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ORGANIC CHEMISTRY TOPICAL:
Molecular Structure of OrganicCompounds
Test 1
Time: 23 Minutes*Number of Questions: 18
* The timing restrictions for the science topical tests are optional. If youare using this test for the sole purpose of content reinforcement, youmay want to disregard the time limit.
MCAT
2 as developed by
DIRECTIONS: Most of the questions in the following testare organized into groups, with a descriptive passagepreceding each group of questions. Study the passage,then select the single best answer to each question in thegroup. Some of the questions are not based on adescriptive passage; you must also select the best answerto these questions. If you are unsure of the best answer,eliminate the choices that you know are incorrect, thenselect an answer from the choices that remain. Indicateyour selection by blackening the corresponding circle onyour answer sheet. A periodic table is provided below foryour 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.
Molecular Structure of Organic Compounds Test 1
KAPLAN 3
Passage I (Questions 1–7)
Until recently, it had been common practice forpharmaceutical companies to manufacture chiral drugs asracemates, not as single enantiomers.
Usually, only one enantiomer of a given compoundpossesses therapeutic value, while the other may have nobeneficial pharmacological properties and may even induceserious physiological side effects.
A typical example of this problem was the use of thedrug Thalidomide in 1961. Administered to reduce nauseaand vomiting during the early stages of pregnancy, thedesired physiological activity lies with the R-isomer. TheS-isomer of Thalidomide is a teratogen (an agent thatproduces physical defects in developing embryos). As aresult, administration of this drug as the racemate causedcongenital malformations in thousands of infants. Thestructure of Thalidomide is shown below.
NO O
NC
O
C
OH
Figure 1
Today, advances in the synthesis of specificenantiomers should severely limit the number of racemicdrugs produced. Traditionally, the enantiomer of a chiraldrug could be synthesized via resolution (conversion of theracemic mixture into diastereomers, separation, and thenreformation of the enantiomers). However, this process iswasteful because of the numerous steps involved in theprocedure. The emergence of a new technique calledenantioselective catalysis may soon be used to avoid thisinefficiency.
The catalysts employed in enantioselective catalysisare usually based on optically active metal complexes andare highly effective in the production of one enantiomerover another. The success of this technique is attributed tothe production of thousands of chiral products from onemolecule of catalyst. A good example is the use of anickel-based catalyst to synthesize the S-isomer of thechiral drug Naproxen (Reaction 1).
MeO
+ HCNL + [Ni]
6-Methoxy-2-vinylnapthalene
CH3CN
H
MeO
CH3CO2H
H
MeO
Naproxen
(L + [Ni] = Metal complex catalyst)
Reaction 1
Another example of enantioselective catalysis is thatof meso breaking. For example, azidotrimethylsilane hasbeen shown to react with cyclohexene oxide in the presenceof the enantioselective catalyst titanium isopropoxide:
O + R3SiN3
chiral
catalyst
OSiR3
N3
Reaction 2
GO ON TO THE NEXT PAGE.
MCAT
4 as developed by
1 . How many chiral centers are present in theThalidomide molecule?
A . 0B . lC . 2D . 4
2 . What is the most likely reaction mechanism betweenHCN and 6-methoxy-2-vinylnapthalene?
A . Electrophilic additionB . Bimolecular nucleophilic substitutionC . Bimolecular eliminationD . Free radical addition
3 . Reaction 2 proceeds through an SN2 mechanism.Consequently, the attacking nucleophile is:
A . the alkyl group.B . the azide ion.C . silicon.D . the chiral catalyst.
4 . The double bond of the vinyl group in 6-methoxy-2-vinylnapthalene contains which of the followingorbitals?
I. sp orbitalsII. sp2 orbitals
III. sp3 orbitalsIV. p orbitals
A . II onlyB . III onlyC . I and IV onlyD . II and IV only
5 . Which of the following statements most accuratelydescribes the properties of Thalidomide?
A . It is a mixture of enantiomers that have the samechemical and physical properties, with theexception of the direction of optical rotation, butdifferent physiological properties.
B . It is a mixture of diastereomers that have the samephysical and chemical properties, but differentphysiological properties.
C . It is a mixture of enantiomers that have differentphysiological, chemical, and physical properties.
D . It is a mixture of diastereomers that have differentphysical, chemical, and physiological properties.
6 . Which of the following is the R-isomer of Naproxen?
A .C
CO2H
H
CH3
MeO MeO
CH3
H
CCO2H
MeO
CO2HCH3
C
H
CO2HCH3
C
H
MeO
C .
B . D .
7 . The term “meso breaking” can be applied to Reaction2 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.
GO ON TO THE NEXT PAGE.
Molecular Structure of Organic Compounds Test 1
KAPLAN 5
Passage II (Questions 8–13)
One of the major problems in using solar energy as apower source is the conversion of the raw material,sunlight, into a usable form of fuel. The transformation ofenergy contained in sunlight to chemical energy viaphotochemically induced isomerization (photo-isomerization) is one method that has been suggested inorder to alleviate this problem. In this process, a simpleorganic molecule is converted to a product that has far moreenergy than the starting material. This excess energy maybe stored in the product as angle, torsional or non-bondedstrain. The reverse reaction, formation of the startingmaterial, provides a means in which energy can be releasedand used in a controlled manner.
A good illustration of the photoisomerization processis the conversion of norbornadiene to quadricyclane(Reaction 1). Quadricyclane has 62 kcal/mol of excessstrain energy relative to norbornadiene and is thereforethermally unfavorable; the reverse reaction, quadricyclane tonorbornadiene, is highly favored in the presence of metalcatalysts.
Since it is readily synthesized in high yield,norbornadiene is a convenient starting material in thelaboratory. However, further research is required before itcan be mass-produced for large scale adaptation.
NorbornadieneStrain energy = 33 kcal/mol
QuadricylaneStrain energy = 95 kcal/mol
hv
Reaction 1
Norbornadiene converts to quadricyclane via aphotochemically allowed [2+2] ring closure, analogous tothe conversion of 1,3-butadiene to cyclobutene (Reaction2).
hv
1,3-Butadiene Cyclobutene
Reaction 2
The alkane equivalent of norbornadiene, norbornane,is shown in Figure 1.
4
2 6
3
7
5
1
Norbornane
Figure 1
8 . The ring strain that arises in small cycloalkanes, suchas cyclopropane or cyclobutane, is mainly attributedto:
I. compression of the bond angles to less than109.5°.
II. interaction between eclipsed hydrogens onadjacent carbons.
III. nonbonding interactions between hydrogenson non-adjacent carbons.
A . I onlyB . II onlyC . I and II onlyD . I, II, and III
9 . Which of the following will most readily undergo anintramolecular [2+2] ring closure?
A .
B .
C .
Ph
C C
CH2
CH2
Ph
Ph Ph
D .
GO ON TO THE NEXT PAGE.
MCAT
6 as developed by
1 0 . Catalytic hydrogenation of norbornadiene releasesapproximately 57 kcal of heat per mole ofnorbornadiene. The product, norbornane, is:
A . more stable, due to the decrease in angle strainaccompanying saturation.
B . more stable, due to the increased number ofhydrogens in the product.
C . less stable, due to the change in hybridization ofthe double bonded carbons.
D . less stable, due to steric effects and an increase inangle strain.
1 1 . For substituted norbornanes, it has been found thatthose substituted in the 7 position undergo SN2reactions with the appropriate nucleophile while thosesubstituted in the 1 position do not. This occursbecause:
A . it is difficult to form a carbocation intermediate atcarbon 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 thenucleophile is difficult because it is stericallyhindered.
D . the carbon at position 7 is more electrophilic thanthe carbon at position 1.
1 2 . Which of the following is true of norbornane?
A . Norbornane is less stable than cyclopentane butmore stable than cyclohexane.
B . Norbornane is more stable than both cyclopentaneand cyclohexane.
C . Norbornane is less stable than chair cyclohexanebut more stable than boat cyclohexane.
D . Norbornane is less stable than both cyclopentaneand cyclohexane.
1 3 . A Diels-Alder process ([4+2] cycloaddition) takes placeaccording to the following scheme:
A Diels-Alder reaction between which of the followingwould result in the formation of norbornadiene?
A . Propene and 1,3-butadieneB . Acetylene and 1,3-cyclopentadieneC . Ethylene and propeneD . Ethylene and 1,3-cyclopentadiene
GO ON TO THE NEXT PAGE.
Molecular Structure of Organic Compounds Test 1
KAPLAN 7
Questions 14 through 18 are NOTbased on a descriptive passage
1 4 . Which of the following compounds will exhibit thegreatest dipole moment?
A . (Z)-1,2-Dichloro-1,2-diphenyletheneB . (E)-1,2-Dichloro-1,2-diphenyletheneC . 1,2-Dichloro-1,2-diphenylethaneD . 1,2-Difluoroethane
1 5 . How many structural isomers of C3H6Br2 are capableof exhibiting optical activity?
A . 1B . 2C . 3D . 4
1 6 . C=C, C=O, N=N, and C=N bonds are quite commonin organic compounds. However, C=S, C=P, C=Si,and other similar bonds are not often found. The mostprobable explanation for this observation is that:
A . carbon does not combine with elements foundbelow the second row of the periodic table.
B . sulfur, phosphorus, and silicon do not form pibonds due to the lack of occupied p orbitals intheir ground state electron configurations.
C . sulfur, phosphorus, and silicon are incapable oforbital hybridization.
D . the comparative sizes of the 2p and 3p atomicorbitals make effective overlap between them lesslikely than between two 2p orbitals.
1 7 . Which of the compounds listed below is linear?
A . Carbon tetrachlorideB . PropyneC . AcetyleneD . 1,3-Hexadiene
1 8 . What is the order of increasing carbon-carbon bondlength in the molecules listed below?
I. AcetyleneII. Benzene
III. Ethylene
A . I, II, IIIB . I, III, IIC . II, III, ID . II, I, III
END OF TEST
MCAT
8 as developed by
ANSWER KEY:1. B 6. C 11. C 16. D2. A 7. D 12. D 17. C3. B 8. C 13. B 18. B4. D 9. D 14. A5. A 10. A 15. A
Molecular Structure of Organic Compounds Test 1
KAPLAN 9
EXPLANATIONS
Passage I
1 . BThe 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.
NO O
NC
O
C
OH
*
A racemic mixture is one that contains equal amounts of two enantiomers, chiral molecules that have oppositeconfigurations from each other at every stereocenter and so are non-superimposable mirror images. In organic compounds, astereocenter or a chiral center is usually defined as a carbon that is attached to four different groups. Thalidomide definitely has atleast one chiral center since the text talks about its administration as a racemate and the R– and S– isomers. Therefore, choice Acan be eliminated. A chiral center is an atom attached to four different functional groups. Starting at the benzene ring, you cansee that there are no chiral centers, as all of the carbons are attached to only three substituents. The same thing applies to thecarbonyl groups. Moving on to the carbon in the nitrogen-containing ring, you can see that this is a chiral center. It is attachedto a hydrogen (not shown explicitly), a nitrogen, a CH2 group, and a carbonyl group. Moving clockwise around the ring, thenext group is the carbonyl functionality. Again, this center is achiral since it is only attached to three substituents; the samething applies to the carbonyl group opposite. Although the nitrogen in the ring is attached to four different groups (the lone pairof electrons not shown is considered a group), it undergoes rapid inversion of configuration that causes it to not be a chiralcenter. Moving along to the last two carbons in the ring, you can see that neither of these are chiral since they are each attachedto two hydrogens atoms. So, there is only one chiral center in the Thalidomide molecule: the carbon in the nitrogen containingring which we discussed earlier.
2 . AIn 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 iselectron rich, it is likely to be attacked by positively polarized molecules or electrophiles. The hydrogen in HCN acts as anelectrophile and, therefore, adds to the least substituted carbon to form the most stable carbocation intermediate. Thisintermediate is formed as electrons are given up from the double bond to form a new bond to the hydrogen, leaving behind apositive 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 productshown in Reaction 1. Obviously from the mechanism I have just described, there is addition across the double bond. Thisaddition 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 multiplebond; 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 inorder 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 whenperoxides are added to the reaction mixture. This reaction occurs by a radical mechanism and the hydrogen adds to the mostsubstituted carbon. This is not the case in this reaction, so choice D is incorrect.
3 . BEpoxides, 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, thenucleophile attacks one of the epoxide carbons, and since carbon cannot form 5 bonds, the bond between it and the oxygen isbroken. The electrons from this bond are taken on by the oxygen forming a negatively charged anion.
In Reaction 2, the R3SiN3 molecule forms R3Si+ and N3–. 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 thecarbon–oxygen bond. The positively polarized silicon-alkyl group can then add to the negatively charged oxygen to form theproduct shown. As a result, the attacking nucleophile is the azide ion and choice B is the correct response.
MCAT
10 as developed by
O
N3–
N3
O–
+ SiR3+
N3
OSiR3
The chiral catalyst is involved in the reaction, but not as a nucleophile. A catalyst serves to speed up the reaction andat the end of the reaction, it remains unchanged. Therefore, it would not behave as a nucleophile and become incorporated intothe product; choice D is wrong.
Choices A and C are wrong since these molecules constitute the electrophilic portion of the R3SiN3 molecule. Sincethey are positively polarized, there is no way they could be nucleophilic.
4 . DCarbon–carbon double bonds are made up of both sp2 hybridized orbitals and p orbitals. The electron configuration of
carbon is 1s22s22p2. The 2s orbital and two 2p orbitals hybridize to form three sp2 hybrid orbitals. This leaves one free porbital which can overlap with an adjacent orbital to form a pi bond. sp2 hybridized orbitals have a geometry of 120°, and theseconstitute the carbon-carbon and carbon-hydrogen sigma bonds in the molecule. Therefore, the double bond in the vinyl group isformed by the overlap of two sp2 orbitals and two p orbitals.
5 . AThalidomide 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 oppositedirections. However, they do behave differently in chiral environments, and so they exhibit different behavior in the humanbody. One enantiomer reduces nausea and vomiting while the other is a teratogen which is an agent that causes physical defectsin the developing embryo. Therefore, Thalidomide is a mixture of enantiomers that have the same chemical and physicalproperties 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 werediastereomers, then they would possess different chemical and physical properties and would form a mixture of two differentcompounds, not a racemic mixture as described in the passage. Choice C is incorrect because enantiomers have the samechemical and all of the same physical properties except in the direction that plane polarized light is rotated.
6 . CFirst, 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 thehydrogen. The other three atoms attached to the stereocenter are carbons, so the atomic weights of the groups attached to thesecarbons now have to be considered. The substituent that will have the next to last priority is the methyl group, since hydrogensare attached to the methyl carbon. In the remaining two substituents, the carbons are attached to two other carbons (in the caseof the substituted naphthalene group) and two oxygens (in the case of the carboxyl group). Another rule you need to know, isthat in double bonds, the atoms have to be duplicated; a carbonyl group would be classed as a carbon bonded to two oxygenatoms. When this is done, the carboxyl group turns out to be of highest priority. So to summarize, the order of increasingpriority is hydrogen, methyl, naphthalene and then carboxyl. In order to assign a configuration, the lowest priority substituenthas to be rotated to the back and then arrows are drawn from the highest priority substituent (numbered 1) to the lowest prioritysubstituent (numbered 3). If this is done for Naproxen in Reaction 1, you should see that the arrows are in an anti–clockwisedirection 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, makingit a non-superimposable mirror image of the S–isomer. Choice C is correct because it has the opposite configuration of the S-isomer. The hydrogen group is already oriented toward the back, and so in drawing arrows from the carboxyl group through tothe 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 fromthe naphthalene ring. These choices are wrong because both groups would be in the plane of the ring. In addition, theorientation around the chiral carbon is incorrect in both responses. Choice B is incorrect because this is the S-isomer ofNaproxen; the configuration around the chiral center is exactly the same as that shown in Reaction 1.
7 . DA 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
Molecular Structure of Organic Compounds Test 1
KAPLAN 11
cyclohexene oxide, shown in Reaction 2. The two carbons attached to the oxygen are in fact chiral centers; each one is attachedto four different groups; oxygen, hydrogen, CH2, and CH. However, intersecting the bond perpendicular to these two chiralcarbons is a plane of symmetry. Therefore, this molecule is superimposable on its mirror image and is optically inactive. Theaddition of azidotrimethylsilane, R3SiN3 to cyclohexene oxide results in the loss of this plane of symmetry. The two chiralcarbon 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 correctanswer. Choice A is incorrect since the chirality of the catalyst is retained in the reaction. Remember: although a catalyst maytake part in a reaction, it comes out the same at the end of the reaction. Choice B is incorrect because the chiral centers are notgenerated in the reaction: they have always been present. Choice C is wrong because the plane of symmetry is lost in theconversion of the meso compound to the optically active product.
Passage II
8 . CYou should be aware that in saturated ring systems (ring systems consisting of all single bonds), the carbons are sp3
hybridized. sp3 hybridized carbons have preferred bond angles of 109.5°. However, in cyclopropane, the bond angles arecompressed 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 anglestrain 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 repulsiveinteraction, so the molecule suffers from torsional strain. The diagram below illustrates one such pair of eclipsed hydrogens:
C
C C
H
H
H
H
H
H
eclipsinginteractions
Cyclobutane also suffers from torsional strain, but to a lesser degree. If the bond angles were 90°, the cyclobutanemolecule would be planar and all eight hydrogens would be eclipsed. To alleviate this, the molecule folds and results is a bondangle of 88°. At the expense of slightly more angle strain, the torsional strain is reduced. Both molecules possess eclipsedhydrogens 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 inspace. The classic example is the repulsion between substituents on opposite ends of cyclohexane in the unfavorable boatconformation. There are no such interactions on cyclopropane or cyclobutane, so statement III is false.
9 . DThe 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. Ringclosure involves the movement of these electrons; in the case of a [2+2] reaction, 2 electrons are used up in forming a newsigma bond, while the other two simply move between carbons. Right away, you can rule out choices A and B. Benzene asshown in choice A, contains 6 electrons and is very stable. A [2+2] ring closure in this molecule is highly unlikely since itwill involve breaking the aromaticity of the ring. Choice B, norbornene, won’t undergo a [2+2] ring closure since it onlycontains 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, thephenyl groups would end up on the same side of the molecule. This would result in a great deal of steric interaction, and, as aresult, the trans isomer is favored and ring closure is extremely difficult. On the other hand, the double bonds in choice D are incloser proximity to each other, making cyclization somewhat easier.
1 0 . ACatalytic 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°. Lookingat norbornadiene, you can see that the sp2 bond angles are much less than 120°, and as a result, the molecule suffers from anglestrain. When catalytic hydrogenation occurs, the double bonded carbons become sp3 hybridized, and the preferred bond anglelowers to 109.5°. The bond angles in norbornane are much closer to this bond angle so saturation of norbornadiene is
MCAT
12 as developed by
accompanied by a decrease in angle strain. In other words, catalytic hydrogenation results in the formation of a more stablemolecule. This is best described by choice A. Further evidence that the product is more stable is provided from the questionstem: it states that 57 kilocalories of heat per mole are released upon hydrogenation; in other words, 1 mole of product is morestable 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 theproduct. Choice C is correct in that the change in stability is due to the change in hybridization. However, this answer choicealso states that the product is less stable. If the product was less stable than the reactant, 57 kcal would be taken in uponformation of norbornane. For this reason, choice D is also incorrect.
1 1 . CYou 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 bythe loss of a good leaving group. The nucleophile then adds to the carbon bearing this positive charge. As a carbocation isformed, the order of reactivity of an alkyl halide, for example, will be tertiary, then secondary, then primary (in fact, primaryalkyl halides do not react via an SN1 mechanism). On the other hand, SN2 reactions depend on the concentration of both thesubstrate and the incoming nucleophile. In a concerted mechanism, the incoming nucleophile attacks the electrophilic carbon180° to the leaving group. No carbocation intermediate is formed, and any factor that effects the incoming nucleophile and thesubstrate 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 thecarbon is primary, nucleophilic attack will be easier since there is less steric hindrance. For this reason, reactivity of alkylhalides 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 atertiary alkyl halide, whereas substitution at the 7 position would result in the formation of a secondary alkyl halide. Tertiaryalkyl halides won’t react by an SN2 mechanism due to steric hindrance, so at position 1, the incoming nucleophile would besterically hindered and would not be able to attack 180° to the leaving group. SN2 would be significantly easier at position 7since 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 achievedat position 1. However, this is relevant only in an SN1 reaction, which is not what we have here. Therefore, choice A isincorrect. Choice B also talks about carbocation formation and so is incorrect. Finally, choice D is wrong because the carbons atpositions 1 and 7 would be more or less equally electrophilic if substituted by the same group.
1 2 . DBreaking 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 ismade up of carbons 1, 2, 3, 4 and 7). The six-membered ring is not in its preferred chair conformation and the bond angles arehighly compressed. Therefore, the molecule will be less stable than cyclohexane, eliminating answer choices A and B. Both theboat 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 bentconformation to alleviate torsional strain at the expense of slightly more angle strain. The cyclopentane rings in norbornane arefar more strained than isolated cyclopentane rings, so the molecule is more unstable than both cyclopentane and cyclohexanemaking choice D the correct response.
1 3 . BThe 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:
Molecular Structure of Organic Compounds Test 1
KAPLAN 13
1, 3-butadiene propene
1, 3-cyclo-pentadiene
acetylene
1, 3-cyclo-pentadiene
ethylene
A.
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 thefact that acetylene contains a triple bond; it still donates only two electrons in the cyclization reaction. The overall processresults in cyclization to form norbornadiene. Two double bonds remain; one from the original diene and one from acetylene (as atriple bond was involved in the reaction, not a double bond). Cyclization involves the formation of a six–membered ring and theformation 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 adouble 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, notnorbornadiene. Therefore, choice A is incorrect.
Independent Questions
1 4 . AThe 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 theelectronegativities of the atoms involved in the bond. Therefore, carbon–carbon bonds have little or no dipole moment, whilecarbon–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 onone 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 isassigned according to molecular weight, so the two chlorines take highest priority). The electronegative chlorines pull electrondensity 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 pullelectron density toward themselves, both dipoles cancel each other out to give a net dipole moment of zero. Therefore, choice Bcan be eliminated.
Choice C contains the same substituents, but there is a single bond in this molecule, not a double bond. The keydifference in this molecule is that the single bond rotates, so any effective dipole moment that is set up will be canceled out byan 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 thecarbon–carbon single bond will destroy any net dipole that is set up.
1 5 . AStructural 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 canattach to carbon 1 of the propane chain forming 1,1–dibromopropane, both bromines can attach to carbon 2 forming2,2–dibromopropane, a bromine can attach to carbons 1 and 2 forming 1,2–dibromopropane or finally, a bromine can attach tocarbons 1 and 3 forming 1,3–dibromopropane. All of these are structural isomers since they have the same molecular formula,
MCAT
14 as developed by
but different atomic connectivity. So, which of these will exhibit optical activity? Recall that in order to possess opticalactivity, a molecule must be chiral. Most chiral molecules are identified as possessing atoms attached to four differentsubstituents. Of the four isomers, only the 1,2–substituted propane has a chiral center. The second carbon is attached to abromine, a hydrogen, a CH2Br group and a CH3 group. Therefore, this isomer may exhibit optical activity. All of the otherisomers don’t exhibit optical activity because they don’t possess any chiral centers, so A is the correct response.
C C C
H
Br
H
H
Br
H
H
H
*
1 6 . DIn the case of carbon–carbon double bonds, they are made up by the overlap of sp2 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. Sigmabond lengths and strengths are largely determined by the size and shape of the atomic orbitals involved in formation of the bondand the ability of these orbitals to overlap effectively. Sigma bonds are stronger than pi bonds because head to head overlap ismore efficient than sideways overlap. Sigma bonds formed from two 2s orbitals are shorter than those formed from two 2porbitals or two 3s orbitals.
The difference between the first four bonds named in the question stem and final three, is the position of theirrespective atoms in the periodic table. Carbon, oxygen and nitrogen can be found in the second period, while sulfur, phosphorusand silicon are found in the third period. Therefore, sulfur, phosphorus and silicon would use 3p orbitals in the formation of pibonds, not 2p orbitals in the case of carbon, nitrogen and oxygen. The key problem, however, is that 3p orbitals are muchlarger than 2p orbitals and so overlap of the 2p orbital of carbon and the 3p orbital of silicon, phosphorus or sulfur is unlikelyto 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. Forexample, carbon can form bonds with the elements in the halogen group. Choice B is also wrong. In their ground state electronconfigurations, silicon, sulfur and phosphorus all have partially occupied p orbitals which can form bonds. Finally, choice Cis incorrect since sulfur, phosphorus and silicon can undergo hybridization. These elements can not only mix s and p orbitals,but d orbitals as well.
1 7 . CIn 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 triplebond 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; twoare sp hybridized, while the remaining carbon is sp3 hybridized. sp3 hybridization results from the combination of the 2s and allof the 2p orbitals. Four sp3 carbons form and the preferred bond angle becomes 109.5°. In other words, sp3 bonds are tetrahedralin their geometry. Therefore, propyne won’t be linear since the carbon not involved in the triple bond will have a tetrahedralarrangement of atoms around it. Choice A is wrong since the carbon in this molecule is sp3 hybridized as well. The carbon isattached to four chlorine substituents, so in order to achieve maximum separation between the bonds, a tetrahedral geometry issought, hence carbon undergoes sp3 hybridization. Finally, choice D is wrong because 1,3–hexadiene contains sp2 hybridizedorbitals (when the double bonds occur) and sp3 orbitals where the single bonds occur. Bond angles of 120° are associated withsp2 orbitals and as we have already discussed, bond angles of 109.5° are associated with sp3 orbitals. Therefore, there is no waythis molecule can be linear and so choice D is incorrect
1 8 . BIf 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 bondin acetylene (ethyne) is a triple bond, and so is the strongest. In ethylene, the carbon-carbon bond is a double bond and so is thenext 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.
