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 ...
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)
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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
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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
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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
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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%)
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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
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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
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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.
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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
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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
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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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