Organic chemistry section test (3)

MCAT Subject Tests
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_______________________________________________________________O R G A N I C C H E M I S T R Y S U B J E C T T E S T 3

Organic Chemistry Subject Test 3

1.

Which of the following alkyl halides would be the least
reactive substrate in a Williamson ether synthesis?

3.

What is the reagent X in the following reaction?

CH3 CH=CHCH
2 CH3

A.

X

CH3 CHCHCH2 CH3
CH2

CH2CH2Br
B.

A.
B.
C.
D.
E.

CH3CHCH3
Br

C.

HCHO, heat
CO2, LiAlH4
CH3OH, H+
CH3COOH, H+
CH2N2, UV light

CH3C(CH3)CH3
4.

Cl

Which of the following is a pyrimidine?
A.

D.

CH3CH2CHCH
2CH3
Br
E.

CH2Cl
B.
2.

Which of the following will yield a racemic product
mixture?
C.

A.

CH3
CH3 CCH3
CH2 OH
B.

N
N

N

HBr
H+

N

N
H

D.

1)CH3MgCl
CH3CH2OCH2CH3
2) H2O

N

C.

CH3CH2CH CH2

Br2 / CCl4

N
E.

D.

CH3CH2CH

CHCH2CH3

KMnO4

N

Heat

E.

HCl
CH2 CHCH3

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5.

Starting with a molecule of benzyl alcohol ΦCH2OH,
we wish to form ΦNH2 (aniline). Which of the
following syntheses should be used? [Φ = phenyl
group]

Addition of Br2 to 2-pentene yields

6.
A.

C2H5

A.
Φ CH2 OH

KMnO4

NH3

PBr3

NaCN

H

Br

H

OBr

B.
Φ CH2 OH

Br
-

CH3

H2 , Ni
B.

C.
Φ CH2 OH

PBr3

NH4 + , H+

Cu

NH3

C2H5
Br

Br

H

H

D.
Φ CH2 OH

LiAlH 4

CH3

E.

HCN
Φ CH2OH

H2O

NH3

C.

C2H5
H

Br

Br

H

CH3

D.

C2H5
H

Br

H

Br
CH3

E. More than one of the above

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7.

What is the major product of the reaction below?

8.

(CH3 )3 C

COOH

Br
NaOH

KMnO4 (conc.)
CH3 CH=CH

Predict the product of the following reaction:

C
?

H

heat

CH2 CH3

CH3

A.

(CH3 )3 C

A.

H

CH3

H3 C

H

H

C

C

?

C
HO

CH2 CH3

CH3

B. (CH3)3CC=CH2
C.

(CH3 )2 C=CCH3

OH OH
B.

COOH

H
D.

(CH3)2CCH2CH3
OH

CH3CH

CH

COOH

E. Either A or B, depending on reaction conditions.

C.

COOH

COOH

COOH

D.

CH3

OH H
H3 C

C

C

H

OH

CH2 CH3

E. No reaction will occur.

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9.

10. Which of the following is an acetal?
A.

C2 H5
KMnO4

SOCl2

HCN

HOCCH3

H2 O
H+

O
B.

CH3 CCH2 CCH3
A.

O

O
C

O

C.

CH3

COOH

OR
C

H
B.

OH

D.

O

C

CH2 OH
NH2

O

OH

OCH3

OH

OH

E.

C.

COOH

CH2 OH
OH
OH

O
OH
OH

D.

CH2 COO-

E.

CN
CH

OH

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11. If the compound pictured below undergoes E2
elimination, what is the most likely product?

Ha

12. Which of the choices below lists a possible sequence via
which the three compounds shown, initially in a
chloroform solution, may be separated?

NH2

Cl Hc
C2 H5

H
Hb Hd

A.

C2 H5

Hc
Hb Hd

B.

Ha
C2 H5
Hd
C.

Ha

H

N

;

; CH NO
3

2

NO2
A. Extraction with a weakly acidic aqueous solution,

followed by extraction with a strongly acidic
aqueous solution, followed by distillation.
B. Extraction with a strongly acidic aqueous solution,
followed by extraction with a weakly acidic
aqueous solution, followed by distillation.
C. Extraction with a strongly acidic aqueous solution,
followed by extraction with an even stronger acidic
aqueous solution, followed by distillation.
D. Extraction with a strongly basic aqueous solution,
followed by extraction with a weakly acidic
aqueous solution, followed by distillation.
E. Extraction with a weakly basic aqueous solution,
followed by extraction with a strongly basic
aqueous solution, followed by distillation.

C2 H5
Hb Hc
D.

C2 H5

Hc
Ha Hd

E.

Ha
C2 H5

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13.

14. Which monosubstitution product(s) would be expected
from the following reaction?

H3C

excess

O

+

CO

CH2MgBr

O
C

H2O

Br2 , FeBr3

H+

A.

OH
C

A.

O

BrCH2

CH2

B.

C

B.

O
C

O

H3C

CH2

O
C

C.

OH
C CH2
CH2

Br
C.

H3C
D.

O
C

OH
C

O

CH2

CH2

Br
D.

H3C
E.

Br

OH
C

O
C

CH2

E. A mixture of B and D.

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15. Which one of the following compounds can be most
easily converted to a Grignard reagent?
A.

17. A general formula for para-substituted benzoic acids is
shown below. The Ka of a particular para-substituted
benzoic acid will be smallest when X is which of the
following?

Cl

X

COOH

B.

COOH
Cl

A.
B.
C.
D.
E.

NH2
OCH
NO2
Cl
CH3

NH2

18.

C.
Cl

O
CH3C
CH3C

OH

D.

Cl

OCH3

Cl

C

O

The products of this reaction are
A.
B.
C.
D.
E.

E.

O + 2NH3

2CH3CONH2 + H2O
CH3COONH2 + HCONH2 + H2
CH3CONH2 + CH3COO–NH4+
CH3COO–NH4+ + CH=CH–O–NH2
CH3CONH2 + CH4

19. In the electrophilic aromatic substitution of phenol,
substituents add predominantly in which position(s)?
A.

B.
C.
D.
E.

ortho to the hydroxyl group
para to the hydroxyl group
meta to the hydroxyl group
A and B
A, B, and C

C
H
16. The base-catalyzed condensation product of two
acetaldehydes is

A.
B.
C.
D.
E.

an α-hydroxyaldehyde.
an α-ketoaldehyde.
a β-hydroxyaldehyde.
a β-keto aldeyhyde.
an α-hydroxyketone.

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20. Which of the following compounds will give a single
proton NMR signal?
I. hexane
II. 1, 2-dichloroethane
III. tert-butyl chloride
IV. 2-methyl-2-butene
A.
B.
C.
D.
E.

III only
II and III
II and IV
I, II, and II
None of the above

22. Which of the following compounds would be expected
to be the most basic?
A.
NH2

Cl

B.
NH2

21. To prepare a primary alcohol with a Grignard reagent,
the Grignard reagent must react only with
A.
B.
C.
D.
E.

CH3COCH3.
CH3CHO.
HCHO.
HCOOH.
CO2.

NO2

.
C.
NH 2

Cl

D.
NH2

E.
NH2

OCH3

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23. Which of the following compounds would not give a
positive result under the iodoform test?
A. CH3CH2OH
B.
O

R1 C

C

OH H
A.
B.
C.
D.
E.

CH3CH

C.
O
CH3CCH2 CH3

D.

O

O
CCH3

heat

R1

H
C

C
H

+ H2O

R1 = R2 = H.
R1 = CH3 and R2 = H.
R1 = CH3 and R2 = Cl.
R1 = CH3 and R2 = CH3.
R1 = Cl and R2 = Cl.

27. FCH2COOH is a stronger acid than BrCH2CH2COOH
because

28. The organic acid pictured below will be most acidic
when X is which of the following?

COOH

24. A compound produces an infrared spectrum with a sharp
peak at approximately 2950 and 1700 cm–1, as well as a
number of smaller peaks between 900 cm–1 and 1460
cm–1. The substance yields a negative result under
Tollens’ test. This substance is most likely a(n)
A.
B.
C.
D.
E.

H

R2

H+

A. the more electronegative F is closer to the carboxyl
group than is the less electronegative Br.
B. Br is more electronegative than F.
C. FCH2COOH has a smaller Ka than does
BrCH2CH2COOH.
D. FCH2COO- is a stronger base than BrCH2CH2COO-.
E. F- is a better leaving group than Br-.

CH

E.

R2 H

ketone.
aldehyde.
alcohol.
alkane.
alkyne.

X
A.
B.
C.
D.
E.

CH3
NH2
CH2Cl
OH
NO2

25. Which of the numbered hydrogens in the molecule
pictured below is most acidic?

O

O

CH2CH2CH2CCH2CH
1 2 3 4 5
A.

B.
C.
D.
E.

1
2
3
4
5

26. The dehydration reaction below proceeds most readily
when
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29. What is the major product of the reaction below?

OH
(CH3)3C

C

CH3

H

H+
heat

?

30. Which of the following aromatic compounds will
undergo electrophilic substitution primarily in the ortho
and para positions?
A.

NO2

A.

(CH3)2C

CH

CH3

CH3

B.

OCCH3

B.

(CH3)2C

C(CH3)2

(CH3)3C

CH2

CH3

(CH3)3C

CH

CH2

O

C.
C.

COOH

D.
E.

O
(CH3)3C

C

CH3

D.

CN

E.

CHO

STOP! END OF TEST.

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THE ANSWER KEY AND EXPLANATIONS BEGIN ON THE FOLLOWING PAGE.

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ORGANIC CHEMISTRY SUBJECT TEST 3

ANSWER KEY
1. C

7. C

13. C

19. D

25. D

2. C

8. E

14. C

20. B

26. D

3. E

9. A

15. D

21. C

27. A

4. D

10. D

16. C

22. E

28. E

5. A

11. C

17. A

23. D

29. B

6. E

12. A

18. C

24. A

30. B

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EXPLANATIONS

1.

C
The Williamson ether synthesis is an SN2 reaction in which an alkoxide ion (OR–) acts as the nucleophile while an
alkyl halide serves as the substrate:

-

RO Na

+

+

NaX

ROR'

R'X

Since this is an SN2 reaction, it follows that those alkyl halides which are most susceptible to nucleophilic attack are
most suitable to serve as the substrate in this synthesis. The order of reactivity toward SN2 reaction is dictated largely by
steric factors and goes 1°>2°>3°. The compound in choice C, as a tertiary alkyl halide, is thus the least suitable of the
compounds offered in the choices. Choices A and E are primary (and benzylic) halides while choices B and D are
secondary alkyl halides; these compounds all undergo SN2 reactions more readily than does the tertiary alkyl halide of
choice C.

2.

C

In order for a reaction to produce a racemic mixture, the product formed must possess at least one chiral center and its
enantiomers must form in a one-to-one ratio. The product of the addition reaction in choice C is 1,2-dibromobutane. This
compound has one chiral center, at carbon number 2, and its R and S enantiomers should be expected to form in equal
amounts in the absence of any special effort to produce either in enantiomeric excess.
H

H
CH 3CH 2C

CH 2

+

Br 2

CH 3CH 2C
Br

Br
CH 2

Br
+

CH 3CH 2C
H

CH 2
Br

As for the other choices, the reaction in choice A should produce a mixture of neopentyl bromide (via substitution of
the protonated hydroxyl group by bromide) and 2-bromo-2-methylbutane (via substitution after a 1,2 methyl shift); neither
product is chiral. Choice B shows a Grignard reagent over the arrow and diethyl ether as the reactant; no reaction will occur
since ethers are generally inert to Grignard reagents and are, in fact, often chosen as solvents for Grignard reactions for that
very reason. The reaction in choice D will produce, upon heating, two equivalents of the achiral compound propionic acid
as the permanganate oxidatively cleaves the alkene double bond, while the reaction in choice E will yield the Markovnikov
addition product 2-chloropropane which is likewise achiral.

3.

E
The synthesis of cyclopropane via addition to the double bond in an alkene is accomplished by reaction with
methylene, H2C:, an example of a class of compounds known as carbenes in which carbon forms two single bonds to make
a neutral molecule. (Notice the lone pair of electrons on the carbon atom.) This reactive intermediate is most commonly
formed by the heat- or light-induced decomposition of diazomethane, as shown in choice E. The lone pair of electrons on
the carbene intermediate are used together with the _ electrons of the alkene to form the two new single bonds in the
cyclopropane product. (An alternative route to cyclopropane synthesis involves reacting diiodomethane and a zinc-copper
couple to generate a carbene-like species called a carbenoid which will add a methylene group to the double bond.)

4.

D
The compound in choice D is pyrimidine itself, a heterocyclic analog of benzene containing two nitrogen atoms,
positioned as shown. Pyrimidine and its derivatives are essential components of nucleic acids such as DNA and RNA.

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Choice A is almost, but not quite, the general structural formula for a steroid; it differs from the standard steroidal ring
structure in that the ring on the far right has six carbons instead of five. Choices B and E, known as quinoline and pyridine
respectively, each have only one nitrogen atom and thus fail to meet the definition of a pyrimidine. Choice C is purine,
another essential component of nucleic acids.

5.

A
Since many synthetic pathways might be available to effect any given transformation, it makes most sense to check
the choices offered. Choice A will yield the desired product: permanganate oxidizes primary or benzylic alcohols to the
corresponding carboxylic acids in the first step. The benzoic acid formed by this oxidation will, in the presence of aqueous
ammonia, form an ammonium carboxylate salt which, upon heating, will produce an amide in the second step. Finally,
treatment of the benzamide formed in the second step with hypobromite will produce an amine with one less carbon. This
reaction is known as a Hofmann rearrangement, and it will convert benzamide into aniline, the desired product, in the third
step.
Choice B is incorrect. Phosphorus tribromide, step one, will convert an alcohol into an alkyl bromide, in this case
benzyl bromide. Treatment of this benzyl halide with sodium cyanide, step two, will produce a nitrile via an SN2 pathway.
Finally, hydrogen over a nickel catalyst will reduce the nitrile to a primary amine; the primary amine formed by this series
of steps will have two carbons between the phenyl ring and the amino group, and thus is not aniline. Incorrect choice C,
like B, produces benzyl bromide after the first step. Benzyl bromide should be inert to ammonium and acid in the absence
of any nucleophilic species, thus no reaction should be expected in the second step. Choices D and E also do not lead to the
formation of aniline.

6.

E
The electrophilic addition of bromine to an alkene proceeds through a cyclic bromonium ion intermediate.
Configuration-wise, the addition is anti. Regiospecificity also dictates that the geminal dihalide shown in choice B cannot
possibly be a product of this reaction. Complete scratchwork would show that choices A and D should form with equal
probability, and therefore that choice E is correct:

H5C2

H
C

H

Br

Br

H5C2

C

+

Br
C

CH3

H
C

H

CH3

H5C2

CH3
C

H

C

Br

H
Br

Br-

OR

H5C2

+

Br
C

H

H5C2
C

C

H

CH3

CH3

Br

H

C

Br
H

BrA faster way to get to the correct answer here is to notice that both reactants are achiral, and because of that the
product(s) as a whole cannot be optically active. Furthermore, since the groups attached to the two stereocenters are not
identical (a methyl group on one and an ethyl group on the other), absence of optical activity cannot be due to the
generation of one lone achiral meso compound. An enantiomeric pair must therefore be the product of the reaction. By this
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faster method of reasoning it is not necessary to figure out which product or products will form, but only that more than one
will.
7.

C
This reaction is an example of benzylic oxidation. Substituted benzene derivatives are easily oxidized by
permanganate or dichromate to benzoic acid derivatives if there is at least one hydrogen attached to a benzylic carbon. (The
first step in the reaction is the abstraction of a benzylic hydrogen to give a benzylic radical, thus making the hydrogen atom
necessary.) The reactant shown has three benzylic carbons, one of which, the carboxylic acid carbon at the top of the
structure, is already completely oxidized. The other two, the benzylic carbon in the alkyl group on the right and that in the
alkenyl group on the left, can both be oxidized to carboxylic acid groups, yielding the tricarboxylic acid compound shown
in choice C. Choices A and D can be dismissed for several reasons, among which is that they suggest reduction of the
COOH group in the reactant to a CH3 group in the choices; permanganate is far and away not a reducing agent. Choices A
and D also show the conversion of the alkene side chain to a diol; this mild oxidation would occur only if the permanganate
used were cold, dilute, and dissolved in basic solution. One might more quickly eliminate choices A and D by noticing that
they differ only by rotation around a carbon-carbon single bond; they are thus equivalent choices. Choice B is wrong
because it shows the selective oxidation of the alkyl side-chain only, when in fact both side-chains will be oxidized under
the conditions indicated in the reaction shown.

8.

E
This question is testing our understanding of fundamental organic reaction mechanisms. The reactant shown is a
secondary alkyl halide, thus it is susceptible to nucleophilic attack via either an SN1 or SN2 pathway. It also has a proton α
to the bromine atom (on the methyl group), so it is capable of undergoing elimination to form an alkene. Sodium hydroxide
is a strong base which can facilitate E2; the hydroxide ion can also behave as a nucleophile and lead to an SN2 reaction
under the proper conditions. The preference for either E2 or SN2 can be enhanced by the appropriate choice of reaction
conditions, justifying choice E as the credited choice.

9.

A
This question is testing our knowledge of organic synthesis, and requires that we work through the steps shown to
predict the final product. (Partial knowledge of the reagents shown would, however, allow for the elimination of some of

the answer choices. For instance, noticing that the final step in the synthesis includes acidification allows us to eliminate
choice D since this product would be protonated in acidic solution.) Let’s go through the steps:
KMnO4 is a strong oxidizing agent which will oxidize most alkylbenzenes, including the ethylbenzene starting
material here, to benzoic acid. After the first step we thus have benzoic acid, which can now react with thionyl chloride,
SOCl2, in the second step. Thionyl chloride is most commonly used to convert a carboxylic acid to the corresponding acyl
chloride; the product of the second step is thus benzoyl chloride. Acyl halides are especially reactive toward nucleophilic
acyl substitution, the reaction by which the halide is replaced by any of a large number of suitable nucleophiles while the
carbonyl C=O bond remains intact. The third step of the synthesis in this question should therefore result in the substitution
of the chloride by cyanide from the HCN and the product of this step should be the α-ketonitrile derivative of benzoic acid.
The cyano group of nitriles is easily hydrolyzed in aqueous acid; the final step of the synthesis should thus be expected to
convert the α-ketonitrile to the α-ketoacid shown in choice A. The sequence of reactions is as follows:
O
C2H5
KMnO4

C

O
OH
SOCl2

C

O
Cl
HCN

C

O

N

C
H+

C

COOH
= Choice A

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10.

D
By definition, an acetal must contain a carbon atom bonded to two -OR groups and a hydrogen atom. Acetals might
thus be thought of as geminal diethers. The carbon atom on the right side of the ring in choice D is bonded to the OCH3
group on the far right and to the O atom in the ring which is, in turn, bonded to the rest of the ring; a third substituent would
be a hydrogen atom which is not shown by convention. Choice D does therefore meet the criteria for classification as an
acetal. Choice A is a carboxylic acid, while choice B is a b-diketone. Choices C and E can be classified as hemiacetals, but
not as acetals. Hemiacetals, by definition, contain a carbon atom which is bonded to one -OR group, one -OH group, and a
hydrogen atom.

11.

C
E2 elimination occurs in the anti configuration, requiring that the proton abstracted be in a position trans to the

leaving group. As drawn, only the proton labeled as Hd is in the proper position for anti elimination, i.e., trans to the
chlorine atom in the reactant, and thus Hd must be absent from the final product. This qualifies choice C as the only
possible choice.

12.

A
This question is testing our understanding of organic separation techniques, specifically extraction, as well as our
knowledge of the relative acidity of the three compounds shown. The first compound, aniline, is the most basic of the three;
it has an amino group in which the nitrogen has a lone pair of electrons available for donation to a suitable Lewis acid. The
amino group nitrogen atom in the second compound, para-nitroaniline, is rendered less basic due to the electronwithdrawing nature of the nitro group; with decreased electron density, it is less eager to donate electrons, i.e., to act as a
Lewis base. Lastly, the third compound, nitromethane, is not basic at all. Applying this relative order of basicity to the
extraction, now, requires the understanding that acidic extraction will make organic bases soluble in the aqueous phase via
protonation of the base to form a soluble ionic salt. Therefore choice A is the best choice. Extraction with a sufficiently
weak aqueous acid will protonate only the most basic compound, aniline, and allow it to transfer, in protonated form, to the
aqueous phase. A second extraction with a stronger acid will then protonate the more basic of the two compounds
remaining in the chloroform layer, para-nitroaniline, and allow it to enter the aqueous phase. The third compound would
then remain alone in the organic layer, and the separation could be completed by distillation.
If a strong enough acid were used first, both aniline and nitroaniline would be protonated and transferred to the
aqueous layer, and the separation would not be effective. Extraction with a base would not effect a separation at all, since
the first two compounds, as bases themselves, are inert to base.

13.

C
The reaction shown is an example of a Grignard reaction. One should predict that the Grignard reagent used, benzyl
magnesium bromide, will act as a nucleophile and attack the carbonyl carbon in the substrate, phenyl benzoate, forming the
tetrahedral intermediate shown below.

OC

O

CH2

After formation of the tetrahedral intermediate, a pair of electrons on the negatively charged oxygen can move down
to the carbon atom of the original carbonyl group, restoring the carbon-oxygen double bond and displacing the phenoxide
leaving group. Once the carbonyl group has been restored, it is susceptible to further nucleophilic attack by a second
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molecule of the Grignard reagent, resulting, after acid workup, in the formation of dibenzylphenylcarbinol, the compound
shown in answer choice C.

14.

C
The reagents shown in this reaction, bromine and ferric bromide, are the reagents of choice to effect the bromination
of most benzene derivatives via EAS (electrophilic aromatic substitution). Both the rate and the regioselectivity of EAS
reactions are influenced by the presence of electron-donating or withdrawing groups already on the ring. Electronwithdrawing groups, such as the acetyl substituent shown here, slow down the substitution by the incoming electrophile
while directing it toward the meta position relative to the position of the original substituent. Thus the major product of this
reaction should be meta bromoacetophenone, the compound shown in choice C. Choices B and D, and therefore choice E,
would be the results if the original compound had an electron-donating, ortho/para directing group on it in place of the
electron-withdrawing acetyl group. Choice A is the product of alpha bromination, which should be expected to result only
via a radical mechanism. Absent an active metal, NBS, hν, peroxides, or extremely high temperatures, radicals are unlikely;
choice A should thus be eliminated.

15.

D
This question tests detailed knowledge of the Grignard reaction. Usually, such reactions are initiated by the addition
of clean magnesium metal to dry ethereal solutions of alkyl halides, though substituted alkyl halides are possible substrates
if no other reactive functional group is present. Absence of water is a requirement, or else the newly formed alkyl
magnesium halide will react with the hydroxy proton in the water, forming an alkane and a halomagnesium hydroxide,
thereby killing the reaction. Similarly, it is necessary that no other acidic protons be present in the solvent system nor in the
substrate, or else these would react with the newly formed alkyl magnesium halide. It is this last fact which accounts for the
elimination of the four wrong answer choices: choices A, B, and C can be eliminated because each contains a proton
bonded to an electronegative element (oxygen in A and C, nitrogen in B), while choice E contains an acetylenic proton.
Elimination of choice E requires that we recall the relatively high acidity of an acetylenic proton; the proton attached to the
sp hybridized carbon is sufficiently acidic to react with strong bases such as hydride or amide anions or, as relevant here, to
protonate a Grignard reagent. Choice D thus remains as the only viable choice.

16.

C
The base-catalyzed condensation of acetaldehyde is an example of an aldol condensation and is outlined below:

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O-

O
CH2

CH

CH2

O
-:CH2

CH

CH

H
H3C

-:OH

CH
O
OH O

O

CH3CHCH
2CH

CH3CHCH
2CH
: OH

OH

The newly formed hydroxyl group thus winds up on the second carbon from the aldehyde carbonyl; the product is
therefore classified as a β-hydroxyaldehyde, choice C. It should be noted that this product can easily undergo further
dehydration (elimination of water to form double bond) to yield a compound where a carbon-carbon double bond is in
conjugation with the carbonyl bond.

17.

A
This question requires us to translate acid strength into Ka. Recall that a smaller Ka corresponds to a weaker acid; do
not confuse Ka with pKa! We can thus translate the question into: "Which substituted benzoic acid will be the weakest
acid?" Recall next that electron-withdrawing groups increase acid strength since they help to stabilize the anion formed
upon dissociation, so a further translation is, "Which of the following is the least electron-withdrawing as a substituent?".
Of the choices offered, B, C, and D are electron-withdrawing groups, while choices A and E are electron-donating. Of the
electron-donating groups listed, NH2 is more so than is CH3 due to the lone pair of electrons on the nitrogen atom. (In
electrophilic aromatic substitutions, the amine group is a strongly activating ortho/para director because of its electrondonating ability.) Choice A is thus the most electron-donating and has the most destabilizing effect on the conjugate base
(the anion), thereby disfavoring dissociation of the proton. The compound para-aminobenzoic acid is therefore the least
acidic, and has the smallest Ka.

18.

C
The reaction is that of an acid anhydride, acetic anhydride to be precise, and a nucleophilic base, ammonia.
Nucleophilic attack on one of the anhydride carbonyl carbon atoms by the ammonia nitrogen should be expected, followed
by the loss of an acetate leaving group concurrent with the restoration of the carbon-oxygen double bond.

O

:O-

O

O

+ :NH3

O
+NH3

O

O

O
+

+NH3

-O

The protonated amide thus formed can then release a proton (to the acetate ion or to another molecule of ammonia)
producing one molecule of acetamide. Choices B and D can be eliminated at this time due to their lack of acetamide. The
remaining species in the product mixture (acetate, ammonia, and proton) would then combine to form ammonium acetate,
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the second product shown in choice C. Note that in the presence of ammonia, carboxylic acid cannot remain protonated and
exists as a salt.

19.

D
Phenol, or hydroxybenzene, has an -OH group on the ring. The hydroxy oxygen has two nonbonded pairs of
electrons, either of which can be donated (via resonance) to the aromatic ring after addition of an electrophile. Recall that
electron-donating groups stabilize most the cations formed upon addition to the ortho and para positions, thus they serve as
ortho/para directing activators. The -OH group is not particularly bulky, and thus there will be no steric hindrance that
prevents substitution at the ortho position. Both ortho and para substituents will result.

20.

B
A single proton NMR signal means that all the hydrogen atoms in the molecule find themselves in identical chemical
environment with respect to the other atoms in the molecule. Hexane has three different types of protons: the terminal
methyl ones, the methylene ones on the adjacent carbon atoms, and the methylene protons on the innermost carbon atoms.
Because they will have approximately similar chemical shifts, the observed spectrum may be quite complex. Compound II,
1, 2-dichloroethane, possesses four equivalent protons and will give one signal:
Cl

Cl
C

H

C

H

H
H

Compound III, tert-butyl chloride, has nine equivalent protons and will also give a single proton NMR signal.

H3C
H3C
H3C

C

Cl

Choice B is therefore correct. The last compound, 2-methyl-2-butene, has four distinct types of protons: the three
methyl groups attached to the double bond are in different chemical environments, plus the vinylic proton.

21.

C
This question is about the reaction of a Grignard reagent, RMgX, with various carbonyl compounds to form alcohols.
Choices D and E can be eliminated first since a carboxylic acid, choice D, will neutralize a Grignard reagent (RMgX +
R'COOH ∅ RH + R'COOMgX) while reaction with carbon dioxide, choice E, results in the formation of a carboxylate
anion (RMgX + CO2 ∅ RCOOMgX). The aldehydes and ketone in the remaining choices do yield alcohols as the major
product. However, only choice C, formaldehyde, will result in the formation of a primary alcohol. Choice A, acetone,
would react with the Grignard reagent to produce a tertiary alcohol: the two methyl groups in the original acetone molecule
will wind up on the same carbon as that attached to the hydroxyl group in the product. Acetaldehyde, choice B, would react
with RMgX to form a secondary alcohol, i.e., the methyl group of the acetaldehyde would be attached to the central carbon
atom. Only in choice C, formaldehyde, are there two hydrogen atoms already present; these two hydrogens are necessary to
form the -CH2OH moiety which would define the end product as a primary alcohol.

22.

E

The less stable an unshared pair of electrons, the likelier it is to extract a proton from other molecules around, and thus
the more basic the compound. This question about the order of basicity can therefore best be approached by examining the
expected stability of the lone electron pair on the nitrogen atom in the amino group of the molecules shown. The stability of

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the electron pair will be affected by the nature of the other substituent group present, specifically in terms of whether it is
electron-withdrawing or electron-donating (either inductively or by resonance) and how effective its influence is based on
its position relative to the amino group. The first thing to notice is that aniline itself (choice D) is a rather weak base
because the electron pair is stabilized via delocalization through the _ cloud of the phenyl group (i.e. by resonance):
(+)NH

: NH2

(+)NH

2

(+)NH

2

2

(-)

(-)

(-)

One would expect that electron-withdrawing groups would lead to a further stabilization (and therefore weakening) of
the base. Choices A and C, which contain the electronegative chlorine, would make the compound less basic, and are
therefore not correct. Choice B, which has the nitro group at the meta position, would also stabilize the electron pair
inductively, even though not as effectively as it would if it were at the o- or p- position, where it could also stabilize
through resonance. Choice E, however, with the methoxyl group at the para position, is destabilizing because of the highly
unfavorable resonance contributions:

: NH2

: NH2
(-)

: OCH3

etc.

(+)OCH

3

The proximity of the electron pair of nitrogen and the electron density released by the methoxyl group causes the
compound to be more basic than aniline itself, overwhelming the inductive effect of the electronegative oxygen atom.
Choice E is the most basic, and hence the correct answer.

23.

D
The iodoform test is useful in identifying the presence of the following groups:

O
C

OH
CH3

,

C

CH3

Compounds containing these groups react with iodine in NaOH to give a yellow precipitate. Choices C and E are
methyl ketones and will give positive results. Choice B, acetaldehyde, also possesses the —COMe group that leads to the
formation of the precipitate. Choice A, ethanol, possesses the group on the right above and will first be oxidized to
acetaldehyde, yielding a positive result. Choice D, benzaldehyde, is the only compound shown that will not react in the
solution.

24.

A
Students are expected to know the characteristic frequencies of certain functional groups in the infrared. The strong
peak at ~1700 cm-1 indicates the presence of a carbon-oxygen double bond, i.e. a carbonyl group. Choices C, D, and E can
thus be eliminated. At this point, one may use either of two pieces of information to distinguish between the two remaining
alternatives: (1) Tollens’ test is also known as the silver mirror test, and gets the name from the fact that silver is deposited
on the walls of a clean test tube or flask as a mirror if an aldehyde is present. The negative result of the test indicates that
the compound is a ketone, choice A. (2) The C-H stretch of the aldehyde group would give two weaker bands in the
infrared spectrum at ~2700 cm–1 and ~2850 cm–1, both of which are absent. This latter approach, however, probably
requires more knowledge about infrared spectroscopy than the student is expected to know.

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25.

D
The most acidic hydrogen is the hydrogen that, upon leaving the molecule, will leave behind the most stable anion.
Alpha-carbons, i.e. carbon atoms adjacent to carbonyl carbons, have protons that are acidic compared to protons attached to
normal carbon atoms because the carbonyl group can stabilize by resonance the conjugate base that results from
deprotonation:

O
C
..

C

C

O

O
C

C

C

H

:OHChoices C and D are both α-hydrogens and will thus be expected to be more acidic than the other choices. In addition,
choice D will be more acidic than choice C because the anion will be stabilized by two carbonyl groups rather than one.

26.

D
Dehydration of an alcohol to yield an alkene under acidic conditions proceeds via the E1 mechanism. The hydroxyl
group is protonated, and departs as a water molecule, leaving behind a carbocation which then loses a proton to the solvent
while a double bond forms between the two carbon atoms:

CH3

CH3
R1

C
R2

OH

H+

R1

C
R2

CH3
+
OH
H

-H2O

R1

C+
R2

CH2
-H+

R1

C
R2

Formation of the carbocation is the rate-determining step in this reaction. Therefore, the more easily it is formed (i.e.
the more stable it is), the faster the reaction will go. The carbocation intermediate will be stabilized by the presence of
electron-donating substituent groups. Alkyl groups are electron-donating and thus will increase the rate of reaction by
facilitating the formation of the intermediate. The more numerous they are, the more stable the cation will be, and the faster
the reaction. Choices C and E are incorrect because the electronegative chlorine will destabilize the carbocation
intermediate instead.

27.

A

Because F is the most electronegative atom, it helps to stabilize the anionic conjugate base, thus increasing the
strength of the corresponding acid, especially when, as in this case, it is close to the acid functional group. Choice B is a
false statement, as F is more electronegative than Br. Choice C is also a false statement, as Ka is a direct measure of acid
strength: the higher the Ka, the stronger the acid. (The lower the pKa, the stronger the acid: keep in mind the difference
between the two.) Choice D is false as well: the stronger the acid, the weaker its conjugate base. As for choice E, it is a
false statement that is irrelevant at best. Only very indirectly can one rationalize acid strength from how good a leaving
group it contains: high electronegativity is one factor that makes a species stable as an anionic leaving group and hence one
may expect to see a correlation between the ability of the species to leave and the acid strength of a compound that contains
this species, but in this case Br- is actually the better leaving group, although Br is the less electronegative atom.

28.

E
The group that will stabilize the anionic conjugate base will yield the most acidic compound. An electron-withdrawing
group stabilizes the negative charge either by induction or by delocalization. Choices A–D are all electron-donating, and

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would tend to lower the acidity of the molecule (compared with X = H). Only E, —NO2, is electron-withdrawing, thereby
increasing the acidity of the molecule.
29.

B
Upon heating under acidic conditions, alcohols will undergo dehydration to yield alkenes. This is an E1 reaction that
proceeds via the formation of a carbocation intermediate. Choices A, C, and E are all incorrect and easily dismissed because
they are not alkenes. Of the two choices B and D, B is correct because it is the most substituted alkene. Choice D is indeed
a product of the reaction as written, but a minor one because it is relatively unsubstituted. The secondary carbocation

formed upon the departure of the protonated hydroxyl group will rearrange (via migration of a methyl group) to generate
the more stable tertiary carbocation which in turn will lead to the more substituted alkene:

CH3

CH3
H3C

C

+
C

CH3 H
30.

CH3

H3C

C+

C

CH3

-H+

(CH3)2C

C(CH
3)2

CH3 H

B
Electron-donating species will activate aromatic rings for ortho/para electrophilic substitution. Only choice B, which
has an ester substituent, is electron-donating. All of the other choices have electron-withdrawing groups, which are
deactivating and meta-directing. (The exceptions to this are the halogen atoms, which are deactivating but ortho/paradirecting.)

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