2Oxygen containing compounds test w solutions

ORGANIC CHEMISTRY TOPICAL:
Oxygen-Containing Compounds
Test 1
Time: 22 Minutes*
Number of Questions: 17

* The timing restrictions for the science topical tests are optional. If you
are using this test for the sole purpose of content reinforcement, you
may want to disregard the time limit.


MCAT

DIRECTIONS: Most of the questions in the following test
are organized into groups, with a descriptive passage
preceding each group of questions. Study the passage,
then select the single best answer to each question in the
group. Some of the questions are not based on a
descriptive passage; you must also 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)

GO ON TO THE NEXT PAGE.

2

as developed by


Organic Chemistry Discretes Test
Passage I (Questions 1–6)

β-keto esters and malonic esters (commonly known
as β-dicarbonyl compounds) are compounds that contain
two carbonyl groups separated by an intervening carbon.
The central carbon of these compounds contains highly

acidic protons, known as α-hydrogens. An α-hydrogen can
be abstracted with an appropriate base, forming an enolate
ion that can be either acylated or alkylated.
Two common laboratory syntheses that rely on this
principle are the malonic ester synthesis (Reaction 1) and
the acetoacetic ester synthesis (Reaction 2). The malonic
ester synthesis is one of the most profitable methods of
preparing substituted carboxylic acids, while the acetoacetic
ester route is often used in the synthesis of methyl ketones.
O

O
OEt

OEt
1) NaOEt
2) RX

1) H3O

R

2)

R

1 . Which of the following best explains the chemistÕs
choice of ethanol as the solvent used in the synthesis
described in the passage?
A . The solvent chosen should have a boiling point

similar to that of the starting material.
B . The solvent chosen should have a boiling point
similar to that of the desired product.
C . Ethanol solvates the ethoxide anions, rendering
them less nucleophilic.
D . Ethanol is a better solvent than polar aprotic
solvents such as DMSO.

COOH
+CO 2(g)

O

O

Figure 1

+

OEt

OEt

COOH

Reaction 1

2 . Which of the following compounds can most readily
be synthesized via the acetoacetic ester pathway shown
in Reaction 2?

O

O

O
1) NaOEt

1) H3O+

R

2)

2) RX

OEt

OEt

O

R

O

+CO2(g)

Reaction 2

A
chemist
attempted
to
synthesize
cyclohexanecarboxylic acid (Figure 1) using a slight
variation of the malonic ester synthesis. Diethyl malonate
(1 equiv) was dissolved in a solution of sodium ethoxide in
ethanol (2 equiv). The resulting solution was then added
dropwise, with constant stirring, to a solution of 1,5dibromopentane (1 equiv). When the reaction was
complete, the reaction mixture was hydrolyzed with 6M
HCl(aq) and heated to remove any CO2 that formed. After
the desired product was isolated, the final yield was
determined to be 75%.

A.
B.
C.
D.

Hexanoic acid
2-Hexanone
3-Hexanone
Ethyl hexanoate

3 . Which of the following best describes the mechanism
of both synthetic pathways shown in the passage?
A . Alkylation followed by hydrogenation and
carboxylation .
B . Deprotonation followed by alkylation and

carboxylation.
C . Alkylation followed by protonation and
decarboxylation.
D . Deprotonation followed by carboxylation and
hydrolysis.

then
then
then
then

GO ON TO THE NEXT PAGE.

KAPLAN

3


MCAT
4 . Which of the following halides would be the most
suitable starting material for a malonic ester synthesis
of dibenzylacetic acid?
A.
B.
C.
D.

Bromobenzene
2-Chlorophenol
4-Chlorophenol

Benzyl bromide

5 . Attempts to produce laboratory quantities of
asymmetrically substituted carboxylic acids such as 2ethylpentanoic acid will result in lower yields than
those usually obtained in the case of symmetric
products. This difference is primarily attributed to:
A . reduced solubility of the participating reactants.
B . competitive alkylation resulting in a mixture of
products.
C . decreased nucleophilicity of the anionic
intermediate.
D . increased difficulty in hydrolyzing the asymmetric
ester.

6 . The two synthetic pathways described in the passage
can best be described as:
A.
B.
C.
D.

nucleophilic substitution reactions.
nucleophilic addition reactions.
alpha substitution reactions.
simultaneous alpha substitution and nucleophilic
substitution reactions.

GO ON TO THE NEXT PAGE.

4

as developed by


Organic Chemistry Discretes Test
Passage II (Questions 7–12)
Triacylglycerols are fats or oils commonly found in
plant and animal systems. These molecules are formed by
the attachment of three long chain carboxylic acids (fatty
acids) to glycerol via an ester linkage. The carbon chain of
the fatty acids tends to be unbranched and even in number
(usually between 8 and 20 carbons). Fatty acids can be
either unsaturated or saturated: Unsaturated fats contain
double bonds, whereas saturated fats do not. Since recent
research has supported the suggested link between dietary
saturated fats and serum cholesterol, the degree of saturation
in various animal fats and vegetable oils is of great interest
to nutritionists.
Table 1 contains a number of vegetable oils consisting of
various fatty acids that differ in both chain length and
degree of saturation.
Table 1
FATTY ACID CONTENT(%)
Palmitic Palmitoleic Steric Oleic Linoleic Linolenic
acid
acid
acid acid
acid
acid
(C16)*

(C16:1)
(C18) (C18:1) (C18:2)
(C18:3)
SOURCE:
Corn

10

1

2

25

62

Ð

Linseed

10

Ð

4

23

57

6

Safflower

7

Ð

2

13

75

3

Soybean

9

Ð

4

43

40

4

Castor Bean

Ð

Ð

1

93

4

*C x:y
x = number of carbons
y = number of double bonds

Ð

The fatty acid content of these oils can be easily
determined by employing the transesterification reaction
(Reaction 1).
O
O

C

O
R'

OH MeO

O
O

C

C

R'

O

H+/OHÐ
R'' + 3MeOH

OH + MeO

O
O

C

C

R''

O
OH MeO

R'''

Triacylglycerol

Methanol

Glycerol

C

R'''

Fatty acids

Reaction 1
A student attempted to determine the fatty acid
composition of a commercially available cooking oil by
using Reaction 1. The procedure involved dissolving the
oil to be analyzed in toluene; when dissolution appeared to
be complete, a solution containing BF3 in methanol was
added. After heating for 30 minutes, the reaction mixture
was cooled to room temperature and extracted with distilled
water. The organic layer was then retained for analysis.

7 . What is the purpose of using BF3 in the experiment
described?
A . To absorb excess water formed
B . To catalyze the transesterification by polarizing
the carbon-oxygen double bond
C . To catalyze the transesterification by acting as the
initial nucleophile

D . To solvate the methanol molecules

8 . If the student analyzes the product mixture by proton
NMR, which of the following observations would
signify that the reaction had reached completion?
A . The presence of several singlets in the region 3.0
to 45 ppm
B . The absence of multiplets in the region 3.5 to 4.5
ppm
C . The presence of peaks in the region 2.0 to 3.0
ppm
D . The absence of a sharp singlet at 9.5 ppm
GO ON TO THE NEXT PAGE.

KAPLAN

5


MCAT
9 . Triacylglycerol is a useful starting material in the
production of soaps (salts of long chain carboxylic
acids) via a process known as saponification. If this
process was carried out by boiling the fat in aqueous
sodium hydroxide and then extraction with toluene,
what products would you expect to see in the aqueous
layer?
A.
B.
C.

D.

A mixture of methyl esters and glycerol
Glycerol and sodium chloride
Both unreacted sodium chloride and triglyceride
A mixture of glycerol and sodium salts of fatty
acids

1 0 . Partial transesterification of triacylglycerol can result
in the formation of ÔdiolsÕ. These compoundscould
also be described as:
A.
B.
C.
D.

methanol dimers.
nonalkylated fatty acids.
monoesters of glycerol
diesters of glycerol

1 1 . According to Table 1, which of the following oils has
the highest polyunsaturated fat content?
A.
B.
C.
D.

Corn oil
Linseed oil

Soybean oil
Castor bean oil

1 2 . In order to determine the degree of unsaturation in the
oil being analyzed, the student treats a portion of the
reaction mixture with a solution of bromine in carbon
tetrachloride. If it is found that 1.50 equivalents of
bromine react, what is the most likely source of the
oil?
A.
B.
C.
D.

Soybean
Safflower
Corn
Castor bean

GO ON TO THE NEXT PAGE.

6

as developed by


Organic Chemistry Discretes Test
Questions 13 through 17 are NOT
based on a descriptive passage.
1 3 . Which of the following compounds is the strongest

acid?
A.
B.
C.
D.

HOOC-CH2Cl
HOOC-CH2OCH3
HOOC-CH2F
HOOC-CH3

1 6 . Reaction of benzoic acid with thionyl chloride
followed by treatment with ammonia will yield which
of the following compounds?
A.
B.
C.
D.

Benzamide
p-Aminobenzaldehyde
p-Chlorobenzamide
3-Chloro-4-aminobenzaldehyde

1 7 . The lH NMR spectrum below is characteristic of
which of the following compounds?
1 4 . Which of the following compounds is the most
susceptible to nucleophilic attack by OHÐ?
A.
B.

C.
D.

Propanoic acid
Ethanal
Benzyl chloride
Propionyl chloride

1 5 . The Williamson synthesis of methoxybenzene is
outlined below

(1)
10

(2) (2) (3)
9

8

7

1) NaOH
2) CH3 Cl

5

4

3

2

1

0

ppm (δ)

OCH3

OH

6

+ NaCl + H2 O

A.
B.
C.
D.

Butanol
Butanone
Butanoic acid
Butanal

Which of the following best describes the mechanism
of this reaction?
A.
B.

C.
D.

Bimolecular nucleophilic substitution
Unimolecular nucleophilic substitution
Bimolecular elimination
Unimolecular elimination

END OF TEST

KAPLAN

7


MCAT
ANSWER KEY:
1.
C
2.
B
3.
B
4.
D
5.
B

8

6.
7.
8.
9.
10.

D
B
B
D
C

11.
12.
13.
14.
15.

B
C
C
D
A

16.
17.

A
D

as developed by


Organic Chemistry Discretes Test
OXYGEN-CONTAINING COMPOUNDS TEST EXPLANATIONS
Passage I (Questions 1Ð6)
1.
For question number 1, the correct answer is C. This question requires an understanding of the malonic
ester synthesis described in the passage and the mechanism by which it occurs, as well as the concept of solvation.
Ethanol is a protic solvent: a solvent whose molecules have a hydrogen attached to an oxygen or a nitrogen. Protic
solvents solvate anions very effectively by hydrogen bonding to them. In this experiment, sodium ethoxide dissociates to
produce an ethoxide anion--which, depending on the solvent, can act as either a base, abstracting an acidic hydrogen, or as a
nucleophile, attacking an electrophile. However, as you can see from the reaction, sodium ethoxide abstracts an acidic alpha
hydrogen from diethyl malonate, and does not attack the electrophilic carbonyl carbon. In using the protic solvent ethanol,
the ethoxide anion becomes hydrogen bonded and consequently solvated. This "solvation sphere" renders the anion much less
nucleophilic and so ensures that alpha deprotonation can proceed unhindered. Therefore, choice C is the correct response.
Now for the other answer choices. Choices A and B discuss the boiling point of the solvent. Of course, ethanol has
to be distilled off at the end of the reaction, but its boiling point was not the reason it was chosen. Ethanol is used for its
protic properties as we discussed, therefore A and B can be discarded. Choice D is wrong, since a polar aprotic solvent would
be just as good a solvent as ethanol. However, aprotic solvents are not suitable for this reaction: they do not possess a
hydrogen attached to an oxygen or a nitrogen. Again, choice C is the correct answer.
2.
The correct answer to question 2 is choice B. This question is testing your understanding of the
syntheses described in the passage, along with a bit of nomenclature. It's mentioned in the passage that the acetoacetic ester
synthesis is commonly used in the production of methyl ketones: ketones of the type CH3COR. So, the question is really
asking: "which of the following is a methyl ketone?". 2-Hexanone is the only answer choice that fits this definition and so is
correct. Hexanoic acid--choice A--and ethyl hexanoate--choice D--would be more easily synthesized by the malonic ester
synthesis, not the acetoacetic ester synthesis. Finally choice C, 3-hexanone, has the carbonyl in the middle of the chain and
would most likely be synthesized by methods other than those mentioned in the passage--such as direct oxidation of 3hexanol. Again, choice B is the correct answer.
3.

The correct answer to question 3 is choice B. Just like the previous question, this one tests your
understanding of the mechanism of both syntheses. According to the reactions shown, both starting materials are reacted with
a base (in this case, an alkoxide) which abstracts a proton, forming a carbanion: deprotonation. Choices A and C can be
eliminated. The carbanion, being a nucleophile, attacks the alkyl halide, displacing the halide: alkylation. Choice D can be
eliminated, leaving choice B as the correct response. (By the way, since carbon dioxide is a product in both of the reactions,
they also undergo decarboxylation.)
4.
The correct answer to question 4 is choice D. This question requires you to have a thorough
understanding of the malonic ester synthesis--probably more so than the other questions. It would probably be useful to drawout the target compound, dibenzylacetic acid, and compare it to the cyclohexane carboxylic acid made by the chemist in the
passage. If you do this, you will get a figure showing an acetic acid molecule in which the alpha carbon is attached to two
benzyl groups--so the overall formula is C6H5CH2 twice and then CHCOOH. The CH carbon is the original middle alpha
carbon of the malonic ester or more specifically, the one that is deprotonated first by sodium ethoxide. So, you can see that
two benzyl groups have been substituted onto this carbon, so the alkyl halide that reacts must contain a benzyl group. The
only answer choice that contains this benzyl group is choice D, so this is the correct response. When the alpha carbon of
diethylmalonate is deprotonated, a benzyl group adds to this carbon, resulting in loss of bromide from the alkyl halide. The
same process occurs with a second benzyl bromide--so overall, two equivalents of benzyl bromide and sodium ethoxide are
needed. Finally, the substituted acetic acid is formed by addition of acid and then heating in order to decarboxylate the
molecule. All the other answer choices are aryl halides and since this reaction does not proceed by nucleophilic aromatic
substitution, we could have quickly eliminated them. Again, the correct answer is choice D.
5.
The correct answer to question 5 is choice B. This question again tests your understanding of the malonic
ester synthesis. The named product--2-ethylpentanoic acid--could be made from diethyl malonate by first deprotonating the
ester with a strong base, then adding one equivalent of propyl halide, deprotonating for a second time and then adding one
equivalent of ethyl halide. If this were done simultaneously, without removing the first alkylated product, we would expect to
wind up with a mixture of 5, 6, and 7 carbon chains attached to the acid groups in the final products, since the propyl and
ethyl halides compete for the same nucleophilic site. To obtain the desired product, there would have to be a separation step
so that the first alkylated product could then be treated independently. In either case the yield of desired product will be
reduced: by formation of undesired side products or by losses during separation. Anyway, back to the question. The primary
reason the yields are low is that in any case we have competitive alkylation of the anionic intermediate. Done in a single
step, the nucleophilic carbanion can attack either alkyl halide. This competitive alkylation can be avoided by doing the

reaction in sequence, resulting in losses from other factors. However, the primary reason that the reaction would be carried
out in sequence is to avoid competitive alkylation. Choice B is the correct response.

KAPLAN

9


MCAT
Solubility problems--choice A--can be avoided by suitable choice of solvents, and are not related to the symmetry of
the final product--this answer choice is incorrect. Choice C tries to distract you with the nucleophilicity of the anion--which
is the deprotonated diethyl malonate molecule. If sodium ethoxide is present, the nucleophilicity will be exactly the same,
whether symmetric or asymmetric products are formed, so choice C is also wrong. Meanwhile, choice D is, like some of the
choices in earlier questions, looking at the wrong part of the reaction. Hydrolysis is basically part of the work-up and so
occurs after the alkylation steps are complete.
6.
The correct answer to question 6 is choice D, simultaneous alpha substitution and nucleophilic
substitution reactions. The alpha position of both reactants is the position where the alkyl group has been substituted for one
of the hydrogens. Now, in order for substitution to occur, the alkoxide must first abstract a proton from the alpha position,
forming a nucleophilic carbanion. This nucleophilic carbanion then attacks the alkyl halide, displacing the halide: a
nucleophilic substitution reaction. Choice D is the correct response. (Even though choices A and C could be considered true,
the question asked you to choose the best description of the pathways, and that is choice D.) Choice B--nucleophilic addition-is a reaction characteristic of carbonyl compounds that have no leaving groups. In these reactions the carbonyl is not
attacked, so choice B is incorrect. Again, choice D is the correct response.
Passage II (Questions 7–12)
7.
The correct answer to question 7 is choice B. This question tests your understanding of Reaction 1. The
most likely mechanism for a reaction like the one illustrated is nucleophilic attack by the oxygen of methanol on the
carbonyl carbon of the ester followed by the loss of a good leaving group--which in this case is an OR group, where R is the
glycerol "backbone". This will result in the formation of glycerol and three fatty acids--the nature of which will depend on
the alkyl chain. Neutral alcohols are only slightly nucleophilic and react slowly with the carbonyl carbon. Boron trifluoride

is a strong Lewis acid and can coordinate to the carbonyl oxygen, drawing even more electron density away from the carbonoxygen double bond thereby making the carbon more susceptible to nucleophilic attack by methanol. This is described best
by answer choice B. As for the wrong choices, A misses the minor detail that water is not a product of the transesterification
reaction. Choice C wants us to believe that a strong Lewis acid like BF3 can act as a nucleophile, which is impossible by
the definition of a Lewis acid as it is an electron pair acceptor, not a donor. Choice D--solvating the methanol molecule--is
incorrect as this would decrease its nucleophilicity, not enhance it. Boron trifluoride has nothing to do with the methanol in
the reaction--as I said before, it simply polarizes the carbon-oxygen double bond in the triacylglycerol molecule. Again,
choice B is the correct response.
8.
The correct answer to question 8 is choice B. This question requires a basic understanding of NMR and
its relationship to the structural analysis of esters. Correctly answering this question also requires a thorough understanding
of the experiment described in the passage. Well, the product mixture retained for analysis was the organic layer.
Presumably, this should be a mixture of fatty acid esters and an NMR spectrum should reflect this if the reaction did indeed go
to completion. If the reaction was not complete then we would expect some water insoluble reactants to be present also.
Water insoluble reactants would have to be the triacylglycerol, and perhaps some mono- and di- esters of glycerol resulting
from the incomplete, though partial, transesterification reaction. With that level of understanding, we would be safe to hit the
choices. Choice A states that 3H singlets are observed. This is evidence that methyl groups are present, but does not signify
that nothing but methyl esters are present in the product mixture. Choice B, on the other hand, indicates that no multiplets-meaning no signals corresponding to the hydrocarbon "backbone" of triacylglycerol--are observed. If this signal were absent,
there would also be no mono- or di-esters produced either. This should make for pretty good evidence that starting material
and partially reacted starting material are absent, and as such, it should make for a pretty good answer choice. As for the rest
of the choices, choice C indicates signals in the region where protons alpha to a carbonyl would be observed. Since the
initial triglyceride and the final fatty acid esters, as well as the partially reacted species, all have CH2 groups next to carbonyl
carbons, this observation would not tell us anything about the progress of the reaction, or lack thereof. Finally, choice D
gives us a singlet at 9.5, or rather its absence to consider. This resonance is characteristic of an aldehyde proton, so its
absence indicates the absence of aldehydes in the product mixture. Since neither reactant nor product is an aldehyde, this
observation says nothing at all about the reaction or its progress. Again, choice B is our answer.
9.
For question 9, the correct answer is choice D. Saponification involves the basic hydrolysis of
triacylglycerol to yield glycerol and sodium carboxylates which are in fact soaps. When triacylglycerol is boiled in aqueous
sodium hydroxide, the hydroxide group attacks and adds to the carbonyl carbon of the ester, forming a tetrahedral intermediate,
which in turn decomposes to give a carboxylic acid and an alkoxide ion. The alkoxide then deprotonates the acid yielding the

sodium salt of the acid and glycerol. Choice D is the correct response.
Because the question states that triacylglycerol is hydrolyzed in aqueous sodium hydroxide and then extracted with
toluene, choices B and C can be eliminated since there is no mention of chloride ions. As far as choice C is concerned, if
there were any unreacted triacylglycerol it would be found in the organic layer, not the aqueous layer. Finally choice A is
wrong because methyl esters would not be formed in the reaction. This answer choice is trying to confuse saponification

10

as developed by


Organic Chemistry Discretes Test
with transesterification. Methyl esters would be formed if say, methanol were present, but again, only sodium hydroxide and
toluene are mentioned, so this answer choice is wrong and again the correct response is choice D.
10.
The correct answer to question 10 is choice C. The diol formed would be the remains of the triglyceride
after two fatty acid residues have been removed. The term "diol" would refer to glycerol with one remaining fatty acid attached
to it and so could be described as a monoester of glycerol. The other choices are not diols. Choice A--methanol dimers-would just be composed of two molecules of methanol bonded to each other. Choice B--nonalkylated fatty acids--are acids not
diols, and finally, choice D--a diester of glycerol--will have only one free OH group. Again, choice C is the correct response.
Questions 11 and 12 both require an understanding of the table provided in the passage. The table lists several
common oils, along with a breakdown of their fatty acid composition by percentage.
11.
The correct answer to question 11 is choice B. This question requires us to determine which is highest in
polyunsaturated fat. Remember that an unsaturated compound is one in which double bonds are present, so a polyunsaturated
compound will have two or more double bonds. This corresponds to the last two columns in the table, and if you add the
percentages up, you can see that linseed oil has the greatest percentage of polyunsaturated fatty acids out of all of the answer
choices at 63%. Of the choices given, corn oil has 62% polyunsaturated fatty acids, soybean oil is 44%, and castor bean is
4%. So choices A, C, and D can be eliminated and again, choice B is the correct answer.
12.
The correct answer here is choice C. This question requires you to do a bit more than just table reading;

here we need to consider the reaction of bromine with the fatty acids isolated in the experiment. Bromine in carbon
tetrachloride is a standard test solution for double bonds. One equivalent of bromine would indicate one equivalent of double
bonds and in this reaction we are told 1.5 equivalents of bromine react so there should be 1.5 double bonds per molecule, or
rather, an average of 1.5 double bonds per molecule. We might translate this mathematically as for every one hundred
molecules there needs to be 150 double bonds. This is a convenient number since the table gives us percentages, so if we
just multiply the given percentage by the number of double bonds in that component we will come up with the number of
double bonds which that component contributes to each one hundred molecules of the sample and hence the equivalents of
bromine. So let's try the answer choices starting with choice C--corn oil. From the table, corn oil has 1% of fatty acids with
16 carbons containing one double bond, 25% C-18 with one double bond for a total of 26% with one double bond. It also
has 62% times two double bonds, or 124 double bonds per one hundred molecules. Adding it all up, we have 26 + 124 = 150
which is the correct answer. Trying the others, safflower oil is 13% mono-, 2 times 75% diunsaturated and 3 times 3%
triunsaturated making a total of 172, so this is more than 150 making choice B incorrect. Soybean oil has 43% mono- plus
80 for the di- and 3 times 4 or 12 for the triunsaturated for a total of 135, so choice A is wrong as well. Finally choice D,
castor bean is 93% mono- and 2 times 4% diunsaturated so it totals up to 101 double bonds per one hundred molecules or
1.01 equivalents. Again then, choice C is the correct answer.
Discrete Questions
13.
The correct answer here is choice C. Remember that acid strength is increased by the inductive effect of
electron withdrawing groups on the neighboring carbon atoms. You should know that: (1) the more electronegative the
substituents, the greater the inductive effect; (2) the closer the electronegative group is to the carbonyl carbon, the greater the
effect; (3) the greater number of electronegative substituents there are, the greater the inductive effect. With this in mind let's
look at the answer choices. Choice A, chloroacetic acid, has a pKa of 2.86. Choice B is methoxyacetic acid and it would be
hard to predict how it's acidity compares with that of chloroacetic acid without experimental evidence. However, choice C,
fluoroacetic acid, would definitely be more acidic than both of them, since it is the most electronegative. Choice D, acetic
acid, would be the least acidic since it has no electronegative substituents at all--choice C is the correct response.
14.
For question 14, the correct answer is choice D. This question deals with the susceptibility of different
compounds to nucleophilic attack. All of the answer choices will undergo nucleophilic attack, but the compound that will
undergo attack the easiest is propionyl chloride. Choice A--propanoic acid--has a carbonyl carbon that is slightly susceptible
to nucleophilic attack, since the double bonded oxygen has an electron-withdrawing effect. However, if you compare

propanoic acid and its functional derivative propionyl chloride, you can see that this molecule not only has withdrawing
effects from the oxygen, but also the chloride. You should be aware that out of all of the carboxylic acid derivatives, acyl
halides are the most reactive towards nucleophiles due to the electron withdrawing effects of oxygen and a halide. Choices B
and C are susceptible to nucleophilic attack since they both have electronegative substituents, but not to the same extent as
propionyl chloride. Choice D is the correct answer.
15.
The correct answer to question 15 is choice A. This question asks you about the mechanism of the
Williamson ether synthesis. If you look at the reaction, and even draw out the mechanism, you can see that this is an SN2
reaction, with a deprotonated alcohol acting as the nucleophile on an alkyl halide. Sodium hydroxide reacts with phenol to
form sodium phenoxide. The phenoxide ion, being a good nucleophile, attacks the alkyl halide, displacing the chlorine. The

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MCAT
fact that the alkyl halide is primary should point you towards bimolecular nucleophilic substitution. SN1--choice B--and E1-choice D definitely won't occur because dissociation of the alkyl halide would leave a pretty unstable carbocation. Always
remember that primary alkyl halides only undergo SN2 and E2 reactions. Choice C is wrong because there are no elimination
products--the acidic proton of phenol has been replaced by a methyl group via the nucleophilic attack of the phenoxide ion on
the alkyl halide, so it must be substitution. Again, choice A is the correct response.
16.
The correct answer to question 16 is choice A. The first step in this reaction involves substitution of the
-OH group in benzoic acid by the -Cl group from thionyl chloride which incidentally has the formula SOCl2. The resulting
compound is benzoyl chloride--a highly reactive acyl halide which is also susceptible to nucleophilic substitution. Treating
this molecule with ammonia will result in substitution of the Cl group by an NH2 group, resulting in formation of an amide.
The molecule is named benzamide, which corresponds to choice A. Avoiding the other choices here requires a bit of
knowledge regarding reagents. As benzoyl chloride is so susceptible to nucleophilic substitution, there is no way that a
chlorine will be found in the final product if another nucleophile, such as ammonia, is around. Therefore, you can discard
answer choices C and D. Choice B, p-aminobenzaldehyde, is wrong as well. p-Aminobenzaldehyde is a benzene ring with an

NH2 substituent para- to an aldehyde functionality. This may be confuse you a bit if you're not sure of the reactions of
carboxylic acids and their derivatives. Again, choice A is the correct response.
17.
The correct answer here is choice D. One thing about chemical shifts before I answer the question: the
chemical shift of the aldehyde proton is 9-10 ppm. Remember it! Of all the chemical shifts it is probably the most
important one to have memorized for the MCAT. All right, looking at the spectrum, what do you see? Low and behold, a
peak at 9.8 ppm--an aldehyde peak! The only aldehyde in the answer choices is D, making it the correct response. All of the
other signals are due to alkyl protons and you don't need to worry about these too much as long as you can identify the
aldehyde proton. Since none of the other answer choices has an aldehyde functionality, none would have a proton that
appeared at 9.8ppm, so they can be eliminated from the answer choices.

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