Chapter 14 - Test Bank Chem 200

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Package title: Solomons Test Bank Course Title: Solomons 11e

Chapter Number: 14

Question type: Multiple choice

1) The correct name for the compound shown below is:

Cl HO a) m-chorophenol b) o-chorophenol c) p-chorophenol d) o-chorobenzyl alcohol e) 2-chlorophenyl alcohol Answer: B

Topic: Nomenclature and Isomerism Section: 14.2

Difficulty Level: Easy

Br NH₂ a) m-bromoaniline b) o-bromoaniline c) p-bromoaniline d) o-bromobenzyl amine e) 3-bromophenyl amine Answer: A

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3) The correct name for the compound shown below is: OCH3 O2N a) o-nitroanisole b) m-nitroanisole c) p-nitroanisole d) p-nitrotoluene e) 4-nitrophenol Answer: C

CH₃ Br a) 3-bromobenzene b) 4-bromobenzene c) m-bromobenzene d) o-bromotoluene e) Bromomethylbenzene Answer: D

5) Which of the following is NOT 2-bromo-5-nitrobenzoic acid?

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IV I II V III NO2 Br HO O Br O2N OH O Br NO2 HO O NO2 Br HO O Br NO2 HO O a) I b) II c) III d) IV e) V Answer: D

6) Toluene is the name commonly assigned to: a) Hydroxybenzene b) Aminobenzene c) Methylbenzene d) Ethylbenzene e) Methoxybenzene Answer: C

7) Anisole is the name commonly assigned to: a) Hydroxybenzene

b) Aminobenzene c) Methylbenzene d) Ethylbenzene

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e) Methoxybenzene Answer: E

Difficulty Level: Medium

8) Phenol is the name commonly assigned to: a) Hydroxybenzene b) Aminobenzene c) Methylbenzene d) Ethylbenzene e) Methoxybenzene Answer: A

9) Aniline is the name commonly assigned to: a) Hydroxybenzene b) Aminobenzene c) Methylbenzene d) Ethylbenzene e) Methoxybenzene Answer: B

10) Xylene is the name commonly assigned to: a) Hydroxybenzene b) Aminobenzene c) Dimethylbenzene d) Ethylbenzene e) Methoxybenzene 4

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Answer: C

11) Cumene is the name commonly assigned to: a) Hydroxybenzene b) Aminobenzene c) Ethylbenzene d) Isopropylbenzene e) Isobutylbenzene Answer: D

Br NH₂ Br a) 3,4-Dibromoaniline b) 2,4-Dibromoaniline c) 2,5-Dibromoaniline d) 3,6-Dibromoaniline e) 2,6-Dibromoaniline Answer: C

13) 2-Bromo-4-nitroaniline is:

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IV I II V III NO₂ Br OCH3 NH₂ NO₂ Br NH₂ Br NO₂ NO₂ H₂N Br NO₂ H₃CO Br a) I b) II c) III d) IV e) V Answer: C

14) 2-Bromo-5-nitroaniline is: IV I II V III NO₂ Br OCH₃ NH₂ NO₂ Br NH₂ Br NO2 NO₂ H₂N Br NO2 H₃CO Br a) I b) II c) III d) IV e) V 6

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Answer: B

15) 2-Bromo-4-nitroaniline is: IV I II V III NO₂ Br OCH₃ NH₂ NO₂ Br NH₂ Br NO₂ NO₂ H₂N Br NO₂ H₃CO Br a) I b) II c) III d) IV e) V Answer: C

16) m-chlorobenzenesulfonic acid is:

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IV I II V III Cl HO3S SO3H Cl Cl SO3H SO3HCl HOOC Cl Cl HOOC a) I b) II c) III d) IV e) V Answer: D

17) 4-Bromo-2-nitroaniline is: IV I II V III NO₂ Br OCH3 NH₂ NO₂ Br NH₂ Br NO₂ NO₂ H2N Br NO₂ H₃CO Br a) I b) II c) III d) IV e) V 8

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Answer: A

18) 4-Amino-2-chlorophenol is: IV I II V III NH₂ Cl OH NO₂ O Cl NH₂ Cl OH O Cl H₂N NO₂ Cl OH a) I b) II c) III d) IV e) V Answer: D

19) 3-bromo-4-nitroacetophenone is: IV I II V III H O Br O2N H O O2N Br O Br O O2N Br O2N _H O Br O2N 9

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a) I b) II c) III d) IV e) V Answer: C

20) 4-Chloro-3-nitro-benzaldehyde is: IV I II V III NH₂ Cl OH NO₂ O Cl NH₂ Cl OH O Cl H₂N NO2 Cl OH a) I b) II c) III d) IV e) V Answer: B

21) 4-Amino-2-chlorobenzaldehyde is:

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IV I II V III NH₂ Cl OH NO₂ O Cl NH₂ Cl OH O Cl H₂N NO₂ Cl OH a) I b) II c) III d) IV e) V Answer: C

22) 4-Amino-2-bromoanisole is: IV I II V III NH2 Br OCH₃ NO₂ Br H₃CO NH₂ Br OCH₃ O Br H2N NO2 Br OCH₃ a) I b) II c) III d) IV e) V Answer: D

Topic: Nomenclature and Isomerism

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Section: 14.2

23) 4-Bromomethyl-2-nitroanisole is: IV I II V III NH₂ Br OCH₃ NO₂ Br H₃CO NH₂ Br OCH₃ O Br H₂N NO₂ Br OCH₃ a) I b) II c) III d) IV e) V Answer: B

24) 3-chloro-4-nitrobenzoic acid is:

IV I II V III H O Cl O2N OH O O2N Cl O Cl O O2N Cl O2N _H O Cl O2N a) I 12

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b) II c) III d) IV e) V Answer: B

25) In theory, a single molecule of this compound will rotate plane-polarized light. a) Butylbenzene

b) Isobutylbenzene c) sec-Butylbenzene d) tert-Butylbenzene e) None of these choices. Answer: C

26) Which dichlorobenzene might theoretically yield three mononitro products? a) o-Dichlorobenzene

b) m-Dichlorobenzene c) p-Dichlorobenzene d) None of these choices. e) All of these choices. Answer: B

Topic: Nomenclature and Isomerism Section: 14.2 and 14.3

27) Which dibromobenzene can yield only one mononitro derivative? a) o-Dibromobenzene

b) m-Dibromobenzene

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c) p-Dibromobenzene

d) More than one of these choices. e) None of these choices.

Answer: C

28) If thiophene is an aromatic molecule and reacts similarly to benzene, how many (neutral) monobromothiophenes could be obtained in the following reaction?

S Br₂ FeBr₃ a) 1 b) 2 c) 3 d) 4 e) 5 Answer: B

29) Which of these is the single best representation for naphthalene?

V III IV II I a) I b) II c) III d) IV e) V 14

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Answer: A

30) How many dichloronaphthalenes are possible? a) 7 b) 8 c) 9 d) 10 e) 12 Answer: D

Difficulty Level: Hard

31) How many monochloronaphthalenes are possible? a) 2 b) 3 c) 4 d) 5 e) 6 Answer: A

32) The complete name for the following compound is:

CH₃ H a) sec-Hexylbenzene b) 2-Phenylhexane c) (R)-2-Phenylhexane 15

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d) (S)-2-Phenylhexane

e) Butylmethylphenylmethane Answer: D

33) The complete name for the following compound is:

CH3 H a) sec-Hexylbenzene b) 2-Phenylhexane c) (R)-2-Phenylhexane d) (S)-2-Phenylhexane e) Butylmethylphenylmethane Answer: C

34) Which dibromobenzene can, in theory, yield three mononitro derivatives? a) o-Dibromobenzene

b) m-Dibromobenzene c) p-Dibromobenzene d) All of these choices. e) None of these choices. Answer: B

35) Which dihydroxybenzene can yield only one mononitro derivative? 16

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a) o-Dihydroxybenzene b) m-Dihydroxybenzene c) p- Dihydroxybenzene

d) More than one of these choices. e) None of these choices.

Answer: C

36) How many different dibromophenols are possible? a) 8 b) 7 c) 6 d) 5 e) 4 Answer: C

37) How many different dichloronitrobenzenes are possible? a) 8 b) 7 c) 6 d) 5 e) 4 Answer: C

38) How many different dihydroxybromobenzenes are possible?

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a) 8 b) 7 c) 6 d) 5 e) 4 Answer: C

39) Which is the only one of these reagents which will react with benzene under the specified conditions? a) Cl2, FeCl3, heat b) H2, 25°C c) Br2/CCl4, 25°C, dark d) KMnO4/H2O, 25°C e) H3O+, heat Answer: A

Topic: Chemical Tests and Reactions Section: 14.3

40) Which reagent(s) would serve as the basis for a simple chemical test that would distinguish between benzene and 1-hexene?

a) NaOH in H2O b) Br2 in CCl4

c) AgNO3 in C2H5OH d) NaHSO3 in H2O e) None of these choices. Answer: B

41) Which reagent(s) would serve as the basis for a simple chemical test that would distinguish 18

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between benzene and cyclohexene? a) NaOH in H2O

b) Br2 in CCl4

c) AgNO3 in C2H5OH d) NaHSO3 in H2O e) None of these choices. Answer: B

42) Which reagent(s) would serve as the basis for a simple chemical test that would distinguish between ethylbenzene and vinylcyclohexane?

a) H2CrO4 b) LAH

c) NaBH4, H2O d) KMnO4, OH−, H2O e) Two of these choices. Answer: D

43) Which reagent(s) would serve as the basis for a simple chemical test that would distinguish between ethylbenzene and vinylcyclohexane?

a) H2CrO4 b) LAH

c) NaBH4, H2O d) O3/Zn, H+

e) Two of these choices. Answer: D

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44) Which reagent(s) would serve as the basis for a simple chemical test that would distinguish between naphthalene and 2,4,6-decatriene?

a) NaOH in H2O b) KMnO4, OH−, H2O c) NaBH4, H2O d) H2CrO4

e) None of these choices. Answer: B

45) Why would 1,3-cyclohexadiene undergo dehydrogenation readily? a) It is easily reduced.

b) Hydrogen is a small molecule.

c) 1,3-Cyclohexadiene has no resonance energy.

d) It would gain considerable stability by becoming benzene. e) It would not undergo dehydrogenation.

Answer: D

Topic: Explanations and Predictions Section: 14.5

46) Recalling that benzene has a resonance energy of 152 kJ mol-1 and naphthalene has a resonance energy of 255 kJ mol-1, predict the positions which would be occupied by bromine when phenanthrene (below) undergoes addition of Br2.

1 2 3 4 5 6 7 8 9 10 a) 1, 2 b) 1, 4 c) 3, 4 d) 7, 8 20

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e) 9, 10 Answer: E

Topic: Explanations and Predictions Section: 14.3 and 14.5

47) Which of the following statements about cyclooctatetraene is NOT true? a) The compound rapidly decolorizes Br2/CCl4 solutions.

b) The compound rapidly decolorizes aqueous solutions of KMnO4. c) The compound readily adds hydrogen.

d) The compound is nonplanar.

e) The compound is comparable to benzene in stability. Answer: E

48) Cyclopentadiene is unusually acidic for a hydrocarbon. An explanation for this is the following statement.

a) The carbon atoms of cyclopentadiene are all sp2-hybridized. b) Cyclopentadiene is aromatic.

c) Removal of a proton from cyclopentadiene yields an aromatic anion.

d) Removal of a hydrogen atom from cyclopentadiene yields a highly stable free radical. e) Removal of a hydride ion from cyclopentadiene produces an aromatic cation.

Answer: C

49) Consider the molecular orbital model of benzene. In the ground state how many molecular orbitals are filled with electrons?

a) 1 b) 2 c) 3

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d) 4 e) 5 Answer: C

Topic: Molecular Orbital Theory Section: 14.6

50) In the molecular orbital model of benzene, how many pi-electrons are delocalized about the ring? a) 2 b) 3 c) 4 d) 5 e) 6 Answer: E

51) In the molecular orbital model of benzene, how many pi-electrons are in bonding molecular orbitals? a) 6 b) 5 c) 4 d) 3 e) 2 Answer: A

52) Which cyclization(s) should occur with a decrease in pi-electron energy?

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I II III IV CH2 CH CH CH2 + H₂ CH2 CH CH2 + H₂ CH2 CH CH2 + H₂ CH2 CH CH2 + H₂ a) I b) II c) III d) IV

e) All of these choices. Answer: B

53) Which cyclization(s) should occur with a decrease in pi-electron energy?

+ H2 + H2 + H₂ + H₂ I II III IV 23

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a) I b) II c) III d) IV

e) All of these choices. Answer: B

54) Which cyclization(s) should occur with an increase in pi-electron energy?

e) All of these choices. Answer: D

55) Which cyclization(s) should occur with an increase in pi-electron energy? 24

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e) All of these choices. Answer: A

56) Application of the polygon-and-circle technique reveals that single electrons occupy each of the two nonbonding orbitals in the molecular orbital diagram of:

a) Cyclobutadiene b) Benzene c) Cyclopropenyl cation d) Cyclopentadienyl anion e) Cycloheptatrienyl cation Answer: A

57) In the molecular orbital model of benzene, the six p-orbitals combine to form how many 25

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molecular orbitals? a) 6 b) 5 c) 4 d) 3 e) 2 Answer: A

Topic: Molecular Orbital Theory Section: 14.6 and 14.7

58) In the molecular orbital model of benzene, how many π-bonding molecular orbitals are there? a) 6 b) 5 c) 4 d) 3 e) 2 Answer: D

59) In the molecular orbital model of benzene, how many π-antibonding molecular orbitals are there? a) 6 b) 5 c) 4 d) 3 e) 2 Answer: D

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60) In the molecular orbital model of benzene, how many pairs of degenerate π-molecular orbitals are there?

a) 6 b) 5 c) 4 d) 3 e) 2 Answer: E

61) In the molecular orbital model of benzene, how many pairs of degenerate π-bonding molecular orbitals are there?

a) 1 b) 2 c) 3 d) 4 e) 5 Answer: A

62) In the molecular orbital model of benzene, how many pairs of degenerate π-antibonding molecular orbitals are there?

a) 1 b) 2 c) 3 d) 4 e) 5 Answer: A

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63) In the molecular orbital model of benzene, how many non-bonding molecular orbitals are there? a) 1 b) 2 c) 3 d) 4 e) 0 Answer: E

64) In the molecular orbital model of the cyclopentadienyl cation, how many π-bonding molecular orbitals are there?

a) 6 b) 5 c) 4 d) 3 e) 2 Answer: D

65) In the molecular orbital model of the cyclopentadienyl cation, how many π-antibonding molecular orbitals are there?

a) 6 b) 5 c) 4 d) 3 e) 2 Answer: E 28

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66) In the molecular orbital model of the cyclopentadienyl cation, how many pairs of degenerate π-molecular orbitals are there?

a) 6 b) 5 c) 4 d) 3 e) 2 Answer: E

67) In the molecular orbital model of the cyclopentadienyl cation, how many pairs of degenerate π-bonding molecular orbitals are there?

a) 1 b) 2 c) 3 d) 4 e) 5 Answer: A

68) In the molecular orbital model of the cyclopentadienyl cation, how many pairs of degenerate π-antibonding molecular orbitals are there?

a) 1 b) 2 c) 3 d) 4 e) 5 29

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Answer: A

69) In the molecular orbital model of the cyclopentadienyl cation, how many non-bonding molecular orbitals are there?

a) 1 b) 2 c) 3 d) 4 e) 0 Answer: E

70) In the molecular orbital model of cyclobutadiene, how many π-bonding molecular orbitals are there? a) 1 b) 2 c) 3 d) 4 e) 5 Answer: A

71) In the molecular orbital model of cyclobutadiene, how many π-antibonding molecular orbitals are there?

a) 5 b) 4 c) 3 d) 2

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e) 1 Answer: E

72) In the molecular orbital model of cyclobutadiene, how many pairs of degenerate π-molecular orbitals are there?

a) 0 b) 1 c) 2 d) 3 e) 4 Answer: A

73) In the molecular orbital model of the cyclobutadiene, how many pairs of degenerate π-bonding molecular orbitals are there?

a) 1 b) 2 c) 3 d) 4 e) 0 Answer: E

74) In the molecular orbital model of cyclobutadiene, how many pairs of degenerate π-antibonding molecular orbitals are there?

a) 1 b) 2

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c) 3 d) 4 e) 0 Answer: E

75) In the molecular orbital model of cyclobutadiene, how many non-bonding molecular orbitals are there? a) 1 b) 2 c) 3 d) 4 e) 0 Answer: B

76) Consider the molecular orbital model of cyclopentadienyl anion. In the ground state how many molecular orbitals are filled with electrons?

a) 1 b) 2 c) 3 d) 4 e) 5 Answer: C

77) In the molecular orbital model of cyclopentadienyl anion, how many pi-electrons are delocalized about the ring?

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a) 2 b) 3 c) 4 d) 5 e) 6 Answer: E

78) In the molecular orbital model of cyclopentadienyl anion, how many pi-electrons are in bonding molecular orbitals?

a) 6 b) 5 c) 4 d) 3 e) 2 Answer: A

79) In the molecular orbital model of cycloheptadienyl cation, how many pi-electrons are in bonding molecular orbitals?

a) 6 b) 5 c) 4 d) 3 e) 2 Answer: A

80) In the molecular orbital model of cycloheptadienyl cation, how many pi-electrons are 33

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delocalized about the ring? a) 6 b) 5 c) 4 d) 3 e) 2 Answer: A

81) In the molecular orbital model of 1,3-cyclobutadiene, how many electrons are in non-bonding molecular orbitals?

a) 6 b) 5 c) 4 d) 3 e) 2 Answer: E

82) In the molecular orbital model of which of the following substance(s) are non-bonding orbitals found?

a) 1,3-Cyclobutadiene b) Cyclopentadienyl anion c) 1,3,5,7-Cyclooctatetraene d) Two of these choices. e) All of these choices. Answer: D

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83) Which of the following is NOT true of benzene?

a) Benzene tends to undergo substitution rather than addition reactions, even though it has a high index of hydrogen deficiency.

b) All of the hydrogen atoms of benzene are equivalent.

c) The carbon-carbon bonds of benzene are alternately short and long around the ring. d) Only one o-dichlorobenzene has ever been found.

e) Benzene is more stable than the hypothetical compound 1,3,5-cyclohexatriene. Answer: C

Topic: Bonding and Resonance Section: 14.4 and 14.6

84) Which of the following is true of benzene?

a) Benzene tends to undergo addition rather than substitution reactions. b) All of the hydrogen atoms of benzene are equivalent.

c) The carbon-carbon bonds of benzene are alternately short and long around the ring. d) The benzene ring is a distorted hexagon.

e) Benzene has the stability expected for cyclohexatriene. Answer: B

85) The carbon-carbon bonds in benzene are:

a) of equal length and are shorter than the double bond of ethene.

b) of equal length and are intermediate between a double bond and a single bond. c) of unequal length and are alternately short and long around the ring.

d) due only to p-orbital overlap.

e) of equal length and intermediate between the carbon-carbon bond lengths in ethene and ethyne.

Answer: B

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86) We now know that the two Kekule structures for benzene are related in the following way: a) They are each equally correct as a structure for benzene.

b) Benzene is sometimes one structure and sometimes the other. c) The two structures are in a state of rapid equilibrium.

d) Neither of the two structures adequately describes benzene; benzene is a resonance hybrid of the two.

e) None of these choices. Answer: D

87) In which of the following compounds would the shortest carbon-carbon bond(s) be found? a) Toluene b) 2-Ethylcyclopentene c) 4-Methyl-1,3-cyclohexadiene d) 3-Methyl-1-hexyne e) 3-Methyl-1,3-hexadiene Answer: D

Topic: Nomenclature, Bonding and Resonance Section: 14.4 and 14.6

88) In which of the following compounds would the longest carbon-carbon bond(s) be found? a) 2-Bromobenzaldehyde b) Vinylbenzene c) 1,3,5-Heptatriene d) 2,4,6-Octatriene e) 2-Ethylbenzoic acid Answer: D

Topic: Nomenclature, Bonding and Resonance Section: 14.4 and 14.6

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89) Which of the following statements regarding the cyclopentadienyl radical is correct? a) It is aromatic.

b) It is not aromatic. c) It obeys Huckel's rule.

d) It undergoes reactions characteristic of benzene. e) It has a closed shell of 6 pi-electrons.

Answer: B

Topic: Aromatic/ Antiaromatic/ Nonaromatic Section: 14.7

90) Which of the following statements regarding the cycloheptatrienyl radical is correct? a) It is aromatic.

b) It is not aromatic. c) It obeys Hückel's rule.

Answer: B

91) Which of the following statements regarding the cyclopentadienyl cation is correct? a) It is aromatic.

b) It is not aromatic. c) It obeys Hückel’s rule.

Answer: B

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92) Which of the following statements regarding the cyclopropenyl anion is correct? a) It is aromatic.

Answer: B

93) Which of the following statements regarding the cyclopropenyl radical is correct? a) It is aromatic.

Answer: B

94) Which of the following statements regarding the cycloheptatrienyl anion is correct? a) It is aromatic.

Answer: B

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95) Of Hückel’s requirements for aromatic character, only this one is waived in the case of certain compounds considered to be aromatic.

a) The ring system must be planar. b) The system must be monocyclic. c) There must be (4n + 2) π electrons.

d) The Hückel number of electrons must be completely delocalized. e) None. All of these rules must apply in every case.

Answer: B

Topic: Aromaticity, Hückel’s Rule Section: 14.7

96) Which of the following would you expect to be aromatic?

I II III IV V H H a) I b) II c) III d) IV e) V Answer: E

Topic: Aromaticity, Huckel’s Rule Section: 14.7

I II III IV H H 2+ V a) I b) II 39

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c) III d) IV e) V Answer: E

IV I II III H H a) I b) II c) III d) IV

99) Which of the following would you expect to show aromatic character?

e) None of these choices.

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Answer: B

IV I II III a) I b) II c) III d) IV

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103) Which of the following structures would be aromatic?

a) I b) II c) III d) IV e) V Answer: B

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104) Which of the following would you expect to be aromatic? IV I II III H H a) I b) II c) III d) IV

e) None of these choices. Answer: C

106) On the basis of molecular orbital theory and Hückel’s rule, which molecules and/or ions should be aromatic?

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IV I II III V H H H H N H H H H VI VII VIII IX a) I and V b) III and VIII c) IV, VII and IX d) IV, VI, VII and IX e) All of the structures, I-IX Answer: D

107) On the basis of molecular orbital theory and Hückel’s rule, which molecules and/or ions should be antiaromatic? IV I II III V H H H H N H H H H VI VII VIII IX a) I and V b) I, II and V c) III and VIII

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d) IV, VII and IX e) IV, VI, VII and IX Answer: B

108) On the basis of molecular orbital theory and Hückel’s rule, which molecules and/or ions should be non-aromatic? IV I II III V H H H H N H H H H VI VII VIII IX a) I and V b) I and II c) III and VIII d) IV, VII and IX e) IV, VI, VII and IX Answer: C

109) On the basis of molecular orbital theory and Hückel’s rule, which of these compounds should be aromatic?

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H H I II III IV V a) I b) II c) III d) IV e) V Answer: E

110) On the basis of molecular orbital theory and Hückel’s rule, which of these compounds should be aromatic? a) I b) II c) III d) IV e) V Answer: E

111) On the basis of molecular orbital theory and Hückel’s rule, which of these compounds should be aromatic?

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a) I b) II c) III d) IV e) V Answer: E

112) Which of these is an aromatic molecule?

I II III IV V N O B S CH₃ CH₃ O N H a) I b) II c) III d) IV e) V Answer: A

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a) I b) II c) III d) IV e) V Answer: A

115) Which of these is an aromatic molecule? 48

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116) Which of these is not an aromatic molecule?

a) I b) II c) III d) IV e) V Answer: D

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117) Of the following C-10 compounds, which is expected to possess the greatest resonance (delocalization) energy? I II III IV V a) I b) II c) III d) IV e) V Answer: C

Topic: Aromaticity, Hückel’s Rule Section: 14.7 and 14.8

118) Which of the following would you expect to be antiaromatic?

IV I II III V a) I b) II c) III d) IV e) V Answer: C

119) Which of these would you expect to have significant resonance stabilization energy?

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I II III N N H a) I b) II c) III

d) All of these choices. e) None of these choices. Answer: D

Topic: Aromaticity, Huckel’s Rule Section: 14.7 and 14.9

O N I II _III a) I b) II c) III

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a) I b) II c) III

S N I II _III a) I b) II c) III

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CH N I II _III a) I b) II c) III

124) Which compound would you NOT expect to be aromatic?

IV I II III N V S N O B H R a) I b) II c) III d) IV e) V Answer: E

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a) I b) II c) III d) IV e) V Answer: E

IV I II III N S N O H V a) I b) II c) III d) IV e) V Answer: E

127) Which of these species is aromatic?

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IV I II III N H V O N H O O O S a) I b) II c) III d) IV e) V Answer: B

128) In which case is the indicated unshared pair of electrons NOT a contributor to the π aromatic system? I II III IV N O H N H a) I b) II c) III d) IV

129) In which case is the indicated unshared pair of electrons NOT a contributor to the π aromatic system?

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S I _II _III _IV O N H a) I b) II c) III d) IV

N H NH O O O O S O I II III IV V a) I b) II c) III d) IV e) V Answer: B

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N H NH O O O S I III IV V O II a) I b) II c) III d) IV e) V Answer: B

N H NH O O O S I III IV V S O II a) I b) II c) III d) IV e) V Answer: B

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N H NH O O O S I III IV V N H II O a) I b) II c) III d) IV e) V Answer: B

N H NH O O O S I III IV V N H N II O NH2 a) I b) II c) III d) IV e) V Answer: E

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O I II III IV V a) I b) II c) III d) IV e) V Answer: C

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O I II _III IV V a) I b) II c) III d) IV e) V Answer: C

138) Which annulene would you NOT expect to be aromatic? a) [6]-Annulene b) [14]-Annulene c) [16]-Annulene d) [18]-Annulene e) [22]-Annulene Answer: C

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N O I CH3 II III IV V O B S CH3 O N H a) I b) II c) III d) IV e) V Answer: A

140) On the basis of molecular orbital theory and Hückel’s rule, which of these compounds should be aromatic? H H I II III IV O V a) I b) II c) III d) IV e) V Answer: E

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141) How many equivalent resonance structures can be written for the cyclopentadienyl anion? a) 3 b) 4 c) 5 d) 6 e) 7 Answer: C

Topic: Resonance Forms Section: 14.7

142) How many equivalent resonance structures can be written for the cycloheptatrienyl cation? a) 3 b) 4 c) 5 d) 6 e) 7 Answer: E

143) How many equivalent resonance structures can be written for the cyclopropenyl cation? a) 1 b) 2 c) 3 d) 4 e) 5 Answer: C

144) How many equivalent resonance structures can be written for the cyclobutadienyl dication? 62

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a) 1 b) 2 c) 3 d) 4

e) No equivalent resonance forms can be drawn. Answer: D

145) How many resonances would be expected in the broad-band decoupled C-13 spectra of:

a) 1 b) 2 c) 3 d) 4 e) 5 Answer: D Topic: Spectroscopy Section: 14.11

a) 1 b) 2 c) 3 d) 4

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e) 5 Answer: C

Topic: Spectroscopy Section: 14.11

a) 1 b) 2 c) 3 d) 4 e) 5 Answer: E Topic: Spectroscopy Section: 14.11

a) 2 b) 3 c) 4 d) 5 e) 6 Answer: E Topic: Spectroscopy 64

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Section: 14.11

149) How many resonances would be expected in the broad-band decoupled C-13 spectra of: CHO Cl a) 2 b) 3 c) 4 d) 5 e) 6 Answer: D Topic: Spectroscopy Section: 14.11

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151) How many resonances would be expected in the broad-band decoupled C-13 spectra of: a) 2 b) 3 c) 4 d) 5 e) 6 Answer: B Topic: Spectroscopy Section: 14.11

153) How many resonances would be expected in the broad-band decoupled C-13 spectra of: 66

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a) 4 b) 5 c) 6 d) 7 e) 8 Answer: A Topic: Spectroscopy Section: 14.11

a) 4

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b) 5 c) 6 d) 7 e) 8 Answer: B Topic: Spectroscopy Section: 14.11

a) 4 b) 5 c) 6 d) 7 e) 8 68

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Answer: E

158) Which of these compounds absorbs at the longest wavelength in the UV-visible region?

I II III IV V a) I b) II c) III d) IV e) V Answer: B Topic: Spectroscopy Section: 14.11

159) A compound has the formula C8H9Br. Its 1H NMR spectrum consists of: doublet, δ 2.0

quartet, δ 5.15 multiplet, δ 7.35

The IR spectrum shows two peaks in the 680-840 cm-1 region; one is between 690 and 710 cm-1 and the other is between 730 and 770 cm-1. Which is a possible structure for the compound?

I Br II Br III Br IV Br V Br a) I b) II 69

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c) III d) IV e) V Answer: D Topic: Spectroscopy Section: 14.11

160) Which of the following substances, formula C8H9Br, might exhibit the following 1H NMR spectrum? triplet, δ 2.8 triplet, δ 4.65 multiplet, δ 7.2 I Br II Br III Br IV Br V Br a) I b) II c) III d) IV e) V Answer: C Topic: Spectroscopy Section: 14.11

161) Which of the following substances, formula C8H9Br, might exhibit the following 1H NMR spectrum?

triplet, δ 1.2 quartet, δ 2.45 multiplet, δ 7.4

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I Br II Br III Br IV Br V Br a) I b) II c) III d) IV e) V Answer: A Topic: Spectroscopy Section: 14.11

162) Which of the following substances, formula C8H9Br, might exhibit the following 1H NMR spectrum? singlet, δ 2.45 singlet, δ 2.5 multiplet, δ 7.35 I Br II Br III Br IV Br V Br a) I b) II c) III d) IV e) V Answer: E 71

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163) Which of the following substances, C8H9Cl, would exhibit five signals in its 13C NMR spectrum? I II Cl III IV Cl V Cl Cl Cl a) I b) II c) III d) IV e) V Answer: C Topic: Spectroscopy Section: 14.11

164) Which of the following substances, C9H10O, would exhibit seven signals in its 13C NMR spectrum? I II III O IV V O O HO O a) I b) II 72

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c) III d) IV e) V Answer: A Topic: Spectroscopy Section: 14.11

165) For which of the following substances, C9H10O, would the aromatic multiplet, in its 1H NMR spectrum, consist of two doublets?

I II III O IV V O O HO O a) I b) II c) III d) IV e) V Answer: A Topic: Spectroscopy Section: 14.11

166) In the 1H NMR spectra of which of the following substances, C9H10O, would you expect to see two singlets in the range of δ 2.0-2.5 ppm?

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I II III O IV V O O HO O a) I and III b) I, II, and III c) IV

d) V

e) All of these substances would exhibit two singlets δ 2.0-2.5 ppm in their 1

H NMR spectra. Answer: B

167) Which isomer of C7H7Cl exhibits strong IR absorbances at 690 and 750 cm1?

I II III IV CH3 Cl CH3 Cl CH3 Cl CH2Cl a) I b) II c) III d) IV e) II and IV Answer: E Topic: Spectroscopy Section: 14.11

168) Which reagent(s)/technique would serve to distinguish between azulene and bicyclo[5.3.0]decane?

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a) Br2, CCl4

b) KMnO4, OH−, H2O c) NMR Spectroscopy d) UV Spectroscopy e) Two of these choices. Answer: E

Topic: Chemical and Spectroscopic Analysis Section: 13.9, 14.7 and 14.11

169) Which reagent(s)/technique would serve to distinguish between azulene and bicyclo[5.3.0]decane?

a) Br2, CCl4

b) KMnO4, OH−, H2O c) NMR Spectroscopy d) NaBH4, H2O

e) Two of these choices. Answer: C

Topic: Chemical and Spectroscopic Analysis Section: 14.7 and 14.11

Question type: fill-in-the-blank

170) A compound that reacts like an alkane, alkene, alkyne, or one of their derivatives, is called an ___ compound, while one that reacts like benzene is called an ___ compound.

Answer: aliphatic, aromatic Topic: General Concepts Section: 14.1

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171) Humans do not have the biochemical ability to synthesize the benzene ring. Therefore, two amino acids, ___ and ___, are essential to the human diet.

Answer: tryptophan, phenylalanine Topic: General Concepts

Section: 14.10

172) Cagelike molecules with the geometry of a truncated icosahedron are called ___. Answer: fullerenes

Topic: General Concepts Section: 14.8

173) Monocyclic compounds with alternating single and double bonds are called ___. Answer: annulenes

174) The difference between the amount of heat actually released upon hydrogenation of benzene and that calculated on the basis of the Kekule structure is called the ___ of benzene. Answer: resonance energy

175) Molecular orbitals of equal energy are referred to as ___ orbitals. Answer: degenerate

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176) When a benzene ring is used as a substituent, it is called a ___ group. Answer: phenyl

Topic: Nomenclature Section: 14.2

Question type: Molecular Drawing

177) Draw the structure corresponding to the following name: 2-bromo-5-nitrophenol

Answer: OH Br O2N Topic: Nomenclature Section: 14.2

178) Draw the structure corresponding to the following name: 3-bromo 5-phenylaniline Answer: NH₂ Br Topic: Nomenclature Section: 14.2

179) Draw the structure corresponding to the following name: 3-nitro-4-iodoanisole

Answer:

O O₂N

I

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Topic: Nomenclature Section: 14.2

Question type: Essay

180) Give the IUPAC name for the following substance:

O OH

Answer: 2-methyl-4-(3-methylbutyl)benzoic acid Topic: Nomenclature

Section: 14.2

181) Draw the structure corresponding to the following name: 2-hydroxy-4-(3-methylcyclopentyl)benzoic acid Answer: HO HO2C Topic: Nomenclature Section: 14.2

Question type: fill-in-the-blank

182) Hückel’s Rule requires ___ π electrons for an aromatic compound. Answer: 4n+2

Topic: Hückel’s rule Section: 14.7

Question type: Essay

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183) The heats of hydrogenation for cyclohexene and benzene are given below. Calculate the resonance stabilization energy of benzene.

C6H10 + H2 Æ C6H12 ΔHo = -120 kJ/mol C6H6 + 3H2 Æ C6H12 ΔHo = -208 kJ/mol

Answer: The resonance stabilization energy of benzene is the difference between the theoretical and observed ΔHo

values.

Hydrogenation of three double bonds should release 3 times the energy released when one double bond is hydrogenated. Thus, the theoretical ΔHo

for benzene can be calculated as [3x (-120)] kJ/mol = -360 kJ/mol

Therefore,

Resonance stabilization energy of benzene = (-208) – (-360) kJ/mol = 152 kJ/mol

Topic: Aromaticity, Resonance Stabilization Energy Section: 14.5

184) Use the “polygon-in-circle” method to draw an MO diagram showing the pi-electron distribution pattern in the cyclopentadienyl anion. How is this diagram useful in explaining the aromatic properties of this anion?

Answer: The following MO diagram can be generated using the polygon-in-circle method. The six pi electrons of the cyclopentadienyl anion in the bonding molecular orbitals form a closed bonding shell, accounting for its aromatic properties.

Bonding MO Antibonding MO

6 pi electrons "closed bonding shell" Topic: Aromaticity, Molecular Orbital Theory Section: 14.7

185) Use the “polygon-in-circle” method to draw an MO diagram showing the pi-electron distribution pattern in the cyclopentadienyl cation. How is this diagram useful in explaining the antiaromatic properties of this anion?

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Answer: This MO diagram can be generated using the polygon-in-circle method. The 4 pi electrons of the cyclopentadienyl cation in the bonding molecular orbitals form a bonding shell which lacks 2 pi electrons from being closed, accounting for its antiaromatic properties.

Topic: Aromaticity, Molecular Orbital T heory Section: 14.7

186) Use the “polygon-in-circle” method to draw an MO diagram showing the pi-electron distribution pattern in the cycloheptatrienyl cation. How is this diagram useful in explaining the antiaromatic properties of this anion?

Answer: This MO diagram can be generated using the polygon-in-circle method. The six pi electrons of the cycloheptatrienyl cation in the bonding molecular orbitals form a closed bonding shell, accounting for its aromatic properties

Topic: Aromaticity, Molecular Orbital Theory Section: 14.7

187) The pka of cyclopentadiene is found to be 16 while, by contrast, that of cycloheptatriene is found to be 36. Explain the large difference found in the two pka values.

Answer: The 20 order of magnitude difference in acidity of cyclopentadiene compared with cycloheptatriene lies in the fact that the conjugate base of cyclopentadiene is the aromatic cyclopentadienyl anion, a very stable conjugate base species. However, in the case of

cycloheptatriene, removal of a proton produces the cyclohe[tatrienyl anion, a cyclic, conjugated, planar system with 8 π-electrons, that is antiaromatic.

H H B:− H .. Aromatic anion 80

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H H

B:

-H

Topic: Aromaticity, Molecular Orbital Theory Section: 14.7

188) Benzene, while unusually unreactive with electrophiles, will react under certain conditions. However, when it does react, it does so by ___ rather than by addition.

Answer: substitution

Topic: Aromaticity, Reactivity Section: 14.3

189) Explain briefly why cyclopentadiene readily reacts with strong bases.

Answer: The acidity of a substance can be often be correlated with the stability of its conjugate base: substances that have exceptionally stable conjugate bases tend to be strong acids. The conjugate base of cyclopentadiene, formed during reaction with a base, is exceptionally stable, by virtue of being an aromatic anion (6 pi electrons, planar ring), thus accounting for the readiness with which cyclopentadiene reacts with bases.

H H B:− H .. Aromatic anion Topic: Aromaticity, Reactivity

Section: 14.7

190) Cyclohepatriene is found to readily undergo a hydride transfer with triphenylmethyl arbocation to generate the cycloheptatrienyl cation and triphenylmethane. Explain why this transformation is facile.

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H H

+ Ph₃C+

H

+ Ph3CH

Answer: This reaction readily proceeds because the product, cycloheptatrienyl cation, is aromatic and much more stable than even the triphenyl substituted carbocation.

191) Pyrrole is not particularly basic, because the lone pair on the nitrogen is ___. Answer: part of the aromatic system

192) The Lewis structures of both pyridine and pyrrole have an electron pair on the nitrogen atom, suggesting that both substances might possess basic properties. However, only pyridine reacts readily with HCl to form pyridinium chloride, while pyrrole is rather unreactive. Explain clearly, using relevant diagrams to illustrate/clarify your answer.

Answer: N .. sp2 + HCl N H + _Cl− pyridinium chloride N H .. p + HCl _X pyrrole electron pair:

part of aromatic pi sextet pyridinium chloride N H H not aromatic +Cl−

Both pyridine and pyrrole are known to exhibit aromatic properties; the nitrogen atom must therefore be sp2 hybridized in both cases.

In the pyridine molecule, this places the electron pair in an sp2 hybrid orbital, and not involved in the aromatic π sextet needed for aromatic stability; the electron pair is thus “available” for

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reacting with HCl, forming pyridinium chloride, while preserving aromatic stability.

By contrast, the electron pair in pyrrole must be placed in the p orbital, since all three sp2 orbitals are used in sigma bonds with neighboring atoms. This allows the electron pair to contribute to aromatic stability by becoming part of the aromatic sextet. Thus, the electron pair on nitrogen is not “available” for reacting with HCl, because doing so would destroy the aromaticity of the ring.

193) Predict the product of the reaction of one equivalent of hydrochloric acid with the organic compound shown: N H N 1 equivalent HCl Answer: N H N 1 equivalent HCl N H N H

194) Consider the nitrogen atom in the molecules of pyrrole and pyrrolidine. Bearing in mind that pyrrole exhibits aromatic properties, compare the hybridization of nitrogen in both species, explaining why it must be different, although at first glance, it may appear identical.

H

N HN

pyrrole pyrrolidine

Answer:

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NH _NH

sp2 _sp3

In pyrrole, the conjugated unsaturated ring system forces the nitrogen to be trigonal planar, that is sp2 hybridized, with the lone pair electrons in an unhybridized p-orbital. This allows for overlap with the conjugate system accounting for the aromatic character. The nitrogen in

pyrrolidine, on the other hand, is closer to tetrahedral, and as a result sp3 hybridized, and the lone pair electrons reside in an sp3 hybrid orbital.

195) Briefly explain why the aromatic hydrocarbon shown possesses a significant dipole moment. Use diagrams as needed to illustrate/clarify your answer.

Answer:

=

+

-The stable, low energy resonance contributor shown creates an aromatic system in both rings. This separation of charge creates the significant dipole observed.

Topic: Aromaticity Section: 14.7

196) Briefly explain why the aromatic hydrocarbon shown possesses a significant dipole moment. Use diagrams as needed to illustrate/clarify your answer.

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Answer:

The stable, low energy resonance contributor shown creates an aromatic system in both rings. This separation of charge creates the significant dipole observed.

Topic: Aromaticity Section: 14.7 and 14.8 Difficulty Level: Hard

197) Briefly explain why the aromatic hydrocarbon azulene, C10H8, possesses a significant dipole moment. Use diagrams as needed to illustrate/clarify your answer.

Answer: Azulene has the bicyclic structure shown below. Since the “non-polar” resonance hybrid meets the Hückel rule for aromaticity, it accounts for the observed aromatic properties of azulene. However, closer scrutiny suggests that a shift in electron density, in which the

7-membered ring is electron deficient and the 5-7-membered ring is electron-rich would be

especially stable: each ring is now independently aromatic, an aromatic cycloheptatrienyl cation fused with an aromatic cyclopentadienyl anion. Thus, the most stable pi-electron distribution would suggest a polar structure for azulene, accounting for this substance having a significant dipole moment.

non-polar _polar

Topic: Aromaticity Section: 14.7

198) Although all bond lengths in benzene are identical, that is not always true for all aromatic compounds. In naphthalene, for instance, the C1-C2 bond is shorter than the C2-C3 bond. Explain, using resonance theory.

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Answer: The major resonance structures for naphthalene are drawn below. The observed bond lengths reflect the contribution of each structure to the overall resonance hybrid:

The C1-C2 bond: two double bonds (I and II), and one single bond (III) The C2-C3 bond: one double bond (III), and two single bonds (I and II) Overall: C1-C2 bond is more double-bond-like, thus shorter.

II III IV I 1 2 3 1 2 3 1 2 3 1 2 3 shorter longer

Topic: Resonance, Bond Lengths, Bond Order Section: 14.7

Question type: Molecular Drawing

199) Draw all significant resonance structures for pyridine, C6H5N

Answer: N.. N..

Topic: Resonance Section: 1.8 and 14.9 Difficulty Level: Hard

200) Draw all significant resonance structures for azulene, C10H8

Answer:

201) Draw all significant resonance structures for the cyclopentadienyl anion. Answer:

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202) Consider the oxygen atom in the molecules of furan and tetrahydrofuran. Bearing in mind that furan exhibits aromatic properties, compare the hybridization of oxygen in both species, explaining why it must be different, although at first glance, it may appear identical.

O.. O.. .. .. THF Furan Answer: O Tetrahydrofuran: sp3 hybridized both electron pairs

in sp3 orbitals

O

.. p

Furan: sp2 hybridized

one electron pair in p orbital: part of aromatic pi sextet

.. sp2 ..

..

sp3 sp3

Oxygen has 2 lone electron pairs in both species and, at first glance, it might appear that the hybridization patterns are identical. In THF, the oxygen atom is sp3 hybridized and both pairs can be placed in sp3 hybrid orbitals. In furan, there are two possibilities that must be considered: i) the oxygen may be sp3 hybridized, with both electron pairs sp3 hybrid orbitals, as in THF, or, ii) oxygen may be sp2 hybridized, with one electron pair in an sp3 hybrid orbital, the other in a p orbital. In the latter case, the electrons in the p orbital can then become part of an aromatic sextet, allowing the molecule to gain aromatic stability. Since furan is known to exhibit markedly aromatic properties, the oxygen must be sp2 hybridized.

203) The compound illustrated below is found to have the following signals in its broadband decoupled C-13 spectra:

190.77, 140.91, 134.82, 130.89 and 129.45 δ

Using resonance theory, assign the C-13 peaks to the appropriate carbons.

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CHO Cl Answer: CHO Cl 190.77 140.91 129.45 130.89 134.82

Topic: Spectroscopy and Resonance Theory Section: 14.11