6Molecular structure of organic compounds test w solutions

ORGANIC CHEMISTRY TOPICAL:
Molecular Structure of Organic
Compounds
Test 1
Time: 23 Minutes*
Number of Questions: 18

* 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


Molecular Structure of Organic Compounds Test 1
Passage I (Questions 1–7)
Until recently, it had been common practice for
pharmaceutical companies to manufacture chiral drugs as
racemates, not as single enantiomers.
Usually, only one enantiomer of a given compound

possesses therapeutic value, while the other may have no
beneficial pharmacological properties and may even induce
serious physiological side effects.
A typical example of this problem was the use of the
drug Thalidomide in 1961. Administered to reduce nausea
and vomiting during the early stages of pregnancy, the
desired physiological activity lies with the R-isomer. The
S-isomer of Thalidomide is a teratogen (an agent that
produces physical defects in developing embryos). As a
result, administration of this drug as the racemate caused
congenital malformations in thousands of infants. The
structure of Thalidomide is shown below.
O
C
N

L + [Ni]
+ HCN
MeO
6-Methoxy-2-vinylnapthalene
CH3
CN
H
MeO
CH3
CO2 H
H
MeO
Naproxen

C
O

N
H

O

O

(L + [Ni] = Metal complex catalyst)

Figure 1
Today, advances in the synthesis of specific
enantiomers should severely limit the number of racemic
drugs produced. Traditionally, the enantiomer of a chiral
drug could be synthesized via resolution (conversion of the
racemic mixture into diastereomers, separation, and then
reformation of the enantiomers). However, this process is
wasteful because of the numerous steps involved in the
procedure. The emergence of a new technique called
enantioselective catalysis may soon be used to avoid this
inefficiency.

Reaction 1
Another example of enantioselective catalysis is that
of meso breaking. For example, azidotrimethylsilane has
been shown to react with cyclohexene oxide in the presence
of the enantioselective catalyst titanium isopropoxide:
chiral

O + R3 SiN 3

OSiR 3

catalyst
N3

Reaction 2

The catalysts employed in enantioselective catalysis
are usually based on optically active metal complexes and
are highly effective in the production of one enantiomer
over another. The success of this technique is attributed to
the production of thousands of chiral products from one
molecule of catalyst. A good example is the use of a
nickel-based catalyst to synthesize the S-isomer of the
chiral drug Naproxen (Reaction 1).

GO ON TO THE NEXT PAGE.

KAPLAN

3


MCAT
1 . How many chiral centers are present
Thalidomide molecule?
A.
B.

C.
D.

in

the

0
l
2
4

2 . What is the most likely reaction mechanism between
HCN and 6-methoxy-2-vinylnapthalene?
A.
B.
C.
D.

Electrophilic addition
Bimolecular nucleophilic substitution
Bimolecular elimination
Free radical addition

5 . Which of the following statements most accurately
describes the properties of Thalidomide?
A . It is a mixture of enantiomers that have the same
chemical and physical properties, with the
exception of the direction of optical rotation, but
different physiological properties.

B . It is a mixture of diastereomers that have the same
physical and chemical properties, but different
physiological properties.
C . It is a mixture of enantiomers that have different
physiological, chemical, and physical properties.
D . It is a mixture of diastereomers that have different
physical, chemical, and physiological properties.

6 . Which of the following is the R-isomer of Naproxen?
A.

CH3
CO2H

C

C.

CH3

H

H

3 . Reaction 2 proceeds through an S N 2 mechanism.
Consequently, the attacking nucleophile is:
A.
B.
C.
D.

the alkyl group.
the azide ion.
silicon.
the chiral catalyst.

4 . The double bond of the vinyl group in 6-methoxy-2vinylnapthalene contains which of the following
orbitals?
I.
II.
III.
IV.
A.
B.
C.
D.

sp orbitals
sp2 orbitals
sp3 orbitals
p orbitals

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

MeO

B.

MeO

D.

CH3
CO2H

C

CH3
CO2H

C

H
MeO

CO2H

C

H
MeO

7 . The term “meso breaking” can be applied to Reaction
2 because:
A . the chirality of the catalyst is destroyed.
B . chiral centers in cyclohexene oxide are created
when the product is formed.

C . symmetry is retained when cyclohexene oxide is
converted to the product.
D . the symmetry of cyclohexene oxide is broken to
form an optically active compound.

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4

as developed by


Molecular Structure of Organic Compounds Test 1
Passage II (Questions 8–13)
One of the major problems in using solar energy as a
power source is the conversion of the raw material,
sunlight, into a usable form of fuel. The transformation of
energy contained in sunlight to chemical energy via
photochemically
induced
isomerization
(photoisomerization) is one method that has been suggested in
order to alleviate this problem. In this process, a simple
organic molecule is converted to a product that has far more
energy than the starting material. This excess energy may
be stored in the product as angle, torsional or non-bonded
strain. The reverse reaction, formation of the starting
material, provides a means in which energy can be released
and used in a controlled manner.

The alkane equivalent of norbornadiene, norbornane,
is shown in Figure 1.
7
4
1
2
6
Norbornane

Figure 1
8 . The ring strain that arises in small cycloalkanes, such
as cyclopropane or cyclobutane, is mainly attributed
to:

A good illustration of the photoisomerization process
is the conversion of norbornadiene to quadricyclane
(Reaction 1). Quadricyclane has 62 kcal/mol of excess
strain energy relative to norbornadiene and is therefore
thermally unfavorable; the reverse reaction, quadricyclane to
norbornadiene, is highly favored in the presence of metal
catalysts.
Since it is readily synthesized in high yield,
norbornadiene is a convenient starting material in the
laboratory. However, further research is required before it
can be mass-produced for large scale adaptation.

5

3

I. compression of the bond angles to less than
109.5°.
II. interaction between eclipsed hydrogens on
adjacent carbons.
III. nonbonding interactions between hydrogens
on non-adjacent carbons.
A.
B.
C.
D.

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

9 . Which of the following will most readily undergo an
intramolecular [2+2] ring closure?
hv
A.

C.
CH 2

Ph

Norbornadiene
Strain energy =
33 kcal/mol

Quadricylane
Strain energy =
95 kcal/mol

C
CH 2
B.

Reaction 1

C
Ph

D.
Ph

Ph

Norbornadiene converts to quadricyclane via a
photochemically allowed [2+2] ring closure, analogous to
the conversion of 1,3-butadiene to cyclobutene (Reaction
2).
hv

1,3-Butadiene

Cyclobutene

Reaction 2
GO ON TO THE NEXT PAGE.

KAPLAN

5


MCAT
1 0 . Catalytic hydrogenation of norbornadiene releases
approximately 57 kcal of heat per mole of
norbornadiene. The product, norbornane, is:
A . more stable, due to the decrease in angle strain
accompanying saturation.
B . more stable, due to the increased number of
hydrogens in the product.
C . less stable, due to the change in hybridization of
the double bonded carbons.
D . less stable, due to steric effects and an increase in
angle strain.

1 3 . A Diels-Alder process ([4+2] cycloaddition) takes place
according to the following scheme:

A Diels-Alder reaction between which of the following
would result in the formation of norbornadiene?
A.
B.
C.
D.

Propene and 1,3-butadiene

Acetylene and 1,3-cyclopentadiene
Ethylene and propene
Ethylene and 1,3-cyclopentadiene

1 1 . For substituted norbornanes, it has been found that
those substituted in the 7 position undergo SN2
reactions with the appropriate nucleophile while those
substituted in the 1 position do not. This occurs
because:
A . it is difficult to form a carbocation intermediate at
carbon 1 in the initial step of the reaction.
B . it is easier to form a carbocation at the 7 position,
making substitution kinetically favorable.
C . back-side attack at the 1 position by the
nucleophile is difficult because it is sterically
hindered.
D . the carbon at position 7 is more electrophilic than
the carbon at position 1.

1 2 . Which of the following is true of norbornane?
A . Norbornane is less stable than cyclopentane but
more stable than cyclohexane.
B . Norbornane is more stable than both cyclopentane
and cyclohexane.
C . Norbornane is less stable than chair cyclohexane
but more stable than boat cyclohexane.
D . Norbornane is less stable than both cyclopentane
and cyclohexane.

GO ON TO THE NEXT PAGE.

6

as developed by


Molecular Structure of Organic Compounds Test 1
Questions 14 through 18 are NOT
based on a descriptive passage
1 4 . Which of the following compounds will exhibit the
greatest dipole moment?
A.
B.
C.
D.

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

1 7 . Which of the compounds listed below is linear?
A.
B.
C.
D.

Carbon tetrachloride
Propyne
Acetylene

1,3-Hexadiene

1 8 . What is the order of increasing carbon-carbon bond
length in the molecules listed below?
1 5 . How many structural isomers of C3H6Br2 are capable
of exhibiting optical activity?
A.
B.
C.
D.

1
2
3
4

I. Acetylene
II. Benzene
III. Ethylene
A.
B.
C.
D.

I, II, III
I, III, II
II, III, I
II, I, III

1 6 . C=C, C=O, N=N, and C=N bonds are quite common

in organic compounds. However, C=S, C=P, C=Si,
and other similar bonds are not often found. The most
probable explanation for this observation is that:
A . carbon does not combine with elements found
below the second row of the periodic table.
B . sulfur, phosphorus, and silicon do not form pi
bonds due to the lack of occupied p orbitals in
their ground state electron configurations.
C . sulfur, phosphorus, and silicon are incapable of
orbital hybridization.
D . the comparative sizes of the 2p and 3p atomic
orbitals make effective overlap between them less
likely than between two 2p orbitals.

END OF TEST

KAPLAN

7


MCAT

ANSWER KEY:
1.
B
2.
A
3.
B

4.
D
5.
A

8

6.
7.
8.
9.
10.

C
D
C
D
A

11.
12.
13.
14.
15.

C
D
B
A
A

16.
17.
18.

D
C
B

as developed by


Molecular Structure of Organic Compounds Test 1
EXPLANATIONS
Passage I
1.

B
The first three paragraphs talk about the problems associated with manufacturing chiral drugs as their racemates. It then
goes on to give Thalidomide as a classic example of this problem.
O
*

C
N
C

O

N

H

O

O

A racemic mixture is one that contains equal amounts of two enantiomers, chiral molecules that have opposite
configurations from each other at every stereocenter and so are non-superimposable mirror images. In organic compounds, a
stereocenter or a chiral center is usually defined as a carbon that is attached to four different groups. Thalidomide definitely has at
least one chiral center since the text talks about its administration as a racemate and the R– and S– isomers. Therefore, choice A
can be eliminated. A chiral center is an atom attached to four different functional groups. Starting at the benzene ring, you can
see that there are no chiral centers, as all of the carbons are attached to only three substituents. The same thing applies to the
carbonyl groups. Moving on to the carbon in the nitrogen-containing ring, you can see that this is a chiral center. It is attached
to a hydrogen (not shown explicitly), a nitrogen, a CH 2 group, and a carbonyl group. Moving clockwise around the ring, the
next group is the carbonyl functionality. Again, this center is achiral since it is only attached to three substituents; the same
thing applies to the carbonyl group opposite. Although the nitrogen in the ring is attached to four different groups (the lone pair
of electrons not shown is considered a group), it undergoes rapid inversion of configuration that causes it to not be a chiral
center. Moving along to the last two carbons in the ring, you can see that neither of these are chiral since they are each attached
to two hydrogens atoms. So, there is only one chiral center in the Thalidomide molecule: the carbon in the nitrogen containing
ring which we discussed earlier.
2.

A
In Reaction 1, you should be able to see that the first step involves the electrophilic addition of hydrogen cyanide to
the double bond in the vinyl group. Let’s look at the mechanism of this reaction in more detail. Since the double bond is
electron rich, it is likely to be attacked by positively polarized molecules or electrophiles. The hydrogen in HCN acts as an
electrophile and, therefore, adds to the least substituted carbon to form the most stable carbocation intermediate. This
intermediate is formed as electrons are given up from the double bond to form a new bond to the hydrogen, leaving behind a
positive charge. The formation of a secondary carbocation is favored over a primary carbocation, and so the intermediate
–CH+CH3 is formed. The cyanide ion (CN–) then adds to the carbon bearing the positive charge to form the nitrile product

shown in Reaction 1. Obviously from the mechanism I have just described, there is addition across the double bond. This
addition is initially electrophilic and so choice B, which describes the reaction as being nucleophilic substitution, is incorrect.
Elimination, as stated in choice C, is also incorrect. Elimination usually involves the removal of fragments to form a multiple
bond; the process that occurs in the first step of Reaction 1 is just the opposite of this. This leaves choices A and D.
Markovnikov’s rule states that in the addition of a HX to an alkene, the hydrogen will add to the least substituted carbon in
order to form the most stable carbocation intermediate. This rule is followed in Reaction 1, since the hydrogen adds to the –CH2
in the vinyl group, not the –CH group so choice A is the correct response. Anti–Markovnikov addition can be obeyed when
peroxides are added to the reaction mixture. This reaction occurs by a radical mechanism and the hydrogen adds to the most
substituted carbon. This is not the case in this reaction, so choice D is incorrect.
3.

B
Epoxides, such as the one shown in Reaction 2, are highly strained and, because of the electron withdrawing nature of
the oxygen, a nucleophile can attack one of the carbons attached to it resulting in a ring opening reaction. Simultaneously, the
nucleophile attacks one of the epoxide carbons, and since carbon cannot form 5 bonds, the bond between it and the oxygen is
broken. The electrons from this bond are taken on by the oxygen forming a negatively charged anion.
In Reaction 2, the R 3SiN 3 molecule forms R3Si+ and N 3 – . The latter is called an azide ion and as it is negatively
charged; it will attack the electrophilic carbon. It does so 180° to the oxygen, which takes on electron density from the
carbon–oxygen bond. The positively polarized silicon-alkyl group can then add to the negatively charged oxygen to form the
product shown. As a result, the attacking nucleophile is the azide ion and choice B is the correct response.

KAPLAN

9


MCAT
O–
+ SiR3+

O

N3

OSiR3

N3

N3 –

The chiral catalyst is involved in the reaction, but not as a nucleophile. A catalyst serves to speed up the reaction and
at the end of the reaction, it remains unchanged. Therefore, it would not behave as a nucleophile and become incorporated into
the product; choice D is wrong.
Choices A and C are wrong since these molecules constitute the electrophilic portion of the R 3SiN 3 molecule. Since
they are positively polarized, there is no way they could be nucleophilic.
4.

D
Carbon–carbon double bonds are made up of both sp 2 hybridized orbitals and p orbitals. The electron configuration of
carbon is 1s22s22p 2. The 2s orbital and two 2p orbitals hybridize to form three sp 2 hybrid orbitals. This leaves one free p
orbital which can overlap with an adjacent orbital to form a pi bond. sp2 hybridized orbitals have a geometry of 120°, and these
constitute the carbon-carbon and carbon-hydrogen sigma bonds in the molecule. Therefore, the double bond in the vinyl group is
formed by the overlap of two sp2 orbitals and two p orbitals.
5.

A
Thalidomide was administered as a racemate which is an equal mixture of two enantiomers. You should know that
enantiomers have identical chemical and physical properties with one exception: they rotate plane polarized light in opposite
directions. However, they do behave differently in chiral environments, and so they exhibit different behavior in the human
body. One enantiomer reduces nausea and vomiting while the other is a teratogen which is an agent that causes physical defects

in the developing embryo. Therefore, Thalidomide is a mixture of enantiomers that have the same chemical and physical
properties but completely different physiological effects; choice A is the correct answer.
Choices B and D are incorrect because the isomers in Thalidomide are enantiomers, not diastereomers. If they were
diastereomers, then they would possess different chemical and physical properties and would form a mixture of two different
compounds, not a racemic mixture as described in the passage. Choice C is incorrect because enantiomers have the same
chemical and all of the same physical properties except in the direction that plane polarized light is rotated.
6.

C
First, let’s look at Naproxen drawn in Reaction 1, which you are told is the S-isomer. Priorities have to be assigned to
each substituent directly attached to the stereocenter. This is done by atomic number, so the lowest priority goes to the
hydrogen. The other three atoms attached to the stereocenter are carbons, so the atomic weights of the groups attached to these
carbons now have to be considered. The substituent that will have the next to last priority is the methyl group, since hydrogens
are attached to the methyl carbon. In the remaining two substituents, the carbons are attached to two other carbons (in the case
of the substituted naphthalene group) and two oxygens (in the case of the carboxyl group). Another rule you need to know, is
that in double bonds, the atoms have to be duplicated; a carbonyl group would be classed as a carbon bonded to two oxygen
atoms. When this is done, the carboxyl group turns out to be of highest priority. So to summarize, the order of increasing
priority is hydrogen, methyl, naphthalene and then carboxyl. In order to assign a configuration, the lowest priority substituent
has to be rotated to the back and then arrows are drawn from the highest priority substituent (numbered 1) to the lowest priority
substituent (numbered 3). If this is done for Naproxen in Reaction 1, you should see that the arrows are in an anti–clockwise
direction so the molecule is an S-isomer.
In order to qualify as an R–isomer, the answer choice must have an opposite configuration at all chiral centers, making
it a non-superimposable mirror image of the S–isomer. Choice C is correct because it has the opposite configuration of the Sisomer. The hydrogen group is already oriented toward the back, and so in drawing arrows from the carboxyl group through to
the methyl group, you can see that the direction is clockwise, or R.
Choices A and D are wrong because they show the chiral carbon and the methoxy group respectively sticking out from
the naphthalene ring. These choices are wrong because both groups would be in the plane of the ring. In addition, the
orientation around the chiral carbon is incorrect in both responses. Choice B is incorrect because this is the S-isomer of
Naproxen; the configuration around the chiral center is exactly the same as that shown in Reaction 1.
7.

D
A meso compound is a molecule that contains chiral centers, so you would expect it to be optically active . However,
meso compounds also contain an internal plane of symmetry, so the molecule is optically inactive. A good example of this is

10

as developed by


Molecular Structure of Organic Compounds Test 1
cyclohexene oxide, shown in Reaction 2. The two carbons attached to the oxygen are in fact chiral centers; each one is attached
to four different groups; oxygen, hydrogen, CH 2, and CH. However, intersecting the bond perpendicular to these two chiral
carbons is a plane of symmetry. Therefore, this molecule is superimposable on its mirror image and is optically inactive. The
addition of azidotrimethylsilane, R 3SiN 3 to cyclohexene oxide results in the loss of this plane of symmetry. The two chiral
carbon centers still remain, but the molecule is now optically active since it is nonsuperimposable on its mirror image.
Therefore, the symmetry of cyclohexene oxide is broken and an optically active compound is formed: choice D is the correct
answer. Choice A is incorrect since the chirality of the catalyst is retained in the reaction. Remember: although a catalyst may
take part in a reaction, it comes out the same at the end of the reaction. Choice B is incorrect because the chiral centers are not
generated in the reaction: they have always been present. Choice C is wrong because the plane of symmetry is lost in the
conversion of the meso compound to the optically active product.
Passage II
8.

C
You should be aware that in saturated ring systems (ring systems consisting of all single bonds), the carbons are sp3
hybridized. sp 3 hybridized carbons have preferred bond angles of 109.5°. However, in cyclopropane, the bond angles are
compressed to 60°, whereas in cyclobutane, the bond angles are about 90° (cyclobutane takes on a ‘folded’ or ‘bent’
conformation). In both cases, the sp3 orbitals cannot assume the ideal bond angle of 109.5°, and so there is a great deal of angle
strain in the molecule. This makes statement I correct and answer choice B wrong.
Because of the rigidity of the cyclopropane ring, all six hydrogens are eclipsed. This results in a great deal of repulsive

interaction, so the molecule suffers from torsional strain. The diagram below illustrates one such pair of eclipsed hydrogens:
eclipsing
interactions

H

H

H
C
H

H

C

C
H

Cyclobutane also suffers from torsional strain, but to a lesser degree. If the bond angles were 90°, the cyclobutane
molecule would be planar and all eight hydrogens would be eclipsed. To alleviate this, the molecule folds and results is a bond
angle of 88°. At the expense of slightly more angle strain, the torsional strain is reduced. Both molecules possess eclipsed
hydrogens though and suffer from torsional strain; statement II is also correct and choice A is out.
Non–bonding or steric interactions occur when atoms bonded to non-adjacent atoms compete for the same position in
space. The classic example is the repulsion between substituents on opposite ends of cyclohexane in the unfavorable boat
conformation. There are no such interactions on cyclopropane or cyclobutane, so statement III is false.
9.

D
The passage gives you an example of a [2+2] ring closure by the conversion of 1,3–butadiene to cyclobutene. It is

pretty evident from this reaction that ring closure involves 4
electrons in relatively close proximity to each other. Ring
closure involves the movement of these electrons; in the case of a [2+2] reaction, 2 electrons are used up in forming a new
sigma bond, while the other two simply move between carbons. Right away, you can rule out choices A and B. Benzene as
shown in choice A, contains 6 electrons and is very stable. A [2+2] ring closure in this molecule is highly unlikely since it
will involve breaking the aromaticity of the ring. Choice B, norbornene, won’t undergo a [2+2] ring closure since it only
contains 2 electrons.
This leaves choices C and D. You may think that ring closure in these molecules is equally likely since both contain 4
electrons. However, in choice C, the double bonds are not in close proximity to each other. In order for cyclization to occur,
the double bonds must be on the same side, in other words, the molecule has to be in the cis configuration. To achieve this, the
phenyl groups would end up on the same side of the molecule. This would result in a great deal of steric interaction, and, as a
result, the trans isomer is favored and ring closure is extremely difficult. On the other hand, the double bonds in choice D are in
closer proximity to each other, making cyclization somewhat easier.
10.

A
Catalytic hydrogenation of norbornadiene involves the addition of two molecules of hydrogen to the double bonds, the
result being the formation of the saturated ring system, norbornane. You should be aware that double bonded carbons are sp2
hybridized so a p orbital is free to overlap, forming a double bond. The bond angle of an sp2 hybridized carbon is 120°. Looking
at norbornadiene, you can see that the sp2 bond angles are much less than 120°, and as a result, the molecule suffers from angle
strain. When catalytic hydrogenation occurs, the double bonded carbons become sp 3 hybridized, and the preferred bond angle
lowers to 109.5°. The bond angles in norbornane are much closer to this bond angle so saturation of norbornadiene is

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11


MCAT
accompanied by a decrease in angle strain. In other words, catalytic hydrogenation results in the formation of a more stable

molecule. This is best described by choice A. Further evidence that the product is more stable is provided from the question
stem: it states that 57 kilocalories of heat per mole are released upon hydrogenation; in other words, 1 mole of product is more
stable than 1 mole of reactant by 57 kilocalories.
Choice B is incorrect because the increase in the number of hydrogens is not the cause of increased stability of the
product. Choice C is correct in that the change in stability is due to the change in hybridization. However, this answer choice
also states that the product is less stable. If the product was less stable than the reactant, 57 kcal would be taken in upon
formation of norbornane. For this reason, choice D is also incorrect.
11.

C
You really need to be familiar with SN1 and SN2 reaction mechanisms in order to answer this question. The rate of an
SN1 reaction is dependent only on the concentration of the substrate. The rate-limiting step is the formation of a carbocation by
the loss of a good leaving group. The nucleophile then adds to the carbon bearing this positive charge. As a carbocation is
formed, the order of reactivity of an alkyl halide, for example, will be tertiary, then secondary, then primary (in fact, primary
alkyl halides do not react via an SN1 mechanism). On the other hand, SN2 reactions depend on the concentration of both the
substrate and the incoming nucleophile. In a concerted mechanism, the incoming nucleophile attacks the electrophilic carbon
180° to the leaving group. No carbocation intermediate is formed, and any factor that effects the incoming nucleophile and the
substrate will affect the rate of reaction. One of these factors is steric hindrance: if the electrophilic carbon is highly substituted,
for instance, if it is a tertiary carbon, then it will be difficult for the nucleophile to get to this carbon. On the other hand, if the
carbon is primary, nucleophilic attack will be easier since there is less steric hindrance. For this reason, reactivity of alkyl
halides toward SN2 decreases in the order primary, then secondary, then tertiary (tertiary won’t react).
Looking at norbornane, you can see that substitution with a halide at the 1 position would result in the formation of a
tertiary alkyl halide, whereas substitution at the 7 position would result in the formation of a secondary alkyl halide. Tertiary
alkyl halides won’t react by an SN2 mechanism due to steric hindrance, so at position 1, the incoming nucleophile would be
sterically hindered and would not be able to attack 180° to the leaving group. SN2 would be significantly easier at position 7
since the incoming nucleophile would be less sterically hindered. This makes choice C the correct response.
Now for the wrong answers. It would be very difficult to form a carbocation at position 1 as stated in choice A.
Remember: carbocations prefer a planar geometry, with a bond angle of 120°. There is no way this geometry could be achieved
at position 1. However, this is relevant only in an SN1 reaction, which is not what we have here. Therefore, choice A is
incorrect. Choice B also talks about carbocation formation and so is incorrect. Finally, choice D is wrong because the carbons at

positions 1 and 7 would be more or less equally electrophilic if substituted by the same group.
12.

D
Breaking norbornane down into its constituent rings, you can see that it is actually composed of a six-membered ring
(numbers 1 through 6) and two cyclopentane rings (the first ring is made up of carbons 4, 5, 6, 1 and 7, and the second ring is
made up of carbons 1, 2, 3, 4 and 7). The six-membered ring is not in its preferred chair conformation and the bond angles are
highly compressed. Therefore, the molecule will be less stable than cyclohexane, eliminating answer choices A and B. Both the
boat and the chair conformations of cyclohexane are more stable than the norbornane molecule, so choice C is wrong as well.
Cyclopentane is almost as stable as cyclohexane. Just like cyclobutane, cyclopentane assumes as slightly bent
conformation to alleviate torsional strain at the expense of slightly more angle strain. The cyclopentane rings in norbornane are
far more strained than isolated cyclopentane rings, so the molecule is more unstable than both cyclopentane and cyclohexane
making choice D the correct response.
13.

B
The reaction given here involves three pairs of
electrons. One pair is pushed to a new position in the product
molecule, while the other two pairs are involved in the formation of new sigma bonds that link the two reactants together.
Choice C is incorrect since there are only a total of 2 pairs of electrons in the reactants and so a Diels-Alder reaction
cannot occur. Among the other choices, only one, choice B, leads to the correct product:

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Molecular Structure of Organic Compounds Test 1
A.

1, 3-butadiene

propene

1, 3-cyclopentadiene

acetylene

1, 3-cyclopentadiene

ethylene

B.

D.

In choice B, 1,3–cyclopentadiene is the diene in the reaction and acetylene is the dienophile. Don’t be put off by the
fact that acetylene contains a triple bond; it still donates only two electrons in the cyclization reaction. The overall process
results in cyclization to form norbornadiene. Two double bonds remain; one from the original diene and one from acetylene (as a
triple bond was involved in the reaction, not a double bond). Cyclization involves the formation of a six–membered ring and the
formation of a bridgehead carbon, originating from carbon 5 of 1,3–cyclopentadiene.
Choice D is different from choice B in that ethylene is used instead of acetylene. This molecule contains a double bond,
not a triple bond, so when this molecule acts as a dienophile, it will give up its electrons to leave behind a single bond, not a
double bond. Therefore, the product will have one double bond less and so norbornene is produced instead of norbornadiene.
Choice A will certainly undergo a [4+2] cycloaddition reaction, but the product will be 4–methylcyclohexene, not
norbornadiene. Therefore, choice A is incorrect.
Independent Questions
14.

A

The dipole moment of a molecule is determined by magnitudes of the individual bond dipoles and the spatial
arrangement of these individual bonds. As for magnitudes, bond dipoles are greatest when there is a large difference in the
electronegativities of the atoms involved in the bond. Therefore, carbon–carbon bonds have little or no dipole moment, while
carbon–halogen bonds have relatively large dipole moments.
Looking at the answer choices starting with A, (Z)–1,2–dichloro–1,2–diphenylethene has two phenyl rings attached on
one side of the double bond and two chlorine groups attached on the other side of the double bond. Remember, the
(Z)–designation is assigned when the two highest priority substituents attached to a double bond are on the same side (priority is
assigned according to molecular weight, so the two chlorines take highest priority). The electronegative chlorines pull electron
density toward themselves, setting up a large net dipole moment in that direction. This makes choice A the correct answer. The
(E)-isomer (choice B) differs in that the two highest priority substituents lie across the double bond. As the chlorines pull
electron density toward themselves, both dipoles cancel each other out to give a net dipole moment of zero. Therefore, choice B
can be eliminated.
Choice C contains the same substituents, but there is a single bond in this molecule, not a double bond. The key
difference in this molecule is that the single bond rotates, so any effective dipole moment that is set up will be canceled out by
an opposite dipole as the molecule rotates. So, just like in choice B, there will be no net dipole moment in this molecule.
Choice D is also incorrect for the same reason as choice C. The carbon–fluorine bonds are highly polar, but rotation about the
carbon–carbon single bond will destroy any net dipole that is set up.
15.

A
Structural isomers have the same molecular formula, but a different atomic connectivity. With a molecular formula of
C3H6Br2, the carbon skeleton obviously consists of a propane chain. Four structural isomers can exist; both bromines can
attach to carbon 1 of the propane chain forming 1,1–dibromopropane, both bromines can attach to carbon 2 forming
2,2–dibromopropane, a bromine can attach to carbons 1 and 2 forming 1,2–dibromopropane or finally, a bromine can attach to
carbons 1 and 3 forming 1,3–dibromopropane. All of these are structural isomers since they have the same molecular formula,

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13

MCAT
but different atomic connectivity. So, which of these will exhibit optical activity? Recall that in order to possess optical
activity, a molecule must be chiral. Most chiral molecules are identified as possessing atoms attached to four different
substituents. Of the four isomers, only the 1,2–substituted propane has a chiral center. The second carbon is attached to a
bromine, a hydrogen, a CH2Br group and a CH 3 group. Therefore, this isomer may exhibit optical activity. All of the other
isomers don’t exhibit optical activity because they don’t possess any chiral centers, so A is the correct response.

H

H

H

H

C

*
C

C

Br

Br

H

H

16.

D
In the case of carbon–carbon double bonds, they are made up by the overlap of sp 2 orbitals and p orbitals on adjacent
carbon atoms. The sp2 orbitals overlap head on to form a sigma bond, whereas the p orbitals overlap to form a bond. Sigma
bond lengths and strengths are largely determined by the size and shape of the atomic orbitals involved in formation of the bond
and the ability of these orbitals to overlap effectively. Sigma bonds are stronger than pi bonds because head to head overlap is
more efficient than sideways overlap. Sigma bonds formed from two 2s orbitals are shorter than those formed from two 2p
orbitals or two 3s orbitals.
The difference between the first four bonds named in the question stem and final three, is the position of their
respective atoms in the periodic table. Carbon, oxygen and nitrogen can be found in the second period, while sulfur, phosphorus
and silicon are found in the third period. Therefore, sulfur, phosphorus and silicon would use 3p orbitals in the formation of pi
bonds, not 2p orbitals in the case of carbon, nitrogen and oxygen. The key problem, however, is that 3p orbitals are much
larger than 2p orbitals and so overlap of the 2p orbital of carbon and the 3p orbital of silicon, phosphorus or sulfur is unlikely
to occur. This makes choice D the correct response.
Choice A is incorrect since carbon does combine with elements below the second row of the periodic table. For
example, carbon can form bonds with the elements in the halogen group. Choice B is also wrong. In their ground state electron
configurations, silicon, sulfur and phosphorus all have partially occupied p orbitals which can form bonds. Finally, choice C
is incorrect since sulfur, phosphorus and silicon can undergo hybridization. These elements can not only mix s and p orbitals,
but d orbitals as well.
17.

C
In order for a molecule to be linear, the bond angles must be 180°. The orbital that is associated with this bond angle
is an sp hybridized orbital. In sp hybridized carbons, the 2s orbital combines with a 2p orbital. You should know that a triple
bond contains 2 sp hybridized carbons. Acetylene, choice C, having a triple bond, contains two carbons that are sp hybridized.
Therefore, choice C is the correct answer.
Now for the wrong answers. Choice B, like choice C is an alkyne. However, there are three carbons in propyne; two
are sp hybridized, while the remaining carbon is sp3 hybridized. sp3 hybridization results from the combination of the 2s and all

of the 2p orbitals. Four sp3 carbons form and the preferred bond angle becomes 109.5°. In other words, sp 3 bonds are tetrahedral
in their geometry. Therefore, propyne won’t be linear since the carbon not involved in the triple bond will have a tetrahedral
arrangement of atoms around it. Choice A is wrong since the carbon in this molecule is sp3 hybridized as well. The carbon is
attached to four chlorine substituents, so in order to achieve maximum separation between the bonds, a tetrahedral geometry is
sought, hence carbon undergoes sp 3 hybridization. Finally, choice D is wrong because 1,3–hexadiene contains sp 2 hybridized
orbitals (when the double bonds occur) and sp 3 orbitals where the single bonds occur. Bond angles of 120° are associated with
sp2 orbitals and as we have already discussed, bond angles of 109.5° are associated with sp 3 orbitals. Therefore, there is no way
this molecule can be linear and so choice D is incorrect
18.

B
If we compare bonds between the same atoms, the longer the bond, the weaker it is. In this case we are comparing all
carbon-carbon bonds, and so essentially we are asked to arrange the bonds in decreasing bond strength. The carbon-carbon bond
in acetylene (ethyne) is a triple bond, and so is the strongest. In ethylene, the carbon-carbon bond is a double bond and so is the
next in strength. The carbon-carbon bonds in benzene are intermediate between single and double bonds, and so are the weakest
(and longest) among the three.

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