4Hydrocarbons test w solutions

ORGANIC CHEMISTRY TOPICAL:
Hydrocarbons
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
Rf
(261)

105
Ha
(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


Hydrocarbons Test 1
–

Passage I (Questions 1–6)
–

In order to investigate the reactivity of carbanions, a

chemist carried out the following reaction:
LiNH2

CH3CH2CH2Br

–

–
–
–

–

+

Figure 1
Product B
80%

Product A
20%

1 . All of the resonance contributors in Figure 1 are
similar in that they are:
I. not aromatic.
II. aromatic.
III. equally stable.

Reaction 1
Product B was then subjected to further alkylation

according to Reaction 2.
LiNH 2

–

CH 3CH 2CH2 Br

+

Product C
80%

Product D
20%

A.
B.
C.
D.

I only
II only
I and III only
II and III only

2 . Which of the following best represents the reaction
profile for the conversion of cycloheptadiene into
Products A and B?
A.

C.

Reaction 2
The heat of hydrogenation of Product C was found to
be 0.5 kcal greater than that of Product D. Likewise, the
heat of hydrogenation of Product B was about 0.5 kcal
greater than that of Product A. The resonance contributors
of the carbanions formed in Reactions 1 and 2 are as
follows:

B
A

B.

A
B

D.

B
A

A
B

GO ON TO THE NEXT PAGE.

KAPLAN

3


MCAT
3 . Based on the heats of hydrogenation given in the
passage, which of the following statements is correct?
A . Product A is more stable than Product B due to Π
electron delocalization.
B . Product A is more stable than Product D because
it has fewer propyl groups.
C . Product C is more stable than Product D because
it has a greater separation of the double bonds.
D . Product B is more stable than Product A because
the propyl group is situated between the double
bonds.

6 . Hydrogenation of the starting material in Reaction 1
would result in the formation of cycloheptane. This
molecule would have greater ring strain in comparison
to:
I. cyclohexane.
II. cyclopropane.
III. cyclobutane.
A.
B.
C.
D.

I only
I and II only

II and III only
I, II, and III

4 . Which of the following accurately represents the
resonance hybrid of the cycloheptadienyl anion formed
in Reaction 1?
A.

C.
–

–

B.

D.
δ–

δ–

δ–
δ–

δ–

5 . What would be the major product formed if Product B
was hydrogenated and then reacted with bromine in the
presence of UV light?
A.
B.

C.
D.

4,5-Dibromo-3-propylcycloheptene
2,3-Dibromo-1-propylcycloheptane
2-Bromo-l- propylcycloheptane
1-Bromo-l- propylcycloheptane

GO ON TO THE NEXT PAGE.

4

as developed by


Hydrocarbons Test 1
Passage II (Questions 7–1 2 )
As Reaction 1 illustrates, alkanes commonly undergo
bromination in the presence of ultraviolet light.
CH 3
H

CH 3

hv
+

Br2

Br

+ HBr

Reaction 1

The reactivity of the halogens toward free radical
substitution decreases from top to bottom within the
group: fluorine being the most reactive and iodine being
the least reactive. It should be noted that even though
iodine will readily form free radicals in the presence of
ultraviolet light, this radical is unable to react with
alkanes. However, molecular iodine will react with an
alkane radical to form the iodo-substituted compound and
another iodine radical.

The mechanism of this reaction is as follows:
Br 2

hv

CH3
Br • +

(1)

2Br •

•

CH3 + HBr

H

(2)

CH3

•

CH3 + Br2

Br + Br •

Br 2

Br • + Br •

(3)

(4)

2

A . it helps propagate the reaction by generating a
dimer.
B . it acts as a chain-initiating step by generating
alkyl radicals.
C . it acts to terminate the reaction by using up one
of the reactive intermediates.
D . it doubles the rate of reaction because there are

two alkyl radicals present.
8 . Which of the following alkyl radicals would be the
most stable?
A.

CH 3
•

7 . Step 5 plays an integral role in the bromination of
methylcyclopentane in that:

(5)

CH3

• CH
3
H3 C

•

CH3
CH3 + Br •

Br

B.
(6)

•

CH2

Figure 1
Since the bromine radical can react with water to
generate hydrogen bromide and a hydroxyl radical, it is
essential that reactants and glassware be as dry as possible.
(This hydroxyl radical can then go on to form hydrogen
peroxide, which decomposes to form water and oxygen.)

C.

•
CH3

D.
•
CH3

GO ON TO THE NEXT PAGE.

KAPLAN

5


MCAT
9 . If the rate law for Reaction 1 is
Rate =

k

1 2 . After the radical-initiation step by UV light, which of
the following would slow down or stop the
halogenation
reaction
if
added
to
the
methylcyclopentane/bromine reaction vessel?

CH 3

what is the rate-determining step in the mechanism
shown in Figure l?
A.
B.
C.
D.

Step 1
Step 2
Step 3
Cannot be determined

A.
B.
C.
D.

Iodine
Water
Chlorine
Methylcyclopentane

1 0 . From the information given in the passage, which of
the following reactions is unlikely to occur?
A.

CH4 + Cl2

hv

CH3Cl + HCl

hv

B.

+ Br 2

Br + HBr

hv
C.

CH3CH2CH 3 + I2

CH3CHCH 3 + HI
I

CH3
D.

CH3CHCH3 + Br2

Br
hv
CH3CCH3 + HBr
Br

1 1 . If propane was reacted with chlorine in the presence of
UV light, what products would you expect to form and
what would their approximate percentages be?
A.
B.
C.
D.

Propyl chloride (95%), isopropyl chloride (5%)
Propyl chloride (75%), isopropyl chloride (25%)
Propyl chloride (45%), isopropyl chloride (55%)
Propyl chloride (100%)

GO ON TO THE NEXT PAGE.

6

as developed by

Hydrocarbons Test 1
Questions 13 through 17 are NOT
based on a descriptive passage.

1 6 . Which of the following is the most stable form of
trans-1-isopropyl-3-methylcyclohexane?
A.

1 3 . Which of the following is NOT an example of a
conjugated system?
A.
B.
C.
D.

Benzene
1,3-Cyclohexadiene
1,2-Butadiene
Cyclobutadiene

CHCH3
CH 3

CH 3CHCH 3
D.

CH 3

Heptane

2,2,3-Trimethylbutane
3-Methylhexane
2,3-Dimethylpentane

CH 3CHCH 3

CH 3

CH 3CHCH 3

1 7 . Which of the following reactions will NOT occur?
A.

CH 3CH 2CH2 CH3 + HBr

CH3 CH2CHCH3
Br

CH3
B.

CH 3CHCH2CH 3 + 8 O 2

C.

CH 3CHCH2CH 3 + Br 2

1 5 . Products A, B, and C of the following reaction are,
respectively:
CH3

CH3 C

H3 C

B.

1 4 . Which of the compounds listed below would have the
lowest boiling point?
A.
B.
C.
D.

C.
CH 3

CH3

∆

hv

5 CO2 + 6 H2O

CH3
CH3 CCH 2CH 3
Br

CHCH 3

O
D.

HBr

O
H 2O2

+

NBr

Br +

NH

Product A

HBr/H 2O2

O

O

Product B
H2 O

Product C

H+ /∆

A.
CH3
BrCH2C

CH3

CH3
CHCH3 CH3 C

CH2 CH3 CH3 C

CHCH3

Br
B.
CH3

CH3

CH3 C

CH2CH3 CH3 C

Br

Br

CH3

CH3

CH3
CH2 CH3 CH3 CH

CHCH 3
OH

C.

CH3 C

CHCH2Br CH3 CH

CH3
CHCH3 CH3 C
Br

CH2CH3

OH

END OF TEST

D.
CH3
CH3 C
Br

KAPLAN

CH3
CH2CH3 CH3 CH

CH3
CHCH3 CH3 C
Br

CH2CH3

OH

7


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

8

6.

7.
8.
9.
10.

A
C
A
B
C

11.
12.
13.
14.
15.

C
B
C
B
D

16.
17.

D
A

as developed by

Hydrocarbons Test 1
HYDROCARBONS TEST 1 EXPLANATIONS
Passage I (Questions 1–6)
1.

A
An aromatic compound has to fit four criteria: 1) it must have (4n + 2) Π electrons, where n = 0,1,2,3 etc. This is
known as the Hückel rule; 2) it must be cyclic; 3) it must be planar; 4) it must be conjugated. These features impart unusual
stability on the molecule.
The resonance forms in Figure 1 do obey Hückel’s rule, but the system is not fully conjugated so they are not
aromatic. Choice B and D can be eliminated. Statement III is wrong because the products formed in Reaction 1 and 2 are of
unequal proportions, indicating that these resonance structures are of different energies. Therefore, choice C is wrong.
2.

B
If you look at the paragraph in the passage that discusses heats of hydrogenation, you can see that the heat of
hydrogenation of product B is 0.5 kcal greater than that of product A. In other words, product A evolves 0.5 kcal less energy
when it is hydrogenated, so it contains 0.5 kcal less energy to start with, making it more stable than B by 0.5 kcal. Looking at
the structures of A and B, you would expect product A to be more stable because it’s a conjugated system. Therefore, choices C
and D can be rejected because they show product B as the more stable. So, you are now left with choices A and B. The key
difference in these graphs is that in choice A, the activation energy to form product A is lower than that required to form product
B, whereas it is higher than that required to form product B in choice B. To decide between these two, you have to look to the
percentages of product formed. Product B constitutes 80% of the product formed and so as it is the major product it must have
the lower activation energy. Therefore choice A is wrong and choice B is the correct response.
3.

A
ince the heat of hydrogenation of product A is 0.5 kcal less than that of product B (see explanation to #2 above), it is

more stable. Product A is more stable because it is a conjugated system. Conjugated systems are those systems in which
double or triple bonds are separated from each other by a single bond. Because of this arrangement, the Π electrons can
delocalize; imparting added stability to the molecule. Product B has the double bonds separated by two single bonds, not one,
so it is not a conjugated system. This makes choice A the correct response. Choice D can be eliminated because product A is
more stable. Choice B can be eliminated because we are not given any information comparing the stabilities of products A and
D. Finally, choice C is wrong because product D is more stable than product C because this system, like product A, is
conjugated.
4.

B
The resonance hybrid of the cycloheptadienyl anion is the combination of the three resonance contributors shown in the
first line of Figure 1. Choices A and C can be eliminated because they are resonance contributors of the anion, NOT hybrids.
So, the choice is between B and D. Both choices are resonance hybrids in that they have electron delocalization indicated by the
dashed curve. However, the resonance forms in Figure 1 show that there are three centers of negative charge: one at both ends
of the delocalized Π electron system and one in the center. Choice D can be eliminated because it shows only two centers of
negative charge. Since choice B shows the three centers of negative charge, it is the correct response.
5.

D
Hydrogenation of compound B will result in the formation of propylcycloheptane, a substituted cycloalkane. This
molecule is then susceptible to free radical substitution: in the presence of UV light, the bromine molecule breaks up into two
free radicals, one of which attacks the cycloalkane, abstracting a hydrogen. This results in the formation of an alkyl radical and
hydrogen bromide. The tricky part of this question, however, is that you have to decide what the major radical will be. In the
case of propylcycloheptane, there are three types of hydrogens: primary (at the end of the propyl chain), secondary (on the
cycloheptane ring and the propyl chain) and a tertiary one (at the point where the propyl chain is attached to the ring). So which
of these hydrogens is most likely to be abstracted? To answer this you have to think of which alkyl radical will be the most
stable. Radical stability decreases in the order: tertiary, secondary, primary. So, in this case, the tertiary hydrogen will be
removed, forming a tertiary alkyl radical. This alkyl radical can then react with another molecule of bromine to produce the
substituted cycloalkane: 1-bromo-1-propylcycloheptane: choice D. Let’s examine the incorrect choices. Choice C is correct in
that it will be one of the substitution products, but it will not be the major product. This product is formed through the

formation of a secondary alkyl radical, which is not as stable as a tertiary radical, and so you can discard this answer choice.
Choice B is also wrong for this reason--since both of the bromines end up on secondary carbons. Finally, choice A is wrong
because it’s an alkene not an alkane. Both double bonds in product B will add hydrogen, saturating the molecule.

KAPLAN

9


MCAT
6.

A
In cyclic molecules, ring strain arises primarily from angle strain. Angle strain results when bonds cannot assume
ideal angles: the sp3 bond angle is 109.5° and the sp2 bond angle is 120°. (You should also know the other strains that exist
in cyclic molecules: steric strain, also called nonbonded strain, is the strain that results when nonbonded atoms or groups are
too close to each other. Torsional strain is the strain that results from eclipsed interactions.) Since cyclohexane can assume
conformations that do not experience any type of strain, a fact that you should have committed to memory, choice C, which
does not contain roman numeral I, can be eliminated. Cycloheptane, having one more carbon than cyclohexane, has only
moderate ring strain. On the other hand, cyclopropane, having only three carbons, has considerable ring strain: with bond
angles of only 60°, a considerable amount of energy is stored in these bonds. Cycloheptane certainly has less ring strain than
cyclopropane, so choices B and D can be eliminated, making choice A the correct response.
Passage II (Questions 7–12)
7.

C
You should know that free radical chain reactions, which this is an example of, have three steps: an initiation step, a
chain-propagation step, and a termination step. The initiation step is one that starts the chain reaction by generating the first
free radicals. These free radicals are usually formed by homolytically breaking a covalent bond using ultraviolet light or heat.
A chain-propagating step is one that carries the reaction forward. A chain-propagating step always produces a product and more

free radicals. A termination step is one that stops a chain reaction by the consumption of reagents or reactive intermediates. In
Figure 1, step I is the initiation step, step 2, and 3 are chain-propagation steps, and steps 4, 5, and 6 are chain termination
steps. Step 5, the step in question, is acting to terminate the reaction by generating a dimer: Choice C is the correct answer.
8.

A
Free radical stability is the same as carbocation stability: tertiary is more stable than secondary which is more stable
than primary. Because radicals, like carbocations, are electron poor species, alkyl groups stabilize the radical by their electron
donating nature. Choice A is the only one that is a tertiary radical, and is, therefore, the most stable and the correct answer.
Choices C and D are secondary radicals and choice B is a primary radical.
9.

B
Since the rate of reaction depends only on the concentration of methylcyclopentane, you need to look for the step that
has this compound as a reactant. Choice A, step 1, can be eliminated because it only involves bromine. Choice C, step 3, can
also be eliminated because bromine and methylcyclopentane radicals are reactants in this step. This leaves step 2: in this step
methylcyclopentane is converted to methylcyclopentane radical by reaction with bromine radical. Since this step has
methylcyclopentane as a reactant this is the step we’re looking for. Choice B is the correct answer. Choice D is wrong because
if you are given the rate equation and the steps in the reaction, you can identify those steps that are rate-limiting.
10.

C
In the presence of ultraviolet light, iodine will certainly break up into radicals (you are told this in the passage). It is,
however, only with great difficulty that iodine radicals can abstract hydrogens. Iodine is often used as a free-radical inhibitor
because it will readily react with other radicals. Remember: iodine inhibits reactions; it rarely initiates them. Choice C, being
the only one with iodine as a reactant, is the correct response. Choices B and D are wrong because bromine will react with
alkanes in the presence of ultraviolet light. In choice D, a tertiary hydrogen is being substituted by a bromine, just like the
reaction in the passage. In choice B, a secondary hydrogen has been substituted by bromine: a reaction that happens with
relative ease. If bromination of alkanes occurs, chlorination certainly will, according to the reactivities that you are given.
Therefore, choice A is wrong.

11.

C
First, let us assume that each hydrogen on propane will be substituted by chlorine with equal probability. To form
propyl chloride, any one of six hydrogens can be substituted (3 on each methyl group on the ends), but to form isopropyl
chloride, the hydrogen extracted has to be of the two on the middle carbon. So based on pure statistical probability, propyl
chloride will form in a 6:2 or 3:1 ratio compared to isopropyl chloride. But before we jump to choice B, remember that we are
assuming that each hydrogen is substituted with equal probability. This assumption is not correct. When we extract a terminal
hydrogen (which will ultimately lead to propyl chloride), we get a primary radical intermediate. When we extract an internal
hydrogen (which will ultimately yield isopropyl chloride), we get a secondary radical intermediate. Since carbon radicals do not
have a full octet, they are electron deficient. Alkyl groups are electron-donating, and so a secondary radical is more stable than a
primary radical. The reaction going through the secondary radical intermediate will therefore proceed faster. In other words, the
reaction leading to the formation of isopropyl chloride proceeds more readily. This has an effect opposite to that of statistical
considerations, and so we expect that we would get more of isopropyl chloride than the predicted 25%. Choice C is the only

10

as developed by


Hydrocarbons Test 1
reasonable answer. Note that we could NOT have predicted exactly what the distribution would be like, but choice C is the only
one that shows isopropyl chloride forming in more than 25%.
12.

B
As stated in the passage, water can react with a bromine radical to produce a hydroxyl radical. The hydroxyl radical can
then go on to form hydrogen peroxide which then breaks down to form water and oxygen. Therefore, water "mops up" the
bromine radical which propagates the second step of Figure 1—the formation of an alkyl radical. This would slow down or
completely stop the chain reaction, so choice B is correct. Choice A is wrong because iodine can react effectively with an alkyl

radical—again you are told this in the passage. Iodine can, and does, compete with bromine in the propagation step, resulting in
a mixture of products: 1-iodo-1-methylcyclopentane and 1-bromo-1-methylcyclopentane. This wouldn’t affect the rate of the
reaction, because as discussed in #9, the slowest step is the formation of the alkyl radical, not the reaction of this radical with
halogen. The same would apply to chlorine—choice C. This would compete with bromine, but in this case you would get 1chloro-1-methylcyclopentane and again, as this is not the rate-limiting step, it will not affect the rate of the reaction, making
choice C incorrect. From question 9, you should know that the rate of the reaction is equal to the rate constant multiplied by the
concentration of alkane. Therefore, if the concentration of methylcyclopentane is increased, so will the rate of the reaction. This
makes choice D incorrect.
Discrete Questions
13.

C
Remember that a conjugated system contains two or more double or triple bonds separated by a single bond. 1,2Butadiene does not fit this definition because the double bonds are adjacent to each other, not separated by a single bond. In this
case, the double bonds are said to be cumulated. Benzene, choice A, is an example of a conjugated system: there are three
double bonds separated from each other by a single bond. Choice B, 1,3-cyclohexadiene, is also a conjugated system: there are
double bonds between carbons 1 and 2 and carbons 3 and 4; therefore, this answer choice can also be eliminated. Finally,
cyclobutadiene is also an example of a conjugated system. This molecule has two double bonds separated by two single bonds,
so this answer is wrong as well.
14.

B
All of the answer choices are related in that they are structural isomers: they have exactly the same molecular formula,
C7H16, but different atomic connections. So, for this question, atomic connectivity, not molecular weight, is what you have to
focus in on. You should know that branching in alkanes lowers the boiling point because branched molecules cannot interact as
effectively with each other as unbranched molecules. You can think of branched molecules as balls and unbranched molecules as
sheets. Certainly, the surface area of interaction is greater for the sheets than the balls. So, the molecule that has the lowest
boiling point is the one that is the most highly branched. That is choice B, 2,2,3-trimethylbutane. All of the other answer
choices are not as highly branched. Choice D is the next highly branched, followed by choice C and choice A, the straight-chain
hydrocarbon heptane is not branched at all.
15.

D
The first reaction tests your knowledge of Markovnikov’s rule: the hydrogen will add to the least substituted carbon,
resulting in the most stable carbocation. The bromine then adds to this carbocation. In this case the hydrogen adds to the
secondary carbon and the bromine to the tertiary carbon. Choices A and C can be eliminated. The second reaction involves free
radical intermediates. You should be able to recognize this by the presence of hydrogen peroxide. In this case, the result is antiMarkovnikov addition: the hydrogen adds to the most substituted carbon and the bromine to the least. Choice D is the correct
response. You didn’t need to know the third reaction to answer this question, but let’s talk about it anyway. The reaction to
form product C also follows Markovninov's rule: the addition of water yields 2-methyl-2-propanol, which corresponds to that
drawn in choice D.
16.

D
In substituted cyclohexanes, the larger groups tend to take equatorial positions. As you have a methyl and an
isopropyl group to choose from, it is pretty evident that the latter will take the equatorial position. This rules out choices A
and B since the isopropyl groups here are in axial positions. In addition, the molecule in choice B is a cis isomer not a trans
isomer. Choice C is wrong because this molecule is a 1,4 disubstituted cyclohexane, not a 1,3 disubstituted cyclohexane as the
question states. This leaves choice D, which is correct because it is a 1,3 disubstituted compound, and it is a trans isomer
having the larger isopropyl group in an equatorial position.
17.

A

KAPLAN

11


MCAT
Alkanes will not be substituted by a bromide from hydrogen bromide. Since they are highly unreactive molecules,
alkanes will only undergo substitution when extreme conditions are employed--like UV light for instance. This initiates highly
reactive free radicals which will now react. This free radical substitution reaction is shown in choice C, therefore this answer

choice can be eliminated. Anyway, back to choice A. No such conditions are present in the reaction in choice A, so this will
not occur, making it the correct answer. At high temperatures, alkanes do undergo combustion to form carbon dioxide and water
so choice B can certainly be eliminated. Finally, N-bromosuccinimide (NBS) will place a bromine atom in the allylic position
of an alkene, assisted by the free radical initiator hydrogen peroxide, so this is a valid reaction too. Therefore, choice D can be
eliminated.

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