Organic chemistry section test (2)

MCAT Subject Tests
Dear Future Doctor,
The following Subject Test and explanations contains questions not in test format and
should be used to practice and to assess your mastery of the foundation content
necessary for success on the MCAT. Simply memorizing facts is not sufficient to achieve
high scores; however, an incomplete understanding of basic science knowledge will limit
your ability to think critically. Think of building your content knowledge as learning the
vocabulary and practicing MCAT-like questions as actually speaking.
All rights are reserved pursuant to the copyright laws and the contract clause in your
enrollment agreement and as printed below. Misdemeanor and felony infractions can
severely limit your ability to be accepted to a medical program and a conviction can
result in the removal of a medical license. We offer this material for your practice in your
own home as a courtesy and privilege. Practice today so that you can perform on test
day; this material was designed to give you every advantage on the MCAT and we wish
you the best of luck in your preparation.
Sincerely,

Albert Chen
Executive Director, Pre-Health Research and Development
Kaplan Test Prep

© 2003 Kaplan, Inc.
All rights reserved. No part of this book may be reproduced in any form, by Photostat, microfilm,
xerography or any other means, or incorporated into any information retrieval system, electronic
or mechanical without the written permission of Kaplan, Inc. This book may not be duplicated,
distributed or resold, pursuant to the terms of your Kaplan Enrollment Agreement.


________________________________________________________________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 2

Organic Chemistry Subject Test 2

Questions 3-6 refer to the following choices:

1.

Which of the following is the most stable carbocation?

A.
B.
C.
D.
E.

A.

H2 C

CH+

H2 C

CH

B.

CH2 +

enantiomers
diastereomers
structural/constitutional isomers
conformational isomers

tautomers

3.

C.

CH3CH2CHCH2+

H
CH3 C

CH3
D.

CH3 andCH3CH2CH2OH

OH
+

4.
E.

CHO

CH2 CH2 +

CHO

H

C

OH

H

C

OH

and

OH

C

H

H

C

OH

CH2OH
2.

CH2OH

Which is correctly paired?

A.

5.

O
R

CH

C

OCH2CH3

amino este

and

C

C

NH2

C

C

B.

R

C

NH

R

amide

O

6.

C.

R

H

D.

R

CHOH
C

α− hydroxy acid

C2H5

CH3

amine

NH2

C

OH and

C2H5

H

C

OH

CH 3

OH

O
E. All of the above

KAPLAN _______________________________________________________________ 1


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 2 ________________________________________________________________

7.

What is the hybridization of the carbonyl carbon in N,Ndimethylformamide?
A.
B.
C.
D.
E.

8.

O

sp3
sp2
sp
sp3d
sp3d2

A.

H 2N

C
O

O
B.

A compound is separated from an ethereal mixture by

extraction with cold, aqueous NaHCO3, then recovered
by addition of HCl to the extract. Which of the
following is a likely identity of the compound?
A.
B.
C.
D.
E.

9.

10. Which of the following is an appropriate resonance
form of the conjugate base of p-aminobenzoic acid?

O
O
C

N

O

H
O
H 2N

D.

E.

1,3-Cyclopentadiene reacts with sodium metal at low
temperatures according to:

196K

C

H
C.

CH3(CH2)7CH2OH
CH3(CH2)7CHO
CH3(CH2)7CH2NH2
CH3(CH2)7COOH
CH3(CH2)7COONa

Na

H 2N

Na

What is the best explanation for this observation ?
A. The reactant is more unstable at reduced
temperatures.
B. The cation formed is stabilized by aromaticity.
C. Sodium metal is highly specific for cycloalkenes.
D. The rehybridization of the saturated carbon atom
provides additional stability to the product.
E. The anion formed is stabilized by aromaticity.

C
O

H

O
N

C

H

O

11. Which of the following statements is true of acetone?
A. It exists as a pair of tautomers, of which the keto
form predominates.
B. It exists as a pair of tautomers, of which the enol
form predominates.
C. It is useful as a solvent due to its unusually high
boiling point.
D. It is useful as a solvent due to its polar, protic
nature.
E. It is useful as a solvent due to its lack of reactivity
toward nucleophiles.
12. The compound pictured below is a member of which
compound class?

COOH

O
A.
B.
C.
D.
E.

2

C(CH3)3

carboxymethoxybenzenes
methoxybenzoic acids
propoxybenzoic acids
t-butoxybenzoic acids
carboxycumene oxides

_______________________________________________________________ KAPLAN


________________________________________________________________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 2

13. Which of the following could be the formula of an ester?
A.
B.
C.
D.
E.

C7H12O
C7H12O2
C7H14O
C7H16O2
C6H12O

14. What is the rate law for the elimination (E2) reaction
shown below?
t-BuBr + t-BuO–K+ →
← H2C=C(CH3)2 +t-BuOH + KBr
A.
B.
C.
D.
E.

rate = k[t-BuBr]
rate = k[t-BuBr][t-BuO–]
rate = k[t-BuBr]2
rate = k[t-BuO–]2
rate = k[t-BuOH][H2C=C(CH3)2]

18. How many distinct organic compounds have the
molecular formula C5H12?
A.
B.
C.
D.
E.

One
Two
Three
Four
Five

19. Which of the structures below is most stable?

A.

CH 3
H 3C

Cl

B.

CH 3

15. Cyclohexane has how many fewer hydrogen atoms per
molecule than does n-hexane?
A.
B.
C.
D.
E.

0
1
2

3
4

Cl

CH 3
D.

16. Which of the following has the most exothermic heat of
hydrogenation?
A.
B.
C.
D.
E.

Cycloheptene
Cyclohexene
Cyclopentene
Cyclobutene
Cyclopropene

17. What are the bonding types and degree measures of the
bond angles normally associated with the hybridizations
sp2 and sp3, respectively?
A.
B.
C.
D.
E.

σ 109.5°, and σ 109.5°
σ 109.5°, and π 109.5°
σ 120°, and π 120°
π 120°, and σ 120°
π 120°, and σ 109.5°

CH 3

C.

CH 3
E.

CH 3
CH 3

20. Which of the choices below best describes the
relationship between Z and E isomers?
A.
B.
C.
D.
E.

Enantiomers
Geometric isomers
Conformational isomers
Structural isomers
Tautomers

KAPLAN _______________________________________________________________ 3


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 2 ________________________________________________________________

21. What is the approximate C-N-C bond angle in
diethylamine?
A.
B.
C.
D.
E.

25. How many different stereoisomers does the following
compound have?
OH OH OH OH

107°
110°
120°
150°
180°

HOOC

22. What are the positions occupied by the bromine atom
and the tert-butyl group respectively in the most stable
conformation of trans-1-bromo-3-tert-butylcyclohexane?
C(CH3)3

H

A.
B.
C.
D.
E.

A.
B.
C.
D.
E.

H

Bromine: axial, t-butyl: axial
Bromine: equatorial, t-butyl: equatorial
Bromine: axial, t-butyl: equatorial
Bromine: equatorial, t-butyl: axial
Bromine: either axial or equatorial, t-butyl:
equatorial

23. What is the IUPAC name of the following compound?
CH 3

CH 3CH 2CHCH 2 CCH 2 CH 2 CH 2 CH 3

CH 3CH 2

A.
B.
C.
D.
E.

CH 3

5-ethyl-3,3-dimethylnonane
3-ethyl-5,5-dimethyloctane
3-ethyl-5,5-dimethylnonane
7-ethyl-5,5-dimethyloctane
7-ethyl-5, 5-dimethylnonane

C

C

C

H

H

H

H

H

2
3
4
8
16

26. Which of the following compounds will exhibit the
greatest dipole moment?

Br

A.
B.
C.
D.
E.

C

27.

(Z)-1,2-Dichloro-1,2-diphenylethene
(E)-1,2-Dichloro-1,2-diphenylethene
1,2-Dichloro-1,2-diphenylethane
1,2-Dichloroethane
1,2-Difluoroethane

How many structural isomers of C3H6Br2 are capable of
exhibiting optical activity?
A.

B.
C.
D.
E.

0
1
2
3
4

28. Which of the compounds below would be the best
starting material for the synthesis shown?

?

A.
B.
C.
D.
E.

SOCl 2

CH 3CH 2MgBr

CH 3(CH 2) 6CH 3

CH3(CH2)4COOH
CH3(CH2)4CH3

CH2=CH(CH2)3CH3
CH3(CH2)4CH2OH
CH3(CH2)2CH2OH

24. What is the product of the reaction below?

CH3Cl + CH3O- Na+
A.
B.
C.
D.
E.
4

CH3CH2OCH2CH3
CH3OCH2CH3
CH3OCH3
CH2=CH2
CH3CH3

_______________________________________________________________ KAPLAN


________________________________________________________________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 2

29. What is the major product of the elimination reaction
below?

H 3C

H

H

CH 3

C

C

C

Cl H

H

KOH

CH 3

alcohol

H

A.

H 3C

CH 3

C

C

C

H

H

CH 3

CH 3
B.

H 3C

C

C

C

CH 3

H

C.

H 3C

C

H

CH 3

C

C

Cl

H

H
D.

H 3C

CH 3

C

C

H

H
H 3C

E.

H 3C

CH 3

C
H

C

C

CH 3

H
C

CH 3

H

30. A mixture of alkyl halides is subjected to elimination
(E1) with one mole of aqueous acid and alcohol. If the
reactant mixture initially contains one mole each of npropyl chloride and isopropyl chloride, what can be
predicted about the product mixture?

RX + R'X

H +/H 2O
CH 3CH 2OH

products

A. A single alkene product is formed, and isopropyl
chloride is present in excess.
B. A single alkene product is formed, and n-propyl
chloride is present in excess.
C. Two distinct alkene products are formed, and npropyl chloride is present in excess.
D. Two distinct alkene products are formed, and
isopropyl chloride is present in excess.
E. No reaction occurs because both alkyl halides used
are inert to elimination (E2).

STOP! END OF TEST.

KAPLAN _______________________________________________________________ 5


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 2 ________________________________________________________________

ORGANIC CHEMISTRY SUBJECT TEST 2

ANSWER KEY

6

1. B

7. B

13. B

19. A

25. D

2. E

8. D

14. B

20. B

26. A

3. C

9. E

15. C

21. A

27. B

4. B

10. C

16. E

22. C

28. D

5. D

11. A

17. E

23. C

29. A

6. A

12. D

18. C

24. C

30. B

_______________________________________________________________ KAPLAN

ORGANIC CHEMISTRY SUBJECT TEST 2

Explanations
1.

B
Cation stability, like that of the free radicals, declines in the order 3°>2°>1°>methyl. Choices C, D, and E, as primary
cations, can therefore be disposed of first. (Each would quickly rearrange via hydride shifts to form tertiary or benzylic
cations.) Of the remaining choices, choice A is a vinylic cation which is rather unstable due to the high s character of the
sp2 orbital. The _ bond and the empty orbital are perpendicular to each other and hence the carbocation cannot be stabilized
through charge delocalization. Choice B, however, is an allylic cation which can be stabilized via resonance (or
equivalently, delocalization of the _ electron cloud):

CH2 CH CH2

2.

CH2

CH

CH2

E
All of the structures listed are correctly paired with their description. Choice A is an ester (RCOOR) that has a amine
substituent. Choice B is an amide (RCONR2), choice C is a simple (in this case primary) amine, and choice D is an acid
with a hydroxy (alcohol) substituent on the α-carbon.

3. C 4. B 5. D 6. A

Both compounds in #3 have the same atoms, but the connectivity among them is different. These are structural or
constitutional isomers by definition.
The two compounds listed in #4 have the same two chiral centers on each and differ in the configuration around only
one of these; they are thus not mirror images of each other. They are therefore diastereomers.
The two compounds in #5 are chemically identical, differing only in their rotation about a single bond, in this case the
central carbon-carbon bond. These are referred to as conformational isomers.
The last pair of compounds are non-superimposable mirror images and are recognizable as such because the two
compounds have two of the groups interchanged about a chiral center. By definition, they are enantiomers. Note that
another switch between any two of the four groups will bring the molecule back to its original configuration.

7.

B
A carbonyl carbon is one that is double bonded to an oxygen. Since the carbon contains one π bond and three σ
bonds (one double bond and two single bonds), it must be sp2 hybridized. Note that we do not need to know the structure of
N, N-dimethylformamide to answer this question.

CH 3

O
N

CH 3
N,N -Dimethylformamide
8.

D
The correct answer choice has to be a compound that is hydrophobic (since it started in the ether layer), but when
reacted with a weak base becomes a water soluble salt. Choice A is an alcohol which is not acidic enough to react with

KAPLAN _______________________________________________________________ 7


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 2 ________________________________________________________________

sodium bicarbonate, a weak base. It would remain dissolved in the ether. Choice B is an aldehyde, which can be
eliminated for the same reason as A. Choices C and E are both bases, so neither will react with another base. Choice D,
nonanoic acid, will react with sodium bicarbonate and form sodium nonanoate, a water soluble salt. Extra Q: How can
you tell you have a carboxylic acid by observing the addition of sodium bicarbonate? A: In addition to the formation of a
salt, carbonic acid is also formed (H2CO3) which can decompose into H2O and CO2--you'll see bubbles!
9.

E
When 1,3-cyclopentadiene reacts with sodium, it goes from a nonaromatic to an aromatic compound. A planar cyclic
compound is aromatic if it satisfies Huckel's rule—that is, it has (4n+2) π electrons, where n is any integer. The product of
the reaction has 6 π electrons, so is therefore aromatic. An aromatic compound is very stable because the π electrons are
delocalized throughout the ring. In this reaction the products are much more stable, so the reaction will be spontaneous.

10.

C
When aminobenzoic acid loses a proton, the conjugate base assumes a charge of –1. Therefore, any resonance
structure must also posses the same net charge. Only answer choices B and C satisfy this requirement, and B can easily be
eliminated because nitrogen is not part of the benzene ring.

11.

A
Acetone is a compound commonly used as a solvent to promote SN2 and E2 reactions because it is aprotic and polar.
Since it cannot form strong intermolecular bonds and is of low molecular weight it has a low boiling point. Acetone exists

in two forms, with the keto form predominating by 99%. This is because the carbon-oxygen double bond (the carbonyl
bond) is much stronger than the carbon-carbon double bond.

OH

O
CH3
12.

C

CH3

CH3

C

CH2

D
Benzoic acid is a benzene ring with a carboxylic acid function attached directly. This eliminates choices A and E.
Meta to the acid is a t-butoxy group, leaving only choice D.

13.

B
An ester must have at least two oxygens; both of which are bonded to the carbonyl carbon (one via a double bond and
one via a single bond). This leaves only choices B and D. The maximum number of hydrogens a seven carbon ester can
have is 14,. This happens when all the other carbon atoms (except for the carbonyl carbon) are saturated. One possibility is
given below:

O

H2
H3C

C

C
H2

H2
C

C
H2

C

O

CH3

Again, this is the maximum number of hydrogen atoms. So choice D, with 16 H’s, cannot be correct. Choice B is the
correct response. The fact that it has two fewer H’s means that the compound is unsaturated. One possibility for the
structure of choice B is as follows.

8

_______________________________________________________________ KAPLAN

ORGANIC CHEMISTRY SUBJECT TEST 2

O

H2
H3C
14.

C

H

C
H2

C

C
H

C

O

CH3

B
For SN2 and E2 reactions, the rate law is first order with respect to both the nucleophile and the substrate. In this

reaction the nucleophile is the t-butoxide anion (t-BuO-) and the substrate is t-butyl bromide (t-BuBr). Thus the rate law
must be B.

15.

C
Cyclic hydrocarbons have the general formula of CxH2x, while straight chain hydrocarbons have the formula CxH2x+2,
giving a difference of two. Cyclohexane has 12 hydrogens, while n-hexane (straight chain) has 14.

H
C
H
H C
H

H H
C

H
C H
C

C
H H

H
H C

H

H

H

H12
cyclohexane,6C
16.

H H
C
C
H H

H H
C
H
C
C H
H H
H

n-hexane, 6CH14

E
Heat of hydrogenation (_HH) is the amount of energy released when π bonds are converted to σ bonds by the addition
of hydrogens. The greater the _HH of a molecule, the greater the potential energy it has. The more potential energy a
molecule has, the less stable it is. By comparing the _HH of different molecules, their relative stabilites can be obtained.
Cyclopropene, a three-membered ring, has the greatest bond strain, and is the most unstable. It will therefore release the
most energy upon hydrogenation, so will have the greatest _HH.

17.

E
sp2 hybridized carbons are involved in a double bond, which consists of one sigma and one pi bond. The bond angle is
about 120°:

120°
120°

C

sp2 hybridized

sp3 hybridized carbons are involved in single (sigma) bonds only, and the bond angles are about 109.5°.

KAPLAN _______________________________________________________________ 9


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 2 ________________________________________________________________

109.5°

C

sp3 hybridized
18.

C
The molecular formula C5H12 corresponds to that of a saturated alkane, since it is of the form CnH2n+2. Distinct
compounds arise because of the different degrees of branching that can occur in the carbon chain:

H
H C

H
C

H
C

H
C

H
C

H

H

H

H

H

H

H
H C

H
C

H

H

H
C
H C H

H
C
H

H
n -pentane

19.

isopentane

H
H C H
H

H
H C
H

C
H C H
H

H
C H
H

neopentane

A
The question is asking you to decide between the different positions that substituents of cyclohexane can assume. The
structure that is most stable will have the least repulsion of the electron clouds of neighboring substituents. It is therefore
more favorable for substituents to occupy equatorial positions since they are more “out of the way.” Choice A has all the
substituents in the equatorial position (designated by a bond that is not directed vertically), while all the other answer
choices have substituents in the axial position (designated by a vertical bond).

20.

B
Z and E isomers (and their cousins cis and trans) describe the relative positions of substituents about a double bond.
They are by definition geometric isomers. Since double bonded carbons are not chiral, they cannot be enantiomers,
eliminating choice A. Conformational isomers can be interconverted by simple rotation about a single bond (e.g.
boat/chair, eclipsed/gauche). Since no rotation can occur about a double bond, C is incorrect. Structural isomers, choice D,
differ about the connections between the atoms in the molecule, but have the same general formula. E and Z isomers have
the same connections, but differ about their arrangement in space (and are therefore classified as stereoisomers). Tautomers
are a specific type of structural isomer.

21.

A

N
3HC2HC

CH2CH3

H
N,N- Diethylamine
The normal bond angle for an sp3 hybridized atom is 109.5o. By looking at the drawing of diethylamine, you will
notice that nitrogen has a free electron pair. Nonbonded (or lone) electrons pairs exert stronger repulsive effects, forcing

10

______________________________________________________________ KAPLAN


ORGANIC CHEMISTRY SUBJECT TEST 2

the atoms attached to nitrogen down, and closer together. VSEPR predicts that the bond angle must be less than 109.5°
leaving only choice A. Extra Q: What is the shape of the molecule? A: Trigonal pyramidal.
22.

C
The most stable conformation will have the least steric hindrance. The t-butyl is the bulkiest substituent, so it must
assume the equatorial position (this eliminates choices A & D). Since the bromine is trans to the t-butyl and is two
positions away, it must assume the opposite configuration as the t-butyl group, so is therefore axial. This leaves us with
choice C. If members of the ring are trans, and are an even number of carbons apart they will have the opposite position
(equatorial vs. axial), and those an odd number apart will have same positions. If two members of the ring are cis, and are

an even number of carbon apart they will have the same position, and odd number apart will have opposite positions. These
descriptions need not be memorized--all you need to do is just visualize the ring, keeping in mind the alternating positions
of members on the ring.

A
3

2

B

5

4

Alternating postions
:
A and B are both above the ring
A is axial and B is equatorial

1

6

CH3

H3C

C

CH3

H3C

CH3
CH3

C

H

H
Br

H

H
Br

(If both groups were in the equatorial position, that would of course be more favorable still. But then the compound
would be cis, not trans.)
23.

C
In following the IUPAC naming procedure outlined in the review notes, we start by identifying the longest carbon
chain, here being 9. This molecule is a hydrocarbon with no double bonds, so we call it nonane. Next, the carbons must be
numbered such that the substituents receive the lowest numbers possible, so this molecule gets numbered left to right.
Finally, attached to carbon 3 is an ethyl group, and to carbon 5 is two methyl groups. Putting this together we get answer
choice C.

24.

C
In order to identify the product, we start by recognizing and classifying the reactants. Methyl chloride is a primary
alkyl halide, and sodium ethoxide is a strong nucleophile. What reaction mechanism does this scream out? SN2 : the
ethoxide attacks while the chloride leaves, forming dimethyl ether, choice C. (The sodium is just a spectator ion here).
This reaction type goes by the name of the Williamson ether synthesis.

KAPLAN ______________________________________________________________ 11


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 2 ________________________________________________________________

H
C

CH 3O

Cl

H3C

O

CH3

+ Cl -

H
H

25.

D
The number of stereoisomers can be calculated from the formula 2n, where n = the number of chiral carbons (since
each center gives two isomers: S and R). This number is a maximum because the presence of meso compounds will reduce
the number of stereoisomers. (This rule holds as long as we do not have double bonds, which may lead to geometric
isomers.) In this case, there is no possibility of meso compound, since no matter how the groups are arranged, there will not
be an internal plane of symmetry in the molecule. This molecule has 3 chiral carbons, and thus 8 stereoisomers.
(Note that if the carbon on the right end had been connected to a carboxylic group (COOH) rather than a second
hydrogen, there would have been 4 chiral centers, but there would also be the possibility of having meso compounds.)

26.

A
(Z)-1,2-Dichloro-1,2-diphenylethene

φ

(E)-1,2-Dichloro-1,2-diphenylethene

φ
C

φ

C

Cl

Cl
C

Cl

C

φ

Cl

net dipole moment

no net dipole moment

As seen above, the Z isomer has a net dipole moment, where in the E isomer the dipole moments cancel one another
out. Dicholoroethane and difluorethane also have no net dipole moment.
27.

B
Four structural isomers exist for the formula C3H6Br2:

H3C

C
H2

H
C Br
Br

1, 1-dibromopropane

H3C

Br
C
Br

CH3

2, 2-dibromopropane

H3C

H
C
Br

H
C H
Br

1, 2-dibromopropane

H
H C
Br

H

C
H

H
C H
Br

1, 3-dibromopropane

Among these, only the third structure, 1, 2-dibromopropane, contains a chiral carbon. The middle carbon is attached to
four different groups: an H, a Br, a methyl group, and a –CH2Br group. It can therefore be optically active.
None of the other choices contains a chiral carbon. The compounds are all achiral and optically inactive.

28.

D
This is a SN2 reaction made to look difficult. SOCl2 reacts with primary and secondary alcohols to generate
something with a very good leaving group, ClSO2–. (OH– is a poor leaving group.)

12

______________________________________________________________ KAPLAN


ORGANIC CHEMISTRY SUBJECT TEST 2

O

O
R

OH

+

Cl

S

Cl

R

O S

Cl

+

HC

good leaving group
The compound formed is reactive in nucleophilic substitution reactions. The Grignard reagent is a good nucleophile.
To get the question correct, you have to identify R. Since our product has 8 carbons, and 2 came from the ethyl magnesium
bromide, our starting compound must have 6 carbons attached to a leaving group; answer D.
29.

A
KOH dissociates in to K+ and OH– ions, the latter of which is a strong base. The elimination will therefore proceed via
the E2 mechanism. Since the hydroxide ion is not a bulky base, and the substrate is a secondary alkyl halide (rather than a

more sterically hindered tertiary alkyl halide), the resulting alkene will be the more thermodynamically stable, more
substituted one depicted in choice A. I.e. the double bond forms between the second and third carbon atoms.

30.

B
When either of the alkyl chlorides undergoes elimination, the same alkene will be formed: propene. Since this is an
E1 reaction, the substrate that can form the more stable carbocation will be most reactive. Isopropyl chloride, a 2o alkyl
halide, can form a more stable carbocation than n-propyl chloride, a 1o alkyl halide. Therefore isopropyl chloride will react
the most and n-propyl chloride will be present in excess.

KAPLAN ______________________________________________________________ 13