Chapter 19
Aromatic Substitution Reactions
Review of Concepts
Fill in the blanks below. To verify that your answers are correct, look in your textbook at
the end of Chapter 19. Each of the sentences below appears verbatim in the section
entitled Review of Concepts and Vocabulary.
• Electrophilic aromatic substitution involves two steps:
o Formation of the __________ complex, or arenium ion.
o Deprotonation, which restores _________________.
• Sulfur trioxide (SO
3
) is a very powerful ____________ that is present in fuming
sulfuric acid. Benzene reacts with SO
3
in a reversible process called __________.
• A mixture of sulfuric acid and nitric acid produces the nitronium ion (NO
2+
).
director.
• All activators are ______-______directors
• A nitro group deactivates an aromatic ring and is a ______director.
• Most deactivators are ______directors.
• Strong activators are characterized by the presence of a __________________
• An elimination-addition reaction occurs via a ____________intermediate.
CHAPTER 19
429
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 19. The answers appear in the section entitled
SkillBuilder Review.
19.1 Identifying the Effects of a Substituent
19.2 Identifying Directing Effects for Disubstituted and Polysubstituted Benzene Rings
19.3 Identifying Steric Effects for Disubstituted and Polysubstituted Aromatic Benzene Rings
19.4 Using Blocking Groups to Control the Regiochemical Outcome
of an Electrophilic Aromatic Substitution Reaction
430
CHAPTER 19
19.5 Proposing a Synthesis for a Disubstituted Benzene Ring
NO
2
IDENTIFY A THREE-STEP PROCESS FOR ACHIEVING THE FOLLOWING TRANSFORMATION:
1)
2)
3)
19.6 Proposing a Synthesis for a Polysubstituted Benzene Ring
19.7 Determining the Mechanism of an Aromatic Substitution Reaction
INDICATE THE MECHANISM THAT OPERATES IN EACH OF THE THREE SCENARIOS SHOWN IN THE FOLLOWING DECISION TREE:
CHAPTER 19
431
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 19. The
answers appear in the section entitled Review of Reactions.
Electrophilic Aromatic Substitution
Nucleophilic Aromatic Substitution
Br
NO
2
OH
NO
2
432
CHAPTER 19
Solutions
19.1.
I
H
HIHIHI
I
Base
CHAPTER 19
433
19.3.
H
D
D O S O D
O
OO SO
19.4.
O
NOH
O
O
O
O S O HH
O
N
O
O
H
H
N
O
O
– H
2
O
nitronium ion
19.5.
434
CHAPTER 19
19.6.
AlCl
3
Cl
Cl
Cl Cl
Cl AlCl
3
H
CHAPTER 19
435
19.7.
H O S
O
O
O H
19.8.
a) It is necessary to perform an acylation followed by a Clemmensen reduction to avoid
carbocation rearrangements.
19.9. It cannot be made via alkylation because the carbocation required would undergo a
methyl shift to give a tertiary carbocation. It cannot be made via acylation followed by a
436
CHAPTER 19
19.10.
R O
O
R
OAlCl
3
AlCl
3
RCO
R O
O
R
O
19.11.
Br
+
B
r
CHAPTER 19
437
b)
O
NOH
O
O
O
O S O HH
O
N
O
O
H
H
N
O
O
– H
2
O
19.13. As show below, attack at C4 or C6 produces a sigma complex in which two of the
resonance structures have a positive charge next to an electron-withdrawing group (NO
2
).
These resonance structures are less contributing to the resonance hybrid, thereby
destabilizing the sigma complex. In contrast, attack at C5 produces a sigma complex for
which none of the resonance structures have a positive charge next to a nitro group.
NO
2
H NO
2
O
2
N
NO
2
N OO
O
2
N
NO
2
H NO
2
O
2
N
NO
2
H NO
2
O
2
N
ATTACK
AT C4
438
CHAPTER 19
19.14. The chlorine atom in chlorobenzene deactivates the ring relative to benzene. If
19.15. Ortho attack and para attack are preferred because each of these pathways
involves a sigma complex with four resonance structures (shown below). Attack at the
meta position involves formation of a sigma complex with only three resonance
structures, which is not as stable as a sigma complex with four resonance structures. The
19.16.
a) The nitro is strongly deactivating and meta-directing.
b) An acyl group is moderately deactivating and meta-directing.
c) A bromine atom weakly deactivating and ortho, para-directing.
CHAPTER 19
439
19.18.
N
O
A
C
19.19.
a)
O
2
N
CH
3
NO
2
b)
CH
3
NO
2
O
2
N
c)
O
OO
O
19.20.
a)
OH OO
HNO
3
OH OO
440
CHAPTER 19
b)
OMeO
Br Br
2
FeBr
3
OMeO
Br
B
r
19.21.
Br Br
2
Fe
Br Br
B
r
Br
19.22.
19.23. All three available positions are sterically hindered.
19.24.
Br
2
FeBr
3
OO
Br
CHAPTER 19
441
19.26.
19.27.
a) The nitro group must be installed in a position that is meta to each of the OH groups.
19.28.
a) Cl
2
, AlCl
3
19.29.
a)
SO
3
H
b)
NO
2
c)
Cl
d)
Et
e)
Br
f)
NH
2
19.30.
a)
1) Br
2
, AlBr
3
442
CHAPTER 19
c)
d)
NH
2
Cl
1) HNO
3
, H
2
SO
4
2) Cl
2
, AlCl
3
3) HCl, Zn
g)
1) AlCl
3,
Cl
O
2) HCl, Zn[Hg], heat
3) Fuming H
2
SO
4
4) CH
3
Cl, AlCl
3
5) Dilute H
2
SO
4
CHAPTER 19
443
j)
1) CH
3
CH
2
COCl, AlCl
3
2) HCl, Zn[Hg], heat
3) CH
3
CH
2
COCl, AlCl
3
4) HCl, Zn[Hg], heat
19.31. The para product will be more strongly favored over the ortho product if the
tert-butyl group is installed first. The steric hindrance provided by a tert-butyl group is
greater than the steric hindrance provided by an isopropyl group. Of the following two
possible pathways, the first should provide a greater yield of the desired product.
19.32.
a) Nitration cannot be achieved effectively in the presence of an amino group.
b) Each of the two alkyl groups is ortho–para directing, but the two groups are meta to
each other. A Friedel-Crafts acylation will not work in this case (see solution to problem
19.9)
19.33.
444
CHAPTER 19
b)
1) CH
3
CH
2
COCl, AlCl
3
c)
NH
2
Cl
Cl
2) HNO
3
, H
2
SO
4
3) HCl, Zn
4) excess Cl
2
1) (CH
3
)
3
Cl, AlCl
3
d)
19.34.
O OH
1) (CH
3
)
2
CHCl, AlCl
3
2) Fuming H
2
SO
4
19.35.
Br
NO
2
NaOCH
3
, heat
Cl
Br
NO
2
H
3
CO
CHAPTER 19
445
19.36.
1) Cl
2
, AlCl
3
19.37.
a) Each additional nitro group serves as a reservoir of electron density and provides for
19.38.
NH
2
+NH
2
446
CHAPTER 19
b)
BrBr FeBr
3
FeBr
3
BrBr
I
N
O O
NH
2
NH
2
N
O O
I
–
CHAPTER 19
447
19.41.
C
l
NaOH HO
H
O
O
H
heat + +
19.42.
a)
NO
2
ONa
19.43.
O
OH
O
Cl
O
AlCl
3
HCl, Zn[Hg], heat
EtCl
AlCl
3
KMnO
4
, NaOH
heat
448
CHAPTER 19
19.44.
Increasing Reactivity toward Electrophilic Aromatic Substitution
19.45.
NO
2
OMeHO
m
o
s
t
a
c
t
i
v
a
t
e
d
l
e
a
s
t
a
c
t
i
v
a
t
e
d
19.46.
19.47.
Cl
OH
HO
OH
NO
2
CHAPTER 19
449
19.48.
a) This group is an activator and an ortho,para-director.
b) This group is an activator and an ortho,para-director.
c) This group is an activator and an ortho,para-director.
19.49.
a)
Cl Cl
+
b) unreactive c) unreactive
h)
O
O
O
O+
19.50.
450
CHAPTER 19
19.51.
OHNO
3
/ H
2
SO
4
O
O
2
N NO
2
19.52.
H OSO
3
H
19.53.
a)
ClCl AlCl
3
ClCl AlCl
3
CHAPTER 19
451
b)
ONOH
O
O
O
O S O HH
ONO
O
H
H
N
O
O
– H
2
O
nitronium ion
c)
SO
O
O
H
O SO
3
HH SOH
O
O
452
CHAPTER 19
H
Al
Cl
Cl
Cl
ClCH
H
H
HH H
AlCl
3
Cl
+ HCl
+ AlCl
3
e)
Br
2
Fe
CHAPTER 19
453
19.54.
a)
ClIAlCl
3
ClIAlCl
3
454
CHAPTER 19
b)
C Cl
H
H
Cl
Al
Cl
Cl
Cl
AlCl
3
Cl
+ HCl
+ AlCl
3
Cl
Al
Cl
Cl
Cl
Cl
AlCl
3
H
CHAPTER 19
455
19.55.
O
Cl
OMe
O
Cl
OMe
O
OMe
b)
1) AlCl
3
,
2) Zn [ Hg], HCl, heat
Cl
O
d)
1) CH
3
Cl, AlCl
3
2) excess NBS Br
3
C
456
CHAPTER 19
b)
H
2
N Br
1) HNO
3
, H
2
SO
4
2) Br
2
, AlBr
3
3) HCl, Zn
19.58.
a) The second step of the synthesis will not work, because a strongly deactivated ring will
not undergo a Friedel-Crafts alkylation. The product of the first step, nitrobenzene, will
19.59.
a)
O
b)
N
O
CHAPTER 19
457
19.60.
Br
HNH
2
Br
NH
2
NH
2
HNH
NH
2
Br
–
H
19.61.
OH
NO
2
O
2
N
19.62.
OSO
3
HH
HO
OH
HO O
HO H
H
OSO
3
HH OH
H
OH
– H
2
O
– H
2
O
CHAPTER 19
459
19.63.
b)
N
H
E
N
H
E
N
H
E
O O O
d) The nitroso group should be ortho–para directing, because attack at the ortho or para
position generates a sigma complex with an additional resonance structure.
e) The nitroso group is a deactivator, yet it is an ortho–para director, just like a chlorine
atom.
19.65.
460
CHAPTER 19
19.66.
a) Toluene is the only compound containing an activated ring, and it is expected to
19.67.
O
O
O
O
Compound A Compound B
c)
NO
2
OHO
1) Fuming H
2
SO
4
2) HNO
3
, H
2
SO
4
3) Dilute H
2
SO
4
4) KMnO
4
, NaOH, heat
5) H
3
O
+
19.69.
a)
O
O
O
O
Br
C
l
d)
HO
NO
2
Br
NO
2
19.70.
462
CHAPTER 19
19.71. Attack at the C2 position proceeds via an intermediate with three resonance
structures:
OEOE
H
OE
H
OE
H
19.72.
a)
OMe
BrBr
NO
2
1) Cl
2
, AlCl
3
2) HNO
3
, H
2
SO
4
3) NaOH
4) CH
3
I
5) Br
2
, AlBr
3
CHAPTER 19
463
e)
NO
2
C
l
O
1) CH
3
CH
2
COCl, AlCl
3
2) HNO
3
, H
2
SO
4
3) Cl
2
, AlCl
3
f)
g)
OEt
Cl
NO
2
1) Cl2, AlCl3
2) HNO3, H2SO4
3) NaOEt, heat
4) Cl2, AlCl3
i)
Br
O
2
N
1)
2) Br
2
, AlBr
3
3) HNO
3
, H
2
SO
4
Cl Cl
AlCl
3
464
CHAPTER 19
19.74. Bromination at the para position occurs more rapidly because ortho attack is
sterically hindered by the ethyl group:
Zn [Hg] , HCl, heat
O
19.75.
a)
Cl
OO
Br
2
Br
O
Br
OH
1) MeMgBr
2) H
2
O
AlCl
3
AlBr
3
19.76.
a)
+
CHAPTER 19
465
19.77.
The OH group activates the ring toward electrophilic aromatic substitution because the
19.78.
O
2
N NO
2
NO
2
2,4,6-trinitrotoluene
19.79.
a) A phenyl group is an ortho–para director, because the sigma complex formed from
ortho attack or para attack is highly stabilized by resonance (the positive charge is spread
over both rings). The ortho position is sterically hindered while the para position is not,
19.80.
466
CHAPTER 19
19.81.
OH
H OSO
3
H
H H
OH
H H
OH
H H
O
H
2
SO
4
OH
OH
HO
OH
O
H
H
OH
H OSO
3
H
– H
2
O
resonance
stabilized
CHAPTER 19
467
19.82.
HO H OSO
3
H
HO
H
HO
H
19.83. The amino group in N,N-dimethylaniline is a strong activator, and therefore, an
ortho-para director. For this reason, bromination occurs at the ortho and para positions.