salt, has a high melting point. The pH at which the zwitterion exists inhighest concentration is referred to as the isoelectric pH. The isoelectric pH isobtained by averaging the two pK^ values that involve the neutral molecule-That is to say that you average the pKu that takes the molecule from -1 to neutralwith the pK6 that takes the molecule from neutral to +1. The molecule is aperfect zwitterion exactly midway between the two pKu values which include theneutral species. The isoelectric point coincides wiih an equivalence point on atitration curve, the first equivalence point of all amino acids except the ones wiilrbasic side chains, where it coincides with the second equivallnce point. Tocalculate the isoelectric pH you must decide if the side chain is acidic, basic, mneutral. For amino acids with neutral or acidic side chains, the isoelectric point bfo"ld by averaging pKal with pKu2. For amino acids with basic side chains, theisoelectric point is found by averaging pKa2 with pKu3. Figure 7-22 shows zpictorial overview of the structure of an amino acid ai it chang:es with increasingpH. The R-group represents any neutral side chain.

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ooo

HROverall .hu.g"' 0

. PLr + PKuzpL=--

HROverall

"hurg", -1-

ooo

H CFLl'#on

Overall charge' *1-

""xOverall .hurg", 0

. PLr + PKazpr=--

";.

Overall chu.ge, -1-

H gFr,

Ih"Overall

"hurg",'2

OH

FigureT-23

Figure 7-24 shows the titration curve associated with full deprotonation of anacidic amino acid, starting from its fully protonated state and proceeding to itsfulIy deprotonated state,like it is shown tnFigureT-23.

1,/z z

Equivdentsbase added

Figure 7-24

-*/oHi'=-#o i+L*-/"HR

Overall churg", *1-

FigureT-22

Figure 7-23 shows a pictorial overview of the structure of an acidic amino acid asit changes with increasing pH. The side group is specific for aspartic acid, butthe charges are the same with any acidic side chain such as glutamic acid,cysteine, or tyrosine. Acidic side chains lose a proton from their neutral state toform an anion. Acidic side chains contain a hydrogen on either oxygen or sulfur.

n,*/oH Pqr-- og,"/os!e**-/"

Isoelectric Point



Figure 7-25 shows a pictorial overview of the structure of a basic amino acid as itchanges with increasing pH. The side group is specific for lysine, but the chargesare the same with any basic side chain such as histidine and argininu. "By

definition, basic side chains gain a proton to form a cation at low "pH

,rut.r"s.Basic side chains contain a lone pair of electrons on nitrogen when neutral.

H (CH,),t-'NHs

Overali charge, *2

H (CHr)ol-'*NH,

Overall charge' *L

H tfuna*NH,

Overall charge: -1-Overall chu.ge, 0

ot=ryfuFigure 7-25

Figure 7-26 shows the titration curve associated with fully deprotonation of anacidic amino acid, like shown in Figure Z-25.

Isoelectric PointpH=pK

pH=pK

HzA*

11 /z ort- /2

Figure 7-26

For calculating the isoelectric point of an amino acid, it is first necessary todetermine the nature of the side chain. The isoelectric point is the averagl ofpK"1 and pKn2 for all amino acids except lysine, histidine, and arginin". f-h"isoelectric point of lysine, histidine, and arginine is an average of pKu2 and pK6.No matter what isoelectric point you are determining, you should remembei thitit's an average of two consecutive pKu values.

Equivalents base added

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Easy Method to Get the Isoelectric point of a proteinFirst assume that the,protein exists at pH = 1, where all sites on the protein areprotonated. Carboxyl groups are uncharged when protonated, while imino sitescarr.y a positive charge when protonated. only the amino terminal and the sidechains of lysine, arginine, and histidine carry a charge (a positive one) at low pH,so the net charge of the protein is the rn- oi the lysires, irginines, and histidinesin the protein plus one (for the N-terminal). consider the p"rotein tnFigareT-27.

o_@@oH"N-Met-Thr-Ala- L;s-As Gly

-His- Glu- S"r4"#Llhil"'i'"V"i'-o"

scH3 cHoH CHa NHro,AoH H \n, oA on

cH2oH

CHs '4\(NH

HN-./

Figure 7-27

The protein in Figure 7-27 canbe abbreviated as H6A3*, because when it is fullyprotonated, it carries a +3 charge, and there are six sites from which a protonmay be lost (the N-terminal, the C-terminal, and the side chains of Lys,Asp, His,and Glu). Now consider the number of times the protein must be deprotonatedlo reach zero charge.

_The charge starts at +3, so the fully protonated proteinnust be deprotonated three times to become neutral. This results in pK^3eading to the zwitterion. Figure 7-28 shows an abbreviated titratior of u6d3i.'"

r{uA3* +I HsAr, +L *rno tro- *L ,ro'- SE* or-

Figure 7-28

1-" p-I is an average of the two pKu values surrounding the neutral species (the:Ku that leads to the zwitterionind the pK2 that leads"from the zwitierion). Inlis c1s9, the pI is found by averaging pKa3 with pKu4. The shortcut derived:rom this example is that the-total .hutg" at pu = t

"o-esponds to the subscript

- i the first pKu averaged to determine pt. This is because the pKu leading to thereutral species corresponds numericaily to the charge wherrfully protJnated.-onsider a protein with a +7 charge at pH = 1. It requires seven dlprotonations:,1 get to the neutral species, so pKuT is the pKu value that leads to the zwitterion.Figure 7-29 surnrnarizes this example and any generic charge.

-) +t

PK^z PK+^:_

-tl-

PKae,: H3APKa+

PKat+q :: +(q-1) T +1

p PKaq+t0

-r- PKaz+PKaaPr-

-

2

pI= PKuq+PKuq*t

2

PKarE...-+/

-*6

Figure 7-29

T: pI rs found by summing the number of basic amino acids plus one, and--aking the pKz with that numerical subscript and averaging it with the nextrK" value sequentially up. so, if a protein has two lyr;i"t,-or,u arginine and::ee histidines, then the pI is_an averige of pKuT and pkug. The sample protein:.s a lysine and histidine, so the pI is an aveiage of pfu3 a.,a pfua.

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The final step is to determine which pKu values are which. For the originalprotein shown, pKul is for the carboxyl terminal (this is always true). The

nffi:T"";1ii""-Tlffi;[ *r::lined rrom pKu inrormation' Figure 7-30

@@ooHrN- N4et-Thr-Ala- Lyr-A:p- Clv

- His- Clu- s"r4

pKa-=e7 # I I # + I I # | 'oHrr scH3 cHoH CHr trlnr _A^ _ H ftr cH2oH pKn=2..1

I " pKo = ro.'so-l -oH

,!. d'' _9H #1

.TJ #6 PKa = 3'9 .// Nrr r Y,.2 - 4.3!'I 13 #2 \ ,r" #3

pKan= 6.1HNJ

Figure 7-30

The assignment of pKu values is sequential with increasing numerical value, sothe pI for this protein is found by averaging pKa3 (4.3) and pKn4 (6.1), whichresults in a pI of 5.2. This means that at pH = 5.2 it exists as a zwitterion, at pH <5.2 it exists as a cation, and at pH > 5.2 it exists as an anion.

^r -PKa3+PKa4 -4.3+6.7 -70.4 - <'t''- 2 - 2 - u -"''

Example7.22\Alhich of the following statements is false about amino acids?

A. The carboxyl terminal always has the lowest pKu.B. A carboxylic acid side chain always deprotonates before the amino terminal.C. The sidechain group of lysine is less acidic when protonated than the amino

terminal of lysine.D. The D-isomer of glycine is the predominant isomer in human beings.

SolutionThe carboxyl terminal has the lowest pKu value for all twenty of the amino acidsfor which we code. This can be verified by looking at the amino acids in Figures7-17 through7-2L. This eliminates choice A. Because the carboxyl terminal has alower pKu value than the amino terminal, the carboxyl terminal is more acidicthan the amino terminal. By being more acidic, the carboxyl terminal loses a

proton more readily than the amino terminal. This means that choice B is valiiand thus is eliminated. Choice C requires that you look at the amino acids irFigure 7-27. The pKu of the amino terminal of lysine is 9.2. The pKu of the sidechain of lysine is 10.8. The lower pKu is associated with the amino terminalwhich implies that the amino terminal is more acidic than the side chain. Thismakes choice C valid, and thus it is eliminated. The only choice left is answer D.There is no D-isomer of glycine, because the side group for glycine is a hydroger-therefore glycine does not contain a chiral center. Without a chiral center, therecan be no D or L configuration associated with the isomer. This confirms that thestatement is false, making choice D the best answer. For the other amino acid-.the naturally occurring form is the L-isomer. This can be recalled by thinking a,i

the phrase "L is for life." It can also be remembered by thinking amino acids arenatural. Either way you choose to recall the meaning of the L label, amino aciisare naturally L. If you are bothered by the word "always" in choice A, sometimesan always is okay.

I

1

J

III

fiM.

,n

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Example 7.23tr\4rat is the isoelectric point for tyrosine?

A. 2.2B. 5.6c. 6.2D. 9.1,

SolutionTvrosine is not a basic amino acid, so itsoKul and pKaz. The isoelectric pointnathematics:

isoelectric point is found by averagingis solved according to the following

pI = PKal +PKa2 =2.2+9J. - 11.3 = 5.65222

This makes choice B the best answer. As a general rule, acidic amino acids havelI values less than 5.5 and basic amino acids have pI values greater than 2.5. Allrther amino acids have pI values within the 5.5 to 7.5 range.

Example7.24

"\hat is the isoelectric point for lysine?

.\. 5.7B. 6.5c. 9.2D. 10.0

5olutionLysine is a basic amino acid, so its isoelectric point is found by averagin g pKaZand pKu3. The isoelectric point is solved according to the following mathemitics:

pI =PKa2+PKa3 =9.2+10.8 =20 = I0222This makes choice D the best answer. As a general rule, basic amino acids have:i values greater than7.5, so choices A and B are eliminated before any math:reeds to be done.

Example 7.25i\hat is the isoelectric point for the dipeptide alanine-arginine?

A. 7.5

B. 9.0c. 11.1

D. 11.5

Solution-{,lanine is a hydrophobic amino acid and arginine is a basic amino acid, so the:-.oelectric point for the dipeptide is found by averaging pKa2 and pKu3. As-.',-ritten, the peptide linkage is made from the C-terminal of alanine and the N-:erminal of arginine, so there is no pKulgOOH; for alanine and no pKuglgr*; forarginine. For the dipeptide, pKu2 is the N-terminal of alanine and pKu3 is the;ide chain of arginine. The isoelectric point is solved as follows:

p1 = PKa2 + PKa3 = 9.9 + 13.2 -'zz 23.1.

2= 11.55

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This makes choice D the best answer. The difficult task in these questions is notdetermining which pKu terms belong in the equation, but rather assigning a

numerical value to each pKu term. If you've grown weary of continually turningback to the amino acids in Figures 7-77 tLrrough 7-21, then start making someapproximations. Knowing that the N-terminal is pKu2 and its value is around 9

should tell you that the pI is greater than 9. This eliminates choices A and B,

which helps to some extent. From here, you have to know the values. TheMCAT test writers will provide pKu values if the question is this specific.

Example 7.26

VVhat is the isoelectric point for the dipeptide histidine-leucine?

A. 5.5B. 6.7

c. 7.6D. 9.2

SolutionHistidine is a basic amino acid and leucine is a hydrophobic amino acid, so theisoelectric point for the dipeptide is found by averaging pKa2 and pKu3. Aswritten, the C-terminal of histidine and the N-terminal of leucine make up thepeptide linkage, so pKu3 is the N-terminal of histidine and pK62 is the side charnof histidine. The presence of the leucine does not alter the pI from what it woulchave been with histidine alone. The isoelectric point is solved according to thefollowing mathematics:

pI= pK62 +pKu3=6.1+9.2 -15.3 =7.6522

This makes choice C the best answer. You should have been able to solve thi-.particular question without consulting any other data. The side chain o:histidine has a pKu close to physiological pH, so you know that the pI is ar.

average of roughly 7 and9, which means that the pI is about B. Only choice C i'close to that number. Another method for determining the best answer is tcconsider that histidine is a basic amino acid. We know that the pI is greater thar7.5,but it must be less than the pKu of the amino terminal, given that pKu:corresponds to the amino terminal. Only choice C fits in the 7.5 to9.2rcnge.

Example7.27At what pH is the tripeptide serine-cysteine-isoleucine perfectly neutral?

A. 5.3

B. 6.1,

c. 8.4D. 9.4

SolutionThe pH at which it is perfectly neutral is the isoelectric point. Of the three amiri;acids, none are basic, so the isoelectric point for the tripeptide is found b',

averaging pKul and pKu2. The tripeptide has three active protons, the N-terminal of serine, the side chain of cysteine, and the C-terminal of isoleucineThe C-terminal is always pKal and in this case pKu2 is the side chain of cysteineThe isoelectric point is found using the following mathematics:

61 =PKal +PKa2

=2.3+8.4 =70.7 = 5.35'222

I

p'

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Organic Chemistry Nitrogen Compounds Biologicat

Example 7.28l4lhich of the following tripeptides has the LOWEST pI value?A. Asp-Lys-AspB. Glu-Lys-GluC. Lys-Asp-LysD. Lys-Glu-Lys

SolutionThe lowest pI value corresponds to the most acidic tripeptide. Choices A and Bhave two acidic amino acids and one basic amino acid, while choices C and Dhave two basic amino acids and one acidic amino acid. This eriminates choices Cand D' Choices A and B each have one lysine,.so both pls are found by averagingpKu2 and pKu3. In each tripeptide, ther! are five active protons (the N_terminal,the C-terminal, and th::: the siae ciains;. Both pKu2 and pKu3 correspond to theside chains of the acidic amino acids. tn choice-l, *," aliaic amino acid isaspartic acid while in choice B the acidic amino acid is gt.rtumi" acid. Asparticacid has a lower side chain pKu than grutamic acid, so tti" t.ipuftiae in choice Ahas a lower pI than the tripeptide ln ctioice B. Choice A is the best answer.

This makes choice A the best answer. Choices C and D should have beeneliminated, because there are no basic amino acids in the tripeptide. Cysteine isactually an acidic amino acid, but because most cysteine side chains are involvedin cross linking, we rarery consider their acid-base propertie, i'prot"ir,r.

Example7.29-\t pH = 5.0, histidine exists with a net charge of;{. +2.B. +1.c. 0.D. -1.

SolutionTo determine the charge of an amino acid at a given pH, first determine whether'he sites are protonatJd or deprotonated. wh"en the pH ;i;", than the pK,,the site is protonatga'

- ^{rren jhe pH is greater than the pKu, the site isdeprotonated. At pH = 5.0, histidine has a dJprotonated carboxyl terminal (pKu- 1.8), a protonated side chain (pKa = 6.1), anja pr;;;";rJ;;ino terminal (pKu= 9..2). The carboxyl terminal .itriur a negative .hurg", while ttre side chain andamino terminal are both carry a positive

"frurg". The"overall charge is therefore apositive one. The best answer isihoice B. Hiitidine at pH = 5.0 is drawn below:

PKa=9'2pH . pK HsIr(

PKa = 1'8o- pH>pK

deprotonated

protonated

PKu = 6.1

PH < pKoprotonated

overall charge = +1 +1 -1 = +1

H

H 9Hz

#-opyright @ by The Berkeley Review ra9 Exclusive MCAT preparation


Example 7.30At pH = 9.5, tyrosine exists with a net charge of:

A. +18.0c. -1D. -2

SolutionAt pH = 9.5, tyrosine has a deprotonated carboxyl terminal (pKa = 2.2), adeprotonated amino terminal (pKa = 9.1), and a protonated side chain (pKu =10.1). The carboxyl terminal canies a negative charge, while the amino terminalis uncharged. The side chain is uncharged when protonated, so the overaltrcharge is a negative one. The best answer is choice C.

ProteinsProteins are biological polymers built from amino acids. They are held togetherby covalent bonds in peptide linkages. A peptide linkage is formed when theamino terminal of one amino acid attacks the carbonyl carbon of the carboxr-lterminal of another amino acid, resulting in the loss of water. Because aminoacids have specific stereochemistry, proteins are highly chiral (on average, ther-have one chiral center for every amino acid in the protein). This large degree ofchirality explains why enzymes are so selective.

Protein Structural Features and LevelsThe structure of a protein can be broken down into levels. The most elementan'level is the primary structure, which is defined as the sequence (connectivity) o:amino acids within the protein. To break down the primary structure, you mus:cleave the peptide bonds. Figure 7-30 shows a tetrapeptide (four- amino aciiprotein).

Tetrapeptide (4 amino acid protein)

*/'

R3

oH

HH

N

I

H

Peptide linkage (amide bond)

Figure 7'30

In this case, the primary structure is the order in which the four amino acids a:earranged from the amino terminal to the carboxyl terminal. The prima;r.structure of a protein or enzyme is determined by sequencing the protein oneamino acid at a time. This will be discussed later. The interactions between ttwamino acids within the protein are responsible for the secondary structure"which is defined as the folding of the amino acids into their natural configuraticmwithin the protein. The chemical reasons for special features in the seconda--r'structure are most often attributed to hydrogen bonding, and sometimes to ctcei:-"

linking through disulfide bridges and kinks and turns caused by the presence rffproline. Some of the structural features of interest in proteins are the s-helix and

Copyright O by The Berkeley Review 19() The Berkeley Revier


G-pleated sheets. In the o-helix, two residues near one another in the protein areheld together by-hydrogen bonds. There are 3.6 amino acids per turn and eachturn is 5.4 A in length. They coil like a phone cord. In B-pieated sheets, theamino acids strands are not coiled. They ire held together by hydrogen bondsfrom the_hydrogen on nitrogen to the carbonyl oxyge"n. Figure 7-31 shows a B-pleated sheet with antiparallel strands

R7

-t_t*r,/

ilo,,,.

H

(^lt -N.

5_q

RzoH

I

N

4 roH lll-rvA1Is\ll"\ Rzs H 9,

A-.42-- I llRgz ,H P

Rzz

\

il9'.

H

I

N.

H

rfo

o

T+i/r&Ht;J

38

YililoH oHR+rH o

H R,"HIts.fiFigure 7-31

{ disulfide bridge is formed when two cysteine resid.ues lose the H from the:hiol group to form a sulfur-sulfur bond. Disulfide bridging is most often:onsidered to be part of the tertiary structure. Because loss of hydrogen resultsrn an increase in the oxidation state of sulfur, this process is oxidatiae, so a cross-jnked protein is considered more oxidized than the protein without cross-

:^khg. To break a cystine cross-linkage, one can add. a reducing agent such as:imercaptoethanol, HSCH2CH2oH. As a side note, cross-litrtitr[ iJresponsible:or an increase in rigidity for the polymer (polypeptide in this case) due to the-oss of both flexibility and entropy. In terms of thermodynamics, the process of=oss-linking is driven by enthalpy and not entropy (recall that AG = lu - us).-igure 7-32 shows the reductive cleavage of a disulfide linkage using theremical reagent B-mercaptoethanol.

H

I

N

N

I

Ho

N

I

H

-opyright O by The Berkeley Review l9l Exclusive MCAT Preparation


, Vasopressin (antidiuretic hormone)

|

-i

Cys-Tyr-Phe- GIn-Asn- Cys-pro-Arg- Gly - NH-

SH

I

Cys

-Tyr -Phe-

I

lr,,rA-tti

s -oHIGln-Asn- CYt

-Pro-Arg - Gly - NH-

l"oA-ttt

SH SHttCys-Tyr Gln-Asn- Cys-pro-A.g -

Gly - NH-+

Horyt-rA.-otFigure 7-32

when any of the molecular interactions that hold together the secondary o:tertiary structure of a protein are broken, the protein is said to be denqtured 1ntlonger in its natural form). The overall folding can be changed by theenvironment in which the protein exists. ln an aqueous environment, a protei::will fold in such a way as to expose the hydrophilic side groups to the iolven:(water) while minimizing the exposure of the hydrophobic (alkyl and phenr.side groups to the solvent (water). in a hydrophobic envirotr-eni, a protein n-i;fold in such a way as to expose the hydrophobic side groups to the solvent (lipicwhile minimizing the exposure of the hydrophilic (hydrogen bonding) itdegroups to the solvent (tipid). This overall folding of the protein is referred to a-.the tertiary structure. The tertiary structure is globular for many proteins.

Example 7.31Taking a protein from a lipid environment and placing it into an aqueousenvironment would most affect which of the following?A. Primary structureB. Secondary structureC. Tertiary structureD. Quaternarystructure

SolutionA protein in a lipid environment has its hydrophobic side chains exposed to thesolvent and its hydrophilic side chains in the inner core. A protein in an aqueonsenvironment has its hydrophilic side chains exposed to the solvent and ishydrophobic side chains in the inner core. The conversion from lipi;environment to aqueous environment breaks apart the internal hydroge:bonding, which denatures the protein. This affects the secondary structure trchanging the structural features. The tertiary and quaternary structures are alsoaffected, but only as a consequence of changes in the secondary structure. Thisquestion calls for the most specific change. The best answer is choice B.

:-nW

rltcnffim&,mgffifrffi

G

m'lburulMflumrfls&d.ft,mfr,$n

&,li@M

Copyright @ by The Berkeley Review 192 The Berkeley Revieu


Example 7.32Turns are caused by which of the following amino acids?

A. CysteineB. HistidineC. ProlineD. Glycine

SolutionBecause proline is cyclic, it cannot freely rotate about its sigma bonds, forcing theprotein chain to turn. Choice C is the best answer. The following pictorialrepresentation shows the impact of the cyclic structure of proline on the turn.*y'

N \*:proline residue .A oturn--d,

NH

"YEvprotein turn

oo

,*.,rJL *^If *tr-

H:'=R

up

oBent peptide containing proline Blown view of protein tum

Example 7.33\Alhich of the following results in a denatured protein?

A. Cleaving disulfide bondsB. Spiitting base pairsC. Breaking alpha bondsD. Forming of beta bonds

SolutionBreaking disulfide bonds requires the reductive cleavage of the S-S bond. Thisdefinitely changes the tertiary structure, thus denaturing the protein. Basepairing is involved in DNA and RNA, not proteins, so choice B is eliminated.Alpha bonds and beta bonds are throw away answers, so choices C and D areeliminated. Choice A is the best answer.

Unique Biological ProcessA common rule in the chemistry of amino acids is that they exist exclusively as L-stereoisomers in biological systems. An exception to this "all amino acids are L"rule is found in the bacterial cell wall. The cell wall of a bacteria is held togetherby a net-like structure that involves a cross-link between D-alanine and glycine.As a point of biological interest, B-Iactams such as penicillin act in a medicinalfashion by breaking this linkage and destroying the cell walls of bacteria. Again,these are not facts to be processed and stored, but they are interesting biologicalanecdotes that have been presented on previous MCATs. Discussing them hereis simply to provide a little background, so if they show up again, there will be asmall air of familiarity to them. Figure 7-33 shows the formation of the cross-linkin the bacterial cell wall.

Copyright O by The Berkeley Review t93 Exclusive MCAT Preparation


O HOHHil til ltR-C-CHz-NH, + R'-N-C-C-N-C-COr- +ll t'

Terminal Residue of the H CHe CHspentaglycinebridge TerminalD-Ala-D_Alaunit

O H HH Hil I I I *lR-C- CH"-N-C-C-N-R' + H.NIC-CO?-' ll | --r^' | 'o CHs CHs

Gly-D-Ala crosslink D-Ala

Figure 7-33

Gel ElectrophoresisL9t us consider gel electrophoresis, a biochemistry lab technique based on theidea that charged particles migrate through eleciric fields. Particles migrateaccording to their charge (which impacts the strength of the electrical forcelandtheir size-(which impacts their resistance to flow ur th"y travel through a gel). Apolyacrylamide gel is chosen to offer resistance ro ihut the particles do notmigrate too rapidly and thus not have the time to separate to a distinguishableamount. Figure 7-34 shows the schematic of an electrophoresis apparatus.

+--_@

o-

li)

!c

o

I(tu

ffi

&

EWm-

Jl"

IL

ffihrqonn

m,imery

t{c.

Figure 7-34

In gel electrophoresis, cations migrate to the cathode and anions migrate to theanode. This allows for separation of cations from anions. To get separation c,flike charges (cations from cations for instance), the particles

"must migrate at

different speeds. The speed depends on the acceleration, which dependJ on tlwelectric force (F = qE) and the resistive force due to drag (Fa.ug). iquation i-Jshows the relationship of the particle's mass and acceleration to ?ts charge, q, tllgstrength of the electrii field, E, and the resistive force due to the drag urlt rrlorothrough the gel.

ma=gE-Fdrag (7Jl

The charge of an amino acid, polypeptide, or protein fragment depends on tlepH of the environment and the isoelectric point, pI, of the species. when pH kgreater than pI, the environment is basic, resulting in a negatively chirgedspecies. when pH is less than pI, the environment is acidic, resulting iir mpositively charged species. The bigger the difference between the pH or *,environment and pi of the species, the greater the magnitude of its charge. Authe magnitude of charge increases, the electric force inireases, and thereioreacceleration increases, This means that the species can be separated by their Ivalues. The other factor that influences the migration rate is the size of tispecies. Larger species experience greater resistance as they travel through 1

medium. As a result, there are two types of electrophoresis tlchniques emp-.-lorto separate protein fragments.

Copyright O by The Berkeley Review 194 The Berkeley


The first type of electrophoresis employs a pH gradient in the gel and thecharged species migrate through the gel until they ieach a pH equalto their pI.At that point they have no net charge, so there is no electric force. This techniqleis referred to as isoelectric focusing. The second type of electrophoresis involvesadding sodium dodecyl sulfate, sDS, to the piotein mixture. The proteinfragments incorporate sDS into their secondary structure, resulting in everyfragment taking on a negative charge. Larger protein fragments can incorporatemore sDS than smaller ones, resulting in a greater magnitude of cirarge.However, larger species are more massive. As a result, the incorporation of sbscreates an environment where all of the protein fragments have the same mass-to-charge ratio. Given that they are experiencing the same electric field, anyseparation is due to differences in size, where smail species migrate faster thanlarger species, because they experience less resistance due to drag.

B' It is the heaviest fragment in the mixture and is most likely to containc. It is the lightest fragment in the mixture and is most likely to contain

Example 7.34What is true of the protein fragment in a mixture that migrates fastest in SDSPAGE and migrates to the highest pH value in a gel with a pH gradient?A. It is the heaviest fragment in the mixture and is most likely to contain

aspartic acid.

lysine.aspartic

acid.D. It is the lightest fragment in the mixture and is most likely to contain lysine

SolutionIn SDS PAGE, the speed at which a species migrates depends on the resistance itexperiences moving through the gel. Larger fragments experience greaterresistance, so a species that migrates rapidly through the gel must be small. Thismeans that the fastest fragment must be the lightest fragment, so choices A and Bare eliminated. Because the species migrated to a high pH value, it must have alarge pI value. The presence of lysine, arginine, and histidine increases the pI, sothe species must have at least one of those three amino acids. Aspartic acidIowers the pI value, so choice c is eliminated and choice D is the best answer.

.{ff inity ChromatographyAnother technique that can separate proteins according to their charge is affinitychromatography. we will look specifically at ion-exchange chromatography.Ion-exchange chromatography separates by the affinity of charged protein- foroppositely charged sites on the polymer in the column (stationary phase). Thestationary phase is made of an insoluble polymer, often polystyrene or cellulose,to which functional groups capable of carrying a charge have been attached.Typical anionic groups include sulfate on sulfonated polystyrene and carboxylateon carboxymethyl-cellulose. Typical cationic groups include diethylaminoethoxyon diethyiaminoethoxy-cellulose. Positively charged columns bind anionicspecies tightly and negatively charged columns bind cationic species tightly. Aprotein carries a positive charge when the solution pH is less than the isoelectricpoint, pI, of the protein. v\4ren a mixture of proteins is added to the column andailowed to migrate down the polymer, selected proteins can be separated fromthe mixture by binding the column. Proteins carrying the same iharge as thecolumn elute. To release a bound protein from an ion-exchange column, thepolymer can be washed with a solution of varying pH or a solution of graduallyrncreasing salt concentration. Both techniques for releasing the protein coulddenature it irreversibly, so the preferred technique is not always obvious.



Cutting and Sequencing (Primary Structure Determination)Determining the primary sequence of a protein is critical in biochemistry. In thesequencing of proteins, peptide bonds must be broken and component aminoacids identified. To sequence a protein, several reagents are involved.Sometimes it is best to break a large protein into smaller, more manageablefragments. Each fragment can be sequenced and then the fragments &n bereassembled to determine the overall sequence of the original protein. Thismeans thai the original protein must be mirked in such u *iy that the first andlast fragments are easy to identify. we will start by determining which aminoacid is first in the protein. The identification of the first amino acid in thesequence can be accomplished by adding 2,4-dinitrofluorobenzene, 2,4-DNFB

llanger's reagent), to the protein, followed by comprete hydrolysis using 6 MHCl. sanger's reagent binds the amino terminal of a protein, thereby labeling thefirst amino acid. Complete acid hydrolysis of the piotein then yields ail o] thecomponent amino acids, of which only one is marked. Figure 7-35 shows bothfree 2,4-DNFB and a complexed amino acid residue

HNq

"b,f -);x:KITO0 Nq Nq

2,4-dinitrofluorobenzene Labeled amino acid residue

Figure 7-35

Sanger's reagent serves to mark the first amino acid in a protein and to ultimatelr-identify it, but that's just the beginning when req.tetrcitrg a large protein wiiirseveral amino acids. To break the protein into smaller fragments,-biochemistsmimic what nature does. we employ enzymes that break a large protein intosmaller fragments by cleaving proteins at specific sites (bondsy. fne enzymechymotrypsin, which is active in the small intestine, cleaves an amino acid cirainat the peptide linkage on the carboxyl side of phenylalanine, tyrosine, ortryptophan, As a point of biological interest, the pancreas actuaily releaseschymotrypsinogen, an inactive species that gets activated once in the smallintestine. chymotrypsin would cleave the twelve amino acid protein, Arg-Cys-Gly-Ala-Phe-Thr-Met-Ala-Tyr-Cys-His-Leu, twice, Ieaving three smalierfragments. The three pieces are: Arg-Cys-Gly-Ala-phe, Thr-Met-Ala-Tyr, andCys-His-Leu. Reassembling the three fragments can be done if the first frigmen:is known from the 2,4-DNFB result. Table 7-r shows various enzymei andchemical reagents used to break peptide iinkages, their specific activity, and forthe digestive enzymes, their corresponding zymogen and sites of release aniactivity.

Copyright @ by The Berkeley Review 196 The Berkeley Reviex 1{L


Enzyme or Agent Cleavage activity Site (if applicable)Chymotrypsin

(digestive enzyme)C-side of Phe, Trp, and Tyr Released as chymotrypsinogen

from pancreas into small intestineClostripain(enzyme)

C-side of Arg

Cyanogen bromide(chemical reagent)

C-side of Met

O-Iodosobenzoate(chemical reagent)

C-side of Trp

Hydroxylamine(chemical reagent)

Asp-Gly bonds

2-Nitro-S-cyanobenzoate(chemical reagent)

N-side of Cys

Pepsin(digestive enzyme)

C-side of Asp, Clu, and Leu Released as pepsinogen from chiefcells into stomach

Staphylococcal protease(digestive enzyme)

C-side of Asp and Glu (Glu onlyunder specific conditions)

Thermolysin(enzyme)

N-side of Leu, Val, and Ile

Trypsin(digestive enzyme)

C-side of Lys and Arg Released as trypsinogen frompancreas into small intestine

TableT-l

sequencing is often done in multiple steps. some methods involve randomhydrolysis followed by sequencing of the fragments and analysis of the overlapof information between fragments. There is also a systematic approach whereihe protein is broken down at specific linkages using different reagents and theoverlap between fragments is again analyzed. Consider the following example.

A twenty-seven amino acid protein is treated with various reagents in fourseparate experiments. The results of each experiment are listed below.1. Treatment of the protein with 2,4-dtnitrofluorobenzene and then 6M HCI

yields a histidine residue bonded to 2,4-dtnitrobenzene.2. Treatment of the protein with chymotrypsin yields the following

fragments:

a) N-His-Val-Ser-Gly-Ala-Ile-Phe-Cb) N-Leu-His-Gly-Thr-Tyr-Cc) N-Val-Asp-Arg-Cys-AIa-Leu-Cd) N-Val-His-Lys-Glu-Trp-Ce) N-Cys-Ile-Met-Phe-C

3. Treatment of the protein with trypsin yields 3 fragments that are 19amino acids, 5 amino acids, and 3 amino acids in length.

4. Treatment of the protein with thermolysin yields a free histidine residueand a free leucine residue along with the following six fragments:a) N-Ile-Phe-Cb) N-Leu-His-Gly-Thr-Tyr-Cys-Cc) N-Ile-Met-Phe-Cd) N-Va1-His-Lys-Glu-Trp-Ce) N-Val-Asp-Arg-Cys-Ala-C0 N-Val-Ser-Gly-Ala-C



Frorn the first experiment, the conclusion you should draw is that histidine is thefirst amino acid in the protein, because of the reaction with sanger's reagent.From the second experiment, we can conclude that leucine is the last amino acidin the protein, because leucine is the only carboxyl terminal amino acid in one ofthe five fragment that would not have been fragmented by chymotrypsin.Combining the conclusion of the first experiment with fragments in the secondexperiment, we know that amino acids 1-7 are His, val, ser, Gly, Ala,Ile, and phesequentially. We also know that amino acids 22-27 are Val, Asp, Arg, Cys, Ala,and Leu sequentially. The free leucine from Experiment 4 further supports thatleucine is the last amino acid in the sequence. From this point, we will label eachfragment from Experiment 2 A through E and use information from the otherexperiments to evaluate the overlap of information and determine the position ofeach amino acid. The five fragments from Experiment 2 are arbitrarily assignedthe letters A through E as follows:

Fragment A: N-His-Val-Ser-Gly-A1a-Ile-phe-CFragment B: N-Leu-His-Gly-Thr-Tyr-CFragment C: N-Va1-Asp-Arg-Cys-Ala-Leu-CFragment D: N-Va1-His-Lys-Glu-Trp-CFragment E: N-Cys-Ile-Met-phe-C

Because we know the first and last amino acids, we know that fragment A is first,and fragment C is last. This leaves six possible combinations for the fragmentsproduced in Experiment 2. The possibilities are: A-B-D-E-C, A-B-E-D-C, A-D-B-E-C, A-D-E-B-C, A-E-B-D-C, and A-E-D-B-C. From Experiment 3, we know thatthe 3-AA fragment must be Cys-Ala-Leu, the last three amino acids. Trypsincleaves the C-side of Lys and Arg, so the Lys-Glu bond is broken. This leads usto conclude that fragment D must be the fourth fragment, because the five aminoacid fragment from Experiment 3 must end with arginine. This narrows thechoices to two: A-B-E-D-C and A-E-B-D-c. From Experiment 4, it canbe inferredthat Phe from fragment A must be bonded at its carboxyl site to either Leu, val,or Ile. This means that fragment E cannot be second in the overall sequence,which narrows the answer down to A-B-E-D-C. This makes the original protein:N-His-Val-Ser-Gly-Ala-Ile-Phe-Leu-His-Gly-Thr-Tyr-Cys-Ile-Met-phe-val-His-Lys-Glu-Trp-va1-Asp-Arg-Cys-Ala-Leu. Be sure you work through the logic ofsequencing.

!T

n

$

Copyright @ by The Berkeley Review l98 The Berkeley Review


Edman's ReagentTo sequence the amino acids in a small polypeptide (twenty to fifty amino acidsmaximum, depending on the amino acids), biochemists employ Edman'sreagent, phenylisothiocyanate (H5C6N=C=s). This reagent r"q,t"t"rtiuily removeseach amino acid one at a time starting from the amino terminal. This leads to theamino acid sequencing equipment used in most biology labs. Edman's reagentand its reactivity is shown in Figure 7-36.

Phenylisothiocyanate

I\-L-b

S

il

1tH

ii-(*" \*HzN

orr,"^ 1l Trransrertt_ry\NH-peptide

H

Figure 7-36

Example 7.35Hydrolysis of a peptide results in the cleavage of:

A. the carbonyl group.B. acarbon-carbonbond.C. the carboxyi terminal.D" the amide bonds.

SolutionPolypeptides are held together by peptide linkages, which are amide bondsbetween amino acids. To break apart a polypeptide, the amide linkage must besevered. This makes choice D the best answer.

J

Peptide + G"

NH-peptide

NH-peptide

/c- wH



Example 7.362,A-Dinitrofluorobenzene reacts with the:

A. nitrogen of the amino terminal.B. carbon of the carboxyl terminal.C. nitrogen of the amide bonds.D. carbon of the side chains.

Solution2,4-Dinitrofluorobenzene, known as sanger's reagent, is an electrophile thatreacts with the nitrogen of the amino terminal of a protein by iay of anucleophilic aromatic substitution reaction. The nitroget of th" amino terminalattacks the benzene ring nucleophilically to displace th"e fluorine on the aromaticring in a two-step reaction. The best answer is choice A.

Example7.37The tripeptide Lys-Arg-Ser would NOT:A. have an overall positive charge atpH = 7.B. be water-soluble.C. from any disulfide bonds.D. have an isoelectric point greater than Asp-Glu_Thr.

SolutionThe tripeptide has an amino terminal on lysine (pKa = 9.2), acarboxyl terminalon serine (pKa = 2.2), an ammonium side chain ori tyrin" (pKa = 10.g), and aguanidinium side chain on arginine (pKa =13.2). At pil = 7,'ih. amino terminalis protonated and thus positively chaiged, the carboxyl terminal is deprotonateda.d thus negatively charged, and thelwo basic side chains are protonated andthus each are positively charged. The overall charge is therefore positive two_This means that "The tripeptide Lys-Arg-ser would, not have an overall positivecharge at pH = 7" is an invalid statemenl, so choice A is eliminated. Because all:{1h." sites on the tripeptide are charged, the compound is ionic and thereforehighly water-soluble. This means that "The tripeptide Lys-Arg-ser would not],r-water-soluble" is an invalid statement, so choice B is eliminatel. Because neithenlysine, arginine, nor serine contain thior groups (only cysteine does) there can beno disulfide cross-linking associated with this tripeptiie. This means that "Thetripeptide Lys-Arg-ser would not have any disuiride bonds formed" is a valirrstatement, so choice C is the best answer. The isoelectric point for the tripeptidcLys-Arg-ser is an average of pKu3 and pKu4 (the two sidl chain pKu vaiues fclysine and arginine), while theisoelectrii point for the tripeptide Asp-GIu-Thrban-average of pKul and pKu2 (the carboxyl terminal pru value and the side chainpKu value for aspartic acid). The average of pKul u"a pru2 for Asp-Glu-Thr iselower number than the average of pKul anapiua ro. LyJ--a.g-sei. This ,,that "The tripeptide Lys-Arg-ser would nothive inisoeiectric"point greaterAsp-Glu-Thr" is an invalid statement, so choice D is eliminaied. Basicacids (lysine, arginine,, and hjstidine) always increase the isoelectric pointacidic amino acids (aspartic acid and glutamic acid) always decreaseisoelectric point. This question is difficult,lecause the language is crrmbe.stMaking use of denotations next to each answer choice tik-e "ialse, because itpositive," may prove highly beneficial. you need to not only recall the subir11!ter when working with these books, but you must also hone your test-taskills.

Copyright O by The Berkeley Review 200 The Berkeley

Organic Chemistry Nitrogen Compounds Section Summary

Key Points for Nitrogen Compounds (Section 7)

Non-biological Nitrogen-Containing Compounds

1. Amines (Contains a central nitrogen with either alkyl groups or hydrogensattached)

a) Basicity (Amines are weak bases)

i. The pK6 for amines is between 3 and 5, the pKu of protonatedamines is between 9 and 11

ii. Typically, 2' > 3'= 1", because of steric hindrance and H-bondingiii. Common reagent in buffers with pH of 10 t 1

b) Nucleophilicityi. Good nucleophiles for 51112 reactionsii. Can undergo multiple additions when treated with alkyl halides

c) Amines can be synthesized in many ways

i. Gabriel synthesis using alkylation of a phthalimide followed byhydrolysis

ii. Reduction of azides, nitriles, imines, and amidesiii. Hofmann rearrangement of an amide

d) Amine reactivity is based on nucleophilic substitutioni. Amines readily displace a leaving group from alkyl halidesii. Amines react with nitrite to form diazonium salts

iii. Quaternary amines can undergo Hofmann elimination to yieldterminal alkenes where three of the alkyl groups are methyl groups

Imines have nitrogen double bonded to a central carbon (R2C=NR')

a) They react in a similar fashion to aldehydes and ketones

i. They are in equilibrium with enaminesii. The carbon is electrophilic and can be attacked by nucleophilesiii. Oximes and hydrazones form in a similar way as imines

Amides have nitrogen bonded to a carbonyl carbon (RC=ONR')

a) Amides are neither acidic nor basic

i. Amides are formed from substitution reactions where an amineattacks an ester, anhydride, or acid halide

ii. Amides are also formed from an oxime via Beckmann rearrangementiii. Amide bonds are referred to as peptide bonds when formed between

amino acids

Amino acids can be synthesized by several methods in aitro

a) The alpha carbon has four things attached, an H, CO2H, NH2, and R-group, so synthesis routes center around adding these groups to carbon

i. Hell-Volhard-Zelinskii synthesis activates the alpha-carbon of a

carboxylic acid and then adds an amineii. Strecker synthesis adds a cyano group to an imine and then

undergoes hydrolysisiii. An alpha keto acid can be converted into an alpha imino acid, which

is reduced into an amino acidiv. Gabriel synthesis, which is used to synthesize a primary amine can

make an amino acid, where the phthalimide is added to an alkylhalide with a carboxylic acid group rather than an alkyl halide

i

1l

I

!

2.

.)-

4.


Organic Chemistry Nitrogen Compounds Section Summary

Biological Nitrogen-Containing Compounds1. Amino acids have an amine group on the alpha carbon of a carboxylic acid

a) They are classified according to their R-groupi. Hydrophobic amino acids have alkyl side chainsii. semi-hydrophobic amino acids have functional groups that are only

slightly water solubleiii. Hydrophilic amino acids have highly water soluble side chains,

usually alcohols and amines

b) Side chains can be acidic, basic, or neutrali. Acidic amino acids contain oxygen, lose a proton from their neutral

state, and have isoelectric points less than 5.5ii. Basic amino acids contain nitrogen, gain a proton from their neutral

state, and have isoelectric points greater than 2.5iii. Neutral amino acids do not gain or lose a proton from their neutral

state and they have isoelectric points between 5.5 and 7.5

c) Isoelectric points, the pH where the charge is 0, are found by averagingthe two pKus leading to and from the zwiiterioni. pI is an average of pKul and pKu2 for all amino acids except Lys,

Arg, His, because "his lys are basic.', For Lys, Arg, and His, pI is anaverage of pKu2 and pKu3.

ii. The pI of a protein is found easily by adding the number of Lys, Arg,and His plus one for the N-terminal, which gives the p*u t".i-,leading to the zwitterion

iii. when the pH of the environment exceeds pI, the species is an anioniv. \A/hen pI exceeds the pH of the environment, the species is an cation

2. Proteins are polymers of amino acidsa) Primary structure is the amino acid sequence and peptide linkagesb) secondary structure involves the interactions of nearby amino acids

i. An alpha-helix is a coil in the backbone with 3.6 amino acids per tumii. Beta-pleated sheets involve H-bonding of amides of nearby amino

acid moieties

c) Tertiary strucfure involves side chain interactions of amino acids that arefar apart from one anotheri. Beta-pleated sheets involve hydrogen bonding of amides of nearbv

amino acid moieties

d) Quaternary structure involves interactions between two differentprimary strands

3. Protein lab techniques center around separation and sequencinga) Gel Electrophoresis

i. Anions migrate to anode and cations migrate to cathod.e in an E fieldii. Buffered gel gets separation by pI: migrates until pl = pH of geliii. SDS-PAGE separates by size: bigger fragments migrate slower

b) sequencing involves analyztng the primary structure of a proteini. First treated with urea to break H-bonds and G-mercaptoethanolii. Edman's reagent, Ph-N=C=s, cuts one AA at a time from N-terminaliii. Enzymes can be used to break large proteins into smaller, more

manageable fragments which can be sequenced and reassembled

Copyright @ by The Berkeley Review 202 The Berkeley Revierr

i

I. Basicity of Amines

II. Amine Basicity and Nucleophilicity

III. Effects of Alkyl Qroups on Amine Nucleophilicity

IV. Imine and Enamine Chemistry

V. Amino Acid Synthesis in vitro

VI. Amino Acid Trivia

VII. Acidity and Basicity of Amino Acids

VIII. Amino Acids and polypeptides

IX. Physical Properties of Amino Acids

X. Protein tlormones

XI. Enzymatic Cleavage and Sequencing

XII. Peptide Sequencing

XIII. Amino Acid Stereospecificity in Synthesis

Questions not Based on a Descriptive passage

Amines and Amino Acids Scoring Scale

Kaw Score MCAT Score

84 - 100 t3- 15

66-85 IO-1247 65 7 -934-46 4-6l-33 L-3

(L -7)

(8 - 14)

(ls - 20)

(2r - 27)

(28 - 35)

(36 - 43)

(44 - 5t)

(52 - 58)

(5e - 66)

(67 - 75)

(74 - Bt)

(82 - 88)

(Be - e6)

(e7 - 1OO)

ffirm

rm

mfr


The trend for the relative basicity of amines in water is:secondary (2") > primary (1") = tertiary (3') > arnmonia. Theorder is explained by competing influences in water. Theelectron donating effect of the alkyl groups supports arelationship of 3' >2" > 1' > ammonia. This prediction issupported by the gas phase basicity of the amines and amineacid/base reactions conducted in an aprotic solvent. Reducedhydrogen bonding caused by the steric hindrance of alkylgroups supports a relationship of ammonia > l' > Z" > 3..Differences in hydrogen bonding capacity are demonstrated bythe increasing sharpness of the N-H absorbance seen ininfrared spectroscopy. The sharpest peak is observed with thesecondary amine and gets progressively more broad withprimary amines and ammonia. The two opposing trendscompete to give the observed trend in water. Table 1 showsthe pKu values for some alkyl ammonium chloride salts.

Compound PKaNHaCI 9.26(H3C)3NHCl 9.79

H3CNH3CI 10.65

H3CH2CNH3CI 10.71

(H3C)2NH2Cl r0.73

(H3CH2C)3NHC| 10.75

(H3CH2C)2NH2C1 10.95

Table 1

The pK6 for the conjugate base of any given acid can betbund by using the equation pKb = 14 - pKu. This can berseful for determining whether a compound is charged oruncharged at a given pH value in an aqueous solution.Figure 1 shows the structure of a diamine at varied pH.

H/

H\

HH

s, -N\,- NH'

pH = 10.8

HI

E.-NVNH,pH = 12.8

pH values

./^ oE, -N\'t NH'

pH = 8.8

Figure 1 Diamine at varying

1 . In the gas phase, which of the following ammoniumcations is the MOST acidic?

A. NH4+

B. H3CH2CNH3+

C. (H3CH2C)2NH2+

D. (H3CH2C)3NH+

2 . What is the approximare pK6 of CF3CH2NH2?

A. pKu > 10.71

B. 10.71 > pK6 2 7C.1>pK6)3.29D. 3.29 > pKu

Copyright O by The Berkeley Review@

3. What can be determined about the equilibrium constant(K) for the following acidlbase reaction?

H3CNHj+ + NH3

A. K > 1 in both water and gas.

B. K> I in water; K< I in gas.

C. K< 1 in water; K> I in gas.

D. K < 1 in both water and gas.

Which of the following amines is the MOST basic?

A. NH3

B. H3CNH2

C. (H3C)2NH

D. (H3C)3N

Charged compounds are found to be more soluble inwater than neutral compounds. At which of thefollowing pH values would a tertiary amine be MOSTsoluble in water?

A. 3

B. 5

c.9D. 11

The pKa for carboxylic acids is between 2 and 5. In apH = 7 buffered aqueous solution, lysine exists inwhich of the following forms?

A'aB.oHzN HzN

4.

5.

6.

OH

H (CH.),I

NHz"

(:"'lo*NH:

H3N1

H (CH"),I

*NHr

the passage of theof an N-H bond of a

D.oC.

"'*.#oH (CHr)a

l-'NHz

o-

7- Why is there no mention inbroadness of the IR absorbancetertiary amine?

A. Steric hindrance hinders N-H bond stretching.B. There is no N-H bond on a tertiary amine.C. Resonance shifts the absorbance to 3000 cm-I.D. Symmetric molecules exhibit no IR absorbances.

205 GO ON TO THE NEXT PAGE.

Passage ll (Questions B - 14)

Amines are classified as weak bases, compounds thatpartially hydrolyze water to give OH- and its conjugate acid.Amines are one of the most common types of organic bases.The relative basicity of various amines can be predicted bycomparing the substituents bonded to the nitrogen of theamine. Electron donating groups, when bonded to nitrogen,increase the basicity of the amine. This is to say that thebasicity is directly dependent on the groups attached to theamine. Amines obey standard rules for aqueous phase acid-base chemistry. Reaction 1 is a proton transfer reactioninvolving a primary ammonium cation and ammonia. Theequilibrium for Reaction 1 as written lies to the left.

H3CNH3+ + NH3 *H3CNH2 + NH4+

Reaction I

The equilibrium constant for Reaction I is less than 1,because the methyl amine (on the product side) is a strongerbase than ammonia (on the reactant side). The equilibriumthus favors the protonation of methyl amine. By knowingthe direction that an equilibrium lies, it is possible todetermine the relative basicity of two compounds bothqualitatively and quantitatively. The pKu value for an acid (orthe pK6 value for a base) can be derived from the equilibriumconstant (K.O) for the reaction and the pKu of the other acid.Table 1 shows the equilibrium constants for a series of acid-base reactions with ammonium, NH4*. All reactions werecarried out at 35'C.

Reactant Base K"O for reaction with NH4+

NH: 1.00

H3CNH2 25.7

H3CH2CNH2 29.s(H3C)2NH 30.9

(H3CH2C)2NH 50.1

(H3C)3N 3.55

(H3CH2C)3N 32.4

Table I

Which of the following compounds is the MOST basic?

A. H3CNH2

B. H3CCH2NH2

C. (H3C)2NH

D. (H3CCHz)zNH

Which of the following reactions has a K"O of 0.030?

A. (I{3C)3NH+ + NFI3 . ^ GIIC):N + NH4*

B. (I{3C)3N + NFI4* T- (HjC)3NH+ + N}I3

C. GI jC)2NH2+ + NHj -i- (I{3C)2NH + NFI4*

D. (I{3C)2NH+N}Ia+ . ^ (H3C)2NH2++N}Ij

8.

9.

Copyright O by The Berkeley Review@ 206 GO ON TO THE NEXT PA

10. To separate an amine from an amide, it would be best touse which of the following for extraction?

A . Ether and water at pH = 3B . Ether and water at pH = 7C . Ether and water at pH = 11

D. Amides cannot be separated from amines.

1 1. Which of the following statements is true about rherelative basicity of amines compared to amides?

A. Amines are more basic than amides, because amide:have their lone pair tied up in resonance.

B. Amines are less basic than amides, because amideshave their lone pair tied up in resonance.

C . Amines are more basic than amides, because amidesare more sterically hindered.

D. Amines are less basic than amides, because amidesare more stericallv hindered.

12. The highest boiling point is associated with which c,fthe following amine compounds?

A. NH3

B. H3CNH2

C. H3CCH2CH2NH2

D. (H3C)3N

13. The pKa of trimethyl ammonium chloride-(H3C)3NHCI, is 9.8. Benzene is electron withdrau'insby resonance when the substituent has a lone pair crfelectrons, so the pK6 of aniline, H5C6NH2, CANNOTbe which of the following?

A. 3.6

B. 4.8

c. 9.2D. 10.4

14. Which of the following changes to a primary arnircwill NOT result in a compound that yields a lower f,*qwhen undergoing proton transfer with ammonium?

A. Replace a hydrogen on the carbon backbone uiffifluorine.

B. Replace the hydrogens on the carbon bondednitrogen with a double bond to oxygen.

C . Replace one of the hydrogens on nitrogen wir.second alkyl group.

D . Replace the alkyl group on nitrogen wi*hydrogen.

io!

flh

mm

m

,md

rffi


The nucleophilicity of an amine closely parallels itsbasicity, where as the base strength of an amine increases, itsnucleophilicity also increases. This holds true as a generaltrend with most deviations attributed to steric factors. TableI shows the pK5 and log K.O for an Syg2-reaction with ethylchloride, H3CCH2CI, for a series of amines.

Amine pKn log K. o

NH: 4.7 1.8

H3CNH2 3.4 2.6

(H3C)2NH 3.2 3.1

(H3C)3N 4.2 0.6

Table 1

One problem with amines as nucleophiles is that theyare capable of undergoing multiple additions. It is difficult toisolate primary amines. To do so, an excess of amine is usedin the reaction. When there is excess electrophile, thereaction proceeds readily to secondary and tertiary amines. Tosynthesize a primary amine, methods other than the reactionof ammonia with an alkyl halide are employed. Figure 1

shows alternative routes for synthesizing a primary amine.

Method 1: Gabriel Phthalimide Synthesis

r. o

NHr#RNH23. RBr

Method 2:

4. NaOF(aq)/A

Reduction of Nitrogen Compounds

Hr(e)/cat.RC=N

-

RCH2NH2

H2(g)/Ni

C2H5OH

LiAlH4

-.-.-+ether

HoNNHT

Pd/cH3oH

Figure 1 Synthetic routes to form a primary amine

1 5. Reduction of all of the following types of compoundswill yield a primary amine, EXCEPT;

A. a nitrile.B. a primary imine.C. asecondaryamide.D. an oxime

RCH=NH

olt

RC- NH2

RCH2N3

RCH2NH2

RCH2NH2

RCH2NH2

Copyright @ by The Berkeley Review@ 207 GO ON TO THE NEXT PAGE.

I 6. The IR absorbance for anapproximately:

A. 3200 cm-1.

B. 2700 cm-|.C. 1600 cm-l.D. 1300 cm-1.

N-H bond is observed at

17. Which of the following statements BEST explains whymethyl amine is a stronger nucleophile than ammonia?

A. The methyl group is electron withdrawing by theinductive effect, making nitrogen electron poor.

B. The methyl group is electron donating by theinductive effect, making nitrogen electron rich.

C . The methyl group is electron withdrawing byresonance, making nitrogen electron poor.

D. The methyl group is electron donating byresonance, making nitrogen electron rich.

18. Methyl amine, when added to (R)-2,chloroburane, wouldgive which of the following major organic products?

A. 1-butene

B. 2-butene

C. (R)-2-butyl methyl amineD. (S)-2-butyl methyl amine

19. Which of the following labeled nitrogen atoms is theMOST nucleophilic?

a

bNHz

HzN

A. Nitrogen a

B. Nitrogen b

C . Nitrogen c

D. Nitrogen d

2 0. Methyl amine can reactmolecules EXCEPT:

with all of the following

A. o

",.A.,C.

H3C- Cl

B. o

",.4o.",D.

nrc/o - at.

.N. $-H

Passage tV (euestions 21 - 27)

When ketones are exposed to amines (such as ammoniaor alkyl amines), they can be converted into an imine andwater. In the presence of a small amount of acid, conversionof roughly 99Vo of the ketone into amine occurs in about anhour for most amines. The reaction has an equilibriumconstant close to one, so the reaction is driven by either theaddition of excess reactant or the removal of a product. Thereaction proceeds with both primary and secondary amines (inaddition to ammonia), but the secondary amine cannot form aneutral imine. Secondary amines instead form compoundsknown as enamine.r, rather than form a cationic imine.Figure I shows the reactions of a ketone with ammonia. aprimary amine, and a secondary amine respectively.

Ho

oA*,+ NH3 =L

N/

A + HoO

R' R"

N-Ro

*A*, + RNH2 =5

o *-R-o

*,4* +R2NH + .A*,,*n'o

Figure 1 Ketone to imine conversions

The ketone can be regenerated by adding water to theimine, as shown by the reverse reaction of the equilibrium.The imine cation formed from the addition of a secondaryamine to a ketone can undergo the following syntheticallyuseful reaction (Figure 2).

*-*-*

I

oA.r*'*-R- *

*A.n*,I

R''

o

*A.n*I

R"

.A*', * t'o

*-R-*

.AcH2R,*-*-*

*Ao"*

oH-

R''-X--_l>

R-.@- RN11 H+raor

nAcH*, =+I

R"

Figure 2 Ketone synthesis via an enamine intermediate


"\) + H2NCH2Cn3 4.

A.HjCH2CN

C.HN

23. What is the product for the following reaction whe:carried out under acidic conditions?

","{o + Hzo

2 4 . What is the product of the fbllowing reaction?

21. Which of the following conversions CANNOTcanied out via the enamine synthesis route?A . 3-pentanone into 2-methyl_3_pentanoneB. Butanal into 2-ethylhexanalC. 3-hexanone into 3-heptanoneD. Acetone into 2-octanone


b'e

B.HjCH2CHN

B.

+"{ d>A . A ketone and a primary amine.B. A ketone and a secondary amine.C . An aldehyde an<l a primary amine.D . An aldehyde and a secondary amine.


25. To convert 3-pentanone into 4-methyl-3-heptanone,which of the following sequences ofreagents should beused?

A. 1. R3NH+ 2. H3CCH2I 3. H3O+(aq)

B. 1. R2NH/H+ 2.H3CCH2I 3. H3O+(aq)

C . l. R3NH+ 2. H3CCH2CH2I 3. H3O+(aq)

D. 1. R2NHIH+ 2. H3CCH2CH2I 3. H3O+(aq)

26. Treatment of an imine with a reducing agent is similarto treatment of a ketone with a reducing agent. What isthe product formed when the imine produced from thereaction of a ketone and primary amine is reduced usingLiAlHa in ether?

A. A primary amide

B. A primary amine

C. AsecondaryamineD. A tertiary amine

2 7. What are the products formed when N-ethyl-3-pentimineis treated with acidic water?

A. Ethanol and 3-pentanone.

B. Ethyl amine and 3-pentanone.

C. Ammonia, ethanol, and 3-pentanone.D. Diethyl ether, ammonia, and 3-pentanone.



There are many naturally occurring amino acids (formallyknown as 2-aminocarboxylic acids) that can be isolated frombiological systems. Amino acids vary only in the alkylgroup attached to carbon 2. Our body can synthesize someamino acids, but others must be consumed in our diet.Amino acids can be synthesized in vitro using one of threecommon methods. The first method involves thesubstitution of the amino group for a bromine functionalgroup on the carbon 2 of an alpha bromo carboxylic acid.This method is known as the Hell-Volhard-Zelinskiisynthesis, and it is shown in Figure 1 below.

ooll t. Br,/PBr, ll***, -E?- *lAon

Br

oo.+*4*Br

Figure 1 Hell-Volhard-Zelinskii

6- + NHaBr

NH:*

synthesis of an amino acid

The second method for synthesizing amino acids in vitroinvolves the addition of a 2-bromodiester to a dicarboxylicimide (Gabriel's phthalimide). The product of this reaction isthen heated in the presence of strong acid in water, whichresults in hydrolysis of both ester sites to form a diacid. Thedicarboxylic imide is completely hydrolyzed under theseconditions, to yield an alkyl ammonium cation. Afterenough time at an elevated temperature, the diacid undergoesdecarboxylation to form a monoacid, which in this case is an

amino acid. This method is known as the Gabriel synthesisand it is drawn in Figure 2 below.

o

ot*B'ovou,*os]"::

\n"HrN*{

Fo"o

\o"HgN*{

Fo"o

\.,"HTN+J

H*(aq)-*A

Figure 2 Gabriel synthesis of an amino acid

The last method involves the addition of ammonia to anaidehvde to form an imine. This is followed by the additionof a nitrile anion (C=N-) to the imine under acidic conditions.The product is then treated with strong aqueous acid at high[emperature to hydrolyze the cyano group into a carboxylica;ri functional group. This synthesis method is known asrh= Srrecker synthesis and is drawn in Figure 3 below.

o

*A'1. NH3

2. NaCN/HCN

3. ll3O+(aq) /A

:8.

Figure 3 Strecker synthesis of an amino acid

In infrared spectroscopy, the O-H absorbances for D_serine appear at which of the fbllowing values?

A . 3200 cm- I and 3400 cm-lB. 2700 cm-l and 3400 cm-lC. 1700 cm-l and 3400 cm-lD. 1700 cm-i and 2i00 cm-l

\\-hich of the following sequence of reagents would1, ield phenylalanine when added to 2-phenylethanal?

o

2-phenylethanal phenylalanine

\ . 1. CrO3/H+(aq) 2. NH3B. 1. NaCN 2. H+(aq)/AC. 1. NH3 2. NaCN 3. H+(aq)/AD. 1. 2-aminoerhanal 2. H+(aq)lA,

The following 1HNUR was run in CDCI3 on an aminoacid synthesizedby the Gabriel method. Which of thetollowing amino acids could it be?

19.

H NHc

Phenylalanine

7654321-{. Glycine (R = H)B, Alanine (R = CH3)C. Serine (R = CHzOH)D. Phenylalanine (R = CH2C6H5)

C-;', 1-q51 @ by The Berkeley Review@

What is the side product formed during the Gabrielsynthesis of valine?

A. O B.

e(-'o

m

altered.

B. It would be increased, because the proton leavesmore readily to make the species neutral.

C. It would be decreased, because the proton leavesmore readily to make the species negativellcharged.

D. It would be decreased, because the proton lear-e*more readily to make the species neutral.

Which of the following amino acids, starting from imfully protonated state, has three equivalence points wheatitrated by a strong base?

A. GlycineB. AlanineC. Leucine

D. Tyrosine

OH

OH

CN

CN

How many units of unsaturation are there in an aminoacid with an aliphatic side chain?

A . 0 units of unsaturalionB. 1 unit of unsaturationC . 3 units of unsaturationD . 4 units of unsaturation

How many signals do you expect in the 1HNURspectrum for L-cysteine, with a side chain of CH2SH-obtained in a solution with deuterochloroform solvenr?4.2B. 3

c. 4

D. 5

The pKu for the side chain of histidine in aqueousmedium is 6.1. What is true of its pKu in ahydrophobic environment?

A. It is the same, because log functions cannot be

31.

c

32.

33.

34.

35.

ft!ll-trLG.

mlmMh

fim

m]{n[

rmrror

TMfi

mffi

dilrtunfrr



In adult humans, there are nine essential amino acids.The term essential is applied to amino acids that humans:nust obtain through diet, because they are not synthesized in:he cytosol. The essential amino acids include the aromaticrmino acids phenylalanine, tryptophan, and histidine, and.ome hydrophobic amino acids like valine, leucine, and.soleucine. The other three essential amino acids arerethionine, threonine, and arginine. Of the essential amino:cids, only histidine, threonine and arginine are hydrophilic.The water-solubility of histidine increases at a low enough:H to protonate the side chain and form a cation.

In neutral water, the carboxylic acid functional group isrble to deprotonate and become anionic, while the amino:unctional group gets protonated and becomes cationic. Thermino and carboxyl terminals of all amino acids are charged.r an aqueous environment at or near a pH of 7, causing most.mino acids to exist as zwitterions in neutral aqueouslolution. A zwitterion is the form of the amino acid in:,hich the overall charge is zero, but the molecule has charged.ites on it. The exceptions to the zwitterion at pH = 7 rulerre the highly acidic and highly basic side chains.

All naturally occurring amino acids have the same,-hirality in vivo. The natural form for humans is the L-.orm, which is a result of an S chiral center at carbon number:"r'o of the amino acid, with the exception of cysteine.Carbon number two is often referred to as the alpha carbon.Frgure 1 shows the generic structure for a typical L-aminorcid where the R refers to the side chain of the amino acid:

o. ll .^ff:1",!'iffi1"",protonaled H:N\ -ZU\--_

','amino terminal 'O-

/zHR

L-Amino acid

Figure 1 Generic Amino Acid in Water

Six of the essential amino acids have a hydrophobic side:hain. A hydrophobic side chain reduces the water-solubilityrf the amino acid. The amino acid still exists as a zwitterionrn water, as long as the pH is within two units of the aminoacid's isoelectric point. The isoelectric point, pI, can bercund by averaging the two pKa values for the amino acidthat include the zwitterion in their reaction. For instance, thepI of glycine is found by averaging pKul and pKu2, becausepKul involves zwitterion formation and pKu2 involveszwitterion consumption.

3 6. There is at least one essential amino acid with each ofthe following properties EXCEPT:

A. an aromatic ring.B. a hydrophilic side chain.

C. an R-steroecenter at the cr-carbon.

D. a cationic side chain at pH = 6.


3 7. Removing a polypeptide from a lipid environment andplacing it in an aqueous environment would:

A. alter its primary structure by cleaving covalentbonds.

alter its primary structure by cleaving hydrogenbonds.

alter its secondary structure by cleaving covalentbonds.

alter its secondary structure by cleaving hydrogenbonds.

3 8. Before determining the amino acid sequence of a largeprotein, an enzyme is often added, so that the protein is:

A . denatured and turned into its straight chain form.B. cleaved at selected amide bonds and broken down

into smaller fragments.

C . coupled with an identical protein in the reverseorder and then transcripted.

D . sliced into a cross-section and mounted on a slidefor a better view of the component amino acids.

3 9. Threonine, shown below, has what chirality?

o HOH

H"N H

A. 25, 35

B. 2R, 35

C. 25, 3R

D. 2R, 3R

4 0. At a pH of 7.4, the structure of phenylalanine is BESTdescribed as:

A. uncharged.

B. an anion.

C . a cation.

D. a zwitterion.

41. What pH would result if you were to mix 10 mL offully protonated alanine with l0 mL of NaOH(aq)solution, where the concentration of the base solution is1.5 times greater than the concentration of the alaninesolution?

L. 2.35

B. 6.11

c. 7.00D. 9.87

B.

C.

D.


4 2. What is the isoelectric point for isoleucine?

PKal = 2.3 pKa2 = 9.7

A. 2.4B. 6.0c.7.0D. 9.1

4 3. Which of the following amino acids would generate thefollowing titration curve when titrated with 0.10 MNaOH(aq) starting from its fully protonated state?

1.0 2.0 3.0

Equivalents of 0.10 M NaOH(aq) added

A. Phenylalanine

B. GlycineC. LysineD. Aspartic acid

Copyright @ by The Berkeley Review@ 212 GO ON TO THE NEXT PAGL


An amino acid is given its name based on its havingboth an amine group and a carboxylic acid group. Thegeneric structure for a natural (chirally correct) amino acid i*s

drawn in Figure I below.

ollHzN\a/t\o,

luHR

Figure L Generic amino acid

The R represents various groups, each one correspondingto a specific amino acid. This is to say that all amino acidsare the same with the exception of the R group. The genericamino acid in Figure 1 is shown as it exists in a hydrophobirenvironment. This is not how it exists in an aqueousenvironment, however. In water, the amino terminal (aminegroup) is protonated and the carboxyl terminal (carboxyliiacid group) is deprotonated. Figure 2 shows the aqueou_rform of a generic amino acid.

olt

luHR

t,*-"-"-o-

Figure 2 Amino acid in aqueous environment

The pK6 values for the protonated amino terminal anC

the carboxylic acid group are approximately 9.5 and 2.5respectively. At physiological pH of '1.4, the pKu of theamino terminal is greater than the pH, so it is protonated br,the solution according to the Henderson-Hasselbalch equation_At physiological pH, the pKu of the carboxyl terminal is iess

than the pH, so it is deprotonated by the solution. Equatiom1 is the Henderson-Hasselbalch equation.

PH = PKa + log [A-]

IHA]Equation 1 Henderson-Hasselbalch Equation

Using the Henderson-Hasselbalch equation, it is possibleto calculate the ratio of the deprotonated form to protonateJform for each site on an amino acid at a specific pH.

44. Which of the following changes to n-propanol wouliiNOT result in a lower pKu than n-propanol?

A. Oxidizing carbon I by four electrons, resulting in an-bond to oxygen instead of two o-bonds tcthydrogen.

B. Replacing the oxygen with a sulfur atom.C . Replacing the propyl group with a phenyl ring.D. Replacing the hydroxyl group with an amine

group.

What charge does the following dipeptide cany at a pHof 5.0?

o

PKa = 2'0

H (cHt4H oI

+ftHt PKu = lo'8

A. -1

B. 0

C. +1

D. +2

When the pH of solution is greater than the pKu by l,what is the ratio of the deprotonated form to theprotonated form?

A. 1:10B. 1:2C.2:1D. 10: 1

How do the three labeled protons on the followingmolecule rank in terms of relative acidity?

oc+ llHiN\

c/ c\

o*1

ll.- bH CF2OH

A. Proton a > Proton c > Proton b

B. Proton b > Proton a > Proton c

C. Proton b > Proton c > Proton a

D. Proton c > Proton a > Proton b

What can be said about the pKu of a hydrogen on sulfur,given that cysteine exists predominantly in thefollowing form at pH = 7.4? Note that the hydrogen onsulfur is the second proton to be lost when cysteine isdeprotonated.

NI

o.llHrN\a/t\n-

l\H CH2SH

A. It is less than 2.5.

B. It is greater than 2.5 but less than 7.0.C . It is greater than 7.0 but less than 9.5.D. It is greater than 9.5.

Irpyright @ by The Berkeley Review@

W.-\ PKa=3'e

4 9. How many times must the fully protonated form ofglutamic acid be deprotonated to form monosodiumglutamate? (For glutamic acid R = CHZCHZCOZH)

A. 0 times

B. I timeC. 2 times

D. 3 times

50. At what pH value will the ratio of the +2 to +l formsof the following molecule be 10J

t \\.V\*"=u,

I

Given the pKu values for each of the followingcompounds, which of the compounds is the most basic?

o

o

oA

PKa= 9.2+

HgN.

H (CHn),'r*lNH3 PKo

A. 3.08.5.rc. 1.1

D. 8.2

OH PKo = 1.3NI

H

= 10.8

51.

c. D.

o

PKa = 10'0 PKa = 8'4 PKa = 5'0

u'ry*"'

ry"

GO ON TO THE NEXT PAGE.

Passage Vlit (euestions 52 - 58)

Amino acids are the biological building blocks ofnaturally occurring polymers known as proteins. The humanbody codes for twenty amino acids. These twenty make upthe majority of the amino acids found in enzymes, proteinhormones, and other polypeptides. polypeptides are formedwhen amino acids link their respective amino and carboxylterminals. Table 1 lists the twenty amino acids that havebase codons associated with them:

Amino acid Code R-Group PKasAlanine A1a CH: 2.3/9.7Arginine Arg (CH2)3NHC(NHz)z+ 2.2/9.5/13.0Aspartic acid Asp CH2CO2 2.1/3.9t9.8Asparagine Asn CH2CONH2 2.0/8.9Cysteine cyt CH2SH r.8/8.4/10.6Glycine Glv H 2.3/9.6Glutamic Acid Glu CH2CH2CO2- 2.U4.3t9.8Glutamine Gln CH2CH2CONH2 2.2/9.1Histidine His CH2C3H3N2 t.8t6.1t9.2Isoleucine Ile CH(CH3)CH2CH3 2.4/9.7Leucine Leu CH2CH(CH3)2 2.4/9.6Lysine Lys (CH2)aNH:+ 2.2/9.2/10.8Methionine Met CH2CH2SCH3 2.3/9.3Phenylalanine Phe CH2C6H5 t.8/9.2Proline Pro

-CH2CH2CH2- 2.0t9."7Serine Ser CH2OH 2.2t9.2Threonine Thr CH(CH3)oH 2.6/r0.3Tryptophan Trp CH2CsH6N 2.4/9.4Tyrosine Tyr CH2C6HaOH 2.1/9.1t10.1Valine Val CH(CH3)2 2.3/9.6

Table I

Each amino acid can be distinguished from the others bythe R-group attached to the alpha carbon, known as its sidechain. The R-groups in Table I are shown as they exist atpH = 7.4. Figure 1 shows the structure for a generic aminoacid with a generic R-group.

o

Figure I Generic amino acid with natural stereochemistry

52. Which of the following polypeptides will carry apositive one charge when placed in gastric acid (anaqueous solution buffered at pH = 1.5)?

A. Lys-Leu-IleB. Arg-Cys-HisC. Asp-Glu-AsnD. Glu-His-Phe


o.ll

"t*-a/"l\.

HR

53. Hydroxyproline, shown below, is found in humar:beings, but it is not coded for in DNA.

What feature is NOT associated with hydroxyproline?A. Hydroxyproline has more than one chiral center.B. Hydroxyproline induces deviations in the seconda,-,

structure of a protein.C . Hydroxyproline is hydrophilic.D . Hydroxyproline has a side chain pKu around 4.0.

Which of the following peptide fragments has thrHIGHEST isoelecrric pH?

A. Met-Lys-ArgB. Gly-Phe-AlaC. Tyr-Asp-GluD. Leu-Val-Ile

Which of these amino acids contains exactly one chi:ruIcenter?

A. GlycineB. HistidineC. Isoleucine

D. Threonine

Which of these dipeptides requires the lowest soluimpH to exist 100% as a zwitterion?

A. Leu-HisB. Lys-AsnC. Asp-CysD. Tyr-Arg

At pH = 7.0, which of the following amino acids umigrate to the cathode in a buffered electrophoresis gai"

A. AlanineB. Aspartic acidC. Leucine

D. Lysine

An isoelectric point less than 5.0 indicates tharamino acid has what type of side chain?

A . An acidic side chain.B. A basic side chain.C. A hydrophilic side chain.D. A hydrophobic side chain.

H N-ot\-**n(,!",

54.

,t

1

55.

(T

I6ilG

&rp

m

56.

2t4

58.

GO ON TO THE NEXT P.{

\mino.\cid

DecompTemp

Water Sol.@ 25'C

(8/too mI-)

- -rqt0lil PKal PKa2 PKa3

G1 233"C 25.2 z.-'\a 9.78

\la 298"C 16.5 +8.5 2.35 9.81\/ai 315'C 8.9 +13.9 2.29 9.12

Leu 293'C 2.4 -10.8 2.33 9.74

Ile 284'C 4.1 +1 1.3 2.32 9.74

\Iet 280"C 3.4 -8.2 2.11 9.21

Pro 220'C t62.0 -85.0 1.95 10.64

?he 283"C 3.1 -35.1 2.58 9.24

T 289"C t.l -3 1.5 2.43 9.44

Ser 228'C 5.0 -6.8 2.19 9.46

Thr 225"C verv sol. -28.3 2.09 9.10

- \'s +6.5 1.88 8.36 10.33

\I 342"C 0.05 -10.6 2.21 9.12 10.07

-\ Sn 234"C 3.5 -5.4 2.02 8.80

lln 185"C '\- l +6.1 2.11 9.t3j.sp 210"C 0.6 +25.0 r.99 3.88 10.01

llu 241'C 0.9 +31.4 2.13 4.32 9.9s

_\S 225'C very sol. +14.6 2.16 9.20 10.80

lao 244'C 15.0 +12.5 t.82 8.99 13.21

:l is 281"C 4.2 -39.1 1.81 6.05 9. 15


There are twenty amino acids for which human DNA;odes. These twenty amino acids form proteins within theiuman body. The physical properties of the twenty amino:cids are listed in Table l, shown below:

Table 1

Amino acids are distinguished by the side chain attached'. :he alpha carbon. Within a linear protein, there is one,. ::no terminal and one carboxyl terminal, so the side chainsr,-iount fbr the behavior of each amino acid within a peptide.

5:i, tyhi"h of the following side chains would MOSTLIKELY be found in the hydrophilic pocket of a

protein?

A. -CH2CH2SCH3B. -CH2C6HaOHC. -CH2CH2CONH2D. -CH2SH

r i, How can the fact that lysine is more water soluble thanleucine best be explained?

A . The side chain of lysine can form hydrogen bonds,while the side chain of leucine cannot.

B. The side chain ofleucine can form hydrogen bonds,

while the side chain of lysine cannot.C. The side chain oflysine is non-polar.D. The side chain ofleucine is polar.

- :i"right O by The Berkeley Review@ 215

66.


61. The GREATEST isoelectric pH is found with:

A. alanine.

B. glutamic acid.

C. histidine.D. phenylalanine.

62. Which of the following structures represents the mostcommon form of cysteine in a pH = 7.0 solution?

A.

HzN

B.

HzN

CH2SH cH2s -

H CH2SH

63. Arginine has a lower pKui than alanine, because the

carboxyl terminal of alanine is:

A. more acidic than the carboxyl terminal of arginine,due to the inductive effect of the anionic side chain.

B. less acidic than the carboxyl terminal of arginine,due to the inductive effect of the cationic side chain.

C . more acidic than the carboxyl terminal of arginine,due to the inductive effect of the amino terminal.

D. less acidic than the carboxyl terminal of arginine,due to the inductive effect of the amino terminal.

64. Tyrosine best separates from cysteine at a pH equal to:

A. the pKa (Cys), when using ether/water extraction.

B. the pKa (Tyr), when using ether/water extraction.

C . 7.0, when using ether/water extraction.

D. either plg6 or pltyr, when using ether/waterextraction.

At pH = 4.0, what is true for histidine?

A . Most exists with the carboxyl terminal protonated.

B. Over 9070 exists as a zwitterion.C . Most exists with the amino terminal deprotonated.D. Over l}Vo of the side chains are protonated.

Which of the following amino acids would bind to a

column filled with DEAE-cellulose at a pH of 6.20?

A. AlanineB. Glutamic acidC. LysineD. Tyrosine

65.

C.

H CHoS -


Parathyroid hormone (pTH) and calcitonin (CT) are twoof the three protein hormones involved in mineralhomeostasis. PTH is initially synthesized as a 115-aminoacid inactive polypeptide preprohormone. preprohornonesare later activated by removal of specific parts of theirprimary structure. The primary structure of parathyroidhormone is shown below:

Figure 1 Parathyroid Hormone

Calcitonin is a 32-amino acid peptide hormone that hasits carboxyl terminal in the form of an amide. Bothcalcitonin and PTH are found in several animals, but theprimary sequence varies from species to species. Thereactivity is sometimes altered by the substitution of differentamino acids in the polypeptide.

67 . In human PTH, amino acid #46 is alanine. In bovinePTH, amino acid #46 is glycine. What is true whencomparing human PTH with bovine pTH?

A . The secondary structure is more disrupted in bovinePTH, because of its bulkier side chain.

B. The secondary structure is more disrupted in humanPTH, because of its bulkier side chain.

C. The tertiary structure is more disrupted in bovinePTH, because of its bulkier side chain.

D. The tertiary structure is more disrupted in humanPTH, because of its bulkier side chain.


68. In PTH, amino acids #1 through #34 make up theactive portion. Amino acids #35 through #94 protecxthe structure from proteolysis. Amino acids #3-sthrough #43 can best be described as:

A. hydrophilic.B. hydrophobic.

C. polar and anionic.D . polar and cationic.

6 9. In the endoplasmic reticulum, prepro-pTH loses tu.c,methionine residues and a 23-amino acid peptide to formpro-PTH. Pro-PTH is transferred into the Golgi regionwhere it is converted into PTH. What occurs in theGolgi region?

A . Gain of an octapeptideB . Gain of a hexapeptide

C. Loss ofan octapeptideD. Loss ofa hexapeptide

70. Chymotrypsin cleaves at the C-terminus of phe, Trg-and Tyr. What fragments form when human pTH xtreated with chymotrypsin?

A . Three fragments total, of length 22 amino acids, i !amino acids, and 5l amino acids.

B . Three fragments total, of length 27 amino acids. tr .amino acids, and46 amino acids.

C. Four fragments total, of length 27 amino acids. lamino acids, 16 amino acids, and 32 amino acids.

D . Three fragments total, of length 23 amino acids. j i.

amino acids, and 50 amino acids.

71. The fact that Metg and Metl g of human pTH can Lrreplaced with nor-leucine (R = -CH2CH 2CH2Cllz t

with no observed change in reactivity implies that:

A. sulfur has no effect on reactivity.B. nitrogen has no effect on reactivity.C. sulfur has a significant effect on reactivity.D. nitrogen has a significant effect on reactivity.

7 2. Analogs of human-PTH should:

A . be structurally diverse between amino acids #l!through #34 compared to human-pTH.

B. be structurally similar between amino acids #frthrough #34 compared to human-pTH.

C . have a different N terminus and C terminus rlmnhuman-PTH.

D . be structurally diverse between amino acids *-

through #84 compared to human-pTH.

7 3 . What can be concluded from the following data?

Fragment Vo Activity in vitro Vo Activity in vivo1-84l-341 - 31

r -28

I00Vo

llVol0Vo

5Vo

l00Vo

32Vo

6lVo

0.3Vo

A. The active site probably contains amino acid # 17.

B. The active site probably contains amino acid # 30.

C. The active site probably contains amino acid # 66.

D. The active site probably contains amino acid # 83.

Copyright @ by The Berkeley Review@ 2t7 GO ON TO THE NEXT PAGE.


A researcher was interested in determining the sequenceof two polypeptide fragments she isolated from partialhydrolysis of an enzyme. The first fragment (Fragment I) isa six amino acid fragment which cannot be deciphered usingmolecular weight studies of the component amino acids. Theprimary structure of the second fragment (Fragment II) wasdetermined using an amino acid sequencer. Fragment II hasthe following amino acid sequence from N-to-C terminals:His-Ser-Val-Phe-Ile-Tyr-Phe. An automated proteinsequencer cleaves, isolates, and identifies the componentamino acids one at a time from the amino terminal to thecarboxyl terminal using phenylisothiocyanate. The researcherused the enzymes listed in the Table I for analysis.

Enzyme Amino acid Cleavage side

Chymotrypsin Phe, Trp, Tyr Carboxyl

Clostripain Arg Carboxyl

PepsinPhe, Trp, TyrAsp, Glu, Leu

Carboxvl

Thermolysin Leu, Ile, Val Amino

Trypsin Lys, Arg Carboxyl

Table I

The term "cleavage side" indicates the point at which theenzyme breaks the peptide bond. For instance, thermolysincleaves the peptide bond on the amino (left) side of leucine,isoleucine, or valine. It should be noted that the enzymeitself is a chain of amino acids, but it is present in such a

small concentration that it does not cleave itself. Theenzyme concentration is low because it is a catalyst with a

substantial turnover rate, where a "substantial" turnover rateis defined to be greater than 10O/second.

Some small polypeptides can be analyzed according totheir isoelectric point or migration rate in an electrophoresisgel. If the isoelectric point of an amino acid is greater thanthe pH, then that amino acid carries a partial positive chargeand exhibits cationic behavior. Cations migrate to thecathode in gel electrophoresis. Figure 1 shows histidine inits fully protonated state.

CHz * pKo2 -- 6.1

PKat = 1'8HO

PKaz= 9.2 H:N H

Figure 1 Fully protonated structure of histidine

7 4. What is the approximate isoelectric point of histidine?

A. 4.0B. 6.1

c.'7.6D. 9.2

{J+\

7 5. In what direction would histidine migrate if it wereadded to an electrophoresis gel buffered at pH = 7.0?

A. It would migrate to the anode.

B. It would migrate to the cathode.C . It would not migrate.

D. It would denature.

7 6. If treating a polypeptide with z, -dinitrofluorobenzenefollowed by 6 M HCl(aq) results in an alanine residuebound to a dinitrobenzene moiety, then what can beconcluded from this information?

A . The first amino acid in the polypeptide is alanine.B. The second amino acid in the polypeptide is

alanine.

C . The fifth amino acid in the polypeptide is alanine.D . The last amino acid in the polypeptide is alanine.

7 7 , Treatment of Fragment II with thermolysin would yield:

A. 1 fragment.B. 2 fragments.

C . 3 fragments.

D. 4 fragments.

78. What is the side group on rhe89 grams/moles?

A. -HB. -CH2OHC. -CH3D. -SCH3

amino acid that weighs

7 9. Which of the following polypeptides would be observedafter Fragment II is treated with clostripain?

A . His-Ser-Va1-Phe-Ile-Tyr-Phe

B. His-Ser-Val-PheC . Ile-Tyr-PheD. His-Ser-Arg

8 0. If the separate treatment of a protein first with pepsinand then with thermolysin generated the samepolypeptides, then which of the following peptidelinkages is present in the protein?

A. Ile-Phe

B. Leu-Leu

C. Tyr-ValD. Val-Met

Copyright O by The Berkeley Review@ 218 GO ON TO TIIE NBXT PA

8 L. Treatment of Fragment I with chymotrypsin yields thefollowing residues: Ser-Val-Phe and Gly-Cys-Gly.Treatment of the six amino acid Fragment I withSanger's reagent (4-dinitrofluorobenzene) binds serine.Which of the following is the sequence for Fragment I?

A. Gly-Cys-Gly-Phe-Val-SerB. Gly-Cys-Gly-Ser-Val-PheC. Ser-Val-Phe-Gly-Cys-GlyD. Phe-Va1-Ser-Gly-Cys-Gly

I

Itr

:iS

f,ffi

mfi

({ffi

mffiflil

{m

fri

Passage Xlt (Questions 82 - 88)

A protein isolated from the saliva of the NortheasternRadish Spider is found to break down muscle fiber in theTuscaloosa Spitting Lizard. Researchers determined theprimary sequence of the protein by carrying out fourexperiments. In Experiment I, the full protein is treated with2,4-dinitrofluorobenzene (2,4-DFNB) followed by hydrolysisof the protein using an acid catalyst. The 2,4-DNFB is astrong electrophile that accepts electrons from the nitrogen ofthe first amino acid, and in doing so labels it.

The amino acids present in the protein are two alanines,two cysteines, and one each of arginine, glycine, histidineifound to be the first amino acid), isoleucine, leucine, lysine,methionine, phenylalanine, serine, and tyrosine. Followingthe hydrolysis experiment, the researcher treated separatesamples of the protein with the following reagents;

Experiment II: The protein is first treated with B_

mercaptoethanol (breaking any disulfide linkages) followedby treatment with trypsin yields these fragments:

N-Ser-Ile-Tyr-Ala-C

N-His-Lys-CN-Met-Cys-Leu-Gly-Ala-phe-Cys-Arg-C

Experiment III: The protein is first treated with B-mercaptoethanol (breaking any disulfide linkages) followedby treatment with thermolysin yields these fragments:

N-Ile-Tyr-Ala-C

N-His-Lys-MetCys-CN-Leu-Gly-Ala-Phe-Cys-Arg-Ser-C

Experiment IV: The protein is first treated with B-nercaptoethanol (breaking any disulfide linkages) followedby treatment with chymotrypsin yields these fragments:

N-Cys-Arg-Ser-Ile-Tyr-C

free alanine

N-His-Lys-Mer-Cys-Leu-Gly-Ala-phe-C

The researchers were able to use the experimental-:Ibrmation to determine the sequence of the amino acids in:i:e protein. Knowing the first amino acid in the sequence

".ong with any two of the experiments, the sequence can

";curately be deduced. All disulfide linkages must first be:roken in order to isolate the fragments in their correctr-QUOnC9.

E 2. What is the role of B-mercaptoethanol ineach experiment?

B-Mercaptoethanol serves to bindacid in the protein.B-Mercaptoethanol serves to bindacid in the protein.

A.

B.

the first part of

the first amino

the last amino

C . B-Mercaptoethanol serves to cleave the proteinfollowing the cysteine residues.

D . B-Mercaptoethanol serves to cleave any disulfidebridges present in the protein.


83. What is the role of 2,4-d,initrofluorobenzene inExperiment I?

A. DNFB serves to bind the first amino acid in theprotein.

B. DNFB serves to bind the last amino acid in theprotein.

C. DNFB serves to cleave the protein following thecysteine residues.

D. DNFB serves to cleave any disulfide bridges presentin the protein.

8 4 . Based on the results of Experiments II, [I, and IV, whatis the LAST amino acid in the protein?

A. AlanineB. Arginine

C. HistidineD. Serine

85. Based on the results of Experimenrs II, III, and IV, allof the following are valid conclusions EXCEpT:A. chymotrypsin cleaves on the carboxyl side of

tyrosine and phenylalanine.

B . trypsin cleaves on the carboxyl side of arginine andlysine.

C . thermolysin cleaves onand leucine.

D . thermolysin cleaves onand alanine,

the amino side of isoleucine

the amino side ol cysteine

86. Why is it a more valid conclusion that thermolysincleaves before leucine and isoleucine, rather than aftercysteine and serine?

A . Thermolysin cannot cleave after molecules involvedin sulfide bridging.

B. Because serine has a protic side chain, it cannotreact with thermolysin.

C. There are two cysteine residues present in theprotein, so if thermolysin cleaved after cysteine, anadditional cut would have been observed.

D. There are two cysteine residues present in theprotein, so if thermolysin cleaved after cysteine,one less cut would have been observed.

87. From the information in Experiments II and III, howmuch of the exact protein sequence is known?

A. It is known only to amino acid two.B. It is known only to amino acid four.C . It is known up to amino acid eleven.D . The entire sequence of amino acids in the protein is

known.

8 8. Based on all of the experimental information, what isthe primary sequence of the unknown protein?

A . N-His-Lys-Met-Cys-Leu-Gly-Ala-Phe-Cy s-Arg-S er-Ile-Tyr-Ala-C.

B . N-His-Lys-Met-Cys-Ile-Tyr-Ala-Leu-Gly-Ala-Phe-Cys-Arg-S er- C.

C . N-I1e-Tyr-Ala-Leu-Gly-Ala-Phe-Cys-Arg-S er-His-Ly s-Met-Cys - C.

D . N-Leu-Gly-Ala-Phe-Cys-Arg-Ser-His-Lys-Met-Cys-I1e-Tyr-Ala- C.

Passage Xlll (Questions 89 - 96)

When amino acids are synthesized invitro, they form a:a racemic mixture of the D- and L-enantiomers. Since the L-enantiomer is desired, resolution is necessary. For resolvingthe mixture, one of three techniques can be applied.

Technique l: The racemate is first treated with benzoylchloride to form an amide at the amino terminal. Theracemate of N-benzoyl-Dl-amino acid is treated with a

brucine salt causing the D-amino acid to precipitate fromsolution. The solid is filtered from solution and is treatedwith acid to dissolve the brucine-amino acid salt into a ne\\'solution. Upon treatment with aqueous hydroxide, a puresample of the D-amino acid is isolated.

Technique 2: The racemate is first treated with benzol'lchloride followed by a strychnine salt, causing the L-aminoacid to precipitate from solution. The solid is filtered fromsolution and is treated with acid to dissolve the strychnine-amino acid salt into a new solution. Aqueous hydroxide isadded to the new solution to isolate the L-amino acid.

Technique 3: The racemate is treated with acetic anhydrideto form N-acetyl-Dl-amino acid. This racemic mixture ofacylated amino acids is treated with hog renal acylase.which removes the acetyl group from the L-enantiomer.forming the zwitterion. The D-enantiomer remainsacylated at the N-terminal, making it anionic (due to thedeprotonation of the C-terminal). The two species are

separated using ether-water extraction.

A major drawback to the first two techniques is that onl1one enantiomer is isolated in pure form. The enantiomer tha;does not precipitate from solution is isolated after the solvenis evaporated, which does not result in a pure sample. Thesecond drawback is that the success of the techniques varieswith different amino acids. For instance, they work weLwith alanine, but are difficult to use with tryptophanTechnique 3 has the most universal application. Presenre;below is an experimental application of Technique 3:

Experiment: A suspension of I2.9 grams of N-acetyl-Dl-isoleucine in 0.80 liters of water is buffered to 7.0 in an

aqueous solution, following which 0.010g hog renalacylase powder is added. The mixture is stirred for l6hours at 37.0'C (physiological temperature). The mixtureis then acidified with 10.0 mL of concentrated acetic acid-filtered through a frit, and evaporated under vacuum to a

volume of roughly 50 mL. Ethanol is added to form L-isoleucine crystals. The crude product is recrystallized frornan ethanol-water mixture to yield roughly 3.5 grams('71.|Vo yield) of optically pure L-isoleucine.

8 9. Why must the first step in Technique 3 be done in alaprotic solvent?

A. A protic solvent is too inert.B. A protic solvent can react with acetic anhydride.C . To ensure that the N-terminal is protonated,

D. To ensure that the C-terminal is deprotonated.

rNr t

$3.

J

l

Coprright @ by The Berkeley Review@ 220 GO ON TO THE NEXT PAGL

90. When the brucine salt is used to resolve a racemicmixture of alanine, what is likely to occur?A. L-alanine is isolated in relatively pure form fromprecipitate, while D_alanine is^ isolated with

impurities from solution.B. L-alanine is isolated with impurities from

precipitate, while D_alanine is isolated in .etatiueifpure form from solution.

C. D-alanine is isolated in relatively pure form fromprecipitate, while L_alanine is isolated withimpurities from solution.

D. D-alanine is isolated with impurities fromprecipitate, while L-alanine is isolatid in relativeiypure form from solution.

9 1. Which procedure involves an increase in entropy?A . Separation of enantiomers from a racemic mixture.B . Racemization of one enantiomer.C . Reaction of a D-amino acid with a diacid.D. Precipitation of an enantiomeric salt.

9 2. Why is the racemate first treated with acetic anhvdride inTechnique 3?

A. To protect the amino terminal.B. To form a bond that will react with hog renal

acylase.

C. To make the amino acid cationic.D. To invert the chiral center.

93. In Technique 3, the filtrate can be treated with HCI(aq)to a pH of 2.0 to crystallize the N_acetyl_D_isoleucine.The crystals can be dissolved and hydrolyzed intosolution of the D-amino acid using 1.0 M HCI(aq).The D-enantiomer can be treated with racemase togenerate a D-L mixture again. Technique 3 can beapplied to this mixture once again to isolate more L_enantiomer. Why does this process become futile afterthree cycles?

A. The percent of the enantiomer obtained from thethird cycle is only l2.5%o of the product mixture.

B. The percent of the enantiomer obtained from thethird cycle is only 6.25Vo of the product mixture.

C. Ygg_

renal acylase is consumed completely by thethird cycle.

D. The acetic anhydride reacts only when the mixturehas a substantial amount of L_enantiomer.


9 4, In the experiment, why is the solution stirred forsixteen hours?

A . To vaporize the pig renal acylase.B. To agitate the mixture and enhance reactivity. ,

C. To _centrifuge the powder to the bottom of theflask.

D. To generate a suspension, and thus decrease thereaction rate.

9 5 . Why is hydroxide anion used at the end of TechniqueA . To remove the benzoyl substituent from the

terminus.

B. To remove the carboxyl and amino terminalprotons.

C. To remove the cationic brucine salt.D. To racemize the isolated amino acid.

9 6. What precipitates with strychnine in Technique 2?A. B.

1?

N-

H3N1

C.

HrN*

o

*/"

RH

HIN

97.

Questions 97 - 100 are NOT based

on a descriptive passage

Which of the following graphs BEST represents the rate

of the corresponding nucleophilic substitution reaction

relative to the pH of the solution?

H3CNH2 + CF|CI € (l{3C)2NH2+Cl

--+

+

pH+

9 8. Following neutralization, the product of a reaction ofH3CNH2 and H3CCI would show which of the

following proton NMR peak patterns?

A. Singlet (3H), singlet (3H), septet (lH)B. Singlet (3H), singlet (3H), quartet (1H)

C. Singlet (6H), sextet (1H)

D. Doublet (6H), septet (lH)

9 9. How many mL of 0.010 M NaOH(aq) need to be added

to take 10 mL 0.010 M lysine from pH = 5.68 (the firstequivalence point) to pH = 10.80 (pK6 for lysine)?

A. 5 mL 0.10 M NaOH(aq)

B. 10 mL 0.10 M NaOH(aq)

C. 15 mL 0.10 M NaOH(aq)

D. 20 mL 0.10 M NaOH(aq)

'1

6zs

^1

Gz.H

&

"1

Gzs&

'1

6

H


100. A1l of the following can be reduce to form aminef,

EXCEPT.

A. Amides with LiAlH4.B. Nitriles withHCIlZnClZ(aq).

C. Imines with H2iPd.

D . A11 of the above reactions yield amines.

.'GETINTOCHEM!"

1.A 2.C6. D ',|. Bll. A 12. c16. A r1. B21. c 22. A26. C 21. B31. B 32. B36. C 3',7. D41. D 42. B46. D 47. A51. D 52. C56. C 51. D61. C 62. C66. B 67. B11. A 12. B16. A 71. C81. C 82. D86. C 87. D91. B 92. B96. D g',t. C

3.D 4.C 5.A8.D 9.C l0.A

13. A 14. C 15. C18. D 19. C 20. D23. A 24. A 25. D28. B 29. C 30. A33. D 34. D 35. D38. B 39. C 40. D43. D 44. D 45. B48. C 49. C 50. B53. D 54. A 55. B58. A 59. C 60. A63. B 64. D 65. D68. B 69. D 70. D13. B 74. C 15. B78. C 79. A 80. C83. A 84. A 85. D88. A 89. B 90. c93. A 94. B 95. A98. D 99. C 100. B

3.

Amines and Amino Acids passage Answers

7.

Choice A is correct' -According to the passage, the most basic amine compound in the gas phase is the tertiaryamine' The weakest base in the gas phase is the base with the least electron donating aliyl groups. This makesammonia the weakest base. Because ammonia is the weakest base, ammonium .u.-tior-r'(th-e conjugate acid ofammonia) must be the strongest acid in the gas phase. This makes choice A correct.

Choice C is correct. If the pKu for CH3CH2NH g+ is 10.71., then the pK6 for CH3CH2NH2 must be 3.29. BecauseCH3CH2NH2 is more basic than CF3CH2NH 2, due to the electron wiihdrawirig ,-rut,rr" of fluorine, the pK6 forCF3CH2NH2 must be greater than 3.29. The pK5 for CF3CH2NH2 should be Jnly a point or two greater thanthe pK6 for CH3CH2NH2, making choice C coirect.

Choice D is correct. In both the gas phase and in water, primary amines are stronger bases than ammonia. Thismeans that the stronger base is on the product side of the equation. A favorable ieaction has the stronger basereacting to yield the weaker base, which is the reverse reaction as written. This means that reactants are moreabundant than products as written. This makes the equilibrium constant (Keq) less than one in both the gasphase and in water. The best answer is choice D.

Choice C is correct. The donating nature of methyl groups increase the basicity of an amine. According to Table1, the highest pKa is associated with the dimethylammonium cation, so it is ihe weakest acid. As an*acid getsweaker, its conjugate base becomes stronger, so ttre secondary amine is the strongest base. This is choice C.

Choice A is correct- An amine of any substitution, tertiary included, will exist in its protonated state in solutionswith a pH value less than its pKu value. When it is protonated, it is cationic and thus more soluble in waterthan it is in its neutral state. This implies that as the pH of the solution is lowered, the amine becomes moresoluble. The lowest pH of the answer choices is choice A, 3, so choice A is the best choice.

Choice D is correct- When the pH is less than the pKu, the site exists in its protonated state. When the pH isgreater than the pKu, the site exists in its deprotonated state. The pH is seven, which is greater than the pKufor the carboxyl terminal so the carboxyl terminal is deprotonated. When the clrboxyl terminal isdeprotonated, it carries a negative charge. This eliminates choice A. The pH is seven, which is less than thepKu for the amino terminal and the amino side chain, so both the amino terminal and the amino side chain areprotonated. When the amino terminal and the amino side chain are protonated, they each carry a positivecharge. This eliminates choices B and C. The best answer is choice D.

Choice B is correct. The passage discusses the N-H absorbance of a secondary amine, but does not mention atertiary amine. This is because all of the hydrogens in a tertiary amine are bonded to carbon. The nitrogen hasthree bonds to carbon, so there is no N-H bond, making choice g the best answer. There is no resonance to speakof, given that there is no n-bond, so choice C is eliminaied. The tertiary amine may or may not be symmetric, butthat is irrelevant, so choice D is eliminated. Steric hindrance

^uy t""- tempting, because a tertiiry amine hasmore steric hindrance than less substituted amines. But such an answer is designed to make you jump at sterichindrance without thinking through the question completely, because it is not the reason for no absorbance.

8' Choice D is correct. The donating nature of alkyl groups increases the basicity of an amine, although it alsocreates steric hindrance. In this case, the amines are primary and secondary. Secondary amines are more basicthan primary amines, so choices A and B are eliminited. the data in Tabie 1 confirmi that secondary aminesare more basic than primary amines, because secondary amines yield a greater equilibrium constant whenreacting with ammonium than primary amines. Methyl groups are less butky than ethyt groups, but ethylgroups are better electron donating substituents than methyl groups, so the most basic .o--po.rna cannot bedetermined using background inform.ation. According to the data ln table 1, (CH3)2NH g"rl"*t", a K"O of 30.9when reacting with ammonium, while (CH3CH2)2NH generates a K"o of 50.1 when reac"ting with amnionium.This means that (CH3CH2)2NH is a stronger base than (CH3)2NH, so cioice D is the best unilr"r.

6.

Copyright @ by The Berkeley Review@ 223 AMINO ACIDS & AMINES EXPLANATIONS

9. Choice C is correct. Table 1 iists the reactions of neutral amines with ammonium, so for each entry, NH+* shoul:be found on the reactant side. All of the entries in Table t have equilibrium constants greater than 1, ,o to hu.'.an equilibrium constant of 0.03, it must be the reverse reaction of what is shown in Table 1. This means tha:\Hi must be found on the reactant side. This eliminates choices B and D. According to the data in Table ,dimethyl amine ((H3C)2NH) is associated with an equilibrium constant of 50 *nl" trimethyl amrr-.rItH3C)3N) is associated with an equilibrium constant of roughly 30. It is in the reverse reaction of trimeth'.-amine that results in an equilibrium constant of 0.03 (about 1,/30), so choice C is the best answer.

Choice A is correct. Because an amine is a weak base while an amide exhibits no acid-base properties l,r,he:ai:led to water, the two compounds can be separated using acid-base extraction. This eliminates choice D. T:.alnLre can be protonated to form a cationic compound (its ammonium conjugate acid), which is more water solui:,.il'-e-n the amide. The extraction therefore will be most successful at a 1olr,' pH (acidic conditions). The lowest p- -ot the choices is 3, which makes choice A the best answer.

Choice A is correct. Because of the electron withdrawing nature of the carbonyl group through resonance, --:.ele:tron pair on the nitrogen in an amide is less available for donating than the electron pair on nitrogen in .--alr-lie' The difference in basicity is attributed to resonance and not steric hindrun.e, so choices C and D a::el-minated. An amine is more basic than an amide, so choice A is the best answer.

Choice C is correct. The boiling point of a compound is influenced by its molecular mass and its intermolecu-:forces. The highest boiling point is associated with the compound with both hydrogen bonding and the grea:a;;r,olecular mass. Ammonia has hydrogen bonding, but it is very light. Choice A is eliminated. Trimeth/iut --*erhibits no hydrogen bonding, so it does not have the highest boiling point of the choices. This elimina:=sci.oice D. Methyl amine and propyl amine have roughly the same degree of hydrogen bondrng, but propyl ar':,ris heavier. As such, the highest boiling point is observed with propyl amine. Choice C is the best answer.

Choice A is correct. Given that the pKu of (H3C)3NHC1 is 9.8, the pK6 of (H3C)3N: must be 4.2. The elec:::rdonating nature of methyl groups increases the basicity of an amine while the electron withdrawing natur. :rbenzene decreases the basicity of an amine. Aniline is therefore less basic than trimethyl amine. Becaus- ,higher pK6 value is associated with weaker bases, the pK6 of aniline can be no less than 4.2. This makes ch---:*A not possible. Choice A is your choice for correctness and subsequent happiness.

Choice C is correct. A decrease in the K"o value for a proton transfer reaction with ammonium indicates tha: --:c

Ittgty of the compound is reduced. Thi question is thus "Which of the following changes to an amine \\-. riNOT result in a weaker base?" or better yet, "Which change would result in u.o*po.tnd MORE basic th-- ;primary amine?" Replacing hydrogen with fluorine would reduce the basicity, because fluorines are elec:::,nrn'ithdrawing. Choice A is thus eliminated. Replacing the adjacent (alpha) hydrogens with a double bor.; iroxygen would convert the compound to an amide and thus reduce its basicity, because oxygen is elec::rr,vithdrawing. Choice B is thus eliminated. Replacing one hydrogen with an alkyl group -orrta increase -:*basicitv, because alkyl groups are electron donating. Choice C is the best answer. Replacing an alkyl group-,l--iThydrogen would reduce the basicity, because alkyl groups are electron donating. Choice D is thus eliminatec

Choice C is correct. Method 2 in Figure 1 shows the reduction of a nitrile, a primary imine, a primary amide ,:,;an azide into a primary amine, so a nitrile and primary imine, choices A and B, can definitely be reduced ir,:: ,primary amine. Choices A and B are eliminated. A secondary amide can be reduced, but R-groups rer---.flrconstant, resulting in the formation of a secondary amine. This means that a secondary amide, choice C, ca:::ribe reduced into a primary amine, so choice C is the best answer. An oxime, R2C=NOH, can be reduced mucL :.*an irnine can be reduced. The resuit is a primary amine and water. Choice D is eliminated.

Choice A is correct. The N-H bond, like the O-H bond, has an infrared absorbance value above 3000 cm-1. ---eabsorbance is broadened due to hydrogen bonding, although not as broad as a hydroxyl absorbance, be.a-:,"rarnines do not form as strong of hydrogen bonds with amines as alcohols form with alcohols, Broadening r-. :;r'',associated with the question, so we address it for edification purposes only. You simply need to know that 1,.--ibonds are simiiar to O-H bonds, so their IR absorbances are close in value. The best choice is answer A.

10.

17.

12.

t3.

1.4.

E::

15.

16.

Copyright O by The Berkeley Review@ 224 AMINO ACIDS & AMINES EXPLANATIO\:

17. Choice B is correct' The methyl group usually donates electrons through what is called hyperconjugation. Thedonating form of the structure is nof truly a resonance form in the traditional sense, so the methyl group isconsidered to be electron donating through the inductive effect. This is another case of controversy, so when youare uncertain of what-information they are basing their answer on, consult the passage. In this case, which willbe true on occasion, the information is not there. So, from background information, resonance is defined as thedelocalization of electrons through the n-bonding network. The inductive effect is defined as the delocalizationof electrons through the o-bonding network. Theie are no n-bonds in an amine, so the best answer cannot involveresonance' This eliminates choices C and D. The best answer must therefore involve the inductive effect. Anelectron donating group would make an amine a better nucleophile, so choice B is the best answer.

Choice D is correct. The real question here is whether the reaction proceeds by an 51111 mechanism or an 5512mechanism. The electrophile is secondary, so there is no easy way to tell. However, on a multiple choice exam,you should let the answers dictate the choices you make. Tire Sry1 reaction would result in a racemic mixture,which is not listed as a choice. The answer choices tell us to decide between an 5612 reaction and an E2 reaction.Methyl amine is not a strong enough base to cause elimination, so the reaction must follow an S1'tr2 m-echanism,eliminating choices A and B. An 5512 reaction results in an inversion of the chiral center. The reactantelectrophile has R-chirality, so the product shoulcl have S-chirality. The best answer is choice D.

Choice C is correct. The most nucleophilic nitrogen is the one with a lone pair that is most readily available forsharing' Electron pair a is the lone piir of ut't u-id" nitrogen, so it is tied up in resonance and not available foruse as a nucleophilic lone pair. Choice A is eliminated. Electron pair b is ieing shared with the benzene (andthe amide carbonyl on the para position of benzene) through resonance, so it is not as available for nucleophilicattack as a standard amine lone pair. Choice B is eliminated. Electron pair c is free from any conjugation(because it has sp2-hybridization ,^and sp2-1-rrbridization has the electrons in an orbital perpendicular to the p-orbital of the n-network). Because of the resonance from nitrogen d, nitrogen c carries a partially negativecharge making it more nucleophilic. Choice C is the best answer.'Electron pal d is involved ln the aromaticityof the five-membered ring, so it is not readily available for nucleophilic atiack. Choice D is eliminated. Thisquestion is analogous to what is observed with histidine. The best choice is electron pair c, so choose C.

Choice D is correct. An amine can react with both an acid halide (choice A) and an ester (choice B) to form anamide, so choices A and B are eliminated. It can undergo an Sp2-reaction with an alkyl halide (ciloice C), sochoice C is eliminated. Ethers are inert and react with very few compounds. Choice D is the best answer.

Choice C is correct- Using the enamine synthesis route, alkyl groups can be added to the alpha carbon. Thesynthesis that cannot be carried out via the enamine procedure laid out in Figure 2 is the one that forms a newcarbon-carbon bond to a carbon other than the alphi carbon. Choice A add"s a methyl to carbon 2 (the alphacarbon of 3-pentanone), choice B adds a butyl g.o.tp to carbon 2 (the alpha carbon of buianal), and choice D addsa pentyl group to carbon 3 (the atpha carbon of acetone). It is in choiie C that the new carbon-carbon bond isformed with a non-alpha carbon (carbon 6 of 3-hexanone is not an alpha carbon). pick choice C for the warm,fuzzy feeTing of correct answer picking.

Choice A is correct. Water is the side product that is generated in the formation of an imine from a ketone andan amine. The amine reactant is a primary amine, so it wiII lose both hydrogens and form a bond to the carbonylcarbon. The best answer is choice A.

Choice A is correct. The treatment of an imine with acidic water results in the hydrolysis of the imine andsubsequent formation of an amine and a carbonyl. The cyclic nature of the reactant causes the two functionalgroups to be present on the same molecule. By cleaving the double bond and adding two hydrogens to the nitrogenand an oxygen (carbonyl oxygen) to the carbon, the pioduct is determined rather easily. This"shown below. Thebest answer is choice A.

18.

19.

H

AMINO ACIDS & AMINES EXPLANATIONS

*'€o -> "-QO- -rpyright @ by The Berkeley Review@ 22s

24' Choice A is correct. The imine bond, when cleaved with acidic water, forms a ketone compound and a prima:-amine' A quick pictorial route for determining the hydrolysis products formed from the reaction is shown belor.,

+"+1"{# }*n':OPrimary amine Ketone

The best answer is choice A.

25' Choice D is correct. From 3-pentanone to 4-methyl-3-heptanone, the structure has increased its number of carbc:,by three. This means that a three carbon chain has been added to carbon 2 (which is equivalent to carbon -1 :,symmetry) of 3-pentanone. This eliminates choices A and B. To carry out this synthesis, i secondary amine rr.,-,,:be added to the carbonyl reactant (3-pentanone) to form an enamine. This eliminates choice C and makes cho::.D the best answer. In the second step of the synthesis, the alkyl halide compound (RX) is a three-carbon ch: -with the halide on the terminal carbon. The final step in the synthesis in all four choices is the addition .,acidic water, so it must be true. The last step converts the nitrogen compound (enamine) irrto an oxygen compo.i:(carbonyl).

o R:.. ^,--R R:p,7n

ll l{g T H3ccH2cH4 -il- t'r':,- fr

HrcH2cAcH2cH3 HrcH2cA.HCFL lr.nr.AcHCFb rircHrcA.HCH| ' J L

I -'

26.

)n

H3CF{2CF{2C H3CFI2CFI2C

choice C is correct. The imine that is formed from a primary amine and a ketone is the imine shown in ,-::second example in Figure 1. The nitrogen of the imine his a single bond to one carbon and a double bond to ::

=other carbon. Upon reduction of the imine, the nitrogen will gain a new o-bond to hydrogen and lose the n-bo::to the carbon (as the carbonyl oxygen does when the ketone is reduced to a secondary atcJnot;. The final produ::is a nitrogen with two alkyl substituents and a hydrogen, which describes a secondaiy amine. The best dfls\4,€r :choice C.

Choice B is correct. Questions of this type are best solved by drawing the reaction out. Acidic water causes i=hydrolysis of N-ethyl-3-pentimine, as shown below:

H3CFI2C- NF!

Ethyl amineThe products are ethyl amine and 3-pentanone, so the best answer is choice B.

28. Choice B is correct. D-Serine has two hydroxyl absorbances, because it has two non-equivalent O-H bonds. l;IR absorbance for the O-H group of a carboxylic acid is broad and it comes between 2500 cm-1 and 3000 cm-1. l:=IR absorbance for the O-H group of a standard alcohol comes between 3200 cm-l and 3600 cm-1 and is also broa:although not as broad the carboxylic acid O-H absorbance. The carboxylic acid hydroxyl and alcohol hydror.are not equivalent rnJR spectroscopy. Whether it be by memory or inductive reasoning, pick B. A carbonyl bon:C=o, absorbs around 1700 cm-1, so choices C and D aie eliminit"d. Urirrg inductive r6aioning, you can conciuj=thai the o-H bond of an acid is weaker than the o-H bond of an alclhoi (which is *h/ii is more acidicWeaker bonds have lower bond energies (weaker spring constants, if you wili), so they are easier to stretch. ,:the bond is easier to stretch, then less energy is required to stretch the bond. Less eneigy can be seen in the j:absorbance as a lower frequency (lower wave number). This means that the absorbance foi an acidic O-H must't=less than a standard alcohol, around 3300 cm-1, so choice A cannot be possible.

HrcH.c--+1"<:*^.:'cF{2cFI3

H_____-l>Hzo

CFI"CH"

cFI2CF{33-Pentanone

Copyright O by The Berkeley Review@ 226 AMINO ACIDS & AMINES EXPLANATIONS

29. Choice C is correct. There are three synthetic routes presented in the passage. This particular synthesis beginswith an aldehyde, therefore it is the Strecker synthesis, as shown in Figure sI upon matching the reagents fiomFigure 3 with the reagents in the answer choices, it is choice C that is a match. If you choie to eva'iuate eachstep in each answer choice, the sequence of products from the steps in choice C are an imide, cyano amine andamino carboxylic acid. Following the reaction step by step also leads to the correct choice. In choice A, thecompound would be oxidized into a carboxylic acid and then deprotonated. In choice B, the cyano group wouldadd to the carbonyl to form a cyano alcohol which can then form an acid with acidic *orknp, but iot an aminoacid' Without adding nitrogen to the compound, it is not possible to form an amino acid. Choice D has too manycarbons in the reactant, so it cannot form tie product. The best (and correct) choice is answer C.

Choice A is correct. Every amino acid in a hydrophobic solvent has a broad signal between 10 and 12 ppm in the1HruvR due to the carboxylic acid hydrogen uni u luiwtn signat of twice tte intensity between 2.5 ppm and4'5 ppm due to the two hydrogens on nitrogen. Deuterochlorofoim, CDCI3, is an aprotic, hydrophobic solvent, sothe signal just above 10 ppm is attributed to the carboxylic acid proton ur-rd tl'," signal around 3-pp* is attributedto the two amino terminal hydrogens. the 1HNMR has three iignals, in a 1 : 2 : 2 ratio, so the amino acid hasthree types of hydrogens. Because there is only one other sigial (the triplet around 2.1 ppm for the alphahydrogens), there is only one other type of hydrogen present ot'r ih" amino acld molecule. This means that the R-grouP must be a hydrogen, because any other R-group would introduce additional signals into the 1gruUnspectrum. The only amino acid with two equivalent hydrogens in the side chain is glyciie (the side chain is H).Glycine has three types of hydrogens, so choice A is the besianswer.

Choice B is correct. _In the Gabriel synthesis, the amino acid is released when the phthalimide complex is

hydrolyzed under acidic aqueous conditions. This means that the side product results from the hydrolysis of thecyclic imide. This makes the side product a diacid, making choice B the best answer. Choice A"is noi possible,because hydrolysis would cleave the amide bonds. Choice i is too reduced and choice D cannot happen, becausehydrolysis would likely cleave the cyano groups if the pH was correct.

Choice B is correct- An amino acid itself has one unit of unsaturation due to the n-bond of the carbonyl in thecarboxylic acid. The questions centers around the units of unsaturation in the side chain. Aliphatic, bydefinition, refers to an alkyl grouP with no rings and no n-bonds. Thus, there is no unit of unsaturation present inan aliphatic side chain. The units of unsaturation for the side chain is zero so the total units of unsatu'ration forthe molecule is one. Pick B, and you can proudly consider yourself a correct answer picker.

Choice D is correct. When L-cysteine is completely uncharged, there are five unique hydrogens present oncysteine- When cysteine exists as a zwitterion, there are only four unique hydrogens pr"rilt, because thecarboxylic acid proton has been lost. The solvent is deuterochloroform, a hyirophobic solvent because thepresence of deuterium on carbon is aprotic, so cysteine exists in its uncharged state.

^Consequently, there are five

different hydrogen signals in the iHNtvtR. Pick D for optimal results. i-Cysteine is shown below with all ofthe unique hydrogens labeled.

30.

31.

?)

33.

HzN\ ,/'n"t L-Cysteine

34' Choice D is correct. Histidine has a positively charged side chain when the pH of the aqueous solution is lessthan the pK^ value of the side chain This means thit the histidine side chain goes from positively charged toneutral upon deprotonation' Because a hydrophobic environment cannot Jabilize a positive charg-e, thehistidine side chain prefers to be neutral in a hydrophobic environment. Histidine will therefore lose the H+more readily in a hydrophobic environment than a hydrophilic environment. Losing H+ more readily results ingreater acidity and thus a lower value for the pKu. Choice A is eliminated, becauie log functions do changewhen the number changes. The pKu decreases, not increases, when it becomes more acidicl so choice B cannot becorrect. Upon deprotonation of the side chain, the side chain becomes neutral, not negatively charged, so choiceD is a better answer than choice C.

tX"u


36.

J/.

38.

35. Choice D is correct. In order for an amino acid to have three equivalence points, when starting in its fullrprotonated state and being titrated by a strong base, the amino acid must have three acidic protons. In order icran amino acid to have three acidic protons, the amino acid must have an active proton on the side chain at lo-,trpH (i.e., have a side chain that is either acidic or basic). Tyrosine has an acidic side group (a phenol proton ha'ra PKa around 10) while the R-group for the other three choices (glycine, alanine, and leucine) are al1 allo-groups (none of which have an acidic proton). The best answer is therefore tyrosine, choice D.

Choice C is correct. According to the passage, there are three aromatic amino acids, phenylalanrnetryptophan, and histidine. This eliminates choice A. According to the passage, threonine is one of the essenti.damino acids. it has a hydroxyl group on its side chain, so there is at least one essential amino acid that lshydrophilic. This eliminates choice B. Essential amino acids are natural amino acids, so they must be L-aminc,acids. With the exception of cysteine, L-amino acids have an S-stereocenter at the cr-carbon. Consideringcysteine is not one of the essential amino acids listed in the passage, there are no essential amino acids with a:R-stereocenter at the d-carbon, so choice C is the best answer. At a pH of 6, arginine carries a positive charge, slchoice D is eliminated. Histidine has a side chain pKu around 6.05, so it will exist in a state where slightrover half is protonated and cationic.

Choice D is correct. Whenever a protein is removed from a hydrophobic environment and introduced into a

hydrophilic environment, the secondary structure of the polypeptide backbone will be altered, because the,hydrogen bonds that are intramolecular in the lipid environment can now be formed with the surrounding watein the hydrophilic environment. The result is that the protein will turn itself inside-out as it passes from orreenvironment into another. Pick D, and get another one correct.

Choice B is correct. Enzymes such as chymotrypsin, trypsin, and thermolysin break the peptide linkages (amiaebonds) of specific amino acid residues within a protein. This breaks the protein into smaller fragments, whic:can then be evaluated with greater ease. This is done with large proteins, because sequencing techniques hanean upper limit to the number of amino acids they accurately determine. Before using Edman's reagent, a prote:!is denatured using B-mercaptoethanol and urea, to denature and elongate the protein for easier analvshHowever, it is not an enzyme that does this, so choice A is eliminated. Enzymes do not cause a protein to coupLewith another protein, and transcription is not associated with protein sequencing, so choice C can be eliminatE;"Choice D is a throw away answer, because there is no mounting of a protein onto slides in sequencing. The be*answer is choice B, because a large protein can be accurately sequenced after it has been cleaved into srnalpieces.

Choice C is correct. H is in the front position at both chiral centers, so whatever arc is drawn when prioritizingthe substituents from highest through lowest must be reversed to determine the chirality. In this case carbon lgets a clockwise arc and carbon 3 gets a counterclockwise arc. Because the molecule should really be viewed frorthe backside, reversing these arcs correctiy identifies carbon 2 as an S-center and carbon 3 as an R-center. Thecompound has chirality of 25, 3R, so choice C is the best answer. The determination is shown below.

39.

HOH

-f4./t\

o

A,HO(t),.*2

HO/,"c\

n

HzN H

For carbon-2, the arc must be reversed forcorrect viewing, so it has S-chirality.

J\HzN H

For carbon-3, the arc must be reversed forcorrect viewing, so it has R-chirality.

cFI3C

HO H,,t

"o,. | "


{0. Choice D is correct. Phenylalanine has an alkyl side chain, so the only active protons it possesses are for theamino terminal and the carboxyl terminal. The carboxyl terminal has a pKu between 2 ani g,so at a pH of 7.4,the pH is greater than the pKu and therefore the carboxyl terminal of fnenytatanine is deprotonated andanionic' The amino terminal has a pKu between 9 and"1.0, so at a pH of 7.4thepH is less than the pKu andtherefore the amino lelmilal of phenylalanine is protonated and cationic. There are no acidlbase propertiesassociated with its side chain, so phenylalanine will always have a neutral side group. The result is thatphenylalanine has one cationic site ind one anionic site, so it exists primarily as a zwitterion. While it has nonet charge, because it has charged sites, it cannot be considered uncharged. Choice A is eliminated. It has nooverall charge, so it cannot be considered to be an anion or a cation, so cioices B and C are eliminated. The bestdescription is a zwitterion, so choice D is the best answer.

Choice D is correct. This question requires that you understand a titration curve in a conceptual manner. Becausethe base is 1'5 times more concentrated than the alanine solution and an equal volume of each solution is mixed,the amount of base that has been added is one-and-one-half equivalents. Starting at the fully protonated state,addition of 1.5 equivalents of titrant base takes the alanine ,tl,rtior, past its firit equivalencl point and half*-uy

!o the second equivalence point of alanine. If you observe the titraiion curve for ilanine, this makes the pHof

-solution equal to pKu2, which lies exactly halfway between the first and second equivalence points. Thevalue of pKu2 lies between 9 and 10 for most amino acids, so choice D is the best answer. The titration curve foralanine titrated by NaOH is shown below:

{1.

Isoelectric Point

PH = PKaz

Alanine startsPH = PKurfully protonated.-

-HzA* .1,| /2

Equivalents base added

12' Choice B is correct. For an amino acid with a hydrophobic, alkyl side group, there are only two pKu values. Todetermine the isoelectric point, you must urr"rug" the two pKu values" thai are associated with the zwitterion,in this case the only pKu values associated withlhe amino icid. For isoleucine, the values for pKul and pKu2are 2'3 and 9.7 , respectively. The average of the pKu values is 6, so choice B is the best answer.

{3' Choice D is correct. Because there are three equivalence points (vertical inflection points) on the titration curve,the amino acid that the titration curve represents must have a side chain with an actlve proton. An activeproton is associated with either an acidic or basic side group. This eliminates choices A and i], which both haveside chains that contain no active protons. Because both of ine first two plateaus in the graph occur below a pHof 7 '0, the first two pKu values for the amino acid must be less than 7.0. The middle oi tn" plateau representsthe point at which the pK^ is equal to the pH of the solution. So, from the graph, a qualitative determinationof the pKu values can be made. Because the value of pKu2 is less than 7.0, ihe side chain is a carboxylic acidgroup rather than an amine group. Aspartic acid has a iarboxylic acid group on the side chain, which has a pKuvalue less than 7.0, so choice D is the best answer. The side chain of ffrin" is an amine, so its pKu is above 2.0.Choice C is eliminated for this reason.

nine finishesthe two eq. points.


44. Choice D is correct. To make the pKu lower, something must be done to the compound to make the proton mc:gacidic' In choice A, the oxidation of carbon 1 by four electrons results in the conversion of a primary alcohol ir:ra carboxylic acid, which increases the acidity of the compound through electron withdrawal via resonarce-This results in a lower pKu, so choice A is eliminated. Because sulfur is Targer than oxygen, exchanging ox\.ge1lfor a sulfur allows the proton to be lost more easily. If the proton is lost moL easily, tnei tne.orr1pJ,r1j is mo:eacidic and thus has a lower pKu value. This can be seen *h"n

"ornparing serine (primary alcohol side group I :o

cysteine (primary thiol side group), where the side chain pKu foi cystline is about 6 less than the side cha-rpKu for serine. This eliminates choice B. Replacing the propyi grorrp with a phenyl group has the same effejas oxidizing carbon

-1 from a primary alcohol into a .uiUo*yti. acii, althougn th" resonance effect is not a$

strong. Regardless, the compound becomes more acidic, so the pKu decreas"r. thi, can be seen when comparrjg:serine (primary alcohol side group) to tyrosine (phenol side group), where the side chain pKu for tyrosine 15about 4 less than the side chain pKu for serine. Choice C is lhus eliminated. The only u.,r-"," choice left uchoice D By replacing the oxygen with nitrogen, the compound becomes less acidic. Amines are less acidic tha:alcohols. Nitrogen is roughly the same size as o"yg".r, ro the effect is not the same as choice B with sultuBecause nitrogen is less_electronegative than oxygen, it "pulls" electrons less than oxygen does. This makes rrenitrogen more basic and less acidic than the oxygen. This raises the pKu value making choice D the choice tl-a:would nof decrease the pKu.

Choice B is correct. When the pH of the solution is greater than the pKu for a proton, that proton is removed brthe solution and the compound exists in its deprotonated state. wne" ine pu l, hlgn, the solution t ruri. u.;thus deprotonates the proton. This same conch-rsion can be formed using the Henderson-Hasselbalch equatio:-Yh:" the pH is greater than the pKu, the log term in the H-H equation niust be positive, so the concentration irrthe base (deprotonated form) must be greater than the concentrition of the acidlprotonated form). This mearsthat when the pH is less than the pKu, the site retains its proton. Taking the siLs one at a time, the carbor;-terminal has a pKu of 2.0, so at pH = 5.0, it is deprotonated and carries i -1 .hurg". The carboxylic acid siiechain on the aspartic acid (right amino acid) his a pKu of 3.9 which is also le"ss than the pFi, so it too sdeprotonated and thus carries a -1 charge. The amino terminal has a pKu value of 9.2, whichis greater than itupH, so it remains protonated and thus carries a +1 charge. The amine side chain on the lysine (ieft amino aciChas a pKu of 10.8, which is also greater than the pH, so if is protonated and thus carries a +1 charge. The sum olthe charges is zero, so the best answer is choice S. The structure at a pH of 5.0 is drawn below.

oYotf ., = r.n

45.

I5'r

46.

17.

Choice D is correct. In order for the pH to be greater than the pKu, the conjugate base (deprotonated form) mr:-.:be in higher concentration than the conjugate acid (protonated fot-;. The quistion asks for the deprotonated-t"-*protonated ratio, which is greater than 1. This eliminates choices A and B. B".urse the pH is one unit abore]he pKu, the log of the base to acid ratio must equal 1, according to the Henderson-Hasselbalch equation. Thelog of 10 is equal to 1, so the ratio of conjugate basJto conjugate acid must be 10 : 1. The best answer is choice D.

Choice A is correct. Because the electron withdrawal associated with resonance is greater than the electror.withdrawal associated with the inductive effect, a carboxylic acid is more acidic thai the alcohol group on acarbon with one fluorine atom bonded to it. From the puriug", we know that the pKu for the carboiytic acid =lower than the pKu for the protonated amine site, so the caiboxylic acid (site a) must be the most acidic site.This means that proton a is the most acidic proton on the molecule and only .hoi"" A lists proton a as most acidicso choice A is the best answer.

;rrC:plrighr O by The Berkeley Review@ 230 AMINO ACIDS & AMINES EXPLANATIONS

48.

49.

51.

52.

53.

Choice C is correct. Because the sulfur is still protonated at pH = 7.4, the pKu for the proton on sulfur must begreater than7.4. Because the proton on sulfur is the second proton to be lost (it is lostbefore the proton on theamino terminal), it must be more acidic than the amino proton. The amino proton has a pKu around 9.5, so the Hon sulfur must have a pKu less than 9.5. The pKu for the proton on sulfur lies somewhere between 7.4 and 9.5, sothe best answer is choice C.

Choice C is correct. The fully protonated form of glutamic acid starts with a positive one charge, because theamino terminal is protonated carrying a positive one charge and the carboxyl terminal and carboxylic acid sidechain are each protonated and neutral. The compound monosodium glutamate has one sodium cation, sotherefore the glutamate portion must carry a negative one charge. To be converted from a positive one chargeinto a negative one charge, glutamic acid must be deprotonated twice. The best answer is therefore choice C.Drawn below is the step by step conversion from the fully protonated form of glutamic acid to the monoanionform of glutamic acid (glutamate).

oH"*- Jt'

X - oH 1st equivalent

H gHz OH_r r

""y?I "nu,.guO +1

"ttY""Tr chargeoo

o.llt'^fo-H CH?t-

HzcY ?", .n",r"

o -1

o

"'fr#o-H CH,l-

2nd equivalent

oFI-

50. Choice B is correct. As drawn, the dipeptide molecule carries a +3 charge when fully protonated. When themolecule is at a pH of 1.8, the two forms of the molecule present in solution are the fully protonated form(carrying a + 3 charge) and the form with the carboxyl terminal deprotonated (carrying a +2 charge). \Atrhen thepH is 1.8, equal to pKu1, these two forms are present in a fifty-fifty ratio. When the molecule is at a pH of 6.1,the two forms of the molecule present in solution are the form with the carboxyl terminal deprotonated (carryinga +2 charge) and the form with both the carboxyl terminal and histidine side chain deprotonated (carrying a +1charge). VVhen the pH is 6.1, equal to pKu2, these two forms are present in a fifty-fifty ratio. In order to have a

ten fold excess of the +2 species, the pH must be less than pKaZby a factor of log 10, which equals pKa2 - 1. Thiseliminates choices C and D. The pH must be one unit less than 6.1 (pKuz), making choice 8,5.7, the best answer.

Choice D is correct. The most basic compound has the lowest pK5 value and the weakest conjugate acid. ThepK6 value can be found by subtracting the pKu for the conjugate acid from 14. Choice A is eliminated, because itis the conjugate acid of choice D. The conjugate acid is always less basic than its conjugate base. The pK6 valuesfor the three remaining choices can be found from the pKu values given for the sample compounds. For choice B,the pK6 value is 74 - 8.4 = 5.6. For choice C, the pK6 value is 14 - 5.0 = 9.0. For choice D, the pK6 value is 14 -10.0 = 4.0. The lowest pK6 value is found with choice D, therefore the most basic compound is choice D.

Choice C is correct. At a pH of 1.5, the tripeptide is fully protonated. The protonated amino terminal carries a

positive 1 charge and the protonated carboxyl terminal carries no charge. If none of the side chains carry a

positive charge, the overall tripeptide carries a positive 1 charge. Only the basic amino acids (histidine,lysine, and arginine) have a positively charged side chain when protonated. This means that the correctanswer choice shall not contain histidine, lysine, or arginine. Only choice C fits this criterion, so the bestanswer is therefore choice C.

Choice D is correct. Hydroxyproline has the normal chiral center associated with an amino acid plus it has anadditional chiral center associated with the hydroxyl group of the side chain. This eliminates choice A. Just asproline induces deviations in the secondary structure of a protein (turns and kinks), hydroxyproline causesstructural deviations as well. This makes choice B valid, thus it is eliminated. Alcohols are hydrophilic, sochoice C is valid and thus can be eliminated. The side chain of hydroxyproline is an alcohol. The pKu of analcohol is approximately that of water (15.5), therefore the pKu cannot be around 4.0. If it were around 4.0, thenserine would be expected to have an active side chain proton as well. The best answer is choice D.

Copyright @ by The Berkeley Review@ 23I AMINO ACIDS & AMINES EXPLANATIONS

54' Choice A is correct. The highest isoelectric pH is found with the tripeptide with the greatest number of bas:;side chains' This is true because the basic side chains retain positive .nurg" at higher pH values than the othe:side chains' Histidine, lysine, and arginine are the three basic amino acidi, and only choice A contains at lea::one of the three. Because choice A contains two basic side chains, it has an isoeiectric pH that is an average .,:pKu3 and PKa4' The four pKu values associated with Met-Lys-Arg are 2.2 (for the carboxyl terminal .:arginine), 9'3 (for the amino terminai of methionine), 10.8 (for tire sid"e chain of lysine), and 13b (for the sic;chain of arginine). The isoelectric pH is an average of 10.8 and 13.0, which is 11.9. This makes choice A tr;tripeptide with the highest isoelectric point.

55' Choice B is correct. Glycine has no chiral center, so choice A is eliminated. Isoleucine and threoni'e each ha..:a chiral center in their side chain resulting in two chiral centers overall, so choices C and D are eliminate:Histidine has all tp2-hybridized carbons and nitrogens in its ring, so there are no chiral centers on its side cha-:Because histidine has only one chiral center, choice B is the best answer.

56' Choice C is correct. This question is a unique way of asking "which dipeptide has the lowest isoelectric poir,:The lowest isoelectric point is found wittr the dipeptide cirrying no-basic side chains and preferentiallr- :*-acidic side chain (such as aspartic acid or glutamic acid). Choice A is eiiminated, because histidine l;1"r,-which increases^the isoelectric pH. Choice B is eliminated, because lysine is basic, which increases:::isoelectric pH. Choice D is eliminated, because arginine is basic, which inlreases the isoelectric pH. Chorce Chas no basic side chain. The isoelectric point is an iverage of the carboxyl terminal pKu of cysteine (1.g) ancl :--:side chain pKu of aspartic acid (3.9) wnlch results in an [oelectric point of 2.85.

57' Choice D is correct. Cations migrate to the cathode, therefore this question is asking for the amino acid -,:exists as a cation, or carries a partially positive charge, at a pH of z.o. To carry a positive charge, partiar _rfull, the isoelectric pH (pI value) must be greater than"the soluiion pH. This question is therefore uit ir,g for ,:'iamino acid with the highest pI value (only one can be the best urlri"r, ,o orliy one can have a pI value grea:--,:than 7'0, which makes it the greatest pI value). In this case, the only basic urr-ri"o acid (which has the high:s;pI value) is lysine, so choice D is the best answer.

58' Choice A is correct. The isoelectric pH is an average of the pKu value leading to the zwitterion and the :!.,,value going from the zwitterion. To hive a pI less th; 5.0, the pl(u values beinjaveraged must tre low num6.:1For amino acids with side chains having no active protons, tnl pr is found by"uverugir-,g the carboxyl terrn ::ipKu and the amino terminal pKu. This means thai the pI is an average of atout 2.5 u.,d 9.5, resulting in a :-around 6'0' This eliminates choices C and D. For amino acids witl basic side chains, the pI is founc --,"averaging the side chain pKu and the amino terminal pKu. This means that the pI is an u.r"rug" of a nLrn:*:between 6 and 72 and a number around 9.5, resulting in a pI around 9.0. This eliminates choice B. For amino a..::with acidic side chains, the pI is found by averaging the carboxyl terminal pKu and the side chain pKu. i:*means that the pI is an average of about 2.5 and a number between 3 ancl 10, resulting in a pI urorr.d 3.0. I:*makes choice A the best answer. To summarize, acidic amino acids have pI values less"than b, basic amino a;.rihave pI values greater than 7, and amino acids with neutral side chains have pI values around 6.

Choice C is correct. The side chain that is most likely to be founcl in a hydrophilic pocket is the side chai: :r,the most water soluble amino acid. The side chains in the answer choicei are methionine, tyrosine, glutan--- *and cysteine. According to the data in Table 1, the most water soluble amino acid of the choices is glitami:- m3'7 grams per 100 mL water. This is predictable, because of the hydrogen bonding of the side chain. The :,:slanswer is choice C.

Choice A is correct. Choices B and C should be eliminated immediately, because the alkyl side chain of leu:rnrdoes not form hydrogen bonds and the charged side chain of lysine is polar. Choice D is true, but the faci -:,adleucine has a nonpolar side chain does not enhance the waier solubility of leucine. Choice D is true :uiirrelevant, which eliminates it. Because lysine can form hydrogen bonds *ith th" hydrogens on nitrogen, ci.:,1,o*A is valid.

Ellhr

59.

60"

Copyright @ by The Berkeley Review@ 232 AMINO ACIDS & AMINES EXPLANATIO\S

61. Choice C is correct. For amino acids with both neutral and acidic side chains, the isoelectric pH is found byaveraging pKul and pKa2. The pI values for the amino acids with both neutral and acidic side chains areconsequently less than 6.0. For basic amino acids, because the pI is an average of pKu2 and pKu3, the value of pIis greater than 7.0. The only basic amino acid in the answer choices is histidine, choice C.

Choice C is correct. For cysteine, the amino terminal is protonated and carries a positive charge. The carboxylterminal is deprotonated and carries a negative charge. This eliminates choices A and B. Because the pH of thesolution is less than the pKu of the side chain, it is protonated and carries no charge. This makes choice C thebest answer and eliminates choice D.

Choice B is correct. The acidity of a compound is measured by its pKn, so the lowest pKu value is associatedwith the most acidic compound. The carboxyl terminal proton is the first to be lost, so it is represented as pKa1,meaning that the question is comparing the relative acidity of the carboxyl terminal protons of arginine andalanine, The carboxyl terminal pKu for arginine is lower than carboxyl terminal pKu alanine (according to thevalues in Table 1), so choices A and C are eliminated. Choice B is the better answer, because it is the side chain,not the amino terminal, that differs between arginine and alanine.

Choice D is correct. To separate two compounds using extraction in an ether/water biphasic solution, it is bestwhen one is charged and the other is neutral. It is at the isoelectric pH that the species carries no net charge,thus when the pH is equal to the pI of lysine, then lysine is neutral and therefore more soluble in ether than inwater. At this same pH, tyrosine is partially negative and thus water-soluble. \.Alhen the pH is equal to the pIof tyrosine, then tyrosine is neutral and therefore more soluble in ether than in water. At this same pH, lysine ispositive and thus water-soluble. At all other pH values, both lysine and tyrosine are charged. The best answeris choice D. Because of the high water-solubility of lysine in water, the extraction is best carried out at theisoelectric pH of tyrosine.

Choice D is correct. At a pH of 4.0, the carboxyl terminal is predominantly deprotonated, because the pH isgreater than the pKu of the carboxyl terminal. This makes choice A false. The isoelectric pH for histidine is7.60, so at a pH of 4.20 (a value less than 6.05), the side chain carries a positive charge, so far less than 90%exists as a zwitterion. This makes choice B false. The pKu of the amino terminal is greater than the pH, so theamino terminal is predominantly protonated. This makes choice C false. Because the pH is far less than thepKu of the side chain, the side chain is mostly protonated (more than ten percent). The best answer is choice D.

Choice B is correct. The background information that you have to know is that DEAE-cellulose carries apositive charge at neutral pH, so at a pH of 6.20, the column must be positively charged. Because the columncarries a positive charge, negatively charged amino acids are the ones that bind to the column. The lower thevalue of the pI, the more anionic the amino acid. This means that the amino acid with the lowest pI value isthe one that is most hindered by the DEAE-cellulose column. Glutamic acid has the lowest pI, because the sidechain has a pKu value of only 4.32. This makes choice B the best answer.

Choice B is correct. The side chain of alanine (CHe) is bulkier than the side chain of glycine (H), thereforehuman PTH has a bulkier R group than bovine PTH. This elirninates choices A and C. The secondary structure ismore affected by the steric hindrance of the side chain than the tertiary structure, so the best answer is choice B.

Choice B is correct. All of the amino acids from number 35, valine, to number 43, proline, have alkyl side chains,and thus are hydrophobic and uncharged. This eliminates choices A, C, and D. The best answer is choice B.

Choice D is correct. The polypeptide pre-pro-PTH is 115 amino acids in length. After losing the twenty-threeamino acid segment and the two methionine residues (twenty-five amino acids in total), the resulting pro-PTHcontains ninety amino acids. In the Golgi region, pro-PTH (ninety AA) is converted into PTH (eighty-four AA),so six amino acids must be lost. This means that a hexapeptide is lost, making choice D the best answer.

70, Choice D is correct. Within PTH, there is a tryptophan, Trp, a phenylalanine, Phe, and no tyrosine residues,Tyr. The Trp is in position 23 and the Phe is in position 34. This means that PTH chymotrypsin cleaves after AA#23 and afler #34, resulting in three fragments consisting of 23 amino acids (AA #1 through AA#23),11 aminoacids (AA #24 through AA#34), and 50 amino acids (AA #35 through AA #84). The best answer is choice D.

copyright @ by The Berkeley Review@ 233 AMINO ACIDS & AMINES EXPLANATIONS

62.

63.

64.

55.

66.

67.

68.

69.

Choice A is correct' Choices B and D can be eliminated immediately, because there is no nitrogen in the sii;chain of methionine' Replacing the side chain of methionine with an'aliphatic alkyl group does not increase r:decrease the number of nitrogen atoms in the molecule. Choice C can be eliminated, tecause no change -reactivity is observed. This makes the best answer choice A. The side chain listed is similar to methionr::except that the suifur has been replaced by a methylene group (CHZ).

Choice B is correct. The active region of human-PTH is given in the passage as being amino acids #1 throus:#34 1o

the analogs to PTH should be similar in this samJregion (the active"region). This eliminates choices -r-and D and makes choice B the best answer so far. The C-terminal and N-terminal do no affect the active s--:reactivity, so choice C is eliminated. The best answer is choice B.

Choice B is correct. Choice A can be eliminated, because reactivity is eliminated when the structure is reduct,:to twenty-eight amino acids. Choices C and D are poor choices, te.u.rs" both are found in the 35 through >=segment that does not directly affect reactivity, accolding to the passage. Based on the in aiuo reactivity da::the reactivity drops severely when going from 31 to 28 imino u"idr, so the active site includes an amino al:between 29 and 31. Choose B for best t"rrlltr.

74' Choice C is correct. Because basic amino acids, like histidine, have a +Z charge when fully protonated, th.;must be deprotonated twice to reach their neutral state (as a zwitterion). Th-e isoelectric'point is found:,averaging the two pKu values that involve the zwitterion. For an amino acid with a basic ria* gro"p, if.r" pl ;..an average of the last two pKu values (PKaz and pKu3). For histidine, the isoelectric point is"(6.1 + 9.2)1) =7'65, maktns choice C the best answer. Basic amino acids all have a pI greater thanT.0, so choices A anc :should be eliminated immediately.

75' Choice B is correct. To determine the migration direction of an amino acid in a buffered gel, the isoelectric pc:::must first be determined. Histidine is a basic amino acid, so the isoelectric point oihirtidir," is found:;averaging its last two pK2 values. For all basic amino acids (histidine, lysine, and arginine), the iro"i.lo,point (pI) is the average of pK^2 and pKu3. For all other amino acids, the isoelectric point is the average ::pKul and pK22. The pI for histidine is (6.1 + 9.2)/2 = 7.65. This value is greater than the pH of the solution, I -so histidine is partially positive (cationic) in a solution with a pH of 2.0." Cations migrate to the cathode in g.,electrophoresis, so choice B is the best answer o----- --

76' Choice A is correct. Sanger's reagent, 2,4-dinitrofluorobenzene, binds the first amino acid in a polypeptiitfragment at its amino terminal. The amino terminal nitrogen substitutes on the aromatic ring for the fluori::atom' This means that the amino acid that binds 2,4-dinitiofluorobenzene must be the first amino acid in ...rpolypeptide. Because 2,4-dinitrofluorobenzene binds serine in the unknown polypeptide, the first amino acicl :the polypeptide must be serine. Choose A if youknow what's good for you.

77 ' Choice C is correct. Thermolysin cleaves a protein on the amino side of leucine, isoleucine and valine. Fragme:.:II contains one valine (amino acid #3) and one isoleucine (amino acid #5), neither of which is the first aminracid' Hence, thermolysin cleaves Fragment II twice, breaking it into three separate pieces. you always chec1,that neither valine, leucine, nor isoleucine are the first amino *la itr the sequence when using thermolysin. tl,rbreak for Fragment II caused by thermolysin is as follows: His-Ser I Val-phe I Iso-Tyr-ph". -pi.t

C to score b:;on this question.

78' Choice C is correct. Without a side group, a generic amino acid weighs 17 grarns (for NH3) + 13 grams (for CH *44gtams (forCo2) =74grams/moleplusthemassof theRgroup. if th"to"tulweightisae*gramimole,thema=.of the side group must be 15 grams/mole. of the answer"selections, only the riethyl girp has a mass oi ,lgrams/mole' For this reason, you just have to pick C. To determine the answe r,youcould*also have gone throu=each answer choice and systematically.o*pnt"d each of their molecutu, *"ijnt, if you know the"st;;:;;;.:,the amino acid.

79' Choice A is correct. According to Table 1, clostripain cleaves at the carboxyl side of arginine. Fragmen: -contains no arginine, so it will not be cleaved by clostripain, leavirrg the fragment in one piec"e. rragment II sta""..:as His-ser-val-Phe-Ile-Tyr-phe, so choice A is the besJ answer.

71.

n.,

73.

Eil_

Copyright O by The Berkeley Review@ 234 AMINO ACIDS & AMINES EXPLANATIO\s

s0. -lChoice C is correct. Pepsin cleaves at the carboxyl side of phenylalanine, tryptophan, tyrosine, aspartic acid, r

g1utamicacidandleucine.Thermolysinc1eavesattheamino.sideofleucine,l,ot",.l,.'",andvaline.^Thi,-"u.',that if the carboxyl side of either phenylalanine, tryptophan, tyrosine, aspartic acid, glutamic acid or leucineis bonded to the amino side of leucine, isoleucine, or valine, then a peptideiinkage exists that will be cleaved byeither pepsin or thermolysin' In such a case, the same fragments would result whether pepsin or thermolysinwere employed. In choice A, the carboxyl side of isoleucine is linked to the amino side oiphenylalanine, whichis the opposite of what is needed to lead to equal cleavage by pepsin and thermolysin. Choice A is eliminated.In choice B, the carboxyl side of leucine is linked to the amino iiae of leucine, and both pepsin and thermolysinreact with proteins containing leucine. The problem with this answer choice is that tirey cleave on differentsides, so pepsin will generate a fragment ending in leucine and a free leucine while thermoiysin generates a freeleucine and a fragment starting with leucine. Choice B is eliminated. Pepsin does not react with either valineor methionine, so no cleavage occurs with choice D when using pepsin. Beiause thermolysin reacts with proteinscontaining valine, a break will occur before the Val-Met linkage, resulting in fragmentation. This eliminateschoice D. Choice C is the best answer, because pepsin cleaves after tyroJin" ur-rJ thermolysin cleaves beforevaline, so the same break occurs for the Try-Val linkige whether pepsin or thermolysin is used.

Choice C is correct. As given in Table 1, chymotrypsin cleaves the carboxyl side of phenylalanine, tryptophanand tyrosine. There are two tripeptide fragments foimed when using chymotrypsin, Lne ending in either Ser orPhe (although by convention we assume it ends in Phe as written) ur-rd tn" other ending in Gly [th"r" is no issuewith the order of the second fragment, because the sequence is the same either backwaris or forwards.) The Ser-Val-Phe fragment must be first and it must start with Ser, because it is the only one with either phe, Trp, or Tyrpresent at the end. This means that the Gly-Cys-Gly fragment is the secona of *re two fragments. puiting thetwo fragments back together yields the hexapeptide sei-val-Phe-Gly-Cys-Gly. This makes choice C the bestanswer' If you didn't solve it that way, there is an easier way. Only .noi." C has serine as the first amino acidin the sequence. You know serine is the first amino acid in Fragment I from the results of the treatment withSanger's reagent.

Choice D is correct. The role of the G-mercaptoethanol is to cleave any disulfide linkages formed between theside chain residues of cysteine. This is statedseveral times in the passage, at the start oJ each experiment. Thebest answer is choice D.

Choice A is correct. As stated in the passage, the purpose for adding 2,4-dinitrofluorobenzene (2,4-DNFB) wasto bind the amino terminal of the first amino acid in the polypeptide"chain. The 2,4-dinitrofluorobenzene onlybinds the first amino acid making it serve as a marker foi the-fiist amino acid. This means that the role ol 2,4-dinitrofluorobenzene is to identify the first amino acid in the sequence. This makes choice A correct.

Choice A is correct. Only histidine appears at the N-terminal in one of the fragments in each of the threeexperiments, Experiments II, III, and IV. Using the data from Experiment II and k]-rowing that histidine is thefirst amino acid (implying that Lys is the second amino acid), the possibilities for the last amino acid in thepolypeptide are either alanine or arginine. The fragments of Experiment III show that the last amino acid iseither serine or alanine (because His is first, and therefore Cys must be fourth). Combining this with theinformation from Experiment II and Experiment III, the last amino acid in the sequence must be alanine. Thismakes choice A your best choice.

Choice D is correct. The fragments formed in Experiment IV show that chymotrypsin either cleaves after Tyrand Phe or before Cys and Ala. Because the last fragment (N-His-Lys-Met-bys-L""-Cty-afa-phe-C) shows ap_eptide linkage to the amino terminal of alanine still in tack, chymotrypsin must cleave after Tyr and phe.Choice A is valid. The fragments formed in Experiment II show that trypsin either cleaves after Lys and Arg orbefore Ser and Met. There is no way to determine which is actually thl-case, so as it is, choice B is not invalid.The fragments formed in Experiment iII show that thermolysin eithlr cleaves after Cys and Ser or before Ile andLeu. Because the last fragment listed from the thermolyiin cleavage (N-Leu-GlyiAlu-ph"-Cys-Arg-Ser-C)shows a peptide linkage to the carboxyl terminal of cysteine still in tick, thermolysin must cleave before Ile andLeu. This makes choice D your best choice, because thermolysin does not cleave the amino terminal of alanineand cysteine. This also eliminates choice C.

i2.

]).

,opyright @ by The Berkeley Review@ 235 AMINO ACIDS & AMINES EXPLANATIONS

86. Choice C is correct- Because the last fragment in Experiment III (resulting from the thermolysin cleavage), N-Leu-Gly-Ala-Phe-Cys-Arg-Ser-C, shows a peptide linkage to the carbo"xyl terminal "i;yJ,;;;" ,tJril" tu.t,thermolysin cannot cleave following cysteine. if it did

"i'uur," after cysteine and serine, rather than beforeisoleucine and leucine, then the Cys-Arg linkage in fragment 3 would have been broken, resulting in anadditional fragment. The cleavage has nothi.tg t do witl disulfide linkages, so choice A is eliminated. We

have no information to assess whether thermolysin has any interactions witi protic side chains, so while choiceB may be true, we cannot choose it. Based on the third fragment, thermolysin'must cleave before isoleucine (Ile)and leucine (Leu)' As it is, thermolysin also cleaves beforeialine (Val). This makes choice C your best choice.

Choice D is correct. Histidine is the only amino acid that appears at the N-terminal of one of the threefragments in both Expe-riment II and Experiment III, so histidine must be the first amino acid. Coupling the factthat histidine is the first amino acid in the polypeptide aiong with comparison of the f.afm"its f.o-.Experiment II and Experim-ent III, the fragments cat-t 6e orretlappeJand matched to sequence the eitire peptide.For instance, Experiment III gives the first four amino acids as His-Lys-Met-Cys. Knowing the third and fburth11ino acids in the polypeptide identifies the first two fragments in Experiment II (N-HIs-Lys-Met-Cys-Leu-Gly-Ala-Phe-Cys-Arg-C), which defines the third fragmeniin Experiment II by default. The cumulative resul:gives the entire poiypeptide, so choice D is the best answer.

Choice A is correct. Histidine is the only amino acid at the start of a fragment in at least one fragment ir.Experiments II, III, and IV. This means that the polypeptide must begin'with histidine, which eliminate:choices C and D. Alanine is the only amino acid at the end of u f.ug*"nt ii at least one fragment in ExperimenisII, III, and IV. This means that the polypeptide must end with alanine, which eliminates choices B and C. Th;onlychoice that begins with histidine and ends with alanine is choice A, making it the best answer. This can beverified by matching the fragments from the three experiments, which shows"that the correct sequence mu::start with His-Lys-Met-Cys and end with Ser-I1e-Tyr-Ala.

87.

88.

89. Choice B is correct. Choices C and D are eliminated, because in an organic solvent, the N-terminal -.deprotonated and the Cterminal is protonated, resulting in both sites being tincharged. For the experiment t.work, the amino terminal must be neutral and nucleophiti.. choice A is JtiminatJ, because u prttl. soh.e..:would be reactive, not inert. A protic solvent (such is ethanol) can attack acetic anhydride and undergc .substitution reaction at the carbonyl carbon. The best answer is choice B.

Choice C is correct. Technique 1 involves the use of a brucine salt, which forms a precipitate with a D-amL-:acid' The precipitate is what is isolated from the mixture in solution in relatively pure form, not the species '..solution, so the best answer is choice C. Because of the precipitation of the O-amino acid, rather than the --amino acid, choices A and B are eliminated.

Choice B is correct. An increase in entropy involves an ilcrease in randomness. This occurs with a phase char,:.that increases the volume occupied by ihe material or a reaction that generates an increase in the number ::molecules. Choice A is eliminated, because it involves purification Jna separation of stereoisomers, r,r,hi::increases the order. The reaction of a D-amino acid with an achiral species r-rlith"t. increases nor decreases f -entropy, so choice C is eliminated. In choice D, a solute is converted lnto a soiid, which involves a decrease _entropy of the system. This eliminates choice D. The only increase in entropy occurs when one enantiomel --.racemized into a mixture of two enantiomers, because they are no longer separate and ordered. This ,,-ldri,:choice B the best answer.

Choice B is correct. The enzyme acylase cleaves an acyl bond (found in an amide). Individual amino acids ha'" =no peptide linkages (amide bonds), so amino acids must first be acylated to form a bond that can react r.,.-:-acylase' Acetic anhydride will acylate aI1 of the amino acids, but acylase will only remove the acyl group frr:molecules with the correct stereochemistry. This describes choice B. Because the amino terminal is tied ur -the amide bond, it cannot be protonated, and thus the acylated amino acid is only charged at the

"urUo.lterminal. This makes the species anionic, not cationic, which eliminates choice C. The ihiral center is :.. :affected, given that the priorities remain the same, so choice D is eliminated. The amino terminal is : ::protected by the acyl group, because there is no reason to protect it. This eliminates choice A, and makes chc :-.B the best answer.

I90.

91.

Copyright O by The Berkeley Review@ 236 AMINO ACIDS & AMINES EXPLANATIO\S

93. Choice A is correct- Choice C should be ruled out, because hog renal acylase is an enzyme, and thus it is notconsumed in the process. Choice D can be eliminated, because acetic anhydride is achiral, so it shows nopreference for one enantiomer over another. The original mixture is fifty percent D and fifty percent L. Afterremoval of one isomer and racemization, twenty-five percent of the originat product is D and twenty-fivepercent of the original product is L. After the removal of one isomer and iacemization again (the thirdpurification), twelve and one half percent of the original product is D and twelve and one half"percent of theoriginal product is L. This means that only twelve and one half percent of the desired enantiomer is isolatedfr.om the third cycle. After this point, the percent isolated is less ihan ten percent, and is not worth the effort.The best answer is choice A.

Choice B is correct. .stirring will not vaporize (sublime) a solid unless it results in extreme heating. Even withsixteen hours of stirring, not enough energy will be introduced into the system to get a notable amount of phasechange. This eliminates choice A. Stirring implies that the solid does not dissolv! weil into the solvent, so therole of agitation is not to centrifuge the solid to the bottom, but to increase the collision frequency and thusincrease the reactivity. This eliminates choice C and makes choice B the best answer. Generating a suspensionincreases the surface area for collision (and thus increases the reaction surface), causing the riaction rate toincrease, not decrease. This eliminates choice D.

Choice A is correct. The addition of strong base removes the carboxyl terminal and amino terminal protons, ifthere are protons on the amino terminal to be removed. Because the nitrogen is involved in the amide bond withthe benzoyi group, it never gains the proton that an amine group *o.rid gain, so there is no hydrogen to bedeprotonated. This eliminates choice B. The brucine salt was broten apart uling a strong acid, noia strlng base,so choice C is eliminated' Treatment with strong base in an aqueous

"nrriron-un"t hydrolyzes the amide bond and

thereby removes the benzoyl group from the N-terminal. This makes choice A thl best answer, an answer youshould pick if you know what is best.

Choice D is correct. In technique 2, the racemate is first treated with benzoyl chloride, which N-acylates all ofthe amino acids in solution, both D and L. The strychnine salt is formed from the selective precipitation of theL-amino acid after it has been N-acylated. Choices A and C can be eliminated, because they contain no benzoylgroup attached to the amino terminal nitrogen. Choice B has an R-stereocenter at the c,-caibon, while choice Dhas an S-stereocenter at the c-carbon. Choice D is the correct answer choice, because S-stereochemistry isassociated with an "L" amino acid. You should recall the mnemonic, "you're either a DoctoR or you LoSe,"where D and R go together and L and S go together.

Choice C is correct. The real question is "what effect does H+ have on the reaction rate when methyl amine isadded to methyl chloride?" The nucleophile is a good nucleophile and the electrophile is small, so tire reactionis an S52 nucleophilic substitution. In S1g2 reactions, the rate of the reaction depends on both the electrophileconcentration and the nucleophile concentration. This means that the rate oi the reaction varies with theconcentration of neutral methyl amine. Methyl amine must have a lone pair available to attack theelectrophile, so if it is protonated, it cannot be a nucleophile. The implies that at lower pH values there is lessneutral amine, so the reaction is slower at lower pH values and fastei at higher pH values. The correct graphshould show greater rates at higher pH values, so choices A and B are eliminited. Because pH is a log scalL andit is located on the x-axis, the relationship of the rate and pH is not linear. Choice C is i better answer thanchoice D.

Choice D is correct. The final product, after it has been neutralized (deprotonated and returned to an uncharged

?lil",l,: (H3C)2NH. _This

molecule has two unique types of hydrogeni, so it should show only two peaks in its'HNMR spectrum. This eliminates choices A and B, which show three signals rather than just two. The twosignals are in a 6:1, ratio, but that is observed in both choice C and choice D, so we are no further along thanbefore. The six like hydrogens split the neighboring signal into a septet (6 neighboring Hs + 1. = 7 peaks) uia tn"one hydrogen on nitrogen splits the neighboring signal into a doublet (t neighb'oring H + t = 2 peaks). This meansthat the iUNltR should hive a 1H seltet unJ u ZFI doublet, which confirms that choice D is a better answerthan choice C, if correct answers are in fact better than wrong answers. Choose D for the warmth and glow ofwisdom.

94.

95.

95.

97.

93.

lopyright O by The Berkeley Review@ 237 AMINO ACIDS & AMINES EXPLANATIONS

99' choice C is correct'- According to the question, the addition of base takes the amino acid from its firstequivalence point to the point at which the pH equals the value ot'-pKu3. rnis process can be observed using atitration curve, as drawn below. To carry oni thi, .or,version-lvoutarequire one and a half equivalents of strongbase' when dealing with equivalents, the conversion is best be viewed conc"pi.rutty from a titration curve. 1.5equivalents of strong base is equal to 15 mL given that the concentrations oi lysine and NaoH are equal andthere are 10 mL of the lysine solution present itritiutty. select choic" c f*;;itml experience.

pH

10.80

2.1

HrAt

PH = pKu

PH = PKaz

PH = PKug

100' Choice B is correct' This is trivial knowledge, so based on memory, you can say A and C are valid methods"Should your memory fail you, the reaction must be reduction of Ure nitrogen containing compound. LiAlHa ardHz/Pd should be familiar reducing agents. In choice B, there is notiing to provide electrons to nitrogen(reduction is the gain of electrons)'- znc has an oxidation state of +2 and Jhlorine will not give up electron*,Choice B is the best answer.

l: i.ii t i

mL 0.10 M NaOH(aq) added

Copyright @ by The Berkeley Review@ 23a AMINO ACIDS & AMINES EXPLANATI

Section VmSeparation Technigues

a) Distillationi. Vapor Pressureii. Distillation Apparatusiii. Simple Distillationiv. Fractional Distillationv. Vacuum Distitlation

b) Chromatographyi. Mobite Phase and Adsorbentsii. Thin Layer Chromatographyiii. Ri valuesiv. Column Chromatographyv. Designer Columnsvi. Gas Chromatography (OC)

c) trxtractioni. AcidlBase trxtractionii. Analyzing Flow Charts

Purifi cation Techniquega) Recrystallization

i. Solvent choice and Kefluxingii. flot and Cold Filtrationiii. Solvent Washes

Identifi cation Techniques

OrganicChemistryLaboratoryTechniques

by Todd tsennett

Mixture: Amine, Carboxylic acid, Fhenol, and Hydrocarbon

Organic solvent I Weak base(aq)

Organic layer Aqueous laYer

Organic layer Aqueous laYer a) Physical Properliesi. Melting and tsoiling Points

Chemical Testsi. Color of Phase Changesii. Limiting Reagents

Derivative Formationi. Melting Point Evidence

Mass Spectroscopyi. Base and Parent Peaksii. Fragmentation

Organic layer Aqueous layer

to"a{nontact

b)

I

YHvdrbcarbon

c)

d)

REKI{EMYl)R-.E,.v.l.E.wt

Speci aLtztng in MCAT Preparation

Laboratory TechniquesSection Goals

understand and be able to applv the different forms of chromatographrYou must be familiar *itn tn sure liauidchromatography, column chromatography, gel c"hrdmitdgraphy, unO 6"rlif.'r.tfi"t6*rr"[". Ali;fll: T:h.tgle:.111Yglvlng chromatography have basically-the s5me funcrion. They ail dip6nd on acompounds.atftnity for a mobile phase versus its affiriity for the stationary phise as if migratesdor,r"n the column.

underqtand and be able to appty the different forms of distillation.There are four forms of distillation to know: simple, fractional, steam, and vacuum. you must knowthe difference in the apparatus and set-up for'each of the four. [n addition, vo, *"rt know thepurpose.of each, the sifriation where each is best apptied, una lr'" ;;;1;;;;"ri;;i;;;""nrages ofeach of the techniques.

understand and be able to apply the different types of extraction.Extractionbasicallyinvol.ygsthgseparationofcompouna'Denavlor ln two tmmtsctble solvents. You must recognize common solvent mixfures that will b'ebiphasic, and their solubility properties. Be able to diaw a flow.nrriloittu"J.iJ **t.r.tion andacid/base extraction, which simpty alters the pH of the aqueous pn"i" l"

" ili"JaiJ extraction.

Recrystallization is applied to purify acompound is assumed'to be prirer thanwhose details vou should understand.

solid and form nice long crystals. The crystal form of athe powder form. The technique is a multi-step process

Understand and be able to lization.

Be able to deduce structural features chemical tests.Chemicaltestsinvolvereagentsthatselectivelyreactwithasma1lnumb";m,r11$ergo,either

a.ph.ase .lt.u"gg or color chanfe upon.doing so. You """a to ,nJ";;i;tiJ1h[l;;l;behind chemical tests and be able to interpretlhe iesults.

Be able to deduce structural features using mass spectroscopy.\bu must have a fundamental idea of how the. inrtrur*r"'t op*rut"s in terms of ion p.odu.ti,or., i*separation, and particle detection. You must know what a base peak and parent rleak are and beable to identify them on a typical mass spectroscopygraph. YoJ mustbe'able to'identifv tvoicalisotopic d!*tributions for chlorine, bromine, hydrolen]arid carbon. You must be able to intl'ipretgraphs Td it doing so identify not onJy the original ipecies, but also be able to identify the fragmbntsthat are formed during the process.

I

I

1l

il

rfi

u

ilfr@

Organic Chemistry Lab Techniques Introduction

once a compound has been synthesized, it must be separated, purified, andidentified. But let's be pragmatic about things. The aveiage student taking theMCAT doesn't care much about organic chemistry lab techniques, otherwis" tn"ywould be aiming for graduate school in chemistry and not -edicitt". So our goalis to present organic chemistry lab techniques in enough depth to ansia,,ermultiple choice questions, and no more. Although there are rnu.ry techniques forseparation and purification, we shall limit the techniques to distiilation,chromatography, extraction, and crystallization. once purified and isolated, nextyou must identify what you have collected. Taking melting points, boilingpoints, and doing identification tests goes only so far in identifying a compoundlIt requires much more once you start dealing with rarger molecules. But on amultiple choice exam such as the MCAT, there is a great deal of narrowing downalready done for you, As far as studying for the MCAT goes, our concern is withthe simple compounds, logical analysis, and the process of eli-itratio.,.Lab procedures are listed as a topic in the MCAT student Manual, they appear inthe. AAMC sample passages and questions, and have appeared frequently ondifferent versions of the MCAT over the past few years. We will rnr,r"y a few labtechniques, the rationale behind each technique, and their effectiveness undervarious conditions. This section is written from a perspective that assumes youdid only enough work in your organic chemistry rib courses to turn in yourlabwrite-ups and didn't spend much energy to learn the theory. The importantinformation to extract here is the theory behind each technique. For instance, theMCAT Student Manual lists boiling points and distillation aslopics, so distillationis a probable passage topic that can be coupled with theoretical boiling pointquestions. Lab techniques can be broken down into three categories' sepirition,purification, and identification.

To separate anything, one component must move away from the other, whetherit be in different directions or at different speeds. There are four separationsperformed in the lab: 1) solid from liquid, 2) solute from liquid, 3) liquid fromliquid, and 4) solid from solid. For each case, there is rnor" thutr one techniquepossible. For solid from a liquid, there is filtration or centrifugation couplea wiUrdecantation. For solute from liquid, there is distillation, precipitation coupledwith filtration, extraction, chromatography, or ion exchange. For liquid fromliquid, there is distillation. For solid from solid, there is recrystallization, densitygradient columns, molecular sieves, acid-base extraction, sublimation, or columnchromatography. we shall address some of these techniques in this section.

To purify a solid, recrystallization is the most common method. The crystallineform of a solid is the most pure. The purification of a liquid is most commonlycarried out via distillation. For purification of hygroscopic organic liquidi,distillation from a drying agent (such as magnesium sulfate) is used to rehoveyatel from the organic solvent. The drying agent binds water and the organicliquid can be boiled away free of any water azeotrope that might be pr6sentotherwise. Specific examples include aromatic solventJsuch as toluene.

once a compound has been isolated, it must be identified by techniques such asIR, UV, or NMR spectroscopy, gas chromatography, thin layer chromatography,chemical tests (reactions with reagents that identify specific functional gio"pr;,and physical properties (optical rotation, melting point, boiling point, indsolubilities). This section will address each of these techniquei

-except the

spectroscopic techniques, which were addressed in section 2.

Copyright @ by The Berkeley Review 24t Exclusive MCAT Preparation

Organic Chemistry Lab Techniques Separation Techniques

$GH uili,ffiGH e5To separate molecules requires that at least one of the molecules be in motion.We shall use the word flow to describe the motion of bulk material. The abilityto flow depends on the state of matter, where solids cannot flow, liquids flowdown, gases flow in all directions (including up), and solutes flow witha solvent.When two materials are in different phases, they can flow in different directionsand therefore can be separated. A common sense view of separation techniquesdictates that if you have a mixture of two more components, you must convertthe system into a state where the components are in diffeient phases. Forinstance, distillation can be used to separate a tiquid from another liquid byheating the system until one component boils and consequently flows Lp ut iaway from the component that remains as a liquid. w" shall look ut uttseparation techniques as an effort to get the materials to flow in different ways.

DistillationDistillation removes a liquid from either another liquid or from solute byexploiting their boiling point differences. Upon healing, the most volatilecomponent converts to a gas more readily than the less volatile components,although there is a small amount of a less volatile cornponent that vaporlzes too.In distillation, after the vapor escapes from the surface of a liquid it cin travel upthe distilling column. The vapor collides with the inner wills of the distillingcolumn, where it condenses. The amount of condensation depends on the walltemperature. Some of the condensed liquid drips into the distilling flask andsome reevaporates. Elery time the vapor condenses and reevaporates, it goesthrough a cycle of purification. The design of the distilling apparatus is such thatby the time it reaches the top of the distitling column, it hai gone through enoughpurification cycles that nearly 100% of the vapor is the more volatile .o*pon"r,I.

Vapor Pressure Diagram and Vapor RatioThe separation of two liquids via distillation is based on the vapor pressure ratioof the components in solution. The vapor pressure of a material dlpends on itsheat of vaporization, LHvaporizalior-r, and its percent composition in Jolution. AsAHvaporizatlorl increases, if is harder for the molecules to escape solution, so thevapor pressure is reduced. As the species' mole fraction in solution decreases, ittakes up a smaller percentage of the surface area, and therefore cannot have asmany molecules escape per unit area. This too reduces the vapor pressure. Toaddress this phenomenon, Figure 8-1 shows the vapor p."srrr" asi function o{temperature for two hypothetical components in a mixture of two liquids.

Vapor Ratio (More : Less)

3:1

27:1init. Sol'n 1st Condensate 2td Condensate 3td Condensate

Solution Ratio (More : less)

Figure 8-1

9:7 27 :7

-t*l

rl9:1

nl6l

Temperature

- - - - - More volatile species

Less volatile species

Copyright O by The Berkeley Review 242 The Berkeley Review


What this graph and calculation demonstrate is that over three evaporation-condensation cycles, the system becomes richer and richer in the more volatilecomponent until the ratio is 27 : 1, which equates to a mixture that has 96.4% ofthe more volatile component. This values gets larger and larger, approaching100%, with each additional evaporation-condensation cycle. As the surface areain the distilling column increases, the number of cycles increases, and thus thepurity increases. For a mixture where the ratio of vapor pressure of the morevolatile component to the vapor pressure of the less volatile component is small,more surface area is required to get good separation. This is what is referred toas fr a c ti o nal dis tillation.

Distillation ApparatusThe heat is added at the lowest point, because heat rises through the solution andhomogenizes the system through thermal circulation. As liquid at the surface ofthe solution in the distilling pot evaporates, the vapor rises up the distillingcolumn. At the top of the distilling column, the vapor can flow down the sidearm, which is enveloped by a cooling jacket, resulting in condensation of thevapor. The condensed liquid drips from the side arm into the collection flask.Figure 8-2 shows the basic components of a distillation apparatus.

The set up shows an opening between the side tube and the collection flask,which allows the system to vent any build up of pressure that takes place as theliquid vaporizes. As the vapor forms, it pushes the air that is initiatly present inthe distilling apparatus out through the venting hole. The system can be sealedand a drying tube can be fitted to the collection flask if it is iced down, so thatmoisture from the air does not enter the collection flask and condense with thedistilled product. An ice bath around the collection flask is used when thedistillate is highly volatile.

Thermometer

Heating Mantle Collection flask

Figure 8-2

Simple Distillation versus Fractional DistillationDistillation can be carried out under conditions that are referred to as eithersimple distillation or fractional distillation, depending on the amount of surfacearea in the distilling column. Fractional distillation employs a distilling columnwith more surface area than simple distillation. Simple distillation is employedwith liquids that have a large difference in boiling points, while fractionaldistillation is employed with liquids that have a small difference in boiling points(generally less than 30'C in difference, although that is not an absolute number.)Simple distillation may also be employed to remove a solvent from a solute.Simple distillation is faster than fractional distillation and it generates a higheryield, while fractional distillation leads to greater distillate purity.

Copyright @ by The Berkeiey Review 243 Exclusive MCAT Preparation


Fractional distillation is the same as simple distillation for the most part with theexception being that the distillation column is filled with an inert solid to providemore surface area on which the vapors can condense and then re-enaporaie. Thisserves to re-distill the mixture as the vapor climbs up the distillation column.Most often, the column is packed with glass beads, because glass is both inertand cheap. The column must be fitted with a trap in the inner core so that theglass beads do not fall through the column into the distillation flask. Fractionaidistillation allows the liquids to be separated more efficiently and thus in a purerfashion. The apparatus in Figure 8-2 shows the set up for simple distillition,because the distillation column is empty.

Example 8.1

which of the following mixtures can be separated using simple distillation?A. Alpha-D-glucopyranose frombeta-D-glucopyranoseB. Butanol from diethyl etherC. 1-Pentanol from 2-methyl-1-butanolD. Acetone from propanal

SolutionSimple distillation separates a solvent from a solute or a liquid from a liquidwhen they have a large difference in boiling point. The largest difference inboiling point is found with choice B, an alcohol and ether of equal mass. choiceA is two diastereomeric solids and choices c and D are of two liquids withroughly equal boiling points. They cannot be separated by simple distillation.

Example 8.2which of the following materials could be used to pack the column in thefractional distillation of acetic acid from trichioroacetic acid?

A. Paraffin waxB. Magnesium turningsC. Aluminum meshD. Glass beads

SolutionThe material chosen must provide surface area, remain a solid at elevatectemperatures, and be inert. Paraffin wax has a iow melting point, so choice A iseliminated, because the wax will melt and drop into the distillation flask. Meta-can oxidize in the presence of acids, resulting in metal oxides that can neutralizethe acids, so choices B and C are eliminated. The best answer is choice Dbecause the glass beads are inert and will remain solid at the elevatectemperatures used in distillation.

Vacuum DistillationVacuum distillation involves attaching the distillation apparatus to a vacuur.pump at a point after the condensate has been collected. The purpose is lcreduce the pressure in the apparatus and in doing so lower the boiling poin:You should recall that the boiling point is defined as the temperature at whicrthe vapor pressure equals the atmospheric pressure, so lowering atmosphen:pressure lowers the vapor pressure required to boil, which results in a lorr'e:temperature needed to boil. This is done when the compounds have a hig:.boiling point, higher than the flash point or decomposition point.

Copyright @ by The Berkeley Review 244 The Berkeley Relier*


ChromatographyChromatography is the separation of two or more components in a mixture byexploiting their differences in solubility in a migrating solvent and their affinityfor a polymer. Both factors play a significant role in the separation process, sothey must both be considered. The various chromatography techniques are allcentered around a sort of swimming race between the components in the mixtureto some arbitrary finish line. What is meant by "race to the finish" is thatdifferent components exhibit different rates of migration across the surface of astationary phase, referred to as the adsorbent, depending on their solubility in themobile phase, (the solvent), and their affinity for the stationary phase. The solventis migrating through the adsorbent and the components are traveling with thesolvent to a varying extent. As the solvent moves, the components are towedalong with it in most cases. The more soluble the component is in the solvent,the further it is pulled by the solvent through the adsorbent. Likewise, thegreater the affinity for the polymer, the slower the component migrates.

Mobile Phase (Solvent) Properties versus Stationary Phase PropertiesBecause separation in chromatography depends on relative affinities, we shouldconsider the nature of typical materials. The mobile phase is one of severalpossible solvents, ranging from nonpolar (hexane) to polar/protic (ethanol), or insome instances a mixture of two solvents. When mixtures are used in columnchromatography, it usually starts with one pure solvent followed by differentmixtures which gradually become richer in a second solvent. Typical solventpairs include ether-hexane and methanol-methylene chloride.

The stationary phase is one of two materials, alumina or silica gel. Both of thesepolymers are polar, so in the absence of any other information, we know thatpolar materials have a higher affinity for the stationary phase than nonpolarmaterials. As a consequence, polar species have slower migration rates thannonpolar species in chromatography. How much slower depends on thematerial and solvent. In addition, the activity of a polymer can be affected by thesolvent, where the solvent binds the silica gel or alumina and thereby reduces theinteraction of the migrating solute with the polymer. For instance, as aluminagains moisture, it is less able to bind other materials, so it is said to be less active.Alumina comes with a Brockmann activity rating from I to V, where V representshighly moist alumina which exhibits low activity. Anhydrous alumina is morepolar than anhydrous silica gel.

The only thing that can be said with certainty in chromatography is that polarspecies in nonpolar solvents have slow migration rates while nonpolar species innonpolar solvents have a fast migration rate. The questions on the MCAT willlikely stress this point, where you can simply refer to the idea "like dissolves like"to answer them.

Thin Layer Chromatography (TLC)Thin layer chromatography, TLC, is carried out on a small scale to identify thenumber of components or type of components in a mixture. It involves spottingsmall aliquots of sample in a line parallel to the base of the plate near the bottomof a vertical, rectangular plate with either silica gel or alumina on its surface.This plate is then placed into a container that can be fitted with a lid and soiventis added to the container until it is just below the levei of the spots. It isimportant to not have the solvent level in the container above the spots,otherwise they can dissolve into solution. Due to capillary action, the solventslowly migrates up the plate, interacting with each of the spots as it travels. Inorder to prevent the solvent from evaporating from the surface of the plate, a lidis employed to maintain a closed system. Some chemists place a piece of filter



paper on the inside wall of the jar, not touching the plate, to provide additionalsurface area from which the solvent can evaporate. This keeps the chambersaturated with vapor. once the solvent is neariy to the top of the plate, the plateis removed from the container and the top of the solvent front is marked ,rrir,g upencil.

The plate can be deaeloped in one of several ways, such as spraying ninhydrin onthe surface, adding it to a closed chamber with iodine ..yrtalr, or shining UVlight on the plate. In some way, the spots must be made visible for evaluation.Once it is determined where the spots ended, one can ascertain such informationas how many components were in the mixture and whether the componentswere nonpolar, semi-polar, or highly polar. By initially spotting a plate with aknow. material you suspect to be the unknowr .o-po.r'rd -in

ihe originalsample, the spots of the unknown composition can be compared to the knlu,nspot to determine identity of one of the unknown compounds in the originalsample. A TLC set trp is shown in Figure g-3.

Thin Layer ChromatographyLid-----+

TLC plate

lar

(with siiica gel or alumina)

Solvent top

Sample(spotted onto plate)

Example 8.3Thin layer chromatography can be used to do all of the following EXCEpT:A. determine the number of components in a mixture.B. distinguish the relative abundance of enantiomers.C. determine the best solvent for column chromatography.D. monitor the progress of a reaction.

SolutionTLC separates the components in a mixture by their relative affinities for thesolvent and the adsorbent. As a result, the number of components in the mixturecan be ascertained from the number of spots on the plate. This eliminates choiceA. Because column chromatography uses the same solvent and adsorbent asTLC, the results of TLC can serve as a preview of column chromatography, andthus be used to test potential solvents. This eliminates choice c. Samples of areaction can be taken and analyzed by TLC to determine the appearance ordisappearance of either a reactant or product. This eliminates choice D. There isno quantitative capability with TLC and enantiomers travel at the same rate, so

-heycann", r" t"t"t"t"a,*


Organic Chemistn"y Lab Techniques Separation Techniques

Example 8.4If the sample on the plate were submerged below the top of the solvent, then thesample would:A. migrate more rapidly than expected.B. dissolve into solution rather than climb the plate.C. react with the plate.D. cause the plate to dissolve into solution.

SolutionIf the sampie on the piate is placed into a flask at a level that is lower than the topof the solvent, then the cornpound is free to dissolve into the solution, and thus itwill not migrate up the plate. This eliminates choice A and makes choice ts thebest answer. The plate has an inert adsorbent, so no reaction takes piace. Thiseliminates choice C. The plate does not dissolve into solvent, and even if it did,the height of the plate wouldn't matter. Choice D is eliminated.

R1 ValuesIn TLC, we often measure an R1 value for the compounds. We can think of an ftgvalue as a "ratio of the fronts," where the distance a spot travels is compared tothe distance that the solvent travels. The distance that the solvent travels ismeasured from the line on which the spots start, where the solvent first interactswith the sample, to the line where the solvent front finishes. The distance thespot traveis is measured from the center of the spot when it starts to the center ofthe spot when it ftnishes. Some spots deform and spread out as they climb, so itis best to use the center of the spot as reference. The distance the componenttravels is divided by the distance the solvent travels to yield its Rl value. Thecomponent with the greatest R1 value is the component that has the least affinityfor the stationary phase (adsorbent) and is most soluble in that solvent. Therange for R; vah-res is: 0 < R; < 1, because there is a chance the spot does not travelat ali or that it travels almost at the solvent front. The R1 value is a physicalconstant associated with a solute,/solvent/adsorbent combination. The basicidea of all forms of chromatography is separation by relative affinity.

Example E.5

What is NOT true about the Rlvalue for a compound?

A. It cannot be greater than 1.0"

B. It has no units.C. It is different in different solvents.D. it is always the same on the same adsorbent.

SolutionThe spot cannot travel faster or farther than the solrrent carrying it, so the llivalue cannot be greater than 1.0, eliminating choice A. Because it is calculated bydividing a distance by a distance, the units cancel out. This means that R1 valueshave no units, which eliminates choice B. Because solubility is different in eachsolvent, the impact of each solvent varies from solvent to solvent. As a result, theR1 vaiue differs with each solvent, which eliminates choice C. Because themigration rate depends on the affinity of the compound for both the adsorbentand the solvent, the distance it travels in a given amount of time depends on theadsorbent and solvent. The R1 value varies with adsorbent, making choice D nottrue about the It1 value for a compound and therefore the best answer.

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Figure 8-4 shows the TLC plates at the start and after completion of the migrationand development with iodine crystals for two separate Tr,C experiments. In onecase the solvent is nonpolar while in the other case, the solvent is polar. Thenature of the compound can be inferred from the relative R6 values obiained for aseries of solvents. Because a solute dissolves best into a solvent of "like" nature,the higher the value of Ri, the more that the solute is like that solvent.

ilBefore

Solvent frontstopped

dsolvent

- d.po,

dsolvent

Solvent frontstopped

dsolte.t

o'=*

,dspot

AfterPolar Solvent

Figure 8-4

Based on R1 values, you should be able to make predictions as to the best solventfor column chromatography. The best separation occurs when the R6 valueshave the greatest multiplicative (not additive) difference. For instance R6 valuesof 0.2 and 0.1 separate better than Rlvalues of 0.3 and 0.2.

Column ChromatographyColumn chromatography is applied to separate bulk quantities of product. Itoperates on the same principles as thin iayer chromatography: the solubility of acompound in the solvent as it flows down the column versus the affinity of thecompound for the adsorbent in the column. If the compound migrates quicklydown the column, we can conclude that it must experience a minimum attiactionto the silica gel or alumina (stationary phase) in the column. In columnchromatography, the "race to the finish line" is to the bottom of the column andthe components in the mixture separate according to how quickly they finish.To set up a column, it is first packed with a plug (glass wool works well)followed by a thin layer of sand (to form a flat surface), then a large layer of silicagel, followed again by a thin layer of sand, to protect the top of the silica gel fromdisruptions in its flatness. The top and bottom of the silica gel must be flat andparallel to one another, so that components all travel the exact same distancethrough the silica gel no matter where they start. we assume that thecompounds have no affinity for the sand or glass. solvent is added to thecolumn to saturate it. Once the top of the solvent is flush with the top of thesand, the sample can be added.

The sample is dissolved into minimal solvent, which is poured gently into thetop of the column. The solvent slowly flows down the silica filled glass columnto an opening at the bottom, taking the compounds at different rates as it goes.Samples of the solution are then collected in increments that are analyzed foi anvcomponents tirey may contain. Faster migrating compounds reach the bottom ofthe column first, so they have shorter elution times. EIuion time is definecl as thetime it takes a component to travel the length of the column and drip out. Likethe Ri value, elution time is a measurement from the experiment. A large R+value corresponds to a low elution time, given that the compound has i fastmigration rate in each case. Each aliquot of solvent collected from the column L.analyzed by sorne technique (usually spotting on a TLC plate and then eitherillumination with a UV lamp or treatment with 12 crystals) for the presence of

:I(

J

T

(I

tud,l@

IM

difr6

AfterNonpoiar Solvent

Copyright @ by The Berkeley Revierv 24a The Berkeley Revien


components. The aliquots containing the component are consolidated and thesolvent is driven off via evaporation. Figure 8-5 shows a column at differentstages, where the bands separate as they migrate down the column.

Solvent is added so itwill flolv down the column

Sand layer

Silica gel

Sand layer

t=0

Less soluble componentMore soluble component

f-_-t

hFFrC[;:':':;:':':']t-..._.-._.-.-l

[.'i......',..j[.:.:.:.:.:.:.]l:.:.:.:.:.:.:{ltttrIruilrl

[.:::':t:::.:i]

[:::::::::::::][.:.:.:.:.:.:-lt.:.:.:.:.:.:.1

ry6

,_ /

f---lhF#t.:.:.;.:.:.:'tL.-.-.-".._'_.1t..-.-..'....-t

Iii,:.:ii.:':.1t-....'.._...-tlrIlt[ililr4

[;,,i,i'i'i,i'il:.:.:.:':.:.:lt.............tl[[ilil[ilra

[::i:i:::::i:ijl.:.:.:.:.:.:.1l:.:.:.:.:.:.:{t-'-'........,t

-t|6

t=3

Second componentstill migrating

First component hasreached the bottom

6t-?

Figure 8-5

Example 8.6Column chromatography relies on which force to separate compounds?A. GravityB. Gas flowC. Magnetic attractionD. Centrifugal

SolutionThe solvent migrates down the column, because gravity is pulting the solventdown. The compounds are pulled along by the solvent as it migrates down thecolumn. The driving force behind the overall flow of solution is gravity. Thismakes choice A the best answer. Gas flow is associated with gaschromatography, not column chromatography, so choice B is eliminated. Thereis no magnetic field or circular motion, so choices C and D cannot be true. youshould make the correct choice and select choice A.

Designer Polymers (Chiral Columns, DEAE-Cellulose, and Affinity Columns)Given that the migration rate through a column depends on the affinity of acompound for the polyrner, there are some columns designed with specificmolecules in mind. while the application of such columns is more common irrbiochernistry than organic chemistry, the basic idea is still rooted in thefundamentals of coiumn chromatography. An extreme case of designing apolymer is observed in biochemistry, where an antigen column is used to bindand isoiate antibodies from a solution. one problem encountered in such anexperiment involves horv to release the antibody from the column. onceeverything else has eluted from the column, the column can be treated with a

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varying pH buffer until the antigen and antibody denature enough to no longerbind one another. one problem with changing the pFi of the column is thechance that the proteins are irreversibly denatured. In such cases, the antibodymay instead be released by adding a salt solution of gradually increasing ionconcentration or by adding free antigen to compete with the bound antigen forthe antibodies.

In organic chemistry, chirsl columns have become popular in recent years,because they are capable of selecting for one enantiomer over another. If thedesired product is a compound with a hydroxyl group on a stereogenic carbonwith S-chirality, then a column with R-alcohois on the polymer will bind andthereby hinder the flow of the S-enantiomer through the column. As a result, theS-enantiomer has a greater elution time than the R-enantiomer, so they separateas they traverse the length of the column. The fundamental difference betweencolumn chromatography in biochemistry and organic chemistry is that in organicchemistry, the aim is to have all of the compounds elute at different times wherein biochemistry the goal is to have all but one compound elute.

The most common designer polymers on the MCAT to date are the adsorbentsused in ion exchange chromatography. Three with which you should be familiarare sulfonated-polystyrene, carboxymethyl-cellulose (CM-cellulose), anddiethylaminoethoxy-cellulose (DEAE-cellulose). The first two carry negativecharges at neutral pH, while DEAE-cellulose carries a positive charge at neutralpH. Proteins with high pI 'u'alues, ones that are rich in lysine, arginine, orhistidine, will bind CM-cellulose columns, ailowing them to be isolated. Proteinswith iow pI values, ones that are rich in aspartic acid or glutamic acid, will bindDEAE-celluiose columns, allowing them to be isolated. Typicaily withsulfonated-polystyrene columns, the protein mixture is added at low pH so all ofthe amino acids and small peptide residues bind the column. The pH of thecoiumn is gradually increased, releasing amino acids according to increasing pIvalues. Figure 8-6 shows DEAE-cellulose and sulfonated-polystyrene.

DEAE-Cellulose

H-O"t'o

CH,t-o

OH

Figure 8-6

CHtt-o

OH OH

o

OH OH

Sulf onated-polystyrene

i-=\.4

Copyright O by Ti-re Berkeley Review 250

OH

The Berkeley R.eview L--:-

Organic Chemistry Lab Techniques Separation Technigues

Gas Chromatography ApparatusCas chromatography, GC, works by first vaporizing a sample (mixture o{components) into the gas phase and then forcing that organic ,rupo, through apacked column using elevated pressure from a ryti.ra"t o"f an ir-,eit gas lusriaityhelium and sometimes argon). The machine measures retention time on thecolumn by recording collisions as gas molecules leave the end of the long, coiledcolumn and strike the detector. The collisions are converted to peaks (like a bargraph), which can be integrated to determine their relative abundance. A greaterarea for the signal indicates a greater quantity of material presentl Thecompounds in the samplemust be highly volatile at 200"C and inert with regardsto silica gel or wax for GC analysis to work. Figure g-7 shows the design"for astandard gas chromatography machine and a standard chart output for a mixtureof three components.

Injection port,sample may be

through which a

added by syringe

Packedcolumn

Carrier.gas is added ",,nt, oJ^,. rn. r"r,*H

gas at this point forces the vaporized sample tir".i!n the packed coiledcolumn and into the dete-ctor. The detectbr reads Iolisions. The chartgraphs intensity of collisions

"eri"J-iim". *=14.,iJ*;";;, -ih;i';;

identification can be obtained directly from gas chromatography. Theretention time depends on the nature of the co]umn and the"compound.

Figure 8-7

The relative quantities of the components can be determined by integrating thesignals associated with the compounds. GC output is continuorrr, ,oih" gr"aphsare actually read from right to left, where Jignals on the right rep"resentcomponents with smaller eiution times than components on the left. Todetermine the purity of a product mixture (and relative abundance ofcomponents), the graph can be read directly. The purity of the product mixturecan be observed directly in the total number of peaks that are ,"L. o., the graph.To ascertain information about one of the components in a mixture, u- p.r.usample of that component can be added to a gas chromatography apparatus andthe retention time for that particular component can

-be determined. Analternative way to identify a component involves the addition of a small sampleof that compound to the mixture to enrich the sample in that component. Thatcomponent corresponds to the signal that increases its relative size. Thistechnique is referred to as spikirzg the sample with a component. Gaschromatography is used as both an identification tool for the reaction mixtureand" as a tooi for measuring the kinetics of a reaction. By measuring the samplemixture at given intervals, the growth of a peak can be monitored lo determinethe rate of formation of a product. Figure g-g shows a typical graph obtainedfrom the GC apparatus. GC peaks are relatively sharp as the graph rho*r.

Copyright @ by The Berkeley Review 251 Dxclusive MCAT Preparation


Retention time

ttt*

Example 8.7Gas chromatography does NOT work with:A. alcohols.B. aldehydes.C. ketones.D. fatty acids.

SolutionIn this case, the component that cannot work is the component that cannotvaporize in the gas chromatographer. The temperature of the vaporizingchamber is often higher than 200"C, a high enough temperature to vaporize mostorganic liquids, but a fatty acid cannot vaporize at that temperature. The fattyacid is likely to melt and decompose at this temperature, so it cannot beidentified using gas chromatography techniques. Choice D is the best answer.

Analysis of GC DataThe MCAT is apt to ask data analysis questions rather than details of operationfor the GC apparatus. It is imperative that you be able to interpret graphs suchas the one in Figure 8-8. GC analysis focuses on quantitative aspects of themixture rather than specifics about the structures. Given that the compoundsvaporize instantaneously upon addition to a GC machine, the boiling point doesnot affect the elution time. However, because many columns are often slightlypolar, affinity for polar species is greater than affinity for nonpolar species. Asecond factor that dictates the migration rate is mass. Fleavier gases move slowerand have longer elution times. Heavier gases and polar gases typically have highboiling points, so although boiling point has nothing to do with elution time, thefactors that raise boiling point also increase the elution time in a GC. This is whymany chemists say that molecules with lower boiling points come off of a GCcolumn first.

ExtractionExtraction works based on solubility and it is typically employed to takeadvantage of drastic differences in the solubilities of components in two different(immiscible) solvents. To start, we first need to define partitioning. A compoundhas a different solubility in every solvent. The ratio of its maximum solubility inone solvent compared to its maximum solubility in another solvent is known as

the partition coefficient. In cases where the compound can go into either solvent, itwill select between the solvents in a ratio equal to the partition coefficient. Whenthe solvent system involves two immiscible liquids, then the solute must split



between the two solvents. some components wil dissolve primarily into one ofthe solvents, while the remaini.g "o*po.,ents

will dissolve primarily into theother solvent. This means. that th-e components of a mixture &n be separated ifthey have differing partition coefficienis. By decanting one solution from theother,_the components are then separated from one ur,oih", in their solute formaccording to their relative solubilities in the two different solvents. From thispoint, you must remove the component from the solvent in which it is nowdissolved. This can be done via disiillation or controlled evaporation.

Extraction does not always involve two immiscible solvents, although that ismost often the case. A sotid component can be removed from a mixture of solidsby adding a solvent r: whif it is highly soluble and the other compounds arerelatively insoluble. This is how many nitural products are isolated from plants.

Acid-Base ExtractionV\4ren one of the two immiscible solvents is water, the properties of the solutescan change when the solution pH changes. Acid-base extraction in essenceseparates compounds by their pKu values, By using a biphasic solvent mixture(heterogeneous system),-selective protonation or dJprotonation (depending onthe compound) will either increise or decrease tire water soiuuitity of"thecompound, depending on whether an ion is generated. Loss of charge results inan increase in preference for the organic layeiwhile a gain of charge results in anincrease in the preference for the iqn"o.,s layer. onJe extracted into the waterlayer, the species can be returned to a neutral state by adjusting the pH, and inmost cases it will crash out of solution. Figure B-9 showi the f]owchart for theacid-base extraction of a mixture of n-iutylamine (CH3CH2CH2CH2NH2),butanoic acid (CH3Cltrr2CH2CO2H), and ethyl benzene 1bU3ICH2C6H5;.

Mixture: Butylamine, Butanoic acid, Ethylbenzene

Layer I (organic) Layer II (aqueous)

toz lHcrluq;

IButanoic acidLayer IV (aqueous)

rozlNaoHlaql

iButylamine

Figure 8-9

In the first extraction, aqueous sodium hydroxide solution deprotonates butanoicacid to form butanoate, a carboxylate anion (which is highly water soluble due toits negative charge). Brrtylamine and ethylbenzene rerialn in the organic layer.The butanoic acid can be "crashed out" oi solution by adding acid tJregeneiatethe carboxylic acid. The carboxylic acid form is .,ot .r"ry ivater soluble, so itprecipitates out of the water. Butyl amine is removed from the ether layer byextracting it with acidic water, which forms butyl ammonium cation, u i,ignrywater soluble species. Butyl ammonium cation can be "crashed out" of solutionby adding strong base to the aqueous rayer. The ether layer contains theremaining ethylbenzene, which can be isolated by evaporating tire ether away.

Layer III (organic)

Ievapprate

tEthylbenzene

50.0 mL ether 50.0 mL 10% NaOH(aq)

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Example 8.8To separate benzoic acid (C6H5CO2H) from toluene (C6H5CH3), it would be bestto use a mixture of an:

A. alcohol and water at pH = 11.

B. alcohol and water at pH = 3.

C. ether and water at pH = 11.

D. ether and water at pH = 3.

SolutionTo carry out an extraction, the two solvents cannot be miscible in one another.Alcohols are typically miscible in water, so choices A and B are eliminated.Benzoic acid can be deprotonated in basic solution, where it becomes its anionicconjugate base (benzoate) which is readily soluble in water. Toluene dissolvesinto the ether rather than water, so the best answer is choice C.

more soluble inether than water

more soluble inwater than ether

insolublein water

Separatory Funnel and Partitioning into the Biphasic LayerThe physical act of separation is most often carried out in a separatory funnel. Aseparatory funnel is an upside-down, pear-shaped flask with a fritted glass capon top and a stopcock controlled valve on the bottom. The two layers naturallr'separate, with the less dense layer assuming the top. The system is mixed andshaken to ensure complete interaction of the two layers. One thing to note is thatthe aqueous layer has a meniscus, so the layer with the most contact against theglass walls is the aqueous layer. To increase the differences between the tn olayers, the aqueous layer is often saturated with salt. Such a solution is referrecto as a brine solution.

Copyright @ by The Berkeley Review 254 The Berkeley Revierr


Flow Chart with an Ether, Amine, phenol, and Carboxylic AcidOn the MCAT, perhaps the most difficult task when it ctmes to extraction will beanalyzing a flow chart. Due to the time limitations of the exam, the chart cannotbe too exhaustive, so we shall consider a four-compound mixtrrre. Figure g-10shows a typical four-compound flow chart. It is a gtod general rule that a weakbase should be added beiore- a strong base, so as to isolite the stronger acid. Ifyou were to add the sodium hydroxide first, it would extract both the"phenoxideand the carboxylate into the aqueous layer.

NHC

d

distill awayfrom residue

dH3

HeC

cH2cH3

d"lacidify andI filtertcH2cH3

d""

cH2cH3

0.10 M NaHCO3(at)

",'d'd",

Figure 8-10

Special Cases including Esters, Anhydrides, and Acid HalidesAs is true with all lab techniques, the conditions must be inert for the techniqueto be effective. Because extraction involves both acidic water and basic water,there is the potential for hydrolysis. As such, extraction cannot be used to isolatewater-sensitive compounds such as acid anhydrides, acid halides, and esters.These compounds can be extracted from a pu$ into an organic sorvent, but theycannot undergo acid-base extraction.

y"lacidify and

{ruter

#""

0.10 M NaOH(aq)

Copyright O by The Berkeley Review 255 Exclusive MCAT preparation

Organic Chemistry Lab Techniques Purification Techniques

F tdbhilifi$d5Manl of the separation techniques we have discussed so far can also be used aspurification techniques. Column chromatography is useful for purification,because the bands, separates nicely from one another. Depending on thecompounds, the isolation of a few grams of highty pure material is common in:?ty1- chromatography. Distillation, when ao"neLultiple times, can generate ahighly pure liquid. But rather than describe chromatography and distillationagain with the attention to detail that makes fot pntili.aiion, we shall onlr-discuss, recrystallization. your number one goal is to zero ir-, on topics ,n"i n"r.a good probability of showing up on the MCAT, so we shall considerrecrystallization in the purification section.

Recrystallization of a Crude product MixtureA solid product may be separated from solid impurities through selecti'eprecipitation, which organic chemists refer io as recryslallizotion.Recrystallization involves dissolvjng the solid into hot solvent, htt filtering outthe insoluble solid impurities, and then slowly cooling the solution to preciiitatethe purified crystals. By adding a minimal amount of-solvent in which the soluteis saturated at high temperature but insoluble at a lower temperature, one ma'first dissolve a product mixture leaving solid impurities in ,olltior-, i, f

" ?iit"r"aout. upon cooling the solution, the product precipitates out of solution in a

crystalline form free of the impurities from be?ore. The crystals are filtered atlow temperature, to prevent any of the material from dissolving back intosolution, to avoid any reduction in the yield.

solvent Choice (Dissolving sorute into Minimar Refluxing solvent)In recrystallization, solvent choice is essential. Because the Jolid is dissolved intorefluxing soivent, it is ideal that the desired material is highly soluble at elevatedtemperatures. The term refluxing describes a system" where the solvent isboiling, but it is condensing on the walls of the fiask and dripping back intosolution. You may recall that acid reflux is a condition where your gastric fluids"Ji-P

up your esophagus only to fall back into your stomach. Reflux is theclimbing and returning process. In order to maximize the amount of crystalcollected, the desired material should be insoluble (or minimally soluble) in thesolvent at lower temperatures. The solvent can be a homogeneous mixture oftwo solvents to meet desired solubility properties. ny mixing solvents, it ispo-ss_ible to change the overall polarity of the solvent and thus chinge the overallsolubility of the solute at both high temperatures and low temperatures.

Example 8.9if a solid is too soluble in water at room temperature, it would be best to addwhich of the following solvents?

A. HexaneB. EthanolC. Diethyl etherD. Tetrahyclrofuran

SolutionThe key to this question is that the solvent you mix with water must reduce theov-e1a,ll polarity and hydrogen bonding capacity of the solvent mixture and besoluble in water. All of the choices areless polar and less protic than water, butof the choices, only ethanol is soruble in water. Choice s is ihe best answer.

I;

-"!

ffi

IL

5_

t-

!

5

:::_l

- -,rvright @ by The Berkeley Review 256 The Berkeley Review

Organic Chemistry Lab Techniques Purifi cation Techniques

Decolorizing (Adding Charcoal to solution to Bind Colored Impurities)In some instances, the purification can be enhanced by adding a material topreferentially bind some or a1l of the impurities. one such .ur" oJ".r6 when youh.ave colored impurities and a colorleis target crystal. By adding activatedcharcoal, charcoal that is free of impurities, the colored species will bepreferentially bound by the surface of the charcoal and can thus be filtered out.Charcoal binds all organic materials, but because of its n-network and thepl-esencg of conjugated n-networks in colored species, charcoal has a higheraffinity for colored organic molecules than colorlesi organic molecules.

Hot Filtering (to Remove Insoluble Impurities, i.e., boiling chips)Filtration is employed to remove solids from liquids. It is i laboratory techniquepatterned after everyday activities, such as collecting spaghetti in a colanderwhile running water over it. In organic chemistry, a fine mi&oscopic net is usedto separate the solid from the liquid by allowing the tiquid to flow through whilehindering the flow of solids, which are larger than the pore size of the filter. Thesolids are thus collected by the filter. Filtration can also be used to separate largesolid particles from small solid particles by setting up two filtering sites ifdifferent size. The first filtration shouid be of a largei pore size than the secondfiltration so that the larger solid may be collected firit and thereby separatedfrom the smaller solid, which will then be collected in the second fiiter. This issometimes referred to as sieae filtering.

Filtering typically depends on gravity to force the liquid to flow through thefilter, although it can be expedited by a pressure difference between theatmosphere above the filter and the atmosphere below the filter. Gravityfiltration allows large insoluble solid particles (crystals) to be removed from thesolvent (mother liquor). As the solution flowi through the filter, the solidparticles are collected in the pores of the filter. Vacuum iiltration uses a filter inconjunction with a vacuum, so that the liquid flows faster. As with all types offiltration, insoluble solid particl"s ate .errroved from a solvent. The liquii flowsthrough the filter due to both gravity and the pressure difference between above

^"a t"t"- ,^" tttr.

Example 8.10which of the foilowing mixtures can be separated using gravity filtration?A. Ethanol from diethyl ether.B. Sodium chloride from water.C. Butane from dichloromethane.D. Oxalic acid from hexane.

SolutionIn this question, you must rely on experience and memory. In choice A, bothcompounds are liquids, so filtering is pointless. In choice B, the salt is completelysoluble in water, so the salt is actually in solute form. It must be a solid, notsolute, to be filtered, so choice B is eliminated. Butane is a liquid as isdichloromethane, as you may recall from lab experiences. This eliminites choiceC, ieaving choice D as the oniy remaining choice, so it better be correct. Hexaneis a liquid and all carboxylic acids, besides formic acid and acetic acid, are solidsat room temperature. Acids do not readily dissolve into hydrocarbon solvents,so oxalic acid exists as a solid and not as a solute in hexane. choice D iscomprised of a solid in a liquid, so it is the best answer.


Organic Chemistry Lab Techniques Purification Techniques

Crystallization, Cold Filtering (to Collect Crystals), and Solvent WashThe final step in a recrystallization procedure is to collect and isolate the targetcrystals. If the solution is slowly cooled from its refluxing state, it will precipitatelong, pure crystals. These crystals are collected using filtration, but with coldsolvent, so as to prevent the loss of product to dissolving. The crystals at thrspoint still have a residue of solvent with soluble impurities on their surface. Thecrystals must lose this "dirty solvent" to be pure. The dirty solvent can beremoved by washing the crystals with a solvent in which the impurities aresoluble, but the target compound is not. It is optimal to use a solvent with a lortboiling point, so it can readily evaporate away from the solid. The rinsingsolvent dissolves the impurities as it passes across the crystals. A commonexample is the removal of water from glass by rinsing the glass with acetone.You no doubt have done this in lab a thousand times (okay, maybe only three orfour times, but 1000 is so much more impressive). The acetone does not dissoh-ethe glass, but it does dissolve the water and rinses it away. The residual amountof acetone left behind on the glass can evaporate away leaving behind drv(anhydrous) glassware. solvent washes are often carried out on the soliccollected from filtration. It's similar to use rinsing off with water once we havelathered up in the shower. If we didn't rinse the lather away, the water wouldevaporate and leave a soap film on our skin. We opt to rinse in a pure, volatilesolvent in which we are insoluble, at least most of us do.

Example 8.11The best choice with which to rinse sodium acetate in order to remove propano-residue is:

A. hydrochloric acid.B. propanol.C. diethyl ether.D. water.

SolutionThe goal of rinsing a product is to remove the dirty solvent (in this casepropanol) from the surface of the desired product (in this case H3CCO2Nawithout reacting with the product nor dissolving the product. Choice A i;eliminated, because HCI will protonate the sodium acetate to form acetic acicChoice B is out, because you cannot add propanol to remove propanol (tha:would be the same as washing a dirty shirt in mud to remove the dirt). Choice Dis eliminated, because sodium acetate can dissolve into the water. The bes:choice is the only choice left, choice C. Diethyl ether will dissolve the propanc-without dissolving the sodium acetate. The propanol can be rinsed away u'if.the ether and only residual ether will be left behind in the sodium acetate. Th.e

ether can easily evaporate away from the sodium acetate crystals leaving behintpure sodium acetate.

E

T],

AsCD

b|:

-J$iir

a

lfftg

LiCopyright O by The Berkeley Review 2Sa The Berkeley Revien

Organic Chemistry Lab Techniques Identification Techniques

I

I

KThe last aspect of lab we shall consider are the identification techniques. once acompound is isolated, it is important to be able to characterize the material.compounds are most easily idfntified by physical properties. IR, uv-visible,and NMR spectroscopl hiyl arready r"ur, alr"Lrsed, so we shall consider mostlychemical diagnostics in this sectioi. nf-rtu-,gerprinting .o-porr'a, with theirphysical properties, it is possibre to distinguish or," frorriur,other. Chemicar testsfor specific functional groups car, urro"b" employed. In some instances, acompound can be converted into a new compound (ierivative), and the physicalproperties of the derivative can be used to help identify the or(inat compound.

Physical PropertiesIt is common to verify a compound or ascertain its purity from its melting point.A rapid change from a solid into a liquid (occurring orrJ, t*o J"grees or less) isindicative of a pure compound. sharp boiling po"int, ur" ,",oi as obvious. Toidentify a solid, one can mix an unknown with the compound they suspect it is.A,sharp melting point confirms that the unknown is in ti"t tnuicompound, whilea broad melting range indicates that the compound is something else. This is themixed melting point tesi. The name comes from mixing the unknown and knownto,f:tr a homogeneous mixture._ other physicar properties to compare includesolubilities and optical activity. \Arhatevir physical diff"r"r,." io*porr.d, *uyhave can be exploited to distinguish one unknown from either a second:19:f or a

:lmfarabte standaid "o*po,r.,d (known). Table 8_1 shows rhe,t.;;il;i;

Substance Density (g/ml) Melting Point Boiling PointAcetone 0.79 -95'C 56'CBenzophenone 1.1s 48'C 306'CChloroform 1.49 -64'C 61.'CCyclohexane 0.78 6'C 81"CMethanol 0.79 -98'C 65'CNaphthalene 1.15 94'C 288'CStearic acid 0.85 70'c 297'C

Table 8-1

Example 8.12According to the data in Table g-1, which of the foilowing compo'nds is a solidat room temperature and sinks to the bottom of the frask #n"', uaa"a to water?A. CyclohexaneB. ChloroformC. NaphthaleneD. Stearic acid

SolutionThe compound is insoluble in water and d.enser than 1.00 grams/mL,because itsinks, and doesn't dissolve, in water. Incorrect density er#inates choices A andD' Chloroform has a merting point of -64'C, soit is a tiquid at room temperature.This eliminates choice B. Niphtharene, choice C, is a ,oua ut room temperature,because it has a melting poinf of 94'C. Choice C is the best answer.

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Example 8.13If a student suspects that the compound they have is L-tyrosine, what shouldthey add to the sample for a mixed melting point?A. L-TyrosineB. D-TyrosineC. XyloseD. Glycine

SolutionTo support the notion that an unknown compound is L-tyrosine using a mixedmelting point study requires mixing the unknown with a pure sample of L-tyrosine. If the unknown compound is in fact L-tyrosine, then the mixture is infact pure L-tyrosine. This would result in a sharp melting point at 114'c, themelting point for L-tyrosine. If the unknown compound were not L-tyrosine, themelting point range would be broad and would fall outside of 114"C. The bestanswer is choice A. D-tyrosine, although having the same melting point as L-tyrosine, won't work, because the enantiomers interact with each otherdifferently than they do with themselves. Pure enantiomers typically do notpack into crystals as well as racemic mixtures, so they have lower melting points.

Example 8.14A melting point range from 31.5"C to 38.5"C supports which conclusion?A. The compound is very pure.B. The compound is highly reactive at low temperatures.C. The compound will boil at a temperature less than 138.5'C.D. The compound has impurities.

SolutionThe melting point range does not indicate anything about the reactivity of a

compound. As such, no conclusion can be made about the compound =

reactivity, so choice B is eliminated. Knowing a compound's melting point doesnot give any hint as to its boiling point. This eliminates choice C. A range of 7'Cis broad in terms of melting point, and a broad melting point range is associatedwith an impure compound. The only reasonable answer is choice D.

Example 8.15To distinguish D-leucine from L-leucine, it is best to compare their:A. specific rotations.B. solubility in ethanol.C. melting points.D. odors.

SolutionD-leucine and L-leucine are enantiomers, so they have opposite optical rotatiorr.of equal magnitude. To distinguish the two enantiomers from one another, it :sbest to compare their specific rotations. Their solubiiity and melting pointsshould be identical, because of their structural similarities (enantiomers have thesame physical properties.) Their odors may or may not be different, because theolfactory lobes are chirally sensitive, but we cannot determine for sure that the','have different odors. we know for sure that they rotate plane-polarized light i:.opposite directions, so the best answer is choice A.

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Example 8.16The difference in solubility between 4-nitrobenzoic acid and glyceraldehyde (2,3-dihydroxypropanal) in water is most dissimilar with their rotnUitltv difference in:A. benzene.B. ether.C. 5% acid solution.D. 5% base solution.

SolutionNitrobenzoic acid gets deprotonated in a basic solution and the anion that isformed is more soluble in water than the original acid. Glyceraldehyde has noacidic or basic sites, so its solubility is not affected by the pli of the soiution. Forthis reason, nitrobenzoic acid shows greater solubility in tne 5% base solutionthan pure water while glyceraldehyde shows roughiy the same solubility inwater as 5% base solution. The best answer is choice D. The important point onthis

.question is that you are looking for differences in solubility in witer andanother solvent, not differences between 4-nitrobenzoic acid and glyceraldehydein their solubility in benzene, ether, acidic solution or basic solution, This wai infact a really poorly worded question, which happens from time to time on theMCAT.

Example 8.17Butanol can be distinguished from butanone by comparing all of the followingEXCEPT their:A. boiling points.B. freezing points.C. densities.D. infrared spectra.

SolutionButanol and butanone have different boiling points and melting points, becausean alcohol is polar and protic while a ketone is simply polar. This eliminateschoices A and B. The infrared spectra for both comporrt ds differ greatly, becausebutanol has a broad peak around 3500 cm-l that butanone does not have andbutanone has a sharp peak aroun d 1700 cm-1 that butanol does not have. Thesetwo peaks can be used as features that will differentiate the two compounds.This eliminates choice D. The two compounds have different densities, but thequestion does not allow for all four choices to be eliminated. You are looking forthe "best" answer. Both compounds have roughly equal masses and are tiqiridsat room temperature, so their densities are not going to be that diffeient.Although all of the choices work, the best answer is choice C.

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Compound Tested For Reagent PositiveMethyl ketones Iz(s)/OH-(aq) Yellow precipitate1'and 2'alcohols, aldehydes KzCrO+/HzSOq Color change: orange to greenAlkenes Brz/CCL+ Color change: brown to clearUnsaturated fats IBr/CClt Color change: purple to clear

Chemical TestsChemical tests involve reactions that are selective for very few functional groups.The ideal chemical test is positive for only one functional group, but thisls rare.Most chemical tests have multiple functional groups that"give positive results.For a chemical test to be effective, a few things iustte true."

1) The reagent must react with very few functional groups.2) The chemical testing reagent must undergo some visible change, such as

a color change or phase change, so it can be detected.3) The testing reagent must be the limiting reagent, so that any excess of the

testing reagent will not interfere with the reading.

We shall focus on the tests associated with compounds other than carbohydrates.Table 8-2 shows four common chemical tests used in organic chemistry.

Table 8-2

Do not memorize any of these tests. Know the concepts behind a chemical test,but expect that the passage will provide the details about a specific chemical test.

Example 8.181-Iodohexane is distinguished from 3-iodo-3-methylpentane Mosr readily by:A. reaction with ammonia.B. molecular mass.C. water-solubility.D. the iodoform test.

Solution1-Iodohexane and 3-iodo-3-methylpentane have the same molecular weight, sochoice B is eliminated. Alkyl halides are not very water soluble, so choiie C iseliminated. The iodoform test is for methyl ketones, not alkyl iodides, so choiceD is eliminated. The difference between them is that 1-iodohexane is a priman-alkyl iodide and 3-iodo-3-methylpentane is a tertiary alkyl iodide. rhe prlmari.

_kyY,"@""Example 8.19\21'/hat makes KMnoa in basic water not useful as a chemical test agent?A. It reacts with too many compounds.

l. It undergoes both a phase change and color change.C. It is too inert and requires a high concentration.D. It is a gas with low solubility in water.

SolutionPermanganate can oxide many compounds, including alkenes, alcohols, andaldehydes. Although it has a distinct color change, lt reacts with too manr-compounds for a conclusion to be drawn. Choice A ii the best answer


Organic Chemistry Lab Techniques Identification Technigues

Derivative FormationDerivatives are products formed from the reaction of a given reagent with anunknown compound. Derivatives are made to be used is a diagnostic for anunknown. The additional physical measurements associated wlth the newlyformed derivative can be.used to identify both the derivative and indirectly thelnkqown' They are used to support and confirm an identity for an unknown.The.functional group on the unknbwn compound must be kno*n to decide whatderivative should be synthesized. co*,'o. derivatives incrude 2,4-dinitrophenylhydrazones for ketones or aldehydes, benzylesters for alcohols, andosazones for sugars. The ideal derivative is a solid thai can easily be identifiedby its melting point. An example of a 2,4-dinitrophenylhydrazine derivative isshown in Figure 8-11.

o

*AH H

HI

..N-

Nq

o.Y*-ig.Phenylhydrazine derivative

NI

HAldehvde

Ketone

R

Phenylhydrazine derivative

Figure 8-11

Example 8.20The limiting reagent of a reaction has a molecular mass of r2g grams per moleand the product has a molecular mass of 160 grams per mole. If 0.20 grams of thelimiting reagent generates 0.20 grams of prodirct, then what is the percent yietd?A. r00%B. 89%c. 80"/"

D. 75%

Solution

lercent yield is found by taking the actual yield (in either grams or moles) anddividing it by the theoreticai yield (in either grams or molJs). In this case, it iseasier to solve using moles rather grams. Either way you choose, you must becertain that the units cancer out. The mathematics is as follows:

0.2 gl%yietd = ; ufrl?lTol", ='-7teo9l^orc ='l* -728 -64- = 8 = s.6rheorericat motes 0 2 Sl. ^^ s,t lrZ, 160 80 10 -'J

l12gol^ob ,Lv

The yield is 80%, choice C. The percent yield for a reaction must be calculatedusing the limiting reagent. Most numbers should result from easy calculations.

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Example 8.21To distinguish 2-hexanone from 3-hexanone, it is best to compare:A. their boiling points.B. their reactivity with an oxidizing agent.C. their alcohol solubility.D. the melting points of their phenylhydrazine derivatives.

SolutionThe two compounds (2-hexanone and 3-hexanone) have different boiling points,but only slightly different, because they are both six carbon ketones. Thiseliminates choice A. Ketones are inert with most oxidizing agents, so choice B iseliminated. Ketones have similar alcohol solubilities, ro

"hoi." C is invalid. To

distinguish compounds, derivatives are made. Phenylhydrazine derivatives of 2-hexanone ar-rd 3-hexanone have different melting points, because they do notpack the same in the solid phase. This makes choice D the best answer. As apoint of interest, the iodoform test could work too, but it is not a choice.

Example 8.22which of the following physical properties of the 2,4-drnitrophenylhydrazinederivative of a ketone is best to use as an identity diagnostic?A. DensityB. Molecular weightC. Boiling pointD. Melting point

SolutionPhenylhydrazones are solids, so it is not feasible to compare their boiling points.This elimrnates choice C. Their densities and moleculaimasses are not going tobe as useful in distinguishing, because most will have similar values foideniitr-and molecular mass, or close enough that no strong conclusion can be drawn.This eliminates choices A and B. Melting point is ih" dirting,rishing physicaltr"t"ttt r",-""" a e best answer choiie D.

-Example 8.23To distinguish an alcohol from an ether of approximately the same molecularmass, all of the following can be employed EXCEPT comparing their:A. boiling points.B. water solubility.C. rate of evaporation.D. index of refraction.

SolutionAn alcohol exhibits hydrogen bonding while ether does not. This increases theboiling point of the alcohol, which also makes it harder to evaporate than anether. An alcohol has a higher boiling point and slower rate of evaporation thanan ether of roughly equal mass, which eliminates choices A and -. Hydrog"nbonding also increases the water solubility of an alcohol of roughly tire sJn"number of carbons as an ether. This eliminates choice B. The index of refractionmay very well be different for an ether and an alcohol, but it is not predictablelike boiling point and water solubility. This makes it hard to distinglish whichcompound is the ether and which is the alcohol, so choice D is the beJt answer.

Copyright O by The Berkeley Review 26'4 The Berkeley Review


Mass SpectroscopyThe last laboratory technique we shall address is mass spectroscopy. Because themechanics of the device are covered in physics, we shall focus on the data andinterpretation only. At the level of the MCAT, there are only three things weshould understand about the data. First, we can ascertain the molecular mass ofa compound with great accuracy from the parent peak (peak of highest mass).This can help us to distinguish alkanes from alkenes and alkynes of equal carboncount. Second, we can determine if the compound has either chlorine or bromine(from the 4istribution of their isotopes). Chlorine has two abundant isotopes,35Cl and 37C1, ir-a ratio of approximately 3 : 1. Therefore, when we see some ofthe higher mass peaks in a 3 : 1 ratio for the M and M+2 signals in a grouping, wecan conclude that the peak combination must be due to chlorine. Bromine hastwo abundant isotopes, 79Br and 81Br, in a ratio of approximately 1 : 0.98. Apeak ratio of 1 : 1 for the M and M+2 peaks in a group generally indicates thepresence of bromine in the compound. Lastly, we can determine the mass of thealkyl groups attached to either a carbonyl carbon or a heteroatom. From thedifference is mass between the parent peak and the base peak (the tallest peak,which is assigned an intensity of 100 on the y-axis of mass spectroscopy graphs),we can ascertain the alkyl groups that have fragmented off of the original cation.

The one fact we need from physics is that the machine detects ions, and in ourcase, cations. In some cases, the cation is also a free radical, but that is irrelevantto the operation of the machine. The basic ideas behind the device are thatcharged particles move in a circular path when traveling perpendicular to linearmagnetic fields and that heavier particles traverse a circular path of greaterradius than lighter particles. The machine accelerates the cations and then causesa 90' turn, which allows particles of different mass to spread out before theystrike a detector. Figure 8-12 shows the mass spectroscopy graph for butanone.

38 40 42 44 46 48 50 52

Molecular Mass (amu)

Figure 8-12

The mass of the most abundant molecule of butanone is 72.0575 g/ rnole, so thepeak at 72 is d:ue to the radical cation formed when a nonbonding (ione pair)electron was ionized from oxygen. In ionizing a hydrocarbon, a bonding electronis lost from a carbon-hydrogen bond. An electron is ionized when a molecule isbombarded with high energy incident particles, usually electrons. Becausecations and free radicals are unstable, the species rearranges and fragments,

100

80

U]

CJ bU

CJ

t40../

20

Base



sheering off pieces of the molecule. The mass spectrophotometer detects anycharged fragments that splinter off, so the peaks that are observed correspond tbthe most stable cationic species that rearrangement and fragmentation can form.The peaks themselves represent fragments and the difference between peakscorrespond to uncharged fragments that sheered off. In Figure g-r2, the peiks ofinterest are found at 72, 57 , 43, and 29 . The peak at 72 is the parent peali, so it isattributed to the cationic compound. The peaks at 57 and 43 correspond to theacylium ions formed when the parent compound loses a methyl group and ethyigroup respectively. The peak at29 is possibly due to an ethyl cation.

Example 8.24In mass spectroscopy, the material being analyzed should be:

A. at low pressure and in the gas phase.B. at high pressure and in the gas phase.C. at low concentration and in the liquid phase.D. at high concentration and in the liquid phase.

SolutionBecause the particles must travel independently of one another in the apparatus,they must be in the gas phase. This eliminates choices c and D. In order tominimize the number of peaks, the material is at low pressure, so that the highlyreactive free radical and cationic particles do not collide and undergo reactions.If there are too many molecules, they can combine, which would lead to morecomplicated data. The pressure should be low, so choice A is the best answer.

Example 8.25which of the following peak combinations would imply that chlorine waspresent on the molecule? .L'"LL

737 732 733 134 135 131 132 133 134 135 737132 133 134 135

B.A.

SolutionAs mentioned earlier, chlorine has two abundant isotopes that come in a ratio ofapproximately 3 : 1 and differ in mass by 2. we shall ignore the peak at 135 tointerpret the ratios, because that peak does not vary from choice to choice.Choice A shows peaks in a 2 :7 : 2: 7 ratio, which in no way can be interpretedas 3 : 1. This eliminates choice A. Because the peaks that differ in mass by 2 areof roughly comparable height, it is more likely that the species contains bromine.Choice B shows peaks in a 1 : 1 : 2:2 rctio, which in no way can be interpreted as3 : 1 in favor of the lighter peak. This eliminates choice B. Choice c shows peaksin a 9 : 4 : 3 : l..3ratio which shows a 3 : 1 ratj.o for the 133-to-131 peaks and a 3 : 1ratio for the 134-to-132 peaks. This makes choice C the best answer. Choice Dshows peaks in a 9 : 3 :2:8 ratio, which in no way can be interpreted as 3 : 1.This eliminates choice D. Although it may seem arbitrary as to how the heightratio was determined, only choice C shows peaks differing by 2 n mass with aheight ratio of 3 : 1 in favor of the lighter one.

137 732 133 134 135


Organic Chemistry Lab Techniques Section Summary

Key Points for Lab Techniques (Section g)

Separation Techniques

1,. Based on different flow directions or different flow speedsa) Distillation

i. Converts the most volatile component into a gas so it can flow up thedistilling column and away from the mixture

ii. Vapor pressure differences are maximized at the boiling point of theleast volatile component

iii. The distillation apparatus is comprised of a distillin g pot, a distillingcolumn, a thermometer, a condensing side arm, and a colection frask

b) simple distillation has minimal surface area in the distilling columni. Maximizes yield at the expense of purityii. It is employed when the boiling points of the components differ by a

large amount.

c) Fractional distillation has significant surface area in the distilling columni. Maximizes purity at the expense of yieldii. It is employed when the boiling points of the components differ by a

small amount (less than 30'C).

d) Vacuum distillation is employed when boiling pointsi. Carried out in a closed system where the internal

are highpressure is actively

reducede) Chromatography separates

stationary phaseby relative affinities for a mobile versus

i. Mobile phase is the solvent which travels across the adsorbentii. Adsorbents (the stationary phase) are polar polymers through which

or across which migration occursiii. Thin layer chromatography provides a flat surface across which the

solvent migrates, taking solute with itiv. Ri values are the ratio of the solute migration distance to the soivent

migration distance in thin layer chromatographyv. column chromatography is a separation and/or purification

technique where a mixture separates as it travels down a columnvi. Elution time is defined as the time it takes from when the solutes first

interact with the adsorbent to when they exit the base of the columnvii' Gas chromatography (GC) involves the vaporization of a mixture

which migrates through a hot column and is pushed along by aninert carrier gas

f) Extraction involves the removal of a component from a mixture byselective solubility in a new solventi. Partitioning describes the distribution of a compound between two

solventsii. Partition coefficient is the numerical ratio of the solubility of a

compound in one solvent versus another solventiii. Acid/base extraction involves partitioning between an organic

solvent and water, where the pH is varied to enhance the solubilityof compounds that can form ions when protonated or deprotonated

iv' The yield is better when carried out in three consecutive extractionsof small volume rather than one extraction of large volume

v. Extraction flow charts involve tracing the pathway of variouscomponents in a mixture when they travel through the steps


Organic Chemistry Lab Techniques Section Summary

Purification Techniques

a) Recrystallizationpurifies a solid by dissolving it and then crystallizing itin a slow, methodical fashioni. The compound

_must be highly soluble at high temperatures andminimally soluble at low teniperatures

ii. The dissolving step is carried out under refluxing conditions tooptimize the solubility

iii. A decolorizing step is carried out when there are colored" impuritiesand a colorless compound. Activated charcoal is typically used

iv. Hot and cold filtration are used depending on whether thecompound must stay in solution or remain crystallized

v. solvent washes are employed to rinse away residual solvent thatmay contain impurities

Identif ication Techniques

a) Physical properties are a quick way to verify the identity of a compoundi. Melting points and boiling points are the common physical

properties used as a diagnostic for a compound's identityb) Chemical tests involve reagents that react with a minimal number of

functional groupsi, Color or phase changes indicate a positive testii. The test agent must be the limiting reagent in the reaction

c) Derivatives are formed to provide additionar evidence as to theof a compoundi. Melting points are the most common physical property measured for

a derivative

d) Mass spectroscopy converts a molecule into a cationic radical and thenaccelerates, deflects, and separates the cationic species that are formedfrom fragmentationi. Base and parent peaks represent the most stable molecular ion and

the ionic form of the original compound respectivelyii' Fragmentation occurs when the unstable species formed upon

ionization undergoes chemical processes in an effort to form a morestable species

iii. Bromine on compound results in a 1 : 1 peak ratio for the high massfragments while chlorine on compound results in a 3 : 1 peik ratiofor the high mass fragments

identitv

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LaboratoryTechniquesPassages

l5 Passages

lOO Questions

r Suggested schedule:I r' After reading this section and attending lecture: passages I, IV, vl, & xQrade passages immediately after completion and log your mistakes.

II: Following Task I: Passages II, VII, x, & xr (2g questions in sz minutes)Time yourself accurately, grade your answers, and review mistakes.III: Review: passages III, v, VIII, xII, XIII, & euestions 94 - loo

Focus on reviewing the concepts. Do not worry about timing.

R. E. V. I . b. \llT@

altztng in MCAT Preparation

$,fi.

I. Distillation and Separation I - Z)

II. Fractional versus simpre Distillation @ - 14)

III. steam Distillation of Natural products es - zl)IV. Caffeine Extraction Experiment e2 - 2g)V. Partition Coefficient Experiment eg _ bS)

VI. Acid/Base Extraction Experiment 66 - 42)

vll. Extraction and Thin Layer chromatography Experiment (4b _ 49)

vlll. Thin Layer and corumn chromatography (5o - 56)

IX. Column Chromatography 6Z - 6Z)

X. Gas Chromatography rc4 - Zl)XI. Recrystaltization e2 - Zg)

xll. Qualitative Anarysis (Bo - 86)

XIII. Synthesis and Extraction @Z - gJ)

Questions not Based on a Descriptive passage (g4 - l oo)

Lab Techniques and Spectroscopy Scoring Scale


84 - loo 15-1566-85 lo-t247 -65 7 -934-46 4-61-33 I -5

llL,;

lli"ir I rr

ivl i.

1l:.tn

['nel

nt;


A student is given a mixture of two liquids to separate.There are three possibilities for the composition of themixture: 1) equal parts by volume of Z-1,2-dichloroethene(b.p. 60'C) and E-1,2-dichloroerhene (b.p. a8'C), 2) equatparts by volume of 3-pentanone (b.p. 102"C) and 2-hexanoneft.p. 128'C), and 3) equal parts by mass of diethyl ether (b.p.34"C) and di-n-propyl ether (b.p. 90'C). The student isuncertain as to which of the three mixtures he may have, sohe uses fractional distillation to separate the mixture.Fractional distillation is employed when the boiling points ofthe components are less than 30'C apart fiom one another.The distillation apparatus is fitted with a collection vial that;an collect the distillate in 10-mL aliquots. The apparatus;ontains the following components: boiling flask, distilling;olumn (with optional packing), thermometer, condenser*'ith cooling sleeve, and collection flask. Figure 1 shows thedistillation apparatus used by the student.

Thermometer

DistillationColumn

Condenser with cooling sleeve

OptionalPacking

Collection flask

Boiling Flask

Figure 1 Distillation apparatus used in the experiment

After approximately thirty minutes, the first drop of.rquid appears in the collection vial. The boiling flask is held.t a constant temperature and the distilling column is,:lsulated with glass wool. A total of three samples (of:oughly 8 mL each) were collected before the distilling flask.,, as removed from the heat source. A small portion of tiquid:emained in the distilling flask after the heat source was:emoved. Once the system cooled back to room temperature,.!ere were approximately 10 mL of solution still remaining:r the distilling flask.

1 . When distilling a liquid with a boiling poinr of 56.6"C,you should use an apparatus with which of thefollowing?

A . Two openings to the atmosphere, one before thecollection flask and one after the collection flask.

B. One opening to the atmosphere before thecondenser.

C . One opening to the atmosphere after the collectionflask.

D. No opening to the atmosphere, so no vapor canescape.

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2. For the experiment. the thermometer read 49.2"C asthefirst aliquot was collected. The compound beingcollected is most likely which of the following?

A. E-1.2-dichloroethene

B. Z-1.2-dichloroethene

C . Diethyi ether

D. 3-pentanone

A boiling chip. when added to the distillation flask,prevents bumping (the sudden pop of vapor in thesolution) by:

A. providing surface area fiom which to boil.B. causing even distribution ofthe heat.

C . reducing intermolecular forces.

D . increasing intermolecular forces.

The majority of the vapor from the boiled liquidcondenses on which of the following areas?

A. Cold glass in the condenser

B. Cold glass in the collection flask

C . Warm glass in the condenser

D. Warm glass in the collection flask

To separate a dissolved solid (solute) from a liquidsolvent in which it is completely soluble, it is best touse:

A. gravity filtration.B. vacuum filtration.

C. simple distillation.D. fractional distillation.

Simple distillation would work best with which of thefollowing mixtures?

A. H3CH2COCH2CH3 and H3CH2CCOCH2CH3

B. H3CH2COCH2CH3 and E-i,2-dichloroethene

C. H3CH2CH2COCH2CH2CH3 and

H3CH2CCOCH2CH3

D. H3CH2CH2CH2CCOCH3 and

H3CH2CCOCH2CH3

Fractional distillation is employed to separate:

A. agasfromaliquid.B. a liquid from a gas.

C. asolidfromaliquid.D. a liquid from a liquid.

3.

4.

5.

6.

7.


Passage ll (Questions 8 - 14)

A student intended to separate 10.0 mL of toluene from10.0 mL of heptane. The boiling poinr of heprane is 9g.4.Cand the boiling point of toluene is 110.6"C. To separate thetwo liquids, the student could use either simple or fractionaldistillation. Simple distillation involves distillation througha short, hollow distilling column and it is chosen when thesubstance to be separated has a boiling point that is at leastthirty degrees lower than any other component in themixture. Simple distillation takes less time than fractionaldistillation and generates a higher yield. In the separation ofheptane from toluene, it would be impractical to use simpledistillation, because their boiling points are too close.

Fractional distillation diff'ers fiorn simple distillation inthat the distillation column has more surface area. This isaccomplished either by using a longer column or by packingthe column with an inert material onto which the vapors cancondense. The advantages of fiactional distillation are thatthe product is purer and the vapor pressure is rnore constant,which reduces bumping in the solution. Fractionaldistillation does not recover all of the component, however,and it can take up to five times as long as simple distillation.Table I lists the boiling points of some organic liquids.

Liquid BoilingPoint Liquid

BoilingPoint

Diethyl ether 34.6"C Octane l25.l"cPentane 36.1'C n-Pentanol 138.0'CTHF 65.4'C Nonane 150.8'cHexane 68.9'C Anisole 158.3'CIsopropanol 82.3"C Decane t74.t'c

Table I8 . Over time, the effectiveness of {iactional distillation is

diminished. How can this best be explained?

A . Over time the distilling column heats up, and asthe temperature increases, the amount ofcondensation and reevaporation decreases.

B. Over time the distilling column heats up, and asthe temperature increases, the amount ofcondensation and reevaporation increases.

C . Over time, the solution becomes poor in the morevolatile component, so azeotropes increase.

D. Over time, the solution becomes poor in the morevolatile component, so azeotropes decrease.

9 . When separating the following moiecule pairs usingdistillation, which pair requires the longesr clistillingcolumn?

A . Diethyl ether and pentanol.

B. Tetrahydrofuran (THF) and anisole.

C . Octane and pentane.

D. Isopropanol and toluene.

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11.

10. Which of the following is NOT an advantage offractional distillation over simple distillation?

A . The components that are collected are purer.B. The fractional distillation procedure is faster than

simple disrillarion.

C . Fractional distillation allows for the mixture tocontinually re-distill throughour the distillationcolumn.

D. Fractional distillation provides a more constanivapor, thus reducing the tendency of the solution to"bump".

Why is distillation bulb (at the base) placecl in a sancbath in a heating mantle rather than placed over a flame lA . The sand serves as a themal insulator.B. The sand is less reactive with the glass than the air.

C. The sand bath keeps the vapor pressure of warerlow by ahsorbing moisturc.

D. The sand bath allows for uniform distribution o-heat around the distillation bulb.

What is the role of copper mesh when it is inserted inla distilling column?

A . The copper mesh serves as a filter to collect ai;solid particles that have become airborne.

B. The copper mesh serves to neutralize any acii.:vapor that forms.

C . The copper mesh allows for an even distribution .

heat due to the electrical resistance of the copper.

D. The copper mesh serves to provide addition.surface area onto which vapors may condense a: _

then re-evaporate in fractional distillation.

Which of the following pairs of compounds is b.-separated using simple distillation?

A. Diethyl ether and heptane.

B. Tetrahydrofuran (THF) and hexane.

C . Octane and pentanol.

D . Nonane and anisole.

Which stereoisomers would MOST likely shou --,:same boiling point?

A . 2R,3S-dichloropentane and 2R,3R-dichloropenr.,.B. 3R-bromocyclopentene and 35-bromocyclopen:: .C . D-Glucose and D-mannose

D . 3R-ethy1-trans-decalin and 3S-ethyl-trans-decali:

t2-

13.

F

!t

lfm'

i[it r.

14.


Because molecules can exist in the vapor state attemperatures below the compound's boiling point, it ispossible to carry out distillation at temperatures below theboiling point of the components in the mixture when themixture does not form an ideal solution. A commontechnique for carrying this out is steam distillation. Steam isdirected to the distilling flask, resulting in a temperature of100'C during the course of distillation. The solution slowlyevaporates, and is collected in the same fashion as standarddistillation. Because of the presence of water vapor in theimmediate atmosphere, the distillate is rich in water. Unlikethe distillate in simple or fractional distillation, the distillatein steam distillation must be further purified. To simplifyfurther separation, steam distillation is often employed withoils that are immiscible in water.

A student was assigned the task of separating citral (b.p.

=229'C) from lemon grass oil. Citral is a l0-carbon terpenethat is a precursor in the commercial synthesis of Vitamin A.It is isolated in conjunction with neral, which varies at thedouble bond between carbons 2 and 3. Citral and neral areshown in Figure 1 below.

Citral

CH: CH3

Neral

Figure 1 Citral and neral, components of lemon grass oil

After steam distillation, the mixture is approximately90Vo water,9Vo cttral, and l7o neral. To isolate the organicproducts, the mixture is extracted using diethyl ether and driedusing anhydrous magnesium sulfate.

15. What compounds are best separated using steamdistillation?

A . Organic liquids with boiling points over 100'C

B. Organic liquids with boiling points under 100'C

C. Organic solids with melting points over 100'C

D. Organic solids with melting points under 100'C

1 6. Neral and citral are best described as:

A. conformational isomers.

B. geometrical isomers.

C . optical isomers.

D. structural isomers.

Copyright O by The Berkeley Review@ 213 GO ON TO THE NEXT PAGE.

17. Standard distillation of lemon grass oil at 229"C couldresult in all of the following EXCEPT:

A . decomposition of citral.

B. hydrogenation of citral.

C. oxidation ofcitral.D . polymerization of citral.

18. When ether is added to the distillate, what is observed?

Water dissolves into the ether layer while citralsinks to the bottom of the flask.

Water dissolves into the ether layer while citralfloats to the top of the solution.

C . An ether layer forms on top of the water layer, andcitral dissolves into it.

D. An aqueous layer forms on top of the ether layer,and citral dissolves into it.

19. What is the purpose of adding anhydrous magnesiumsulfate to the ether extract?

A . To isomerize neral into citralB . To reduce the aldehyde into a primary alcohol

C . To remove any residual water in the ether layer

D. To selectively bind neral, leaving behind pure citral

2 0. What is the approximate boiling point of neral?

A. 78"C

B. 100'c

c. t41'cD. 211"C

21 . What property is expected for Vitamin A?

A. A lower melting point than citralB. A high water solubility

C . A high vapor pressure at room temperature

D. A high affinity lbr lipids

A.

B.


A student attempts to remove caffeine from tea leaves bya series of extraction procedures. Initially, 2.0 grams of tealeaves are added to 10 mL of boiling water for 20 minutes.The aqueous solution is then extracted three times withexactly 5.0 mL of merhylene chloride (CH2C!) during eachof the three extractions. The 15.0 mL of methvlene chloridesolution is collected and consolidated into one flask. Thesolvent is removed through evaporation by flushing vaporfrom the flask using a constant stream of nitrogen gas. Thetemperature of the solution is maintained at 60'C by partiallyimmersing the flask in a water bath.

After the solvent is completely removed from the flask, awhite residue remains at the bottom. This white residue issublimed under vacuum and 28 milligrams of a white powderare isolated. The melting point and infrared spectroscopy dataof the white powder marches that of caff'eine (m.p. = 238 'C).The student concludes that the isolated compound is in factpure caffeine. Following sublimation, the bottom of theflask has no white powder remaining, just a waxy residue.The structure of caffeine is shown in Figure 1 below.

o

":I5 CHrI

N

N

I

CH:

Figure 1

The caifeine could have been removed from the organicsolvent by treatment with a strong acicl such as hydrochloricacid. The acid can protonate the caf'fcine to form the ionicsalt complex, which is relatively insoluble in organicsolvent.

22. From the values presented in the passage, what is theapproximate percentage by mass of caffeine in tea?

A. 1.470

B " 2.87n

C. 11Vo

D.28Vo

23. Based on the information in the passage, caffeine haswhich of the following solubility properties?

A. It is soluble in both water and CH2CI2, but moresoluble in CH2C12.

B . It is soluble in both water and CH2C|2, but moresoluble in water.

C . It is soluble in water, but insoluble in CH2C|2.D. It is soluble in CH2C12, but insoluble in warer.

Copyright @ by The Berkeley Review@ 214

28.


24. The flow of nitrogen gas serves to:

A. reduce the partial pressure of the solvent vapor somore solvent can evaporate.

B. increase the partial pressure of the solvent vapor somore solvent can evaporate.

C. reduce the partial pressure of the solvent vapor so

less solvent can evaporate.

D . increase the partial pressure of the solvenl vapor so

less solvent can evaporate.

Which procedure would you use to determine the purit..of the caffeine?

A. Adding it to methylene chloride to test itssolubility.

B. Adding it to water to test its solubility.C . Mixing the compound with caffeine and observins

the melting point for a sharp range.

D. Mixing the compound with an organic solid othe:than caffeine and observing the melting point for a

sharp range.

Why is the methylene chloride added in three 5-mLportions rather than one 15-mL portions?

A . Methylene chloride is too unstable to be added in E

big portion.

B. 15 mL of methylene chloride weighs more than 10-mL of water, so the organic layer would sink to theborrom ol the flask.

C . Less caffeine can be extracted by three smal-extractions than one large extraction.

D . More caffeine can be extracted by three smal1extractions than one large extraction.

Methylene chloride was added to the aqueous solution[o:

A . extract all of the caffeine while leaving behind th;water-so1ub1e impurities.

B. extract most of the caffeine while leaving behincthe water-soluhle impurities.

C . extract all of the caffeine along with the water-soluble impurities.

D. extract some of the caffeine along with the water-soluble impurities.

Sublimation is employed to purify a:

A. solid from other solids.

B. solid from liquids.

C. solid from gases.

D. liquid from solids.

t<

26-

:ti

!-

27.


A student wishes to determine the partition coefficient of4-methylaniline between the water layer and diethyl etherlayer in a biphasic mixture. The student decides to dissolve aknown mass of 4-methylaniline into a l:l mixture byvolume of water and anhydrous diethyl ether. In the firsttrial, the student uses l0 mL of water, l0 mL of anhydrousdiethyl ether, and 1.0 gram of 4-methylaniline. This mixtureis placed into a separatory funnel and shaken. The lowerlayer is removed, without removing any of the upper layer.A drop of water is added to the lower layer, and it is found todissolve completely into the solution. The student concludesfrom this that the upper layer, still in the separatory funnel,is the organic (diethyl ether) layer. The student removes theupper layer into a 25 mL Erlenmeyer flask and adds to this0.5 grams of anhydrous calcium chloride, which forms a layerof powder on the bottom of the flask.

The flask is swirled gently for two minutes and then sitsfor five minutes. The ether layer is removed from the flaskby decantation using a warm pipette and is added to a 10.50gram vial. The calcium chloride remaining in the Erlenmeyerflask is rinsed with 5.0 mL of anhydrous diethyl ether. This5.0 mL portion is removed by decantation using a warmpipette and added to the vial. The vial is heated mildly underan exhaust fan until the diethyl ether evaporates awaycompletely. The mass of the vial and dried 4-methylanilineis 11.25 grams. The student assumes that the missing 4-methylaniline from the original 1.0 g sample must havedissolved into the water layer. By dividing the grams whichdissolved into the ether layer by the grams that dissolved intothe water layer, a partition coefficient is calculated.

To verify that the answer is valid, the student consults abook of physical constants and finds that the solubility at25'C for 4-methylaniline in water is 3.8 grams/100 mL andthe solubility at25'C for 4-methylaniline in diethyl ether is22.8 grams/100 mL. From these values, the studentconcludes that the experimental partition coefficient is toolow. This deviation is most likely attributed to the loss ofsolid in the transfer steps.

29. From the experimental values using 1.0 gram in amixture of l0 mL each of water and diethyl ether, whatis the partition coefficient?

A. 2.5

B. 3.0

c. 4.0

D. 6.0

31 . A second student attempted this same experiment using3.0 grams 4-methylaniline in a mixture of l0 mL eachof water and diethyl ether. Which of the followingstatements predicts the results?

A. The partition coefficient was found to be the sameas with the 1.O-gram experiment.

B. The partition coefficient increased from the valuedetermined in the 1.O-gram experiment.

C. The partition coefficient decreased from the valuedetermined in the 1.O-gram experiment.

D. The partition coefficient could not be determinedbecause the 3-methylaniline didn't dissolvecompletely.

3 2. How can the 4-methylaniline that adhered to the pipettewall during transfer be recovered?

A . The pipette could be rinsed with water.B. The piperte could be rinsed with diethyl ether.

C . The pipette could be heated to a high temperature.D . The pipette could be cooled to a low temperature.

3 3 . If the ether used was NOT anhydrous, how would thisaffect the calculated partition coefficient?

A. It would lower the value, because initially therewould be more than 10 mL of water and less than10 mL of ether.

B. It would raise the value, because initially therewould be more than l0 mL of water and less than10 mL of ether.

C. It would lower the value, because initially therewould be more than l0 mL of ether and less than10 mL of water.

D. It would raise the value, because initially therewould be more than 10 mL of ether and less than10 mL of water.

3 4 . All of the following solvents be used in place of diethylether EXCEPT:

A. tetrahydrofuran, THF.B. ethanol.

C. cyclohexane.

D . methylene chloride.

3 5. Which of the following is a desired property of theorganic solvent?

A. Highly miscible in warer

B. Highly dense

C. Solute is highly soluble in the solventD. Its boiling point is less than room temperature

30" From the literature valuespartition coefficient?

A. 2.5

B. 3.0

c. 4.0

D. 6.0

for solubility, whar is rhe



Two students were each presented with a mixture of equalparts benzoic acid, resorcinol, and meta dinitrobenzene. Tohelp in separating the three compounds into their pureindividual components, the students had availabl e 70Vosodium bicarbonate solution, 5% hydrochloric acid solution,l)Vo acetic acid solution , and 5Va sodium hydroxide solution.Each student mapped out a scheme in the form of a flow chartto separate the three compounds. Student I chose first to addsodium hydroxide solution and ether to separate thecompounds. Student II chose first to add sodium bicarbonatesolution and ether to separate the compounds. Each scheme,along with the structures of each compound and their pKuvalues, are shown in Figure I below.

oY?l= - ,r !H nK" = o se

OG*NO"

meta-DinitrobenzeneBenzoic Acid Resorcinol

Proposed Scheme for Student I

Tube 2

Ether layer

Aqueous sodiumbicarbonate layer Tuhe I

Tube 3

Proposed Scheme for Student II

Ether layer

Aqueous sodiumhydroxide Iayer

Aqueous sodiumbicarbonate layer

Aqueous sodiumEther layer roxide layer Tube 4

Tube 5 Tube 6

Figure 1 Reaction schemes lor Students I and II

The ether layer and aqueous layer are immiscible in oneanother, so they can be separated using decanting techniques.To isolate the solute from the ether layer, the ether solvent isallowed to evaporate" To isolate the components from theaqueous layers, the soiutions are neutralized and filtered. Thetwo students did not get the same resuits. The melting pointranges for the three solids isolated by Student II are sharperthan the meiting point ranges for the three solids isolated byStudent L


3 6. Which of the following organic compounds could beused in place ofether in this acid/base extraction?

A. Ethanol (b.p. = 79'C)B . 1,4-Dichlorobenzene (m.p. = 53"C)C. Naphthalene (m.p. =94"C)D. Chloroform (b.p. = 6l'C)

3 7. What solution should be added to Tube 4 to cause a

precipitate to fall out of solution?

A. 10Vo sodium bicarbonate solution.B. 5Vo hydrochloric acid solution.

C. I}Vo acetic acid solution.D. 5Vo sodium hydroxide solution.

3 8. What can be said about Tube 2 and Tube 5?

A . Tube 2 contains dinitrobenzene, while Tube -<

contains resorcinol.

B. Tube 2 contains dinitrobenzene, while Tube _<

contains both resorcinol and benzoic acid.C. Tube 2 contains resorcinol, while Tube 5 contains

dinitrobenzene.

D. Tube 2 and Tube 5 both contain dinitrobenzene.

39. What is true about the pK values for the conjugat:bases of the benzoic acid and resorcinol?

A . pKb(sodium benzoate) > pKb2(sodium resorcinoxid:and pKullr"rorcinol) > pKa2(resorcinol)

B. pKUZ(soOium resorcinoxide) > pKb(sodium benzoa::and PKal (resorcinol) > pKa2(resorcinol)

C . PKI(soaium benzoate) > pKb2(sodium resorcinoxi::and pKu2lt.rorcinol) ) PKal(resorcinol)

D . pKUZ(soaium resorcinoxide) > pKb(sodium benzoa::and pKu2lr.rorcinol) > PKai (resorcinol)

4 0. What can be concluded about the two schemes preseni3:in the passage?

A . The scheme for Student II worked while the scher:-tbr Student I did not work.

B. The scheme for Student I worked while the scher.for Student II did not work.

C. Both the scheme for Student I and the scheme i::Student II worked.

D. Both the scheme for Student I and the scheme:.Student II did not work.

216 GO ON TO THE NBXT PAGL

41. If 3-methylaniline were present in the mixture forStudent II, where it be found?

A. Exclusively in Tube 4

B. Exclusively in Tube 6

C . Evenly split between Tubes 5 and 6

D. Predominantly in Tube 5

4 2. Which of the following compounds cannot be isolatedusing acid-base extraction techniques?

A. 4-Ethylcyclohexanone

B. 3-Bromophenol

C. 3-Nitrobenzoic acid

D. Ethyl benzoate



Organic compounds can be isolated from plants througha process known as extraction, a common industrial practiceused to isolate natural products for medicine, preservatives,dyes, and flavoring agents. Trimyristin, for instance, can beisolated from nutmeg beans by grinding the beans into a pulpand then adding diethyl ether to the ground pulp. Trimyristindissolves readily into diethyl ether. In ordinary extractionprocedures, the solution is isolated by either decanting itaway from the suspension or by filtering the solid impuritiesout of the suspension by passing it through cotton. In bothprocedures, the desired compound remains dissolved in rheorganic solvent. When left to sit, the solvent will slowlyevaporate, leaving behind residue of the compound. In thecase of trimyristin, precipitation can be expedited by addingmethanol to the solution. The purity of the extract may betested by one of several spectroscopic techniques (includingIIINMR and IR), by thin layer chromatography, or byevaluating the melting point of the powder.

Trimyristin is a fatty acid triglyceride, derived fromglycerol and three fourteen-carbon fatty acids. Like all esters,it is sensitive to acid-base extraction techniques. In anexperiment, students attempt to extract trimyristin fromnutmeg beans. The solvent chosen for extraction wastetrahydrofuran, THF. After filtering through cotton andadding methanol to solution, a white powder formed on thebottom of the flask. It was isolated by passing it through a

Biichner funnel fitted with filter paper. A small sample ofthe powder was dissolved into three drops of dichloromethane,and the new solution was spotted onto a TLC (thin layerchromatography) plate in three places. The plate was placedinto a beaker with minimal dichloromethane at the bottom.The beaker was sealed and the solvent proceeded to climb theplate. Before the solvent reached the top of the plate, theplate was removed from the solvent and the top of the solventfront was marked. The plate was set into an iodine chamberfor developing. Spots appeared and were analyzed. Figure Ishows a rough sketch of what appeared:

Solventfront

stopped

dsolvent

Before After

Dichloromethane Solvent*r=;*

Figure 1 Thin layer chromatography results of extract

The rhree spots resulted in R1 values of 0.19, 0.59, and0.83. The compound with an R1 = 0.59 (referred to asCompound II) ri'as the most abundant species of the threecompounds rhat were isolated from the crude product mixtureusing column chromatography techniques. The other twoproducrs \', ere present in smaller quantities.

{3. -\ spot that shows an R1 value of 0.j2 in a nonpolarsolr'ent like hexane will likely show what R1 value in apoiar solvent like ethyl acetate?

A . 1.84

B. 0.12

c . 0.13

D. -0.21

44. What solvent should be added to the beans to removeany naturally occurring sugar? A typical sugar has acarbonyl group and all other carbons have an alcoholgroup (cH2oHCH(oH)cH(oH)cH(oH)cH(oH)cHo)

A. A long-chain alkane.

B. A cyclic ether.

C. A short-chain alcohol.D. A long-chain ketone.

45. Why should the initial height of the solvenr in rhebeaker NOT be above the initial height of the spots onthe TLC plare?

A. The spots would not migrate up as fast if they aresubmerged below the level of the solvent.

B. The spots would become saturated and thus migratetoo rapidly up the plate.

C. The spots would migrate in a radial manner ratherthan a linear manner, if submerged below the top ofthe solvent.

D. The spots would dissolve into solution ifsubmerged below the top of the solvent.

46. If two compounds in a mixture have respective R1values of 0.58 and 0.29 in hexane, and 0.07 and 0.21 inethanol, how are they best separated?

A. By column chromatography, using hexane in ashort column.

B. By coiumn chromatography, using hexane in along column.

C . By column chromatography, using ethanol in ashort column.

D . By column chromatography, using ethanol in along column.

Copyright O by The Berkeley Review@ 218 GO ON TO THE NEXT PAGE"

4 7. Which of the following solvents would show a similarresult as a TLC solvent as carbon tetrachloride (CCl+)?

A. Acetone (CH3COCH3)

B. Propanal (CH3CH2CH=O)

C. Ethanol (CH3CH2OH)

D. Pentane (CH3CH2CH2CH2CH3)

48. Which of the following values is NOT possible for rheR1 value in a TLC analysis?

A. 0.00

B. 0.35

c. 0.70

D. 1.05

4 9. How can THF and trimyristin best be described?

A . THF is a nonpolar solvent and trimyristin is ahydrophobic compound.

B. THF is a nonpolar solvent and trimyristin is ahydrophilic compound.

C . THF is a polar solvent and trimyristin is ahydrophobic compound.

D. THF is a polar solvent and trimyristin is ahydrophilic compound.

Passage Vllt (euestions 50 - 56)

Chromatography operates based on the partitioning of a;ompound between a mobile phase and a stationary phase. If:he compound has a higher affinity for the eluting solventmobile phase) than the adsorbent (stationary phase), then it

'.i ill travel quickly. One type of chromatography is thin layerchromatography, which is done on a small scale to analyzethe components of a mixture. It can also be done on a largerscale in a column, where the length of the column is at leastten times the diameter of the column. Columns are capableof separating up to approximately l0 grams of material,rihether it is a liquid or solid.

In thin layer chromatography, the adsorbent is bound to arectangular plate made of either glass or plastic, depending onthe adsorbent. The sample is spotted near the bottom of theplate and solvent is allowed to climb up the plate throughcapillary action. The solutes in the spots migrate with thesolvent up the plate, but at varying rates depending on theirrelative partitioning between eluting solvent and adsorbent.In column chromatography, the same material is added to thetop of a column packed with adsorbent and a thin layer ofclean sand at the top. The sand prevents the adsorbent fromdistorting when solvent is poured into the column.

There are two common adsorbents, alumina (Al2O3) andsilica gel (SiOz). Alumina comes is three forms, acidic,basic, and neutral. The neutral form is given a BrockmannActivity rating from I to V, where I is the most active interms of binding a solute. The activity can be reduced by theaddition of water, because water binds the alumina, therebyreducing the surf'ace area to which the solute can bind. ABrockmann Activity rating of III refers to alumina that is 6Vor.vater by mass.

Solvents are rated according to their polarity, from leastpolar to most polar. As a solvent becomes more polar, it hasa greater affinity for a polar solute. Table 1 shows solventsaccording to elution rate for a nonpolar solute from fastest toslowest across alumina. Often, the affinity of polar solutesfor the adsorbent exceeds all other affinities.

Rank Solvent Rank Solvenl

1 n-Hexane 8 Tetrahydrofuran2 Petroleum ether 9 Dioxane.l Cyclohexane 10 Ethyl acetate4 Ligroin 11 Dimethyl sulfoxide5 Carbon disuiflde l2 2-Propanol6 Ethyl ether t3 Ethanol1 Dichloromethane t4 Methanol

Table 1

The general elution order for solutes in nonpolar solventsis: alkanes > alkenes > dienes > aromatic hydrocarbons>ethers > esters > ketones > aldehydes > amines > alcohols >phenols > carboxylic acids. R1 values can be predicted fromthe relative elution rate and rhe solvent rankings by applyingthe simple principle: "like dissolves like."

Copyright @ by The Berkelev Review@

5 0. Which combination has the fastest elution time througha column filled with alumina?

A. Benzoic acid solute in n-pentaneB. Glycerol solute in petroleum etherC. o-Xylene solute in ligroinD. Glycine solute in ethanol

51 . For two compounds with R1 values that are different bya factor of 2, what is true in column chromatography?

The elution time is 4 times greater for thecompound with the larger R1 value than thecompound with the smaller Rp value.

The elution time is 2 times greater for thecompound with the larger R1 value than thecompound with the smaller R1 value.

The elution time is 4 times greater for thecompound with the smaller R1 value than thecompound with the larger R1 value.

The elution time is 2 times greater fbr thecompound with the smaller Rp value than thecompound with the larger Rp value.

Using TLC, which solvent would give salicylic acid thegreatest R1 value?

A. n-Hexane

B. Ligroin

C. Methanol

D. 2-Pentanol

What is true of alumina with a Brockmann Activitvrating of IV?

A . It has 4Vo water by mass.

B. It has a greater affinity fbr solute than alumina witha Brockmann Activity rating of III.

C. It is rich in silicon aroms.

D. It can be formed by adding more than 6 grams ofwater to 94 grams of alumina with a BrockmannActiviry raring ol I.

How does a sudden change in solvent from n-pentane tomethanol in an alumina column harm the ei'fectivenessof the column?

A. The methanol endothermically binds the alumina,forming vapor pockets in the column.

B. The methanol exothermically binds the alumina,forming vapor pockets in the column.

C. Methanol is immiscible in hexane, so it dropsthrough the column too fast.

D. Methanol is immiscible in hexane, so it dropsthrough the column too slow.

A.

B.

C.

D.

52.

53.

54.


5 5 . Why does adding water to alumina raise its BrockmannActivity rating?

^4. . After water binds the surface of alumina, there arefewer sites on the alumina for solute to bind.

B , Water causes the alumina to contract, reducing itssurface area for the binding of solute.

C . Water makes the nonpolar alumina more polar.D . Water oxidizes the alumina into a new species that

is more hydrophobic than alumina.

56. .\iumina would likely have the highest affinity for',', hich of the following solutes?

\, 1,3-Cyclohexadiene

B. CyclohexanolC . Ethyl butanoate

D. 3-Methylpentanal

Copyright @ by The Berkeley Review@ 280 GO ON TO THE NEXT PAGE. if,iE

Passage lX (Question 57 - 63)

Benzylic hydrogens are unusually reactive for hydrogenson an spj-hybridized carbon. They have pKus around 23-25.considerably lower than 45-50 for alkyl hydrogens. They arealso more susceptible to oxidation than alkyl hydrogens.Figure 1 shows the oxidation of toluene into benzoic acid. a

typical oxidation reaction for benzylic protons.

CH: [o] -

Figure 1 Oxidation of the benzylic hydrogens of toluene

A chemist decided to oxidize fluorene into fluorenone bradding sodium dichromate, Na2Cr2O7, in acetic acid solvenrThe reaction is shown in Figure 2 below.

Figure 2 Oxidation of fluorene into fluorenone

Because both fluorene and fluorenone are solids at roomtemperature, separation cannot be done by distillation. Foiseparating solids, extraction or chromatography are typicallyemployed. If the sample is 10 grams or less, then columrchromatography works well. The chemist chose to separarethe product mixture using column chromatography. prior tocarrying out the column chromatography experiment, she rana series of thin layer chromatography trials to determinewhich elution solvent and adsorbent to use in the column.

The adsorbent in each trial is silica gel. Two spots areobserved in every trial. One spot is yellow in color, and bothare UV active. Table I shows the R1 values for each of thetwo compounds in various solvents used in the TLC trials.

Solvent R1 for YellowCompound

R1 for Colorlesscompound

Liquid I 0.18 0.31Liquid II 0.24 0.33Liquid III 0.19 0.51Liquid IV 0.41 0.55

THF 0.31 0.42

Table 1

5 7. Fluorenone should have the greatest affinity for whichof the following solvents?

A. Acetone

B. Cyclohexane

C. Dichloromethane

D. Methanol

wo

tol+

58. Why does fluorene elute from the column beforefluorenone when cyclohexane is the eluting solvent?

A. Fluorene has a higher affinity for silica gel thanfluorenone.

B. Fluorene is lighter than fluorenone, so it travelsfaster.

C. Fluorene has a higher affinity for cyclohexane thanfluorenone.

D. Fluorene is smailer than fluorenone, so it can passmore easily through the pores in silica gel.

According to the data in Table 1, what eluting solventis best for separating fluorenone from fluorene?

A . Liquid IIB. Liquid ItrC . Liquid IVD. THF

When eluting a mixture of biphenyl, ethyl benzoate,and toluic acid from an alumina column using ligroin asthe eluting solvent, the sequence off the column is:

A. biphenyl first, ethyl benzoate second, and toluicacid last.

B. toluic acid first, ethyl benzoate second, andbiphenyl last.

C. ethyl benzoate first, biphenyl second, and toluicacid last.

D. biphenyl first, toluic acid second, and ethylbenzoate last.

Once the silica gel is set in the column, why shouldyou wait to add the sample until the solvent is flushwith the top of rhe silica gel?

A. So that the sample does not float on the solventand never get to the silica gel

B. So that the sample does not adhere to the glass ofthe column

C. So that all of the sample starts traveling down thecolumn at the same time

D. So that top of the column can dry out before thesample interacts with the silica gel

6 2. Why is a thin layer of sand added to the top of the silicagel?

A . To prevent the sample from reaching the silica gelB . To reduce the poiarity of the gel

C . To keep the top of the silica gel flat when moresolvent is poured into the column

D. To bind sandphilic solures

59.

60.

61.

Copyright @ by The Berkele-v- Rer.iew@ 281 GO ON TO THE NEXT PAGE.

63. Why do nonpolar solutes elute before polar solutes fromcolumns containing silica gel, even when a polarsolvent is used?

A. Polar compounds do not follow the ',like dissolveslike" rule.

B. The silica gel is highly polar, which hinders rhemigration of polar solutes.

C . Nonpolar species become more polar when thesolvent is polar.

D. Polar compounds aggregate and form largemolecules that do not migrate quickly because ofsteric hindrance.

Passage X tQuestions 64 - Z1)

-{ researcher carried out a two-step reaction using theketone 3,3-dimethylcyclopentanone as the reactant. Thereaction sequence is shown in Figure 1 below.

CH: CH:Reaction I

CHr

cHsCHr CH:

Reaction 2

Figure 1 Two-step reaction of 3,3-dimethylcyclopentanone

The reaction sequence involves deprotonating an alphahydrogen and subsequently adding an electrophile to form abond to the alpha carbon. Because the reactant contains twoalpha carbons, there are two possible structural isomerproducts. The two methyl groups on carbon three offer sterichindrance, so one alpha carbon is less sterically hindered thanthe other alpha carbon.

The reaction was carried out at different temperatures. Atlower temperatures, the product mixture favored the lesssterically hindered product. This is often referred to as thekinetic product of the reaction. The thermodynamic productresults from the formation of the more stable bond. In thereaction of Figure 1, there is no distinct thermodynamicproduct. Table I shows the percentage of each species insoiution following the reaction at various temperatures.

o

A* R-x

H-.", -

Temp 7o Reactant Vo Product A Z, p.oduct BTrial I: -33"C 25 66 g

TrialII: 0"C 20 60 20Trial III: 15'C 16 56 28Trial IV: 30'C t3 53 34Trial V: 50'C l0 50 40Trial Vl: 78"C j 48 4.5

ooo

Q.f-q;,ft.,

Table 1

Figure 2 is the gas chromatography chart produced whenthe product mixture tiom Trial III is analyzed. Thecompounds were passed through a Carbowax_ 12 column thatis partially polar. The identity of each peak was obtained byspiking the product mixture with a sample of one of the purecomponents. Spiking is the term given to adding a smallportion of one of the components to the product mixture.The peak that increases is deducecl to beiong to the compoundthat has been spiked.

Copyright O by The Berkeley Review@ 282 GO ON TO THE NEXT PAGE. .ir

Figure 2 Gas chromatography results for Trial III

Gas chromatography can be used to quantitativelyana\yze the composition of mixtures by evaluating the areaunder each peak. The relatives areas under each curve can beused to determine the percent composition of the mixture.

64. If the product mixture from Trial III were spiked withthe reactant, which of the peaks in the example wouldgrow?

A. Peak 1

B. Peak 2

C. Peak 3

D. Without knowing the composition of the column,it cannot be determined.

6 5. If the column in the gas chromatography machine werepacked with a nonpolar polymer, what would be trueabout the retention times of polar and nonpolarcompounds?

A. Nonpolar compounds would have longer retentiontimes, because they bind the column less tightl1,than polar compounds.

B. Nonpolar compounds would have longer retentiontimes, because they bind the column more tightll,than polar compounds.

C . Nonpolar compounds would have shorter retentiontimes, because they bind the column less tightl;.than polar compounds.

D. Nonpolar compounds would have shorter retentiontimes, because they bind the column more tightil.than polar compounds.

6 6. Which of the following relationships accurately depictsthe relative retention times of the components?

A. Product A > ProductB > ReactantB. Reactant > Product A > product BC. Reactant > Product B > productA

D. Product B > Product A > Reactant

6 7. According to Table 1, what can be concluded aboutincreasing the temperature of the reaction?

A . It makes the reaction more favorable.B. It favors the formation of the kinetic product.C. It favors the retention ofreactants.D . It makes the reaction less favorable.

6 8. For the compounds associated with the crude productmixture from Trial IV, gas chromatography separatesprimarily by;

A . the size of molecule.B. the mass of the heaviest atom in the molecule.C. the charge ofthe compound.D. the affinity for the column.

69. The best gas to use as a carrier gas (to propel thevaporized components of the mixture thiough thecolumn) is:

A. carbon dioxide.

B. hydrogen.

C. helium.

D. water.

7 0. The major product from the reaction in Trial II can beisolated using which of the following techniques?

A. Column chromatographyB. Simple distillationC. Vacuum sublimationD . Acid-base extraction

71. Why must a strong base be used in Reaction I inFigure 1?

A . B_ecause alpha hydrogens are highly acidic, withpKu values around 2 _ 5.

B. B,ecause alpha hydrogens are highly acidic, withpKu values around lj _20.

C. B-ecause alpha hydrogens are weakly acidic, withpK. values around 2 _ 5.

D. B_e-cause alpha hydrogens are weakly acidic, withpKu values around 17 _ 20.

Passage Xl (euestions 72 - 79)

Recrystallization is the laboratory process ofdissolving asolid into a minimal amount of solvent, and thenprecipitating the solute from solution in a highly purecrystalline form. The procedure includes some filtering stepsto remove solid impurities (such as a boiling ctrips frcharcoal that may have been added) and ultimately to iiolatethe desired crystals. The general procedure for therecrystallization process is listed below:

I. Add the impure solid mixture into a minimalamount of solvent at room temperature to dissolvethe compound partially.

II. Heat the solvent- to its boiling point in a refluxingsystem. If not all of the solid has dissolved, add hols-olvent drop by drop until the solid is completelydissolved.

III. Once the solid is completely dissolved, hot_filter thesolution, avoiding any precipitation of the solute.

IV. Slowly cool the filtered solution until it reachesroom temperature, and then place the container intoan ice bath. Crystals should form slowly duringthis step of the procedure.

V. Filter the collected crystals from the solvent atreduced temperature.

VI. Rinse the crystals with a volatile solvent in whichthey are insoluble and allow them to air dry.

_ The procedure is generic and can be applied to any solidthat is soluble in an organic solvent. If the solid is toosoluble in the solvent, the quantity of solute that isrecrystallized is minimized. If the solid is not solubleenough in the solvent, the quantity of solvent is so large thatimpurities are dissolved as well. The ideal solvent has thesolid barely soluble at room temperature and fully soluble atthe boiling point of the solvent. The solvent is refluxed sothat it exists at its highest temperature (boiling point), butdoes not evaporate away.

Charcoal is a decolorizing agent that is sometimes addedto the first solution. Charcoal has a high affinity for coloredorganic molecules, because both charcoal and organicchromophores are rich in n-bonds. Charcoal is only added tothe solution when the desired compound is colorless and thesolution is colored.

The purity of the crystals can be quickly tested bylooking at the melting point range. A narrow rangecorresponds to pure crystals. As a general rule, larger crystJsare purer than smaller crystals.

7 2 . Which of the following crystals are the purest?

A. Small cubic crystals formed by rapid cooling.B. Small rectangular crystals formed by slow cooling.C. Long cubic crystals formed by slow cooling.D. Long rectangular crystals formed by rapid cooling.


73. The ideal solvent for recrystallization should have whichof the following properties?

I. A low affinity for the solid lattice

II. The solid should be highly soluble in the solvent atall temperatures

m. Impurities should be highly soluble in the solvent

A. I only

B. II only

C . I and II only

D. I and III only

7 4, The purity of the crystal CANNOT be verified by whichof the following?

A . Density of the crystal

B. Melting point of the crystal

C . Mass of the crystal

D. Index ofrefraction ofthe crystal

7 5. What is the role of horfiltering the solution in Step III?

A . To filter out any soluble impurities

B. To filter out any insoluble impurities

C . To filter out the compound

D . To reduce the amount of solvent in the flask

7 6. What is the purpose of using the ice bath in Step V?

A. To increase the amount of crystals formed byincreasing the solubility of the solid

B. To decrease the amount of crystals fbrmed byincreasing the solubility of the solid

C . To increase the amount of crystals formed bydecreasing the solubility ofthe solid

D. To decrease the amount of crystals formed bydecreasing the solubility ofthe solid

7 7. What is a potential problem if the crystals are formedtoo rapidly?

A. They grow to be too large.

B. They trap impurities from solution in their latticestructure.

C . It does not allow enough time for impurities toform in the lattice.

D. It prevents solvent from being incorporated into the

lattice structure.


7 8. What is NOT true about recrystallization?

A. The final crystals are purer than the starting solid.

B. The process requires knowing the boiling point ofthe solvent.

C. It is best carried out in a highly volatile solvent.

D . It results in less than 100Vo recovery of the originalsample.

7 9 . To which solution should activated charcoal be added

during a recrystallization experiment?

A. A solution with colorless impurities and a colorless

target crystal

B. A solution with colorless impurities and a colored

target crystal

C . A solution with colored impurities and a colorless

target crystal

D. A solution with colored impurities and a coloredtarget crystal

Passage Xll (Questions B0 - 86)

A researcher has chemicals organized according tomolecular mass. In one section there are three compoundswhich say only 100 grams/mole on their respective bottles.In an attempt to identify the three unknowns, the researcherlabels the bottles Compound A, Compound B, andCompound C. 1.00 gram of each compound is isolated andpurified.

A 1.00 mg sample of each compound was completelyoxidized. For all three compounds, 2.64 mg CO2 gas and1.08 mg H2O liquid was collected after complete oxidation.No nitrogen, sulfur or halides were found in any of thecompounds. No analysis for the oxygen content wasconducted due to the difficult nature of analyzing oxygencontent. Calculations show that there are exactly six carbonsand twelve hydrogens in each of the three compounds,implying that the three unknowns are isomers. After theformula analysis was complete, the researcher subjected eachof the three compounds to standard chemical tests. as

summarized in Table 1.

CrO3/H+ B12(CCla) (NO2)2C6H3COCl

Compound A turnedgreen

remainedbrown

precipitate formed

Compound B remainedofange

remainedbrown

no precipitateformed

Compound C remainedorange

turnedclear

no precipitatefbrmed

Table I

It should be noted that the organic product from thereaction of Compound A with CrO3/H+ does not turn litmuspaper red. Compound B, when treated with 12 and KOH,forms a yellow precipitate and acidic solution (a positiveresult for the iodoform test for methyl ketones). From themolecular mass and carbon count, it can be inferred that thecompounds contain one oxygen. The chemical tests indicatethat each compound is either an alcohol, an ether or a

carbonyl compound. A positive test with bromine in carbontetrachloride indicates the presence of an alkene. Each ofthese three types of functional group reacts differently, sopredictions can then be made about further reactions that theisomers can undergo, in order to further support their identity.

8 0. Which of the following IR absorbances would be thekey peaks in Compound A and Compound B?

A. Compound A: broad absorbance at 3500 cm-1;Compound B: sharp absorbance at 1710 cm-l

B, Compound A: sharp absorbance at 1710 cm-i;Compound B: broad absorbance at 3500 cm-1

C. Compound A: sharp absorbance at 2980 cm-i;Compound B: sharp absorbance at 1710 cm-1

D. Compound A: sharp absorbance at 1710 cm-l;Compound B: sharp absorbance at 2980 cm-1


81. A corrpound that turns clear with Br2lCCl4 and has one

degree of unsaturation CANNOT be:

A. a ketone.

B. an alkene.

C. an ether.

D. an alcohol.

Which of the lollowing types of compounds would losea peak in its 1HNMR when D2O is added?

A. Alcohol (ROH)

B. Ketone (RCOR)

C. Ether (ROR)

D. Aldehyde (RCHO)

Which of the three compounds would undergo a colorchange when treated with KMn04. considering KMnO4can oxidize alkenes, selected alcohols, and selectedcarbonyl compounds'/

A. Compound AB. Compound C

C. Compounds A and C

D. Compounds B and C

(NO2)2C6H3COC1 is NOT a good reactant to tesr fbrwhich of the following?

A. Alcohol (ROH)

B. Amine (RNHz)

C. Ether (ROR)

D. Thiol (RSH)

Compound A, by turning green with CrO3/H+ and notforming a compound that turns litmus red, is mostprobably:

A. aprimaryalcohol.B. a secondary alcohol.

C. a tertiary alcohol.

f). an ether.

A positive iodoform test can be associated with whichof the following?

A. A yellow precipitate confirming a methyl ketone.

B. A yellow precipitate contirming an aldehyde.

C . A black precipitate confirming a methyl ketone.

D . A black precipitate confirming an aldehyde.

82.

83.

84.

85.

86.

28s GO ON TO THE NEXT PAGE.

Passage Xlll (Question 87 - 93)

A student starts with 138 mg of 1,4-dimethoxybenzene(138.166 grams/mole) and treats it with 1.32 mL anhydrousHNO3 (1.50 g/ml, 63.012 g/mole) dissolved into 2.61 mLanhydrous H2SOa. This mixture is then heated to 35'C fortwenty minutes and then left to cool to room temperature.As the mixture is being heated, a brown gas appears in thereaction flask. To this mixture is added 5.0 mL of waterpreviously cooled to 0'C. This results in a white crystallineprecipitate forming instantly in the solution upon addition ofthe water.

The white solid is filtered from solution using a Hirschfunnel and washed with three aliquots of 1.0 mL of coldwater. The product is then isolated, dried and weighed. Themass collected for the isolated crude product (assumed to be2,S-dimethoxynitrobenzene (183 g/mole)) is 91.5 mg. Themelting point range is measured to be from 12.5'C to i4'C.

A small sample of the product is then pulverized and adrop of mineral oil is added to the pulverized sample. Themull (mixture of the mineral oil drop and pulverized solid) isstirred and added to the face of a salt plate for analysis usinginfiared spectroscopy. Key peaks in the IR show that thecompound contains a nitro group, a C-O single bond, and abenzene ring. Based on the infrared data and the meltingpoint of the crude product, the student concludes that thecrude product is 2,5-dimethoxynitrobenzene. The literaturevalue for the melting point of 2,5-dimethoxynitrobenzene is74 - 15"C. The error in the observed melting point isattributed to thermometer error and not an error in thereaction.

8 7. What is the percent yield for this reaction?

A. 25Vo

B.50VoC. 15Vo

D. 1007o

8 8. What can be deduced from the melting point range andthe IR data for the crude product?

A. The product is still wet with solvent.

B. The product is free of any significant impurities.

C . The product is impure.

D. The starting material was isolated rather than theproduct.

89. Filtering is used to separate a:

A. solid from other solids.

B. solid from liquids.

C. solid from gases.

D. liquid from other liquids.


9 0. What is the identity of the brown gas formed in the firstpart of the reaction?

A. Nz

B. CO

C. NO2

D. SOo

91. Which of the following is a possible cause for rhe

thermometer error?

I. Air has leaked into the thermometer resulting ilreduced vacuum within the thermometer.

II. A divot exists on the inside wall of thethermometer between the 45 and 55 'C mark.

m. A divot exists on the inside wall of thethermometer between the 85 and 95 'C mark.

A. I only

B. I and II only

C . I and III only

D. I, II, and III all explain the themometer error.

92. To further purify the product mixture, what type tr:laboratory technique should be carried out?

A . The crude product mixture could be dissolved inlrhot solvent and then recrystallized from the sol\ei-as it cools.

B. The crude product mixture could be filtered a secon:time with the same solvent.

C . The crude product mixture could be heated to liqur*form and then distilled via fractional distillation.

D. The crude product mixture could be treated rl;r:nitric acid and sulfuric acid again to react a:-iunreacted starting material.

93. Why is the addition of water carried out at a lc';temperature?

A . To increase the solubility of the product.

B. To decrease the solubility ofthe product.

C . To increase the acidity of the solution.D. To terminate the reaction by protonating the nir:

group.

GO ON TO THE NEXT PAGL

Questions 94 through 100 are NOT based on adescriptive passage.

9 4 . If a pure sample of a compound with S-stereochemistryhas a specific rotation of -55", then what is thecomposition of a mixture of that compound and itsenantiomer that exhibits a specific rotation of +22.?

A , 50Vo S-enantiomer with S\Ea R-enantiomerB . 33Vo S-enantiomer with 6lVo R-enantiomerC. 30Vo S-enantiomer wtth70% R-enantiomerD, 20Vo S-enantiomer withS}Vo R-enantiomer

9 5. The following TLC plate and column correspond to thesame components within the mixture and the samemobile phase (solvent).

Which correctly correlates the spot to the band?

A. BandA = o; BandB =0 ; Band C = oB. BandA = o; BandB = $; Bandc = oC. BandR = 0; BandB = o; Band C = oD. BandA = o; BandB = O; Band C = Q

Fractional distillation differs from simple distillation inall of the following ways EXCEpT:

A . The distilling column in fractional distillation hasmore surface area than the distilling column insimple distillation.

B. Fractional distillation is chosen over simpledistillation when the boiling points are close forthe components in the mixture.

C . Fractional distillation generates a higher purity thansimple distillarion.

D. Fractional distillation generates a greater yield thansimple distillation.

96.

Column TLC Plate

Copyright O by The Berkeley Review@ 281

9 7. Which of the following compounds, during an acid_baseextraction, would be extracted into 0. l0 M NaOH(aq),?

A. 4-Methyl benzoic acid

B. 3-Ethyl anisole

C. Ethyl benzoate

D. N-Merhylanilinc

98. In which of the following solvents or solutions is 4_hydroxybenzoic acid MOST soluble?

A. 0.10 M HCI(aq)

B. 0.10 M KOH(aq)

C. Water

D. Diethyl ether

9 9. Which of the following lab techniques will NOT workfor the task listed?

A. Using distillation ro separare two liquids.B. Using chromatography to separate two solutes.C. Using extraction to separate two gases.

D. Using crystallization to separate two solids.

100. Butanone shows all of the fbllowing EXCEpT:

A . high solubility in acetone.

B. a negative Jones test for oxidation.C. a mass spcctroscopy peak at 57 amu.D . a specific rotation of op = {1.2'.

l.c 2.A 3.A 4.A 5.C 6.41.D 8.A 9.D l0.B l1.D 12.D13. A 14. B 15. A 16. B 11. B 18. C19. C 20. D 21. D 22. A 23. A 24. A25. C 26. D 21. B 28. A 29. B 30. D31. D 32. B 33. A 34. B 35. C 36. D31. B 38. D 39. C 40. A 41. D 42. D43. C 44. C 45. D 46. C 41. D 48. D49. A -50. C 51. D 52, C 53. D 54. Bs5. A s6. B 51. A 58. C s9. B 60. A6r. c 62. C 63. B 64. A 65. B 66. D61. A 68. D 69. C 70. A 11. D 12. C13. D 14. C 15. B 16. C 71. B 78. C'19. c 80. A 81. A 82. A 83. C 84. C85. B 86. A 87. B 88. B 89. B 90. C91. B 92. A 93. B 94. C 95. A 96. D91. A 98. B 99. C 100. D

STOP! START GRADING.

Laboratory Techniques Passage Answers

2.

Choice C is correct. No matter what the boiling point of the liquid, no distillation should ever be done in aclosed svstem, otherwise pressure will build up (as T increases, so does P if there is constant V) and the systemwill er-entually explode. This eliminates choice D. The venting of the system must occur after the vapor hasbeen condensed and collected, otherwise the vapor will escape and nothing will be collected. This eliminateschoices A and B and makes choice C the best answer.

Choice A is correct. A temperature of 49.2"C indicates that the boiling point of the compound being collected isnear 49.2"C. The compound has to be one of the six given in the passage, so you must scan the first paragraph forthe compound with a boiling point closes t to 49.2 "C. The best choice for the identity of the comporlnd is t-t,Z-dichloroethene, which has a boiling point of 48"C. This makes choice A correct. This question might seem sostraight forward that you get nervous and think you're missing something. Keep in mind that the MCAT has arange of question difficulties and that straight forward questions appear.

Choice A is correct. When a boiling chip is added to the distillation flask, it sinks to the bottom, which is thehottest point in the solution (heat is added at the bottom of the solution.) Because vapor must escape from asurface and there is minimal surface area at the point where the solution is hottest, the solution can iuperheat(reach a temperature above its boiling point). Because heat rises (hotter solutions are less dense and thereforemore buoyant than colder solutions), the superheated solution will migrate to the top of the solution, at whichpoht it has surface area from which to evaporate. Because the solution is beyond i1s boiling point, it rapidlr'evaporates (bumps), causing a splash. To avoid this problem, the system needs more surfac" ur"u at the bottomof the boiling flask. A boiling chip is added to provide a rough surface on which vapor can collect and build upnear the bottom of the flask. This is the kinetic theory explanation of bumping. The best explanation is choiceA. Choices C and D should have been eliminated, because the boiling chip does not interact with the moleculesirr solution other than providing a surface area for them to collect on. Intermolecular forces are neither enhancednor diminished by a boiling chip. The boiling chip is stationary, so it does not help in the homogenization of thesolution. This eliminates choice B.

Choice A is correct. The majority of the vapor collects on cold glass, as opposed to warm glass, which eliminateschoices C and D. The condenser is cooled so as to condense the vapor. The condensed vapor then collects andtrickles into the collection flask in droplets. The actual condensation occurs in the condenser, however, whichmakes choice A the best answer.

Choice C is correct. Filtration can not be used when there is a solute, because the compound is fully dissoived intosolution and cannot be collected in the filter. This eliminates choices A and B. A liquid may be separated from a

dissolve solute by distilling (evaporating) the liquid away. This makes choice C or choice D the best answer.Because the boiling points are very different (the solute is normally a solid at room temperature, so it has tomelt before it can boil), there is no need to employ fractional distillation. The sample is best distilled usingsimple distillation, because the boiling points are drastically different. Choice C is your choice.

Choice A is correct. Simple distillation works best when the two compounds have the greatest difference rnboiling points. Choice A is composed of two compounds which have boiling points of 34'C and 102"C, a differenceof 68'C. Choice B is composed of two compounds which have boiling points of 34"C and 48"C, a difference o.'14'C. Choice C is composed of two compounds which have boiling points of 90"C and 102"C, a difference of 12"C.Choice D is comprised of two compounds which have boiling points of 128"C and 102'C, a difference of 26"C. Thelargest difference in boiling points between the components is observed with the compounds in choice A. Thisquestion is reaily a nomenclature question. You must translate from formula in the question into the IUPAC nameand boiling point from the text of paragraph 1.

Choice D is correct. Fractional distillation is employed to separate two liquids with relatively close boilingpoints. This means that it is employed to separate a liquid from a liquid. This eliminates choices A, B, and Cand makes choice D the best answer.

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5.

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Copyright O by The Berkeley Review@ LAB & SPECTROSCOPY EXPLANATIONS

9.

8' Choice A is correct' Fractional distillation depends on the condensation and re-evaporation of the more volatilecomponent in the mixture' Each cycle of condensation/re-evaporation further purifies the vapor so that the firstc ond ens a te o rr o r th;' ;;i;;; ilh ;" *; #;il:ii ;; il:TT;T#ll": ffi ff ili " EX"'J ;,1'J 1t'1 ","1T;surface of the distillation apparatus heats up so that the amount of condensation'on the inner walls is reduced,and the components condenie and re-evaporate less. This does not segregate the less volatile component as wellas when the walls are cooler. The net result is that over time, the distil'iate becomes less pure in terms of themore volatile component present in the vapor. The best answer is choice A. Azeotropes should have an equalbearing on the mixture whether the distillation is simple or fractional. The surface area does not affect theazeotropic mixture.

Choice D is correct. A long distilling column is used in fractional distillation, which is chosen when the boilingpoints of the components in the mixtlre are close in value. Close is defined as boiling points within 30.C of oneanother' In this question, you are asked for the best choice, which for fractionui dirtillution is the pair ofcomponents with the closest boiling points. In choice A, diethyl ether (b.p. = 34.6.C) and pentanol (b.p. =138'0"c) have a difference in boiling points of over 100'c. simple distillation'works with choice A. In choice B,tetrahydrofuran (b'p' = 65'4'c) and anisole (b.p. = 158.3'c) have a difference in boiling points of nearly 100"c.Simple distillation will work fine for choice B. In choice C, octane (b.p. = 725.7"C) and pentane (b.p. = 36.1.C)have a difference in boiling points of nearly 90"C. Simple distillatiorr *itt work j# fine in choice C as well, Inchoice D, isopropanol (b.p, = 82.3"C) and ioluene (b.p. = 110.6"C) have a difference in boiling points of under30"C' simple distillation will not work well, so fractional distillation should be used with choice D.

Choice B is correct' The major advantage of fractional distillation over simple distillation is that thefractional distillation procedure allows components to be distilled and collected in a purer manner. Fractionaldistillation can always be used no matter how close the boiling points of two substances may be. choice A isvalid' Fractional distillation works by providing surface area from which the vapor can continuously condenseand re-evaporate' Choice C is thus vaiid. The drawback to fractional distillation is the time required to carryout the distillation and the loss of the vapor that condenses throughout th; appurutrr. Fractional distillationtakes a larger amount of time than simple distillation, thus choice B i, un incorrect statement. The best answeris choice B.

Choice D is correct' Just as a water bath is used for better heat transfer (in both cooling and heating), a sandbath is chosen to transfer heat into the flask most efficiently. The sand bath covers the outside surface of theglass' and transfers heat-into the system through the glass. The sand serves to conduct heat (not insulate), sochoice A is eliminated. sand is not reactirr" *ith glasl, but neither is air. Choice B may or may not be a truestatement' but either way, it is not the best answer. sand is not particularly hygroscopic, so it will not affect thevapor pressure of water, eliminating choice C. The best answer involves heai t"ransfer, so choose D for optimalperformance.

'l'2' Choice D is correct' The role of the copper mesh in the distilling column is to increase the surface area ontowhich the vapor may condense so as to collect more condensate in ttre coiumn. The condensate can evaporate onceagain from the copper mesh, further distilling the condensed liquid- The copper mesh does not filter anything,it is inert, and it does not distribute heat anyLetter than the vaptr does. copper mesh just allows for fractionaldistillation to take place. The best answer is choice D.

13' Choice A is correct' Simple distillation is chosen when the boiling points of the components in the mixture arefar apart' Far apart is defined as boiling points that are *oru Ihun 30"C apart from one another. For thisquestion, you are asked for the best choice. The best choice for simple distillition will be the two componentswith the greatest boiling point difference. In choice A, diethyl ether'(b.p. = gi..e"C) and heptane (b.p. = 9g.4.C)have a difference in boiling points of over 60"C. simple distiilation will work fine with choice A, but you mustlook at the other choices to iee if there is a difference greater than 63.g'c. In choice B, tetrahydrofuran (b.p. =65'4'c) and hexane (b'P' = 68'9"c) have a difference in"boiling points or or"rty sl'c. simple distillation will notwork with choice B, thus fractional distillation must be used. Ctoice B is eliminated. In choice C, octane (b.p. =725'7'c) and pentanol (b'p' = 138.0"C) have a difference in boiling points or auo.rt 12"C. Simple distillation willnot work with choice C. Choice C is thus eliminated. In choice"d,lonlne 6; = 1S0.g"C) and anisole (b.p. =158'3"c) have a difference in boiling points of only 7.5"C. simple distillation witt not work in this case, thuschoice D is eliminated. Choice A is lhe best answei.

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Copyright @ by The Berkeley Review@ 2a9 LAB & SPECTROSCOPY EXPLANATIONS

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17.

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'19.

'l'4' Choice B is correct. The same boiling point for two stereoisomers is observed when the two stereoisomers areenantiomers, as they have the same functional groups and the same steric hindrance. They show the sameintermolecular forces, which results in the same boiling point. The two molecules in both choice A and choice Dare diastereomeric pairs, therefore both choice A and choice D can be eliminated. Diastereomers do not sho*-the same intermolecular forces, so they likely show different boiling points. Glucose and mannose are C-lepimers of one another, which is another way of saying that they are diastereomers. The only pair oienantiomers in the choices are 3R-bromocyclopentene and 3S-bromocyclopentene, making choice B the bestanswer. When a compound has only one chiral center, it cannot have a diastereomer, beciuse diastereomersrequire at least two chiral centers. Choices A, C, and D have multiple chiral centers.

15. Choice A is correct. Steam distillation involves boiling a mixture of components in the presence of water, so theboiling point is more important than the melting point. This eliminates ihoices C and b. Water boils at 100"C.so adding steam to help distill a mixture of organic compounds is best done when the boiling point of eachcomponent is greater than 100'C. This makes choice A the best answer.

Choice B is correct. Both compounds contain ten carbons and have three units of unsaturation, so they areisomers. All of the answer choices are isomers of some kind, so that didn't help much. They have the san.econnectivity (and therefore the same root for their IUPAC names), so they are not structural isomers. Thiseliminates choice D. Neither citral nor neral has a chiral center, so they are not optical isomers. Thi:eliminates choice C. They cannot readily contort or rotate about a bond to inter-convert between one another, scchoice A is eliminated. Conformational isomers must be able to rotate or flip about sigma bonds to interchangebetween one another, like the gauche and anti conformations of a compound lik" brtur-r". Citral and neral var' aithe positioning around the double bond between carbons 2 and3, so they are cis-trans isomers. Cis-trans isomer-.are more correctly defined as geometrical isomers, so the best answer is choice B.

Choice B is correct. Standard distillation of lemon grass oil at 229' exposes the compounds in lemon grass oil tcextreme heat, which can decompose, destroy, or denature a compound. Decomposition can certainly oc-cur at hig:temperatures, as is seen with most biological molecules, so choice A is eliminated. High heat in the pr"r"rrc" otoxygen gas results in the oxidation of most organic compounds, so choice C is valid, ind therefore LhminatecCitrai can react with itself at the n-bond carbons, so at high temperature, polymerization is a possibility. ThL.eliminates choice D. Without a source of hydrogen, hydrogenation is not a vlible possibility, so hydrogenatioris not a iikely result when heating a citral-neral mixture. This makes choice B the best answer.

Choice C is correct. Ether is chosen to extract the organic components from the aqueous layer, because it i.immiscible in water and it has a high affinity for organic molecules. The immiscibility of water in ethe:eiiminates choices A and B. Citral dissolves into the ether layer, which favors choice C over choice D. Thewording is not perfect, so we must also consider that ethers float on water, given that their densities are les.than 1.00 g/nrL. This supports choice C as the best answer.

Choice C is correct. The term anhydrous provides a big clue here, because it refers to something without wate:which implies that water must have been removed from it, otherwise the fact that it is anhydrous is irrelevant.The best answer is that an anhydrous salt is added to the ether layer to absorb, and thereby remove wateiChoice C is the best answer. Isomerization of neral into citral is done with either light or heat, but an insolublesalt has no bearing on the isomerization. Choice A is eliminated. Sulfate, SO42-, is rich in oxygen an;magnesium dication, Mg2*, is poor in electrons, so neither could be a reducing agent. Neither

"u. i"drr." *,

aldehyde into an alcohol, so choice B is eliminated. Selectively binding a compound such as neral might souniiike something a chemist would want to do, but there is no reason to believe that magnesium sulfate could dithat. Choice D is eliminated.

Choice D is correct. Neral only varies from citral in the positioning of one double bond, so in solution, it exhibi-similar intermolecular forces as citrai. \A4rile the melting point may be impacted by the cis versus trans positiorof the double bond (given that packing depends on steric hindrance), the boiling point is less impact"a. Th.boiling point of citral is 229"C, so the boiling point of neral should be close to thit value. Choice b is the bestanswer. Given that steam distiilation is employed, the boiling point of neral is above 100"C, so at the t,er.,-least, choices A and B should have been eliminated.

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Copyright @ by The Berkeley Review@ 29o LAB & SPECTROSCOPY BXPLANATIO\S

21' Choice D is correct. Vitamin A is made from citral, so it is likely a hydrophobic aldehyde too. It is larger thancitral, so it will have a higher melting point, eliminating choice A.

" It d hydrophobic, so it has a low watersolubility, eliminating choice B. Citral is a liquid with i high boiling point, ,o it hu, a low vapor pressure.vitamin A is likely the same, so choice C is eliminated. Being a tl*pia, Vitamin A has a higir afiinity forlipids, which makes choice D the best answer. Vitamin A is a lipil-soluble drug.

22' Choice A is correct._ The percentage of caffeine in the tea leaf is the mass of the extracted caffeine divided bythe mass of the tea leaves multiplied by 100%. The amount extracted is 0.028 grams (28 milligrams) and theoriginal sample of tea leaves weighed 2.0 grams, so the value is 0.028/2x 100% = 0.014 xI0O"/, = \.4"h. Choice Ais the best answer.

23, Choice A is correct. Because caffeine is extracted with both hot water and methylene chloride, it must besoluble in both hot water and methylene chloride. This eliminates choices C and o. tn the second extraction,methylene chloride is used to remove the caffeine from the water, so caffeine must be more soluble in methylenechloride than water. This eliminates choice B and makes choice A the best al1swer.

24' Choice A is correct. A stream of nitrogen, when flowing across the surface of a liquid, will force the gases presentabove the surface away' thus reducing the partial pt"s.rr" of the vapor above soiution. If the air above thesolution is saturated with vapor before the nitrogen gas stream, lt witl not be saturated afterwards, aliowingnew vapor to form' The solution continues to vaporize (evaporate) to equilibrate with the vapor present abovesolution, but because the vapor is continualty being ,e*ou"d, it can never reach equilibrium. Eventually, theliquid evaporates away completely. This is best explained as choice A.

25' Choice C is correct. A quick and easy test for the identity of a solid is the mixed melting point test. By adding asmall amount of the compound you believe the sample to be, it is possible to observe lhl melting point for themixture of the known and unknown compounds. If they are the ru-" .o-pound, then the meltin! point will besharp and exact. If_the two compounds are not the same, then the melting point wili be broad. TIii means thatto prove that an unknown is caffeine, the unknown is best mixed with cafieine for a melting point measurement.This makes choice C the best answer.

26' Choice D is correct. Methylene chloride is common as a solvent, so it must be a stable compound. Even if it wereunstable, the quantity should not be the factor that triggers its instability. Choice A is eliminated. Choice Baims for the test taker who is asleep at the wheel, so io speak. The top and bottom layers have to do withrelative densities, not relative masses. Less dense species float in more dense mediums, no matter how massivethey are. Choice B is eliminated. The goal of extriction is to isolate the product, so hopefully any method orprocedure would aim to maximize what is isolated. This makes choice D a better ir-rr*"r" thin choice C,although that logic does not explain why it is true. This is where you have to keep in mind that it's astandardized exam, and not an exam of what you know. Using test logic to choose D in this case is perfect,because you save time' A mathematical example is shown here Io r.rppo"rt the answer, but it is unnecessary toconsider during the actual exam.

Consider a solute that splits evenly between two solvents. In one case, 15 mL of Solvent A is used with 10 mLsolvent B. This would result in 60% of the solute being extracted into Solvent A. In a second approach, the 15 mLof Solvent A is added in three 5-mL portions. In portion 7,33.3"/' goes into Solvent A and 66.6% remains inSolvent B. In portion 2, based on the percentage oi th" original matIrial, the partitioning ratio of Solvent A toSolvent B is 22.2"h,to 44'4"/,. In portiorr 3, based on the percentage of the original material] the partitioning ratioof Solvent A to Solvent B is 14.8% to29.6"k. Adding up the percentages extracted into Solvent A from the threeseparate extractions, we get: g3.g% + 22.2% + 74.g% = 70.3"/".

Three separate smaller extractions yields 70.3% while one larger extraction yields 60%. Three smallextractions do in fact generate a better result than one big extraction, so choice D is the best answer.

27 ' Choice B is correct. Initial treatment of the tea leaves with hot water dissolved the water-soluble products,including caffeine' Methylene chloride is added to this solution to remove organic substances thai are onlypartially soluble in water at lower temperatures. Because caffeine is water solui'le at elevated temperatures, itis still partially soluble at a lower temperature, so not all of the caffeine can be extracted into the methvlenechloride layer. The best answer is choice B.

Copyright O by The Berkeley Review@ 29I LAB & SPECTROSCOPY EXPLANATIONS

28' Choice A is correct' Sublimation is the physical process of converting a solid directty into a gas, so choices C andD are not valid answers. As a laboratory technique, it is used to convert one solid into a gas, while leavingbehind the other solids' The gas can migrate u*uy f.o- the other solids and then be collected on a cold surfaceas a pure sample of the solid. Purifying a solid from a liquid does not work, because the liquid can vaporize,resulting in more than one compound ln the gas phase. ThiJ ehminates choice B. The best answer choice is A.

29' Choice B is correct.- Exactly 0'75 grams of the original 1.0 gram sample was isolated from the organic layer. It isassumed that the other 0.25 grams of sample were dissolrrld ir-rto thl aqueous layer. The partition coefficient isthe amount that dissolves into the organic layer divided by the u-or;r,t that dissolves into the aqueous layer,which in this case is O.zs/0.2s = 3.0. This makes choice B the best answer.

30' Choice D is correct. As stated in the passage, the solubility of 4-methylaniline in water at ZS"Cis 3.g grams/100mL and in diethyl ether at 25'C is 22.8 grams/ 100 mL. The partition coefficient is the solubility of the sample indiethyl ether divided by the solubility or tn" sampie in water. The partition coefficient is found as follows:2? Rl

f;=H=#=#=#=uChoice D is the best answer.

31' Choice D is correct. As stated in the passage, the solubility of 4-methylaniline at 25'C in water is 3.g grams/100mL and the solubility of 4-methylaniline itzs'c in diethyl ether is )z.s grams1100 mL. This means that themaximum amount of 4-methylaniline that can dissolve inio 10 mL diethyi ether is 2.2g grams. The maximumamount of 4-methylaniline that can dissolve into 10 mL water is 0.38 grams. The maximum total mass betweenthe two solutions is 2.66 grams. Adding 3.00 grams exceeds this alount, so not all of the 3.00 grams of .1-methylaniline can dissolve ilto the two solutions combined. The undissolved portion will not partition betweenthe two solvents, but instead forms a precipitate at the bottom of the flask. This precipitate makes the lowerlayer appear to ha_ve_more 4-methylaniline than it actually has dissolved into solution. This makes choice Dthe best choice' Had the compound been fuliy soiuble, ihe partition coefficient should have been the same.assuming any errors remains the same in magnitude.

32' Choice B is correct. Because 4-methylaniline is more soiuble in the diethyl ether than water, the pipette coulcibe rinsed with diethyl ether to wash off any residue of 4-methylaniline. Heating and cooling the pipette ispointless once the compound has adhered to the walls of the pipette. Water is J poor choice"to remove theadhered soiid, because 4-methylaniline is relatively insoluble irrwater. The residue on the glass was lost frorr.the diethyl ether layer, so if it is recovered, it is best recovered into the diethyl ether layerf pick choice B forbest resuits.

-i3' Choice A is correct. If the ether were not anhydrous, it would contain some water impurity. This implies that irrthe 10 mL of ether, there is actually a smali portion of water and therefore less than 10 mL of ether. Thi-.eliminates choices C and D, because they indiiate an excess (a value greater than 10 mL) of ether. With lessthan 10 mL of ether, less 4-methylaniline than expected wouid dissolve" into the ether layer. This means that asmaller amount of 4-methylaniline would be isoiated from the ether layer. This would lower the numeratorand thus lower the value for the calculated partition coefficient. This makes choice A the best answer.

Sl Choice B is correct. Diethyl ether is used because it is immiscibte in water and the organic compounds are solubl.'n it' Any solvent that_is used in place of diethyt ether must have the same properties. tetrihydrofuran, THf:s a cyclic ether, so it has the same properties as diethyl ether. This eliminates choice A. iyclohexane an;nethylene chloride are both organic solvents that are immiscible in water, so choices C and D are eliminateiEthanol is miscible in water, so it cannot be used in lieu of diethyl ether. This makes choice B the best answer.

I r Choice C is correct. For two layers to {orm, the organic solvent must be immiscibie in water, eiiminating choi;e\' The density is not important, as long as its not the same density as water. As long as it has a densi:..,lifferent than water, higher or lower, the system will have a layer on top and on bottom. rnls eiimrr1ate.:hoice B. The solute should be soluble in the organic layer, so that it can be removed from the aqueous lar-e:Choice C is the best answer. If its boiling point iJ less than room temperature, then it will exist u, u gu5, at roor:-,:emperature. To be useful as a solvent, it needs to be a liquid at room temperature. This eliminates choice D.

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Choice D is correct. Ether is chosen as a solvent, because it is immiscible in water and it is a liquid at roomtemperature. The organic solvent used in lieu of ether must also be immiscible in water and a liquid at roomtemperature. Choices B and C are eliminated, because they are not liquids. Choice A (ethanol) dissolves intowater, so choice A is eliminated. Only choice D remains. Chloroform is an immiscible liquid, so choose D.

Choice B is correct. To extract the contents of Tube 4 from the organic layer, a basic aqueous solution was used, soTube 4 must be at a high pH. This means that the solute is likely in its anionic form. To remove the solute(dissolved anion) from solution, the solution must be neutralized. To neutralize a basic solution, an acid isadded. This eliminates choices A and D. The acid should be strong, so it can fully react with the base insolution. The best answer is choice B, hydrochloric acid.

Choice D is correct. Tubes 2 and 5 contain the component that does not react with either the strong base or theweak base to become more water soluble. Dinitrobenzene has no acidic protons, given that all of its hydrogensare bonded to carbon, so it will remain neutral and organic soluble during the entire separation. It is not possibleto convert dinitrobenzene into a charged species with just acid and base reactions. The best answer isdinitrobenzene for both Tube 2 and Tube 5, so choice D is the best answer.

Choice C is correct. The second proton is by definition less acidic than the first proton, so the second pKu isgreater than the first pKu. Choices A and B are eliminated. The value of pK6 can be found by subtracting thepKu for the conjugate acid from 14. Because the pKu for benzoic acid is lower than the pKu for the first proton ofresorcinol, the pK6 for the conjugate base of benzoic acid (benzoate) must be greater than the pK6 for theconjugate base of the first acidic proton of resorcinot (pK62). The first proton lost by a diprotic acid is the secondone gained by the fully deprotonated form of the conjugate base. This makes choice C the best answer.

Choice A is correct. Because the melting points for the compounds isolated by Student II showed a narrowerrange than the compounds isolated by Student I, the products from the scheme of Student II must have been purerthan the compounds from the scheme of Student I. Thus, the scheme for Student II must have worked better thanthe scheme for Student I. The scheme for Student II isolated the stronger acid first (Tube 4), the weaker acidsecond (Tube 6), and the neutral species in the organic layer (Tube 5). It can be concluded that the scheme forStudent II worked while the scheme for Student I did not. The best answer is choice A.

Choice D is correct. 3-Methylaniline is an amine, making it a basic compound. Basic compounds dissolve intowater under acidic conditions. F{owever, in the reaction scheme for Student II, only base is added to the waterand not acid. As such, the amine is not protonated, and remains in the organic layer for both extraction steps.The result is that it starts and finishes in the organic layer, which finishes in Tube 5. A small amount maydissolve into water, so the best answer is choice D, predominantly in Tube 5.

Choice D is correct. A compound can be separated using extraction techniques as long as it is not reactive in any ofthe mediums. 4-Ethylcyclohexanone is a large ketone, which is not reactive with the ether layer, acidic water,or basic water. 4-Ethylcyclohexanone can be isolated using acid-base extraction techniques, so choice A iseliminated. 3-Bromophenol is a mild organic acid, which is not reactive with the ether layer or acidic watet,and undergoes reversible deprotonation in basic water. 3-Bromophenol can be isolated using acid-base extractiontechniques, so choice B is elimirated. 3-Nitrobenzoic acid is a mild organic acid, which is not reactive with theether layer or acidic water, and undergoes reversible deprotonation in basic water. 3-Nitrobenzoic acid can beisolated using acid-base extraction techniques, so choice C is eliminated. Ethyl benzoate is an ester. Esters canundergo hydrolysis under acidic aqueous conditions or saponification (base catalyzed hydrolysis) under basicaqueous conditions. Because esters react under the conditions used in acid-base extraction, ethyl benzoate cannotbe isolated using acid-base extraction techniques. Choice D is the best answer.

43. Choice C is correct. An Ri value of 0.92 implies that the compound is highly soluble in the solvent and that thecompound has a very low affinity for the surface of the TLC plate. A change in solvent from nonpolar to polarwould show a reduction in solubility, and thus a reduction in the Rp value. The Ri cannot be negative, so choice Dis eliminated. The R6 vaiue cannot be greater than 1, so choice A is eliminated. The only value greater than zerobut less than 0.92 is choice C, making it the best answer.

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Copyright O by The Berkeley Review@ 293 LAB & SPECTROSCOPY EXPLANATIONS

++. choice C is correct. Because a sugar contains a large number of hydroxyl groups, it will be most soluble in asolvent with a hydroxyl group, such as an alcohol. This is baseclon the idea'that,,like dissolves like.,, Analkane, ether, and ketone do not have a hydroxyl group, so the best answer is choice C.

Choice D is correct. The basic principle behind thin layer chromatography is having the spots migrate up theriate at different rates by exploiting their differences in attraction to the solvent and affinity to the TLC plate.Spotting should be done in a wav that allows the spots to migrate. If the spots are placed Uetow the top of tn"'oh ent (submerged into the solvent), they can dissolve into the solvent. This makes choice D the best answer.Choice A should be eliminated, because tire rate of migration does not depend on the placing of the solvent a.d:t'ots' The spots are not going to become_ saturated, * tn"y will not rise up the plate too rapidly. Choice B ist-imrnated' The spots will dissolve in al1 directions, so radial migration ii an ii-rcor.ect answer. Choice C isrliminated. Choice D is better than the other poor answers.

Choice C is correct. For column chromatography, the best separation results from the solvent that shows thei:eatest difference in ft1 values in terms of ratio. In hexane, the R1 values are different by a factor of 2, while ine:hanoi, the R6 values are different by a factor of 3. Do not subtrict the R1 values to deiermine the separation,'re R1 values must be divided. A better separation is observed with ethanol than hexane, so choices A and Bsrould be eliminated. Because the R1 valnes are small, the column should be short, so that less solvent is:equired and the comPonents do not remain on the column for any unnecessary time after separation has been=stablished' A mixed-solvent of hexane and ethanol wouid not work well, because the solutes will travel faster::rd the ratio of the R1 values will decrease only slightly. Choice C is the best answer.

Choice D is correct. Carbon tetrachloride is a nonpolar solvent, so it shows similar solubility properties (and::'lus similar TLC results) as other nonpolar solvents. Acetone and propanal both contain the carbonyl functional:roL1p/ which makes them polar. This eliminates both choices a ana

-B. Ethanol is not only polar, tut it also is

:ap.able of forming hydrogen bonds with solute particles. Choice C can thus be eliminated. 'Because

pentane is a1''-''drocarbon, it is nonpolar. Pentane will behavl most like carbon tetrachloride, so the best answer is choice D.

Choice D is correct. The Ri value is defined as the distance that a spot travels up a TLC plate (droo1) divided by:he distance that the solvent travels up a TLC plate (ds61rr"rl1). Because the spoi migrates due to its affinity forlne solvent, the spot can never migrate a greater aiiiance'ttran the solveni migrates. This means that the;5oh'sn1-wi11 always be a greater value than the droo1. This consequently means ttat the R; value can never be{reater than or equal to 1.00. Choice D, 1.05, is nof'possible. An R1'value of 0.0 simply -"urls that the spot doesnot dissolve into that particular solvent, and therefore it does not move. An R1 ,rat.re of 0.0 is possible.

Choice A is correct- Trimyristin is a fatty acid triglyceride, meaning it is a triester. Long chain esters arenvdrophobic, because they cannot form hydrogen 6onds and they aie reiatively nonpolar. This eliminatesihoices B and D. THF is an ether, making it nydropnobic. Because of the oryg"n in the ring, THF is a mildlyp'o1ar solvent, although it is aprotic. The best answer is choice A.

50' Choice C is correct. Alumina is a polar adsorbent, so it hinders polar solutes more than nonpolar solutes.\onpolar solutes come off the column first, resulting in a fast elution time. Benzoic acid is hyirophilic andcolar, eliminating choice A. Glycerol is also polar, so choice B is eliminated. Glycine is polar and hydrophilic,so despite the presence of ethanol as the eiuting solvent, it is stilt hindered by the alumina. Choice D iselimi-r-rated. A nonpolar compound like o-xylene exhibits _little affinity for the alumina, so it travels quicklyi-lorvn the column. Using a nonpolar solvent liice ligroin speeds it up even"more, so choice C is the best answer.

51' Choice D is correct. The compound with the iarger R1 value_ climbs a TLC plate faster, so it travels through acoiumn faster as well' This means that it should nu,r" i smaller elution tlme. This eliminates choices A and B.The R1 values differ by a factor of two,, so the relative migration distance in a given time differs by a factor oftrr-o This means that their rates also differ by a factor oi t-o, meaning that their times for traveiing a givendistance shall also differ by a factor of two. This makes choice D the besianswer.

52' Choice C is correct. Salicylic acid is highly polar, as stated in the passage. The greatest R1 value corresponds tothe greatest migration, which is observed inlhe most polar solveni. H"iun" andhgroin are nonpolar, so choicesA and B are eliminated' Methanol is more polar than pentanol, so choice C is the best answer.

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Copyright O by The Berkeley Review@ 294 LAB & SPECTROSCOPY EXPLANATIONS lLfiWr

5J. Choice D is correct. The BrockmarLn Activity rating for alumina has to do with the amount of water bound to thecolumn. According to the passage, a rating of III refers to a column that is 6ok water by mass. A rating of IV hasmore water than a column with a rating of III, so choice A is eliminated. Because there is so much watei bound toalumina, there is less binding of solute to the column. Having more water bound with a rating of IV than a ratingof III, a rating of IV represents a lower affinity than a rating of III. This eliminates choice B. Alumina is madeof aluminum oxide, so there is no silicon present. This elimrnates choice C. A Brockmann Activity rating of III iscaused by 6% water by mass, so a rating greater than III results from more than six percent water by *uri. Gi,r"r-,that a rating of I is perfectly anhydrous, a rating of IV requires more than 6 grams of water p"i f OO grams ofcolumn, so choice D is the best answer.

Choice B is correct. Methanol has a high affinity for alumina, so the binding of methanol to the alumina isexothermic. This eliminates choice A. When methanol is added to the column, heat pockets can form, whichresults in the formation of vapor pockets in the column. These vapor pockets can ctaik the gel, resulting in abreak in uniformity for the pathway of solute through the column. Choices B is the best answer. Methanol ismiscible in hexane, but even without knowing such a fact, there is no reason to assume that rate at whichmethanol migrates through the column has any dependence on the miscibility of the two solvents. Thiseliminates choices C and D.

Choice A is correct. A higher Brockmann Activity rating represents a greater amount of water on the alumina,which corresponds to a lower affinity for polar molecules. The lower affinity is due to the fact that water istaking up space that the polar molecule would bind if it was open. This makes choice A the best answer. Waterdoes not cause the contraction of alumina. If anything, it may dissolve alumina with such a high affinity forwater. Choice B is eliminated. Alumina is polar, so although the addition of water may affect the polirity,choice C is incorrect, because alumina is polar to begin with, not nonpolar. Water does not oxidize ilumina,given that aluminum is fully oxidized in At2o3. This eliminates choice D.

Choice B is correct. Alumina is polar, so the solute for which it has the highest affinity must also be polar.Choice A, 1,3-cyclohexadiene, is nonpolar and hydrophobic, so it is eliminated. Choice C, ethyl butanoite, isan ester, so while it has a dipole, it is classified as nonpolar species and hydrophobic. Choice C is eliminated.Choice B, cyclohexanol, is polar and able to form hydrogen bonds, so it has a high affinity for alumina. ChoiceD, 3-methylpentanal, is an aldehyde, so it polar and should also exhibit an affinity for alumina. A proticspecies has a greater affinity for alumina than an aprotic species, so choice B is a better answer than choice D.

Choice A is correct. Because like dissolves like, fluorenone should be most soluble in another ketone. Assolubility increases in the solvent, the affinity is said to be higher. The only ketone in the answer choices isacetone, so choice A is the best answer.

Choice C is correct. Fluorenone is more polar than fluorene, so independent of solvent, fluorene should elute fromthe column first. The fact that a nonpolar, hydrophobic solvent is being used makes the difference in migrationrate even greater. Fluorene is a nonpolar species, so it has a low affinity for silica gel. This eliminates choice A.Fluorene is in fact lighter than fluorenone, but in a column, much like free fall, the mass of the species does notfactor into its migration rate except for the steric hindrance associated with larger, more massive molecules.Choice B is eliminated. Fluorene does in fact have a higher affinity for cyclohexane then fluorenone does, sochoice C is a valid statement. Fluorene is smaller than fluorenone, but there are no pores in silica gel thatseparate by molecular size (molecular sieves do that), so choice D is eliminated. Choice C is the best answer.

59. Choice B is correct. The best solvent is the one that allows the two species to separate by the most. This isdetermined by comparing the R1 values of fluorene and fluorenone in each of the iolvents. fne greatest ratiorepresents the best solvent. For Liquid II, the ratio is 0.33 to 0.24, which is less than 2. For Liquid III, the ratiois 0.51 to 0.19, which is slightly greater than 2.5. For Liquid IV, the ratio is 0.55 to 0.41, whiah is less than 2.For THF, the ratio is 0.42 to 0.31, which is slightly over 1. Liquid II is best, so choice B is the best answer.

60. Choice A is correct. Aiumina is a polar adsorbent and ligroin is a nonpolar solvent, so nonpolar species will exitthe column before polar species. Of the three compounds, biphenyl is nonpolar and hydrophoblc, so it will elutefrom the column first. This eliminates choices B and C. Toluic acid is polar and hydrophilic, so it will elutefrom the column last. This makes choice A the best answer.

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Choice c is correct' \4rhen the solvent is flush with the top of the silica gel, all of the solute is aiso flush withthe top of the siiica gel. This means that the solute will enter the gel ui th" same time, resulting in sharperbands gorng down the column. Choice C is the best answer. The sJmple must dissolve into the solvent to beeffective, so it cannot float, making choice A incorrect. Whether the solvent is flush with the top of the silicagel does not impact whether or not the solute adheres to the glass. If that is an issue, it will result in problemsas.the sample migrates down the column, no matter where the solvent level starts. choice B is eliminated. Thesilica ge1 should not dry out, so choice D is incorrect. The best answer is in fact choice C.

Choice C is correct. Although the world miglt be a better place with words like "sandophilic,, in it, there is nosuch term' This unfortunately makes choice D incorrect. tfr" sample must reach the silica gel, so choice A iseliminated' The sand does not interact with the silica gel, so it has no impact on the polarity of the gel. ChoiceB is eliminated. The role of the sand at the top of the;ilica gel is ensure that the top of the silica gi stays flat,so all solute travels the same distance through the column no*matter where it starts.

^Without the sXnd, pouringsolvent into the system could disturb the flat surface. Choice C is the best answer.

Choice B is correct. Silica gel is a polar adsorbent, so its affinity for polar species is high while its affinity fornonpolar species is low. The question states that when using siiica git, ,"ronpotar species elute first, no matterwhat solvent is used. This simply means that the affinity of"the adsorbent *.rrt ort*"igh the affinity for thesolvent' Polar compounds do follow the "like dissolves like" rule, so choice A is eliminated. Nonpolar speciesremain nonpolar when polar solvent is added, so choice C is eliminated. polar compounds do not form largemolecular networks any 1nor9 than nonpolar molecules would, so choice D is eliminated. Choice B is the bestanswer/ because silica gel is highly polar and it thereby hinders polar species by binding them as the traveldorvn a column.

Choice A is correct. According to the data in Table 1, the reactant is in the lowest concentration in Trial III(76?''), so Peak 1 must be due to reactant. In GC graphs, the peak with the smallest area correlates to the lowestconcentration. Upon adding reactant, spiking the milture wiih reactant, peak 1 should be the one to grow. Thisis done to confirm the identity of a peak. The best answer is choice A.

Choice B is correct. If-the column is packed with a nonpolar material, then nonpolar compounds will bind to thecolumn more so than the polar compounds. This allows the polar compounds to travel ttuough the column withless resistance. Polar compounds will th,rs have shorter elutiln times. The best answer is choice B.

Choice D is correct._ Figure 1 represents the output for Trail III at 15'C. According to Table 1, the relativeconcentrations at 15"C are Product A > Product B > Reactant. Peak 2 is the largest of area, so peak 2 must beassociated with Product A. Peak 3 is the second largest, so Peak 3 is associat"a #itn product B. peak 1 must bedue to the reactant. The first peak to appear haslhe shortest retention time, so the reactant has the lowestretention time' This eliminates choices B and C. Product B comes off last, so Product B has the longest retentiontime. The best answer is choice D.

Choice A is correct. By definition, an increase in temperature favors the formation of the thermodynamicproduct, so choice B is eliminated. Choices C and D are the same answer, but worded differently. There cannotbe two best answer choices, so choices C and D are both eliminated. The MCAT writers love to place the sameanswer twice with slightly different wording. From Table 1 in the passage, it can be seen that the amount ofreactant decreases with increasing temperature, so the reaction ir proceJding more towards products. Thisimplies that the reaction is more favorable with increasing temperature. The best answer is choice A.

Choice D is correct. Gas chromatography separates by differences in migratory rates of vaporized organiccompounds' Although size and mass are variables ihat affect the migiato ry rcte of a gas, the primaryiifferences in retention times in this example can be attributed to the atlractivl forces between the packing::rateriai in the column and the compounds. If mass and size were the primary factors, then good separation" ' ould not be possible for the two products from the reaction in the passage. the'charge of the compo'nd woutd- ave an effect if charged .o^porrt ds could easily vaporize. Because charf"d .o-poun"ds do not reaiily vaporize,:j i gas chromatography apparatus does not distinguish charged species.- atthough the passage doesn,t mention': :ackground knowledge should tell you that ionic compoirnds (1ike NaCl) do"not vaporize easily. The best;:-i'.\-er is choice D.

I72

64.

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^:--: 3 by The Berkeley Review@ 296 LAB & SPECTROSCOPY EXPLANATIONS

69.

70.

Choice C is correct. The best gas to propel through the gas chromatography apparatus is an inert gas that doesnot react with any of the compounds. Nucleophiles can react with carbon dioxide, so choice A is eliminated.Hydrogen gas can react with air, so hydrogen gas should not be used. This eliminates choice B. Helium is theonly inert gas of the choices, so helium should be chosen. The best answer is choice C. Water is too reactive.

Choice A is correct. The major product from Trial II is the less hindered compound. It has no acid-baseproperties, so acid-base extraction is not viable. Choice D is eliminated. The boiling point of the major productis likely to be close to the boiling point of the minor product, so only fractional distillation would work, notsimple or vacuum distillation. Choices B and C are eliminated. Choice A is the best answer by default.

7'1,. Choice D is correct. A strong base is used to deprotonate a very weak acid. This eliminates choices A and B. ThepKu for a hydrogen on a carbon that is alpha to a carbonyl is around 19, so the best answer is choice D.Sometimes trivia is necessary to solve a problem.

Choice C is correct. Longer crystals are indicative of higher purity. Impurities disrupt the crystal lattice, socrystals cannot grow as large. This eliminates choices A and B. Slow gradual formation of crystals yields thefinest, purest crystals. This makes choice C the best answer.

Choice D is correct. The ideal solvent should not bind to the lattice structure of the crystals, otherwise it couldincorporate as an impurity in the crystal lattice as it precipitates from solution. This makes Statement I valid.The ideal solvent for recrystallization should have the solid insoluble at low temperatures and highly solubleat elevated temperatures. This eliminates Statement IL The impurities should be highly soluble in the solventso that they do not precipitate out and possibly get trapped in the crystals. This makes Statement III valid.Statements I and III are both valid, so choice D is the best answer.

Choice C is correct. Physical properties can be used to determine the purity of a compound. If the density of thecrystal is precise, it implies that there are no impurities in the crystal. The melting point of the compound(crystalline or not) can be used to determine the purity. A broad melting point range implies that the compoundis not pure. The mass of the crystal has nothing to do with its shape and purity. This makes choice C an invalidstatement, and therefore the best answer. The index of refraction is another of the physical properties thatdepends on the lattice structure. The index of refraction varies with impurities. The best answer is choice C.

Choice B is correct. By keeping the solvent and glassware hot, the original solid remains in solution during thefiltration. The purpose of filtering at this point is to remove any solid (insoluble) impurities that are present, sothe target compound should remain in solution for highest yield. This makes choice B the best answer choice.

Choice C is correct. The ice bath serves to cool the solution to 0'C. The solubility of a solute decreases as thetemperature of the solution decreases, causing more solid to crystallize (precipitate) out of solution. Overall,the purpose of the ice bath is to cool the solvent to decrease the solubility and thus increase the amount ofcrystals formed. Pick C for correctness.

Choice B is correct. If crystals form too rapidly, then impurities (such as solvent molecules) can get trapped inthe crystal lattice. If an impurity happens to collide with the surface of the crystal, it has time to escape if thecrystals are formed slowly. However, if the next layer of the crystal forms too quickly, the impurity cannotescape. The best answer is choice B.

78. Choice C is correct. Ideally, the crystals formed from recrystallization are purer than the original solid,otherwise recrystallization would be pointless. Choice A is a valid statement. Considering that the solventmust be heated to its boiling point to maximize solubility (and thus minimize the amount of solvent required), itis important to know the boiling point of the solvent. This makes choice B valid. If the solvent is highlyvolatile, then it cannot be raised to a very high temperature (the more volatile the liquid, the lower its boilingpoint), thus more solvent is required to fully dissolve the solid. The more solvent required at maximumtemperature, the more solvent present when the solution is cooled. The more solvent present whenrecrystallizing, the less solid that will crystallize out from solution. This reduces the percent recovery, makingchoice C the incorrect statement. It is never possible to recover 100% of a material after it has been dissolvedinto solution. A small portion always remains soluble. The best answer is choice C.

copyright @ by The Berkeley Review@ 297 LAB & SPECTROSCOPY EXPLANATIONS

72.

73.

74.

/5.

/o.

77.

80.

79. Choice C is correct. Charcoal binds colored organic molecules. If impurities are colored, they can bind thecharcoal and the charcoal can be filtered awayt taking the bound impurities with it. This means that charcoalis employed when there are colored impurities. This eliminates choices A and B. If the target molecule iscolored, then charcoal can bind it, and using charcoal would decrease the yield. This means that the desiredcompound must be colorless, making choice C the best answer. To be a chemistry warrior supreme, pick C.

Choice A is correct. Because Compound A is oxidized by chromic acid, it must be either an alcohol or analdehyde' Dinitrobenzoylchloride reacts readily with nucleophiles (such as alcohols) to form a derivativethat is a solid at room temperature. The formation of a precipitate therefore confirms that Compound A is anucleophile, so the aldehyde choice is not possible. Although you may not know what these reagents dospecifically, you should be able to deduce the purpose of each test. The molecular formula for Compound A(C6H12O) indicates that the compound must be either a carbonyl, ether, or alcohol. The ether ihoice iseliminated, because it would not be oxidized by the chromic acid. The carbonyl choice is eliminated for thereasons stated above. The compound must be an alcohol. An O-H stretch is found at 3500 cm-1. The peak is broaddue to hydrogen bonding. Because Compound B does not change color in CrOg/H+, it must either be an ether orketone. Because it shows a positive iodoform test, it is a methyl ketone. If you did not know that, you candeduce that the ether possibility is out, because ethers are inert and will not test positive with 12 and KOH.The key peak in a ketone would be the C=O peak at 17L0 cm-1. This makes choice A the best answer.

Choice A is correct. This question has more depth than meets the eye. According to the formula, the compoundhas one degree of unsaturation, which can be either a ketone n-bond, an alkene n-bond, or a ring. The reagentadded (Br2 in CCl4) reacts with an alkene n-bond, therefore an alkene functionality must be present in themolecule. The ketone is not possible for the structure, because the C=O of a ketone would require aiecond degreeof unsaturation and given that the one unit of unsaturation present in the compound has already been used (bythe alkene). There can be no carbonyl present in the compound. The positive test confirms that there is analkene present, so it does not show that it cannot be an alkene. Choice B is thus eliminated. The ether andalcohol functionalities are possible, because the molecule could be an alkene (for which the bromine in CC14 testis positive) with an ether or alcohol functionality. The positive test only shows that the compound cannot be aketone, thus you must pick A you correctness trooper youl

Choice A is correct. An acidic hydrogen (protic hydrogen) will exchange a proton for deuterium in the presenceof D2O. This results in the loss of one signal (broad peak) in the proton NMR spectrum. The only compound witha protic hydrogen is the alcohol, choice A. An ether should have been eliminated immediately, because ethershave no acidic protons. A ketone and aldehyde have alpha hydrogens that can be removed by strong base, butthey do not readily exchange in water without the presence of a strong base or strong acid. This question is bestsolved by selecting the most acidic compound of the choices. A proton on the alcohol oxygen is more acidic thanthe hydrogen on an alpha carbon. No matter what your path to wisdom, finish off beautifully by choosinganswer choice A.

Choice C is correct. Because Compound A was oxidized by the chromic acid, it can be oxidized by the KMnO4 aswell. Both chromic acid and potassium permanganate are strong oxidizing agents. The other compound ofinterest is Compound C, which turned the bromine solution clear (confirming the presence an alkene double bond).Alkenes also react with permanganate solution, forming a vicinal diol with syn orientation. This is stated tnthe passage. Both Compound A and Compound C can react with aqueous potassium permanganate (KMnO4(aq)),resulting in the purple permanganate solution changing color upon reaction. This makes choice C the bestanswer.

Choice C is correct" Acid chlorides are highly reactive electrophiles (because the chloride is a good leavinggroup and the carbonyl carbon is electron poor). Acid chlorides can react with even the weakest of nucleophiles.Alcohols, amines, and thiols are all nucleophiles, so they all wiII react with the dinitrobenzoyl chioride((NO2)2C6H3COC1) kicking out the chloride leaving group. Thus the best choice for a compound that will rzr:react with (NO2)2C6H3COCI is the ether. The best answer is choice C. Based on the inert nature of ethers rr,general, it is safe to pick the ether, even if you know nothing about acid halide reactivity. In other words, fo:questions that ask "Which will not react?", they are really asking 'Which is the least reactive species?"

81.

82.

83.

84.


85.

86.

Choice B is correct. Choices C and D can be eliminated, because tertiary alcohols and ethers cannot be oxidizedby chromic acid (CrO3 in H+). The color change observed from the treatment of Compound A with chromic acidshows that the compound can be oxidized. A primary alcohol (choice A) can be eliminated, because whenprimary alcohols are oxidized, they form carboxylic acids which would turn litmus paper red when added tothe paper. For this reason, the secondary alcohol is the best choice. FeeI correctness euphoria with choice B. Beeuphoric and pick it.

Choice A is correct. This is perhaps a question of trivial knowledge to a point. If you have your chemical testsmemorized, then this question is a no brainer. The MCAT provides a great deal of information, therefore werecommend that you focus on extracting information via critical analysis of the passages and tables, rather thanmemorization. The color of the precipitate is given in the passage as yellow. This eliminates the choices C andD which list the precipitate as being black. The passage also states that the test confirms the presence of amethyl ketone, which outright gives choice A as the correct answer. This is a free question for just reading thepassage. If you did not read the passage, the aldehyde should be eliminated, because aldehydes can beoxidized, and Compound B did not react with chromic acid. If Compound B were an aldehyde, it would haveshowed a positive test with the chromic acid (a color change from orange to green). From the information listedin Table 1 and knowledge that the precipitate is yellow, the best choice is answer A.

87. Choice B is correct. The limiting reagent is 1,4-dimethoxybenzene, not nitric acid, so the percent yield should becalculated based on 1,4-dimethoxybenzene. The percent yield for the reaction is the actual yield (97.5 mg 2,5-dimethoxynitrobenzene) divided by the theoretical yield (183 mg 2,5-dimethoxynitrobenzene) based on the 1.0mmole of the limiting reagent (1,4-dimethoxybenzene). Because the actual yield is only half of the theoreticalyield, the percent yield is fifty percent. The best answer is thus choice B. The math is shown below.

%yield - molesproduct -91"5*glfigglmole - 0.5 mmoles product

= 50o/"moles limiting reagent 1s-g/rss

g/mole1.0 mmoles limiting reagent

Choice B is correct. The melting point of the product has a range of 1.5 "C, which is a small enough range todeduce that the product is relatively pure. Because the melting point is sharp (a sharp melting point range isgenerally regarded as two degrees or less) and close to that of the literature value for the product, the bestanswer is choice B. Be a B supporter! There is no mention of a broad OH peak in the passage, so it is unlikelythat it is wet. Choice A is eliminated.

Choice B is correct. Filtering collects (and thus removes) the solid particles from a solution or mixture. Filteringis most commonly used in organic lab for the separation of solids from Liquids. The best answer is choice B.

Choice C is correct. Both nitrogen gas and carbon monoxide gas are colorless, so choices A and B are eliminated.You should know the color of these two gases from your everyday experiences in life. Nitrogen gas makes up themajority of our air, and air is transparent. Carbon monoxide is given off with the exhaust from combustionengines. If you are aware of your trivia, sulfur dioxide is colorless to faint yellow in color, which eliminateschoice D. If you didn't know this, you hopefully thought of smog, a brown gas containing a nitrogen/oxygenmolecule of some kind (nitrogen dioxide). Because smog is comprised of a nitrogen / oxygen compound, and smog is

brown, it is a logical conclusion to select NO2, choice C. Choice C is the best answer.

Choice B is correct. The error in the thermometer resulted in a reading that was too low (the experimental valueof 72.5 to 74 is less than the literature value of 74 to 75). This means that the mercury did not climb as high as itshould have in the inner column. The presence of some air above the mercury column (resulting from a reducedvacuum) hinders the ability of the mercury column to rise, thus Statement I results in a reading lower thanexpected. A divot present between 45'C and 55'C results in the mercury column filling the divot at temperatureswhere the top of the mercury column is above the divot (temperatures greater than 55'C in Statement II). Thiscauses the mercury column to not rise as high. This results in a reading that is too low. If the divot is above themercury level, then it will have no effect, because the mercury has not risen high enough to fill the divot. Thismeans that Statement II is valid but Statement III is invalid. Choice B is the best answer.

88.

89.

90.

9't.


92. Choice A is correct. Filtering the compound again would not purify the crystals, it would simply reduce theyield, so choice B is eliminated. Distilling the liquid (formed frorn melting) would be impraciical given thehigh temperature and.probability of deposition of ihe gas into solid throughJut the distilling.oi.r-r,,"ro choiceC is eliminated' Addition of more sulfuric acid and niiric acid could furtlier react with the iroduct rather thanthe starting material, so choice D is eliminated. Recrystallization is the ideal laboratory technique to employin the purification of a solid. The best answer is choice A, so you better pick that A.

Choice B is correct. The addition of cold water causes the crashing out (rapid crystailization) of product fromsolution. This means that solubility decreased, not increased, eliminiting cnoic" A and making choice B the bestanswer. The solution already has two strong acids, a mixture of nitric and sulfuric, so the additlon of water willnot increase its acidity of the solution. This eliminates choice C. Water will not protonate a species that isdeprotonated in the presence of strong acid, so choice D is eliminated. you should ,""uil thut lowertemperatures result in reduced solubility, so choice B is the best answer choice.

94. Choice C is correct. If the S-enantiomer has a specific rotation of -55', then the R-enantiomer has a specificrotation of +55'. The net rotation is positive, so there must be more of the R-enantiomer than the S-enantiomer.This eliminates choice A. Soiving this is going to require math, so it is easier to start with an answer choice andwork back to the specific rotation. For plugging in the numbers, choice C is best, because only one calculationneeds to be done. The value is either going to be too high (implying that choice B is correct), iigt-,t on the mark(making choice C correct), or too low (implying that choice D is coriect). A mixture of 30% of the S-enantiomerand 70"/, of the R-enantiomer has a 40"h excess of the R-enantiomer. A 100% R-enantiomer solution has a specificrotation of +55', so 40"k excess of R yields a rotation of 40"/o of +55", which is 22". Choice C is the best answer.

95' Choice A is correct. The solute that elutes first from the column is the fastest. In a TLC experiment, the fastestsol-rte climbs the plate the most, resulting in the highest spot. This means that the dark clrcle corresponds toBand A. This eliminates choices B and C. By the same reasoning used to ascertain the dark spot, thl hollon,spot corresponds to Band C, making choice A the best answer and thus, the most pickable answer.

96, Choice D is correct. Fractional distillation is used to separate liquids with boiling points that are close to oneanother. This is achieved by increasing the surface area in the distllling column. This makes choice A a validstatement, thereby eliminating it. The boiling points are close to one inother in fractional distillation, whichmakes choice B a valid statement, thereby eliminating it. Fractional distiliation is more selective, so thepurity is high, but a great deal of residue is left behind in the distilling column, resulting in a lower yield withfractional distillation. This makes choice D the best answer.

97, Choice A is correct. The compound that gets extracted into sodium hydroxide solution is the organic compoundthat becomes an ion when treated with hydroxide. Neutral acids become anions when they are ieprotonated, sothe answer is an organic acid. The only organic acid is choice A, a carboxylic acid. Choice A is the^best answer.

98. Choice B is correct. Carboxylic acids are hydrophilic, although the aromatic ring helps with its lipidsolubility. It should be somewhat solubie in diethyi ether. Despite the two nJ,atopniUc groups onhydroxybenzoic acid, it is only slightly soluble in water, because of the benzene rlng. CaiUoxylic acids arerather easily deprotonated, so they can be converted into an anion and made to be m-ore water soluble. Thismakes a compound like benzoic acid highly soluble in a basic environment. The best answer is choice B.

99' Choice C is correct. Distillation involves boiling a liquid away from a mixture, so it works with two liquidsthat have different boiling points and don't form an-azeotrope. Choice A is eliminated. Chromatographvinvolves soiutes rnigrating down a column at different rates, so it works for solutes. Choice B is eliminated.Crl'siallization dissolves and selectively precipitates a solid from a mixture of solids. This eliminates choiceD. Extraction separates solids and liquids, but not gases. Choice C is the best answer.

100' Choice D is correct. Butanone should be highiy soluble in other ketones such as acetone, so choice A iseliminated. Ketones don't oxidize with chromium (VI), so they do show a negative Jone's test. Choice B iseliminated. Butanone has a molecular mass of T2 grams per mole, so if it losesl methyl group during, it canform a fragment with a molecular mass of 57 amu. Choice C is eliminated. There rr" ,ro chirat centers inbutanone, so it cannot have optical activity. This makes choice D the best answer.

93.

Copvright @ by The Berkeley Review@ 500 LAB & SPECTROSCOPY EXPLANATIONS

Writing a book of this nature ended up being a more difficult project than I first imagined back in 1990.For every one hundred questions you see in this book, about three hundred had to be lr-ritten. A g:reatdeal of time went into selecting which questions served the purpose of conceptual analysis, developrn*test-taking skills, honing table-reading skills, testing memory, and building confidence. But aschallenging as it was at times, it has been a wonderful experience. This particular edition is theculmination of fourteen years of writing and rewriting. Along the path I have incorporated thefeedback and input of roughly 5000 students. I have tried my best to rnclude many of the examples thathave proven useful in lecture. I have tried to keep in mind the sole purpose of this book in your life: tohelp you achieve a higher MCAT score in the biological sciences section and offer test-taking tips thatare universally applicable. It is imperative that you keep that in mind while working through ourmaterials. Your goal is to learn information and strategies that allow you to quickly pick the best offour choices when asked a multiple-choice question. Completing this book would have never beenpossible without the support of many special people io -y life. I would like to thank:

Cecile Santos: You have been an inspiration to my teaching. I see the love you have in your eyes for thestudents you teach and it is awe-inspiring. You have been my sounding board for teaching ideas andmuch of the text in this book. Without you, I could never have done this.

Cambria and Madison: You fill my heart with so much love. You keep me balanced at times I need itmost. Your laughter and smiles keep my spirits high and I am so lucky you came into my life. You arethe greatest children a father could ever want.

Our Wonderful Teachers: Over the years I have gleaned so many great ideas from you. It has been anhonor working alongside so many of you. \Atrhile so many of you have been helpful in my path, I wouldbe remiss if I didn't specifically mention six very special teachers to me (although we have had manymore.) Anthony Gopal showed me about the power of enthusiasm and how demonstrations and real lifeexamples can drive classroom learning. Moin Vera showed me how simple genius and fundamentalconcepts can make even the hardest topic seem easy. Derek Welsbie showed me how simple logic,when applied with a little bit of information, can make every question easier to understand. SamBiggers showed me how bonding with the class and empathizing with their anxieties can build a bondthat helps the entire team do better. Kim Gordy showed me the importance of silliness in gaining thestudents' interest and trust and how analogies can power learning. Cecile Santos showed me theimportance of engagement and open dialogue with her open forum discussion in lecture where everystudent is involved. You are all incredible teachers and I learn something from you every time I watchyou in action. Thank you!

Dale Schmidt: You have been incredibly supportive over the years, picking up the slack when it came.Your work ethic and attention to detail is second to none. I am always sure that if you did it, it wasdefinitely done correctly.

Kimberly Gordy and Doug Coe: You spent so much time combing through my books to help them flowbetter" I can't thank you enough, because I am not an easy person to work with when it comes to editing.You managed to not only make this book better, but you made me realize the need for a different styleand voice at many junctures.

Our Incredible Students: For the books I have written, I owe my biggest thank you to the thousands ofstudents who have used them to help prepare for their MCAT. Some of you are well into your practiceand this is a distant memory. Others of you have just recently finished your exam and are working yourbutt off to make it into medical school. To each of you who offered some feedback, told me about a typo,asked me why a question was worded a certain way, or just simply said what you thought, I am deeplygrateful. This book, and the others I have written, is as much a part of you as it is a part of me. This isour project and I am so happy you were a part of its development.


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