Vollhardt  Capítulo 15 (Benzenos e Aromaticidade)

Vollhardt Capítulo 15 (Benzenos e Aromaticidade)

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to
convert benzene into each of the following compounds.

(a)

C(CH3)3
A

 (b)

Cl
A

 (c) (d)

NO2
A

(e)

C
O

A
EN CH3

 (f)

CH2CH3
A

 (g)

SO3H
A

 (h)

Br
A

48. Give the expected major product of addition of each of the following reagent mixtures to benzene.
(Hint: Look for analogies to the reactions presented in this chapter.)

(a) Cl2 1 AlCl3
(b) T2O 1 T2SO4 (T 5 tritium,

3H)
(c) (CH3)3COH 1 H3PO4
(d) N2O5 (which tends to dissociate into NO2

1 and NO3
2)

(e) (CH3)2C P CH2 1 H3PO4 (f ) (CH3)3CCH2CH2Cl 1 AlCl3

(g)
A

(CH3)2CCH2CH2C(CH3)2 AlBr3
Br

A
Br

� (h) H3C COCl AlCl3�

49. Write mechanisms for reactions (c) and (f) in Problem 48.

50. Hexadeuteriobenzene, C6D6, is a very useful solvent for
1H NMR spectroscopy because it dis-

solves a wide variety of organic compounds and, being aromatic, is very stable. Suggest a method
for the preparation of C6D6.

51. Propose a mechanism for the sulfonation of benzene using chlorosulfuric acid, ClSO3H (in the
margin).

�

�
HCl

SO3H

S

O

O

OHClO O
B

B

P r o b l e m s

726 C h a p t e r 1 5 B e n z e n e a n d A r o m a t i c i t y

52. Benzene reacts with sulfur dichloride, SCl2, in the presence of AlCl3 to give diphenyl sulfi de,
C6H5 – S – C6H5. Propose a mechanism for this process.

53. (a) 3-Phenylpropanoyl chloride, C6H5CH2CH2COCl, reacts with AlCl3 to give a single product
with the formula C9H8O and an

1H NMR spectrum with signals at d 5 2.53 (t, J 5 8 Hz, 2 H),
3.02 (t, J 5 8 Hz, 2 H), and 7.2 – 7.7 (m, 4 H) ppm. Propose a structure and a mechanism for
the formation of this product.

 (b) The product of the process described in (a) is subjected to the following reaction sequence:
(1) NaBH4, CH3CH2OH; (2) Conc. H2SO4, 100 8C; (3) H2, Pd – C, CH3CH2OH. The fi nal product
exhibits fi ve resonance lines in its 13C NMR spectrum. What is the structure of the substance
formed after each of the steps in this sequence?

54. The text states that alkylated benzenes are more susceptible to electrophilic attack than is benzene
itself. Draw a graph like that in Figure 15-20 to show how the energy profi le of electrophilic
substitution of methylbenzene (toluene) would differ quantitatively from that of benzene.

55. Like haloalkanes, haloarenes are readily converted into organometallic reagents, which are sources
of nucleophilic carbon.

Br

Phenylmagnesium bromide
Grignard
reagents

MgBr
��

��Mg, (CH3CH2)2O, 25°C

Cl

Phenylmagnesium chloride

MgCl
��

��Mg, THF, 50°C

 The chemical behavior of these reagents is very similar to that of their alkyl counterparts. Write
the main product of each of the following sequences.

(a) C6H5Br

1. Li, (CH3CH2)2O
2. CH3CHO
3. H�, H2O (b) C6H5Cl

1. Mg, THF

2. H2C
O

CH2
3. H�, H2O

O
GD

56. Give effi cient syntheses of the following compounds, beginning with benzene. (a) 1-Phenyl-1-
heptanol; (b) 2-phenyl-2-butanol; (c) 1-phenyloctane. (Hint: Use a method from Section 15-14.
Why will Friedel-Crafts alkylation not work?)

57. Vanillin, whose structure is shown in the margin and is the subject of the Chapter Opening, is a
benzene derivative with several functional groups, each one of which displays its characteristic
reactivity. What would you expect to be the products of reaction of vanillin with each of the
following reagents?

(a) NaBH4, CH3CH2OH
(b) NaOH, then CH3I

 Vanillin is obtained by extraction from the seed pods of plants in the Vanilla genus in a
process that dates back at least 500 years: The Mexican Aztecs used it to fl avor xocoatl, a
chocolate drink. Cortez discovered it in the court of Montezuma and was responsible for introduc-
ing it to Europe. Increasing demand for vanillin necessitated development of synthetic procedures,
which involve extraction of related compounds from other plant-derived sources. One of the most
important of these sources is the wood-derived waste from the manufacture of paper. The conver-
sion of eugenol (an extract from cloves) into vanillin is very similar chemically. First, treatment

Vanillin

CH3O
OH

H O

Vanillin

CH3O
OH

H O

 C h a p t e r 1 5 727

of eugenol with KOH at 150 8C in a high-boiling solvent causes positional isomerization of its
side-chain double bond:

Eugenol

CH2CHPCH2

CH3O

KOH, 150°C, 1.5 h

OH

CHPCHCH3

CH3O
OH

 Subsequently, oxidative cleavage, (see Section 12-12) completes the synthesis of vanillin.

(c) Propose a mechanism for the isomerization depicted in the scheme.

58. Because of cyclic delocalization, structures A and B shown here for o-dimethylbenzene (o-xylene)
are simply two resonance forms of the same molecule. Can the same be said for the two dimethyl-
cyclooctatetraene structures C and D? Explain.

CH3

A

CH3
CH3

B

CH3 CH3

CH3

C

CH3

CH3

D

59. The energy levels of the 2-propenyl (allyl) and cyclopropenyl p systems (see margin) are com-
pared qualitatively in the diagram below. (a) Draw the three molecular orbitals of each system,
using plus and minus signs and dotted lines to indicate bonding overlap and nodes, as in Fig-
ure 15-4. Does either of these systems possess degenerate molecular orbitals? (b) How many
p electrons would give rise to the maximum stabilization of the cyclopropenyl system, relative
to 2-propenyl (allyl)? (Compare Figure 15-5, for benzene.) Draw Lewis structures for both sys-
tems with this number of p electrons and any appropriate atomic charges. (c) Could the cyclo-
propenyl system drawn in (b) qualify as being “aromatic”? Explain.

E

Antibonding molecular orbital

Bonding molecular orbital

Nonbonding molecular orbital

 Cyclopropenyl 2-Propenyl (Allyl)

60. 2,3-Diphenylcyclopropenone (see structure in the margin) forms an addition product
with HBr that exhibits the properties of an ionic salt. Suggest a structure for this product and a
reason for its existence as a stable entity.

61. Is cyclobutadiene dication (C4H4
21) aromatic according to Hückel’s rule? Sketch its

p molecular orbital diagram to illustrate.

62. All the molecules shown below are examples of “fulvenes,” or methylenecyclopentadienes.

5-Methylene-
1,3-cyclopentadiene

“Fulvene”

6-Dimethylamino-
fulvene

N(CH3)2

6,6-Dimethyl-
fulvene

CH3H3C

6,6-Diphenyl-
fulvene

C6H5C6H5

2-Propenyl
(Allyl)

Cyclopropenyl

2-Propenyl
(Allyl)

Cyclopropenyl

O

2,3-Diphenylcyclopropenone
C6H5 C6H5

O

2,3-Diphenylcyclopropenone
C6H5 C6H5

P r o b l e m s

728 C h a p t e r 1 5 B e n z e n e a n d A r o m a t i c i t y

 (a) One of these structures is considerably more acidic than the others, with a pKa around 20.
Identify it and its most acidic hydrogen(s), and explain why it is an unusually strong acid for a
molecule with only carbon-hydrogen bonds.

 (b) None of the 7-methylene-1,3,5-cycloheptatrienes that correspond to the fulvene structures
above show any unusual acidity. Explain.

63. A characteristic reaction of fulvenes is nucleophilic addition. To which carbon in a fulvene would
you expect nucleophiles to add, and why?

64. (a) The 1H NMR spectrum of [18]annulene shows two signals, at d 5 9.28 (12 H)
and 22.99 (6 H) ppm. The negative chemical shift value refers to a resonance upfi eld (to the
right) of (CH3)4Si. Explain this spectrum. (Hint: Consult Figure 15-9.) (b) The unusual molecule
1,6-methano[10]annulene (shown in the margin) exhibits two sets of signals in the 1H NMR
spectrum at d 5 7.10 (8 H) and 20.50 (2 H) ppm. Is this result a sign of aromatic character?

65. The 1H NMR spectrum of the most stable isomer of [14]annulene shows two signals, at
d 5 20.61 (4 H) and 7.88 (10 H) ppm. Two possible structures for [14]annulene are shown
here. How do they differ? Which one corresponds to