EAS-Notes

ELECTROPHILIC SUBSTITUTION

Benzene nucleus is composed of six sp2 hybridised carbon atoms linked together to form a hexagonal

planar structure with one H atom bonded to each carbon. All the carbon atoms have one unhybridized

porbital with one electron, which forms a sextet cloud of electrons above and below the plane of the benzene

ring. It means that there is high density of electronic charge in benzene. Hence, a strong electrophile (electron

deficient group) can now attack this electron cloud i.e. benzene readily undergoes electrophilic substitution.

Electrophilic substitution in benzene ring is believed to be a bimolecular (SE2) reaction, which involves the

following pathways. As the electrophile comes closer to the benzene nucleus, a complex is formed.

complex rearranges to give complex.

electrophilic

+ E attack

E rearranges

E

H

complex (loose association)

complex

complex formation results in the loss of aromaticity (breaking of the sextet) which is compensated by

resonance.

+ E+

H

E +

H

E +

H

E

+

Resonance stabilised

E

+ H+

In the last step, intermediate carbonium ion, in the presence of a proton acceptor, loses a proton to give the

substituted product with complete sextet.Above E+ is an electrophile and may be X+ (Cl+, Br+, I+) in

halogenation, 2NO in nitration, SO3 in sulphonation and R+ or RCO+ in Friedel Craft’s reaction.

MECHANISM OF ELECTROPHILIC AROMATIC SUBSTITUTION REACTIONS Substitution reactions like halogenation, nitration, sulphonation, Friedel Craft’s reactions have been referred to

as electrophilic aromatic substitution reactions. The mechanism of these reactions is described as below.

1. NITRATION

Nitration is brought about by the action of concentrated nitric acid or a mixture of concentrated nitric acid and sulphuric acid often called nitrating mixture. HNO3 alone is a weak nitrating agent whereas the mixture is

strong nitrating mixture.The reaction involves electrophilic attack by the nitronium ion, 2NO which is produced

under the conditions of reaction. When HNO3 alone is the nitrating agent. HONO2 + HNO3 H2O +

23 NONO when concentrated HNO3 and concentrated H2SO4 is the nitrating mixture, 2NO (Nitronium ion) is

produced as follows:

HONO2 H2ONO2 H3O + HSO4 + NO2

H2SO4 ..

H2SO4

i.e. HNO3 + 2H2SO4 NO2 + H3O + 2HSO4

+

HSO4

when HNO3 alone is the nitrating agent,

generation of water dilutes the acid and generation of 2NO ion is slowed down. Hence it is a mild nitrating

agent.But when mixture of concentrated HNO3 and concentrated H2SO4 is the nitrating agent. concentrated

sulphuric acid helps to speed up the generation of nitronium ion by absorbing water molecule and producing

H3O+ in the medium.Now, the 2NO ion attacks the benzene nucleus and forms an intermediate cation, a

benzenonium ion, which loses a proton to yield the nitro derivative.

+ NO2

complex

Electrophilic

attack NO2

nitronium ion NO2

H

Nitrobenzene

NO2

H

NO2

H NO2

H+ + 4HSO + H+

H2SO4The reaction by which nitronium ion is produced is simply an acid base equilibrium in which sulphuric

acid serves as the acid and the much weaker nitric acid serves as a base. The very strong acid, sulphuric acid

causes nitric acid to ionise in the sense, HO NO2+, rather than in the usual way,

H+ ONO2.

The

nitronium ion is well known existing in salts such as 42ClONO and

42BFNO which smoothly nitrate benzene at

room temperature. It supports the mechanism, in which the electrophile species attacking the aromatic

compound is nitronium ion, .NO 2 Highly reactive aromatic compounds, such as phenol, are found to undergo

ready nitration even in dilute nitric acid and at a far more rapid rate than can be explained on the basis of the

concentration of NO2

that is present in the mixture. This has been shown to be due to the presence of nitrous

acid in the system which nitrosates the reactive nucleus via the nitrosonium ion, NO

.

HNO2 + 2HNO3 H3O + 2NO3 + NO

NO

H

OH

NO3

OH OH

NO NO

HNO3

OH

NO2

+ HNO2

2. SULPHONATION

Sulphonation is done by heating the substrate with conc. sulphuric acid or fuming sulphuric acid containing

varying proportions of sulphur trioxide.Experimental work based on kinetic studies in concentrated sulphuric acid and in oleum (H2SO4, SO3) strongly favours the theory that sulphur trioxide is the active species. SO3 is

bonded to three more electronegativity oxygen atoms. Hence S is electron difficient centre acting as electrophile.

S

O

O O S

O

O O

+

S

O

O O

+

S

O

O O

+

S

O

O O

+

Now the electrophilic reagent, SO3, attacks the benzene ring to form the intermediate carbocation.

2H2SO4 SO3 + H3O+ + 4HSO

S

O

O O

+

SO3

H

SO3

H

SO3

H

SO3H Sulphonation, like

iodination, is reversible and is believed to take place in concentrated sulphuric acid via the pathway,

+ SO3 k1

SO3

k1

H

SO3

k2

+H fast

SO3H

In oleum, the complex is believed to undergo protonation of the SO3 before undergoing CH fission

to yield the SO3H analogue. Like iodination, sulphonation exhibits a kinetic isotope effect, indicating that CH

bondbreaking is involved in the ratelimiting step of the reaction, i.e. that k1 ~ > k2.

3. HALOGENATION

In halogenation, electrophiles are Cl+, Br

+ and I

+. As halogens are neutral (nonpolar) covalent molecules, so

to generate polarity these react with halogen carriers i.e. Lewis acids

(transition metal halides). The Lewis acids polarizes the halogen molecule forming complexes with negative ion, leaving positive ion to react with benzene nucleus. The positively charged or electrophile species then

attacks the nucleophilic aromatic substrate to give an intermediate carbonium ion which is resonance stabilized. It abstracts a proton and electron pair binding the hydrogen to the ring moves to restore the highly

stable benzene ring system.

(i) FeCl3 + Cl2 Cl Cl FeCl3 Cl+ + FeCl4

+

(ii)

+ Cl+

+ H

Cl

H

Cl +

H

Cl +

Cl

+ H+

H

(iii) HClFeClHFeCl 34 or FeCl3 + Cl 2 Cl Cl FeCl 3

+

+ Cl Cl FeCl3 Cl+ Cl FeCl3

+ +

+ H

Cl

H

Cl +

Cl

+ H+

H

Cl +

+ 4FeCl

4FeCl + H+ FeCl3 + HCl

Role of Lewis acid

The loosely held electrons of the double bond in alkenes polarize the halogen molecule even in the absence

of a Lewis acid catalyst. The electrons cloud in benzene is relatively

less available and consequently the presence of a Lewis acid catalyst is necessary to polarize the halogen molecule, at least in the case of less reactive aromatic compounds

(benzene, chlorobenzene etc.). The fact that halogenation of more reactive aromatic compounds (phenol, aniline etc.) where electrons are more available proceeds smoothly even in the absence of Lewis acid

catalyst establishes the role of Lewis acid in the above mechanism for halogenation of aromatic compounds.Note: A similar dual mechanism can also operate when halogenation is carried out with

hypochlorous or hypobromous acid. This reaction is acid catalyased. HOCl + H+ H2O+Cl

+ H2O+Cl

H2O

+

H

Cl +

H

Cl

+ H

Cl + H+

Cl

Instead of a protonated hypohalous acid, the attacking reagent can be a positive halogen cation, without affecting the course of reaction.H2O+Cl H2O + Cl+. Kinetic isotope effects have not been

observed for chlorination and only rarely for bromination, i.e. the reactions normally follow pathway like nitration. In iodination, which only takes place with iodine itself on activated species, kinetic isotope effects are

the rule. This presumably arises because the reaction is readily reversible (unlike other halogenations), loss of

I occurring more often from the complex than loss of H, i.e. k1 ~ > k2.

1k

H

OH OH OH

I

+ I2 I 1k

2k

I

+ HI

Thus kH/kD for the iodination of phenol and 2, 4, 6trideuteriophenol is found to be 4. Iodination is

often assisted by the presence of bases or of oxidising agents, which remove HI and thus displace the above

equilibrium to the right. Oxidising agents also tend to produce I

, or a complex containing positively polarized iodine, from I2, thus providing a more effective electrophile. Halogenation may also be carried out by use of

interhalogen compounds +

BrCl, ICl, +

etc., attack occurring through the less electronegative halogen as this will

constitute the ‘electrophilic’ end of the molecule.

4. FRIEDEL CRAFT’S REACTON

Alkylation or acylation of aromatic ring with alkyl halides or acyl halides in presence of a Lewis acid,

generally anhydrous AlCl3 (Friedel craft catalyst) is called Friedel crafts reaction.The aromatic ring, to which the side chain is attached, may be that of benzene itself, certain substituted benzenes or more complicated

aromatic ring systems like naphthalene and anthracene. e.g.

RX / AlX3

R

+ HX

where X = Cl, Br, I, but the most effective catalyst is anhydrous AlCl3: R = CH3 , CH3CH2 ,

C6H5CH2 substituted alkyl groups, or acyl group

CH3C

O

or

C6H5C etc.

O

Because of low reactivity of

halogen attached to an aromatic ring, aryl halides cannot be used, in place of alkyl halides. Also, vinyl halides

cannot be used in stead of alkyl halides. MECHANISM

It has been observed that rate of reaction [substrate] [RX] [AlX3]. It means the reaction is 3rd order,

which suggests that AlX3 is involved in the formation of transition complex.

4.1 ALKYLATION

The carbon atom of alkyl halides, R X, +

is electrophile, but rarely is it sufficiently effective so, to affect

the substitution of aromatic species. So, the presence of a Lewis acid catalyst is also required. Anhydrous aluminium chloride, AlCl3, being a Lewis acid, accepts a lone pair of electrons from halogen (Chlorine atom) of

R Cl:

..

.. . This makes R (alkyl) group to be sufficiently polar so as to act as an electrophile. Now, two

mechanisms are possible for Friedel Craft’s alkylation. Both involve electrophilic substitution, but they differ as to the nature of the electrophile. One of the mechanism for Friedel Craft’s reaction involves the following steps.

(i) R Cl:

..

.. + AlCl3 R Cl AlCl3 R+ +

4AlCl Carbocation

+

(ii)

+ R H +

R

complex

(iii)

H +

R

+ 4AlCl

R

+ HCl + AlCl3

In the above mechanism, the electrophile is a carbocation. The function of the aluminium chloride is to generate this carbocation by abstracting the halogen from alkyl halide.On the other hand, the another

mechanism is carried out by the electrophile which is an acid base complex of alkyl halide and Lewis acid, from which the alkyl group is transferred in one step from

halogen to the aromatic ring.

(i) R Cl:

..

.. + AlCl3 R Cl AlCl3 +

.

(ii)

+ R Cl AlCl3 H +

R

complex

+ + +

4AlCl

(iii)

+ AlCl3 + HCl +

R

H + 4AlCl

R

Product

(i) Nature of alkyl groups

If the alkyl group is simple CH3 or CH3CH2, then a complex between alkyl halide and Lewis acid is

the electrophile as shown in second mechanism. But because of the relative stability of s and t carbonium

ions, the adducts with s ant t alkyl halides ionise and it is now the carbonium ion that is predominantly the

active species. e.g.Me3CCl + AlCl3 Me3C +

4AlCl

(ii) Temperature

Not only nature of the alkyl group, but also temperature determines the nature of electrophile. e.g.

nalkyl group can be introduced to a fair extent without rearrangement at low temperatures, because ionisation

of the adduct is retarded. But at higher temperatures, carbonium ion is formed which rearranges and the product is rearranged alkyl benzene. Thus npropylchloride gives isopropyl benzene.

CH3CH2CH2Cl: + AlCl3 +

CH3CHCH2 + 4AlCl

H (ii)

(i) rearranged

CH3CH2CH2 Cl AlCl3 + CH3CHCH3

+

..

..

CH2CH2CH3 + AlCl3 + HCl

npropyl benzene

2° Carbocation (more stable)

CH

Isopropyl benzene

CH3

CH3 + HCl + AlCl3

In the same way, isobutyl chloride gives tbutyl benzene.(iii) Nature of Lewis acid as catalyst

The order of effectiveness of Lewis acid catalysts has been shown to be

AlCl3 > FeCl3 > BF3 > TiCl3 > ZnCl2 > SnCl4

The action of Me3CCH2Cl/AlCl3 on benzene is found to yield almost completely the rearranged product,

PhCMe2CH2Me, which can be explained on the basis of the initial electrophilic complex being polarized enough to allow the rearrangement of [Me3CCH2]+ClAlCl3

to the more stable [Me2CCH2Me]+

ClAlCl3. By contrast Me3CCH2Cl/FeCl3 on benzene is found to yield almost completely the

unrearranged product, Me3CCH2Ph. This is due to the fact that the complex with the weaker Lewis acid, FeCl3,

is not now polarized enough to allow of rearrangement.

4.2 ACYLATION

Acylation of benzene may be brought about with acid chlorides or anhydrides in presence of Lewis acids. e.g.

+ CH3COCl

Acetophenone

+ HCl AlCl3

COCH3

Mechanism:

Friedel crafts acylation is found to follow the same general rate law as alkylation Rate [Substrate]

[RCOCl] [AlX3]In case of acylation, the nature of electrophile will be as follows:

(i)

R C Cl: + AlCl3 R C = O + 4AlCl

O ..

..

(ii)

R C Cl + AlCl3 R C Cl

O

+

O AlCl3

(iii)

R C Cl: + AlCl3 R C Cl AlCl3

O

+

O ..

..

Due to the presence of lone pair of electron on both chlorine and oxygen two intermediates are

possible and both are capable of electrophilic attack on benzene.

Acylium ions have been detected in a number of solid complexes, in the liquid complex between CH3COCl and AlCl3 (by spectroscopy), in solution in polar solvents and in a number of cases where R is very

bulky. In less polar solvents the polarized complex R C O AlCl3

+

Cl

has been detected to act as the

electrophile. Direct chemical evidence clearly indicates that both can be involved depending upon the circumstances.Mechanism may be represented as follows:

(i)

R C Cl + AlCl3 R C = O + 4AlCl

O +

or

R C Cl + AlCl3 R C

O

Cl

O AlCl3 +

(ii)

+

C R H Cl

+

O AlCl3

R C O AlCl3

Cl

R C O + AlCl4 +

C R H

O

AlCl4

COR

+ AlCl3 + HCl

Product

Comparison between alkylation and acylation

(i) A comparison of the electrophilic nature of both alkyl and acyl group indicates that

acyl group is a better electrophile on account of two electron withdrawing atoms attached to C.

+ H

R C Cl

H alkyl halide

O

R C Cl

Effective +ve charge density on carbonyl carbon is greater.

(ii) Acylation requires more catalyst than alkylation because much of the catalyst is removed by the formation of a complex with the product (ketone)

+ AlCl3

COCH3 H3CC O AlCl3

+

and is removed from further participation in the reaction.

(iii) Unlike polyalkylation, polyacylation does not take place as the product ketone is much less reactive

than the original hydrocarbon.

(iv) Rearrangement of R does not take place, as in alkylation, but decarbonylation can take place,

especially where R would form a stable carbonium ion, so that the end result is then alkylation rather than the expected acylation.

Me3C C = O

CO + Me3C

+ Me3C+

CMe3

5. DIRECTIVE INFLUENCE OF SUBSTITUTENTS IN BENZENE NUCLEUS

The first substituent may occupy any position in benzene ring i.e. one and only one monosubstituted

benzene is obtained. The next group may go to ortho, meta or para position.

It is the group already present in the benzene nucleus that determines low readily the attack occurs and at what position of the ring it occurs. In other words, the group attached to the ring not only affects the reactivity

but also determines the orientation of substitution. This is called directive influence of substituents in benzene nucleus. The substituent group is able to activate or deactivate the ring due to a number of factors like

inductive effect, electromeric effect, resonance effect and hyperconjugative effect. Depending on their directive

influence the substituent groups, except halogens, are divided into two different classes:

Class I : R, OH, OR, NH2, NHR, NR2, NHCOCH3, OCOCH3, Cl, Br, I, F, CH2Cl, SH,

Ph, etc. These groups direct the incoming electrophile mainly to the o/ppositions.

Class II : NO2, CHO, CO2H, COCl, CONH2, CO2R, SO3H, SO2Cl, COCH3, CN, CCl3, NH3, NR3, OR2

etc. These groups direct the incoming electrophile mainly to the m position.

5.1 ACTIVATING GROUPS OR ELECTRON RELEASING GROUPS

All groups having one or more lone pair of electrons are activating groups because they release

electrons towards the nucleus increasing electron density and hence energy of the system. Reaction rate is increased due to low energy of activation. Examples:

.. O: , NH2, NHR, NR2, OH, OR, NHCOR, R, Ar, X: ..

.. .. .. .. .. ..

..

..

decreasing o & pdirecting strength

Activating groups make the electrons more readily available to an attacking electrophile by increasing

the electron density at o & ppositions and the product is always a mixture of

o & p isomers.

From an examination of the electronic structure and polar characteristics of o and

p directing groups, it is evident that with the exception of alkyl groups all of them possess at least one lone

pair of electrons at the atom adjacent to the benzene ring known as key atom. Also that their polar

characteristics are I effect and + R types, except alkyl group which is +I type and halogens which have strong

+ E effect in addition.

This lone pair of electrons is in conjugation with the electrons of the ring and exhibits a strong +R

effect, thus increasing the over all electron density in the benzene ring. Although the

I effect opposes the +R effect but latter predominates. Such groups. e.g. OH, OR, NH2, NHR, NR2 etc.

activate the benzene ring towards electrophilic substitution. However the relative increase of electron density is

great at o & p position and due to the nature of conjugation and hence the substitution occurs at these

positions.

Examples: Let us consider monosubstituted benzene, C6H5S where S is a substituent.

5.1.1 WHEN THE SUBSTITUENT(S) IS ELECTRONDONATING

When the substituent S present in the ring, has one or more lone pairs of electrons on the atom

adjacent to the ring, it interacts with electron system of the ring. This gives rise to the following five

resonance forms.

S : S

+

..

S +

S +

:

..

S S

+

I II III IV V Resonance Hybrid

The structure of the monosubstituted benzene, C6H5

S , is in fact represented by the resonance hybrid.

In the hybrid structure the overall electron density of the benzene ring is enhanced compared to unsubstituted benzene ring and more so at the o & p positions of the ring. Therefore the presence of an electron donating

group such as

S causes further electrophilic substitution in o & p positions and also activates the ring to

electrophilic attack.

Let us take the example of phenol, ,HOHC 56

and aniline 256 HNHC

, which have available

electron pair on the key atoms of the substituents. Thus phenol and aniline exhibit resonance and can be

represented as hybrid of the fine canonical forms.

:OH ..

+

.. :

..

+


OH ..

+ OH ..

+ OH ..

:OH ..

:OH ..

NH2 :

+

.. :

..

+


NH2 + NH2

+ NH2 NH2 : NH2

The overall electron density of the ring in each hybrid is increased and the benzene ring system as a

whole is activated to electrophilic attack. Of course, the electrophile (E+] would attack the ring preferentially at o and p positions where the electron density is relatively greater as compared to the meta positions.

Thus, all the groups which are electron donating HO

, RO

, 2HN

, NHR, NR2, NR3, etc. are orthopara

directing and facilitate electrophilic substitution in the benzene ring. 5.1.2 EFFECT OF ALKYL GROUP AS A SUBSTITUENT

Since alkyl group has no lone pair of electrons on key atom i.e. carbon atom, yet it is o & p directing.

First it is because of the fact, that alkyl group releases electron due to its +I (inductive) effect and hence tends to stabilise the carbocation by dispersal of its positive charge.Secondly it

is because the group electronegativity of methyl group is less than that of phenol group. At the same time, c of methyl group is sp3

hybridised whereas that of phenyl group is sp2 hybridised. So due to greater scharacter

of carbon atom of phenyl group, electron is withdrawn towards benzene ring. Third, methyl group repels

electrons towards the ring by hyperconjugation.

H C H

H

H C H+

H

H C H+

H

H C H+

H

H C H

H

All the nine hyperconjugative resonating structures reveal that electron density is enhanced (i.e. ve

charge is developed) at o & p positions simultaneously. Therefore, further electrophilic substitution occurs at

these positions.The electrophilic substitution of tert butyl benzene yields almost exclusively the para isomer.

This is so because the electrophile approach to the ortho position is impossible on account of the steric

influence of the substituent C(CH3)3.

H3C C CH3

CH3

2NO

Nitration + H+

tertbutyl benzene

H3C C CH3

CH3

NO2

+

It has been observed that the larger the size of the alkyl group already present, the smaller is the amount of

the o isomer formed in an aromatic electrophilic substitution reaction. In other words, the bulky alkyl groups

lead to steric hindrance to the introduction of new substituents in

o positions. 5.1.3 EFFECT OF HALOGENS

Halogens are unusual in their effect on electrophilic aromatic substitution. They are deactivating yet ortho, para directing.A halogen substituted benzene (C6H5X) by virtue of the presence of unshared electron pair on the

halogen, exhibits resonance. Thus it can be represented by the resonance hybrid of the following canonical

forms:

:X: ..

..

..

..

+:X ..

+:X ..

+:X ..

:X: ..

X +

Halogens withdraw electrons through its inductive effect (I) and releases electrons through it

resonance or mesomeric effect (+M). Thus for halogen, the two effects are more evenly balanced and we

observe the operation of both.

Through its inductive effect halogen tends to withdraw electrons and thus destabilise the intermediate carbocation. This effect is felt for attack at all positions, but particularly for attack at the positions ortho and

para to the halogen.

Through its resonance effect halogen, tends to release electrons. This electron release is effective only

for attack at the positions ortho and para to the halogen. The combination of the two effects makes the halogenated benzene deactivated. This is so because the inductive effect is stronger than the resonance

effect. Thus halogen, though, o and p directing, yet deactivates the benzene ring.

5.2 DEACTIVATING GROUP OR ELECTRON WITHDRAWING GROUP OR META DIRECTORS Such groups have tendency to withdraw electrons from the benzene nucleus and thus decreasing its

electron density are known as deactivating groups. Due to decrease in electron density of the ring, the rate of electrophilic substitution is retarded. That’s

why these group are called deactivating group. A look on the resonating structure reveal that these groups develop positive charge at ortho and para

positions leaving the metapositions as the point of relatively high electron density and hence the electrophilic

substitution occurs at mposition, not at o and ppositions. Deactivating groups have bonds with one more

electronegative atom.

Examples:

NR3, N , CN, SOH, CH, CR, COH, COR, CNH2, NH3 + O

O

O

O

O O O O O

+

decreasing meta directing strength

Predict the reactivity and orientation in (a) PhN=O and (b) PhS+Me2.

ELECTRONIC EXPLANATION OF mDIRECTIVE INFLUENCE

The polar characteristic of these groups is I and R with the exception of 3NR and CCl3 groups

which exhibit only I effect. Hence

3NR and CCl3 groups deactivate the benzene ring in general by

decreasing electron density due to I effect. However, the withdrawal of electron from ortho and para positions

is as compared to mposition. Thus, meta position remains the point of comparatively high electron density

and electrophilic substitution occurs preferentially at metaposition.

Examples: (i) When the substituent has at least one strongly electronegative atom and a multiple bond in conjugation

with benzene ring: Let A = B represents the group in which B is more electronegative than A. The highly electronegative

atom pulls the electron pair of the multiple bond which in turn withdraws electrons from benzene ring (

M effect) giving rise to the following five canonical forms:

A =B ..

+

+

+


A B ..

A B ..

A B A =B A B

The structure of the benzene derivative C6H5A = B is, in fact, represented by the resonance hybrid

shown above. Evidently electrons are withdrawn by the substituent group from the ring and more so

from the o p positions where the electron density declines. Also, the meta positions have relatively

more electron density and therefore electrophile substitution take place at the meta positions.

The nitro and sulphonic acid groups offer examples of the type of electron withdrawing substituents

which are characterized by the presence of a strongly electronegative atom attached to another more

electronegative atom by a multiple bond.

ortho, para meta

VSA SA A D SD VSD

OH, NH2, NHCOR, R, Ar I, Br, Cl, SO3H, CN, CHO NO2, CF3,

NHR, NR2 OCOR, OR C C

F, CH2X COR 3RN

O S , NO

COOH, COOR,

COCl

Rank each of these species in order of decreasing reactivity to electrophilic substitution: PhMe, PhNMe2, PhN+Me3, PhCH2N

+Me3.

5.3 ORIENTATION IN BENZENE RING WITH TWO SUBSTITUENTS

The position taken up by a third electrophile entering the ring depends on the nature of the two groups

already present. It is often possible in such cases to predict the correct isomer; but remember following

generalizations in order to do so.

1. When the two groups direct differently, i.e. they belong to the classes I and II, then

class I group takes precedence.

In the following examples, the number of arrowheads indicates (qualitatively) the amount of substitution

and the encircled number below indicates the number of isomers.

For example,

2

Cl

NO2

3

OH

NO2

2. When both groups belong to class I, then introduction of third group is very easy and

the group enters in accordance with higher activating group. For this purpose, we can

arrange the groups in the following order: NH2, OH, NR2, O > OR, OCOR, NHCOR

> R, Ar > halogen. For example,

OH

Cl

1

3. When both groups belong to class II, then it is difficult to introduce a third group and the deactivating

power of groups is in the order:

Me3N+ > NO2 > CN > SO3H > CHO > COMe > CO2H

The more deactivating group controls the orientation because the arenium ion formed will be less

destabilized. For example,

NO2

CHO

NO2 CHO

1 2

4. All other things being equal, a third group is least likely to enter between two groups in the meta

position. This is the result of steric hindrance and increases in importance with the size of the groups

on the ring and with the size of the attacking electrophile. For example,

OCH3

SO3H

3

Most favoured attack

Least favoured attack

Now, using these generalizations, we can predict the preferable attack by the incoming group in

disubstituted benzene. We can have following possibilities:

(a) When the two groups present are such that they belong to different categories and oppose each other.

For example, in mhydroxy benzaldehyde, the incoming electrophile is oriented by OH group (and not

by CHO group). As OH group is o/pdirecting, thus attack of the electrophile can take place at three

sites. Attack at position2 is least favoured according to rule 3, but attack at position4 is the most

preferred.

OH

CHO

3

1

2 3

4

5

6

(b) When the two groups present are such that they belong to different categories but they reinforce each

other, then the third group enters almost entirely at one position.

For example, the incoming group in pchlorobenzoic acid goes to the position ortho to the chloro group

and meta to the carboxyl group.

CO2H

Cl 1

Some other examples are:

Me

NO2

OH

NO2 1 1 2

OH

NO2

(c) When the two groups belong to category I (having large difference in their activating ability) and are

placed such that they oppose each other, then the third group enters in accordance with the group

having higher activating ability.

For example, in ocresol (omethyl phenol), the incoming electrophile could go at position4 and 6 but

attack at position4 is greatly favoured due to less steric repulsion.

OH

2

2

3

6

5

4

CH3 1

(d) When the two groups belong to category I and are placed such that they reinforce each other, the

incoming electrophile enters almost entirely at one position. For example, 1,3dimethylbenzene is

substituted at the position4 (ortho to one CH3 group and para to the other CH3 group), but not at the

position5 (meta to both) because of rule 4.

CH3

CH3

1

3

4

6

5

2

2

(e) When the groups are such that they belong to category I with almost identical directing abilities and

oppose each other, then predictions are more difficult. In a case such as,

NHCOCH3

OCH3

where two groups of about equal directing ability are in competing positions, all four products can be

expected and it is not easy to predict the proportions, except that steric hindrance should probably

reduce the yield of substitution ortho to the acetamido group, especially for large electrophiles. Mixtures

of about equal proportions are obtained in such cases.

Few other examples are

Me

4

Cl

Me

Cl

2

NH2

OH

Cl

Br

2 2

(f) When the groups are such that they belong to category I with almost identical directing ability and they

reinforce each other, then also the predictions are difficult.

For example, in mmethoxy acetamide, the attack at position2 is least favoured according to rule 4

but attack at position4 and at 6 is equally feasible.

NHCOCH3

OCH3

3

1 2

3

4

5

6

(g) When both the groups belong to class II and are present such that they oppose each other, then the

more deactivating group decides the orientation of incoming electrophile.

For example, in onitro benzaldehyde, the incoming group attacks at position3 and at position5, of

which position5 is more favoured because of lesser steric repulsion.

CHO

2

2

3

6

5 4

NO2 1

(h) When both the groups belong to class II and are present such that they reinforce each other, then

attack takes place entirely at one position.

For example, 3nitro benzaldehyde is substituted at position5 as it is meta to both

the groups.

NO2

CHO 1

2

3

4

6

5

1

6. SUBSTITUTION IN OTHER AROMATIC SYSTEMS

6.1 NAPHTHALENE AND ANTHRACENE RINGS

The delocalization of the positive charge in the transition states of electrophilic substitutions is

increased by the fusion of two or more benzene rings and the polycyclic hydrocarbons are therefore all more reactive than benzene.

Substitution in naphthalene is illustrative. The transition states for the reaction at 1 and 2positions

may be represented as follows:

H

E

*

H E

*

1substitution:

H

E

H E H E

H

E *

H E

2substitution:

H E

H

E H

E

In each of the two transition states, the positive charge is more extensively delocalized than in reaction

with benzene, leading to lower activation energies. Further, the three starred structures are of benzenoid type

and therefore of lower energy content than the remainder in which the benzenoid nature of the second ring has

been destroyed. Since there are two such lowenergy contributors for 1substitution as compared with one for

2substitution, it is understandable that the 1position should be the more reactive than 2position.

The presence of an electronattracting group in naphthalene reduces the reactivity and causes

substitution to occur in the unsubstituted ring, mainly at the 5 and 8positions (i.e. the two 1positions of that

ring). An electronreleasing group activates the molecule further and reaction occurs in the substituted ring. If the group is in the 1position, substitution occurs at the 2 and 4positions (i.e. ortho and para to the

electronreleasing group), but a 2substituent directs almost entirely to the 1position, although the 3position

is also an ortho position. The reason is that the stabilization of the transition state which is provided by the

substituent is more effective when the appropriate resonance structure is benzenoid (1substitution) than when it is not (3substitution). For

instance,

H E

OH

H

E

OH

..

..

*

(Benzenoid) (Nonbenzenoid)

In anthracene, the electrophile attacks preferentially at the 9 or 10 positions since

the arenium ion formed by the electrophilic attack at any of these positions can have two intact

benzene rings in its canonical forms, while attack of electrophile at any other position (1 or 2) would give arenium ion having a naphthalene ring in its canonical forms.

The resonance energy of 2 benzene rings is more than the resonance energy of a naphthalene ring.

(Position 1,4,5 and 8 are identical, positions 2,3,6 and 7 are also identical and positions 9 and 10 are same)

1 2

3 4 5

6

7 8 9

10

6.2 FIVEMEMBERED HETEROCYCLIC RINGS

The principles that govern the electrophilic substitutions of this group of heteroaromatic compounds will

be illustrated by reference to pyrrole.

Pyrrole is highly reactive at both the 2 and 3positions. The reason is that the transition state for

substitution at each position is strongly stabilized by the accommodation of the positive charge by

nitrogen, in just the way that aniline owes its reactivity to the exocyclic nitrogen. 2substitution

predominates because the positive charge in the transition state is delocalized over a total of three

atoms, compared with two for 3substitution.

The similarity of pyrrole and aniline is particularly apparent in their reactions with bromine: each reacts

at all its activated carbon atoms, pyrrole giving tetrabromopyrrole and aniline giving

2,4,6tribromoaniline. In fact, pyrrole is even more strongly activated than aniline and should perhaps

be compared with the phenoxide ion: each undergoes the ReimerTiemann reaction, unlike other

benzenoid compounds. In addition, pyrrole undergoes FriedelCrafts acylation in the absence of a

catalyst.

H

2substitution:

N

H

N H

E H

N H

E

H

E

.. ..

(Stable contributoroctet of every atom is complete)

H

3substitution:

N

H

N

H

E

H

E

..

(Stable contributoroctet of every atom is complete)

Furan and thiophen are also activated towards electrophiles and react predominantly at the 2position.

The underlying theory is similar to that for pyrrole, namely, that the heteroatom is able to delocalize the

positive charge on the transition state. Since oxygen accommodates a positive charge less readily than

nitrogen, furan is less reactive than pyrrole, just as phenol is less reactive than aniline. The +M effect of sulphur is smaller than that of oxygen because the overlap of the differently sized porbitals of carbon

and sulphur is less than in the case of carbon and oxygen so that, understandably, thiophen is less

reactive than furan. Thus, the reactivity order of 5membered heterocyclics towards electrophilic

substitution would be

pyrrole > furan > thiophene.

6.3 SIXMEMBERED HETEROCYCLIC RINGS

The principles governing the reactivity of these compounds are illustrated by reference to pyridine. The

transition states for substitution at the 3 and 4positions can be represented as the hybrids.

3substitution:

N

H

E

N

H

E

N

H

E

4substitution:

N

H E

N

H E

N

H E

In each case, the positive charge is less well accommodated than in reactions on benzene because

nitrogen is more electronegative than carbon. Hence, both the 3 and 4positions are deactivated, the

latter more strongly because of the high energy of the contributing structure which contains divalent

positive nitrogen. The 2position resembles

the 4position, as reference to the appropriate resonance structures will show.

REACTIONS OF BENZENE AND ITS HOMOLOGUES

(i) Benzene on vigorous oxidation (combustion) gives CO2 and H2O.

C6H6(l) + )g(O2

152 6CO2(g) + 3H2O(l)

Benzene on oxidation by air at 723 K in presence of V2O5 gives maleic anhydride.

+ 2

9O2 3

V2O5 at 723 K

CH

CH

C

C

O

O

O

+ 2H2O + 2CO2

(ii) Side chain oxidation of alkyl benzenes

All alkyl benzenes (irrespective of the chain length) on oxidation by acidified or alkaline KMnO4

(followed by acidification) are degraded to benzoic acid provided they have a benzylic hydrogen.

KMnO4, H+

CH3 CO2H

KMnO4, H+

CH2CH3 CO2H

KMnO4, H+

CH CO2H

CH3 H3C

KMnO4, H+

CH2CH2CH2R CO2H

Even if the side chain contains some functional groups, whether they are electron pushing or electron

withdrawing, are also degraded to benzoic acid on oxidation.

KMnO4, H+

CH2CH2CHCH3 CO2H

NO2

CH=CH2 CCH

and are also oxidised by KMnO4, H+ or KMnO4, OH to give benzoic acid due to

oxidative cleavage of C=C and CC.

Compounds that do not contain a benzylic hydrogen will not get oxidised to benzoic acid.

KMnO4, H+

RCR

No reaction

R

Compounds which contain, two alkyl carbons bearing benzylic hydrogen are oxidised to give two CO2H

groups on the bezene ring.

KMnO4, H

+ CH2R

CH2R

CO2H

CO2H

Phthalic acid

KMnO4, H

+ CH2R

CH2R

CO2H

CO2H

Isophthalic acid

KMnO4, H+

CH2R

CH2R

CO2H

CO2H

Terephthalic acid

KMnO4, H

+ CO2H

CO2H

KMnO4, H

+ CO2H

CO2H

(iii) Addition reactions:

Benzene is so stable that it gives addition reactions much less readily. It can be reduced by catalytic

hydrogenation, when three molecules of hydrogen add to it in presence of Pt as a catalyst.

Pt

+ 3H2 423 K, 100 atm

Addition of 3 molecules of chlorine also takes place to one molecule of benzene in presence of

sunlight. The product formed is 1,2,3,4,5,6hexachlorocyclohexane

(also called as benzene hexachloride). This compound is used as an important insecticide and is sold commercially in the name of BHC or gammaxene.

Sunlight + 3Cl2

Cl

Cl

Cl

Cl

Cl

Cl

PROFICIENCY TEST

The following 10 questions deal with the basic concepts of this section. Answer the following briefly. Go to the next section only if your score is greater than or equal to 8.

Do not consult the study material while attempting the questions.

1. True/False. An electron donating substituent in benzene orients the incoming electrophilic group to the

meta position.

2. True/False. The electrophile in the sulphonation of aromatic molecule may be either SO3 or .HSO 3

3. True/False. In FriedelCrafts alkylation the electrophile is a carbocation.

4. True/False. In FriedelCrafts alkylation, AlCl3 acts as a Lewis base.

5. True/False. A substituent, which stabilises the intermediate carbocation in the electrophilic substitution

in benzene, activates the latter.

6. True/False. An activating group activates only ortho and para positions in benzene ring.

7. True/False. Benzene in the presence of AlCl3 reacts with isobutyl bromide to give tertbutylbenzene.

8. True/False. The monobromination of acetanilide proceeds slower than benzene. 9. True/False. A tribromobenzene gives two isomers when it is mononitrated. The compound is 1, 2,

4tribromobenzene.

10. True/False. Monobromination of ptoluenesulphonic acid followed by treatment with acid and

superheated steam gives mbromotoluene.

ANSWERS TO PROFICIENCY TEST

1. False

2. True

3. True

4. False

5. True

6. False

7. True

8. False

9. False

10. False

SOLVED OBJECTIVE EXAMPLES

Example 1:

In the following reaction,

N

H

O conc. HNO3

conc. H2SO4 X

the structure of the major product ‘X’ is

(a)

N

H

O

NO2 (b)

N

H

O O2N

(c)

N

H

O

NO2

(d)

N

H

O

O2N

Solution:

The ring to which –NH group is attached is activated due to the lone-pair on N; while the ring to which

C

O

is

attached is deactivated. Hence, the electrophile would go to the para-position of the activated ring.

(b)

SOLVED SUBJECTIVE EXAMPLES

Example 1:

Compare the products of the reactions of PhCH2CH3 with Br2 and Cl2 in light and explain any differences.

Solution:

Bromination gives bromoethylbenzene exclusively. The 2° benzylic position is only slightly favored over the 1°

alkyl position in chlorination, the products being PhCHClCH3 (56%) and PhCH2CH2Cl (44%). The less reactive Br

is more selective than the Cl, which abstracts an H in a more random fashion (ReactivitySelectivity principle).

Example 2:

Irradiation of an equimolar mixture of cyclohexane and PhCH3 gives mostly cyclohexyl chloride with Cl2 and PhCH2Br with Br2. Explain.

Solution:

In these competitive reactions the reactivities of cyclohexane and toluene are compared, Cl , being more reactive

and less selective than Br, reacts with the kind of H present in greatest number, which in this case is one of the

twelve equivalent H’s of cyclohexane. The less reactive and more selective Br reacts with the most reactive H, in this case one of the three alkyl H’s of PhCH3.

Example 3:

Deduce the structure of compound A, C9H8, from the following experimental data: A decolorizes Br2 in CCl4 and adds one eq. of H2 under mild conditions, forming B, C9H10. At high temperature and pressure, A adds four eq. of H2. Vigorous oxidation of A yields phthalic acid, 1,

2C6H4(COOH)2.

Solution:

A has 6° of unsaturation, four of which often signal the presence of a benzene ring, as confirmed by isolation of phthalic acid on vigorous oxidation. Reaction with Br2 and one eq. of H2 indicate there is a C=C. Addition of three more eq. of H2 further indicates the presence of a benzene ring. So far 5° of unsaturation have been accounted

for the sixth degree resists reduction and must be a ring with C=C. Oxidation to the orthodicarboxylic acid indicates the ring is fused to the benzene ring. The structure is indene.

COOH [O]

COOH

H2 3H2

Phthalic acid Indene, (A) Indane, (B) Bicyclo[4.3.0] nonane

Example 4:

Identify (a) the chiral compound C, C10H14, that is oxidized with alkaline KMnO4 to PhCOOH and (b) the achiral compound D, C10H14, inert to oxidation under the same conditions.

Solution:

(a) C is a monoalkyl substituted benzene with four C’s in the side chain. The only R with four C’s, one of which is

a chiral center, is CH(CH3)CH2CH3; C is secbutylbenzene, PhCH(CH3)CH2CH3. (b) Any R attached to the

benzene ring cannot have a benzylic H (no oxidation); D is tertbutylbenzene, PhCMe3

Example 5:

(a) Give the structures for all possible chiral compounds, C10H12, that do not decolorize Br2 and that can be oxidized to phthalic acid. (b) Identify E, also chiral, with the same formula, but which is oxidized to PhCOOH.

Solution:

(a) The formula reveals a fifth degree of unsaturation in addition to the four of the benzene ring. This fifth degree of unsaturation must be a ring, not C=C, because the Br2 test is negative. Production of phthalic acid means the ring is fused to the benzene ring. This fused ring has the chiral carbon and must be a

monoRsubstituted fivemembered or diRsubstituted fourmembered ring. Only in this way can we account for the additional four C’s of the formula.

1Methylindane trans1, 2Dimethylbenzcyclobutane

H Me

H

Me

H Me

H

Et

1Ethylbenzcyclobutane

(b) The extra unsaturation is in the single side chain. E is 3phenyl1butene, CH3CHCH=CH2.

Ph

EXERCISE – I

AIEEE-SINGLE CHOICE CORRECT

1. Oxidation of benzene with air at 723 K in the presence of V2O5 gives

(a) Malic acid (b) Maleic acid

(c) Malonic acid (d) Maleic anhydride

2. nButyl benzene on oxidation will give

(a) Benzoic acid (b) Butanoic acid

(c) 4Phenyl butanoic acid (d) Benzaldehyde

3. Which reaction sequence would be best to prepare 3chloroaniline from benzene?

(a) Chlorination, nitration, reduction.

(b) Nitration, chlorination, reduction.

(c) Nitration, reduction, chlorination.

(d) Nitration, reduction, acylation, chlorination, hydrolysis.

4. All the following groups are activating ortho, para directors when attached to a benzene ring except

(a) OCH3 (b)

NHCCH3

O

(c) Cl (d) N(CH3)2

5. Rank the following compounds in terms of increasing reactivity towards nitration with conc. HNO 3/

conc. H2SO4.

(1) (2)

Cl

(3)

NH–CH3

(a) (1) < (2) < (3) (b) (2) < (1) < (3)

(c) (3) < (1) < (2) (d) (3) < (2) < (1)

6. For the reaction,

Br

NO2

?

the best reactants are

(a) C6H5Br + HNO3, H2SO4 (b) C6H5Br + H2SO4,

(c) C6H5NO2 + Br2, FeBr3 (d) C6H5NO2 + HBr

7. Consider the following statements concerning the effect of a trifluoroemethyl group, CF3, on an

electrophilic aromatic substitution.

1. The CF3 group will activate the ring.

2. The CF3 group will deactivate the ring.

3. The CF3 group will be a meta director.

4. The CF3 group will be an ortho, para director.

Which of these statements are correct?

(a) 1,3 (b) 1,4

(c) 2,3 (d) 2,4 8. Among the following, the compound that can be most readily sulphonated is

(a) benzene (b) nitrobenzene

(c) toluene (d) chlorobenzene 9. Among the following statements on the nitration of aromatic compounds, the false one is

(a) The rate of nitration of benzene is almost the same as that of hexadeuterobenzene.

(b) The rate of nitration of toluene is greater than that of benzene.

(c) The rate of nitration of benzene is greater than that of hexadeuterobenzene.

(d) Nitration is an electrophilic substitution reaction. 10. Which of the following reactions is not an example of electrophilic substitution

(a) 3AlCl3266 CHCHCHHC C6H5CH(CH3)2

(b) ClCHHCClCHHC 256AlCl

22663

(c) 666lightUV

266 ClHCCl3HC

(d) 3AlCl

56 HClCOOHHC

OH

CHO

11. Which of the following is meta directing group?

(a) COOH (b) OH

(c) NH2 (d) Cl

12. In the nitration of benzene with a mixture of conc. HNO3 and conc. H2SO4, the active species involved

is

(a) 3NO (b) NO2

(c) 2NO (d)

2NO

13. In the sulphonation of benzene, the active species involved is

(a) 4HSO (b) SO3

(c) SO2 (d) 24SO

14. Benzene reacts with acetyl chloride in presence of anhydrous aluminium chloride to form

(a) acetophenone (b) phenyl acetate

(c) chlorobenzene (d) benzoic acid

15. The direct iodination of benzene is not possible because

(a) I2 is an oxidising agent. (b) resulting C6H5I is reduced to C6H6 by HI (c) HI is unstable (d) the ring gets deactivated

16. Identify the correct order of reactivity in electrophilic substitution reactions of the following compounds:

CH3 Cl NO2

(I) (II)

(III)

(IV)

(a) (I) > (II) > (III) > (IV) (b) (IV) > (III) > (II) > (I)

(c) (II) > (I) > (III) > (IV) (d) (II) > (III) > (I) > (IV)

17. Sulphonation of benzene differs from most of the electrophilic aromatic substitution reactions. Which

one of the following statement is correct? (a) is reversible.

(b) requires the presence of Lewis acid as catalyst. (c) takes place with explosive violence.

(d) requires elevated temperature.

18. Benzene reacts with npropyl chloride in the presence of anhydrous AlCl3 to give predominantly

(a) npropyl benzene (b) isopropyl benzene

(c) 3propyl1chloro benzene (d) no reaction

19. Which one of the following will undergo meta substitution on mono chlorination?

(a) chloro benzene (b) phenol

(c) ethyl benzoate (d) ethoxy benzene

20. NH2, Cl, OH and CH3 groups when attached to benzene ring activate/deactivate it for electrophilic

substitution reaction. Their decreasing order of activation is (a) NH2 > OH > Cl > CH3 (b) NH2 < Cl > OH > CH3

(c) NH2 > OH > CH3 > Cl (d) OH > NH2 > Cl > CH3

21. The function of anhydrous aluminium chloride in the Friedel-Craft’s reaction is

(a) to absorb water. (b) to absorb hydrochloric acid.

(c) to produce an electrophile. (d) to produce nucleophile.

22. Isopropylbenezene can be obtained by

(a)

H+ + CH3–CH=CH2 (b)

AlCl3

+ CH3 CH2CH2Cl

(c)

AlCl3 +

CH3 CHCl

CH3

(d) all of these

23. A particular form of tribromobenzene (A) melts at 44°C. On nitration it forms one possible mononitro

tribromo benzene. The structure of compound (A) is

(a)

Br Br

Br

(b)

Br

Br

Br

(c)

Br

Br

Br

(d) Both (a) and (b)

24. To prevent the formation of carbocation in Friedal craft alkylation the best possible condition

(a) AlCl3 + high temperature (b) SnCl4 + high temperature

(c) AlCl3 + low temperature (d) SnCl4 + low temperature

25. To prepare

NO2

Et from

, which of the following is the correct sequence of steps?

(a) Nitration followed by Friedel Crafts ethylation. (b) Friedel Crafts ethylation followed by nitration.

(c) Friedel Crafts acylation (CH3COCl), nitration followed by Clemenson’s reduction. (d) Friedel Crafts acylation (CH3COCl), Clemenson’s reduction followed by nitration.

EXERCISE – II

IIT-JEE- SINGLE CHOICE CORRECT 1. Chlorination of toluene in the presence of light and heat followed by treatment with aqueous NaOH

gives (a) ocresol (b) pcresol

(c) 2, 4dihydroxytoluene (d) benzoic acid

2. For the electrophilic substitution reaction involving nitration, which of the following sequence regarding

the rate of reaction is true?

(a) 666666 TCDCHC kkk (b)

666666 TCDCHC kkk

(c) 666666 TCDCHC kkk (d)

666666 TCDCHC kkk

3. Which of the following statement made on FriedelCrafts reaction is false?

(a) Friedel Crafts alkylation and acylation leads to new carboncarbon bond formation.

(b) Alcohol/acid combination can be used as reagent for FriedelCrafts alkylation.

(c) When benzene and nitrobenzene, both liquids, are mixed and treated with CH3COCl/AlCl3 under proper reaction conditions, the products obtained are macetyl nitrobenzene

and acetophenone. (d) npropylbenzene can be made in good yield from benzene using npropyl chloride and anhydrous

AlCl3 combination as reagent at low temperature.

4. The presence of which one of the following groups on benzene nucleus activates it towards

electrophilic substitution? (a) CN (b) CHO

(c) COOR (d) OCOR

5. The halide that will not react with benzene in the presence of anhydrous AlCl3 is

(a) CH3CHClCH3 (b) C6H5CH2Cl (c) C6H5Cl (d) CH3CH2CH2Cl

6. Which of the following species is expected to have maximum enthalpy in an electrophilic aromatic

substitution reaction?

+ E+

(I) (II) (III) (IV) (V)

E H E+ H E +H E

+

+

+

(a) Species (II) (b) Species (III)

(c) Species (IV) (d) Species (V)

7. The relative reactivities of benzene, aniline, nitrobenzene and toluene are

(a) benzene > toluene > aniline > nitrobenzene

(b) nitrobenzene > aniline > toluene > benzene

(c) toluene > aniline > nitrobenzene > benzene

(d) aniline > toluene > benzene > nitrobenzene

8. During electrophilic substitution the major product in the following reaction will be

FeCl3

+ N

Cl2 ?

H

(a)

Cl N

H

(b)

Cl

N

H

(c)

Cl

N

(d) both (a) & (b)

9. Electrophile 2NO attacks the following aromatic species.

CCl3

(I)

NO2

(II)

(III)

3HN

O

(IV)

In which cases 2NO will be at meta position?

(a) (II) and (IV) (b) (I), (II) and (III)

(c) (II) and (III) only (d) (I) only

10. Which one of the following aromatic compounds fails to undergo FriedelCrafts reactions?

(a) C6H5CH3 (b) C6D6

(c) C6H5NO2 (d) C6H5Cl

11. Phenol reacts with bromine in carbon disulphide at low temperature to give

(a) mbromophenol (b) o and pbromophenol

(c) pbromophenol (d) 2,4,6tribromophenol

12. Benzene on reaction with conc. HNO3 in presence of conc. H2SO4 followed by the treatment of Cl2 in

presence of FeCl3, gives

(a) 2chloro1nitrobenzene

(b) 3chloro1nitrobenzene

(c) 4chloro1nitrobenzene

(d) a mixture of 2chloro and 4chloro1nitro benzene

13. Which of the following substituted benzenes would furnish three isomeric compounds when one more

substituent is introduced?

(1)

Cl

(2)

Cl

Cl

(3)

Cl

Cl (4)

Cl

Cl

Select the correct answer using the codes given below:

(a) (1), (2) and (3) (b) only (1)

(c) (2) and (4) (d) (1) and (3)

14. Toluene when treated with Br2/Fe gives pbromotoluene as the major product because the CH3 group

(a) has I effect. (b) is meta directing.

(c) activates the ring by hyperconjugation. (d) deactivates the ring.

15. The major product formed on nitration of N, Ndimethylaniline with conc. H2SO4/HNO3 mixture is

(a)

NMe2

NO2

(b)

NMe2

NO2

(c)

NMe2

NO2

NO2

(d)

NMe2

NO2

16. Which of the following reaction is wrong?

(a)

Anhydrous AlCl3

+ CH3Cl

CH3

(b)

Anhydrous AlCl3

+ CH3CH2CH2Cl

CH2CH2CH3

Room Temperature

(c)

Anhydrous AlCl3 + CH3CH2CH2Cl

CH(CH3)2

Room Temperature

(d)

Anhydrous AlCl3

+ CH3COCl

CH3C

O

Room Temperature

17. Which of the following triad of group activates the benzene ring and directs the electrophile to o and

pposition for substitution?

(a) NO2, CHO, COOH (b) OH, O, CH3

(c) OH, SO2OH, NO2 (d) NH2, CHO, SO2OH

18.

Anhydrous

AlCl3 + CH3CH2CCl

O

(X)

The structure of (X) would be

(a)

C

O

(b)

(c)

O

(d)

O

OH

19.

Cl2, Fe

CCl3

X. What is X as a main product?

(a)

CCl3

Cl

(b)

CCl3

Cl

(c)

CCl3

Cl

(d)

CCl3

Cl

Cl Cl

20. pNitrotoluene on further nitration gives

(a)

CH3

NO2

NO2

(b)

CH3

NO2

NO2

(c)

CH3

NO2

NO2

(d)

CH3

O2N

NO2

21. Which of the following statements is correct?

(a) Bromination of toluene occurs faster than that of benzene. (b) Nitration of toluene is difficult than that of nitrobenzene.

(c) The bromonium ion is a good nucleophile. (d) Effective nitrating agent is nitrate ion.

22. The order of reactivity of following compounds

CH3

(I)

CH2CH3

(II) (III)

CH2(CH3)2

(IV)

C(CH3)3

towards electrophilic substitution will be-[where φ = C6H5]

(a) (I) > (II) > (III) > (IV) (b) (IV) > (III) > (II) > (I)

(c) (II) > (I) > (III) > (IV) (d) (III) > (II) > (I) > (IV)

23. NO2+BF4

–

O2N

(X). What is the product (X) formed in the reaction. CH2

(a)

CH2

NO2

O2N

(b)

CH2

NO2

O2N

(c)

CH2 NO2

O2N

(d) both (b) and (c)

24.

+

Anhydrous AlCl3

COCl

COCl

(X). The major product

(a)

C=O

C=O

H

H

(b)

O

C

O

C

(c)

O

C

O

C

(d) all of these

25.

Br2/Fe(1eq) C

N

O

(a)

N

O Br

(b)

N

O Br

(c)

N

O

Br (d)

N

O Br

C +

EXERCISE – III

MORE THAN ONE CHOICE CORRECT

1. Which of the following groups is/are deactivating but o/pdirecting?

(a) Cl (b) CH2Cl

(c) NO2 (d) –NO

2. Which of the following compound will direct the incoming electrophile to the meta position?

(a)

OCH3

(b)

CF3

(a)

CHO

(d)

CH2OH

3. In the given sequence of reaction,

+ methyl succinic anhydride Anhydrous

(A) Zn / Hg

Conc. HCl (B)

SOCl2 (C)

AlCl3 (D)

AlCl3

(a) (A) is

CH CH2

CH2

CO2H

Me

(b) (A) is

CH2

CHCH3

CO2H

O

(c) (D) is

Me

O

(d) (D) is

Me

O

4. Which of the following is not a suitable method to prepare, PhCH2CH2Ph?

(a) 2PhH + CH2Cl2 3AlCl

Anhydrous

(b) PhH + ClCH2CH2Ph .templow,AlCl

Anhydrous

3

(c) PhH + CH3CHPh

Cl

3AlCl

Anhydrous

(d) PhH + PhCH2CCl

O

3AlCl

Anhydrous

HCl.Conc

Hg/Zn

5. A chiral compound (A) with molecular formula C10H12 do not decolourize Br2 water but can be oxidised

to phthalic acid. The possible structure of (A) could be

(a)

Me H

(b) Me

H

(c)

Me H Me

H

(d)

Me H

Me H

6. What is the correct about the given compound,

SMe2

?

(a) The compound is more reactive than benzene towards electrophilic substitution.

(b) The compound is less reactive than benzene towards electrophilic substitution. (c) The incoming electrophile will be directed towards orthopara position.

(d) The incoming electrophile will be directed towards meta position.

7. Which of the following compounds are meta directors?

(a)

CH2Cl

(b)

CCl3

(c)

NMe3

(d)

CH2NMe3

8.

+ CH3CH=CH2 (A) HF, 0°C

+ CH2=CHCH2Cl (B) Anhydrous

ZnCl2

Which is the correct about (A) and (B)?

(a)

CH

CH3

CH3 (A): (b)

CH2CH=CH2

(B):

(c)

CH2CH=CH2

(A): (d)

CH

CH3

CH3 (B):

9. The groups that are activating and orthopara directing would be

(a)

OCCH3

O

(b)

SCH3

O

(c) CH2CH2NMe3

(d) SCH2Cl

10. Which of the following combination of reagents or sequence of reactions can give linear alkyl benzene

(LAB)?

(a)

+ CH3CH2CH2Cl ? Anhydrous

AlCl3,

(b)

+ CH3CH2CH2Cl ? FeCl3

(c)

+ CH3CCH2Cl ? Anhydrous

AlCl3

O

Zn / Hg

Conc. HCl

(d)

+ CH3CH2CCl ? Anhydrous

AlCl3

O

Zn / Hg

Conc. HCl

EXERCISE – IV

MATCH THE FOLLOWING

1. Compare the rates of bromination of given compounds, relative to benzene taken

as standard.

Column I

Compounds

Column II

Rate of Bromination

I. C6H5OCH3 (A) 1012

II. C6H5N(CH3)2 (B) 10

III. C6H5CH3 (C) 105

IV. C6H5NO2 (D) 1

(E) 109

2.

Column I

Groups attached to benzene

Column II

Effect of the group

I. 33 )CH(N

(A) I, +R

II. CH=CH2 (B) I, R

III. OCH3 (C) I only

IV. NO2 (D) +I, +R

(E) I, +R, –R

3.

Column I

Groups attached to benzene

Column II

Nature of the group

I. OH (A) strongly deactivating, ortho para directing

II. Br (B) strongly activating, ortho-para directing

III. NO2 (C) weakly deactivating, ortho-para directing

IV. NHCOCH3 (D) strongly deactivating, meta directing

(E) Moderately activating, ortho-para directing

ASSERTION AND REASON

Direction: Read the following questions and choose

(A) If both Assertion and Reason are true and Reason is the correct explanation of the assertion

(B) If both Assertion and Reason are true but Reason is not correct explanation of the Assertion.

(C) If Assertion is true but Reason is false.

(D) If Assertion is false but Reason is true.

1. Assertion: methyl nitrobenzene should be prepared by the nitration of benzene followed by its

ethylation of nitrobenzene.

Reason: nitro group is meta directing group.

(a) (A) (b) (B) (c) (C) (d) (D)

2. Assertion: Alkylation of benzene by Friedel-Crafts reaction gives polyalkylated benzene.

Reason: The ring gets activated for further substitution after the introduction of one alkyl group.

(a) (A) (b) (B) (c) (C) (d) (D)

3. Assertion: Nitration of aniline can be done by protecting —NH2 group by acetylation.

Reason: Aniline ring is oxidized by concentrated HNO3.

(a) (A) (b) (B) (c) (C) (d) (D)

4. Assertion: In strongly acidic solution, aniline becomes more reactive toward electrophilic reagents.

Reason: The amino group is completely protonated in strongly acidic medium. The lone pair of electrons

on the nitrogen, therefore, is no longer available for resonance.

(a) (A) (b) (B) (c) (C) (d) (D)

5. Assertion: A mixture of concentrated nitric acid and concentrated sulphuric acid is better than

concentrated nitric acid for electrophilic nitration of benzene.

Reason: H2SO4 is a stronger acid than HNO3 which accelerates formation of 2ON

electrophile.

(a) (A) (b) (B) (c) (C) (d) (D)

PASSAGE BASED PROBLEMS

Electrophilic substitutions are characteristic reactions of aromatic compounds. In this reaction hydrogen

of the aromatic ring is easily replaced by an electrophile because hydrogen in the form of H+ is a very good

leaving electrophile. CO2 and SO3 are also good leaving electrophiles when –COOH and –SO3H groups are

present either at ortho or at para position to the highly activating groups like –NH2 and –OH etc. Electrophilic substitution reactions become difficult in presence of electron withdrawing groups. Position of incoming

electrophile depends on several factors like electronic and steric effect of already present group, intramolecular hydrogen bonding and the temperature at which reaction is being done. All electron donating

groups also activate oxidation of aromatic compound beside activating electrophilic aromatic substitution.

1.

excess

.)aq(Br2 SO3H

OH

(X)

The compound (X) is

(a)

SO3H

OH

Br

Br

(b)

SO3H

OH

Br

(c)

OH

Br

Br Br

(d)

OH

Br

2. Which of the following compound will not give electrophilic aromatic substitution?

(a)

CH3

H3C CH3

(b)

NO2

O2N OH

(c)

NO2

OH

(d)

NO2

O2N NO2

3. 2, 4, 6Trinitrophenol (picric acid) can be prepared by treating phenol with cold concentrated H2SO4

followed by treatment with hot concentrated HNO3. In this process:

(a) Cold concentrated H2SO4 makes sulphonation at ortho position which accelerates oxidation of aromatic ring by concentrated HNO3.

(b) –SO3H group incorporated at ortho position is also substituted by –NO2 group on treatment with HNO3.

(c) –SO3H group at ortho position further activates the ring towards electrophilic substitution.

(d) all the above statements are false.

4. What is the correct order of rate of electrophilic aromatic substitution among following compounds?

(1)

CH3

(2)

CH3

(3)

CH3

CH3

CH3 (4)

(a) (2) > (3) > (1) > (4) (b) (1) > (2) > (3) > (4)

(c) (3) > (2) > (1) > (4) (d) (2) > (1) > (3) > (4)

EXERCISE – V

SUBJECTIVE PROBLEMS

1. Using resonance contributors, answer the following. Which is a more stable carbocation intermediate?

OH

NO2 H

+ or

CH3

NO2 H

+

2. Give the products of the following reactions.

(a)

OCCH3

O

+ HNO3 H2SO4 ?

(b)

CH3

1. Mg/Et2O

Br 2. D2O ?

(c)

OCH3

+

O

O

O

AlCl3 ?

(d)

+ Br2

N

? Fe

(e)

CH3 1. NBS/

3. ethylene oxide

2. Mg/Et2O

4. H+

? (f)

CF3

FeCl3 + Cl2 ?

3. Give the product(s) obtained from the reaction of each of the following compounds with Br2/FeCl3.

(a)

OC

O

(b)

CH2O

(c)

C

O

H3C COCH3

O

(d)

CH3O

NO2

4. What products would be obtained from the reaction of the following compounds with hot and acidified

KMnO4?

(a)

CH2CH3

CH3

(b)

CH3

C(CH3)3

(c)

CH2CH2CH2CH3

CHCH3

CH3

5. Propose a mechanism for each of the following reactions.

(a)

CH2CH2CHCH=CH2

H+

H3C CH2CH3

CH3

(b)

CH=CH2

H+

CH3

6. Show how the following compounds could be prepared from benzene.

(a)

CH3

CH3C=CH2

(b)

SO3H

CH2CH(CH3)2

CH3C

O

(c)

OCH3

NO2

Br

7. Determine the major product for each of the following reactions.

(a)

OH

CO2H Br2 in

H2O ?

(b) Anhydrous

AlCl3 + O ?

(c)

Cl2/Fe

N

H

O

CH3

?

(d)

OEt

Conc. HNO3 CH2O

HCl ZnCl2

? ?

8. Compare the rate of nitration under similar conditions of PhOMe and PhSMe. Explain.

9. PhNH2 reacts with Br2 in H2O to give more than 90% yield of 2,4,6tribromoaniline, while PhNMe2 is

mononitrated at the meta position with the more powerfully electrophilic reagent HNO3/H2SO4. Explain. 10. Explain the following percentages of meta electrophilic substitutions:

(a) ArCH3, ArCH2Cl, ArCHCl2, ArCCl3

4.4% 15.5% 33.8% 64.6%

(b) ArN+Me3, ArCH2N+Me3, ArCH2CH2N+Me3

100% 88% 19%

ANSWERS

EXERCISE –I

AIEEE-SINGLE CHOICE CORRECT

1. (d) 2. (a) 3. (b) 4. (c) 5. (d)

6. (a) 7. (c) 8. (c) 9. (c) 10. (c)

11. (a) 12. (d) 13. (b) 14. (a) 15. (b)

16. (c) 17. (a) 18. (b) 19. (c) 20. (c)

21. (c) 22. (d) 23. (a) 24. (d) 25. (c)

EXERCISE – II

IIT-JEE SINGLE CHOICE CORRECT

1. (d) 2. (c) 3. (c) 4. (d) 5. (c)

6. (a) 7. (d) 8. (d) 9. (b) 10. (c)

11. (b) 12. (b) 13. (d) 14. (c) 15. (a)

16. (b) 17. (b) 18. (a) 19. (b) 20. (a)

21. (a) 22. (a) 23. (d) 24. (c) 25. (c)

EXERCISE – III

MORE THAN ONE CHOICE CORRECT

1. (a, d) 2. (b, c) 3. (b, d) 4. (a, c) 5. (a, d)

6. (b, d) 7. (b, c, d) 8. (a, b) 9. (a, c, d) 10. (b,c,d)

EXERCISE – IV

MATCH THE FOLLOWING

1. I (E) ; II (A) ; III (B) ; IV (C)

2. I (C) ; II (E) ; III (A) ; IV (B)

3. I (B) ; II (C) ; III (D) ; IV (E)

ASSERTION AND REASON

1. (d) 2. (a) 3. (a) 4. (d) 5. (a)

PASSAGE BASED PROBLEMS

1. (c) 2. (d) 3. (b) 4. (c)

EXERCISE – V

SUBJECTIVE PROBLEMS

1.

OH

NO2 H

is the most stable Intermediate.

OH

NO2 H

OH

NO2 H

OH

NO2 H

OH

NO2 H

etc

Here OH stabilises adjacent carbocation by +R effect but in the other one methyl stabilizes the adjacent

carbocation only by +I effect.

2. (a)

OCOCH3

NO2

+

OCOCH3

NO2

(Major) (Minor)

(b)

CH3

Br

(1) Mg/Et2O

(2) D2O

CH3

D

(c)

(Major)

(Minor)

OCH3

C=O

CH2CH2COOH

+

OCH3

CO(CH2)2COOH (d)

(Major)

(Minor)

+

Br

Br N N

(e)

CH2CH2CH2OH

(f)

CF3

Cl

3. (a)

Br O C

O

(b)

CH2 O Br

(c)

CH3 C

O

Br

COCH3

O

(d)

OCH3

Br

NO2

4. (a)

COOH

COOH

(b)

COOH

C(CH3)3

(c)

COOH

COOH

5. (a)

(CH2)2CHCH=CH2

H

(CH2)2CHCHCH3

CH3

1, 2 hydride shift

(CH2)2CCH2CH3

CH3

H

CH3

H

(b)

CH=CH2

H

CHCH3

CH=CH2 CHCH3

CH3

CH3

H

CH3

H

6. (a)

CH3Cl,

AlCl3,

Heat

CH3

CH3CH=CH2

H

CH3

CH

H3C

CH3

CCl (4)

alc KOH,

Heat

1 eqv. Cl2 + h

H3C

H3C

H3C

(b)

(CH3)2CHCH2Cl + AlCl3

Cold

(No rearrangement)

CH2CHMe2

CH3COCl,

CH2CH(CH3)2

COCH3

Conc. H2SO4

CH2CH(CH3)2

COCH3

SO3H AlCl3 +

(c)

NO2

Conc. HNO3

+ conc.H2SO4

NaOCH3,

NO2

OCH3

Br2/Fe

NO2

Br

Br2/Fe

NO2

Br

Br

Br

7. (a)

OH

Br

Br

Br (b)

OH

(c)

Cl

N

H

O

CH3

(d)

Cl

OEt

NO2

OEt

NO2

8. PhOMe > PhSMe. The bond from O (which uses a 2porbital) to the ring C is shorter and stronger

than the comparable bond from S (which uses a 3porbital) to the ring C.

9. The tribromoderivative forms because NH2 is a strongly activating o,porienting substituent (as is

NMe2). Recall that amines are also proton acceptors, so in strong acid they are completely protonated.

Consequently, the only species present in HNO3/H2SO4, is the conjugate acid PhNMe2H+ and it is the

species undergoing nitration. The NR2H+ group, with a full + charge, is a deactivator that orients meta. 10. (a) Successive replacement of H’s in electronreleasing CH3 by electronegative Cl’s makes G

increasingly electronattracting and morienting.

(b) The + charge on N in +NMe3 makes the substituent electronattracting and mdirecting by an

inductive effect. Its influence wanes with increasing CH2’s between +N and Ar. With two CH2’s, the

electronreleasing effect of the alkyl group is more significant and o,porientation dominates.

MIND MAP

ELECTROPHILIC AROMATIC

SUBSTITUTION

O, NH2, NHR, NR2, OH,

OR, NHCOR, R, Ar, COO F,Cl, Br, I

+NR3, NO2, CN, SO3H,

CHO, COR, CO2H, CO2R,

CONH2,

1. PhH + RX PhR.

2. PhH + RCOX PhCOR

Carbocation can also be formed from

ROH, Alkene or CO in presence of acid to give electrophilic substituted product.

Alkylation may involve rearrangement. With phenol excess AlCl3 and with aniline excess AlCl3 is required for alkylation. In case of aniline unexpected meta alkylated product is

obtained.

PhH + conc. H2SO4 PhSO3H

Deprotonation is rate determining step. PhH + conc. HNO3 + H2SO4

PhNO2

Electrophilic attack to benzene ring is rate determing step. Other nitrating

species are N2O5, NO2+ BF4

and

NO2+ ClO4

.

PhH + X2 PhX.

Activating Deactivating

o, p directing groups mdirecting groups

Directing groups

Friedel Craft’s

reaction

Nitr

atio

n

Sulphonation

Halogenation

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