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Chapter 18
Aromatic Compounds
Review of Concepts
Fill in the blanks below. To verify that your answers are correct, look in your textbook at
the end of Chapter 18. Each of the sentences below appears verbatim in the section
entitled Review of Concepts and Vocabulary.
• The stability of benzene can be explained with MO theory. The six π electrons all
occupy ___________ MOs.
• The presence of a fully conjugated ring of π electrons is not the sole requirement
for aromaticity. The requirement for an odd number of electron pairs is called
___________ rule.
• Frost circles accurately predict the relative energy levels of the ________ in a
conjugated ring system.
• Alkyl benzenes are oxidized at the benzylic position by _____________ or
______________________.
• In a Birch reduction, the aromatic moiety is reduced to give a nonconjugated
diene. The carbon atom connected to ___________ is not reduced, while the
carbon atom connected to _________________________ is reduced.
CHAPTER 18
411
Review of Skills
Fill in the blanks and empty boxes below. To verify that your answers are correct, look
in your textbook at the end of Chapter 18. The answers appear in the section entitled
SkillBuilder Review.
18.1 Naming a Polysubstituted Benzene
OH
Cl
Br
PROVIDE A SYSTEMATIC NAME FOR THE FOLLOWING COMPOUND
1) IDENTIFY THE PARENT
2) IDENTIFY AND NAME SUBSTITUENTS
3) ASSIGN LOCANTS TO EACH SUBSTITUENT
4) ALPHABETIZE
18.2 Determining Whether a Compound is Aromatic, Nonaromatic, or Antiaromatic
18.3 Determining Whether a Lone Pair Participates in Aromaticity
N
N
H
NH
H
IN THE FOLLOWING COMPOUND, IDENTIFY WETHER EACH LONE PAIR PARTICPATES IN AROMATICITY:
18.4 Manipulating the Side Chain of an Aromatic Compound
412
CHAPTER 18
18.5 Predicting the Product of a Birch Reduction
Review of Reactions
Identify the reagents necessary to achieve each of the following transformations. To
verify that your answers are correct, look in your textbook at the end of Chapter 18. The
answers appear in the section entitled Review of Reactions.
OH
O
Br
Br + EtOH + NaBr
CHAPTER 18
413
Solutions
18.1.
a) 3-isopropylbenzaldehyde or meta-isopropylbenzaldehyde
18.2.
18.3.
a)
OMe
C
l
BrBr
b)
OH
NO
2
18.4.
a) meta-xylene
414
CHAPTER 18
18.6.
Compound A
(C
8
H
8
)
Compound B
(C
8
H
8
Br
2
)
Br
Br
18.7.
18.8.
18.9. The cyclopropenyl cation is expected to exhibit aromatic stabilization.
Antibonding
Bonding
CHAPTER 18
415
18.12. Cyclopentadiene is more acidic because its conjugate base is highly stabilized.
Deprotonation of cyclopentadiene generates an anion that is aromatic, because it is a
continuous system of overlapping p orbitals containing 6 π electrons. In contrast,
deprotonation of cycloheptatriene gives an anion with 8 π electrons.
18.13. The first step of an S
N
1 process is loss of a leaving group, forming a carbocation,
so we compare the carbocations that would be formed.
18.14. The first compound is more acidic because deprotonation of the first compound
generates a new (second) aromatic ring. Deprotonation of the second compound does not
introduce a new aromatic ring:
18.15.
a) One of the lone pairs on oxygen
b) One of the lone pairs on sulfur
416
CHAPTER 18
18.16.
N
OH OH
OH
O
F
O
NH
N
H
N
S
N
N
Lipitor
TM
Zyprexa
TM
18.17. The first compound is expected to be more acidic (has a lower pK
a
), because
deprotonation restores aromaticity to the ring. The second compound is already aromatic,
even before deprotonation.
18.18.
a) Yes, it has the required pharmacophore (two aromatic rings separated by one
18.19.
a)
OH
O
b)
OH
O
HO
O
c)
OH
O
HO
O
CHAPTER 18
417
b)
NBS
Br
NaOEt
heat
18.21.
NBS
heat
Br Mg MgBr
18.22.
NBS Br NaOEt O
heat
1) O
3
2) DMS
418
CHAPTER 18
18.24.
18.25.
OCH
3
OCH
3
OCH
3
OCH
3
OCH
3
18.26.
a)
O
b) acetophenone c)
O
18.27.
18.28.
a) 4-ethylbenzoic acid or para-ethylbenzoic acid
CHAPTER 18
419
18.29.
OH
18.30.
420
CHAPTER 18
18.32.
Cl
Cl Cl
Cl
Cl
Cl
18.33.
O
2
N
O
2
N
NO
2
NO
2
O
2
N NO
2
NO
2
NO
2
O
2
N
O
2
N
O
2
N
NO
2
O
2
N
O
2
N
NO
2
2,3,4-
trinitrotoluene
2,3,5-
trinitrotoluene
2,3,6-
trinitrotoluene
2,4,5-
trinitrotoluene
3,4,5-
trinitrotoluene
18.35.
a) benzene b) benzene c) benzene
CHAPTER 18
421
18.36. a) Yes b) No c) No d) Yes e) No
18.37.
O
aromatic
18.38.
a) Nonaromatic. The lone pairs on the oxygen atom will remain in sp
3
hybridized
orbitals in order to avoid anti-aromaticity.
b) Nonaromatic. The lone pair on the nitrogen atom will remain in an sp
3
hybridized
orbital in order to avoid anti-aromaticity.
18.39.
a)
O
Cl
Loss of the leaving group generates an aromatic cation.
b)
Cl
Loss of the leaving group generates an antiaromatic cation.
422
CHAPTER 18
18.41. The second compound is a stronger base, because the lone pair on the nitrogen
18.42. Six π electrons are required in order to achieve aromaticity. This cation only has
four electrons.
18.45. Steric hindrance forces the rings out of coplanarity.
18.46. Benzene does not have three C-C single bond and three C-C double bonds. In
fact, all six C-C bonds of the ring have the same bond order are the same length.
18.47.
a)
NBS
heat or light
B
r
b)
Na
2
Cr
2
O
7
H
2
SO
4
, H
2
O
O
H
O
CHAPTER 18
423
18.48.
a) 6 b) 5 c) 3 d) 9
18.49.
Me
Me
Me
H
HH
Me
H
H
O
H
3
C
18.50. meta-Xylene.
18.51.
a) The first compound would lack C-H stretching signals just above 3000 cm
-1
, while the
second compound will have C-H stretching signals just above 3000 cm
-1
.
18.52. When either compound is deprotonated, an aromatic anion is generated, which
can be drawn with five resonance structures. The resulting anion is the same in either
case.
18.53. In cycloheptatrienone, the resonance structures with C+ and O- contribute
424
CHAPTER 18
18.54.
a) Each of the rings in the following resonance structure is aromatic.
18.55.
Br 1) Mg
2)
H
O
3) H
2
O
H
2
SO
4
, heat
OH
18.56.
OH
Cl
18.57.
O
O
CHAPTER 18
425
18.58.
H
O
1) NBS, heat
b)
O
1) NBS, heat
2) NaOEt
18.59.
OH
O
426
CHAPTER 18
18.61.
a) The second compound holds greater promise as a potential antihistamine, because it
18.62. No, this compound possesses an allene moiety (C=C=C). The p orbitals of one
18.63.
Compound A Compound B Compound C Compound D
18.64. The nitrogen atom in compound A is localized and is not participating in
resonance. The nitrogen atom in compound B is delocalized, and some of the resonance
18.65.
O
1) NBS, heat
2) NaOEt
1) O
3
2) DMS
CHAPTER 18
427
18.66.
NBS
heat
Br HC CNa
