Organic Chemistry Notes by Jim Maxka
CH 16: Reactions of Aromatic Compounds
jim.maxka@nau.edu
Topics:
Aromatic compounds are less reactive than many other molecules because of their stability
Formation of very HOT electrophiles can lead to EAS – Electrophilic Aromatic Substitution
EWG and EDG can influence the reactivity and regiochemistry of benzene reactions
Benzene can react through a benzyne intermediate with strong base and a good LG
Benzene can react towards a nucleophile as long as there is an EWG in the o or p position
Benzene has special reactivity at the benzyl position
Connections:
O
+
N
O
O
NH2
-
OH
OH
S
O
O
CH3
OH
O
CH3
CH3
NH2
Br
OH
Br
Nu
Nu = OH, NH 2, NHNH2, Cl, I, F
O 2N
O 2N
O 2N
CH16-1
Organic Chemistry Notes by Jim Maxka
jim.maxka@nau.edu
Reactions of Aromatics
There are 3 main ways that Aromatic Compounds react:
(1) Addition onto Benzene: Electrophilic Aromatic Substitution (EAS)
(2) Benzene with LG: E2 (Benzyne Intermediate) or Nucleophilic Aromatic Substitution (NAS)
(3) Reactions at the benzyl C: oxidation, reduction and radical bromination.
Electrophilic Aromatic Substitution (EAS)
Benzene is stablized by
energy and so benzene is less reactive than an alkene.
For example, an alkene reacts with a variety of electrophiles. What are the products of the following
reactions?
Br
H3C
+
CH2
Br
+
Br
Br
-
H3C
+
CH
H3C
+
CH2
H
H3C
H
Br
+
CH2
Br
CH3
H3C
H
+
CH
H3C
-
CH3
+
O
+
H
H2O
H2O
H3C
CH3
However, the reactions with benzene are different.
Benzene + Br2 Î
Benzene + HBr Î
Benzene + H2O (H+) Î
Benzene can react, but we need a very strong electrophile: A super-electrophile! When benzene does
react, we call it Electrophilic Aromatic Substitution (EAS).
Electrophilic Aromatic Substitution is really divided in parts:
(1) Creation of the “super”-electrophile, E+.
(2) Addition of the E+ to the Aromatic ring
(3) E1 Elimination of H to return to aromaticity
Draw arrows to accomplish the following steps in the reaction.
H
E
E
E
Addition
+
S
Elimination
+
SH
Final product = substitution
All of the reactions proceed the same except the nature of the super-electrophile (E+) can vary.
Formation of the super-electrophile:
Only certain electrophiles are strong enough to react with benzene. They are always formed by reaction
with a strong protic acid or Lewis Acid. The types of the reactions can be seen below. These reactions are
so useful synthetically, that we need to look at followup reactions as well.
Reaction
Reagents
E+
Product
Follow-up Rxn
Nitration
HNO3/H2SO4
NO2+
Nitrobenzene
Reduction
Bromination
Br2/FeBr3
Br+
Bromobenzene
Benzyne or NAS
Sulfonation
fuming sulfuric acid SO3H+ Phenylsulfonic acid Sodium Fusion
Friedel-Crafts Alkylation RCl/AlCl3
R+
Alkylbenzene
Oxidation of NBS
Friedel-Crafts Acylation RC=OCl/AlCl3
RC=O+ Acylbenzene
Reduction
CH16-2
Organic Chemistry Notes by Jim Maxka
jim.maxka@nau.edu
Nitration
Nitration takes place via the formation of NO2+ (nitronium ion). NO2+ is strong enough to react with
benzene.
Draw arrows for the following steps to make NO2+.
The official reagents are HNO3 + H2SO4: Draw the necessary arrows.
O
O
H+
N
O
O
H
N
OH
O
N
O
H
nitric acid
+ H2O
O
Overall, the reaction is
H2SO4
+ HNO3
The followup reaction for nitration is the reduction of Nitrobenzene to form aniline. Since reduction is loss
of bonds to O and gain of bonds to H. The reaction involves the exchange of the oxygens on N with
hydrogens. This reaction is accomplished by oxidizing a metal, either Sn or Zn, or SnCl2 (aq) and adding
HCl to provide the hydrogens. Since aniline is a base, to get aniline, we need basic workup. Formally, the
reaction goes like this:
+
NO2
NH3
Sn
+
HCl
NH2
NaOH
SnO2
+
workup
H2O
Bromination
The formation of the super-electrophile Br+ is from the reaction of Br2 with the Lewis Acid FeBr3. Draw
arrows:
Br+ (bromonium) ion is strong enough to react with benzene.
Br
Br
FeBr3
Br
Br
Br+
FeBr3
FeBr4
Overall reaction is written:
FeBr3
+ Br
Br
The followup reaction for bromination involves the reaction of benzene with a LG. Benzene can react to
eliminate the LG to give a very reactive benzyne intermediate that usually reacts with the solvent or a
substitution reaction when there is a strong EWG in the ortho or para position. More later.
Sulfonation
The formation of the super-electrophile SO3H+ is from fuming sulfuric acid. Fuming sulfuric acid is more
acidic than sulfuric acid. Below is the “acid rain” reaction. The inverse can dehydrate sulfuric acid. When
ulfuric acid reacts with itself to protonate an OH group on another sulfuric aicd and then eliminate it.
SO3H+ is a good enough E+ to react with benzene..
O
O
O
-
S
+
(g)
O
O
HO
S
O
+
HO
H2 O
OH
O
O
H+
OH
S
HO
S
O
O
H
O
H
HO
S
+ H2O
O
CH16-3
Organic Chemistry Notes by Jim Maxka
Overall
Benzene + H2SO4 (fuming) Î
jim.maxka@nau.edu
The followup reaction for sulfonation is Na fusion. Na fusion involves the reaction with NaOH(g) at 3000C.
The NaOH must be dry or other reactions occur. Phenoxide (phenol anion) is produced. The reaction
must be acidified to make phenol. The reaction can be written:
-
SO3H
O dilute aq acid
NaOH (dry)
heat
OH
workup
Friedel-Crafts alkylation:
The Friedel-Crafts reaction is the Lewis acid catalyzed ionization of the C-Cl bond (like in Br-Br).
The catalyst is AlCl3. The super-electrophile is
. A poor example of this is below. Arrows:
H3C
H3 C
AlCl3
Cl
Cl
CH3+
AlCl3
AlCl4
Since there is a C+, watch out for rearrangements. Consider the problem of reacting propyl chloride with
benzene in AlCl3. You don’t get much propyl benzene from this reaction. We’ll learn how to make propyl
benzene soon.
H3C
H3C
AlCl3
CH
CH H
CH2
Cl
CH2
CH2
H3C
CH2
H
1,2-hydride shift
Two products are produced. Which is the major?
Which is the minor?
There are other issues with this reaction that will be discussed in the next section.
The followup reaction of alkylation is reactions at the newly formed benzyl Carbon. You already learned
the NBS can selectively replace an H in the allyl or benzyl positons with a Br. We will learn some
oxidation chemistry soon.
Friedel-Crafts acylation
Acylation reaction does not rearrange because of resonance stabilization of the acylonium ion.
O
H3C
C
AlCl3
Cl
H3C
O
O
O
C
C
C
CH3
CH3
Cl
AlCl3
AlCl4
Overall with an acyl chloride reagent:
AlCl3
O
+
H3C
C
Cl
The followup reaction for acylation is reduction of the the benzyl carbonyl. This position is easy to reduce
with either Zn, HCl (as with nitro) or with H2, Pt. Note that benzene is not reduced with H2 and Pt. Why?
O
CH3
Zn, HCl
Or
CH2
CH3
Pt, H 2
CH16-4
Organic Chemistry Notes by Jim Maxka
jim.maxka@nau.edu
Practice:
Predict the products of the following reactoins:
Benzene + Br2 (FeBr3) Î
Benzene + HNO3 (H2SO4) Î
Benzene + H2SO4 (fuming) Î
Benzene + H2O (H2SO4) Î
Benzene + (CH3)3C-Cl (AlCl3) Î
The Second Substitution
What is the effect of a previously substituted benzene?
It depends if we have an EWG or EDG. Since the reaction is the pi bond in the benzene ring reaching out
for the E+, then adding electron density to ring will make the reaction go better; withdrawing electron
density from the ring will make reaction go worse. There are two ways to justify of this effect,
Reactivity and Directing Effects
The case for EDG:
Anisole and aniline have a strong EDG because the lone pair electrons on the O or N can interact with the
aromatic ring by resonance. Draw the 4 other resonance contributors for anisole:
O R
O
R
R
O
O
R
?
Would anisole react better than benzene with E+?
In what positions would the E+ most like go?
Anisole is so reactive that it reacts with Br2 without iron catalyst.
Be sure to review the EDG’s.
Now for the case of EWG:
As we saw before, reacting with benzene and EWG is more difficult than benzene by itself, because the
ring has less electron density and will be less able to react with E+. Let’s look at Nitrobenzene. Draw the
4 additional resonance structures:
O
O
N
O
O
N
O
O
N
O
O
N
?
Would nitrobenzene be less reactive than benzene?
Why? Remember you are reacting E+.
What positions would be less likely to be substituted?
Why? Remember you are reacting E+.
Which position(s) would be more likely to be substituted?
Nitrobenzene reacts very slowly with a second mole of NO2+ (from HNO3//H2SO4). The reaction requires
reflux and long reaction times. Why is this fact a good thing for the synthesis of nitrobenzene?
The second nitro group goes in the meta position.
CH16-5
Organic Chemistry Notes by Jim Maxka
jim.maxka@nau.edu
Reactivity of Aromatic Compounds is related to the pKa of the p-substituted benzoic acids
One way to understand the activation/deactivation effects is to look at a chart that plots the pKa’s of
substituted benzoic acids. What we see is that the electronic effects that influence the electron density
between the O-H of the acids is an indication of the electronics that influence activation or deactivation.
The chart on the next page plots pKa from 3 on the right (very unreactive) to 5 on the left (very reactive)
with benzene in the middle at 4.19. We use this chart to compare reactivities of different substituted
aromatic compounds.
less acidic
more reactive towards E+
4.46
OMe
5.03 OH
O
3.55
3.96 OH
CH3
3.41
3
OH
O
C
Br
3.43
NMe3+
OH
O
O
OH
O
3.52
SO2CH3
4
H
O
N(CH3)2
O
3.75
4.19
4.34 OH
O
OH
O
O
O
5
OH
OH
OH
NO2
N
more acidic
less reactive towards E+
Example of activation/deactivation
Below are the results of a student’s nitration of toluene: The student had previously nitrated benzene.
The nitration of benzene took about 30 minutes. The student noted that toluene reacted in just about one
minute.
CH3
CH3
CH3
CH3
NO2
HNO3
H2SO4
O2N
NO2
57%
3%
40%
Justify the difference in reaction time:
Justify the ortho/para to meta ratio.
CH16-6
Organic Chemistry Notes by Jim Maxka
jim.maxka@nau.edu
Resonance vs Inductive Effects:
In the chart above, it is not hard to see that the on the far right, benzene is substituted by E_G and on the
far left by E_G. The hard one to explain is the effect of the halogens, F, Cl, Br, I = X. X is very
electronegative (EWD) but has LP that can be EDG. We say that X is resonance donating but inductively
electron withdrawing. The inductive effect is another way of saying through sigma bonds, and resonance
is through pi bonds. Therefore, X directs to the o,p positions but is a little less reactive than benzene. In
general, which effect is strongest?
The following chart might help explain inductive vs resonance effects.
Type
Reaction
Favored next
Example
Note
Influence
position
EDG
Resonance
Activating
o,p
anisole
LP electrons
EDG
Inductive
Activating
o,p
toluene
No LP
electrons
EWG
Resonance*
Deactivating
M
nitrobenzene,
Resonance
benzaldehyde
structures
EWG
Inductive*
Deactivating
M
R3N+-C6H5
No resonance
structures
Halides
EDG Res-/
Slightly
o,p
bromobenzene Effects cancel
EWG Inductdeactivating
each other out.
* Friedel Crafts reaction does not occur.
Practice: Classify the following as inductive or resonance EWG/EDG or both.
O
NH3
+
+
O
N
O
CH3
-
Br
CH3
Practice
Order the following in terms of reactivity: Most
Least
Medium
Methoxybenzene (CH3OC6H5)
benzene
benzaldehyde (C6H5C(=O)H)
N,N-dimethylaniline
Toluene
Nitrobenzene
Bromobenzene is only slightly less reactive than benzene. Why?
Nitrobenzene is much more unreactive than benzene. Why?
Methoxybenzene (anisole) is much more reactive than benzene. Why?
Why can isopropyl chloride react with AlCl3 and bromobenzene but not nitrobenzene?
CH16-7
Organic Chemistry Notes by Jim Maxka
More Resonance Structures to explain directing effects:
Ortho, para directors
Aniline + E+
NH2
NH2
NH2
H
E
H
H
E
NH2
NH2
NH2
NH2
H
E+
+
jim.maxka@nau.edu
E
E
NH2
NH2
NH2
E+
+
E
E
H
H
E
E
H
H
para like ortho
NH2
NH2
NH2
NH2
NH2
E+
+
H
E
H
E
H
E
H
resonance structures for o,p and
Notice there are
E
for meta.
Meta directors
O
O
O
O
N
N
E
N
+
O
O
N
N
E
E
O
O
H
H
E
N
O
N
+
O
O
O
O
H
+
O
O
O
N
E+
E
E
E
H
H
H
para like ortho
O
O
O
O
O
N
N
+
O
O
N
N
E+
H
Notice there are
O
E
H
E
+ charges next to each other for o,p and
H
E
for meta.
CH16-8
Organic Chemistry Notes by Jim Maxka
jim.maxka@nau.edu
Back to the Friedel-Crafts reaction:
The reaction of benzene with (CH3)2CHCl (AlCl3) goes to isopropylbenzene very easily. But there is a
problem. What about the reactivity of isopropylbenzene: more or less reactive than benzene?
. A
practical consequence is multi-substitution:
R
AlCl3
R
AlCl3
Cl
R
Cl
more or less reactive
than Benzene?
Most chemists prefer to perform the Friedel-Crafts acylation. Why?
Consider the reaction for the synthesis of acetophenone.
O
C
O
AlCl3
CH3
C
H3C
Cl
more or less reactive
than Benzene?
If we react with H2 (Pt) we can reduce the carbonyl in the benzyl position. Draw the complete reaction to
make propyl benzene.
It’s not the same products that we would get if we reacted benzene with CH3CH2CH2Cl (AlCl3) Î
Draw the major product of this reaction:
Summary of EAS:
1. Benzene needs super E+ or cannot react.
2. EDG-Substituted Benzenes (ACTIVATED) react faster than benzene towards super E+ and even
reacts with normal E+.
3. EWG-Substituted Benzenes (DEACTIVATED) react slower than benzene towards E+; regular E+, are
you kidding?
4. Activated aromatics direct o,p. Deactivated aromatics direct m. Halogens are deactivated o,p
directors.
5. Halogens are resonance EDG, inductive EWG. You need to know the difference between EDG/EWG
resonance/inductive
Benzene with Leaving Groups: Benzyne and Nucleophilic Substituion (NAS)
Benzyne
Benzyne intermediates. Benzene with a LG could conceivably go through elimination with a strong base,
creating an alkyne intermediate. Why is the benzyne so unstable?
What is this mechanism called?
Base takes a proton on the C next to C with LG
Br
NH2
H
CH16-9
Organic Chemistry Notes by Jim Maxka
jim.maxka@nau.edu
Because benzyne is so hard to form, we need an extremely strong base or we need harsh conditions.
NaOH/H2O can be used with high temperatures and pressure. Why is benzyne so hard to form?
CH3
CH3
Hi T/P
Cl
OH
H
OH
H
CH3
CH3
Draw possible structures
OH
OH
Consider the results of the following reaction NaOH/H2O must react >3000C and 150 atm (ortho
substituted LG):
Now consider the results of meta-chlorotoluene reacting with NaOH/H2O >3000C and 150 atm:
CH3
CH3
Hi T/P
H
OH
H
OH
Cl
CH3
H
CH3
Draw possible structures
OH
OH
CH3
CH 3
H
Hi T/P
H
OH
Cl
OH
H
You can also consider the results of the para-chlorotoluene reaction with strong base. Two products can
form.
CH3
CH3
H
Hi T/P
OH
H
OH
H
Cl
CH16-10
Organic Chemistry Notes by Jim Maxka
jim.maxka@nau.edu
Finally benzyne can react with a diene to form a Diels Alder adduct. Which Diene was used to make this
molecule? Write out the reaction conditions.
Br
MgBr
Mg
H2C
+
-MgFBr
F
F
H2C
Nucleophilic Aromatic Substitution:
We just studied the
reaction, in which a strong base reacts with benzene substituted
with a
group. Now, let’s look at another reaction in which benzene has a Leaving Group and
reacts by nucleophilic aromatic substitution. We already studied
aromatic substitution.
NEW REACTION: Start with a nucleophile. Remember a good nucleophile has strong
charge
like
or it has a reactive
of electrons like:NH3 or an amine. A good nucleophile
can also be classified as a good
.
What do you expect for the result of the reaction of bromobenzene with OH- at low pressures?
Br
Br
Nu
+ Nu:
???????
H
Well, doesn’t look too good. Can you list some problems with this intermediate????
However, if there is an EWG in the ortho or para position, then we can stabilize the negative charge from
the intermediate. Try this! This is called the Meisenheimer Intermediate.
Br
O
N
Br
O
Nu
Nu
+ Nu:
O
N
O
Br
O
N
O
O
N
+ Br-
O
Nu
This is how we can get away with nucleophilic aromatic substitution. Not exactly like Sn2, but similar.
So now, you have to watch for two reactions with benzene+LG: if there is strong base or OH- with high
pressures and heat the reaction goes by benzyne; if there is an EWG in the ortho or para position,
then the reaction will go by Nucleophilic Aromatic Substitution at normal temperatures and pressure, with
moderate nucleophiles. This is hard because all benzenes in this situation have LGs.
Practice:
Can you justify why NO2 in the meta position does not react through the Meisenheimer intermediate?
CH16-11
Organic Chemistry Notes by Jim Maxka
jim.maxka@nau.edu
More Practice: Predict the main products for the reactions below. Indicate the mechanism.
Br
NH2 / NH3
Br
+ OH-
reflux
NO2
Br
+ OH-
350oC
1000 psi
What is the main difference between the two mechanisms?
CH3
What would happen if we reacted meta-bromonitrobenzene with aqueous NaOH solution at normal
temperatures?
Benzylic C Chemistry: Bromination (NBS), Oxidation, and Reduction
Finally, we have chemistry at the benzylic position. Now, we always need practice. Finish the resonance
structures for the benzyl cation (like the allyl cation).
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
The benzylic position is reactive to three types of reagents: radicals (review allylic bromination), oxidation
and H2 reduction.
Benzylic Bromination
Toluene reacts with NBS, light or FRI to make benzylbromide.
CH3
NBS, FRI
or
Br2 light hv
Predict the product of the reaction of ethylbenzene + NBS (FRI) Î
What is the difference between the following reaction conditions:
(a) toluene + Br2 (FeBr3) (DARK) Æ o-bromotoluene + p-bromotoluene
(b) toluene + Br2 (FeBr3) (LIGHT) Æ o-bromotoluene + p-bromotoluene + benzyl bromide.
CH16-12
Organic Chemistry Notes by Jim Maxka
jim.maxka@nau.edu
Benzylic Oxidation. The benzyl position is oxidized by MnO4-(permanganate) or CrO3 in sulfuric acid
(Jones). The result is a carboxylic acid in the benzyl posiiton no matter how many carbons are hanging off
benzene, as long as there are H’s to lose.
Oxidation/Reduction in organic chemistry: Simply put
Oxidation is gain of bonds to O. Loss of bonds to H.
Reduction is gain of bonds to H. Loss of bonds ot O.
Predict the following oxidation products.
Toluene + permanganate Î
Propylbenzene + Jones reagent Î
tert-Butylbenzene + Jones reagent Î
Benzylic Reduction: We saw that a C=O in the benzyl position can be easily reduced by catalytic
hydrogenation (H2/Pt). What is the result of the following sequence?
Br
O
+ CH3CH2
H2/Pt
AlCl3
Cl
Synthetic Strategy
Aromatic Synthetic strategy. See book- 16.12.
Synthesis of p-bromobenzoic acid vs. m-bromobenzoic acid from benzene.
Targets
O
O
Br
OH
OH
Br
You must start with benzene and react with any reagent that will give the exact aromatic compound. Since
some groups are o,p-directors and others are m-directors, it is important which order you add the groups
and transform them. Retrosynthetic Analysis
Friedel Crafts Alkylation
O
OH
Br
Br
CH3
Benzylic oxidation
+
CH3Cl AlCl 3
+
Br2 FeBr3
+
Br2 FeBr3
o,p-director
Br
o,p-director
para
+
KMnO4
aromatic bromination
H3C
+
CH3Cl AlCl 3
CH16-13
Organic Chemistry Notes by Jim Maxka
o,p-director
jim.maxka@nau.edu
O
aromatic bromination
Friedel Crafts Alkylation
O
Benzylic oxidation
Br
OH
OH
m-director
CH3
+
Br2 FeBr 3
+
meta
KMnO4
+
CH3Cl AlCl 3
Now make p-bromonitrobenzene
NO2
Br
To make m-bromonitrobenzene:
Br
NO2
Case Study: Synthesis of Sulfanilamide, a Sulfa drug
O
O
NH2
HN
CH3
HN
O
CH3
HN
CH3
NH2
2 steps
O 2S
NH2
O 2S
O 2S
NH2
Cl
sulfanilamide
Summaries
1. Understand how aromatics react differently than alkenes.
2. Know the reagents and products of these EAS reactions
a. Aromatic nitration
b. Aromatic bromination
c. Aromatic sulfonation
d. Friedel-Crafts alkylation
e. Friedel-Crafts acylation
3. Know the reagents and products of the followup reactions
a. Aromatic nitration Æ aniline
b. Aromatic bromination Æ reactions with LG
c. Aromatic sulfonation Æ sodium fusion
d. Friedel-Crafts alkylation
e. Friedel-Crafts acylation Æ reduction at the benzyl position.
4. Know the reagents and conditions of the benzyne reaction.
5. Know the reagents and conditions of the NAS reaction.
6. Understand how the benzylic position is selectively brominated, oxidized or reduced.
7. Understand how the 2nd substituent EAS (o,m or p) is influenced by EWG/EDG.
8. Understand how the reactivity of the benzene ring is influenced by EWG/EDG.
9. Understand the principles of retrosynthetic design and how to decide what steps are required.
10. Use retrosynthetic-thinking to design 3 step syntheses of aromatics.
CH16-14