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CHAPTER 16 - CHEMISTRY OF BENZENE: ELECTROPHILIC AROMATIC SUBSTITUTION

As stated in the previous chapter, benzene and other aromatic rings do notundergo electrophilic addition reactions of the simple alkenes but ratherundergo electrophilic substitution reactions to preserve the stable aromaticring system. The general reaction is:

+ E+E

+ H+H

In each of the cases we will study, the reaction mechanism is the same - whatis different in each is the attacking electrophile, E+, and the reagents thatproduce that electrophile and the subsequent products. The general reactionmechanism for all electrophilic aromatic substitutions (EAS) is:

EH

E E E

H H H

Resonance stabilized intermedite

+ H++ E+

In each of the cases we will study, there will be an overall reaction thatshould show starting material (e.g. benzene), the reagents used for thatreaction and the product obtained.

There will also be a mechanism for the particular reaction, which is the samein all cases except for the attacking electrophile and how it is generated fromthe reagents. In writing a mechanism you must show how the E+ isgenerated, all resonance forms of the intermediate and the product.

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Below is a summary of the five electrophilic substitution reactions covered inthis chapter, showing

Reagents Generation of E+ Product

1. Halogenation:

A. Bromination and Chlorination

X XX

X2/FeX3 FeX3+

"Polarized halogen"X = Cl, Br

X3Fe----Br----Brd d

B. Iodination

I2/CuCl2 I2 + 2Cu2+ 2Cu+ + I+I

2. Nitration

HNO3/H2SO4 HOHO2 +2H2SO4

H3O + 2HSO4 + NO2

NO2

3. Sulfonation

SO3/H2SO4S

O O

O+H2SO4 S

O O

OHHSO4 +

SO3H

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4. Friedel-Crafts Alkylation

RX, AlCl3 R X + AlCl3 AlCl3X + RR

R = alkyl

When R = methyl or a primary halide, the carbocation does not form as aseparate entity but rather is a polarized complex with the aluminumtetrahalide anion. Secondary and tertiary carbocations form butremember with any carbocation (except methyl and ethyl)rearrangement is possible and will occur if a more stable cation canbe formed.

H+ CH3CH2CH2Cl AlCl3

In the reaction -

CH3CH2CH2- rearranges to H3C

CHH3C

5. Friedel-Crafts Acylation

CO

XR

CO

XR

R C O

R C O

AlCl3 + AlCl3 AlCl3X +

OR

R = alkyl,aryl, vinyl

Now let's look at some reactions:

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CH+ AlCl3

O

CH3

NO2H

BrH

CO

ClH3C

HNO3

H2SO4

Br2

FeBr3

Remember that when asked for a mechanism, you must show generation ofthe attacking electrophile, then the attack on the benzene ring showing allresonance forms of the intermediate and finally the formation of product withloss of H+. (See beginning of chapter notes).

Show complete mechanism for the nitration of benzene:

Directional and Activation/deactivating Properties of Ring Substituents

What happens when there is already a substituent on the ring? In whichposition, relative to that substituent, will the incoming groups go - ortho,meta or para????

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First group on the ring determines the reactivity of the ring toward furthersubstitution relative to benzene itself, and also determines the where thesecond incoming group will go. These are two separate aspects to consider infurther substitution on the ring.

An activating group makes the ring more reactive than benzene to furthersubstitution. In some cases the activation is so great that polysubstitutionoccurs and cannot be controlled.

A deactivating group makes the ring less reactive than benzene to furthersubstitution. In some cases the ring is deactivated sufficiently to preventcertain reactions from occurring.

With one group already on the ring, a second substituent can come on thering in ortho, meta or para positions.

Xortho

meta

para

meta

orthoThere are two ortho positions relativeto group X, two meta positions andone para position that are available for subsequent substitution.

The group, X, already on the ring determines where the incoming group willgo......regardless of what the second reaction is.

Experimentally we find that some groups give a mixture of ortho and paraproducts almost exclusively with little or no meta product. Some groups givemostly meta product with some ortho and para product in the mixture. (SeeTable 16.1 on p. 606 of your text).

When we examine these cases we find three groups - those that direct ortho,para and activate the ring, those that direct o, p and deactivate the ring(halogens) and those that direct meta and deactivate the ring.

Below is a chart summarizing these effects. The vertical double line separatesthe ortho/para directors from the meta directors. This simply means thatwhen one of the o-p directors is on the ring further substitution will result inproducts where the new group is ortho or para to that first group. When a

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meta director is on the ring, further substitution will produce primarily themeta product. Mixtures will occur is almost all cases.

Note that with the exceptions of phenyl, vinyl and alkyl, all of the o-pdirectors have a lone pair of electrons on the atom attached to the ring. In themeta director, with the exception of -CF3, the atoms attached to the ringhave a multiple bond to an electronegative element. All of the meta directorsare electron-withdrawing groups.

ORTHO,PARA DIRECTORS META DIRECTORS

-NH2-NHR-NR2-OH

-OCH3-OR-NHCOCH3

Phenyl--CH=CH2Alkyl-

-F-Cl-Br-I

-CHO-COOH-COOR-COCl-CONH2-SO3H-CN

-CF3-NO2

strong moderate weak strong very strong

ACTIVATORS DEACTIVATORS

weak

The bold vertical line separates the activating groups from the deactivatinggroups. Note that of the deactivating groups, only the halogens are o-pdirectors.

Two effects are working here. One is induction, which is an effect thatoccurs through the sigma bond system. Electron withdrawing groups will"pull" electron density away from the ring making the electrons of the ringless available for attack by an electrophile. Electron donating groups dothe opposite. Induction also plays a role in some of the directional effects.

The other effect is resonance, which occurs through the pi system of bondsin the molecule. Resonance plays an important role in the stability ofintermediates of the reactions.

Sometimes these two effects are opposite to one another. One effect may bestronger and exert greater influence. We will point these out as we discuss the

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reactions. As an example, look at phenol (benzene with a hydroxyl group onring). OH is an electron withdrawing group (induction) but OH is a verystrong activator and o, p director. Resonance plays a more important role ascan be seen by resonance structures of phenol shown below.

OH OH OH OH

Let's examine some general reactions with one group already on the ring.

1. Electrophilic substitution of toluene (methylbenzene). Alkyl groups areelectron donators so we can understand why these groups activate the ring.Why are alkyl groups o, p directors?

If we examine toluene under attack by some electrophile and examine theintermediate if the electrophile goes in the para position we get:

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CH3 CH3 CH3 CH3

H E H E H EH

E+

resonance forms of intermediate

In the circled contributing resonance structure the positive charge is directlystabilized by the electron donating effect of the alkyl group. If you examineattack in the ortho position, you will find a similar phenomenon.DRAW THIS FOR PRACTICE!

Now examine meta attack on toluene:

H

E

H

E

H

EH

CH3 CH3 CH3CH3

E+

There is no one very stable form of the intermediate. The intermediate inortho or para attack is more stable, more easily formed and reaction goesin that direction.

2. Electrophilic substitution of Phenol (hydroxy benzene)

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Examine ortho attack on phenol:

E+

OH OH OH OH OHH

EHE

H

EH

E

Why is the circled resonance contributing structure very stable?

Draw the resonance forms of the intermediate in meta attack on phenol.Convince yourself that the lone pair of electrons on the oxygen cannot bepart of resonance with the ring.

3. Electrophilic substitution of Nitrobenzene

If the incoming group goes in the para position

H E H E H EH

E+

NO O

NO2 N NO2O O

The circled intermediate resonance form is especially UNSTABLE. WHY?

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Meta attack on nitrobenzene:

H

E

H

E

H

EH

E+

NO O

NO2NO2 NO2

In meta attack there is no very unstable form, nor is there an especially stableform. Both substitutions are difficult, but meta is more likely.

EXCEPTIONS!!!!!!!!!!!!!

1. Hydroxy and amine are very strong activators. In bromination of phenoland aniline, reaction can be accomplished with Br2 in H20 (no iron catalystneeded) and does not stop with monosubstitution. A trisubstituted product isobtained.

OH OHBrBr

Br

Br2

H2O

Solution: If the OH or the NH2 is first acylated to give -OCOCH3 or -NHCOCH3, the ring is not so active (resonance forms that draw electrons toside group rather than back into ring can be drawn) Monobromination can beaccomplished, usually in the para position because the ortho positions are nowstericly hindered. The acyl groups can be hydrolyzed off in a later reaction.

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NH2 HN NHCOCH3 NH2

Br Br

Br2

HOAc

CH3COCl

CH3

O

H2OHO_

2. The NH2 group on aniline is a strong base. In a Friedel-Crafts reaction, theNH2 group will react with the Lewis acid, AlCl3 to form a complex. Thenthe Lewis acid is not available for the F-C reaction. Again acylation of theamine group will help in eliminating this problem

NH2 H2N AlCl3

CH3CH2ClAlCl3

Not alkylation product.

3. Friedel-Craft electrophiles are not strong electrophiles. If any deactivatinggroup stronger than halogen is already on the ring, F-C will not occur.

Trisubstituted Benzenes:

If two groups are already on the ring, where will a third group go? Sameinductive and resonance effects govern the reaction.

1. If directing effects of the two groups reinforce each other, no problem.

OH

NO2

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2. If the two groups oppose each other, the stronger activating group willdirect the incoming group. (Mixtures often result).

OH

CH3NOTNOT

3. Substitution rarely occurs in the position between two groups that are 1,3-(meta) to each other because of steric hindrance.

Cl

CH3

NOT

Side Chain Reactions:

1. Oxidation of Alkylbenzene Side Chains

The aromatic ring is not oxidized by strong oxidizing agents such andNa2Cr2O7 of KmnO4 which will cleave normal alkenes. But the ringactivated alkyl groups to oxidization (Remember alkanes themselves do notreact with these reagents).

CH3 COOHKMnO4H2O, 95˚C

As long as the carbon directly adjacent to the ring has at least one hydrogenon it, that carbon is subject to oxidation. Regardless how long the alkyl groupis, the product is always benzoic acid.

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CH2CH(CH3)2 COOHKMnO4

H2O, 95˚C

t-Butyl benzene is not subject to oxidation because there is no hydrogen onthe carbon adjacent to the ring.

CCH3

CH3

CH3KMnO4

H2O, 95˚C N.R.

2. Bromination of Alkyl Side Chains:

If an alkyl benzene is treated with NBS (N-Bromosuccinimide), in thepresence of benzoyl peroxide (a radical stimulator) bromination occursexclusively on the benzylic C - that is the carbon adjacent to the ring.The intermediate is stabilized by resonance with the ring.

CH2CH2CH3 CHCH2CH3

BrNBSROOR

Benzylic carbon

CH-CH2CH3 CH-CH2CH3 ETC.

3.Reduction of Aromatic compounds:

A. Reduction of benzene ring

Under normal reduction conditions of H2/Pd in ethanol for example, thebenzene ring cannot be reduced. In the reaction below notice that thecarbonyl (ketone) and ring are not reduced, only the alkene.

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CH CH CO

CH3 CH2 CH2 CO

CH3H2,Pdethanol

Excessive conditions are necessary to reduce the ring to a cyclohexane. a. H2 with a Platinum catalyst at 2000 psi pressure will reduce the ring.

b. Normal pressure but using a more powerful catalyst such as rhodium on carbon will reduce the ring.

CH3

CH3

CH3

CH3

H2,Pt,ethanol2000 psi, 25˚C

Or:H2,Rh/C,ethanol.1 atm. 25˚C

B. Reduction of Aryl Alkyl Ketones

The aromatic ring activates adjacent carbonyl groups toward reduction.

CCH3

OCH2CH3H2,Pd

ethanol

This gives us a way to overcome the problem in F-C alkylation ofrearrangement of alkyl carbocations.

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CCH2CH3

OCH2CH2CH3

CH2CH2CH3CHCH3

CH3

H2,PdethanolCH3CH2COCl

AlCl3

AlCl3

CH3CH2CH2Cl+

C. Reduction of Other Side Groups

Nitro groups can be reduced with H2, Pd/C in ethanol. Remember this alsoreduces adjacent ketones and alkenes anywhere in the molecule.

CH3

NO2

OCH2CH3

NH2

H2, Pd/Cethanol

Nitro groups can also be reduced with tin and hydrochloric acid (SnCl2/HCl),which will not reduce the carbonyl or the alkene. This gives us a way to putan amine group on the ring.

CH3

NO2

O

CH3

NH2

O

1.SnCl2, H3O+

2. H2O, HO-

SYNTHESIS: In all preparations (syntheses) assume that ortho and paraproducts can be separated. Show only the one you need for your synthesis.

Prepare p-bromobenzoic acid from benzene

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Prepare 4-chloro-1-nitro-2-propylbenzene from benzene

Prepare 2-bromo-4-nitrotoluene from benzene

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PrepareCH2OH

CH2OH

from benzene.

Omit Sections 16.8 & 16.9