1.
Organic Chemistry
Discretes Test
Time: 30 Minutes
Number of Questions: 30
This test consists of 30 discrete
questions—questions that are NOT based on a
descriptive passage. These discretes comprise
15 of the 77 questions on the Physical
Sciences and Biological Sciences sections of
the MCAT.

2.
MCAT
2 as developed by
ORGANIC CHEMISTRY
DISCRETES TEST
DIRECTIONS: The following questions are not
based on a descriptive passage; you must select
the best answer to these questions. If you are
unsure of the best answer, eliminate the choices
that you know are incorrect, then select an
answer from the choices that remain. Indicate
your selection by blackening the corresponding
circle on your answer sheet. A periodic table is
provided below for your use with the questions.
PERIODIC TABLE OF THE ELEMENTS
1
H
1.0
2
He
4.0
3
Li
6.9
4
Be
9.0
5
B
10.8
6
C
12.0
7
N
14.0
8
O
16.0
9
F
19.0
10
Ne
20.2
11
Na
23.0
12
Mg
24.3
13
Al
27.0
14
Si
28.1
15
P
31.0
16
S
32.1
17
Cl
35.5
18
Ar
39.9
19
K
39.1
20
Ca
40.1
21
Sc
45.0
22
Ti
47.9
23
V
50.9
24
Cr
52.0
25
Mn
54.9
26
Fe
55.8
27
Co
58.9
28
Ni
58.7
29
Cu
63.5
30
Zn
65.4
31
Ga
69.7
32
Ge
72.6
33
As
74.9
34
Se
79.0
35
Br
79.9
36
Kr
83.8
37
Rb
85.5
38
Sr
87.6
39
Y
88.9
40
Zr
91.2
41
Nb
92.9
42
Mo
95.9
43
Tc
(98)
44
Ru
101.1
45
Rh
102.9
46
Pd
106.4
47
Ag
107.9
48
Cd
112.4
49
In
114.8
50
Sn
118.7
51
Sb
121.8
52
Te
127.6
53
I
126.9
54
Xe
131.3
55
Cs
132.9
56
Ba
137.3
57
La *
138.9
72
Hf
178.5
73
Ta
180.9
74
W
183.9
75
Re
186.2
76
Os
190.2
77
Ir
192.2
78
Pt
195.1
79
Au
197.0
80
Hg
200.6
81
Tl
204.4
82
Pb
207.2
83
Bi
209.0
84
Po
(209)
85
At
(210)
86
Rn
(222)
87
Fr
(223)
88
Ra
226.0
89
Ac †
227.0
104
Unq
(261)
105
Unp
(262)
106
Unh
(263)
107
Uns
(262)
108
Uno
(265)
109
Une
(267)
*
58
Ce
140.1
59
Pr
140.9
60
Nd
144.2
61
Pm
(145)
62
Sm
150.4
63
Eu
152.0
64
Gd
157.3
65
Tb
158.9
66
Dy
162.5
67
Ho
164.9
68
Er
167.3
69
Tm
168.9
70
Yb
173.0
71
Lu
175.0
†
90
Th
232.0
91
Pa
(231)
92
U
238.0
93
Np
(237)
94
Pu
(244)
95
Am
(243)
96
Cm
(247)
97
Bk
(247)
98
Cf
(251)
99
Es
(252)
100
Fm
(257)
101
Md
(258)
102
No
(259)
103
Lr
(260)

3.
Organic Chemistry Discretes Test
GO ON TO THE NEXT PAGE.
KAPLAN 3
1 . On the graph below, which is the titration curve
for an amino acid, what does point A represent?
H
nits of base
A . The isoelectric point
B . pH = pKb2
C . pH = 14 – pKb2
D . pH = pKb2
– 14
2 . The diagram below shows a step in which of the
following processes?
OH
O
CH2OH
H
O
CH2OH
OH
H
O
H
OH
CH2OH
A . Aldehyde formation
B . Hemiketal formation
C . Mutarotation
D . Anomerization
3 . Which of the following products might be
formed if benzoyl chloride was treated with
excess CH3MgBr?
A . C6H5COOH
B . C6H5COOCH3
C . C6H5CHO
D . C6H5C(CH3)2OH and C6H5COCH3
4 . Which of the following reactions would produce
an ester?
A . 2CH3OH
H2SO4
B . CH3CO2H + SOCl2
C . CH3CO2H + C2H5OH
H2SO4
D . C6H5OH + CH3CH2Br
NaOH
H2O
5 . Which of the following reactions will NOT yield
C6H5CH2COCH3?
A . C6H5CH2C CH + HgSO4 / H2O
B . C6H5CH2CH2CH2OH + cold, dil. KClO 3
C . C6H5CH2(CH3)C CH2 + 1. O3
2. Zn / H2O
D . C6H5CH2(CH3)CHOH + cold, dil. Na 2Cr2O7
6 . Which of the following reactions will NOT
result in the formation of a carboxylic acid?
A . Carbonation of Grignard reagents
B . Hydrolysis of nitrites
C . Reduction of aldehydes
D . Oxidation of primary alcohols
7 . The following reaction is an example of:
O OH
H2C
A . esterification.
B . tautomerism.
C . elimination.
D . dehydration.

4.
MCAT
GO ON TO THE NEXT PAGE.
4 as developed by
8 . Which of the following compounds is the most
basic?
A . H2N OCH3
B . H2N CHO
C . H2N NO2
D . H2N Br
9 . Which of the following can be synthesized from
an arenediazonium salt?
I. C6H5Br.
II. C6H5CN.
III. C6H5OH.
A . I only
B . II and III
C . I and III
D . I, II, and III
1 0 . In the following reaction, what is the final
product?
R2C=O + H2NNHC6H5 → R2COHNHNHC6H5
→ R2C=NNHC6H5
A . An oxime
B . A phenaylhydrazone
C . A semicarbazone
D . An aromatic nitrile
1 1 . Which of the following processes describes the
reaction below?
A . Alkylation
B . Hoffman elimination
C . Salt formation
D . Conversion into amides
1 2 . The reaction below will primarily yield which of
the following products?
A . CH3
NO2
B . CH3
NO2
C . CH3
NO2
C6H5CH3
HNO3
H2SO4
D . A and B
1 3 . Which of the following describes the reaction
below?
H5C2 C C CH3
H2, Pd
H H
CH3H5C2
A . Catalytic hydration
B . Substitution
C . Stereospecific reduction
D . Racemization
1 4 . Which of the following reactions is NOT an
electrophilic aromatic substitution reaction?
A . C6H6 + CH3CH2COCl/AlC13 →
B . C6H6 + H2, Rh/C →
C . C6H6 + C6H5CH2Cl/AlCl3 →
D . C6H6 + Br2/FeBr3 →
1 5 . Proteins can be characterized by the fact that
they:
A . are composed of a single peptide chain.
B . have a primary structure formed by covalent
linkages.
C . retain their conformation above 40°C.
D . always have quaternary structures.
R2CHCR2 – N
+
R3 O
–
H
heat
 → R2C = CR2 + R3N:+H2O

5.
Organic Chemistry Discretes Test
GO ON TO THE NEXT PAGE.
KAPLAN 5
1 6 . Halogenation of alkanes proceeds partly by the
mechanism depicted below
I. X2
hv
→ 2X•
II. X • + RH → HX + R •
III. R • + X2 → RX + X •
Which of these steps can be described as chain
propagation?
A . I, II, and III
B . III only
C . II and III
D . I and II
1 7 . Which of the following compounds is least
susceptible to electrophilic aromatic
substitution?
A . C6H5CH3
B . p-Br– C6H4– NH3
+
C . p-O2N–C6H4–NH2
D . p-H3CO–C6H4–OCH3
1 8 . How many different stereoisomers can the
following compound have?
COOH
H
H OH
H OH
HHO
H OH
A . 2
B . 3
C . 4
D . 8
1 9 . In thin-layer chromatography, which of the
following best describes the behavior of the
solvent?
A . It moves downward because of gravity.
B . It moves upward because of capillary action.
C . It moves downward because of capillary
action.
D . It does not move.
2 0 . Which of the following statements is correct?
I. CH2OH
CH2
OH
H OH
HHO
III. COOH
COOH
OHH3C
OHH3
C
II. CH2OH
CH2
OH
HHO
HHO
IV. CHO
CH2
OH
HHO
H OH
A . Compounds I and II are diastereomers .
B . Compounds I, II, and IV are meso structures.
C . Compounds II and IV are optically inactive.
D . Compounds III and IV are optically active.
2 1 . Which two of the following compounds are
geometric isomers?
I. III.
II. IV.
A . I and III
B . I and IV
C . II and III
D . II and IV
2 2 . Which of the following is a result of the reaction
below?
Cl + Br–
H
H
C2H5
CH3
A . Inversion of configuration
B . Retention of optical activity
C . Mutarotation
D . Loss of optical activity

6.
MCAT
GO ON TO THE NEXT PAGE.
6 as developed by
2 3 . When placed in an electric field, which of the
following compounds will migrate toward the
cathode at pH 7.0?
A . H2NCH2COOH
B . H2NCH2CHNH2COOH
C . HOOCCH2CHNH2COOH
D . HOOCCHNH2CH2–S–S–CH2CHNH2COOH
2 4 . At atmospheric pressure, a certain organic liquid
has a boiling point of 185°C, and decomposes at
180°C, while its isomer, also a liquid, boils at
215°C and decomposes at 190°C. These isomers
can be separated by:
A . vacuum distillation.
B . simple distillation.
C . sublimation.
D . fractional distillation.
2 5 . Infrared spectroscopy provides a chemist with
information about:
A . functional groups.
B . conjugated bonds.
C . molecular weights.
D . distribution of protons.
2 6 . A compound with the molecular formula C8Hl8,
produces a single NMR signal. What is its
structural formula?
A . CH3(CH2)6CH3
B . (CH3)2CHCH2(CH2)3CH3
C . (CH3)2CHCH2CH(CH3)CH2CH3
D . (CH3)3CC(CH3)3
2 7 . Which of the following compounds will give the
greatest number of proton NMR peaks?
A .
B .
C .
D .
CH3CH3
(CH3)2C=CH2
H2C CH2
CHCH3
CH3CHBrCH2CH3
2 8 . A clear liquid is subjected to infrared
spectroscopy and produces a spectrum with
prominent, sharp peaks at approximately 2950
cm–1 and 1700 cm–1, as well as a number of
smaller peaks between 1460 cm–1 and 900 cm–1.
This substance is most likely:
A . a ketone.
B . an aldehyde.
C . an alcohol.
D . an alkane.
2 9 . Which of the following compounds form a
racemic mixture?
I . CHO
CHO
HO H
HHO
III. CHO
CHO
OHH
OHH
II. CHO
CHO
OHH
HHO
IV. CHO
CHO
HHO
H OH
A . I and III
B . I and IV
C . II and III
D . II and IV
3 0 . What is the most likely mechanism for the
reaction between 1-chloropropane and sodium
cyanide?
A . S N2
B . E2
C . SN1
D . El
END OF TEST

7.
Organic Chemistry Discretes Test
KAPLAN 7
THE ANSWER KEY IS ON THE NEXT PAGE

8.
MCAT
8 as developed by
ANSWER KEY:
1. C 11. B 21. C
2. C 12. D 22. D
3. D 13. C 23. B
4. C 14. B 24. A
5. B 15. B 25. A
6. C 16. C 26. D
7. B 17. B 27. D
8. A 18. D 28. A
9. D 19. B 29. D
10. B 20. A 30. A

9.
Organic Chemistry Discretes Test
KAPLAN 9
ORGANIC CHEMISTRY DISCRETES TEST EXPLANATIONS
1. The correct answer for question 1 is choice C. This problem deals with the titration of amino acids. The
titration given begins with a strongly acidic solution . As we add base, the pH rises at first. But then as the pH
approaches the pKa1, protons begin to dissociate from the carboxyl group. The value of the pH then levels off
because the carboxyl group buffers the solution when the pH is close to its pKa1. Therefore, point A on the graph
represents the value of pH equal to the pKa1. That is, point A shows deprotonation of the amino acid carboxyl
group. But the reverse reaction—protonation of the amino acid, is represented by pKb2. And by the definition of
pK, the pH at point A is equal to 14 minus the pKb2. Therefore, choice C is correct. Choice A--the isoelectric
point--is in fact point B. When we continue to add base after point A, the pH gradually rises until all the molecules
of the amino acid are in the neutral form.
2. The correct answer for question 2 is again choice C. The question calls upon your understanding of the
phenomenon of mutarotation. Cyclic hemiacetal forms of monosaccharides with different configurations around the
first carbon--that is, anomers--are easily broken in aqueous solutions. In such reactions either the alpha or beta
anomer becomes an open chain. In an aqueous solution, especially if it is slightly acidic, this open chain is easily
recyclized, forming a mixture containing both anomers in their equilibrium concentrations. Thus the initial opening
and the subsequent closing of the chain results in a mixture of anomers. This phenonenon is called mutarotation and
so choice C is correct. Choice A, aldehyde formation , is wrong because the open chain form has a carbonyl group
and therefore an aldehyde is already formed. Choice B--hemiketals--are formed as a result of the nucleophilic addition
of a hydroxyl donated by an alcohol to a ketone carbonyl group. Since in this question you have an aldehyde
carbonyl group, this choice is also wrong. Although this reaction is an example of anomerization, choice D is
wrong since mutarotation is a more accurate description of this process. Again then, choice C is the correct answer.
3. Now for question 3. The correct answer here is choice D. The reagent here--benzoyl chloride--is an acyl
halide which will form a ketone on reaction with an equimolar quantity of a Grignard reagent like this one in the
question. With excess of the Grignard reagent however, acyl halides react further to produce primarily tertiary
alcohols along with a small amount of ketone. Therefore, the product of this reaction will be a mixture of a tertiary
alcohol and acetophenone. This corresponds to choice D and so this is the correct answer. As for the other choices,
Choice A--benzoic acid--would be formed by hydrolysis of benzoyl chloride. Choice B--methyl benzoate--would be
formed by reaction of benzoyl chloride with methanol. Finally, choice C--benzaldehyde--would be formed by
reduction of benzoyl chloride. Again then, the correct answer is choice D.
4. The correct answer here is choice C. Esterification is a process in which a carboxylic acid reacts with an
alcohol in the presence of an acidic or basic catalyst to form an ester plus water. Therefore choice B--formation of an
acyl halide--can be rejected at once. Choice A illustrates the formation of dimethyl ether from two molecules of
methanol, therefore this choice is also wrong. Choice D is wrong as this is an example of the Williamson Ether
synthesis which involves reaction of a phenoxide ion with an alkyl halide to produce an ether. So the only choice
that results in ester formation is choice C.
5. Now for question 5. The correct choice is B. This question asks you to find the correct answer choice that
will NOT lead to the formation of a particular compound, which happens to be benzyl methyl ketone. Choice A
shows an alkyne being treated with an acidic mercury sulfate solution. This leads to Markownikov addition of
water—that is, hydrogen adds to the less substituted triple bonded carbon and hydroxyl adds to the other triple bonded
carbon. This results in the formation of an enol which will then spontaneously rearrange to form the ketone we are
looking for. So, choice A DOES yield benzyl methyl ketone and therefore it's NOT the answer we're looking for.
In choice B, a primary alcohol is oxidized with cold, dilute potassium chlorate. Although ketones can be synthesized
by oxidizing secondary alcohols, oxidation of primary alcohols results in the formation of aldehydes NOT ketones.
Therefore, B is the correct answer. Choice C--ozonolysis of an alkene--will yield our ketone plus formaldehyde and
choice D--oxidation of a secondary alcohol--also gives our desired ketone, so both of these choices are wrong.
Again, B is the correct answer choice.
6. Choice C is correct. This question deals with common methods of preparing carboxylic acids. Let's go
through the choices given. The carbonation of a Grignard reagent leads to the formation of the magnesium salt of a
carboxylic acid. When treated with mineral acid, the magnesium salt is then converted to a carboxylic acid.
Therefore choice A CAN be used to prepare carboxylic acids, so this is wrong. The acidic or basic hydrolysis of
nitriles also yields carboxylic acids, so choice B is also wrong. The oxidation of primary alcohols by various strong

10.
MCAT
10 as developed by
oxidizing agents such as potassium permanganate yields carboxylic acids, so choice D can be rejected. Choice C is
correct because carboxylic acids are formed by the oxidation of aldehydes, not their reduction.
7. For question 7, the correct choice is B. This diagram represents the tautomeric transformations of a ketone.
Ketones exist in two spontaneously-interconvertible forms--the keto form and the enol form. These differ in the
placement of a hydrogen atom and a double bond. The keto structure is more stable because a carbon-oxygen double
bond is more stable thermodynamically than a carbon-carbon double bond with a hydroxyl group attached and the
chemical properties of ketones are determined mostly by the keto form.
None of the other choices describes this process. Choice A--esterification--is a reaction between a
carboxylic acid and an alcohol resulting in the formation of an ester. Choice C--elimination--is a reaction
characteristic of alkyl halides and leads to the formation of a stable double bond. Choice D--dehydration--is the loss
of a hydroxyl group and a proton to yield water and usually a stable double or single bond is left behind. So again,
the correct choice is B.
8. Now for question 8. The correct answer is choice A. This question deals with the effect of substituents on
the relative basicities of aromatic amines. The basicity of aromatic amines or anilines manifests itself in
protonation of the amino-group nitrogen atom to make positively charged anilinium ions. In general, electron
withdrawing substituents decrease the basicity of anilines because the withdrawal of electron density increases the
positive charge on the anilinium ion, making it less stable. On the other hand, electron-donating substituents
decrease the positive charge of the anilinium ion and stabilize it. Now look at the choices. B, C, and D have
electron withdrawing substituents, while A has an electron donating methoxy substituent. Since electron donating
substituents increase basicity, choice A is the most basic among the compounds shown and therefore this choice is
correct.
9. Let's now look at question 9. The correct choice is D. Arenediazonium salts, which are synthesized from
primary aromatic amines, are compounds with an N2+ group attached to the aromatic ring. They are useful for
synthesizing a wide variety of compounds due to the fact that they can easily be made to undergo replacement
reactions in which molecular nitrogen is released and a nucleophilic substituent attaches to the aromatic ring in its
place. For example, in the presence of cuprous halides, diazonium salts release molecular nitrogen and form
halogen-substituted arenes--often called the Sandmeyer reaction. Similarly, in the presence of cuprous cyanide,
nitrogen is released and replaced by the cyanide ion, thus forming aromatic nitriles. In the same way, in cold
aqueous solutions, the hydroxyl group replaces the nitrogen to form phenol. Therefore, compounds I, II, and III can
all be obtained by the replacement of nitrogen in diazonium salts, and so the correct choice is D. As all three
choices are correct, you can discard A, B, and C which state that only one or two of the compounds will be formed.
So again, D is the right answer.
10. For question 10, the correct answer is choice B. This question requires you to know the most important
reactions between ammonia derivatives and aldehydes. In these reactions, the nucleophilic nitrogen of an ammonia
derivative attacks the electrophilic carbonyl carbon, while hydrogen released from this nitrogen attacks the
nucleophilic carbonyl oxygen. This reaction results in the formation of a single carbon-nitrogen bond, as well as a
hydroxyl group on the formerly carbonyl carbon. Since this system is quite unstable, the addition is always
followed by dehydration--that is, a water molecule is formed from the carbons hydroxyl and the neighboring
nitrogen's hydrogen. The release of the water molecule results in the formation of a carbon-nitrogen double bond. In
this case, the final product of the reaction between the carbonyl compound and phenylhydrazine is phenylhydrazone
and therefore choice B is correct. Choices A and C are synthesized in reactions with other ammonia derivatives.
Choice A is wrong because oximes are synthesized by reactions between carbonyl compounds and hydroxylamine.
Choice C is wrong because semicarbazones are synthesized by reactions between carbonyl compounds and
semicarbazides. Finally, aromatic nitriles are obtained by reacions between diazonium salts and cuprous cyanide, and
so choice D is also wrong. Again, the correct choice is B.
11. Now look at question 11. The right answer here is choice B. The reaction given is an example of Hofman
elimination. When a basic solution of a quaternary ammonium hydroxide is heated, it undergoes decomposition
forming an alkene, a tertiary amine and water. This reaction proceeds by an E2 mechanism. The base abstracts a
hydrogen ion from carbon and the subsequent release of the molecule of tertiary amine results in the formation of the
double bond. So, choice B is correct. Let's now look at the wrong answers. Choice A--alkylation--is the insertion
of an alkyl group into a molecule. Choice C--formation of ammonium salts--is possible only in acidic solutions
because it requires protonation of of nitrogen. Moreover, in this case nitrogen doesn't have unshared electron pairs,
therefore protonation and subsequent formation of a salt may occur only upon the breakdown of a carbon-nitrogen

11.
Organic Chemistry Discretes Test
KAPLAN 11
bond. Finally, amides are formed in reactions between primary and secondary amines with acyl halides. Therefore,
choice D is also incorrect and choice B is again the correct answer.
12. For question 12, the correct answer is choice D. The methyl group attached to an aromatic ring has an
activating and ortho/para directing effect on the ring--that is, it makes these positions very reactive towards
electrophilic aromatic substitution. Therefore, nitration is most likely to occur as shown in choice A which is ortho
and choice B which is para. This makes choice D the correct answer. The meta position, as shown in choice C, is
deactivated by the methyl substituent and so nitration here is negligible in comparison to the other two positions.
As a result, choice C is wrong and again, D is the correct answer.
13. In question 13, the correct answer is choice C. This is an example of stereospecific reduction or
hydrogenation. This reaction, catalyzed by palladium, results in the addtion of two hydrogen atoms on the same side
of the molecule--refered to as syn-addition. This puts the methyl and ethyl substituents on the same side of the
double bond and so the cis isomer is formed. Therefore, choice C is correct. Metallic sodium in ammonia solution
would cause a reaction with the opposite stereospecificity, yielding the trans isomer. This is called anti-addition.
Let's go through the remaining answer choices. Choice A is wrong because a hydration reaction is one in which
water is added to a molecule; don't confuse this with hydrogenation in which hydrogen is added. Choice B is also
incorrect because this is definately an addition reaction, not a substitution reaction. Finally, choice D is wrong
because racemization means loss of optical activity. Since the reactant is achiral and therefore optically inactive, it
is incapable of being racemized. Again, the correct answer is choice C.
14. The correct answer to question 14 is choice B. This question deals with different mechanisms of
substitution on aromatic rings. An aromatic ring is especially susceptible to electrophilic attack, but in some cases,
nucleophilic aromatic substitution is also possible. Let's go through the choices. Choice A is an example of
Friedel-Crafts acylation. In this reaction, benzene reacts with an acyl chloride in the presence of the Lewis acid
aluminum trichloride. The Lewis acid removes chloride from the acyl halide thus transforming it into a strongly
electrophilic oxycarbonium ion, which immediately attacks the benzene ring. Thus choice A is wrong because it IS
an example of electrophilic attack. Choice C is an example of Friedel-Crafts alkylation, which proceeds similarly to
acylation, therefore C is also wrong. Choice D can be rejected as it is yet another example of electrophilic attack.
But in choice B, the benzene ring is being catalytically hydrogenated to form cyclohexane, so this answer choice is
NOT an example of electrophilic aromatic substitution making it the correct answer.
15. Now look at question 15. The correct choice here is B. The primary structure of a protein is the amino
acid sequence, which is formed by covalent peptide linkages. Choice A is incorrect because many proteins contain
more than one peptide chain. Choice C is incorrect because proteins are denatured by heating and so they LOSE
their conformation at 40°C not retain it. Choice D is incorrect because only those proteins containing more than
one peptide subunit have quaternary structure.
16. The correct answer to question 16 is choice C. The reaction equations shown are 3 steps in the free radical
halogenation of an alkane. Free radical halogenation is one of the few reactions that alkanes will undergo and occurs
by the reaction of the alkane with a highly reactive halogen radical. A halogen radical is a single, neutrally charged
halogen atom, which has an unfilled valence shell containing seven electrons. Since it has an odd number of
electrons, the halogen radical also has a half filled orbital, which is denoted by Xo. Specifically, this might be
denoted as Bro, Clo, or another halogen free radical. In the halogenation reaction, the step that starts the whole
process is the formation of the free radical from a diatomic molecule. This process, which is shown in the first
reaction equation, is usually catalyzed by ultraviolet light although it can also be catalyzed via attack by another free
radical. Step I is followed by step II, in which the halogen free radical attacks the alkane, producing an alkyl halide
and another highly reactive free radical--an alkyl radical. In step III, this alkyl radical attacks a halogen molecule, X2,
picking up one of the halogen atoms to become another molecule of alkyl halide and at the same time producing
another halogen radical. This new halogen radical will then start the process again thus causing a chain reaction.
Anyway, getting back to the question, step I—UV light-induced generation of two free radicals—is called
the chain initiating step. As a result, you can already rule out answer choices A and D. Steps II and III are both
chain propagating steps because they serve to push the reaction chain onward by generating a molecule of product
plus a molecule of free radical. This means that choice C is the correct answer. Choice B is wrong as it fails to
recognize that step II is a chain propagating step, so again, choice C is correct.
17. Now look at question 17. The correct choice here is B. The reactivity of an aromatic compound toward
electrophilic substitution depends on the nature of the ring's substituents. Electron donating substituents enrich the

12.
MCAT
12 as developed by
electron system of a ring and so activate it towards electrophilic attack. Electron withdrawing substituents, on the
other hand, deactivate the ring making it less susceptible to electrophilic attack. In choice B, the ring is severely
deactivated by the combined electron withdrawing effects of the bromine and ammonia substituents. Choice B is
therefore almost totally unreactive towards electrophilic aromatic substitution and so it is the correct answer. As for
the other choices, choice A contains an electron-donating, activating methyl substituent, so this compound is very
reactive towards electrophilic attack. The ring in choice D is even more highly activated because it contains two
electron-donating methoxy substituents. Finally, in choice C, the strong electron withdrawing effect of the nitro
group is compensated by the strong electron donating effect of the amino group, so this compound is moderately
reactive towards electrophiles. Again, the correct answer is choice B.
18. For question 18, the correct answer is choice D. The number of different stereoisomers a molecule can have
depends on the number of chiral centers it has. If the number of chiral centers in n, then the number of different
stereoisomers is 2n, where a chiral center is defined as a carbon bonded to 4 different substituent groups.
For convenience, let's number all the carbons in this compound, beginning with the carboxyl carbon. You
can see that carbons 1 and 5 are not chiral centers, because carbon 1 is attached to only three substituents while
carbon 5 has two identical substituents--namely hydrogen. On the other hand, carbons 2, 3, and 4 are chiral centers,
since each of them is bonded to 4 different groups. Therefore, the molecule has 3 chiral centers and the number of
different stereoisomers is 2 to the power of 3--that is 8. So, the correct choice is D.
19. Let's now take a look at question 19. The correct answer here is choice B. This question tests your
understanding of how thin-layer chromatography works. A mixture is spotted onto the thin layer of adsorbant that
covers a plastic or glass sheet. The sheet is then placed upright in a developing chamber that contains a solvent,
called the eluant. Capillary action forces the eluant up the plate, and with it, portions of the mixture that have been
dissolved by the eluant. Since the different compounds move at different rates based on their size and how well they
dissolve in the eluant, the mixture gradually separates. When the plate is dried, the spots of different compounds can
be visualized, or made visible, by UV light. Okay, I said that the mixture moves by capillary action, so choices A
and D are wrong. Also, the capillary action forces the mixture up, not down, so choice B is correct and C is wrong.
Incidentally, the downward movement of solvent under the influence of gravity is utilized in chromatography, but in
column chromatography, not thin-layer chromatography. Again then, the correct choice is B.
20. For question 20, the correct choice is A. This question describes the terminology used in describing
stereochemistry. It can best be answered by going through the answer choices. Choice A states that compounds I
and II are diastereoisomers. These are structural isomers that are not mirror images. Looking at compounds I and II,
you should be able to see that that these compounds are indeed structural isomers, but not mirror images. Therefore,
choice A is correct.
Choice B states that compounds I, II, and IV are meso compounds. Meso compounds have chiral carbons,
but also have planes of symmetry--that is they are their own mirror images and are thus optically inactive. You can
see from the structures that compound II is definately meso but compounds I and IV are not. Therefore, choice B is
incorrect. Choice C states that compounds II and IV are optically inactive. As we just said, compound II is meso so
its optically inactive, but compound IV has chiral carbons but no plane of symmetry and so IS optically active.
Therefore, choice C is incorrect. Compound III is meso and so it is optically inactive making choice D wrong. To
repeat then, the correct choice is A.
21. Now for question 21. The correct choice here is C. Geometric isomers are isomers that differ in the
orientation of their substituents around a carbon-carbon double bond. In cis isomers, the substituents are on the
same side of the double bond, while in trans isomers, the substituents are on opposite sides. Compounds I and IV
have no double bonds, so they can't possibly be geometric isomers. Therefore, if you knew the definition of
geometric isomers, you could immediately have eliminated every choice except C. You can easily see that
compound II is the trans isomer of compound III, which is the cis isomer. The proper term for compounds I and IV
is structural isomers, because they have the same molecular formula but their atoms are connected differently.
Compounds II and IV have different molecular formulas as do compounds I and III , so these pairs are not isomers at
all. Again, the correct answer is choice C.
22. The correct choice for this one is D. This question tests your familiarity with the stereochemistry of
substitution reactions. In this case, we're clearly dealing with an SN1 reaction, because tertiary alkyl halides can't
undergo SN2 reactions because of steric hinderance. The central carbon of the alkyl halide reactant is bonded to 4
different substituents, so this compound is chiral. It's also asymmetrical, so it will be optically active. In the first
step of the reaction that's given here, the alkyl chloride will dissociate to form a stable tertiary carbocation. This

13.
Organic Chemistry Discretes Test
KAPLAN 13
will result in the loss of optical activity, which always happens in SN1 reactions. Let's see why. In any
carbocation, the positively charged carbon is always achiral because it has only three substituents. This means that
all carbocations lie pretty much within a plane. Therefore, in the second step, the bromide amion can attack the
carbocation from either side of the plane with equal probability. As a result, the reaction will yield equal amounts of
two chiral products, which are enantiomers--non-superimposable mirror images of each other. These enantiomers
rotate the plane of polarized light to the same extent but in opposite directions, so the product will be an optically
inactive racemic mixture. Getting back to the question, since this reaction results in the loss of optical activity,
choice D is correct and choice B is pretty clearly wrong. Choice A is also wrong because inversion of absolute
configuration only occurs in SN2 reactions, where a nucleophile is attached from the back in a one-step reaction.
Since this is an SN1 reaction, that won't happen. Some of the individual molecules will wind up inverted, but
overall, the absolute configuration of the sample will be lost. Finally, choice C will be irrelevant to this question,
because mutarotation concerns the equilibrium between open-chain forms and cyclic hemiacetal forms of
monosaccharides in aqueous solutions. Again, the correct answer is choice D.
23. For question 23, the correct answer is choice B. This question deals with the amphoteric properties of
amino acids. At neutral or alkaline pH values, the COOH group of an amino acid dissociates to become a
negatively-charged carboxylate. At a neutral or acidic pH, the NH2 group is protonated and becomes positively
charged. Thus at neutral pH an amino acid is a dipolar ion. The overall charge of an amino acid depends on the ratio
of positively charged amino groups to negatively charged carboxylates in the molecule. In order for a molecule to
migrate to the cathode, its overall charge must be positive. Of the choices given, choice B is the only one in which
the positively charged amino groups would outnumber the negatively charged carboxylate groups. Therefore, only
choice B would be positively charged at pH 7. Again then, the correct answer is B.
24. Now let's look at question 24. The correct answer here is choice A. To figure out the best way of
separating two compounds, you have to take into effect their chemical and physical properties. What you're told
about these compounds is that at atmospheric pressure, they both decompose before they reach their boiling point--
the temperatures of which are given. Okay, three of the answer choices here are types of distillation. In simple
distillation, a mixture is heated until each component boils off in turn, and can be collected separately. Since both
of these compounds decompose before they reach their boiling point at one atmosphere of pressure, you can't
separate them by simple distillation, since they'll just decompose before they are separated. So choice B is wrong.
Instead, you should use vacuum distillation, choice A, which is actually distillation at very low pressure. This
makes compounds boil at lower temperatures, which will make it possible for these compounds here to to reach their
boiling point before they decompose. Choice D, fractional distillation, is a procedure used to separate compounds
that have boiling points very close together. It's generally done at atmospheric pressure, so as with simple
distillation, these compounds would decompose instead of distilling, so D is also wrong. Finally, choice C,
sublimation, is the direct conversion of a solid into a gas. This is likely to require lower pressures and/or higher
temperatures than vacuum distillation, meaning it would be more difficult to carry out and would also carry more
risk of the compounds decomposing, so choice C is wrong too. Again then, the correct answer is choice A.
25. The correct answer to question 25 is choice A. Information about functional groups is provided by IR
spectroscopy. Choice B is wrong because information about conjugated double bonds is obtained by UV
spectroscopy, not IR. Choice C is also wrong because information about molecular weight is obtained by mass
spectrometry. Information about protons is provided by nuclear magnetic resonance spectroscopy, therefore choice D
can also be rejected. Again, choiec A is the correct answer.
26. Now for question 26. The correct answer to this one is choice D. This question requires you to understand
the main principals of NMR spectroscopy. This technique provides a chemist with information about the proton
enviroment. Equivalent hydrogens--that is, hydrogens whose positions are identical with respect to the surrounding
atoms--give a single NMR signal, whereas nonequivalent hydrogens give separate NMR signals. Since the
compound given shows just one NMR signal, you can assume that all its hydrogens are equivalent. So let's now go
through the choices given. Choice A is wrong because it will give three different signals, one from the two terminal
methyl hydrogens, one from the hydrogens on the CH2 groups next to them and one for the hydrogens on the two
innermost CH2 groups. Choice B gives 6 different signals: one from the two methyls on the far left end of the
molecule as it's shown, one from the CH group next to that one, 3 from the four CH2 groups in the chain, and 1
from the methyl group on the right end. Thus choice B is also wrong. Using the same principle, you can easily see
that choice C will give 7 signals. However, in choice D, all the methyl groups are equivalent and so this compound
will give one NMR signal. It is interesting to note that in choices B and C, the signals probably won't be as simple

14.
MCAT
14 as developed by
as we have described. Coupling effects would mean that these NMR patterns would be highly complex due to
overlapping and split signals. Again, the correct answer is choice D.
27. Now for question 27. The correct answer here is choice D. Let's go through the answer choices. All the
hydrogens in choice A are equivalent, so this compound would produce one NMR proton peak. In choice B, the 6
methyl group hydrogens are identical so they would produce one NMR signal. The two hydrogens of the CH2 group
are also identical and so would produce another NMR peak. Therefore, choice B would give two peaks. By the same
principle, choice C produces three NMR peaks--one from the two CH2 groups, one from the CH group and one from
the CH3 group. Finally, choice D has a chiral carbon, the one with the bromine group attached to it. The three
hydrogen atoms in each methyl group are equivalent, but the two methyl groups are not, so each methyl group will
produce a peak. The two hydrogens of the CH2 peak will produce another peak and the single hydrogen bonded to
the brominated carbon will produce a fourth. Thus, choice D gives 4 different NMR signals--most of any of the
choices--and therefore it's the correct answer.
28. The correct answer is choice A. This question tests your knowledge of IR spectroscopy and what sorts of
spectra different kinds of compounds produce. In the spectrum that's described, the various peaks between 1600 and
900 cm–1, which aren't described in detail, are in the fingerprint region. The tall peak at 1700 cm–1 indicates the
carbon-oxygen double bond of a carbonyl group. Well, knowing that, we can eliminate choices C and D since
neither of them have a carbonyl group in them. That leaves either a ketone or an aldehyde. Well, the peak at 2950
cm–1, which lies in the range that describes the hydrogen stretching patterns of a molecule, doesn't show the
characteristic double bond absorption at about 2800 cm–1 of the carbonyl hydrogen in aldehydes. The absorption is
more like the normal absorptions of hydrocarbon hydrogens. That means that since the characteristic hydrogen
pattern of aldehydes is missing but we know we have a carbonyl group, the compound must be a ketone, which is
choice A.
29. For question 29, the correct choice is D. This question deals with optical activity in stereoisomers. A
racemic mixture contains equal quantities of two enantiomers--that is, isomers that are non-superimposable mirror
images of one another. Lets go through the choices that we're given. Compounds I and III are mirror images, but if
you rotate one of them by 180 degrees, you will see that you can superimpose it on the other. That's because each
compound has a plane of symmetry. So even though these molecules contain chiral carbons, they are optically
inactive, and would be called meso compounds, not enantiomers. In fact, compounds I and III are the same
compound so choice A is wrong.
It is somewhat easier to eliminate choice B, because isomeric compounds I and IV are clearly not mirror
images at all. In fact, they are diastereoisomers. The same is true of compounds II and III in choice C, which is
also wrong. The correct answer is choice D, because, although compounds II and IV are mirror images, they can't be
superimposed even if you rotate one of them by 180 degrees. Again then, the correct choice is D.
30. For question 30, the correct choice is A. To answer this question, you need to know under what conditions
SN2 reactions will occur.
Well, you can see that the reactant is a primary alkyl halide and chloride is also a good leaving group. You
should also know that CN– is an extremely strong nucleophile, so you should be able to narrow the choices down to
at least A and B, since SN1 and E1 mechanisms are highly unlikely in primary alkyl halides. So now, how do we
know that SN2 will occur over E2? Well, remember we said that CN– is an extremely strong nucleophile.
Therefore, substitution will occur exclusively over elimination. Only when strong bulky bases are used is E2
favored over SN2. So again, the correct answer is choice A.