Richard F. Daley and Sally J. Daley
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Organic
Chemistry
Chapter 18
Aromatic Substitution Reactions
18.1 Mechanism of Aromatic Electrophilic Substitution
18.2 The Nitration of Benzene
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18.3 Halogenation and Sulfonation of Benzene
18.4 Friedel-Crafts Alkylation and Acylation
18.5 Effects of Monosubstituted Arenes on Substitution
18.6 Rate Effects with Monosubstituted Arenes
18.7 Classification of Substituents
935
18.8 Friedel-Crafts Acylation
943
Synthesis of o-Benzoylbenzoic Acid
947
18.9 Multiple Substituent Effects
948
18.10 Substitution on Polycyclic Arenes
951
18.11 Diazotization
954
Synthesis of Methyl Orange
957
Sidebar - Sulfa Drugs
958
18.12 Other Diazonium Salt Reactions
961
18.13 Nucleophilic Aromatic Substitution
963
18.14 Benzyne
965
Synthesis of Trypticene
968
18.15 Synthesis Examples
969
Key Ideas from Chapter 18
975
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920
924
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Organic Chemistry - Ch 18
912
Daley & Daley
Copyright 1996-2005 by Richard F. Daley & Sally J. Daley
All Rights Reserved.
No part of this publication may be reproduced, stored in a retrieval system, or
transmitted in any form or by any means, electronic, mechanical, photocopying,
recording, or otherwise, without the prior written permission of the copyright
holder.
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Chapter 18
Aromatic Substitution
Reactions
Chapter Outline
18.1
Mechanism of Electrophilic Aromatic Substitution
The mechanism of electrophilic substitution of benzene
18.2
The Nitration of Benzene
18.3
Halogenation and Sulfonation of Benzene
A case study of aromatic electrophilic substitution
The mechanism of chlorination, bromination, and sulfonation of
benzene
18.4
Friedel-Crafts Alkylation
Formation of alkyl benzenes
18.5
Effects of Monosubstituted Arenes on Substitution
The effects of one substituent on the position of reaction by a second
substituent
18.6
Rate Effects with Monosubstituted Arenes
The effect of one substituent on the rate of reaction by a second
substituent
18.7
Classification of Substituents
A listing of common substituents showing their directive and rate
controlling effects
18.8
Friedel-Crafts Acylation
Formation of acyl benzenes
18.9
Multiple Substituent Effects
Predicting the position of substitution when two or more substituents
are on the ring
18.10
Substitution on Polycyclic Arenes
Aromatic electrophilic substitution on polycyclic aromatic compounds
18.11
Diazotization
Diazotization and the use of the diazonium ion as an electrophile
18.12
Other Diazonium Salt Reactions
Replacement of the diazonium ion with a variety of groups
18.13
Nucleophilic Aromatic Substitution
Nucleophilic substitution on an aromatic ring
18.14
Benzyne
The formation and reaction of the reactive benzyne intermediate
18.15
Synthesis Examples
Organic synthesis using aromatic electrophilic substitution reactions
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Objectives
✔ Understand the mechanism for aromatic electrophilic substitution
reactions
✔ Recognize appropriate electrophiles that will substitute on an
aromatic ring
✔ Be able to predict the position of a new substitution on an aromatic
ring with one or more existing substituents
✔ Know how the structure of one substituent affects the rate of
reaction for adding a second substituent on the ring
✔ Know the diazotization reaction and how the diazonium salts react
✔ Understand the nucleophilic substitution reaction and its
mechanism
✔ Be able to use the reactions in this chapter in synthesis
‘Tis true; there’s magic in the web of it. . .
—Shakespeare
C
hapter 17 presents the characteristics that make a compound
aromatic. Understanding those characteristics is the
foundation for this chapter, as it examines the various types of
reactions that occur with aromatic hydrocarbons. The chapter first
discusses electrophilic aromatic substitution—a major aromatic
hydrocarbon mechanistic type. Electrophilic substitution allows you to
directly introduce a variety of functional groups onto the aromatic
ring. The chapter then looks at several examples of electrophilic
substitutions on benzene and its derivatives. Much of the rest of the
chapter discusses how the substituents already on the ring affect the
placement of additional substituents.
18.1 Mechanism of Electrophilic Aromatic Substitution
Chapter 14 discusses electrophilic addition reactions to the π
bond of an alkene. The result of an electrophilic addition reaction is
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that a new atom or group of atoms adds to both carbons involved in
the double bond. The mechanism proceeds in two steps: The electron
cloud of the π bond reacts with the electrophile to form an
intermediate carbocation. A nucleophile adds to the intermediate
carbocation, thus replacing the π bond with two new σ bonds.
E
E
Nu:
Nu
E
The AdE2 mechanism.
Benzenoid aromatic
compounds contain
benzene rings or fused
benzene rings.
Benzenoid aromatic compounds also have an electron-rich
π bond cloud that is susceptible to attack by an electrophile. As with
an alkene, the π electron cloud of the benzenoid aromatic compound
reacts with the electrophile and adds the electrophile to one of the
carbons in the ring. This reaction produces a carbocation
intermediate.
Step 1
The
A σ complex is a
resonance-stabilized
carbocation
intermediate.
complex formed by the attack of the electrophile
The carbocation intermediate, called a σ complex, is not aromatic.
The σ complex is written as follows to show the delocalization of the
positive charge:
H
+
E
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The carbon receiving the electrophile becomes sp3 hybridized. Having
an sp3 hybrid carbon in the ring interrupts the continuous overlap of
the p orbitals, which prevents continuous electron delocalization all
the way around the ring and causes the benzene ring to lose its
aromaticity.
At this point, the electrophilic addition reaction with benzene
ends its similarity to an electrophilic addition with an alkene. As you
may recall from Chapter 17, compounds that can become aromatic
will. So it is with the σ complex. The closed shell of benzene is so
stable that the σ complex rapidly loses a proton to a base, enabling the
reaction to regenerate the aromatic benzene ring. If the nucleophile
attacked the aromatic ring the reaction would have formed a
substituted cyclohexadiene, and the benzene ring would have
permanently lost its aromaticity. However, because the aromatic ring
is more stable than the cyclohexadiene ring, an attack of a nucleophile
on the ring would require an endothermic process rather than an
exothermic process. Thus, the lower energy pathway is the loss of a
proton to regenerate the now substituted benzene ring.
Step 2
Base:
Electrophilic aromatic
substitution is a
reaction in which an
electrophile displaces
another atom.
Because an electrophile substitutes itself for a hydrogen on the ring,
the reaction’s overall type is an electrophilic substitution instead of an
electrophilic addition. This reaction type is called an electrophilic
aromatic substitution.
Exercise 18.1
The previous illustration of the second step of an electrophilic
aromatic substitution shows the reaction of only one of the three
resonance structures of the σ complex formed in step one. Show the
second step from each of the other two resonance structures of the σ
complex.
In an electrophilic aromatic substitution reaction, the σ
complex is an intermediate, not a transition state. The reaction is not
concerted; the bond to the electrophile forms before the bond to the
proton breaks. Chemists have gathered much experimental evidence
to verify this process. Thus, when drawing a reaction progress
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diagram, such as Figure 18.1, place the σ complex ion in a valley
between the two transition states.
H
H
+
E
H
+
+
E
E
‡
G2
G1‡
+ H
+ E
Go
Reaction Progress
Figure 18.1. Reaction progress diagram for electrophilic aromatic substitution.
The reaction progress diagram in Figure 18.1 shows that the
first step in the mechanism of an electrophilic substitution reaction is
the rate-determining step. Note that ∆G1‡ , the free energy of
activation for the reaction of benzene and the electrophile to form the
σ complex, is greater than ∆G2‡ , the free energy of activation for the
reaction of the σ complex with the nucleophile to form the final
product. Because the resonance energy in the benzene ring is lost
when forming the σ complex, the reaction of the electrophile with
benzene to form the σ complex is endothermic, and it proceeds slowly.
As the σ complex loses the proton and the benzene ring regains its
resonance energy, the reaction is exothermic and proceeds rapidly.
Electrophilic aromatic substitution is the most common method
used to synthesize substituted aromatic compounds. The reaction
directly introduces functional groups onto the benzene ring and works
with a variety of electrophilic reagents. Sections 18.2, 18.3, and 18.4
discuss the major electrophilic aromatic substitution reactions. All
these reactions follow the mechanism presented in this section.
18.2 The Nitration of Benzene
Benzene reacts slowly with hot concentrated nitric acid in an
electrophilic aromatic substitution reaction to form nitrobenzene. This
reaction is potentially dangerous, however, because nitric acid is a
strong oxidizing agent that often explodes in the presence of any
material that readily oxidizes. A safer, faster, and more convenient
synthesis employs a mixture of concentrated nitric acid and
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Organic Chemistry - Ch 18
The nitration of
benzene is the reaction
of a benzenoid
compound with the
⊕NO electrophile.
2
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concentrated sulfuric acid. The concentrated sulfuric acid acts as a
catalyst allowing nitration to take place more readily at more
moderate temperatures.
NO2
HNO3
H2SO4
50 - 55oC
Dehydration of
alcohols is discussed in
Section 13.9, page 000.
Daley & Daley
Nitrobenzene
(85%)
The nitronium ion (⊕NO2) is the electrophile in the nitration of
benzene to form nitrobenzene. Although concentrated nitric acid
produces the nitronium ion by itself, the equilibrium is so far to the
left that the process is slow. Adding concentrated sulfuric acid to the
reaction mixture increases the concentration of the nitronium ion,
thereby increasing the rate of the nitration reaction. The nitronium
ion forms via a pathway similar to the first step in the dehydration of
an alcohol.
••
HO
••
NO2
H
OSO3H
••
HO
NO2
NO2
H
After the nitronium ion forms, it reacts with benzene to form
the σ complex, the first step of the electrophilic aromatic substitution
reaction. This step is slow because the σ complex is not aromatic.
Additionally, the σ complex is higher in energy than the benzene and
the nitronium ion.
H
NO2
+
NO2
σ complex
In the next step of the mechanism, the σ complex loses a proton to
form nitrobenzene. This step is rapid because the loss of a proton
allows the molecule to become aromatic again.
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H
NO2
NO2
As you may recall,
wavenumbers are
inversely proportional
to wavelength. Thus,
higher energy = lower
wavenumbers. See
Section 9.2, page 000
for more details.
Chemists tested whether the loss of a proton is the fast step or
the slow step of an electrophilic aromatic substitution by replacing the
hydrogens in benzene with deuterium and then running the reaction.
Deuterium (2H abbreviated as D) is an isotope of hydrogen (1H) that
contains not only one proton in its nucleus but also one neutron. Thus,
deuterium has twice the mass of hydrogen. Because the bond energy
between a pair of atoms changes in proportion to the masses of the
isotopes involved in that bond, the C—D bond is higher in energy than
the C—H bond. This isotope effect is observable in the IR spectrum.
The IR absorption of the C—H bond in benzene is approximately 3050
cm–1; whereas the C—D bond of deuteriobenzene is about 2150 cm–1.
Because breaking a C—D bond requires more energy than
breaking a C—H bond, a reaction whose rate-determining step
involves breaking a C—H bond proceeds more slowly when deuterium
is present. Thus, replacing C6H6 with C6D6 results in a reduction of
the nitration rate if the breaking of a C—H bond is the ratedetermining step. With the electrophilic aromatic substitution
reaction, chemists measured no difference in the rate of reaction
between C6D6 and C6H6. This shows that the rate-determining step is
the formation of the σ complex, not the step that breaks the C—H
bond.
Solved Exercise 18.1
Show a mechanism for the following reaction.
D
D2SO4
Solution
In this reaction, a deuterium replaces one of the hydrogens on the ring. The
D2SO4 is the source of D⊕ electrophile. The formation of the σ complex
involves reaction of the ring with the electrophile.
D
D
OSO3D
H
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- anion removes the proton to form the final product.
The DSO4c
D
H
D
OSO3D
Exercise 18.2
Write a mechanism for the formation of a nitronium ion from nitric
acid alone.
18.3 Halogenation and Sulfonation of Benzene
The conversion of benzene to bromobenzene or chlorobenzene
by electrophilic aromatic substitution requires the presence of a Lewis
acid in a mixture of benzene and the halogen (usually Cl2 or Br2). The
most common laboratory procedure involves adding the halogen to
benzene in the presence of some metallic iron. Chemists frequently
add this iron by tossing a few carpet tacks into the reaction mixture.
To provide the iron in an industrial setting, the reaction is run in iron
reaction vessels.
The iron itself is not actually the reaction catalyst. The iron
reacts with the halogen to form a small amount of iron(III) chloride or
iron(III) bromide. These iron halides are the catalysts:
2Fe + 3Cl2
2Fe + 3Br2
2FeCl3
2FeBr3
By themselves, bromine and chlorine are weakly electrophilic.
However, neither halogen is electrophilic enough to react with
benzene. The iron halides are Lewis acids and form complexes with
the halogen atoms.
••
•
•
Cl
••
•
•
Br
••
••
••
•
•
Cl
••
••
•
•
Br
••
FeCl3
FeBr3
•
•
•
•
••
••
••
••
Cl
Cl
••
••
••
••
Br
Br
FeCl3
FeBr3
The formation of the bromine–iron(III) bromide complex
increases the electrophilicity of the bromine to the point that it can
attack the benzene ring and form a σ complex. The next step, in which
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- FeBr ion removes the proton from the σ complex producing
the c
4
bromobenzene, HBr, and FeBr3, is quick.
•
•
••
Br
••
••
Br
••
H
•
•
FeBr3
Br
••
Br
••
Br
FeBr3
Figure 18.2 shows the reaction progress diagram for the bromination
of benzene. Note that ∆G2‡ < ∆G1‡ , so formation of the σ complex is
the rate-determining step.
H
+
H
Br
Br
H
+
+
Br
Br
Br
‡
G2
G1
‡
Br
+ HBr
Go
+ Br2
Reaction Progress
Figure 18.2. The reaction progress diagram for the bromination of benzene.
Chlorination of benzene follows a similar route to the
bromination of benzene. The catalyst, FeCl3, forms from the reaction
of iron and chlorine. On some occasions, chemists also use AlCl3 as a
catalyst. Section 18.4 discusses the uses of AlCl3 as a strong Lewis
acid and its importance in other electrophilic substitutions.
Cl
Cl2
FeCl3
Chlorobenzene
(83%)
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Section 18.12, page
000, describes an
indirect method of
monofluorinating a
benzene by using an
intermediate
diazonium compound.
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Fluorination of benzene is a very exothermic reaction. It reacts
so rapidly that it requires special conditions and reaction vessels.
Even then, the reaction rarely stops with a monofluorination product.
Iodine is so unreactive that it requires special techniques to
iodinate directly. These techniques involve adding iodine to the
reaction mixture in the presence of a strong oxidizing reagent such as
nitric acid.
Iodine may also be
introduced via the
diazonium compound
(Section 18.12, page
000).
I
I2
HNO3
Iodobenzene
(86%)
Exercise 18.3
An important technique for the introduction of fluorine onto an
aromatic ring is via a two-step thallation procedure. Benzene reacts
with thallium trifluoracetate (Tl(OOCCF3)3) to form an
organothallium intermediate. This intermediate reacts with KF and
BF3 to form the fluorobenzene.
O
O
Tl(OCCF3)2
Tl(OCCF3)3
F
KF, BF3
Propose a mechanism for the first step of this reaction.
In a sulfonation reaction, benzene reacts with sulfuric acid to
produce benzenesulfonic acid.
SO3H
H2SO4
Benzenesulfonic acid
(95%)
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Organic Chemistry - Ch 18
Azeotropic distillation
is discussed in Section
8.2, page 000. Benzene
and water form an
azeotrope boiling at
69.4oC having a
composition of 91%
benzene and 9% water.
Chemists form fuming
sulfuric acid by adding
sulfur trioxide to
concentrated sulfuric
acid.
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A sulfonation reaction is readily reversible and yields only low
amounts of benzenesulfonic acid because the equilibrium constant
favors the substrate. However, good yields of benzenesulfonic acids are
obtained by removing the water from the reaction mixture. For
example, an azeotropic distillation removes the water along with the
unreacted benzene, leaving the benzenesulfonic acid.
Another way to remove the water and increase the yield of
benzenesulfonic acid is to use fuming sulfuric acid as the source of
the electrophile. The actual electrophile, or sulfonating agent, is sulfur
trioxide regardless of whether you use sulfuric acid or fuming sulfuric
acid. When you use fuming sulfuric acid, you increase the amount of
sulfur trioxide available to act as the electrophile. Because the sulfur
trioxide reacts with the water to form sulfuric acid, the sulfur trioxide
also removes the water as soon as it forms.
H2SO4
SO3 + H2O
An advantage of the sulfonation reaction is its reversibility.
Simply heating benzenesulfonic acid with an aqueous acid removes
the sulfonic acid group. Reversibility is a very useful synthetic tool as
you can synthesize a benzenesulfonic acid, use it as a synthetic
intermediate, and then remove the sulfonic acid group when it is no
longer needed.
The mechanism for the formation of benzenesulfonic acid
follows the same general mechanism as do all electrophilic aromatic
substitution reactions. The first step is the attack of benzene on the
sulfur of sulfur trioxide, followed by the loss of a proton, which allows
the ring to regain its aromaticity.
OSO3H
O
S
O
O
H
H
O
S
OH
O
SO3H
Exercise 18.4
Write a mechanism to explain the presence of SO3 in sulfuric acid.
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18.4 Friedel-Crafts Alkylation
The Friedel-Crafts
reaction uses a
carbocation or acylium
ion as the electrophile
with benzenoid
aromatic compounds.
In 1877, Charles Friedel at the Sorbonne in Paris and James
M. Crafts of the Massachusetts Institute of Technology collaborated on
the development of a method to produce alkylbenzenes and
acylbenzenes. Their method, the Friedel-Crafts reaction, is one of the
most useful synthetic methods in organic chemistry because it allows
the introduction of a carbon side chain to benzene.
The formation of an alkyl benzene in a Friedel-Crafts
alkylation involves benzene, an alkyl halide, and a Lewis acid.
Frequently, the Lewis acid used is aluminum chloride.
CH2CH3
CH3CH2Cl
AlCl3
Ethylbenzene
(84%)
Carbocation
rearrangements are
introduced in Section
12.4, page 000.
The mechanism for an alkylation reaction begins as the Lewis acid, in
this case AlCl3, reacts with the alkyl halide to form a complex. This
complex is the reacting species for a primary alkyl halide. For a
tertiary alkyl halide, the complex tends to dissociate forming a free
carbocation. In addition, if a carbocation-like rearrangement of the
alkyl group can occur, it will.
CH3CH2
••
•
•
Cl
••
AlCl3
CH3CH2
••
Cl
••
AlCl3
The benzene ring then undergoes electrophilic attack by the complex
to form a σ complex—completing the first step of the electrophilic
aromatic substitution reaction. Immediately following the first step,
the σ complex undergoes the second step and loses a proton to form
the alkyl benzene.
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CH3CH2
••
Cl
••
Daley & Daley
CH2CH3
AlCl3
AlCl4
H
CH2CH3
Rearrangement of the alkyl group can occur during a FriedelCrafts alkylation, so the product of the alkylation is almost never
exclusively a primary alkyl benzene, unless the primary carbocation
cannot rearrange. For example, the reaction of 1-chloropropane with
benzene in the presence of aluminum chloride produces
isopropylbenzene and propyl benzene in a 2:1 ratio.
CH(CH3)2
67%
CH3CH2CH2Cl
Isopropylbenzene
AlCl3
CH2CH2CH3
33%
Propylbenzene
The electrophile for the isopropylbenzene product is an
isopropyl complex that formed when the alkyl halide lost the chlorine
ion and a hydride shift occurred.
CH3CH2CH2
••
•
•
Cl
••
AlCl3
CH3CHCH2
••
Cl
••
AlCl3
H
•
•
CH3CHCH3
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••
Cl
••
AlCl3
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Organic Chemistry - Ch 18
As you may recall from
Section 12.5 (page 000),
more highly
substituted
carbocations are more
stable.
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The driving force for the rearrangement is the stability of the
carbocations (3o > 2o > 1o). The rearrangement of the carbocation is a
major limitation of the synthetic utility of the Friedel-Crafts reaction.
The reaction generally works well if the desired product comes from
the more stable carbocation.
Solved Exercise 18.2
What product forms in the following reaction?
CH2Cl
AlCl3
Solution
In the presence of a Lewis acid catalyst like AlCl3, benzyl chloride forms the
benzyl cation. The benzyl cation is the electrophile that reacts with the
benzene to form diphenylmethane.
CH2Cl
CH2
AlCl3
Diphenylmethane
Exercise 18.5
Propose a step-by-step mechanism for the synthesis of
cyclopentylbenzene from benzene, chlorocyclopentane, and aluminum
chloride.
Lewis acids other than AlCl3 catalyze Friedel-Crafts reactions.
These catalysts include most any Lewis acid that can form a
carbocation from an alkyl halide. Some examples are FeCl3, ZnCl2,
and BF3.
Chapters 12 through 14 cover a number of reactions involving
carbocations. Electrophilic attack on benzene by a carbocation is
simply another of those reactions. Any reagent that forms a
carbocation, not just the ones specifically mentioned in this chapter,
can catalyze a Friedel-Crafts alkylation. For example, a mixture of
propene and liquid hydrogen fluoride, being used as both solvent and
proton donor, reacts with benzene to produce isopropylbenzene.
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CH3CH
Daley & Daley
CH(CH3)2
CH2
HF
Isopropylbenzene
(75%)
Exercise 18.6
Write a step-by-step mechanism for the reaction of cyclopentene with
benzene in liquid HF.
Alcohols in acidic media undergo reactions that are comparable
to the reactions of alkyl halides with Lewis acids. Just like the alkyl
halides, alcohol substrates also readily rearrange as they lose the —
OH group to form the carbocation. For example, isobutyl alcohol with
boron trifluoride (BF3) as a catalyst, reacts with benzene to produce
tert-butylbenzene. The reaction produces tert-butylbenzene instead of
isobutylbenzene because a hydride shift in the isobutyl alcohol occurs
at the same time that it loses the —OH group, thus producing the
more stable carbocation intermediate.
C(CH3)3
(CH3)2CHCH2OH
BF3
tert-Butylbenzene
(64%)
18.5 Effects of Monosubstituted Arenes on Substitution
So far, you have investigated only those reactions that involve
various electrophilic reagents with benzene. This section begins the
discussion of electrophilic aromatic substitution reactions that occur
with benzene rings already bearing a substituent. An electrophilic
substitution with a monosubstituted benzene greatly affects the
outcome of the reaction compared to an electrophilic substitution on a
nonsubstituted benzene. The substituent already on the benzene ring
affects the reaction in two important ways: it affects the
regiochemistry of the incoming electrophile and the rate of the
reaction.
For example, the nitration of toluene produces three products
in differing amounts. Only 3% of the product forms with the nitro
group in the position meta to the methyl group, whereas the
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Daley & Daley
remaining 97% forms in the ortho and para positions (63% ortho and
34% para.
CH3
CH3
CH3
CH3
NO2
HNO3
+
H2SO4
+
NO2
NO2
o-Nitrotoluene
63%
An ortho, para
directing group guides
an incoming
electrophile to either
the ortho or para
position on the ring.
m-Nitrotoluene
3%
p-Nitrotoluene
34%
Because toluene produces predominantly ortho and para substitution
products, the methyl substituent is called an ortho, para directing
group.
Exercise 18.7
Statistically what percentage of each nitrotoluene product would you
expect to get?
In the nitration of (trifluoromethyl)benzene, on the other hand,
91% of the product forms with the nitro group in the position meta to
the trifluoromethyl group. The other 9% of the product is divided with
6% in the ortho position and 3% in the para position.
CF3
CF3
HNO3
CF3
CF3
NO2
+
H2SO4
+
NO2
NO2
(Trifluoromethyl)
benzene
A meta directing group
guides an incoming
electrophile to the meta
position on the ring.
o-Nitro(trifluoromethyl)benzene
6%
m-Nitro(trifluoro- p-Nitro(trifluoromethyl)benzene
methyl)benzene
91%
3%
Because substitution in (trifluoromethyl)benzene occurs primarily at
the meta position, the trifluoromethyl group is called a meta
director.
The nitration of toluene and (trifluoromethyl)benzene
illustrates how the substituent already present on the benzene ring
affects the regioselectivity of an electrophile in a substitution reaction.
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The existing substituent gains its influence over the location of an
incoming group by either donating electron density to or withdrawing
electron density from the π electron cloud. Electron donating
substituents (e.g. CH3—) direct incoming electrophiles primarily to
the ortho and para positions. Electron-withdrawing substituents (e.g.
CF3—) direct incoming electrophiles to the meta position.
The directing effect of the existing substituent group on the
arene is based on the interaction of that group with the positive
charge of the σ complex. When the substituent group stabilizes the σ
complex by its electron-donating qualities, it directs the incoming
group primarily to the ortho and para positions. When the substituent
group destabilizes the σ complex by withdrawing electron density from
the benzene ring, the electrophile reacts mostly to the meta position.
The following discussion of toluene gives a practical
explanation as to why the ortho- and para-substituted σ complexes are
more stable than the meta-substituted σ complex.
Ortho Substitution
CH3
CH3
NO2
CH3
NO2
H
H
NO2
H
Para Substitution
CH3
H
NO2
CH3
CH3
H
NO2
H
NO2
As you look at the three resonance contributors of the σ complex for
the ortho and para substitutions, notice that both complexes contain
one 3o carbocation (the one with positive charge on the carbon bearing
the methyl group) and two 2o carbocations. The tertiary carbocation is
much more stable than either of the two secondary carbocations, so
the tertiary carbocation adds overall stability to the ortho- and parasubstituted σ complexes.
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Organic Chemistry - Ch 18
CH3
930
Daley & Daley
CH3
NO2
Site of the more stable
tertiary carbocation.
H
H
Ortho
complex
Para
NO2
complex
On the other hand, the three resonance structures with the
meta-substituted σ complex are all secondary carbocations.
Meta Substitution
CH3
The donating character
of alkyl groups is
introduced in Section
7.5, page 000.
CH3
CH3
H
H
H
NO2
NO2
NO2
The meta-substituted σ complex is much less stable than the orthoand para-substituted σ complexes.
A methyl group is an electron-donating group, and although it
activates all three positions relative to benzene, it activates the ortho
and para positions more than the meta positions. This increased
reactivity at the ortho and para sites directs the incoming
electrophiles primarily to those positions. All alkyl groups are electron
donating; thus, they are all ortho, para directing groups.
In contrast to the methyl group, the trifluoromethyl group is
strongly electron withdrawing. Because of the high electronegativity of
the fluorines, the C—F bond is quite polar with the positive end of the
dipole at the carbon.
F
F
C
F
When a resonance contributor from a trifluoromethyl σ complex has
the positive charge on the carbon bearing the trifluoromethyl group,
the positive charge from the trifluoromethyl dipole and the positive
charge from the resonance contributor repel each other. This repelling
behavior destabilizes that carbocation, and thus destabilizes the whole
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σ complex. Both the ortho and the para σ complexes have one
resonance contributor with a positive charge on the carbon bearing the
CF3 group; whereas, none of the resonance contributors in the meta σ
complex has a positive charge on the carbon bearing the CF3 group.
Ortho Substitution
CF3
CF3
NO2
NO2
H
H
CF3
NO2
H
Para Substitution
CF3
H
CF3
CF3
NO2
H
H
NO2
NO2
Meta Substitution
CF3
CF3
CF3
H
H
H
NO2
NO2
NO2
The positive charge on the carbon of the ring and the partial
positive charge on the trifluoromethyl group strongly destabilize the
ortho and para σ complexes.
+
CF3
+
NO2
CF3
These positive sites
repel one another.
H
H
NO2
Para complex
Ortho complex
An electrophilic attack at the meta position leads to a more
stable intermediate σ complex than does an electrophilic attack at
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either the ortho or para positions. Although the trifluoromethyl group
deactivates all three positions in comparison to benzene, it deactivates
the meta position less when compared to either the ortho or para
positions.
18.6 Rate Effects with Monosubstituted Arenes
Formation of the σ complex is the rate-determining step in
electrophilic substitution reactions. Factors that affect the stability of
the σ complex also affect the rate at which it forms. Thus, substituents
on a benzene ring not only affect the position of an incoming
electrophile takes on that ring, but also affect the rate of reaction.
Some substituents speed up the rate of reaction in comparison to
benzene’s rate, and some slow it down. As with regioselectivity, the
more stable the σ complex, the easier the σ complex forms, and the
faster it forms.
The data in Table 18.1 shows the relative rates for the
nitration of benzene, toluene, and (trifluoromethyl)benzene. Note that
the rate of reaction for the ortho and para positions of toluene is about
one million times faster than for the ortho and para positions of
(trifluoromethyl)benzene. Note also that the meta position of toluene
is over 30,000 times as reactive as the meta position of
(trifluoromethyl)benzene. Observe that the rates for reaction at the
ortho position are lower than for the para position. However,
experimentally there is twice as much ortho product as para product.
This is due to the fact that there are two ortho hydrogens and only one
para hydrogen.
Relative Rates
Benzene Toluene
(Trifluoromethyl)benzene
Ortho
1
44
Meta
1
2.4
Para
1
59
Table 18.1. Relative rates
(trifluoromethyl)benzene.
for
the
4.6 x 10–6
69. x 10–6
4.5 x 10–6
nitration
of
benzene,
toluene,
and
Recall from Section 18.5 that, when one resonance contributor
in a σ complex possesses more stability than the rest of the resonance
contributors, the more stable contributor increases the overall
stability of the entire σ complex. In fact, it increases the stability of
the whole chemical species. Looking again at the example of toluene,
both the ortho σ complex and the para σ complex have one resonance
contributor that is more stable than the other contributors. None of
An activating ortho,
para directing group
reacts faster than
benzene and directs an
incoming electrophile
to the ortho and para
positions.
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the meta σ complex contributors are more stable than the others.
Nevertheless the meta σ complex benefits from the electron-donating
ability of the methyl group. Thus, the σ complex from toluene is more
stable than the σ complex formed from benzene, and it reacts faster than
benzene in electrophilic aromatic substitution reactions. The presence
of the methyl group increases the rate of reaction of toluene at all
three sites, especially the ortho and para positions. Because of this
increased rate of reaction in comparison to benzene, the methyl group
is called an activating ortho, para director.
The greater the stability of the product of a reaction in a
family of related reactions, in this case the intermediate σ complex,
the less energy of activation required to form it. Figure 18.3 shows the
relationship of the energies of activation for the formation of benzene’s
one σ complex and toluene’s three. All three σ complexes from the
reaction of toluene are more stable than the σ complex from benzene;
therefore, they need less energy to form.
Figure 18.3 also shows the energy relationship among the
various σ complexes of toluene. Recall from Section 18.5 that 63% of
the product forms in the ortho position and 34% in the para position.
Thus, the ortho position has nearly twice as much substitution as does
the para position. Because there are two ortho sites and only one para
site, you would expect exactly a 2:1 ratio of product if both had
identical reactivity. However, the methyl group sterically hinders the
ortho position slightly, which causes the reaction to require more
energy to place the electrophile in an ortho position.
H
+
NO2
‡
G1
Meta
Ortho
Para
CH3
+
NO2
+
NO2
Reaction Progress
Figure 18.3. The relative activation energies for the formation of the σ complexes of
benzene and ortho, meta, and para substitutions in toluene.
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A deactivating meta
directing group reacts
slower than benzene
and directs an
incoming electrophile
to the meta position.
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Daley & Daley
As you saw in Section 18.5, the trifluoromethyl group is
electron-withdrawing in electrophilic aromatic substitution reactions
and destabilizes the σ complexes in comparison to the σ complex of
benzene. The trifluoromethyl group destabilizes the ortho σ complex
and para σ complex the most because both have a resonance
contributor with the positive charge on the carbon bearing the CF3
group. Although the meta σ complex is the most stable, it is less stable
than benzene. The CF3 electron-withdrawing group destabilizes the
meta σ complex because the electron-withdrawing group is only one
carbon away from the destabilized ortho and para resonance
contributors. All three σ complexes are less stable than benzene and
therefore take longer to react. Thus, the CF3 group is called a
deactivating meta director.
Figure 18.4 shows the relationship of the activation energies
and the σ complexes of benzene and the ortho, meta, and para
positions of (trifluoromethyl)benzene.
H
+
NO2
Ortho
Para
Meta
‡
G1
+
NO2
CF3
+
NO2
Reaction Progress
Figure 18.4. The relative activation energies for the formation of the σ complexes of
benzene and ortho, meta, and para substitutions in (trifluoromethyl)benzene.
Exercise 18.8
The rate of nitration of 1,3-dimethylbenzene (m-xylene) is 100 times
as fast as 1,4-dimethylbenzene (p-xylene). Predict the product(s) of
nitration for both xylenes. If there are more than one isomer, which
would you expect to be the major product? Explain the difference in
the relative rates.
18.7 Classification of Substituents
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Methyl and trifluoromethyl groups are only two of the many
substituents that affect the way electrophiles react with benzene in
substitution reactions. In regard to regioselectivity, all substituent
groups are either ortho, para directors or meta directors. Each group
varies as to how it affects the rate of electrophilic substitution. In
general, ortho, para directors activate the aromatic ring with respect
to benzene, and meta directors deactivate the ring.
The ortho, para directors all share the ability to stabilize the σ
complex. For example, the following four resonance contributors can
be written for para substitution on anisole.
••
•
•
H
••
OCH3
E
•
•
H
•
•
OCH3
H
E
••
OCH3
E
•
•
H
OCH3
E
The first three resonance structures are identical to the ones drawn
for para-substituted toluene in Section 18.5. The fourth resonance
structure, however, is particularly stable because the nonbonding pair
of electrons on the oxygen atom helps stabilize the positive charge. In
fact, this is the major resonance contributor because all the atoms
have filled orbitals in their valence shells.
All meta directors inductively destabilize the σ complex
because they have a partial positive or a full positive charge on the
atom attached to the ring thereby discouraging substitution at sites
ortho or para to the substituent. By default, the reactive site is the
meta position because it is not directly destabilized by the substituent.
For example, the least stable resonance contributor for nitrobenzene
has a positive charge on the carbon bearing the nitrogen. The nitro
group nitrogen also has a formal positive charge.
•
•
••
••
O••
O
••
N
H
E
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Table 18.2 summarizes the rate and orientation effects of the
most common substituents. The list places the substituents in order of
decreasing rate of substitution. Amines are the most activating group,
and nitro groups are the strongest deactivating group.
Effect on Rate
Substituent
Very strongly activating
Product
Orientation
••
NH2
Ortho, Para
••
NHR
Ortho, Para
••
NR2
Ortho, Para
••
OH
Ortho, Para
••
Strongly activating
••
O
•
•
Ortho, Para
••
NHCR
••
OR
Ortho, Para
••
•
•
••
O
Ortho, Para
••
OCR
••
Activating
Reference
R
Ortho, Para
Ar
Ortho, Para
CH=CR2
Ortho, Para
H
Deactivating
••
X••
Ortho, Para
••
(X=F,Cl,Br,I)
••
CH2••
X••
Strongly deactivating
•
•
Ortho, Para
••
O
Meta
CR
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Effect on Rate
Daley & Daley
Substituent
•
•
Product
Orientation
••
O
Meta
••
COH
••
•
•
••
O
Meta
••
COR
••
•
•
••
O
Meta
CH
•
•
••
O
Meta
••
CCl
••
Very strongly deactivating
•
•
C N ••
Meta
SO3H
Meta
CF3
Meta
NH3
Meta
NO2
Meta
NR 3
Meta
Table 18.2. Classification of substituents in electrophilic aromatic substitution
reactions.
The following five rules summarize Table 18.2:
1) Activating substituents are ortho, para directors.
2) Ortho, para directors, except for alkyl, aryl, and vinyl
groups, have nonbonding electrons on the atom attached to
the aromatic ring.
3) Deactivating substituents are meta directors.
4) Meta directing groups have at least a partial positive charge
on the atom that bonds to the ring carbons.
5) Halogens are an exception to the above rules. They are
deactivating, but are ortho, para directing groups, and they
have nonbonding electrons.
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Phenol is an example of an ortho, para director with
nonbonding electrons. In a nitration substitution reaction of phenol,
both the ortho σ complex and the para σ complex have four resonance
contributors. The first three are identical to the three shown in
Section 18.5 for either toluene or (trifluoromethyl)benzene. In the
fourth one, the oxygen contributes a pair of nonbonding electrons to
the electron deficient carbon to form a carbon—oxygen double bond.
All the atoms in this resonance contributor have an octet of electrons,
so this contributor is more stable than the other three.
Ortho Substitution
••
•
•
OH
•
•
••
••
•
•
OH
OH
•
•
NO2
NO2
NO2
H
H
H
OH
NO2
H
Major contributor
Para Substitution
••
•
•
OH
H
NO2
•
•
H
••
OH
NO2
•
•
H
••
OH
NO2
••
OH
H
NO2
Major contributor
The stabilization of the positive charge of the σ complex by the oxygen
makes it the major resonance contributor. This stability provided by
the major contributor permits the σ complex to form rapidly. The rate
of formation of the σ complex from phenol in an electrophilic
substitution is much faster than the reaction of the σ complex of
benzene. In fact, the nitration of phenol is approximately 50,000 times
faster than the nitration of benzene.
The nonbonding pairs of electrons on the oxygen cannot
stabilize the meta σ complex of phenol because none of the resonance
contributors have a positive charge on the carbon bearing the OH
group. Thus, the meta σ complex has only three resonance
contributors. The rate of reaction for the meta position, although
slower than the ortho or para positions, is still 10.4 times faster than
the rate of reaction with benzene.
Meta Substitution
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Organic Chemistry - Ch 18
•
•
939
••
OH
•
•
Daley & Daley
••
•
•
••
OH
OH
H
H
H
NO2
NO2
NO2
Beyond what Section 18.6 presents about substitution reactions
on a ring containing an electron-withdrawing substituent, you need to
know that the greater the partial positive charge on the substituent,
the more deactivating the group.
The halogens are an exception to the general rules of how
substituents on benzene behave. The halogen substituents are
electronegative; thus, they are electron withdrawing. They induce a
partial positive charge on the carbon bearing the halogen and thereby
deactivate the ring. However, unlike other deactivating groups,
halogens direct the electrophile to the ortho and para positions
because they have nonbonding pairs of electrons available to stabilize
the σ complex. The stabilization is less significant than with a phenol
or aniline, even though oxygen and nitrogen have electronegativities
equal to or greater than the halogens. The lower stabilization by the
halogens is because the nonbonding electrons are further from the
ring and thus less likely to be donated to stabilize the σ complex.
In the nitration of bromobenzene for example, the bromine
deactivates the ring causing the reaction to be 20 times slower than
the nitration of benzene. The primary products are obromonitrobenzene and p-bromonitrobenzene.
Br
Br
HNO3
H2SO4
Br
Br
+
+
NO2
NO2
NO2
39%
1%
60%
The halogens direct the incoming electrophile to the ortho and
para positions because their nonbonding electrons stabilize the ortho
and para σ complexes but not the meta σ complex. The major
resonance contributor for the halogen-bearing σ complex has the
positive charge on the halogen atom. All atoms in this structure have
an octet of electrons; thus, it is the most stable resonance contributor.
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••
••
•
Br
•
••
Br••
H
H
NO2
NO2
••
Both resonance
contributors have
full octets on all
atoms.
••
Br••
Br••
••
NO2
NO2
H
H
Exercise 18.9
Aniline (C6H5NH2) is more reactive towards electrophilic substitution
than is acetanilide (C6H5NHCOCH3). Explain this difference in
reactivity.
When using highly activating substituent groups, limiting the
number of incoming substituents to only one is difficult. For example,
phenol reacts rapidly with bromine in water to form a quantitative
yield of 2,4,6-tribromophenol.
OH
OH
Br
Br
Br2
H2O
Br
2,4,6-Tribromophenol
(100%)
Solved Exercise 18.3
Complete the following reactions showing the major monosubstitution
product(s) from each reaction.
a)
Br
Br2
Fe
Solution
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Bromine is a weakly deactivating ortho, para directing group. Thus, you will
get a mixture of ortho and para substitution products. Because bromine is
large, there will be a larger fraction of para than ortho product.
Br
Br
Br2
Br
+
Fe
Br
Br
b)
OCH3
CH3CH2CH2Cl
AlCl3
Solution
The methoxy group is an activating ortho, para directing group. Thus, you
will get a mixture of ortho and para substitution. Because the oxygen is
small, the ortho product will likely predominate. In this case, the electrophile
will rearrange to form a secondary carbocation. Thus, the product is an
isopropyl-substituted anisole.
OCH3
CH3CH2CH2Cl
OCH3
OCH3
+
AlCl3
CHCH3 CH3CH
CH3
CH3
c)
O
COCH3
HNO3
H2SO4
Solution
The ester functional group is a deactivating meta directing group. Thus, the
product will have a nitro group substituted meta to the ester group.
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O
Daley & Daley
O
COCH3
COCH3
HNO3
H2SO4
NO2
18.8 Friedel-Crafts Acylation
Section 18.4 describes the Friedel-Crafts alkylation reaction.
As you may recall, the reaction has a limitation in that, if possible, an
alkyl group will rearrange to form the more stable carbocation. A
second limitation of the Friedel-Crafts alkylation is that it does not
reliably produce a monoalkyl benzene. The reaction produces a
slightly activated aromatic ring, which then undergoes a second
alkylation at a higher rate than does benzene.
CH3
CH3
CH3Cl
CH3Cl
AlCl3
AlCl3
The product is more reactive
than the substrate, so it readily
undergoes further reaction.
CH3
CH3
+
CH3
A third limitation of the Friedel-Crafts reaction is that it does
not proceed well with deactivated benzene rings. Any group more
deactivating than the halogens normally gives a low yield or no
reaction at all in the Friedel-Crafts reaction.
A variation of the Friedel-Crafts reaction is the acylation
reaction. A Friedel-Crafts acylation reaction involves the reaction of
an acyl halide or an acid anhydride and a Lewis acid with benzene to
yield an acylbenzene. Both the acyl halide and the anhydride work
well in this reaction because each possesses a good leaving group. The
acyl halide has a halide ion leaving group and the acid anhydride has
a carboxylate ion leaving group. The Friedel-Crafts acylation produces
a deactivated ring and, because deactivated rings cannot undergo a
second acylation, the reaction stops at that point.
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O
O
CCH3
CH3CCl
AlCl3
The product is less reactive
than the substrate, so it does
not undergo further reaction.
Unlike an alkylation reaction, which needs only catalytic
amounts of AlCl3 to react, an acylation reaction requires one mole of
AlCl3 per mole of acyl halide. Acylation reactions need this
stoichiometric amount of aluminum chloride because the aluminum
chloride first forms an acid/base complex with the carbonyl group of
the acyl halide. When running an acylation reaction, chemists usually
allow the acid/base complex to form before they add the benzene to the
reaction mixture.
•
•
••
O
AlCl3
CH3CH2CCl
•
•
AlCl3
O
CH3CH2CCl
Acid-base complex
The acid/base complex is relatively stable in low polarity solvents. In
higher polarity solvents, however, it ionizes and forms a resonancestabilized acyl (or acylium) cation.
•
•
•
•
AlCl3
O
••
O
CH3CH2C
CH3CH2CCl
CH3CH2C
••
O
Acyl cation
An acyl cation reacts with benzene in much the same way as any other
electrophile.
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•
•
944
•
•
••
O
Daley & Daley
••
O
•
•
CCH2CH3
CH3CH2C
••
O
CCH2CH3
AlCl4
H
Chemists use anhydrides to generate the acylium ion less
frequently than they use acyl halides because only a few anhydrides
are commonly available. In addition, acid anhydrides are expensive
and only half the material is available for reaction. The other half is
thrown away. The following mechanism shows the use of an anhydride
in a Friedel-Crafts acylation reaction.
O
O
RC
OCR
O
AlCl3
RC
O
+
Cl3Al
OCR
Exercise 18.10
In some low polarity solvents, the acid/base complex does not readily
form the acylium ion. Thus, the acid/base complex is the reacting
species. Write a mechanism for this reaction. (Hint: Refer to Chapter 8
for a starter.)
The acylationreduction sequence
involves the synthesis
of a 1-phenyl ketone
followed by the
conversion of the
carbonyl to a —CH2—
group.
In contrast to Friedel-Crafts alkylation reactions, the cations in
Friedel-Crafts acylation reactions do not rearrange. Thus, the acyl
group in the acyl benzene product has the same structure as the acyl
group in the acyl halide substrate. Rearrangement does not occur
because the acylium ion is resonance stabilized. Thus, the acylium ion
is much more stable than a carbocation.
The lack of rearrangement of the carbon skeleton of the acyl
group makes the Friedel-Crafts acylation a useful synthesis of alkyl
benzenes. In addition, the acylation reaction does not bring about a
second substitution reaction, whereas most alkylation reactions give
significant di- and trisubstitution products. When chemists want to
form a compound with a —CH2— adjacent to the benzene ring, they
first do an acylation reaction to form a ketone, then they reduce the
carbonyl of the ketone to a —CH2— group. This synthetic strategy is
called an acylation-reduction sequence.
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O
O
CR
RCCl
CH2R
reduce
AlCl3
The Clemmensen
reduction is a method
used to convert the
carbonyl group to a
—CH2— group.
The most common reaction that chemists use to reduce the
carbonyl group to an alkyl group is the Clemmensen reduction. A
Clemmensen reduction reaction involves heating the ketone with a
zinc-mercury amalgam in concentrated HCl. Many ketones that are
stable in hot acids are reduced by the Clemmensen reduction.
O
O
CCH2CH3
Zn(Hg)
HCl
CH3CH2CCl
AlCl3
CH2CH2CH3
Propylbenzene
(71% overall)
The Wolff-Kishner
reduction is another
method used to convert
the carbonyl group to a
—CH2— group.
An alternative procedure to a Clemmensen reduction is the
Wolff-Kishner reduction. The procedure for a Wolff-Kishner
reduction involves heating the ketone with hydrazine (NH2NH2) and
potassium hydroxide in a high boiling alcohol solvent. The WolffKishner reduction works with a variety of ketones and aldehydes that
are stable in hot, concentrated base.
O
(CH3)3CCCl
AlCl3
O
CC(CH3)3
NH2NH2, KOH
CH2C(CH3)3
Diethylene glycol
reflux
(2,2-Dimethyl-1-propyl)benzene
(76% overall)
Both the Clemmensen reduction and the Wolff-Kishner
reduction are very specific for reducing an aldehyde or a ketone to a
methylene group. Neither reaction reduces the carbonyl group of
carboxylic acid derivatives C—O π bonds. The choice between the two
reduction processes depends on other groups present in the molecule.
If there are acid sensitive functional groups, then use the WolffKishner reaction. For base sensitive functional groups, use the
Clemmensen reaction.
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Exercise 18.11
Propose a synthesis for each of the following substituted aromatics.
a) tert-Butylbenzene b) 2-Methyl-1-phenylpropane
c) Butylbenzene
d) Toluene
e) Neopentylbenzene [PhCH2C(CH3)3]
Sample solution
c)
O
O
CCH2CH2CH3
CH3CH2CH2CCl
AlCl3
NH2NH2, KOH
Diethylene glycol
reflux
CH2CH2CH2CH3
Synthesis of o-Benzoylbenzoic Acid
O
O
O
AlCl3
O
Phthalic anhydride
OH
O
o-Benzoylbenzoic acid
(55%)
To a dry 50 mL round-bottom flask, add 1.65 g (0.011 mol) of phthalic anhydride, 7
mL of dry benzene, and 3.4 g (0.025 mol) of anhydrous aluminum chloride. The
apparatus must be dry because aluminum chloride reacts rapidly with water to
release hydrogen chloride, thus, destroying the aluminum chloride. Fit the flask with
a trap to collect the hydrogen chloride produced by the reaction. Stir at room
temperature for 15 minutes. Warm to 50-60oC for 5-10 minutes then reflux for an
hour. Cool the flask in an ice bath then add 15 g of crushed ice. Next, add
concentrated hydrochloric acid until the solution clears (about 3 mL). Place the entire
mixture in a 200 mL flask and distill until the distillate becomes clear. Cool the
residue in the flask in ice, filter the product, and wash it with 5 mL of cold water.
Dissolve the solid in 10 mL of 10% sodium carbonate. Filter any insoluble residue.
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Cautiously acidify with concentrated hydrochloric acid (until a pH of 1-2 is reached),
adding a little acid at a time with stirring. Continue stirring until any oil that
separates becomes crystalline. The yield of the monohydrate of the acid is 2.7 g (55%),
mp 94oC (loses water).
Discussion Questions
1. The Friedel-Crafts acylation does not work well with a carboxylic acid because an
acylium ion cannot readily form from the acid. Explain why the acylium ion cannot
readily form.
2. What is the purpose of dissolving the crude product in sodium carbonate solution?
18.9 Multiple Substituent Effects
When a benzene ring has two or more substituents, all the
substituents exert their combined effects on the reactivity of the ring
and in the placement of any incoming electrophiles. In most cases,
multiple substituents affect an electrophilic aromatic substitution
reaction in one of the following four ways.
1) All available sites are equivalent. This means that a
substitution at any one of these sites gives the same product.
CH3
CH3
Br
Br2
Fe
CH3
CH3
All sites are
equivalent
2) All sites have comparable reactivity, but one site is more
sterically hindered than the other. The reaction then takes place at
the less sterically hindered site.
Less sterically hindered
CH3
CH3
Br
Br2
Fe
C(CH3)3
C(CH3)3
More sterically hindered
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3) The directing effects of the groups reinforce one another. For
example, in p-nitrotoluene the nitro group is a meta directing group,
the methyl group is an ortho, para directing group, and the two groups
are para to each other. Thus, both groups direct to the same pair of
carbon atoms as the preferred site of reaction.
CH3
CH3
Br
Br2
Fe
NO2
NO2
Ortho to the CH3
and meta to the NO2
4) The substituents have directing influences that oppose one
another. When this occurs, the substituent with the greatest influence
controls the outcome of the reaction. If the difference between the two
groups is small, then the reaction does not select one site in preference
to the other. When the difference is large, the more activating
substituent provides the greater influence. An activating group always
directs the reaction in preference to a deactivating group. This is true
even if the activating group is weak and the deactivating group is
strong because there is a higher rate of reaction for the activating
group.
Activated by NH2
NH2
NH2
Br
Br2
Fe
Cl
Deactivated by Cl
Cl
Solved Exercise 18.4
Predict the major product(s) from the bromination of the following
compounds.
a)
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O
NHCCH3
Cl
Solution
Because of the nonbonding electrons on the nitrogen attached to the ring, the
amide group is a strongly activating ortho, para director. The chlorine is a
weakly deactivating ortho, para director. Thus, the sites ortho to the amide
group are more reactive than the sites ortho to the chlorine.
O
O
NHCCH3
NHCCH3
Br2
Fe
Br
Cl
Cl
b)
O
O2N
COCH3
Both the ester and the nitro groups are deactivating meta directors. Because
they are meta to each other, they both direct towards the same site on the
ring.
O
O
O2N
O2N
COCH3
COCH3
Br2
Fe
Br
Exercise 18.12
Predict the major mononitration product(s) for each of the following
aromatic compounds.
a) m-Dichlorobenzene
d) 4-Ethylbenzoic acid
b) p-Methylphenol
e) 1,3-Dimethylbenzene
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c) m-Nitroanisole
f) 2,6-Dichloroanisole
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Sample solution
b) The —OH group is a much stronger activating group than
the —CH3 group, so the —OH group directs the position of the nitro
group.
OH
OH
NO2
HNO3
H2SO4
CH3
CH3
18.10 Substitution on Polycyclic Arenes
Polycyclic aromatic hydrocarbons also undergo electrophilic
aromatic substitution reactions with the same reagents that react
with benzene and the benzene derivatives. In terms of relative rate of
reaction, most polycyclic aromatics react more rapidly than benzene.
Because they lack the symmetry of benzene, the product mixture from
reactions with polycyclic aromatic hydrocarbons is usually more
complex than the product mixture from a similar reaction with
benzene.
Naphthalene, a ten carbon bicyclic aromatic hydrocarbon, has
two positions available for attack by the electrophile in a nitration
reaction. These positions are on C1 and C2. For both positions,
naphthalene has four equivalent sites around the two rings.
Naphthalene needs much milder conditions for the reaction to proceed
than does benzene. A weakly acidic mixture of acetic acid/acetic
anhydride acts well as a solvent for the reaction. The reaction forms
two products in a 91:9 ratio.
NO2
1
8
9
7
6
2
3
10
5
4
Naphthalene
NO2
HNO3
CH3COOH
(CH3CO)2O
warm
+
1-Nitronaphthalene
91%
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2-Nitronaphthalene
9%
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To understand why C1 is so much more reactive than C2,
examine the resonance contributors for the σ complexes that lead to
each of the two products. Attack at C1 produces five resonance
contributors.
Reaction at C1
H
H
NO2
H
NO2
NO2
H
H
NO2
NO2
The first two contributors possess two factors that give them more
stability than the other three resonance contributors and the
resonance contributors in the C2 σ complex. Both resonance
contributors have one ring in which the aromaticity is undisturbed,
and each of the two aromatic contributors in the other ring also has an
allylic resonance. Thus, these two contributors have a special stability
due to the delocalization of the positive charge between C2 and C4.
On the other hand, only one of the resonance contributors for
the reaction at C2 has aromatic character in either ring. The
remaining resonance contributors are not aromatic. Formation of the
C2 σ complex is a higher energy pathway than electrophilic attack at
C1.
Reaction at C2
NO2
NO2
H
H
NO2
H
NO2
NO2
H
H
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Thermodynamic versus
kinetic control of a
reaction is discussed in
Section 16.4, page 000.
952
Daley & Daley
Due to steric hindrance at the C1 position, larger substituents
tend to be more stable at the C2 position. For example, formation of
naphthalenesulfonic acid under mild conditions gives naphthalene-1sulfonic acid, but at high temperatures, the product formed is
naphthalene-2-sulfonic acid. This reaction is another example of
kinetic versus thermodynamic control of a reaction. Because of its size,
a sulfonic acid group on C1 comes within the van der Waals radius of
the hydrogen at C8. But, when the sulfonic acid group is on C2, it has
no such steric interference because the molecule has sufficient room
for the sulfonic acid group between the hydrogens at C1 and C3. Thus,
the product of the higher energy pathway is actually more stable than
the product of the lower energy pathway.
Steric interference
H
Less interference
H
SO3H
SO3H
H
Naphthalene-1-sulfonic acid
Naphthalene-2-sulfonic acid
Using 100% sulfuric acid at a reaction temperature of less than
the yield of the 1-sulfonic acid is 98%. When the reaction is
heated above 165oC, the yield of the 2-sulfonic acid is 88%.
75oC,
Exercise 18.13
Following an analysis similar to that above, find the most reactive site
for electrophilic substitution on anthracene.
Anthracene
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18.11 Diazotization
An aryl diazonium salt
has the formula
ArN2⊕. An aryl
diazonium salt is made
from an aryl amine.
Chemists synthesize aryl diazonium salts from aryl amines.
NH2
N
NaNO2, HCl
N••
Cl
H2O, 0-5oC
Benzene diazonium chloride
The electrophile used to react with the amine to form the aryl
diazonium salt is the mild ⊕NO electrophile. The following
mechanism shows how the aryl diazonium salt forms from HCl and
NaNO2. The ⊕NO ion is resonance-stabilized, so it forms relatively
easily.
••
HO
••
NaNO2 + HCl
••
HO
••
N
••
O •• +
••
N
O•• + NaCl
••
HO
H
N
••
O ••
H
•
•
••
HO
H
Diazotization is the
name of the process
that forms a diazonium
salt.
N
••
N
O••
••
O••
+ H2O
N
•
•
O••
After the ⊕NO electrophile forms, it reacts in acid/base fashion with
the amine nitrogen. The intermediate then loses water and forms the
diazonium salt; thus, completing the diazotization process. Compare
the following diazotization process with the substitution and
elimination reactions of Chapters 12 and 13.
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H
••
NH2
•
•
N
••
O
•
•
N
••
N
••
••
N
O••
N
••
••
•
•
••
••
O
H
H
H
N
N ••
••
N
••
N
••
••
OH2
H
Azo coupling is the
electrophilic
substitution reaction of
a diazonium salt with
an activated aromatic
ring to form a product
with the structure
Ar–N=N–Ar.
N
N
OH
••
Once formed, diazonium salts can be used as synthetic
intermediates for a number of substituted aromatic compounds. For
example, nucleophiles displace the —N2⊕ group, so aryl diazonium
salts can be used in reactions to prepare a number of substituted
aromatic compounds. Section 18.12 (page 000) discusses the reaction
of diazonium salts with nucleophiles.
Diazonium salts can also be used as electrophiles to substitute
onto another aromatic system. This type of electrophilic aromatic
substitution reaction is called azo coupling. Although diazonium
salts are relatively weak electrophiles, they do react well with strongly
activated aromatic rings. Chemists most frequently use the —OH and
the —NR2 groups to activate the ring. The products formed in azo
coupling reactions contain the azo functional group (–N=N–) and have
the general structure of Ar–N=N–Ar. The general mechanism for the
reaction follows a similar mechanism to the other electrophilic
substitution reactions studied in this chapter.
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••
H2O••
N
••
N••
••
N
••
OH
N
H
••
OH
••
••
N
••
N
••
OH
••
4-Phenylazophenol
An azo compound
Most azo compounds are highly colored and make excellent
dyes. These dyes, called azo dyes, are usually more stable and retain
their bright colors better than most other dyes. Following are a few
examples of azo dyes.
N
N••
OH
+
O2N
COOH
N
N
OH
COOH
O2N
Alizarin yellow
(82%)
N
N••
N(CH3)2
+
HO3S
N
N
N(CH3)2
HO3S
Methyl orange
(70%)
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HO
N
N••
HO
+
O2N
N
N
O2N
Para red
(76%)
Synthesis of Methyl Orange
N(CH3)2
NH2
1) Na2CO3
2) NaNO2, HCl < 5oC
HO3S
N
N
Sulfanilic acid
HO3S
N(CH3)2
Methyl orange
(70%)
Dissolve 0.6 g of anhydrous sodium carbonate in 50 mL of water. Add 1.75 g (10 mmol)
of sulfanilic acid to the solution and heat until it dissolves. Cool the solution to room
temperature and add 0.8 g of sodium nitrite. Cool in an ice bath until the temperature
is below 10oC then add 2.5 mL of concentrated hydrochloric acid. A precipitate of
diazonium salt will form. Keep this reaction mixture cold. In a separate flask, dissolve
1.3 mL (10 mmol) of N,N-dimethyl aniline in 1.0 mL of glacial acetic acid. Add this
solution to the reaction mixture. Keep cold for about 15 minutes. During this time a
red precipitate will form. Add 15 mL of 10% sodium hydroxide. Check to see if the
solution is basic. If not, add more 10% sodium hydroxide until the solution is basic.
Heat this solution to boiling and boil until most of the solid is dissolved. Carefully add
5 g of sodium chloride to the boiling solution. Allow the mixture to cool, then cool it
further in ice. Filter the solution. Wash the flask with two ice-cold portions of
saturated sodium chloride solution and pour these through the methyl orange in the
filter. Dissolve the solid in 150 mL of boiling water. All of the solid will not dissolve,
but the sodium chloride contaminant will dissolve. Boil the solution for 5-10 minutes.
Cool the solution to room temperature then place in an ice bath. When cold, filter the
solid and allow it to dry. Yield of methyl orange is 2.15 g (70%), m.p. decomposes.
Discussion Questions
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1. The pH of the reaction mixture is important to the course of the coupling reaction
with N,N-dimethylaniline. The rate of the coupling reaction increases with
increasing pH. Explain this behavior.
2. Methyl orange, and all other azo compounds, is brightly colored. What molecular
features of azo compounds explain the colors of these molecules?
Exercise 18.14
Propose a synthesis for each of the following compounds.
a)
b)
N
CH3
N
N
N
SO2NH2
(CH3)2N
CH3
OCH3
c)
d)
N
N
N
N
NH2
N(CH3)2
NO2
Cl
Sample Solution
c)
NH2
NaNO2, H2SO4
N
H2O, 0 - 5oC
NO2
N
NH2
NH2
NO2
[SIDEBAR]
Sulfa Drugs
The introduction of sulfa drugs in the 1930s hailed the
beginning of modern drug therapy. Before their introduction, even a
minor bacterial infection could become potentially life-threatening.
Because at first no one understood how sulfa drugs worked, even
physicians considered them as almost magical.
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Researchers knew that many bacteria absorbed a number of
dyes. With some dyes, the entire bacteria absorbed the dye; with other
dyes, only specific portions of the bacteria absorbed it. Because of this
behavior, researchers used dyes as a standard method to study the
structure of bacteria. They began to wonder if they might find a dye
that was toxic to the bacteria.
To solve this puzzle, a group of researchers at the German
dyestuff manufacturer I. G. Farbenindustrie investigated the various
dyes in their collection. They screened thousands of dyes and found
that a number of them were active both in vitro and in vivo,
respectively meaning “in glass” and “in life.” They did the test tube or
petri dish (in vitro) screening first. Any promising substances found
there, they tested using living organisms—laboratory animals or
human volunteers (in vivo).
The researchers next decided to determine why some dyes were
active both in vitro and in vivo, some were active in vitro only, and
some were inactive in both cases. As they conducted this work, they
made an unexpected discovery. They discovered that p-[(2,4diaminophenyl)azo]benzenesulfonamide, which was inactive in vitro,
was very active in vivo!
NH2
N
N
H2N
SO2NH2
p-[(2,4-Diaminophenyl)azo]benzenesulfonamide
In
1932,
Gerhard
Domagk
tested
p-[(2,4diaminophenyl)azo]benzenesulfonamide, later marketed under the
trade name Prontosil, on a ten-month-old boy sick with a dangerous
staphylococcal infection. The boy rapidly recovered. Domagk was
awarded the 1939 Nobel Prize in medicine or physiology for this work,
although he was forced to decline the Prize by the Nazi German
Government.
So, Prontosil worked. It had saved a child’s life. But how?
Research had already shown that it was inactive in vitro. Why then
was it active in vivo? Identifying the mechanism by which Prontosil
combats bacterial infections was a major step in the development of
pharmacology.
The goal of the I. G. Farben research group was to find a dye
toxic to bacteria. Prontosil had healed a child with a bacterial
infection, yet, obviously it was not toxic to the bacteria. In vitro
studies had proven that fact. In conclusion, the researchers decided
that Prontosil’s toxicity observed in vivo must be from something
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else—something that had nothing to do with it being a dye. As they
tracked what happened to Prontosil in the body, they found that
Prontosil undergoes a cleavage that produces sulfanilamide.
Sulfanilamide was the substance responsible for Prontosil’s biological
activity. Animals have the necessary enzymes to do this sort of
cleavage, but bacteria do not, thus the explanation for Prontosil’s
inactivity in vitro and activity in vivo.
NH2
N
H2N
N
in vivo
H2N
SO2NH2
SO2NH2
Sulfanilamide
The explanation found by researchers for how sulfanilamide
works was in its structure. Sulfanilamide and p-aminobenzoic acid
(PABA) are so similar that the bacteria mistake sulfanilamide for
PABA. Bacteria require PABA to biosynthesize folic acid, a coenzyme
involved in the transfer of single carbon units in biosynthesis. In the
body, sulfanilamide slows the biosynthesis of folic acid, which, in turn,
slows the bacterial growth, thus allowing the body’s natural defenses
to affect a cure.
H2N
H2N
SO2NH2
Sulfanilamide
COOH
p-Aminobenzoic acid
Once researchers recognized that sulfanilamide was the active
agent against bacteria, they began synthesizing analogs. Between
1935 and 1946, they synthesized over 5,000 compounds. Of these
thousands of compounds, they found two that were more effective than
the others. These were sulfathiazole and sulfadiazine.
H2N
H2N
S
SO2NH
N
N
SO2NH
N
Sulfadiazine
Sulfathiazole
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As researchers found more effective medications, such as
penicillins and tetracyclines, physicians moved away from sulfa drugs,
although veterinarians still widely use them. Sulfa drugs made a large
impact on the lives of people. In the United States alone, the estimate
is that sulfa drugs reduced the death rate from pneumonia by at least
25,000-30,000 deaths per year in the 1940s and 1950s.
18.12 Other Diazonium Salt Reactions
The Sandmeyer
reaction converts a
diazonium salt to an
aryl chloride, bromide,
or nitrile.
In addition to the azo coupling reaction, diazonium salts
undergo a substitution reaction at the carbon bearing the diazo group.
This reaction follows a mechanism outside the scope of this book. The
most common example of the azo substitution reaction is the
Sandmeyer reaction. In a Sandmeyer reaction, the diazonium salt
reacts with copper(I) chloride, bromide, or cyanide to replace the —
N2⊕ group on the ring. Chemists often choose the Sandmeyer reaction
as the best method to prepare aryl chlorides, bromides, or nitriles in
part because it permits precise control of where the substituent will be
on the product.
NH2
NaNO2, H2SO4
X
CuX
H2O, 0 - 5oC (X = Cl, Br or CN)
The following are some examples of the Sandmeyer reaction:
NH2
NaNO2, H2SO4
H2O, 0 - 5oC
Br
CuBr
NO2
NO2
1-Bromo-3-nitrobenzene
(87%)
NH2
NaNO2, H2SO4
Cl
CN
CuCN
H2O, 0 - 5oC
Cl
2-Chlorobenzonitrile
(75%)
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NH2
Cl
NaNO2, H2SO4
H2O, 0 - 5oC
NH2O2S
CuCl
NH2O2S
4-Chlorobenzenesulfonamide
(71%)
Chemists use
Schiemann reactions to
produce aryl fluorides
that are otherwise
difficult to produce.
Another example of an azo substitution reaction is the
Schiemann reaction. To run a Schiemann reaction, fluoboric acid
(HBF4) is added to the diazotization reaction medium. Heating the
reaction mixture then forms the aryl fluoride.
NH2
NaNO2, H2SO4
H2O, 0 - 5oC
F
1) HBF4
2) heat
Fluorobenzene
(83%)
To prepare an aryl iodide, simply stir the diazonium salt with
KI at room temperature until the aryl iodide forms.
CH3
NH2
NaNO2, H2SO4
H2O, 0 - 5oC
KI
CH3
I
3-Iodotoluene
(82%)
Another important reaction involving diazonium salts is the
formation of phenols. In this reaction, you need only to warm the
diazonium salt reaction mixture.
NH2
NaNO2, H2SO4
H2O, 0 - 5oC
warm
CH3
OH
CH3
3-Methylphenol
(74%)
The final substitution reaction discussed here that uses a
diazonium salt is the replacement of the —N2⊕ group with a
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hydrogen. In this reaction, the diazonium salt is mixed with
hypophosphorus acid (H3PO2).
Br
Br
NH2
Br
Br
NaNO2, H2SO4
H2O, 0 - 5oC
H
H3PO2
Br
Br
2,4,6-Tribromobenzene
(68%)
Using deuteriohypophosphorus acid (D3PO2) instead of H3PO2, allows
the selective placement of a deuterium on the ring.
NH2
NaNO2, H2SO4
H2O, 0 - 5o
D
D3PO2
Deuterobenzene
(81%)
Exercise 18.15
Propose a synthesis of 3-chloroethylbenzene from benzene.
18.13 Nucleophilic Aromatic Substitution
Aryl halides that bear one or more strongly electronwithdrawing groups in the ortho or para positions to the halide readily
undergo nucleophilic substitution reactions. These electronwithdrawing groups must be groups that withdraw electron density
due to resonance rather than inductively. Inductive withdrawing
groups include the halogens and the —CF3 group. Resonance electronwithdrawing groups include those groups with a positive charge on the
atom attached to the ring; for example, carbonyl groups and the —
NO2 group. The most commonly used electron-withdrawing group is
the nitro group. The rate of reaction depends on how strongly electron
withdrawing the ortho and para groups are.
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Cl
OH
NO2
NaOH
80oC
NO2
H
NO2
NO2
2,4-Dinitrophenol
(95%)
The mechanism for nucleophilic substitution reactions is
analogous to the mechanism for electrophilic substitution reactions.
The hydroxide ion attacks the carbon bearing the halogen, which
causes the ring to lose its aromaticity. The electron-withdrawing
substituents then stabilize the negative charge by their partial or full
positive charge adjacent to the ring. Finally, the ring loses the
halogen, restoring aromaticity to the ring and forming a phenol.
•
•
••
Cl••
•
•
••
O••
N
O ••
••
•
•
•
•
••
Cl••
••
Cl••
•
•
••
OH
•
•
••
•
•
••
OH
O••
•
•
•
O
•• •
•
•
••
Cl ••
••
O••
N
•
•
••
O••
•
•
N
OH
••
O••
O••
••
••
NO2
N
•
•
OH
••
NO2
•
•
••
•
•
NO2
••
•
•
OH
••
•
•
••
O••
•
•
••
••
•
•
Cl••
OH
••
• •
• •
O
N
N
O••
•
O
•
••
••
••
NO2
NO2
NO2
The rate-determining step in nucleophilic aromatic
substitution reactions is the step that forms the carbon—nucleophile
bond. The rate of reaction depends on which halogen is present on the
ring and the number and strength of the electron withdrawing groups
that are ortho or para to that halogen. The relative rate of reaction for
the halides in nucleophilic aromatic substitution reactions is the
opposite of their rates in aliphatic nucleophilic substitutions. Thus,
aryl fluorides are the most reactive halide leaving groups and aryl
iodides the least reactive. Aryl fluorides are the most reactive because
fluorine is the most electronegative, so it is the best leaving group.
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Fluorine is also the smallest of the halides, thus giving less steric
hindrance to the reaction of the nucleophile than the other halides.
Table 18.3 shows the relative rates of reaction for the halogens.
Aryl Halide
Ar—F
Ar—Cl
Ar—Br
Ar—I
Rate
310
1.0
0.8
0.4
Table 18.3. Relative rates of reactions of 4-halonitrobenzenes with sodium methoxide
in methanol at 50oC.
By increasing the number of electron-withdrawing groups on
the benzene ring, you dramatically increase the rate of reaction. The
increase in reaction rate occurs because each electron-withdrawing
group helps stabilize the negative charge. Table 18.4 shows how
increasing the number of nitro groups affects the rate.
Substitution
None
4-Nitro
2,4-Dinitro
2,4,6-Trinitro
Rate
1.4 x 10–8
1
2.9 x 105
Too fast
to measure
Table 18.4. Relative rate of reaction for various chloronitrobenzenes with sodium
methoxide in methanol at 50oC.
Exercise 18.16
Reaction of 1,2-dichloro-3,5-dinitrobenzene with sodium methoxide in
methanol produces a single product, C7H5ClN2O5. What is the
structure of the product?
18.14 Benzyne
Although
aryl
halides
without
electron-withdrawing
substituents also undergo nucleophilic substitution reactions, they
require extreme conditions or very strong bases. For example the
commercial “Dow process” used for making phenol involves heating
chlorobenzene with sodium hydroxide at 350oC.
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Cl
OH
H
NaOH
350oC
Phenol
(98%)
An example of a nucleophilic substitution reaction that
proceeds with a very strong base is the reaction of chlorobenzene with
sodium amide (NaNH2) in liquid ammonia to produce aniline. Because
the amide ion is such a strong base, the reaction does not require high
temperatures.
Cl
NH2
NaNH2
NH3, -33oC
Aniline
(100%)
The mechanism for a nucleophilic substitution reaction on an
aryl halide without an electron-withdrawing group is different from
the mechanism described in Section 18.13 for a nucleophilic
substitution reaction on an aryl halide with an electron-withdrawing
group. Chemists had no reason to suspect a different mechanism until
they discovered that the substitution takes place not only at the
carbon bearing the halide but also at the carbon adjacent to the halide
bearing carbon. For example, the reaction of 4-bromotoluene with
sodium amide yields a 50:50 mixture of 3-methylaniline and 4methylaniline.
CH3
CH3
NaNH2
+
NH3, –33oC
Br
CH3
NH2
NH2
4-Methylaniline
(p-Toluidine)
50%
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3-Methylaniline
(m-Toluidine)
50%
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A benzene ring that has
undergone an
elimination reaction
that results in an
additional weak π bond
is called a benzyne.
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This nucleophilic substitution reaction follows a similar
pathway to an elimination-addition reaction. The amide ion, acting as
a base, first removes a proton from one of the carbons adjacent to the
halide. Then the halide departs. Both the halide and a proton leave
from adjacent carbons causing the compound to become a neutral
intermediate called benzyne. Chemists use the name benzyne
because the structure is written with a triple bond.
CH3
a Benzyne
The two carbons involved in the “triple” bond of benzyne are
sp2 hybridized. The overlap of their bonding orbitals is ineffective
because the shape of the benzene ring directs them 60o apart rather
than parallel to each other, which is best for an sp2 hybridized bond.
Thus, benzyne is unstable and reactive. In fact, benzyne is so reactive,
it cannot be isolated.
CH3
Poor overlap
After benzyne forms, the amide ion can attack it on either end
of its weak, reactive triple bond. This part of the reaction produces a
benzene anion. The benzene anion then removes a proton from an
ammonia molecule giving the final products, 3-methylaniline and 4methylaniline.
CH3
CH3
CH3
H
••
•
•
NH2
••
NH2
••
••
••
NH2
NH2
H
3-Methylaniline
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CH3
CH3
CH3
•
•
H
••
NH2
••
NH2
•
•
H
•
•
NH2
•
•
NH2
4-Methylaniline
See Section 16.6, page
000 for the Diels-Alder
reaction.
Benzyne is also a very reactive dienophile in a Diels-Alder
reaction. The benzyne is prepared via a Grignard reagent, although an
organolithium works as well. o-Bromofluorobenzene, when treated
with magnesium, first forms the Grignard from the bromide. The
bromine is more reactive than the fluorine in a Grignard reaction. The
Grignard then loses FMgBr to form the benzyne. Because benzyne is
so reactive, the benzyne must be generated in the presence of a diene,
so it can undergo a Diels-Alder reaction.
Br
Mg, THF
F
MgBr
F
o-Bromofluorobenzene
(78%)
Synthesis of Trypticene
NH2
(CH3)2CHCH2CH2NO2
COOH
Anthranilic acid
Trypticene
(70%)
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Dissolve 355 mg (2 mmol) of anthracene in 2.5 mL of 1,2-dimethoxyethane. Add 0.35
mL of isoamyl nitrite. Bring this mixture to a boil. Add a solution of 400 mg (2.9
mmol) of anthranilic acid dissolved in 2 mL of 1,2-dimethoxyethane dropwise over a
period of about 30 minutes. Then add another 0.35 mL of isoamyl nitrite to the
reaction mixture. Add a second 400 mg (2.9 mmol) of anthranilic acid dissolved in 2
mL of 1,2-dimethoxyethane dropwise. Reflux for an additional 15 minutes. Add 1.5
mL of ethanol and a solution of 0.5g of sodium hydroxide in water. Filter the resulting
mixture and wash the solid with an ice-cold mixture of 4:1 methanol and water.
Collect the solid and dry in an oven to a constant weight.
Dissolve the solid product in 4 mL of triglyme. Add 200 mg of maleic anhydride and
reflux the mixture for 5 minutes. Cool the solution to about 100oC and add 2 mL of
ethanol and 500 mg of sodium hydroxide in 5 mL of water. Cool the solution in ice.
Filter the crystals that form and wash with 4 mL of ice cold 4:1 methanol and water
mixture. Dry the product in an oven. The yield of trypticene is 356 mg (70%), mp
255oC.
Discussion Questions
1. The reaction of isoamyl nitrite with anthranilic acid is a diazotization reaction.
Propose a mechanism for formation of benzyne from diazotized anthranilic acid.
2. The purpose for reaction of the crude solid material with maleic anhydride is to
remove unwanted starting material. What is the structure of the product of this
reaction?
Exercise 18.17
What structure would you expect from the formation of a Grignard
reagent from 1-bromo-2-fluoro-4,5-dimethylbenzene in the presence of
furan?
18.15 Synthesis Examples
Planning a synthesis to prepare a benzene compound with
several substituents from benzene requires that you take into account
the directive influences of each group you want to place on the ring.
These directive influences determine the reaction sequence. One
sequence may produce a low yield or even stop the synthesis before it
is complete, whereas another sequence produces the desired product
in good yield. This section works through three syntheses of
substituted benzenes.
To propose a synthesis for p-bromoacetophenone from benzene,
use the principles of retrosynthetic analysis. First ask yourself, “What
is the immediate precursor of this target molecule?”
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O
CCH3
?
Br
p-Bromoacetophenone
Your answer, “A benzene compound that contains one of the
desired substituents. But which one?”
To determine which one, consider each group as the only
substituent on the benzene ring. Then think how that substituent
would direct an incoming group. If the substituent were the ketone
group, it would direct any incoming groups to the meta position. Thus,
you cannot brominate acetophenone because you would get the meta
isomer of bromoacetophenone. If the only substituent were the bromo
group, it would direct an incoming group to the ortho and para
positions. Thus, the immediate precursor is bromobenzene.
O
O
CCH3
CH3CCl
AlCl3
Br
Br
4-Bromoacetophenone
Your first step in the reaction process is to brominate the benzene.
Your second step is to add the ketone group. The following reaction
sequence illustrates the complete synthesis of p-bromoacetophenone:
O
O
Br2
Fe
CCH3
CH3CCl
AlCl3
Br
Br
The second synthesis is of 1,3,5-tribromobenzene from benzene.
Bromine is an ortho, para director, but in 1,3,5-tribromobenzene the
three bromine groups are meta to each other. Your problem in this
synthesis is to find a functional group that will both force the
bromines to substitute onto the benzene ring meta to each other and
then allow you to remove it from the molecule. Two possible groups
are the amine group and the sulfonic acid group, as a hydrogen ion
electrophile easily replaces them on a benzene ring. Recall from Table
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18.2 (page 000), however, that the sulfonic acid group brominates meta
to itself, which would be in the wrong position. An amine group, on the
other hand, is an ortho, para director, so it would place the bromine in
the correct position. In fact, an amine group is such a strong activating
group that it usually causes trisubstitution on a benzene ring.
NH2
NH2
Br
Br
Br2
H2O, NaHCO3
Br
2,4,6-Tribromobenzene
To remove the amine, form the diazonium salt then treat it
with H3PO2. To form the aniline, reduce the nitrobenzene. The
reduction of nitrobenzene to form aniline is a reaction that you have
not previously studied. This reaction is quite simple in that zinc or tin
metal with acid reduces the —NO2 group.
NO2
NH2
Zn (or Sn)
HCl
The complete synthesis of 1,3,5-tribromobenzene is as follows:
NO2
NH2
Zn
HCl
HNO3
H2SO4
Br2
H2O, NaHCO3
NH2
Br
Br
H3PO2
Br
NaNO2, H2SO4
Br
H2O, 0 - 10oC
Br
Br
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The third target molecule for you to synthesize is 4-nitro-3propylphenol from benzene.
OH
?
O2N
4-Nitro-3-propylphenol
4-Nitro-3-propylphenol
has
three
possible
immediate
precursors: 4-nitrophenol, 2-propylnitrobenzene, and 3-propylphenol.
However, 4-nitrophenol will not work because it would give 4-nitro-2propylphenol in a Friedel-Crafts reaction. A Friedel-Crafts reaction is
the reaction you use to add the propyl group. 2-Propylnitrobenzene
will not work either because the propyl and nitro group would direct
the phenol to C5 instead of para to the nitro group. Nitration of 3propylphenol, on the other hand, places the nitro group in the correct
position—the para position. This is the precursor that you must use,
even though the reaction also gives some nitration ortho to the –OH
group and para to the propyl group.
OH
O2N
OH
O2N
OH
O2N
Next, you must determine the precursor to 3-propylphenol. The
only method covered so far that synthesizes a phenol is via a
diazonium salt from an aniline, in this case, 3-propylaniline. To
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convert 3-propylaniline to 3-propylphenol, warm the diazotization
reaction mixture.
NH2
OH
NaNO2, H2SO4
warm
H2O, 0 - 10oC
3-Propylphenol
3-Propylaniline has two substituents that are meta to each
other, but both substituents are ortho, para directing groups. Thus,
you cannot start with the desired substituent. Instead, the first
substituent on the ring must be a meta directing group to direct the
second substitution at its proper position. It must also be a group that
will then react to form the desired group. The two possibilities are a
nitro group, which readily converts to an aniline, and an acyl group,
which readily converts to an alkyl group. Since Friedel-Crafts
acylations do not take place on deactivated aromatic rings, the
sequence you must use is an acylation followed by a nitration.
To do the Friedel-Crafts acylation, start with benzene and
make propiophenone. Then nitrate the propiophenone to produce 3nitropropiophenone.
O
CH3CH2CCl
HNO3
AlCl3
H2SO4
O
NO2
O
1-(3-Nitrophenyl)-1-propanone
At this point, you must convert the nitro group to an aniline
and the carbonyl to a —CH2— group. The conversion from the nitro
compound to an aniline is similar to the Clemmensen reduction of a
ketone. Thus, you can combine the two reductions into one step.
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NO2
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NH2
Zn(Hg)
HCl
O
3-Propylaniline
The following reaction sequence shows the complete synthesis
of 4-nitro-3-propylphenol:
O
CH3CH2CCl
HNO3
AlCl3
H2SO4
NO2
O
O
Zn(Hg)
HCl
OH
warm
OH
HNO3
H2O, 0 - 10oC
CH3COOH
O2N
NH2
NaNO2, H2SO4
Exercise 18.18
When planning a synthesis, knowing what not to do is as important as
knowing what to do. The following syntheses have a flaw. What
product would you expect to obtain from each of the following
schemes? Additionally, propose an alternative synthesis that would
prepare the products shown. You may use any stable monosubstituted
benzene molecule as a starting material.
a)
Cl
Br
CH3CH2CH2Cl
Cl2
AlCl3
Fe
Br
CH3CH2CH2
b)
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OCH3
OCH3
HNO3
Br2
H2SO4
Fe
NO2
Br
c)
COOH
COOH
O
CH3CH2CCl
HNO3
AlCl3
H2SO4
O2N
CCH2CH3
O
Key Ideas from Chapter 18
❑
Aromatic molecules are electron-rich and susceptible to
electrophilic attack by positively charged reagents. Unlike
alkenes, which add electrophiles, aromatic molecules undergo
electrophilic substitution.
❑
The intermediate in the mechanism for electrophilic aromatic
substitution is called a σ complex. The σ complex is a
resonance-stabilized carbocation.
❑
The mechanism for an electrophilic aromatic substitution
involves an attack by the electrophile followed by the loss of a
proton to restore the aromaticity of the ring.
❑
Substitution of the aromatic ring of benzene and its derivatives
is possible by treatment with HNO3 and another strong acid
(usually H2SO4). This reaction generates the nitronium ion
(⊕NO2) electrophile and produces an aromatic nitrate (ArNO2.)
❑
To substitute a bromine or chlorine atom, heat benzene with
the halogen in the presence of a Lewis acid. In most cases this
reaction must be run in situ with elemental Fe, which produces
small amounts of FeX3, the Lewis Acid catalyst.
❑
The methods used to place bromine and chlorine on a benzene
ring do not usually work for fluorine and iodine.
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❑
Benzenesulfonic acids are synthesized by reaction of an arene
with fuming sulfuric acid. Fuming sulfuric acid is a solution of
SO3 in concentrated H2SO4.
❑
The Friedel-Crafts reaction either alkylates or acylates an
aromatic ring. Alkylation occurs when an alkyl halide reacts
with an arene in the presence of a Lewis acid. Acylation
requires an acyl halide instead of an alkyl halide.
❑
Any method for generating a carbocation works well for the
Friedel-Crafts reaction.
❑
Because the Friedel-Crafts reaction generates a carbocation,
the carbon skeleton tends to rearrange whenever possible.
❑
The Friedel-Crafts reaction does not occur if there is an
electron-withdrawing group on the ring. Halogens are the
exceptions because they have nonbonding electrons to stabilize
the σ complex.
❑
The Clemmensen reduction uses zinc amalgam and HCl to
convert many ketone or aldehyde carbonyl groups to CH2
groups.
❑
The Wolff-Kishner reduction also converts many ketone or
aldehyde groups to CH2 groups using NH2NH2 and a strong
base such as KOH.
❑
The presence of a substituent on the ring determines the
regiochemistry of the second substituent being placed on the
ring.
❑
Electron-donating substituents direct incoming electrophiles to
the ortho and/or para positions on the ring.
❑
Electron-withdrawing
substituents
direct
electrophiles to the meta position on the ring.
❑
Electron-donating substituents increase the rate of reaction for
the incoming electrophile.
❑
Electron-withdrawing substituents reduce the rate of reaction
for the incoming electrophile.
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❑
Halide substituents are an exception to the rules for how
current substituents affect reactions. Halide substituents are
electron withdrawing and slow the rate of reaction. But they
direct new substituents to the ortho and para positions,
however, because the nonbonding electrons on the halogen
stabilize the resonance structures.
❑
When a benzene ring contains two or more substituents, you
must consider their combined influences to determine where an
incoming electrophile will go. The two groups either direct to
the same location or they compete to direct the incoming
electrophile to different locations. Generally, an electrondonating group exerts a greater influence than does an
electron-withdrawing group.
❑
Naphthalene has two possible positions for substitution: C1
and C2. Of the two positions, C1 is the more reactive because
two of the resonance contributors do not remove the
aromaticity of the second ring. Substitution at C2 has only one
such resonance contributor.
❑
Diazotization produces a diazonium salt. A diazonium salt is an
arene bearing an —N2⊕ substituent.
❑
Azo coupling is the electrophilic substitution of a diazonium
salt on an aromatic ring bearing an electron-donating group.
Many azo compounds, such as methyl orange and para red, are
used as dyes.
❑
The Sandmeyer reaction reacts copper(I) bromide, chloride, or
cyanide with a diazonium salt to form an aryl bromide,
chloride, or nitrile.
❑
When a diazonium salt reacts with HBF4, it produces an aryl
fluoride. This reaction is known as the Schiemann reaction.
❑
A reaction of a diazonium salt with potassium iodide produces
an aryl iodide.
❑
Warming an aqueous reaction solution containing a diazonium
salt to room temperature or slightly warmer produces a phenol.
❑
To replace a diazo group with hydrogen, react the salt with
hypophosphorus acid (H3PO2).
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❑
Aryl halides bearing strongly electron-withdrawing groups
undergo nucleophilic aromatic substitution with strong
nucleophiles or with weaker nucleophiles only under extreme
reaction conditions.
❑
Benzyne is a benzene ring with an additional π bond between
two adjacent carbons. This additional π bond is very reactive
and can undergo an addition reaction, or it can act as a
dienophile in a Diels-Alder reaction.
❑
Aromatic synthesis requires that you take into account the
directive influences of any substituents already on the ring.
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