Topic 30: Electrophilic Aromatic Substitution
Note to students: This is a single chapter (or a portion of one) from a
textbook that is under construction. Therefore you can ignore
references to other textbook sections.
30.1 Why Should I Study This?
In this topic we will explore a series of related reactions in which a
hydrogen atom of an aromatic ring is substituted (replaced) by another
atom or group. The mechanistic entity that brings about this
substitution process is an electrophile (abbreviated as Elec), so we
refer to this reaction as electrophilic aromatic substitution (EAS).
Elec
H
Elec
EAS is useful in to make new aromatic molecules, and therefore is
commonly used in industrial-scale synthesis. For example, EAS is a
key step in the conversion of toluene into 2,4,6-trinitrotoulene (better
known as TNT, an explosive) and naphthalene into FD&C Red No. 40
(a red dye). Toluene and naphthalene are both available in huge
quantities from petroleum.
CH3
CH3
O2N
NO2
EAS
NO2
Toluene
2,4,6-Trinitrotoluene (TNT)
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Electrophilic Aromatic Substitution
1
Electrophilic aromatic
substitution: A reaction in
which an electrophile adds
to an aromatic ring, ultimately replacing an aromatic
ring hydrogen atom.
Na O3S
OH
N
EAS
N
FD&C Red No. 40
OCH3
Naphthalene
CH3
SO3 Na
We will explore these reactions, as well as many others, in more detail
as this topic unfolds. Although there are few (if any) examples of
biochemical EAS reactions, this topic is worthy of our time and effort
because through its study we will uncover some interesting lessons
about the influence of molecular structure on a reaction’s mechanism,
rate and products.
30.2 Why is Benzene Different? Substitution versus
Addition
Note that the electrophile
is just a protonated leaving
group, and that the leaving
group functions as a base
in a later mechanism step.
We saw this pattern other
mechanisms: SN1 (section
25.xx), E1 (section 27.xx)
and some pi bond addition
reactions (sections 28.xx),
and we will see it again.
Let’s start with a very basic but very important observation about EAS
reactions. A benzene ring has pi electrons, so in analogy to what we
learned in Topic 28, we expect it to react with electrophiles. The
reaction of an alkene such as cyclohexene (a nucleophile) with strong
aqueous acid (H3O+, an electrophile) gives cyclohexanol, an addition
product.
H2O
H
OH2
OH2
H
OH
OH
Benzene has six pi electrons, so by analogy, we expect benzene to
behave much like cyclohexene.
Think Ahead Question 30.1
By working through the reaction mechanism, predict the product
formed by the reaction of benzene with H3O+.
Answer: The predicted mechanism looks very much like the
cyclohexene hydration reaction above, but the nucleophile has two
more double bonds.
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Electrophilic Aromatic Substitution
2
H2O
H
OH2
H
OH
OH2
OH
Arenium ion
The carbocation derived from electrophilic attack on the benzene ring
is called an arenium ion. (Arene is an old term for aromatic molecules
like benzene.) An arenium ion is also called a σ complex or a Wheland
intermediate.
Based on this mechanism, we predict the reaction product is
cyclohexa-2,4-dien-1-ol. However, this is just a prediction, and as we
have seen many times before, Mother Nature sometimes has her own
ideas. When we run the reaction in the lab in order to test the
prediction, we discover the reaction gives substitution products, and
not addition products. For example, heating benzene with a mixture of
aqueous sulfuric and nitric acids gives nitrobenzene (a substitution
product) and not the alcohol product expected by an addition
mechanism.
OH
aq. HNO3/H2SO4
NO2
or
Addition product
expected but not observed
Substitution product
actual product
Lets explore the reason why benzene undergoes substitution instead of
addition by examining the mechanism of a very similar reaction. The
reaction involves introduction of deuterium (2H or D), a hydrogen
isotope that behaves almost exactly like the more familiar hydrogen
isotope 1H in a reaction mechanism, yet it can be differentiated from
1
H in a molecule.* For example, D3O+ is the deuterium analog of H3O+.
Deuterium can help us track proton transfers during a reaction, and
draw conclusions about the reaction mechanism.
Think Ahead Question 30.2
Select the major product when an excess of benzene is reacted with
D3O+, formed in a mixture of D2SO4 and D2O.
*
There is one important difference in the mechanism behavior of hydrogen and
deuterium. Bonds to deuterium are stronger than bonds to hydrogen, so breaking a
bond to deuterium has a larger ΔG‡. This influences the reaction rate, and is called a
kinetic isotope effect.
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Electrophilic Aromatic Substitution
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Arenium ion: A carbocation formed by electrophilic attack on a benzene ring.
D
OD
D2O/D2SO4
or
Answer: In the first mechanism step, D3O+ (the electrophile) attacks
the benzene ring to give an arenium ion.
D
H
OD2
H
D
H
D
D
An arenium ion is just another type of carbocation, so now we
consider carbocation fates. The carbocation can capture a nucleophile,
be deprotonated or rearrange. Recall that carbocations are generally
highly unstable and very reactive, and so they are not very selective
with which carbocation pathway they use to achieve stability. (Review
section 24.xx if you don’t feel confident with carbocation fates.)
However, in this case deprotonation restores aromaticity whereas
capture of water does not. Deprotonation has an extra thermodynamic
incentive over the other carbocation fates. (Recall from section 29.xx
that for benzene, aromaticity is worth 36 kcal mol-1 of extra stability.)
Therefore deuterobenzene is the major product.
Capture nucleophile:
D2O
D
D
OD
O
D2O
H
H
H
D
D
D
+ D2O
D
Aromaticity not restored - Disfavored pathway
Deprotonation:
H
OD2
D
D
+ H
OD2
Aromaticity restored - Favored pathway
(Deprotonation of the deuterium instead of the hydrogen also restores
aromaticity, so it is also favored. However, because this pathway
regenerates benzene we cannot detect its occurrence.)
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Electrophilic Aromatic Substitution
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Because aromatic molecules have significantly less free energy than
similar molecules that are not aromatic, there is a strong driving force
to gain aromaticity whenever possible. For example, cyclohexa-2,4dien-1-ol quickly loses water to form benzene in order to gain
aromaticity.
OH
+ H2O
30.3 A General Mechanism for Electrophilic Aromatic
Substitution
Electrophilic aromatic substitution reactions all achieve the same
change in molecular structure: some other atom or group replaces
hydrogen atom on an aromatic ring. So you might reasonably assume
that the mechanisms for all these reactions also have quite a bit in
common. (We have encountered this general principle many times
before. Similar functional group transformations often have similar
reaction mechanisms.)
Think Ahead Question 30.3
write a general mechanism for a generic EAS reaction of an
electrophile (Elec) with benzene (first reaction in this topic), based on
the reactions in section 30.2.
Answer: As we have seen in section 30.2, the EAS mechanism is very
much like the addition of certain electrophiles to alkenes or alkynes.
The first step is electrophilic attack on the pi electrons, to form a
carbocation. Now we ask ourselves the usual carbocation question:
Which carbocation fate happens? As discussed in section 30.2,
deprotonation of the arenium ion has a great thermodynamic
advantage over nucleophile capture or rearrangement because
deprotonation restores aromaticity (worth an extra 36 kcal mol-1 in
benzene!) whereas the other carbocation fates do not.
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Electrophilic Aromatic Substitution
5
General Mechanism: Electrophilic Aromatic Substitution
Resonance contributors
H
B
Elec
Elec
Elec
Electrophile attacks benzene
ring forming resonance
- stabilized arenium ion.
Carbocation fate: deprotonation.
Only this carbocation fate restores
aromaticity.
The formal charges on the electrophile and base are not specified. This
is because our mechanism is a generalization, and we do not know the
exact structure of the electrophile or base. Without knowing their
structures we cannot assign formal charges. (Like all electrophiles,
Elec carries a δ+ or formal +1 charge. Similarly, B carries a δ- or
formal –1 charge.)
Ipso hydrogen: In an
electrophilic aromatic
substitution, the hydrogen attached to the carbon that is attacked by the
incoming electrophile.
Avoid This Common Problem
Hydrogen atoms are often not shown in skeletal structures (section
1.xx), but the hydrogen on the sp3 carbon atom that bears the
electrophile (called the ipso hydrogen) is shown explicitly. Why does
this hydrogen get special treatment? Sometimes students forget about
this hydrogen and think the carbon has an open octet. This leads to all
sorts of mechanism mistakes. Always drawing the ipso hydrogen will
help you avoid this common pitfall.
Ipso hydrogen
H
Elec
The arenium ion has resonance. As we shall see this resonance plays a
key role when an EAS reaction can give more than one product.
Think Ahead Question 30.4
Draw all of the important resonance contributors, plus the resonance
hybrid for the arenium ion in our general EAS mechanism. How does
this resonance influence the choice of major product and the reaction
rate? (Resonance is an important issue in EAS, so if you don’t feel
confident in your ability to draw these resonance contributors, review
section 5.4.)
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Electrophilic Aromatic Substitution
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Answer: The three important resonance contributors plus the hybrid
are shown.
H
H
Elec
H
Elec
Resonance contributors
H
Elec
Elec
Resonance
hybrid
How does resonance influence the choice of major product? Recall the
useful assumption that the more stable product or reaction intermediate
is the major one (thermodynamic control; section 20.xx). So if more
than one product is possible, the major product will be derived from
the more stable arenium ion (unless the reaction is under kinetic
control).
How does resonance influence the rate of the reaction? Arenium ion
resonance delocalizes the formal positive charge. This stabilizes the
structure and lowers the energy of activation required to form it.
Lower energy of activation results in a faster rate for that step of the
mechanism. But we cannot also say that it will result in a faster rate for
the entire reaction, because the general EAS mechanism has two steps
(or more if additional steps are required to form the electrophile) and
we do not know which is the rate-determining step (rds). As we have
seen numerous times before, it can be useful to explore the features
that influence a reaction’s rate, and EAS is no exception. Therefore we
need to determine which mechanism step is the rds.
Think Ahead Question 30.5
Which step of the general EAS mechanism is the rate-determining
step? Draw an energy profile for the general EAS reaction. Assume the
reaction is slightly exergonic.
Answer: The general mechanism includes two steps (electrophilic
attack to form the arenium ion and subsequent deprotonation of the
arenium ion), each with its own transition state and rate, so the energy
profile has two hills and two free energies of activation (ΔG‡). The
rate-determining (slowest) mechanism step has the largest ΔG ‡.
(Review section 20.xx if you are a bit fuzzy about any of these
concepts.)
In most EAS reactions, several mechanism steps are necessary to
produce the electrophile that actually attacks the benzene ring. For this
discussion we will assume these steps are not rate determining.
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Electrophilic Aromatic Substitution
7
(Kinetic studies suggest this assumption is probably true in many, if
not all EAS reactions.)
When we consider the thermodynamics of the two mechanism steps,
one feature is obvious. Electrophilic attack on the benzene ring (first
step) disrupts aromaticity. Deprotonation of the arenium ion (second
step) restores aromaticity. Aromaticity is a significant stabilizing
feature, and its loss is energetically expensive. Thus we expect ΔG‡ for
the electrophilic attack step to be higher than Δ G ‡ for the
deprotonation step. By this logic we predict that electrophilic attack is
the rate-determining step. (This analysis ignores other energy changes
such as loss and formation of bonds. Experimental kinetic evidence
confirms that the electrophilic attack step is rate determining. This
suggests that either aromaticity is in fact the deciding thermodynamic
factor or the bond energy analysis gives the same result.)
In General...
Compare the rate-determining step for EAS, SN1 and E1. Do you see a
pattern? In each case, the mechanism step in which the carbocation is
formed is rate determining. This pattern applies to many other
reactions as well. So as a general assumption we can say that
carbocation formation is often the rate-determining mechanism step.
Now that we know the rate-determining step we can draw the energy
profile.
H
Elec
δ+
H
B
Elec
Energy
δ+
H
ΔG1 rds
Elec
ΔG2
Elec
Reaction coordinate
Now that we have developed a general EAS mechanism and have an
idea of its kinetics, we can explore specific cases. Because EAS is
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Electrophilic Aromatic Substitution
8
such a valuable reaction for introducing new substituents to a benzene
ring, many different EAS reactions have been developed. We will
examine a few of the more important reactions. We will also see
patterns of reactivity, regioselectivity and mechanism that are common
to all of these EAS reactions. Studying these reactions from the
perspective of these patterns will ease your journey through this topic.
30.4 Halogenation
A. Bromination
EAS is an effective method to replace a benzene ring hydrogen with a
halogen such as bromine.
Br
EAS
Using what we know about electrophiles and EAS, we can deduce the
how this reaction might be achieved.
Think Ahead Question 30.6
What electrophile is needed to convert benzene to bromobenzene by
the EAS mechanism? How might this electrophile be generated? Write
an EAS mechanism for the conversion of benzene into bromobenzene.
Answer: This EAS involves replacement of a hydrogen atom by a
bromine atom, so by analogy to the general EAS mechanism (section
30.3) the electrophile is an electron-deficient bromine atom. This
could be bromine bearing a leaving group (Br-LG). In fact, we have
already seen Br2 react this way, when it adds to an alkene or alkyne
(section 29.xx). So the reaction of Br2 with benzene might proceed this
way:
H
Br
Br
Br
Br
Br
So with great anticipation, we go into the lab and mix Br2 with
benzene, expecting an instantaneous reaction (just like an Br2 with an
alkene or alkyne). But nothing happens!
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Electrophilic Aromatic Substitution
9
Why is Br2 less reactive with benzene than with an alkene? The
answer is aromaticity. As we have seen before (section 30.3), addition
of an electrophile to benzene is energetically expensive because
aromaticity is lost. Bromine is sufficiently electrophilic to add to an
alkene or alkyne (where aromaticity is not lost) but is just not powerful
enough to overcome the higher activation energy needed to disrupt
aromaticity.
How do we make Br2 more electrophilic? One way is to make the
leaving group depart more easily. Another way is to magnify the δ+
charge on one of the bromine atoms. (In other words, we want Br2 to
have a permanent bond dipole instead of an induced bond dipole.)
Both of these goals can be achieved by adding a strong Lewis acid
such FeBr3. Being electron-deficient, the Lewis acid borrows a
bromine lone pair to form a new bond, causing the bromine to have a
formal positive charge. Bromine with a formal positive charge is more
electronegative than neutral bromine, so a Br-Br bond dipole results.
(You can also think of this as an inductive effect.) The Br-Br bond is
also weakened.
δ+
Br
Br
FeBr3
Lewis acid;
accepts electrons
Br
Br
FeBr3
δ+ due to Br+
With a larger δ+ on Br and a weakened Br-Br bond, the Br2-FeBr3
complex is sufficiently electrophilic to overcome the high activation
energy caused by the benzene ring’s loss of aromaticity.
FeBr3 is deliquescent (it absorbs enough water from the air to form a
solution). Moist FeBr3 is not effective as an EAS catalyst. However,
FeBr3 is easily generated in situ by reacting Fe and Br2, so we can
write the overall reaction this way:
Br
Fe, Br2
Think Ahead Question 30.7
Write the EAS mechanism for the reaction of the Br2-FeBr3 complex
with benzene.
Answer: The reaction follows the general EAS mechanism we worked
out before.
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Electrophilic Aromatic Substitution
10
H
Br
FeBr3
Br
Br
Br
δ+
+ HBr + FeBr3
BrFeBr3
The FeBr4- functions as a base in order to neutralize the formal
negative charge on iron. Recall that carbocations are unstable and
highly reactive, so they can be deprotonated even by weak bases
(section 20.xx). Compared to other carbocations, arenium ion
deprotonation is exceptionally easy because aromaticity is restored.
Note that FeBr3 is regenerated in the last mechanism step. It can react
with another Br2 molecule, so it can be considered as a catalyst in this
reaction.
We can often draw two or more similar mechanisms for a reaction that
vary somewhat in the sequence of bond breaking and bond forming.
Your answer to Think Ahead Question 30.7 might be a bit different
that the mechanism just presented. For example, the electrophile that
attacked the benzene ring could be Br+ instead of the Br2-FeBr3
complex.
Br
Br + FeBr4
BrFeBr3
H
Br
Br
Or you may have used Br- instead of FeBr4- to deprotonate the arenium
ion.
Br
FeBr3
Br + FeBr3
H
Br
Br
Br
Both of these are reasonable mechanism alternatives. (The actual
mechanism is dependent on the reactants that are used as well as other
reaction variables such as solvent, temperature, etc. On paper we can
accept any mechanism that is reasonable and consistent with the facts
at hand.)
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Electrophilic Aromatic Substitution
11
Concept Focus Question 30.1
Write the products and mechanism for this reaction.
H3 C
CH3
Br2, Fe
Concept Focus Question 30.2
(a) Why is FeBr3 a Lewis acid?
(b) Which one of these substances is not a Lewis acid: FeCl3, AlCl3,
BF3, CCl4.
B. Chlorination
Now that we know how to introduce bromine onto a benzene ring by
EAS, let’s consider the same transformation with other halogens.
Think Ahead Question 30.8
What reagents might be used to convert benzene into chlorobenzene?
Write a mechanism.
Answer: We used Br2 and Fe to convert benzene into bromobenzene,
so by analogy, Cl2 and Fe might useful to convert benzene into
bromobenzene. The mechanism is also similar.
Cl
Fe, Cl2
Mechanism: EAS Chlorination of Benzene
FeCl3
Fe + Cl2
δ+
Cl
Cl
FeCl3
Cl
Cl
FeCl3
H
Cl
FeCl3
Cl
Cl
Cl
δ+
ClFeCl3
+ HCl + FeCl3
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Electrophilic Aromatic Substitution
12
In parallel to the bromination mechanism, the electrophile could be Cl+
instead of the Cl2-FeCl3 complex, and the base could be Cl- instead of
FeCl4-.
C. Iodination and Fluorination
Iodination of a benzene ring can also be accomplished by EAS. The
electrophile is I+, which is conveniently generated from I2 by oxidation
with HgO or HNO3. (An I2-Lewis acid complex is not sufficiently
electrophilic.)
I
I2, HgO
We might expect a benzene ring to be fluorinated by F+, but the high
electronegativity of fluorine makes this impractical. Reagents for
fluorination in which a fluorine atom bears a leaving group (F-LG)
have been developed. However, the most effect strategy to introduce
fluorine onto a benzene ring is to replace an existing benzene ring
substituent (such as NH2) with fluorine (section 40.xx).
Concept Focus Question 30.3
Give the product of this reaction:
CH3
I2, HgO
H3C
Concept Focus Question 30.4
There is some evidence that conversion of benzene to iodobenzene
with I2 and SO3 may involve I3+ instead of I+. Write a mechanism for
the I3+ version of this EAS reaction.
I
I2, SO3
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Electrophilic Aromatic Substitution
13
Fluorine is considered the
most reactive element,
exothermically attacking
almost everything it encounters, including glass.
Only certain metal alloys
or substances that are
already highly fluorinated
(such as Teflon) resist its
aggressive nature. This
aggressiveness makes F2
impractical for use in
normal organic synthesis.
30.5 Substituent Effects: Electron-Donating Groups
A. Directing Effects
So far we have explored EAS reactions in cases where only a single
reaction product is possible. In the vast majority of cases, however,
more than one product might be formed.
Think Ahead Question 30.9
EAS bromination of toluene with Br2 and Fe leads to more than one
bromotoluene isomer. How many isomers are possible and what are
their structures? (It may be useful at this point to review the
nomenclature of substituted benzenes in section 6.xx.)
Answer: There are three bromotoluene products, namely the ortho,
meta and para isomers.
CH3
CH3
CH3
CH3
Br
Br2, Fe
+
+
Br
ortho-bromotoluene
meta-bromotoluene
Br
para-bromotoluene
(You might think that because there are five benzene ring hydrogens
that there are also five possible products. Compare models of all five
possibilities to determine which are in fact identical.)
Which isomer is the major one? Several factors contribute to this.
Think Ahead Question 30.10
Toluene has two hydrogen atoms that are ortho to the methyl group,
two that are meta and just one that is para. How does this fact
influence the ortho/meta/para product ratio?
Answer: EAS involves replacement of a benzene ring hydrogen with
some other atom or group. Ignoring all other factors, there is a greater
chance that an ortho or meta hydrogen is replaced because there are
two of each of these, but only one para hydrogen. Based just on this
statistical analysis, the product will consist of 40% of the ortho isomer,
40% of the meat isomer and 20% of the para isomer.
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Electrophilic Aromatic Substitution
14
CH3
Two ortho hydrogens
H
H
Two meta hydrogens
H
H
H
One para hydrogen
Think Ahead Question 30.11
How does steric hindrance influence the product ratio during the EAS
bromination of toluene? It may be useful to examine a model of
toluene.
Answer: First, let’s review the basics of steric hindrance. For a new
bond to form, the atoms in question must come within bond distance.
This can be slowed or even prevented if the reactants have group(s) of
atoms that block this close approach. For example, SN2 reactions do
not occur at tertiary carbons because the alkyl groups prevent approach
of the nucleophile. (Read section 22.xx if necessary to review the
concept of steric hindrance.)
Now we can rephrase Think Ahead Question 30.11. Which of the
benzene ring hydrogen of toluene is the most crowded (or in the words
of one student, “has the most shrubbery”)? Refer to your model of
toluene to help you through this discussion. The para and meta
hydrogens are each flanked by two hydrogens. The ortho hydrogens
are flanked by a hydrogen atom and a methyl group. Hydrogen atoms
are small and offer no significant hindrance as the electrophile
approaches. However, a methyl group is larger than a hydrogen atom,
and thus offers some steric hindrance to approach by the electrophile.
Therefore we can conclude, based just on steric hindrance, that attack
at the meta and para positions is favored over attack at the ortho
position. This can be seen in the following space filling model.
Ortho attack
more hindered
Meta attack
less hindered
Para attack
less hindered
We now have two factors that might influence the EAS product ratio.
Based upon your organic chemistry experience so far, neither of these
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Electrophilic Aromatic Substitution
15
factors is obviously dominant, so any conclusion must consider them
both. Meta attack appears most favorable because there are two meta
hydrogens and no steric hindrance. Whether attack at the ortho or para
position is more favorable is a toss-up. The para position is less
hindered but there is only one para hydrogen. The methyl group
hinders the ortho position but there are two ortho hydrogens.
However, we have yet to consider how the details of the reaction
mechanism might also influence the product distribution.
Think Ahead Question 30.12
Write the complete mechanism, including all significant resonance
contributors, for the three products formed when toluene is reacted
with Br2 and Fe. How do the differences in these mechanisms
influence the product distribution?
Answer: Contrasting the three mechanisms requires us to focus on
their differences. Mechanism details that are the same for all three
bromotoluene isomers will not help us here. Formation of the Br2FeBr3 complex does not vary, so we can ignore it for this discussion.
The next portion of the mechanism, electrophilic attack on the benzene
ring leads to three different arenium ions, so lets start our exploration
at that point.
Ortho attack:
CH3
δ+
Br Br
CH3
FeBr3
CH3
H
CH3
H
H
Br
Br
Br
Meta attack:
CH3
δ+
Br Br
CH3
CH3
CH3
FeBr3
Br
Br
Br
H
H
H
Para attack:
CH3
CH3
δ+
Br Br
FeBr3
H
Br
CH3
CH3
H
Br
H
Br
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Electrophilic Aromatic Substitution
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Recall that in general, the formation of the more stable mechanism
intermediate (the arenium ion, in this case) has a lower energy of
activation, a greater reaction rate for the mechanism pathway, and
therefore a greater abundance of the corresponding reaction product.
(Review section 20.xx if this concept isn’t fresh in your mind.)
So which arenium ion is more stable? The arenium ion derived from
ortho attack has three significant resonance contributors, two of which
are secondary carbocations and one of which is tertiary. Meta attack
gives an arenium ion that also has three resonance contributors, all of
which are secondary carbocations. Para attack, like ortho attack, also
gives an arenium ion with three significant resonance contributors, two
of which are secondary carbocations and one of which is tertiary.
Therefore the arenium ion from meta attack is less stable than the
arenium ion from ortho or para attack. Based only on arenium ion
stability, we predict that the ortho and para products will be formed in
greater amounts than the meta product.
So let’s summarize the predictions based on the three factors discussed
above.
Factor
Steric hindrance
Probability
Arenium ion stability
Product Ratio Prediction
meta, para > ortho
ortho, meta > para
ortho, para > meta
These factors are not synergistic, and no obvious conclusion can be
drawn, unless we can decide which factor is dominant. To do this we
need some actual experimental data, which is given in Table 30.1.
(The mechanisms for each of these reactions are discussed later on in
this topic.)
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Electrophilic Aromatic Substitution
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Remember that a methyl
group stabilizes an adjacent
carbocation by electron donation. Review section 20.xx
on carbocation stability if
necessary.
Table 30.1: Toluene EAS Product Ratios
CH3
CH3
CH3
CH3
X
EAS
X
X
Ortho
Meta
Para
60
59
<1
4
40
37
57
3
40
42
19
39
9
1
89
Br2, Fe (X = Br)
Section 30.4A
Cl2, Fe (X = Cl)
Section 30.4B
HNO3, H2SO4 (X = NO2)
Section 30.9
CH3CH2Cl, AlCl3 (X =
CH2CH3)
Section 30.11
O
O
PhCCl, AlCl3 (X = PhC )
Section 30.12
This data is most consistent with arenium ion stability as the dominant
factor. Furthermore, probability carries more weight than steric effects,
unless the steric effects are severe, as illustrated by the data in Table
30.2. This order of importance applies to the vast majority of EAS
reactions.
Table 30.2 Ortho : Para Product Ratios for Nitration of Alkylbenzenes
R
R
R
NO2
HNO3
H2SO4
NO2
R = CH3
R = CH2CH3
R = C(CH3)3
59
50
18
41
50
82
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Electrophilic Aromatic Substitution
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In General...
Relative Importance of Factors Influencing Regioselectivity in EAS
Reactions:
Arenium ion stability > number of hydrogens > steric effects
The methyl group controls the regioselectivity of toluene EAS
reactions. It serves the same role as the director in a movie, telling the
actors where to go, so we call the methyl group an ortho/para
director. The methyl group exerts this directing effect because the
arenium ions from electrophilic attack to the ortho and para positions
are more stable than the arenium ion that comes from meta attack.
There are many different substituents that a benzene ring might have.
Can we make a generalization about which ones are ortho/para
directors?
Think Ahead Question 30.13
A methyl group is an ortho/para director because it stabilizes an
adjacent carbocation. What other benzene ring substituents might also
be ortho/para directors? It may be useful to review the list of
functional groups in section 1.xx and discussion of carbocation
stability in section 24.xx.
Answer: A methyl group exerts its directing effect by releasing
electron density to an adjacent carbocation, thereby increasing arenium
ion stability. Therefore any atom or group that can release electron
density should also be an ortho/para director.
What features must the director have in order to release electron
density? We have already visited this issue when we explored
carbocation stability (section 24.xx). We found that alkyl groups
release electron density by inductive effects. We also found that
carbocations enjoy significant resonance stabilization from adjacent
carbon-carbon pi bonds (alkenes, alkynes and aromatic rings) and
atoms with lone pairs (oxygen, nitrogen and the halogens being most
important). Therefore we predict that alkyl groups, and any group that
has a pi bond or lone pair adjacent to the benzene ring will be an
ortho/para director.
_________________________________________________________
Electrophilic Aromatic Substitution
19
Ortho/para director: In
EAS, a benzene ring
substituent that causes
attack at the ortho and
para positions of the
benzene ring.
R
Alkyl groups stabilize arenium
ions by inductive effects.
Pi bonds stabilize arenium
ions by resonance.
X
X
Lone pairs stabilize arenium
ions by resonance.
Of course all of these groups are not as equally effective or influential.
The effectiveness of resonance stabilization varies with different atoms
and groups. For example, atoms that are more electronegative or
groups with strong inductive electron-withdrawing effects provide less
resonance stabilization. Table 30.3 summarizes the relative strength of
various ortho/para directors. As you examine the substituents in Table
30.3 try to understand how each entry exerts its ortho/para directing
effect.
_________________________________________________________
Electrophilic Aromatic Substitution
20
Table 30.3 Relative Influence of Ortho/Para Directors
Substituent
Directing Strength
H
Directing Cause
Lone pair resonance
strongly activating
N
H
R
N
H
R
N
R
OH
OR
-
O
N
H
R
O
O
R
Ar
H
weakly activating
R
Inductive effects
Aromatic ring resonance
None (reference point)
In General
Aromatic ring substituents that release electrons, either through
resonance or inductive effects, are ortho/para directors.
Concept Focus Question 30.5
Draw all of the important resonance contributors for the arenium ions
formed from each of these molecules when reacted with Br2 and Fe.
Without referring to Table 30.3 decide whether of not the substituent is
an ortho/para director.
OH
(a)
H
N(CH3)2
(b)
(c)
N
O
H3C
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Electrophilic Aromatic Substitution
21
The “Ar” abbreviation
means “any aromatic
ring” (not “argon”!).
O
N(CH3)3
(d)
(e)
N
H
H3C
Concept Focus Question 30.6
Briefly explain.
(a) NH2 is a stronger director than OH.
(b) An amine is a stronger director than an amide.
(c) NH2 is a stronger than N(CH3)2 (Hint: Examine models of the
arenium ions.)
Concept Focus Question 30.7
Suggest the major product of each reaction.
(a)
OCH3
Br2, Fe
(d)
CH3O
O Cl2, FeCl3
(e)
H3C
CH3
Cl2, Fe
CH3CH2
(b)
C(CH3)3
N
H
(c)
I2, HgO
B. Activating Effects
We have discussed how electron-releasing substituents influence the
regioselectivity of an EAS reaction: they are ortho/para directors. But
how do they influence the rate?
Think Ahead Question 30.14
Rank these molecules in order of increasing reaction rate with Br2 and
Fe: benzene, toluene and phenol.
Answer: Remember that for a reaction mechanism with more than one
step, it is the step with the highest energy of activation that has the
greatest influence on the reaction rate. This is the rate-determining step
(rds). It is the rds that controls the overall reaction rate, so we need
only consider this step when exploring the effects of some variable
(such as the identity of benzene ring substituents) on reaction rate.
_________________________________________________________
Electrophilic Aromatic Substitution
22
Cl2, Fe
So what is the rds for EAS? It is electrophilic attack on the benzene
ring (section 30.3). In this mechanism step, the benzene ring
(including its substituents) is the nucleophile, and an arenium ion is
produced.
How can a substituent influence nucleophilicity? As we have in many
cases before (sections 21.xx and 22.xx, for example) enhancing the
nucleophile’s electron density makes the reaction faster. Therefore any
substituents on the benzene ring that release electron density enhance
the reaction rate. This electron release might occur by inductive effects
(methyl and other alkyl groups do this) or by resonance (groups like
OH with an adjacent lone pair or pi bond). Therefore toluene and
phenol are more nucleophilic than benzene, so toluene and phenol
undergo nucleophilic attack more rapidly than benzene. Furthermore,
resonance is generally more effective at increasing nucleophilicity than
inductive effects (sections 21.xx and 22.xx), so phenol undergoes
electrophilic attack faster than toluene.
The electrostatic potential maps are useful to visualize a substituent’s
influence on benzene ring electron density. The center of the benzene
ring is red, indicating this is an area of lower electron density. In
toluene, where the methyl group is a weak electron donor (inductive
effect), the red area is diminished slightly. In phenol, where the
hydroxyl is a strong electron donor (resonance), the red area is
significantly diminished.
Molecule
Electrostatic Potential
Map
Comments
H
H
H
H
H
A hydrogen atom does not
release electron density
into the ring.
H
CH3
H
H
H
H
An alkyl group weakly
releases electron density
by inductive effects.
H
_________________________________________________________
Electrophilic Aromatic Substitution
23
OH
H
H
H
H
Lone pairs strongly release
electron density by
resonance.
H
How can a substituent influence arenium ion stability? We have
already explored this point in section 30.5A. The methyl group of
toluene enhances arenium ion stability by electron release through an
inductive effect. The hydroxyl group of phenol enhances arenium ion
stability by resonance. The formation of an arenium ion that is more
stable has a lower activation energy and faster reaction rate. Therefore
toluene and phenol accept an electrophile faster than benzene. Once
again, resonance is generally more effective for carbocation
stabilization than an inductive effect, so phenols undergo electrophilic
attack faster than toluene.
So to answer Think Ahead Question 30.14, the EAS reaction rates are:
phenol (fastest) > toluene > benzene (slowest).
A c t i v a t o r : In EAS, an
aromatic ring substituent
that enhances nucleophilicity and arenium ion
stability, thereby accelerating the reaction.
It is no coincidence that our exploration of substituent effects on
nucleophilicity and arenium ion stability reached the same conclusion.
This is because the same factor – electron density – is at work in both
cases. Electron-releasing substituents lead to faster reactions, and are
called activators. We also predict that activators are also ortho/para
directors, because electron density influences reaction rate and
regioselectivity in the same way. Experimental data agrees with this
prediction (although there is one important exception, discussed
section 30.7), so we can make a general statement.
In General
In EAS reactions, substituents that release electron density into the
benzene ring by resonance or inductive effects are ortho/para directors
and activators. (For an important exception to this generalization, see
section 30.7).
If the benzene ring has sufficient electron density, because the
activator (or combination of activators) is strong enough, then a
weaker electrophile can be used. For example, toluene has a weak
activator (CH3), and is inert to Br2 without FeBr3 or I2 without HgO.
On the other hand, phenol has a moderate activator (OH), so it has
enough electron density to react with Br2 or I2 without the need for
additional reactants. In fact, phenol is so strongly activated towards
_________________________________________________________
Electrophilic Aromatic Substitution
24
EAS that bromination does not stop until all of the ortho and para
positions have been brominated. Aniline (PhNH2) behaves similarly.
CH3
CH3
CH3
Br
Br2, FeBr3
+
Fails without FeBr3
Br
OH
OH
excess Br2
Br
Br
No FeBr3 necessary
Br
Concept Focus Question 30.8
In each pair of molecules, select the one that reacts fastest with Br2/Fe.
Draw the products of the faster reaction.
(a) Toluene or benzene
(c) Toluene or p-xylene
(b) Toluene or aniline
(d) Phenol or anisole
Concept Focus Question 30.9
Write a complete mechanism for the reaction of phenol with excess
Br2 to give 2,4,6-tribromophenol.
30.6 Substituent Effects: Electron-Withdrawing Groups
In section 30.5 we learned that electron-releasing groups are ortho/para
directors and activators. Not all benzene ring substituents are electronreleasing groups, however. What is the effect of electron-withdrawing
groups on EAS regioselectivity and reaction rate?
Think Ahead Question 30.15
Predict the EAS product(s) when nitrobenzene is reacted with Br2 and
Fe. Is this reaction faster or slower than the same reaction with
benzene? Hint: Consider the Lewis structure and resonance of the nitro
group.
Answer: We learned in the previous section that a ring substituent
controls EAS reaction rate and regioselectivity by its influence on the
electron density of the benzene ring. This electron density controls the
nucleophilicity of the ring and the stability of the arenium ion.
_________________________________________________________
Electrophilic Aromatic Substitution
25
Review section 1.xx if you
have trouble figuring out
the Lewis structure of the
nitro group.
The nitro group is not a functional group we have encountered
frequently, so let us begin our search to answer Think Ahead Question
30.15 by reviewing its structure. The Lewis structure shows the
nitrogen atom is bonded to the benzene ring and two oxygen atoms.
The nitro group has two significant resonance contributors, each with a
+1 charge on the nitrogen and a -1 charge on one oxygen.
O
O
N
N
O
D e a c t i v a t o r : In EAS, an
aromatic ring substituent that
decreases nucleophilicity and
arenium ion stability, thereby
slowing the reaction.
O
What does this Lewis structure suggest about the nitro group’s effect
on the electron density of the benzene ring? A nitrogen atom with a
formal positive charge is highly electronegative (more so than a
nitrogen without a formal charge), so it is drawing electron density
from the ring toward itself. This effect is amplified by the inductive
effect (section 21.xx) of the two oxygen atoms. Taken together these
observations suggest the nitro group to be a powerful electronwithdrawing group. It reduces benzene ring nucleophilicity and
decreases arenium ion stability, causing a slower EAS reaction.
Therefore the nitro group is a deactivator. By extension, any electronwithdrawing group is also a deactivator.
What about the regioselectivity for EAS bromination of nitrobenzene?
As before, we need to consider the arenium ions derived from attack at
the ortho, meta and para positions. Are all of these contributors
important? Do all of the contributors increase arenium ion stability?
Ortho attack:
O
N
O
O
δ+
Br Br
FeBr3
N
O
O
N
O
O
Br
O
H
H
H
N
Br
Br
Unimportant
contributor
_________________________________________________________
Electrophilic Aromatic Substitution
26
Meta attack:
O
O
N
O
δ+
Br Br
N
O
O
N
O
O
N
O
FeBr3
Br
Br
Br
H
H
H
Para attack:
O
N
O
O
δ+
Br Br
FeBr3
H
N
O
Br
O
N
O
H Br
Unimportant
contributor
O
H
N
O
Br
The arenium ion derived from ortho attack has three resonance
contributors, all of which are secondary carbocations. However the
third contributor is not very significant because it has a formal positive
charge on adjacent atoms. So this arenium ion is better described as
having just two resonance contributors. (We can also say that it is not
significant because the strong electron-withdrawing group causes
significant destabilization of the adjacent electron-poor carbocation.)
The arenium ion derived from para attack also has just two significant
resonance contributors for the same reason. In contrast, none of the
resonance contributors for the arenium ion from meta attack have a
problem with adjacent positive charges, so all three are significant. In
summary, the arenium ions from ortho and para attack each have two
significant resonance contributors whereas the arenium ion from meta
attack has three. Therefore meta attack leads to a more stable arenium
ion, so it is the preferred pathway. In other words, the nitro group is a
meta director.
Think Ahead Question 30.16
A nitro group is a meta director because it destabilizes an adjacent
carbocation. What other benzene ring substituents might also be meta
directors? It may be useful to review the list of functional groups in
section 1.xx and carbocation stability in section 24.xx.
The nitro group exerts its meta directing effect because it is electron
withdrawing. Based on this we can predict that any electronwithdrawing substituent will be a meta director. Experimental data
shows this prediction to be true. Table 30.4 lists some commonly
encountered meta directors. Like ortho/para directors, we can also rank
meta directors according to the strength of the influence on the rate of
_________________________________________________________
Electrophilic Aromatic Substitution
27
Meta director: In EAS, a
benzene ring substituent
that causes attack at the
meta positions of the
benzene ring.
EAS reactions. However, because the interplay of electronegativity,
inductive effects and other factors, it is not always readily apparent
why one meta director should be weaker or stronger than another.
In General
In EAS reactions, substituents that withdraw electron density from the
benzene ring are meta directors and deactivators.
Table 30.4 EAS Meta Directors
Substituent
Directing Strength
N
O
NH2R,
NH3,
NHR2,
NR3
CF3
strongly deactivating
O
O
S
OH
O
C
N
O
Cl
O
X
O
R
O
weakly deactivating
X = OH, OR, NH2, NHR, NR2
H
_________________________________________________________
Electrophilic Aromatic Substitution
28
CAUTION!!!
You may think that benzene ring substituents that feature an atom of
high electronegativity directly attached to the ring (such as OH or F)
are meta directors because they destabilize an adjacent carbocation by
an electron withdrawing inductive effect. However, these atoms
usually have one or lone pairs, which stabilize an adjacent carbocation
by resonance, causing ortho/para direction. Using our usual
assumption that resonance dominates other factors suggests (correctly
so) that the substituents in question are all ortho/para directors.
X
X has high electronegativity:
Inductive electron-withdrawing group
Less influential directing effect
X
X has lone pair:
Resonance electron-donating group
More influential directing effect
Concept Focus Question 30.10
A methyl group is an ortho/para director but a trifluoromethyl group
(CF3) is a meta director. Explain. Illustrate this point by writing the
reaction products formed when toluene and trifluoromethylbenzene are
treated with Br2 and Fe.
Concept Focus Question 30.11
(a) Explain why all functional groups in which a carbonyl is
directly bonded to the benzene ring (such as PhCO2H) are meta
directors.
(b) Explain why an acid chloride is a stronger meta director than a
ketone.
Concept Focus Question 30.12
Write the major product of each reaction.
(a)
NO2
(b)
CO2H
Br2, Fe
Cl2, FeCl3
(c)
CF3
(d)
I2, HgO
NO2
Br2, Fe
O2N
_________________________________________________________
Electrophilic Aromatic Substitution
29
30.7 Substituent Effects: The Halogens
You may have noticed that we have avoided discussing the directing
and activating effects of halogens, even though these are common and
important benzene ring substituents. This is because their behavior is a
bit quirky, as we will see in the following section.
Think Ahead Question 30.17
Based upon what we have already learned about EAS directing and
activating effects, write the reaction mechanism (including all arenium
ion resonance contributors) and major product(s) for the reaction of
fluorobenzene with Cl2 and Fe.
F
Cl2, Fe
???
Answer: As we have seen so many times before EAS regioselectivity
is controlled by arenium ion stability. Fluorine has three lone pairs that
can stabilize an adjacent carbocation by resonance (just like oxygen
and nitrogen) so this suggests it is an ortho/para director. The lone
pairs provide an extra, very significant resonance contributor in which
all atoms have full octets.
F
F
δ+
Cl Cl
FeCl3
H
F
F
Cl
H
Cl
H
F
Cl
H Cl
Most important
contributor
On the other hand, fluorine’s high electronegativity destabilizes an
adjacent carbocation, suggesting it to be a meta director. However (as
we were reminded in the previous Caution box) resonance usually
outweighs inductive effects, so we predict the net effect of fluorine to
be an ortho/para director.
We also predict fluorine to be an activator for the same reason it is an
ortho/para director. The lone pairs increase the nucleophilicity of the
benzene ring and provide extra stability for the arenium ion. This
reduces the energy of activation for electrophilic attack, and results in
a faster reaction.
_________________________________________________________
Electrophilic Aromatic Substitution
30
So is our prediction accurate? To test it, we turn to experimental data,
which is shown below.
F
F
Cl2, Fe
F
Cl
+
F
+
Cl
30%
1%
Cl
69%
The data reveals that our regioselectivity prediction was accurate, and
fluorine is an ortho/para director.
However, our rate prediction does not agree with the experimental
data. We predicted reaction of fluorobenzene to be faster than similar
reaction of benzene, but the opposite is true. If the experimental rate
for chlorination of benzene is taken as 1.0 then the rate for chlorination
of fluorobenzene is 0.74, a little bit slower. (In other words,
fluorobenzene is almost as reactive as benzene in EAS, despite the
presence of the most electronegative atom in the periodic table!) A
similar rate effect is observed for EAS reactions with chlorobenzene,
bromobenzene and iodobenzene. Their rates are all slower than
benzene itself, even though all halogens have lone pairs for resonance
donation.
The explanation for the unexpected EAS behavior of halogens (weakly
deactivating ortho/para directors) has been debated, but the general
agreement is that there is a subtle balance between their inductive
electron-withdrawing properties and their resonance electron-donating
properties, resulting in the observed reactivity.
In General
In EAS reactions, a halogen benzene ring substituent is a weakly
deactivating ortho/para director.
_________________________________________________________
Electrophilic Aromatic Substitution
31
CAUTION!!!
The regioselectivity of an EAS reaction is controlled by the
substituent(s) that are attached to the benzene ring at the start of the
reaction, and not the incoming electrophile (the group that is added to
the ring in the course of the reaction). For example, nitro is a meta
director and bromine is an ortho/para director. Bromination of
nitrobenzene gives m -bromonitrobenzene whereas nitration of
bromobenzene gives a mixture of o- and p-bromonitrobenzene.
NO2
NO2
Meta director
Br2, Fe
Br
Br
Br
aq. HNO3
ortho/para director
NO2
aq. H2SO4
Br
+
NO2
Concept Focus Question 30.13
Rank these molecules by their reaction rates in a typical EAS reaction:
benzene, toluene, fluorobenzene, and p-difluorobenzene.
Concept Focus Question 30.14
Write the major products of these reactions.
F
Br
Br2, Fe
(a)
Cl
(b)
I2, HNO3
(c)
F
Cl2, FeCl3
Br2, Fe
(d)
F
30.8 Substituent Effects: Effect of Multiple Substituents
Most of the EAS cases we have examined up to this point have
involved a benzene ring bearing a single substituent. In some of the
Concept Focus Questions we have explored cases with two equal
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Electrophilic Aromatic Substitution
32
substituents. In these cases there was little question as to which
product is formed because the substituent directing effects are
cooperative. What happens if the directing effects are not cooperative?
Lets look at some experimental data then derive a general statement.
Think Ahead Question 30.18
Based on the reactions shown below, formulate a general rule
concerning the regioselectivity of EAS on benzene rings bearing
multiple substituents. Only the major product is shown in each case.
OCH3
OCH3
Br
Br2
Reaction A:
CH3
CH3
OCH3
OCH3
Br2, Fe
Reaction B:
Br
NO2
CO2H
Reaction C:
NO2
Br2, Fe
CO2H
Br
NO2
NO2
Answer: Lets analyze each case individually and then try to find a
pattern. Check the substituent directing effect lists (Tables 30.3 and
30.4) as needed, but only after you have tried to deduce the effects you
have forgotten.
Reaction A: Methoxy is a strongly activating ortho/para director and
methyl is a weakly activating ortho/para director. (This benzene is so
electron rich that Br2 does not have to be activated with FeBr3.) The
product of this reaction tells us that the stronger ortho/para director has
more influence than the weaker ortho/para director.
stronger o/p director
OCH3
OCH3
Br
Br2
CH3
weaker o/p director
CH3
_________________________________________________________
Electrophilic Aromatic Substitution
33
Reaction B: Methoxy is an activating ortho/para director and nitro is a
deactivating meta director. The product of this reaction tells us that the
activating ortho/para director wins out over the deactivating meta
director. The relative strength of the directing effects even overrides
the greater steric hindrance of the methoxy group.
strong o/p director
OCH3
OCH3
Br2, Fe
Br
NO2
NO2
strong m director
Reaction C: Nitro is a strongly deactivating meta director and a
carboxylic acid is a weakly deactivating meta director. The product of
this reaction tells us that the weaker deactivating group controls the
regioselectivity.
weaker m director
CO2H
Br2, Fe
NO2
Br
CO2H
NO2
stronger m director
What is the general ranking of the relative influence of two
substituents?
In General
Stronger activators dominate over weaker activators (reaction A).
Weaker deactivators dominate over stronger deactivators (reaction C).
Activators dominate over deactivators (reaction B).
Many reactions obey this general ranking, regardless of the number of
substituents, although there are exceptions. When the substituents are
very similar in activating or deactivating strength, such as OCH3
versus NHCOCH3, all products are formed in roughly equal amounts,
although steric effects can influence this ratio.
Concept Focus Question 30.15
Rank each set of substituents in order of decreasing influence on EAS
reactions.
(a) -OH, -CH2CH3, -NO2
(b) -OH, -OCH3, -F
(c) -F, -CF3, -NO2
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Electrophilic Aromatic Substitution
34
Concept Focus Question 30.16
Give the major product of each reaction.
CH2CH3
Cl
Br2, Fe
Br2
(a)
(c)
OH
NH2
NO2
CO2H
Cl2, Fe
(b)
(d)
CH3
Cl2, Fe
CO2CH3
O
Up to this point we have only discussed EAS reactions that introduce
halogens. There are many other substituents that we can introduce
using EAS. All of these reactions follow the same general pattern we
have seen so far: an electrophile (Elec) attacks the aromatic ring,
ultimately replacing one of its hydrogens (Ar-H becomes Ar-Elec). For
each new EAS reaction we explore we will deduce the nature of the
electrophile by examining EAS reactions that are (or have been)
important on an industrial scale.
30.9 Nitration
Think Ahead Question 30.19
TNT (2,4,6-trinitrotoluene) was once prepared by reacting toluene
with an aqueous mixture of nitric and sulfuric acids until three nitro
groups were added. (TNT is no longer manufactured on a large scale.)
What is the electrophile in this reaction? What is the mechanism for
the conversion of toluene into p-nitrotoluene? Hint: You will need the
correct Lewis structure for nitric acid.
CH3
CH3
aq. HNO3
+
aq. H2SO4
NO2
CH3
CH3
NO2
CH3
NO2
aq. HNO3
aq. H2SO4
aq. HNO3
NO2
O2N
aq. H2SO4
NO2
NO2
2,4,6-trinitrotoluene
(TNT)
_________________________________________________________
Electrophilic Aromatic Substitution
35
Answer: Let us draw on what we already know about the EAS
mechanism. In the case of halogenation, the electrophile is a naked
halogen atom such (e.g., Br+) or a halogen cation bearing a leaving
group (e.g., the Br2-FeBr3 complex).
Br
comes from
+ Br
New bond forms here
Nitronium cation:
O
N
O
Applying the same idea to nitration, the electrophile is +NO2 (the
nitronium cation) or NO2 bearing a leaving group. Experimental
evidence supports the nitronium cation as the actual electrophile.
NO2
comes from
+ NO2
How is +NO2 formed? Nitric acid could lose an HO group, but
hydroxide is not a leaving group under these conditions. The OH can
be converted to water (a moderate leaving group) by protonation, a
feat easily achieved by H3O+ (from aqueous H2SO4). Once generated,
+
NO2 achieves EAS by the same general mechanism we developed in
section 30.3.
Mechanism: Electrophilic Aromatic Nitration
O
H2O
H
HO
H2O
O
N
N
O
Protonation makes
HO a better LG
+NO
2
Nitronium ion
(an electrophile)
O
Water leaves
CH3
CH3
O
O
O
N
N
O
adds to give most
stable arenium ion
O2N
H
CH3
OH2
NO2
Deprotonation
restores aromaticity
Why does nitric acid accept the proton at the OH oxygen instead of the
oxygen with a negative formal charge? Protonation of the OH oxygen
_________________________________________________________
Electrophilic Aromatic Substitution
36
does not cause a significant loss of resonance whereas protonation of
the negatively charged oxygen does.
EAS nitration is a fairly general reaction but it does have some
limitations. Nitric acid oxidizes primary and secondary amines, so the
reaction gives other products if the ring has an –NH2 or –NHR group.
Tertiary amines (-NR2) are not oxidized by nitric acid. Another reagent
that has been used is nitronium tetrafluoroborate (+NO2 -BF4). This is a
milder, much less acidic way to achieve EAS nitration.
Concept Focus Question 30.17
Suggest the major product(s) of these reactions.
CH3
OCH3
(a)
aq. HNO3
Ph
O
F3C
aq. HNO3
CH3
NO2 -BF4
(c)
aq. H2SO4
(b)
+
CF3
(d)
aq. HNO3
aq. H2SO4
aq. H2SO4
30.10 Sulfonation
Benzene sulfonic acids (Ar-SO3H) are present in several classes of
organic molecules that are manufactured on multi-ton scales, including
dyes and synthetic detergents. The sulfonic acid group can be
introduced onto the benzene ring by EAS, illustrated here as one step
in the synthesis of a detergent.
R
SO3
H2SO4
R = linear alkyl chain,
e.g., C12H25
Sulfonic acid:
O
S
OH
O
O
R
S
OH + R
O
HO3S
Think Ahead Question 30.20
What is the electrophile in the EAS sulfonation reaction? What is the
overall mechanism?
Answer: As in the previous section, we deduced the electrophile by
working backwards. The new substituent is SO3H so the electrophile
_________________________________________________________
Electrophilic Aromatic Substitution
37
A mixture of SO3 and
H2SO4 is called fuming
sulfuric acid or oleum.
might be +SO3H. Working out the correct Lewis structure for +SO3H
suggests it might come from protonation of SO3 by H2SO4.
O
O
S
comes from
OH
+
S
O
O
O
+
OH
S
O
OH
S
comes from
O + "H "
O
O
Now we can write the complete mechanism for electrophilic aromatic
sulfonation.
Mechanism: Electrophilic Aromatic Sulfonation
O
O
O
S
O
H
S
OSO3H
OH
O
O
O
S
OH
Protonation of SO3 makes
it a stronger electrophile
R
R
R
O
O
S
HO3S
H
OSO3H
SO3H
OH
+
SO3H adds to give most
stable arenium ion
Deprotonation
restores aromaticity
A slightly different mechanism is also possible in which SO3 itself is
the electrophile. The sulfur atom carries a large δ+ charge due to the
three highly electronegative oxygen atoms. If SO3 is the electrophile
the mechanism varies only in the timing of the proton transfer steps
(see Concept Focus Question 30.19 at the end of this section).
Experimental evidence suggests that concentration, temperature,
aromatic substrate, and other variables determine whether the actual
electrophile is SO3 or +SO3H. In this textbook we will accept either
mechanism for this reaction. (Other related electrophiles such as
protonated sulfuric acid, H3SO4+, have also been implicated.)
Unlike other EAS reactions, sulfonation is easily reversible, so the
SO3H group can be replaced with a hydrogen atom. The
_________________________________________________________
Electrophilic Aromatic Substitution
38
benzenesulfonic acid need only be heated with dilute aqueous acid to
remove the sulfonic acid group.
SO3, H2SO4
SO3H
H3O+
So a sulfonic acid group is a removable meta director. Its addition and
later removal can be a useful strategy to control regioselectivity of
other EAS reactions, and to synthesize substituted benzene derivatives
that may not be easily made by other routes. For example, consider the
synthesis of m-dibromobenzene from benzene. Treatment of benzene
with excess Br2 and Fe gives ortho- and para-dibromobenzene, but not
the meta isomer.
Br
Br
Br
Br
Fe, Br2
Fe, Br2
+
Br
However, bromination of benzenesulfonic acid followed by
desulfonation gives the desired isomer.
SO3H
SO3
SO3H
H3O+
Fe, Br2
H2SO4
Br
Br
Br
Br
Concept Focus Question 30.18
Give the major products of these reactions.
(a)
C(CH3)3
O
(b)
SO3
H2SO4
SO3
H2SO4
(c)
(d)
SO3H
HO3S
SO3
H2SO4
I
H3O+
heat
Concept Focus Question 30.19
Write the mechanism for the sulfonation reaction of Concept Focus
Question 30.18(a), using SO3 (not +SO3H) as the electrophile.
_________________________________________________________
Electrophilic Aromatic Substitution
39
Concept Focus Question 30.20
Write the mechanism for the desulfonation reaction of Concept Focus
Question 30.18(d).
30.11 Friedel-Crafts Alkylation
In sections 30.4, 30.9 and 30.10 we explored reactions for introducing
halogens, nitro groups and sulfonic acid groups onto a benzene ring.
Synthesis of complex molecules often involves building a more
complex carbon skeleton, so can we use EAS to add alkyl groups to a
benzene ring?
Friedel-Crafts alkylation:
An EAS reaction in which
an alkyl carbocation or
related structure adds an
alkyl group to the aromatic
ring.
Think Ahead Question 30.21
In 1877, Charles Friedel and James Craft published a new reaction for
making bonds between sp3 carbons and benzene ring carbons. This
reaction came to be called the Friedel-Crafts alkylation. For
example, reaction of tert-butyl chloride, benzene and aluminum
chloride gives a mixture of ortho- and para-tert-butyltoluene. What is
the electrophile in this EAS reaction? Write a complete mechanism for
the EAS reaction.
CH3
CH3
(CH3)3CCl
CH3
C(CH3)3
+
AlCl3
C(CH3)3
Answer: Using the same EAS analysis strategy we have used in the
past several sections of this topic, we can disconnect the electrophile
from the benzene ring, revealing it to be a tert-butyl carbocation.
CH3
CH3
comes from
+ C(CH3)3
C(CH3)3
We have produced carbocations before by ionization of a carbonleaving group bond (SN1 and E1 reactions) or in some pi bond addition
reactions. Either of these carbocation sources could also be used in the
_________________________________________________________
Electrophilic Aromatic Substitution
40
EAS reaction. Historically, an SN1-like reaction was developed first, so
that is where we shall start.
The tert-butyl carbocation can be formed by ionization of tert-butyl
chloride, but as we saw before (sections 25.xx and 27.xx) this is an
energetically expensive process. The role of the aluminum chloride is
to assist this ionization by first forming a Lewis acid-halogen bond
(just as we saw in EAS bromination and chlorination, section 30.4).
With these issues resolved we can write the mechanism.
Mechanism: Friedel-Crafts Alkylation
(CH3)3C
Cl
AlCl3
Lone pair from chlorine
fills open octet of aluminum
(CH3)3C
Electrophile adds to
benzene ring
(CH3)3C + ClAlCl3
AlCl3
Chlorine leaving group ability
improved; bond ionizes
CH3
CH3
CH3
C(CH3)3
Cl
(H3C)3C
H
Cl
AlCl3
C(CH3)3
Deprotonation restores
aromaticity
The Friedel-Crafts reaction can be used to add a wide variety of alkyl
groups to a benzene ring, but the mechanism cannot be exactly the
same for all cases.
Think Ahead Question 30.22
Write the reaction and mechanism for the conversion of cumene
(isopropylbenzene) into p-methyl isopropylbenzene by a Friedel-Crafts
reaction.
Answer: Addition of a methyl group suggests the alkyl halide
component of the reaction is CH3Cl. Thus the reaction is:
CH3Cl
AlCl3
CH3
_________________________________________________________
Electrophilic Aromatic Substitution
41
What is the electrophile? Methyl chloride reacts with AlCl3 to form a
CH3Cl-AlCl3 complex. Ionization of this complex is difficult because
it would form a very unstable methyl carbocation. This suggests the
electrophile must be something other than a carbocation. Since the
CH3Cl-AlCl3 complex has a leaving group, it can fulfill this role, as
shown in the following mechanism.
H3C
Cl
H3C
AlCl3
Cl
X
AlCl3
H3C + ClAlCl3
CH3+ too unstable; no ionization
H3C
Incipient carbocation: A
molecule in which carbon
bears a large _+ charge
and a good leaving group.
Behaves in most ways like
a carbocation.
ClAlCl3
H3C
H
Cl
AlCl3
CH3
The carbon-chlorine bond of the CH3Cl-AlCl3 complex is very weak
and the carbon bears a large δ+ charge, more so than the carbon of
CH3Cl itself. It is as if the carbon-chlorine bond is almost completely
broken and the carbon is almost (but not quite yet) a carbocation. This
is called an incipient carbocation. Because the carbon bearing the
leaving group is secondary, the electrophile that attacks the benzene
ring could be either a carbocation or an incipient carbocation.
An incipient carbocation behaves very much like a normal “naked”
carbocation. For example, it is prone to rearrangement as shown in the
following example.
CH3CH2CH2Cl
AlCl3
You might ask why the Friedel-Crafts alkylation reaction of Think
Ahead Question 30.21 involving (CH3)3C+ did not avoid the
energetically expensive carbon-chlorine bond ionization and instead
used the (CH3)3CCl-AlCl3 complex directly as the electrophile. The
answer lies in steric effects: the carbon bearing the chlorine is tertiary,
and thus too hindered to undergo nucleophilic attack.
δ+
(CH3)3C
Cl
AlCl3
X
No nucleophilic attack at 3o carbon
_________________________________________________________
Electrophilic Aromatic Substitution
42
In General
The electrophile in a Friedel-Crafts alkylation reaction can be a
carbocation or an incipient carbocation (RCl-AlCl3 complex). The
more stable the carbocation is, the greater chance that RCl-AlCl3
complex ionizes before attacking the benzene ring. When R is methyl
or 1o the electrophile is the complex. When R is 2o either the complex
or the carbocation can be the electrophile. When R is 3o the
electrophile is most probably the carbocation. (This is analogous to the
choice of SN2 versus SN1 mechanism for an ionic substitution reaction,
section 25.xx.)
Here are a few limitations of the Friedel-Crafts alkylation reaction.
• The reaction fails if the benzene ring is too electron deficient
due to the presence of one or more strong deactivating groups.
(Several weak deactivating groups could combine to have a
similar effect.) For example, the reaction does not occur if the
benzene ring has a nitro, sulfonic acid, ketone or aldehyde
substituent. This reduced reactivity is a combination of
decreased benzene ring nucleophilicity plus reduced arenium
ion stability. For this reason nitrobenzene is sometimes used as
a solvent for Friedel-Crafts reactions. This problem can be
resolved in some cases by running the reaction with an
electron-rich benzene ring, then changing the substituents in
subsequent reactions.
•
The product of a Friedel-Crafts alkylation is more activated
that the starting material due to the presence of an additional
electron-donating alkyl group. Thus the reaction product may
alkylate faster than the starting material, and multiple
alkylations can result. This problem can be addressed by using
a large excess of the starting material, so that the electrophile
has a great chance of colliding with a molecule of starting
material before it finds a molecule of product. This problem
can be avoided with the Friedel-Crafts acylation (section
30.12).
CH3
CH3Cl
+
AlCl3
Major product only
when benzene is
used in large excess
dimethylbenzenes
+
trimethylbenzenes
+
etc.
Can be significant
products if benzene
is not in large excess
_________________________________________________________
Electrophilic Aromatic Substitution
43
Most any alkyl halide can be used, including fluorides,
chlorides, bromides and iodides. However aryl or vinyl halides
cannot be used because the intermediate carbocations are too
unstable to form and direct “SN2” attack at an sp2 carbon is not
possible (section 22.xx).
•
Concept Focus Question 30.21
Write the major organic product(s) of these reactions. Assume that a
large excess of the aromatic starting material is used in each case.
(a)
CH3CH2Cl
(d)
AlCl3
(b)
O
AlCl3
(e)
CH3Cl
ClCH2CPh3
CH3Cl
AlCl3
AlCl3
O
(c)
CH3(CH2)3Cl
AlCl3
Concept Focus Question 30.22
Write a mechanism for the reaction of Concept Focus Question
30.21(c).
In the previous section we learned that in some Friedel-Crafts
reactions, the electrophile is a carbocation. Does this suggest that any
carbocation source can be used to alkylate a benzene ring?
Think Ahead Question 30.23
Give a mechanism for the reaction of benzene and propene in the
presence of sulfuric acid to give cumene.
CH3CH=CH2
H2SO4
Answer: This appears to be an EAS reaction similar to a Friedel-Crafts
alkylation because an alkyl group has replaced a benzene ring
hydrogen. However, the lack of alkyl halide and AlCl3 suggest the
mechanism is something different. So we start the mechanism analysis
by disconnecting the alkyl group to determine the electrophile.
_________________________________________________________
Electrophilic Aromatic Substitution
44
CH3
comes from
and
C
H
CH3
Aha! We know how to generate carbocations from alkenes by
protonation (sections 24.xx and 28.xx). Protonation of propene by
sulfuric acid gives a 2-propyl carbocation. So now we can write the
mechanism.
H
OSO3H
H
OSO3H
The use of only a small (catalytic) amount of H2SO4 and a large excess
of benzene helps ensure the carbocation is captured by the benzene
ring and not by –OSO3H. Liquid HF or other specialized acid catalysts
can also be employed. Other carbocation sources such as alcohols are
also useful. For example, 2-propanol could be used instead of propene
in the previous reaction.
Despite their limitations, the Friedel-Crafts alkylation reactions we
have explored in this section are key steps in the industrial syntheses
of several important products. These include ethylbenzene (converted
to styrene for polymers), cumene (converted into phenol and acetone),
long-chain alkylbenzenes (converted into sodium alkylbenzene
sulfonates for detergents) and short-chain alkylbenzenes (blended in
gasoline for higher octane ratings).
30.12 Friedel-Crafts Acylation
Friedel-Crafts reactions can also be used for acylation, the addition of
a new carbonyl-containing substituent to a benzene ring. For example,
the reaction of thioanisole (PhSCH3) with acetyl chloride and
aluminum chloride was part of the industrial-scale synthesis of the
nonsteroidal anti-inflammatory analgesic (pain killer) Vioxx.
CH3S
CH3
CH3S
O
O
S
O
more steps
Cl
O
O
AlCl3
Vioxx
O
_________________________________________________________
Electrophilic Aromatic Substitution
45
Vioxx was a very
profitable drug for
Merck, netting over $1
billion in annual sales,
until it was implicated
in increased risk of
heart disease and
stroke, and voluntarily
withdrawn from the
market in later 2003.
Think Ahead Question 30.24
Determine the electrophile that is formed in the reaction of thioanisole,
acetyl chloride and aluminum chloride. Write the complete
mechanism.
Friedel-Crafts acylation:
An EAS reaction in which
an acylium ion adds a
ketone group to the
aromatic ring.
Answer: This reaction bears a striking resemblance to the FriedelCrafts alkylation (section 30.11), but an acyl group is added instead of
an alkyl group. In fact was discovered by the same chemists, so it is
called the Friedel-Crafts acylation reaction.
Because of its similarity to the alkylation reaction of the same name,
we might assume the mechanism to be similar as well. Lets start by
applying the same disconnection analysis as we did in section 30.11.
comes from
CH3S
CH3S
O
Acylium ion: R-C≡O
+
and
O
The electrophile is a carbocation, just like in the Friedel-Craft
alkylation. The carbocation is formed by Lewis acid-assisted
ionization of the carbon-chlorine bond, just like the Friedel-Crafts
alkylation reaction. How easy is this ionization? The carbocation has
resonance stabilization, so the ionization occurs readily. The resultant
carbocation is called an acylium ion, because of its structural
similarity to an acyl group. The acylium ion resonance contributor
with a triple bond may look strange, but we have seen plenty of
structures with positively charged oxygens atoms having three bonds,
such as H3O+. Recall also that the resonance contributor with the
maximum number of full octets contributes most to the hybrid (section
5.xx).
Now we can write the complete mechanism.
_________________________________________________________
Electrophilic Aromatic Substitution
46
Mechanism: Friedel-Crafts Acylation
O
O
O
O
C
H3C
Cl
H3C
AlCl3
Cl
O
O
CH3S
H
Cl
AlCl3
CH3
Acylium ion adds to form
most stable arenium ion
H3C
Acylium ion
resonance contributors
CH3S
CH3S
C
H3C
Ionization of weakened
carbon - chlorine bond
Lone pair from chlorine fills
open octet on aluminum
H3C
AlCl3
O
CH3
Deprotonation restores
aromaticity
Note how the mechanism is very similar to the Friedel-Crafts
alkylation.
The Friedel-Crafts acylation reaction has fewer limitations than the
corresponding alkylation reaction. The most significant limitation is
the reaction fails when the benzene ring is too electron deficient due to
the presence of one or more strong deactivating groups (or several
weak deactivating groups could combine to have a similar effect), just
like the Friedel-Crafts alkylation. The same restriction overcomes the
multiple alkylation problem of the Friedel-Crafts alkylation reaction.
Acylation adds a new carbonyl substituent, which serves as a
deactivator and prevents further alkylation. For example, reaction of
benzene with methyl chloride and aluminum chloride gives a mixture
of products bearing one, two and possibly more methyl groups (unless
a large excess of benzene is used). On the other hand, reaction of
benzene with acetyl chloride and aluminum chloride stops after just
one acetyl group has been introduced.
_________________________________________________________
Electrophilic Aromatic Substitution
47
CH3
CH3
CH3
CH3
CH3Cl
+
AlCl3
+
CH3
O
+ other products
CH3
O
Cl
only product formed
AlCl3
The Friedel-Crafts alkylation reaction may also involve carbocation
rearrangements. This may or may not be a problem, depending upon
the product that is desired. However, acylium ions do not rearrange
(rearrangement would sacrifice significant resonance stabilization), so
the carbon skeleton of the acid chloride starting material is retained in
the aryl ketone product.
Cl
Incipient 1o carbocation
rearranges to 3o carbocation
AlCl3
O
O
Cl
AlCl3
Acylium ion does
not rearrange
Concept Focus Question 30.23
Give the major organic product(s) when the following molecules are
reacted with acetyl chloride and aluminum chloride. If no reaction
occurs write “NR.”
(a) Ethylbenzene (b) p-Methoxyisopropylbenzene (c) Nitrobenzene
_________________________________________________________
Electrophilic Aromatic Substitution
48
30.13 The Clemmensen and Wolff-Kishner Reductions
A Friedel-Crafts reaction is a very versatile method to make a new
carbon-carbon bond to an aromatic ring. Friedel-Crafts acylation
provides ketones. Friedel-Crafts alkylation allows addition of a new
methyl group (ArH → ArCH3), as well as a bond to a new secondary
carbon (ArH → ArCHR2) or tertiary carbon (ArH → ArCR3).
However, because of carbocation rearrangements, making a bond to a
primary carbon (ArH → ArCH2R) is not very efficient. How then
might we achieve this very useful transformation?
Think Ahead Question 30.26
What sequence of reactions can be used to convert benzene into
butylbenzene?
Answer: The conversion of benzene into butylbenzene
(PhCH2CH2CH2CH3) requires formation of a new carbon-carbon bond
between the benzene ring and a butyl group. We know that in general,
a Friedel-Crafts reaction can be used to make this carbon-carbon bond.
CH2CH2CH2CH3
comes from
and CH2CH2CH2CH3
However, 1-chlorobutane is not an efficient source of the butyl
carbocation because most of the 1o incipient carbocation rearranges to
a more stable 2o carbocation.
Cl
+
AlCl3
minor
major
How can we add a four-carbon chain without rearrangement? This
calls for a Friedel-Crafts acylation with butyryl chloride. This reaction
solves half the problem by constructing the correct carbon skeleton
(Ph-C-C-C-C) but it creates a new problem: an undesired ketone
group. If we had an effective method to convert the ketone to a
O
PhCH2R) this might be a useful way to get
methylene group (PhCR
around the Friedel-Crafts limitation.
_________________________________________________________
Electrophilic Aromatic Substitution
49
Remember that a reduction reaction is one in
which the number of bonds
between carbon and elements less electronegative
than carbon (usually
hydrogen) is increased
(section xx.xx).
O
O
How to remove C=O?
Cl
AlCl3
As you may have guessed by now, there are at least a dozen reactions
that have been specifically developed (or discovered by accident) for
reduction of a ketone to a methylene group. (The same reactions also
reduce an aldehyde to a methyl group.) Two of these reduction
reactions are discussed here.
A. Clemmensen Reduction
Clemmensen reduction:
Reduction of a ketone to
a CH2 group (or aldehyde
to CH3) with Zn/Hg and
aqueous HCl.
An amalgam is an alloy
with mercury. Gold amalgam was once used to
make gold feelings for
teeth, until concerns
about mercury toxicity
caused this method to be
abandoned.
Reduction of a ketone or aldehyde with zinc in the presence of
mercury (as an amalgam) and hydrochloric acid is called the
Clemmensen reduction. (The Zn/Hg amalgam reacts with acid more
slowly than pure Zn, so the slower Zn/carbonyl reaction can compete
with the faster Zn/acid reaction.) The reaction is most efficient with
aryl ketones, but other aldehydes and ketones can also be reduced. The
mechanism of the reaction is not completely understood. The
Clemmensen reduction solution to Think Ahead Question 30.25 is
shown here.
O
O
Cl
AlCl3
Zn/Hg
aq. HCl
Clemmensen reduction requires strongly acidic conditions, which may
not be tolerated by acid-sensitive molecules. For these cases we can
use the Wolff-Kishner reduction.
B. Wolff-Kishner Reduction
Wolff-Fischer reduction:
Reduction of a ketone to a
CH2 group (or aldehyde to
CH3) with hydrazine and
KOH.
The Wolff-Kishner reduction is an alternate method to convert a
ketone into a methylene group. (This reaction can also be used to
reduce an aldehyde to a methyl group.) It involves heating the
carbonyl compound with hydrazine (H2NNH2) and strong base (KOH)
in a protic solvent with a high boiling point (such as ethylene glycol,
HOCH2CH2OH, bp 197oC). The Wolff-Kishner reduction solution to
Think Ahead Question 30.25 is shown here.
_________________________________________________________
Electrophilic Aromatic Substitution
50
O
O
H2NNH2
Cl
AlCl3
KOH
The mechanism for the Wolff-Kishner reduction is discussed in
section 34.xx.
Wolff-Kishner reduction requires strongly basic conditions, which
may not be tolerated by base-sensitive molecules. For these cases we
can use the Clemmensen reduction. The Wolff-Kishner reduction may
not work if the ketone has too much steric hindrance.
Concept Focus Question 30.24
Give the major organic product(s) of the following reactions.
O
(a)
Zn/Hg
aq. HCl
O
(b)
H2NNH2
KOH
H3C
C(CH3)3
(c)
O
Cl , AlCl3
1.
2. Zn/Hg, aq. HCl
1. AlCl3
(d)
2. H2NNH2, KOH
O
Cl
Concept Focus Question 30.25
Suggest a series of reactions to achieve each of these transformations.
Select the most efficient route possible, using the least number of
reactions. Avoid using only a minor product in the next step.
(a) Benzene into 4-tert-butylacetophenone
(b) Tert-butylbenzene into 5-tert-butyl-2-ethylnitrobenzene
(c) Benzene into m-bromobutylbenzene
_________________________________________________________
Electrophilic Aromatic Substitution
51
30.14 Diazo Coupling
Diazo compound: Any
molecule with the general
formula R-N=N-R.
Molecules containing a nitrogen-nitrogen double bond are called diazo
compounds. If the diazo group links two aromatic rings, a highly
conjugated molecule results. This conjugation causes the molecule to
be strongly colored, often red, orange or yellow. For example, butter
yellow (p-dimethylaminoazobenzene) has an intense yellow color. Its
name comes from its old application for dyeing of white margarine to
give it a more appealing, buttery yellow color. (For a more detailed
discussion of how structure controls color, see section 17.xx.)
N
Azobenzene
(orange-red)
N
(CH3)2N
N
N
(CH3)2N
SO3 Na
N
Methyl orange
(orange)
Butter yellow
(yellow)
N
Lets explore the synthesis of butter yellow, using EAS chemistry. (We
will examine a more complex case, the industrial scale synthesis of
FD&C Red No. 40, in section 30.16, In The Real World.)
Think Ahead Question 30.26
Butter yellow is synthesized by an electrophilic aromatic substitution
reaction using N, N-dimethylaniline is the nucleophile. What is the
electrophile?
Answer: The diazo group may look alien to you, but no need to panic!
We know an EAS reaction is involved, so we apply the same logic to
determine the electrophile as we used previously (sections 30.9-30.12).
(CH3)2N
N
N
from nucleophile
from electrophile
comes from
_________________________________________________________
Electrophilic Aromatic Substitution
52
and
(CH3)2N
N
N
N,N-Dimethylaniline
the EAS nucleophile
Benzenediazonium cation
the EAS electrophile
N
N
This analysis suggests the electrophile has a nitrogen-nitrogen triple
bond bearing a formal positive charge. A molecule of the general
structure RN≡N+ is called a diazonium cation. For the synthesis of
butter yellow we need the benzenediazonium cation, PhN2+. (Of course
the benzenediazonium cation is not a naked cation, but instead is
accompanied by an anion, usually chloride ion. So diazonium
compounds are best described as salts. However the anion is not
particularly important in this process so we will ignore it.)
Diazonium cations are unstable, because nitrogen gas is an
exceptionally good leaving group. (Their high instability may lead to
an explosion!) Diazonium salts are not the kinds of things we can
make and keep around in a bottle until needed. Instead, diazonium
salts are generally made when needed and used right away. Even
though they are unstable, diazonium salts are easy to make: Reaction
of a primary amine (such as aniline) with sodium nitrite (NaNO2) in
the presence of aqueous acid does the trick. The conversion of an
amine to a diazonium salt is called diazotization.
Think Ahead Question 30.27
Work out the reaction mechanism for diazotization of aniline (i.e., its
conversion into benzenediazonium chloride) with NaNO2 and aqueous
HCl. Hint #1: The mechanism involves NO+, the nitrosonium cation.
Hint #2: The mechanism is one of the longest you have seen so far, but
it involves the kinds of steps you have seen before, such as proton
transfers, leaving group departure and resonance. Make logical
decisions based upon normal rules of molecular structure, while
keeping in mind the structure of final product, and you should have
little trouble working this one out.
NH2
Na O
N
aq. HCl
O
N
N Cl
Answer: We’ll use the Ph abbreviation here, because the benzene ring
structure does not change during the course of mechanism. The overall
_________________________________________________________
Electrophilic Aromatic Substitution
53
Diazonium cation:
R
N
N
Diazotization: The reaction
in which a primary amine is
converted into a diazonium
salt.
+
Nitrosonium cation: N=O
structure change requires addition of a new nitrogen atom (probably
from NO+) and loss of two hydrogen atoms.
Mechanism: Diazotization of a Primary Amine
H2O
O
H
N
H2O
O
O
H
N
OH
O
Protonate O- twice to make
leaving group (water)
N
Ph
NH2
N
O
Ph
H
OH2
N
N
O
H2O
H
N
N
Ph
Ph
N
N
H
OH2
Ph
H
H2O
H
N
N
Ph
OH
Ph
N
N
N
N
OH2
Departure of leaving group
gives diazonium cation
Protonation converts OH
into H2O, a leaving group
Deprotonation gives N=N
O
Protonate oxygen to make
water (leaving group)
Remove one N-H bond
OH
N
Nitrosonium cation
(an electrophile)
Water leaves to form
nitrosonium cation
H
Form N-N bond
O
OH2
Ph
N
N
Diazonium cation resonance contributors
Note that water acts as a proton shuttle in this mechanism, serving to
both deliver and take away a proton at various times. If you have
already studied carbonyl chemistry (topics 33-35) you are probably
familiar with the idea of a proton shuttle molecule.
The details of your mechanism may not be the same as shown above,
the difference being the timing of the various steps. Other
diazotization mechanisms are acceptable as long as no fundamental
mechanism or structure rules are violated. It is even possible that
several similar mechanisms, all giving the same product, are operating
in the same reaction vessel simultaneously.
Diazo coupling: The EAS
reaction of a diazonium cation
with another aromatic ring to
form a diazo compound.
Now we are ready to consider the EAS portion of the butter yellow
synthesis. A reaction in which a benzenediazonium cation and another
aromatic ring couple to form a diazo compound is called diazo
coupling. What is the mechanism?
_________________________________________________________
Electrophilic Aromatic Substitution
54
Think Ahead Question 30.28
Write the mechanism for the diazo coupling of phenyldiazonium
chloride and N,N-dimethylaniline to give butter yellow.
N(CH3)2
N2 Cl
N
+
H2O
N
N(CH3)2
Answer: The reaction involves replacement of a benzene ring
hydrogen by a diazonium cation (an electrophile), so the standard EAS
mechanism is likely.
N(CH3)2
N
N
N(CH3)2
N(CH3)2
H
Ph
N
OH2
Diazonium cation adds to
form most stable arenium ion
N
N
N
Ph
Ph
Deprotonation
restores aromaticity
Diazonium cations are weak electrophiles, so the nucleophilic benzene
ring component of the diazo coupling reaction must have at least one
strong activator (OH, OR, NH2, NR2, etc.). (Review activators in
section 30.5B if necessary.)
Concept Focus Question 30.26
Give the major product(s) of these diazo coupling reactions.
(a) Phenol with benzenediazonium chloride
(b) Nitrobenzene with benzenediazonium chloride
OCH3
(c)
NH2
+
NaNO2
aq. HCl
NH2
(d)
HO
NaNO2
+
aq. HCl
N(CH3)2
_________________________________________________________
Electrophilic Aromatic Substitution
55
30.15 Electrophilic Aromatic Substitution Reactions with
Aromatics Other Than Benzene
A. Polycyclic Aromatics
So far in this topic we have seen numerous examples of EAS reactions
on benzene rings. However benzene is not the only aromatic molecule.
(Review sections 10.xx and 10.xx if needed.) EAS reactions can
involve aromatic systems other than benzene, but what are the
mechanisms and products? For example, consider the case of EAS
nitration of naphthalene.
aq. HNO3
mononitration product(s)
aq. H2SO4
Naphthalene
Think Ahead Question 30.30
How many mononitronaphthalene isomers are possible when
naphthalene is reacted with an aqueous mixture of HNO3 and H2SO4?
Which isomer is the major product?
Answer: EAS involves replacement of hydrogen atoms, so we need to
determine how many different hydrogen atoms there are. Models are a
very useful way to do this. Build two models, label the hydrogens in
question then see if the models are superposable. Coincident
hydrogens are equivalent; non-coincident hydrogens are not
equivalent. If they cannot, the hydrogens are not equivalent. (You have
already encountered this skill if you studied NMR spectroscopy in
topic 14 or 15.) Using this comparison we find that the eight hydrogen
atoms of naphthalene can be divided into two sets. H1, H4, H5 and H8
are all equivalent. H2, H3, H5 and H6 are a different equivalent set.
H8
H1
H7
H2
H6
H3
H5
Equivalent hydrogens
H1, H4, H5 and H8
H2, H3, H6 and H7
H4
Naphthalene has only two different hydrogen types, so EAS nitration
can produce only two mononitro isomers.
_________________________________________________________
Electrophilic Aromatic Substitution
56
NO2
NO2
aq. HNO3
+
aq. H2SO4
1-nitronaphthalene
2-nitronaphthalene
Which nitronaphthalene isomer is the major product? Just like all of
the previous EAS reactions we have studied so far, the regioselectivity
of this reaction is controlled by the stability of the arenium ion
intermediates. Start by drawing all the resonance contributors for the
arenium ions derived from electrophilic attack at the 1- and 2positions. (Recall from section 30.9 that the nitronium cation, NO2+, is
the electrophile responsible for EAS nitration.)
Attack at the 1-position:
H
NO2
H
H
NO2
NO2
H
NO2
H
H
NO2
NO2
NO2
H
NO2
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Electrophilic Aromatic Substitution
57
Attack at the 2-position:
NO2
H
H
NO2
NO2
H
H
H
NO2
NO2
NO2
H
NO2
Attack at the 1-position yields an arenium ion with seven resonance
contributors, four of which retain an aromatic sextet. Attack at the 2position gives an arenium ion that only six resonance contributors, and
only two of these retains an aromatic sextet.
The carbocation with the greatest number of significant resonance
contributors is most stable, so the arenium ion derived from attack at
the 1-position is formed more quickly than the arenium ion derived
from attack at the 2-position. Therefore we predict that 1nitronaphthalene is the major product. This prediction is verified by
experimental results.
Concept Focus Question 30.27
Give the major product(s) of the following EAS reactions.
O
(a)
+
Cl
Ph
AlCl3
aq. HNO3
(b)
aq. H2SO4
(c)
aq. HNO3
aq. H2SO4
NO2
(d)
aq. HNO3
aq. H2SO4
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Electrophilic Aromatic Substitution
58
B. Heteroaromatics
Now lets consider EAS reactions on aromatic molecules whose
aromatic ring structure includes atoms other than carbon. Furan, an
aromatic molecule with an oxygen atom in the ring, is a good place to
start.
Think Ahead Question 30.30
Predict the major product when furan is treated with acetyl chloride
and aluminum chloride.
O
O
+
+
O
AlCl3
Cl
O
O
O
2-Acetylfuran
Furan
3-Acetylfuran
Answer: This is another Friedel-Crafts acylation reaction, a typical
EAS reaction whose major product is controlled by the stability of the
carbocation intermediate. (The reaction does not occur on a benzene
ring so the intermediate is not really an arenium ion, but rather a
heteroarenium ion.) So we need to determine the resonance
contributors for the two heteroarenium ions.
Read about the numbering
system used to label the
positions of the furan ring in
section 43.xx.
Attack at the 2-position:
CH3
H
C
O
H
O
O
O
O
H
Cl
O
AlCl3
O
O
O
O
Attack at the 3-position:
O
H
O
Cl3Al
Cl
O
H
CH3
C
O
O
O
O
O
_________________________________________________________
Electrophilic Aromatic Substitution
59
Which heteroarenium ion is more stable, and therefore leads to the
major product? Electrophilic addition at the 2-position gives a
heteroarenium ion with three resonance contributors. The
heteroarenium ion derived from addition at the 3-position has only two
resonance contributors. In both cases the most significant resonance
contributor is the oxonium ion, with its full octet on every atom.
Everything else being equal, the cation with the greatest number of
significant resonance contributors is most stable, so we predict that
attack at the 2-position is favored. Once again, experimental results
agree with our prediction.
Take a moment to consider what we have just accomplished in this
section of this topic. Before this point in our studies, we have paid
little attention to naphthalene or furan, yet we successfully predicted
the products of their reactions, simply by applying the principles of
reactivity and mechanism we have learned previously. This is a
hallmark of understanding organic reactions from a conceptual,
mechanistic viewpoint: Much of the unknown can be predicted with a
fair degree of certainty. Now apply what you have learned to some
additional heteroaromatic EAS examples.
Concept Focus Question 30.28
Provide the major products of each reaction.
I2, HgO
(a)
O
O
(b)
Cl
CH3
AlCl3
O
O
O
(c)
O
N
BF3
H
Hint: Acetic anhydride reacts with BF3 to generate an
acylium cation, just like acetyl chloride plus AlCl3.
aq. HNO3
(d)
aq. H2SO4
N
N(CH3)2
(e)
Cl2
FeCl3
N
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Electrophilic Aromatic Substitution
60
(f)
N
CH3
N
Br2
FeBr3
30.16 In the Real World: Industrial Synthesis of FD&C
Red #40
Look around you and see how colorful your world is. Clothing, food,
cosmetics and even the cover of this textbook are alive with various
shades of vivid color. Some of this coloration is natural, but most of it
(especially in clothing) is added during manufacture. The coloration is
provided by dyes, a broad group of molecules (and mixtures) whose
structures are as varied as the colors they produce.
The first dye was Tyrian purple, isolated by the ancient Greeks and
Romans in small quantities from a gastropod mollusk (M u r e x
brandaris). About 12,000 snails yielded about 1.5 grams of dye.
Because of the labor involved in its production, Tyrian purple was
reserved for royalty. Indigo, used to dye denim, is similar in structure
to Tyrian purple. Note extensive conjugation for both dye molecules,
and the how the addition of two bromine atoms causes a change in
color. (Review section 9.xx on the relationship between color and
molecular structure.)
O
H
N
X
X
X = Br, Tyrian purple
X = H, Indigo
N
H
O
The first synthetic dye was made by accident, but it was an accident
with exceptional consequences. In 1856, William Perkin was trying to
make the anti-malarial drug quinine by oxidation of p-toluidine
(cheaply available in large quantities from coal tar, an industrial waste
product). However the fundamental concepts of organic structure and
complex molecule synthesis were (to be polite) naive at the time,
Perkin’s synthesis yielded no quinine. This does not mean his
synthesis failed completely, because what he did produce turned out to
be the first synthetic red dye, called mauveine (or Perkin’s mauve).
This synthetic dye made Perkin a very rich man at a young age, and
launched the modern dye industry, and arguably all of industrial-scale
organic chemistry. As you examine the structure of the major
component of mauveine, think about its conjugation as well as the
complexity of the reaction mechanism that produces it.
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Electrophilic Aromatic Substitution
61
OCH3
OH
X
N
NH2
N
Quinine
K2Cr2O7
Coal tar
aq. H2SO4
H3C
N
H2N
N
CH3
CH3
p-toluidine
N
H
Major component of mauveine
Modern dyes provide a rainbow of color but are based on just a few
fundamental molecular skeletons. The bright colors of these dyes
result from structures with extensive conjugation. For example, diazo
dyes are usually red, orange or yellow. Adding or changing
substituents (typically phenol and amine groups) changes the color.
Sulfonic acid groups may also be added to control water solubility and
dying properties.
A fundamental concept that we have encountered time and time again
throughout this topic is that EAS is a useful reaction to replace an
aromatic or heteroaromatic hydrogen atom with some other
substituent. This transformation is important for small-scale laboratory
work, but also of great significance for synthesis on an industrial scale
to prepare hundreds of pounds (or hundreds of tons) of some organic
material. EAS is especially important to the dye industry, where diazo
coupling is a key step in the synthesis of many diazo dyes.
Na O3S
OH
N
N
Diazo link provides extended
conjugation and red color
OCH3
CH3
SO3 Na
_________________________________________________________
Electrophilic Aromatic Substitution
62
Some food, drug and cosmetic products owe their bright colors to
diazo dyes. For example, the vivid red color of maraschino cherries (in
canned fruit cocktail) is due to FD&C Red No. 40 (also called Allura
Red AC). Lets practice what we have learned about EAS by thinking
about the industrial scale synthesis of this substance.
Think Ahead Question 30.31
FD&C Red No. 40 could be made by two different diazo coupling
reactions. Give the structures of the starting materials for both routes.
Briefly justify why both of these diazo coupling reactions will give the
desired regioisomer.
Answer: Don’t get discouraged by the complexity of the target
molecule. An EAS reaction (diazo coupling, in this case) follows the
same rules on molecules simple and complex. Start with what you
know about diazo coupling.
As we saw in section 30.14, the diazo coupling reaction links a
diazonium salt (ArN2+) with a nucleophilic aromatic ring by the EAS
mechanism. Diazonium cations are weak electrophiles, so the
nucleophilic aromatic ring must have one or more strongly activating
groups. Both the benzene and naphthalene rings that are attached to the
diazo group of FD&C Red No. 40 have strong activators (OH and
OCH3), so either (in principle) can be the nucleophile in the EAS
reaction.
First we consider the case in which the diazonium cation is derived
from a naphthalene amine.
HO3S
Na O3S
OH
OH
N
+
comes from
N
NH2
OCH3
OCH3
CH3
CH3
SO3 Na
SO3H
Why does this reaction give the diazo regioisomer that corresponds to
FD&C Red No. 40? Recall that when directing groups compete, the
strongest activator dominates (section 30.8). Methoxy is a strong
activator, methyl is a weak activator and the sulfonic acid group is a
strong deactivator. The methoxy activates the two positions ortho to it,
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Electrophilic Aromatic Substitution
63
but the position between OCH3 and SO3H is more sterically hindered.
(The position para to the OCH3 is blocked). Therefore the diazo
coupling should give the desired regioisomer.
Methoxy directs here
OCH3
X
sterically hindered
CH3
SO3H
In the second diazo coupling possibility, the diazonium salt is derived
from 4-amino-5-methoxy-2-methylbenzenesulfonic acid.
HO3S
Na O3S
OH
OH
+
N
comes from
N
NH2
OCH3
OCH3
CH3
CH3
SO3H
SO3 Na
Why does this reaction give the diazo regioisomer that corresponds to
FD&C Red No. 40? Any electrophile would preferentially attack the
most electron rich of the two naphthalene rings, especially a weak
electrophile such as a diazonium cation. Hydroxy is a strong activator
(electron donor) whereas the sulfonic acid group is a deactivator so the
naphthalene ring carrying the OH group is attacked. This ring has three
possible sites for EAS, so which is preferred? We saw in section
30.15A that electrophilic attack at the 1-position of naphthalene gives
an arenium ion that is more stable than attack at the 2-position. In
addition, OH is an ortho/para director, which reinforces attack at the
desired position. Therefore we have faith that the diazo coupling
reaction we have designed will give the desired product as the major
one.
HO3S
Attack on this ring
disfavored by SO3H
Naphthalene favors attack here
OH
HO directs attack here
_________________________________________________________
Electrophilic Aromatic Substitution
64
We end this topic by asking you to apply your new knowledge of EAS
chemistry to work even further backwards in the synthesis of FD&C
Red No. 40, designing route to the two molecules used in the diazo
coupling from structurally simpler starting materials such as anisole
and naphthalene.
Summary
30.1 Why Should I Study This?
Electrophilic aromatic substitution (EAS) is the general name
for a reaction in which a hydrogen atom of an aromatic ring is
replaced by some other substituent. The new substituent enters as
an electrophile. The reaction is the most important way to add new
groups to a benzene ring, and has wide application in organic
synthesis.
30.2 Why is Benzene Different: Substitution Versus Addition
Electrophilic attack on alkenes and alkynes generally leads to
addition products whereas electrophilic attack on benzene gives
substitution products (Ar-H → Ar-Elec).
30.3 A General Mechanism for Electrophilic Aromatic
Substitution
The general EAS mechanism can be divided into three portions.
First an electrophile is generated (if necessary). The electrophile
adds to the aromatic ring to form a resonance-stabilized
carbocation (arenium ion). This is the rate-determining step of the
EAS mechanism. The final step is deprotonation of the arenium
ion to restore aromaticity.
Resonance contributors
H
B
Elec
Elec
Elec
Arenium ion
Electrophile attacks benzene
ring forming resonance
- stabilized arenium ion.
Carbocation fate: deprotonation.
Only this carbocation fate restores
aromaticity.
30.4 Halogenation
EAS bromination and chlorination are achieved by reaction with
Br2 or Cl2 and a Lewis acid such as FeBr3 or FeCl3. The Lewis acid
increases the electrophilicity of Br2 or Cl2, but may not be
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Electrophilic Aromatic Substitution
65
necessary if the benzene ring has a strong activator such as OH.
Iodine can be introduced by reaction with I2 and an oxidant such as
HgO or HNO3. There is no EAS reaction for direct fluorination of
a benzene ring.
Br
Cl
Br2, FeBr3
Cl2, FeCl3
I2, HgO
I
30.5 Substituent Effects: Electron-Donating Groups
Electron-donating groups stabilize adjacent carbocations by
resonance and inductive effects, so they are ortho/para directors.
Because they accelerate an EAS reaction by carbocation
stabilization and by enhancing benzene ring nucleophilicity, they
are also activators. Ortho/para directors include alkyl groups, and
any group that can stabilize a carbocation by resonance with lone
pairs or pi bonds. All ortho/para directors are activators, except the
halogens, which are ortho/para directing deactivators. Table 30.3
lists common ortho/para directors.
X (ortho/para director)
X
X
X
Elec
Elec
+
+
Elec
Elec
Ortho
major product
Para
major product
Meta
minor product
30.6 Substituent Effects: Electron-Withdrawing Groups
Electron-withdrawing groups destabilize adjacent carbocations by
resonance and inductive effects, so they are meta directors.
Because they decelerate an EAS reaction by carbocation
destabilization and by decreasing aromatic ring nucleophilicity,
they are also deactivators. All electron-withdrawing groups are
meta directors. Table 30.4 lists common meta directors.
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Electrophilic Aromatic Substitution
66
X (meta director)
X
X
X
Elec
Elec
+
+
Elec
Elec
Meta
major product
Ortho
minor product
Para
minor product
30.7 Substituent Effects: The Halogens
Halogen substituents are the only deactivating ortho/para directors.
30.8 Substituent Effects: Effect of Multiple Substituents
If the aromatic ring undergoing EAS has more than one
substituent, the order of influence is: Strong activator (most
influential) > weak activator > weak deactivator > strong
deactivator (least influential).
30.9 Nitration
A nitro group (-NO2) can be introduced by reaction with aqueous
HNO3 and aqueous H2SO4. The electrophile is the nitronium
cation, NO2+.
NO2
aq. HNO3
aq. H2SO4
30.10 Sulfonation
A sulfonic acid group (-SO3H) can be introduced by reaction with
SO3 in H2SO4. The electrophile is either SO3 or HSO3+.
Desulfonation is accomplished by heating with aqueous acid.
SO3H
SO3, H2SO4
H3O+
30.11 Friedel-Crafts Alkylation
An alkyl group (-R) can be introduced by reaction of the
corresponding alkyl chloride (R-Cl) with AlCl3. This is the
Friedel-Crafts alkylation reaction. Other carbocation sources
such as alkenes and alcohols can also be used. The carbon bearing
the chlorine must be sp3. The intermediate is a carbocation or
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Electrophilic Aromatic Substitution
67
incipient carbocation (RCl-AlCl3 complex). Carbocation
rearrangement is possible. The reaction fails if the aromatic ring is
strongly deactivated.
R
RCl
AlCl3
30.12 Friedel-Crafts Acylation
O
CR) can be introduced by reaction with an acid
An acyl group (
chloride and AlCl3. This is the Friedel-Crafts acylation reaction.
The electrophile is an acylium ion (R-C≡O +). Carbocation
rearrangement does not occur. The reaction fails if the aromatic
ring is strongly deactivated.
O
R
O
R
Cl
AlCl3
30.13 The Clemmensen and Wolff-Kishner Reductions
A ketone can be reduced to a methylene group, or an aldehyde to a
methyl group by treatment with Zn/Hg in aqueous HCl
(Clemmensen reduction) or with H2NNH2 and KOH (WolffKishner reduction).
O
R
R
Zn/Hg, aq. HCl
or
H2NNH2, KOH
30.14 Diazo Coupling
Diazo compounds (Ar-N=N-Ar) can be synthesized by reacting a
diazonium cation (ArN2+; the electrophile) with another aromatic
ring (ArH). The nucleophilic aromatic ring must be strongly
activated. The diazonium cation is produced by diazotization, the
reaction of a primary aromatic amine (ArNH2) with the nitronium
cation (+N=O), produced from NaNO2 and aqueous acid. Because
of their extended conjugation, diazo compounds are colored red,
orange or yellow, and some have found use as dyes.
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Electrophilic Aromatic Substitution
68
NH2
N2
X
NaNO2, aq. HCl
N
N
(X = strong activator)
X
30.15 Electrophilic Aromatic Substitution Reactions With
Aromatics Other Than Benzene
Aromatic rings other than benzene can also participate in EAS
reactions. The same features that influence EAS of substituted
benzenes also influence these cases.
30.16 In The Real World: Synthesis of FD&C Red No. 40
FD&C Red No. 40 is a diazo dye whose structure may appear
complex, but its synthesis can be readily dissected by applying the
mechanistic concepts and logic learned in this topic.
New Terms
Electrophilic aromatic substitution (page 1)
Arenium ion (page 3)
Ipso hydrogen (page 6)
Ortho/para director (page 19)
Activator (page 24)
Deactivator (page 26)
Meta director (page 27)
Nitronium cation (page 36)
Sulfonic acid (page 37)
Friedel-Crafts alkylation (page 40)
Incipient carbocation (page 42)
Acylium ion (page 46)
Friedel-Crafts acylation (page 46)
Clemmensen reduction (page 50)
Wolff-Kishner reduction (page 50)
Diazo compound (page 53)
Diazonium cation (page 53)
Nitrosonium cation (page 53)
Diazotization (page 53)
Diazo coupling (page 54)
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Electrophilic Aromatic Substitution
69