1

TABLE OF CONTENTS

OVERVIEW 2

ELCB (BIMOLECULAR REACTION) 21

COMPARISON BETWEEN EL, E2 AND ELCB REACTIONS 23

ELIMINATION VERSUS SUBSTITUTION 26

SN2 / E2 CONDITIONS 27

COMPARISON BETWEEN E2 AND SN2 29

SOME COMMON SOLVENTS FOR SUBSTITUTION REACTION 33

HALOARENES 37

ELECTROPHILIC SUBSTITUTION MECHANISM (ARENIUM ION MECHANISM) 38

AROMATIC COMPOUNDS NEVER UNDERGO ADDITION REACTION 40

THE EFFECT OF SUBSTITUENTS ON THE REACTIVITY 50

HALOGENATION 51

GATTERMANN KOCH REACTION 58

FRIEDAL CRAFT ALKYLATION 60

FRIEDEL-CRAFTS ACYLATION 62

2

Overview Few organic reactions give a 100% yield of a single product. The reason for this is that it

is uncommon to find a set of reaction conditions which will permit one reaction to occur

to the total exclusion of another. Nucleophilic substitution reactions are no exception.

They compete for starting material with elimination reactions. Indeed proper selection

of reagents and the reaction conditions may actually result in more starting material

undergoing elimination than substitution.

The most common type of elimination reaction is one in which two fragments are

removed from a substrate to produce a modified substrate and two small units. One of

these fragments is usually the leaving group of the substrate. Eliminations usually

produce a new pi-bond in the modified substrate (1).

(1)

In some elimination reactions (but very uncommon) a new sigma-bond is produced

instead of a pi-bond.

(2)

Thus, elimination reaction is given by those compounds which have nucleophilic group

as a leaving group. The leaving groups which are responsible for elimination reaction are

23232 RSandRN,OH,N,N,OR,OH,X

Classification: One means of classifying elimination reactions uses the Greek alphabet to

identify the skeletal atoms which bear the leaving group of the substrate. The skeletal

atom bearing the typical leaving group is lettered C. The one bearing the other group

or atom which leaves (in most of the cases it is hydrogen atom) is lettered sequentially

3

from C. Since C is always involved, elimination reactions are grouped according to the

higher letter designation. Using this approach, the reaction would be called -, - and -

elimination.

If both departing fragments are bonded to the same skeletal atom, the process is called

as

-elimination or 1, 1-elimination. When the departing groups are on adjacent atoms,

the product is always an alkene. This kind of elimination is called elimination. It is also

called a 1, 2 -elimination because the atoms being removed are on adjacent atoms, i.e.,

atom 1 and 2.

-elimination reactions are the most common elimination reactions. The -elimination

includes solvolysis and base induced elimination reactions of sulphonates, alkyl halides

and quaternary ammonium hydroxide and acid catalyzed dehydration of alcohols.

-elimination reactions can occur by a variety of mechanisms and three mechanistic

pathways are normally distinct routes. These are E1, E2 and Elcb mechanism.

E1 (UNIMOLECULAR ELIMINATION) REACTION (Elimination1)

Mechanism:

C C

L

C C+

= Leaving Group

Slow Step

Rate Determining Step

H H

Base-

CC

Order of reactivity of compounds tertiary > secondary > primary

Rate = k[Substrate or Reactant]

Unimolecular and follows first order kinetics.

4

The reaction of tert-bütyl bromide with sodium hydroxide is a first order elimination

reaction because the rate of the reaction depends only on the concentration of the alkyl

halide. It is called an El reaction E for elimination and I for unimolecular.

Since reaction is unimolecular reaction, the reaction must be at least two step reactions.

An El reaction is a two step reaction. In the first step the alkyl halide dissociates

heterolytically. This step is the rate-determining step of the reaction. In the second step

of the reaction, the base forms an elimination product by removing a proton from a

carbon adjacent to the positively charged carbon. Because the first step of the reaction

is the rate-determining step, increasing the concentration of the base which comes into

play only in the second step of the reaction has no effect on the rate of the reaction.

Thus, the following mechanism agrees with the observed first order kinetics.

Mechanism of the El reaction

5

Regioselectivity: When two elimination products can be formed, the major product is

always the one which is obtained according to the Saytzeff rule. (The hydrogen is

removed from the -carbon bonded to the lowest hydrogen.)

The El reaction of tert-butyl bromide with NaOH

(11)

Relative reactivity's of alkyl halides: The rate of El reaction depends on both the ease

with which the leaving group leaves and the stability of the carbocation that is formed

because the first step is the rate determining step. Thus, the relative reactivity's of a

series of alkyl halides with the same leaving group parallel the relative stabilities of

carbocations.

Relative reactivity's of alkyl halides in El reaction

3° benzyl halide 30 allyl halide > 20 benzyl halide 20 allyl halide = 30 alkyl halide

> 10 benzyl halide = 1° allyl halide = 2° alkyl halide > 10 alkyl halide > vinyl halide.

6

Decreasing reactivity

Rearrangement: Because the El reaction involves the formation of a carbocation

intermediate, rearrangement of the carbon skeleton can occur before the proton is lost.

For example, 3-bromo-2-methyl-2-phenylbutane gives mixture of two alkenes. One

alkene is normal and the other is rearranged product.

In the following example, secondary carbocation undergoes a 1, 2 hydride ion shift to

form more! stable secondary allylic carbocation.

In an El reaction the reaction intermediate is a carbocation which has planar geometry.

In planar geometry the electrons from departing hydrogen can move toward the

positively charged carbon from either side. Therefore, both syn and anti-elimination can

occur.

7

If several -hydrogen's (at least two) are available, the one producing the most stable

alkene usually is that which is lost; i.e., Saytzeff rule is generally followed in El reactions.

Thus, in this case products of El reaction will be similar to those of an E2 reaction.

(12)

The El reaction of chloro 2, 4, 4-tnmethylpefltafle is interesting because it produces an

excess of the less substituted alkene. However, this still goes through the more stable

activated complex (TS). In this case the steric repulsions are greatest in the transition

state leading to the most highly substituted product. This is an example of a steric effect

which leads to Hofmann elimination in E1 reaction.

8

Effect of the structure of the substrate, strength of base, nature of leaving N group

and solvent on the rate of the El reaction

Structure of Substrate: El reaction takes place by the formation of carbocation as

reaction intermediate. Thus, the rate of the El reaction depends on the stability of the

carbocation. Since +I effect, hyperconjugation and conjugative effect stabilize

carbocation, any structure which forms stable carbocation will be reactive. Steric strain

around leaving group also favours the formation of carbocation. Thus, the order of the

reactivity in decreasing order is as follows

30 > 20 > 10 alkyl halide

Thus, alkyl and aryl substitutions on and -carbons with respect to the leaving group

increase the rate of El reactions. As the strain increases the yield of the El product

increases.

Strength and concentration of the base: Since El reactions do not usually require any

base (the solvent molecules serve the purpose which themselves behave as base), the

strength and concentration of the base have nothing to do with the rate of El reactions.

Nature of the leaving group: Reactivity of the substrate depends mainly on the nature

of the leaving group. The best leaving groups are those which are least basic and more

polarizable. Thus, the decreasing order of the leaving group reactivity is

FlCrBI

Nature of Solvents

Since, El reactions involve ionic intermediate, the carbocations, and the rate of the El

reaction increases with increasing polarity of the solvent.

9

E2 (ELIMINATION BIMOLECULAR) REACTION

Mechanism:

C C

L = Leaving Group

Rate Determining Step

H

Base-

CC

Order of reactivity of compounds primary > secondary > tertiary

Rate = k[Substrate][Base]

Bimolecular and follows second order kinetics

The reaction of ethyl bromide with ethoxide ion is an example of an E2 reaction. It is a

second order reaction because the rate of the reaction depends on the concentration of

both alkyl halide and base. It is called E2 reaction: E for elimination and 2 for

bimolecular.

rBOHHCCHCHOHCBrCHCH52225223

(5)

Rate = K [alkyl halide] [base]

The rate law tells us that ethyl bromide and ethoxide ion is both involved in the

transition state of the rate determining step of the reaction. The following mechanism

agrees with the observed second-order kinetics.

10

Mechanism of E2 reaction

The E2 reaction is a concerted, one step reaction, the proton and the bromide ion are

removed in the same step, i.e., in the rate determining step.

An energy profile for the reaction is given in Figure given below

Progress of the reaction

E2 reaction. The reaction of ethyl halide with ethoxide ion

Experimental evidence that is helpful in determining the mechanism of a reaction is a

kinetic isotope effect. The kinetic isotope effect can be used to demonstrate that a

carbon-hydrogen bond is broken in the rate determining step.

11

When the rate constant (KH) for elimination of

H and

Br from l-bromo-2-phenylethane

at 40°C is compared with the rate constant (KD) for elimination of

D and

Br from 1-

bromo-2, 2-dideuterio-2-phenyl ethane determined at the same temperature, KH is

found to be 7 times KD. So, the deuterium kinetic isotope effect is 7.1 which is primary

kinetic isotope effect. The fact that the hydrogen-containing compound undergoes

elimination faster than the deuterium-containing compound tells us that the carbon-

hydrogen (or C–D) bond is broken in the rate determining step. The fact that the C–H

bond is broken in the rate determining step agrees with the mechanism proposed for an

E2 reaction.

OHHCBrCHCHCHHCOHCBrCHCHHC52256

Ka

522256

(6)

BrOHHCCHCDHCOHCBrCHCDHC52256

K

522256D (7)

1.7K

K

D

H

12

Direction of Elimination

When only one type of n-hydrogen is present, the direction of elimination is certain.

For- example all the hydrogen's adjacent to the carbon bearing the bromine in isopropyl

bromide are equivalent. Therefore, the loss of HBr from isopropyl bromide can produce

only one compound, propene.

(8)

When all -hydrogen's cannot be equal, the alkene normally produced is the most highly

substituted one, i.e., the one with the largest number of alkyl groups bonded to the

double bond. This statement is known as Saytzeff rule. Saytzeff pointed out that the

most sub -

carbon that is bonded to the fewest hydrogen's. Or in other word during the E2

elimination of alkyl halide and alcohol, the negative part (i.e., Br or OH) eliminates with

the “H” of the adjacent “C” which has minimum number of hydrogen. Or in terms of

hydrogen Saytzeff rule is “POORER BECAME POORER’ Most dehydrohalogenations

follow this rule. Thus, an E2 reaction is regioselective, more of one constitutional isomer

is formed than the other.

(9)

In the above example, two alkenes are formed. The difference in the rate of formation

of the two alkenes is not very great. Consequently, both products are formed but the

more stable alkene is the major product of the elimination reaction.

Alkyl halides have the following relative reactivity's in an E2 reaction, because

elimination from a tertiary alkyl halide typically leads to a more highly substituted

alkene than elimination from a secondary alkyl halide, and elimination from a secondary

alkyl halide leads to a more highly substituted alkene than elimination from a primary

alkyl halide.

13

Relative reactivity's of alkyl halides in an E2 reaction

Tertiary alkyl halides > see alkyl halides > primary alkyl halides

In most of the dehydrohalogenation reactions, the major products are the Saytzeff

product. But the most substituted alkene is not always the major product. In some cases

the major alkene product is the least substituted alkene. Elimination, in which least

substituted alkene is the major product, is known as Hofmann elimination and the rule

is known as Hofmann rule. Thus, if a reaction produces more of the less highly

substituted alkene, it is following Hofmann rule.

Hofmann elimination reaction takes place in the following four cases:

(i) When the base is bulky.

(ii) When the leaving group is a poor leaving group such as F, 3

RN

and2

RS

.

(iii) Steric hindrance at y-carbon

(iv) -carbon

(i)

(ii)

(iii)

14

(iv)

Formation of Hofmann Product

On the basis of these results one can very easily conclude that the transition state of the

E2 reactions depends on the structure of the substrate and the strength of the base

used for the reaction. One may thus, define the transition state for E2 reaction as a

hybrid of structures (II-IV). These structures gain individual importance depending on

several factors.

Resonating structures of transition states

(i) Structure (II) of the transition state becomes significant when L is good leaving

group and base is strong. This TS gives Saytzeff as well as Hofmann products.

Nature of product depends on the stability of transition state.

(ii) Structure (III) of the transition state becomes significant when L is poor leaving

group or base is bulky. This transition state always gives Hofmann product.

(iii) Structure (IV) of the transition state becomes significant when L is a good leaving

group and base is weak. These TS has some carbocation character. This transition

state always gives Saytzeff products.

15

Stereochemistry of the E2 Reaction

An E2 reaction involves the removal of two substituent's (say H and X). The bonds to the

eliminating groups (H and X) must be in the same plane because the sp3-hybrid orbital of

the carbon bonded to hydrogen and the sp3 hybrid orbital of the carbon bonded to X

become overlapping p-orbital in the alkene product. Therefore, the orbitals must

overlap in the transition state. This overlap can occur only if the orbitals are parallel, and

to be parallel the bonds to the eliminating groups must be in the same plane.

There are two ways in which the C–H and C–X bonds can be in the same plane. They can

be parallel to one another either on the same side of the molecule (syn-periplanar) or

on the opposite sides of the molecule (anti-periplanar).

Eclipsed Newman and sawhorse conformations in which H and X are syn-periplanar

Staggered Newman and sawhorse conformations in which H and X are anti-periplanar

to each other

16

If an elimination reaction removes two substituents from the same side of the molecule,

the reaction is a syn-elimination. If the substituents are removed from opposite sides of

the molecule, the reaction is anti-elimination.

Anti elimination Syn elimination

Optical active alkyl halides, in which

well as stereo specific dehalogenation reaction. Geometry of alkene depends on the

configuration of the substrate as follows:

(i) Erythro form (or meso form) gives E or-trans alkene.

(ii) Threo form (or d or 1 form) gives Z or cis alkene.

Consider the case of 2-bromo-3-phenylbutane:

17

In this form both H and Br is not anti to each other. To get H and Br anti, rotate

Br → H → CH3 clock-wise at carbon- -elimination.

Stereochemistry of elimination can be known by the following two words:

(i) CAR where:

C means cis (or Z) alkene

A means anti elimination

R means racemic mixture (or threo form)

If racemic mixture gives cis alkene then meso form (or erythro form) will give trans (or E)

alkene

(ii) CSM where:

C means cis alkene

S means syn elimination

M means meso form (or erythro form)

If meso form gives cis alkene then racemic mixture (or threo form) will give trans alkene.

18

Effects of the structure of the substrate, strength and concentration of the base and

effect of leaving group on the rate of the E2 reaction

Structure of Substrate

It has been found that branching at and β-carbons increase the rate of the E2

reaction. This is because as the number of substituents increases on the carbon atoms

of the developing double bond, the stability of the transition state increases (Table 9.1).

Substrate structure and rate of E2 reactions

Substrate % Yield Rate of 250C

BrCHCH23 0.9 5100.1

80.3 2.3 105

97 4.7 105

8.9 5.3 105

59.5 8.5 105

Shows that branching at or carbon increases the yield as well as rate of the E2

reaction this table also shows that 3°-halides are more reactive than secondary alkyl

halides which are more than the primary alkyl halides.

30 > 20 > 10

19

Since the transition state of an SN2 react

branches slow down the SN2 reaction rate while they speed up the E2 reaction rate.

Thus, with increasing branching at and -carbons, the E2/SN2 ratio increases.

It has also been found that the electron-withdrawing groups on the -carbon increase

the rate of the E2 reaction. This is because the –I group on -carbon increases the

acidity of the -hydrogens and / stabilizes the carbanion character of the transition

state.

The nature of the leaving group

In gener -

phenyl halides treated with sodium ethoxide in ethyl alcohol, the rate of the E2 reaction

increases with the leaving power of the halogen atom

Leaving groups and rate of chemical reaction

Substrate Rate at 250C

ClCHCHHC2256 0.007 103

BrCHCHHC2256 4.2 103

ICHCHHC2256 27 103

It has also been found that with the increasing size of the halogen atom E2/SN2 ratio

increases but to a minimum extent.

Strength of the base: With the increasing basicity of the added base, the rates of the £2

reactions have been found to increase. The order of basicity among

252HNandOHC,HO

is:

HOOHCHN522

20

The rate of the E2 reactions on a given substrate with the above bases under identical

conditions is also found to be

HOHCOHN522

Thus, the order of the basicity is also the order of the rate.

When the base is weak but strongly nucleophilic toward carbon, E2/SN2 ratio is low but

in the presence of a strong base the E2/SN2 ratio increases.

E2 Elimination of vinyl halides

Vinyl halides in which H and X are anti to each other gives E2 reaction with strong base

(LDA).

The effect of the temperature: As the temperature of the reaction increases, the rate of

the reaction increases.

Thus, the E2/SN2 ratio will also increase with the increasing temperature. For example,

E2/SN2 ratio of the reaction of 2-bromopropane with sodium hydroxide in 62% ethyl

alcohol is found to rise from 1.15 to 2.45 as the temperature is raised from 45°C to

100°C.

The nature of the solvent: The yield of the product in E2 reaction increases with the

decrease in solvent polarity because this favours the formation of the transition state of

the reaction. Thus, the E2/SN2 ratio in the reaction of 2-bromopropane at 50°C with

sodium hydroxide has been found to increase as the solvent polarity decreases.

21

Solvent E2/SN2 ratio

60% C2H5OH and 40% HOH 1.17

80% C2H5OH and 20% HOH 1.44

100% C2H5OH 2.45

El CB (BIMOLECULAR REACTION)

Mechanism:

C C

L = Leaving Group

Rate Determining Step

H

Base-

C C

L = Leaving Group

-

CB = Conjugate base

CCSlow Step

It one begins with the rapid loss of a proton to base. Loss of proton leads to a

stabilized carbanion. The carbanion is then converted into an alkene. Conversion

of carbanion to an alkene is the rate determining step. This elimination, since it

proceeds through the conjugate base of the starting material, is known as Elcb.

The Elcb reaction competes with the E2 reactions. However, Elcb reactions are

much less common than are E2 reactions because of the greater instability of

carbanion. Indeed, only a very small percentage of eliminations follow this path

way.

Rate = k [Conjugate Base] the Elcb reaction is first order in base and first order in

substrate just like the E2 reaction. Thus, the reaction is second order reaction,

first order with respect to substrate and first order with the base. But the

reaction is unimolecular reaction because reaction velocity depends only on the

concentration of the conjugate base of the substrate.

Bimolecular and follows second order kinetics

Important notes:

22

(i) Elcb mechanism is limited to substrates which can stabilize the carbanion

intermediate, i.e., -carbon should contain carbonyl, nitro, cyano, sulphonyl or

other carbanion stabilizing group.

(ii) Product formation generally takes place by Hofmann rule.

(iii) Leaving group of the substrate is generally poor leaving group.

(iv) -Hydrogen should be highly acidic so that the carbanion formation may take

place easily, an

(v) Reaction takes place in the presence of a strong base.

Note: -Halocarbonyl compounds always give Elcb reaction even in the presence of

water and alcohol as base.

Elimination reaction is shown by alcohols and haloalkanes.

Examples of elimination reactions

1. Dehydration of alcohols

In alcohols OH group is not a good leaving group so first of all it is made a good leaving

group by protonation which is done in the presence of acid for instance sulphuric acid,

phosphoric acid etc. after protonation OH group forms OH2+ group which is a good

leaving group. Then OH2+ group eliminates depending upon E1 or E2 mechanism.

Order of reactivity of alcohols for E1 3o > 2o > 1o on the basis of the stability of

carbocation

Order of reactivity of alcohols for E1 3o > 2o > 1o on the basis of strength and

concentration of base

Basicity is the affinity for the proton and nucleophilicity is the tendency to form bonds

with the carbon.

For instance tert-butoxide ion is a strong base then its nucleophilicity because of steric

hindrance.

23

Steric hindrance decreases nucleophilicity and hence nucleophilicity and basicity has

inverse relationship.

Then carbocation will be formed in the E1 mechanism and then think about the

rearrangement to stabilize the carbocation. Then Base comes and picks up the acidic

hydrogen and forms the alkene.

2. Dehydrohalogenation of haloalkanes

Comparison between El, E2 And Elcb Reactions

Comparison between El, E2 and Elcb reactions

El E2 Elcb

1. Two step reaction One step reaction Two step reaction

2. Carbocation as reaction

intermediate

No intermediate forms

Product formation takes

place by formation of

Transition State

Carbanion as reaction

intermediate

3. In the second step the

hydrogen leaves as

proton from the

carbocation

– The leaving group leaves

the carbanion either as an

anion or as a neutral group

4. Weak base is needed in

second step to remove

hydrogen as proton

Strong base is needed to

remove hydrogen as

proton

Strong base is needed to

remove hydrogen as

proton in the first step

5. Follows the first order

kinetics Rate = K

[Substrate]

Follows the second order

kinetics. Race = K

[Substrate] [base]

Follows the second order

kinetics. Rate = K

[Substrate] [base]

6. Main product of the

reaction is Saytzeff

In some cases product is

Saytzeff and in other cases

The Hofmann product

24

product the product is Hofmann

7. Non-stereo specific and

non-stereo-selective

Stereo specific and stereo

selective

Non-stereo specific and

non-stereo selective

8. Reactivity order 30 > 20 >

10

Reactivity order 30 > 20 > 10 Reactivity order 30 > 20 > 10

9. Any structural feature

that stabilizes the

carbocation will

increase the rate. Thus,

branching at and -

carbons increases the

rate.

Branching at and -

carbons and the presence

of electron-withdrawing

group on the a carbon

increases the rate

Any structural feature that

stabilizes the intermediate

carbanion will increase the

rate. The presence of a -

aryl group shifts a -

elimination toward Elcb

pathway.

10. Polar protic solvents

increase rate

In a polar solvent the rate

decreases

Polar aprotic solvent

increases the rate

11. The concentration and

the basicity of the

solvent have no effect

on the rate

With increasing

concentration and basicity

of the added base, the rate

increases With increasing

concentration and basicity

of the added base, the rate

increases

12. As the temperature

increases, the rate

increases

As the temperature

increases, the rate

increases

As the temperature

increases, the rate

increases

13. El reactions compete

with SN1 reactions

E2 reactions compete with

SN2 reactions

–

14. Common reaction The most common

reaction

Very rare

25

15. Favours in the presence

of weak base

Favours in the presence of

strong base

------------------

The use of deuterium labeling can help to distinguish the Elcb reaction from the E2

pathway. If an E2 reaction is carried out in a solvent which could act as a deuterium

source, and the reaction is interrupted before it has gone to completion, recovered

starting material is free from any deuterium. This happens because there is no way by

which deuterium could become incorporated into the starting material.

E2 pathway:

Any reaction which involves a carbanion as an intermediate, on the other hand, should

produce recovered starting material labelled with deuterium. It is precisely such

deuterium incorporation which occurs in those reactions presumed to be Elcb.

Elcb pathway:

This incorporation also confirms that the first step is reversible step.

The reaction of 1, 1-dichioro-1-deuterio-2, 2, 2-trifluoroethane illustrates a -

elimination which follows the Elcb mechanism. It also shows how, by starting with a

deuterated substrate, the carbanion can be trapped using

O H/HOH rather than O

D

/D2O.

26

Elimination Versus Substitution You have seen that alkyl halides can undergo four types of reaction: SN2, SN1, E2, and E1.

The first thing you must decide is whether the reaction conditions favour SN2/E2 or

SN1/El reactions. Fortunately, the conditions that favour an SN2 reaction also favour an

E2 reaction, and the conditions that favour an SN1 reaction also favour an El reaction.

This means that you do not need to worry about SN2/El or SN1/E2 combinations; you just

decide whether the combination is going to be SN2 / E2 or SN1 / E1.

Your decision is easy if the reactant is a primary alkyl halide because, without the ability

to form carbocations, primary alkyl halides undergo only SN2/E2 reactions.

If the reactant is a secondary or tertiary alkyl halide, whether it undergoes SN2 / E2 or

SN1 / El reaction depends on the reaction conditions. SN2/E2 reactions are favoured by

a high concentration of a good nucleophile/strong base, whereas SN1 / El reactions are

favoured by a poor nucleophile/weak base. In addition, the solvent in-which the

reaction is carried out, can affect the choice of mechanism.

Once you have decided whether the conditions will lead to SN2/E2 reactions or to SN1 /

El reactions, you must decide how much of the product will be the substitution product

and how much will be the elimination product. The relative amounts of substitution and

elimination products depend on whether the alkyl halide is primary, secondary or

tertiary and of the nucleophile/base.

27

SN2 / E2 Conditions Let’s first consider the situation in which you have decided that the conditions lead to

SN2 / E2 reactions (a high concentration of a good nucleophile/strong base). The

negatively charged species can act as a nucleophile and hit the back side ofcarbon to

form the substitution product, or it can act as a base and remove a -hydrogen, leading

to the elimination product.

The relative reactivities of alkyl halides in SN2 and E2 reactions are given below.

in an SN2 reaction 10 > 20 > 30 in an SN1 reaction 30 > 20 > 10

in an E2 reaction 30 > 20 > 10 in an E1 reaction 30 > 20 > 10n

Because a primary alkyl halide is the most reactive of the alkyl halides in an SN2 reaction

and the least reactive in en E2 reaction, you can predict that a primary alkyl halide will

primarily form the substitution product in a reaction carried out under conditions that

favour SN2 / E2 reactions.

However, if either the primary alkyl halide or the nucleophile/base is sterically hindered,

the nucleophile will have difficulty getting to the back side of the a-carbon and, as a

result, the elimination product will predominate.

28

The primary alkyl halide is sterically hindered

The nucleophile is stericaily hindered

A secondary alkyl halide can form both substitution and elimination products under SN2

/ E2 conditions. The relative amounts of the two products depend on the base strength

(and the bulk) of the nucleophile. The stronger and bulkier the base, the greater the per

cent of the elimination product for example, acetic acid is a stronger acid than ethanol,

which means that acetate ion is a weaker base than ethoxide ion. The elimination

product is the main product formed from the reaction of 2-ch1oropropane~ with the

strongly basic ethoxide ion, whereas no elimination product is formed with the weakly

basic acetate ion.

29

Increasing the temperature at which the reaction is carried out increases the rates of

both the substitution and elimination reactions, but the rate of the elimination reaction

increases more. Therefore, if the substitution product is the desired product, the

reaction should be carried out at a low temperature High temperatures promote

formation of the elimination product

A tertiary alkyl halide is the least reactive of the alkyl halides in an SN2 reaction and the

most reactive in an E2 reaction. Consequently, only the elimination product is formed

when a tertiary alkyl halide reacts with a nucleophile under SN / E2 conditions.

Comparison between E2 and SN2 reactions: The comparison between E2 and SN2

reactions is as follows:

Comparison between E2 and SN2 SN2 E2

1. Step One One

2. Reagent Strong nucleophile Strong, bulky, Bronsted

base

3. Solvent Aprotic polar solvent

favours SN2 reaction

Aprotic polar solvents

strongly favour E2 reaction

4. Phase transfer catalyst Favour Strongly favour

5. Structure and reactivity

nature of the alkyl group

10 > 20 > 30 30 > 20 > 10

30

6. Kinetics Rate = K [substrate] [

uN

]

Rate = K [substrate] [Base]

7. Stereochemistry Stereospecific: inversion

of configuration

Stereospecific

(i) usually anti-elimination

(ii) in pyrolytic elimination

it is syn.

SN 1 / El Conditions

Let’s now look at what happens when conditions lead to SN1/El reaction (a poor

nucleophile/weak base is used). In SN1 / El reactions, the alkyl halide dissociates to form

a carbocation. The carbocation can either combine with the nucleophile to form the

substitution product or lose a proton to form the elimination product.

Alkyl halides have the same order of reactivity in SN1 and El reactions because they have

the same rate-determining step–ionization of the alkyl halide. This means that all alkyl

halides that react under SN1 / El conditions will give both substitution and elimination

products. Substitution is favoured over elimination at lower temperatures; increasing

the temperature increases the percentage of elimination product. Remember that

primary alkyl halides do not undergo SN1 / El reaction because primary carbocations are

too unstable to be formed.

The Table summarizes the general pattern of reactivity expected from various structural

classes of alkyl halides in reactions with a representative range of nucleophiles (which

may behave as bases).

31

Comparison between SN1 and El: The comparison between SN1 and El reactions is

given in.

Comparison between SN1 and El

SN1 El

1. Step two steps two steps

2. Reagent favoured by solvents and weakly

basic reagents of low

concentration

favoured by weak

base

3. Solvent favoured by polar protic solvent favoured by polar

protic solvent

4. Structure and reactivity

nature of the alkyl groups

30 > 20 > 10 30 > 20 > 10

5. Kinetics first order first order

6. Stereochemistry racemisation and partial

inversion

non-stereospecific

Over all summary of SNl, SN2, El, E2 and E1–CB reaction

Alkyl Halide Poor

Nucleophile

(e.g. H2O,

ROH)

Weak Basic

Nucleophilc

(e.g. I, RS)

Strongly Basic

Unhindered

Nucleophile

(e.g. RO)

Strongly Basic

Hindered

Nucleophile

(e.g. t-BuO-)

Methyl CH3X no reaction SN2 SN2 SN2

Primary(unhindered)

RCH2X

no reaction SN2 SN2 E2

32

Alkyl Halide Poor

Nucleophile

(e.g. H2O,

ROH)

Weak Basic

Nucleophilc

(e.g. I, RS)

Strongly Basic

Unhindered

Nucleophile

(e.g. RO)

Strongly Basic

Hindered

Nucleophile

(e.g. t-BuO-)

Primary

no reaction SN2 E2 E2

Secondary

SN1, E1 (slow) SN2 E2 E2

Tertiary

E1 or SN2 SN1, E1 E2 E2

E1-CB E1-CB E1-CB E1-CB

Effect of Solvent

SN1 reaction is faster in the more polar solvents. .

SN2 reaction involving a negative nucleophile is slower in more polar solvent and

that involving a neutral nucleophile is faster in more polar solvent. .

In addition to these polarity effects, the ability of certain solvents to form

hydrogen bonds to the nucleophile also affects the rate of the 2 reaction. Such

solvents are termed protic solvents and have a hydrogen bonded to nitrogen or

oxygen (H2O, ROH, RCOOH are some example of protic solvents). Nucleophiles,

especially smaller anionic ones, are strongly hydrogen bonded to the solvent

molecules in protic solvents. This makes these nucleophiles less reactive because

33

the solvent molecules block the approach of the electrophile. A protic solvents

(like acetone), which can’t have hydrogen bond to the nucleophile (because they

do not have hydrogen bonded to nitrogen or oxygen), increase the reactivity of

the nucleophile and are especially favourable for 2 reactions.

Table given below some common or organic solvents in order of decreasing “ionizing

power” (or ability to stabilize ions)

Table given below shows some distinctive features of SN1 and SN2 reactions and Table

10 represents preferred substitution mechanism for various carbon structures.

Some common solvents for substitution reaction trifluoroacetic acid

There are polar, protic

solvents. They are quite

good as stabilizing ions

and are especially

favorable for SN1

reactions, although they

can also be used for SN2

reaction with favorable

substrates

Water H2O

Methanol CH3OH

acetic acid

Ethanol CH3CH2OH

dimethyl sulphoxide

(DMSO)

These aprotic solvents are

still relatively polar, so that

they can dissolve both the

nucleophile and the

substrate. They are

especially favourable for

dimethyl formamide

(DMF)

O

CF3COH

O

CH3COH

O

CH3SCH2

O

HCN(CH3)2

34

Acetone

SN2 reactions because

nucleophiles are more

reactive in these solvents

than protic ones.

Table: Summary of the SN1 and SN2 reactions

SN1 SN2

Mechanism two step

R – L R0 R – No

one step

LNuRNuLR

Kinetics first order

rate = k[R – L]

second order

rate = ]Nu][LR[k

Effect of Nucleophile no effect on rate stronger nucleophiles cause

faster rate

Effect of carbon structure tertiary > secondary

resonance stabilization of R important

methyl > primary > secondary

Stereochemistry racemisation (possible

excess inversion)

inversion

Effect of solvent favoured by polar

solvents

favoured by aprotic solvents

Competing reaction elimination,

rearrangement

elimination

O

CH3CCH3

35

Preferred substitution mechanism for various carbon structure

Type Representative structure

(L is a leaving group)

Preferred Mechanism

Methyl CH3 – L SN2

Primary RCH2 – L SN2

Secondary

SN2 (with good

nucleophiles) or SN1

(with poor nucleophiles

and polar solvent)

tertiary

SN1

Allylic CH2 = CHCH2 – L SN2 (with good

nucleophiles) or SN1

(with poor nucleophiles

and polar solvents)

Benzylic

Neopentyl

very slow SN1 and SN2 (SN2

in aprotic solvents, gives

acceptable yields)

Vinylic

Inert to both SN1 and SN2

under nor mal reaction

conditions.

Aromatic

36

Reaction with magnesium. Alkyl halides react with magnesium metal in the presence of

anhydrous ether to form alkyl magnesium halides also called Grignard's reagents. For

example

R X + MgAnhydrous ether

R XMg

Alkyl. Mag. halide(Grignard Reagent)

Reaction with lithium. With lithium alkyl halides form alkyl lithium in the presence of

anhydrous ether. For example

2LiR X +Anhydrous ether

R Li LiX+

Alkyl Lithium

Question: The treatment of alkyl halides with aq.KOH leads to formation of alcohols but

in the presence of alc.KOH, it leads to formation of alkenes.

Answer: With alcoholic KOH alkyl halides forms alkenes because in alcoholic solvent

(aprotic solvent) KOH functions as base and therefore alkenes are being formed as

product.

With Aqueous KOH alkyl halides forms alcohols because in aqueous solvent (protic

solvent) KOH functions as nucleophile and therefore alcohols are being formed as

product.

37

Haloarenes Haloarenes or nuclear Substituted halogen derivatives are the compounds formed by

replacing one or more hydrogen atoms, in an aromatic ring with halogen atoms. For

example,

The compounds in which one or more hydrogen atoms of the alkyl side chains (arenes)

are replaced by halogen atoms are known as side chain Substituted halogen derivatives

or aryl alkyl halides. For example,

It may be noted that the side chain halogen derivatives are quite different in

characteristic compared to haloarenes. They resemble alkyl halides in their behaviour.

For example, these are highly reactive and take part in nucleophilic substitution

reactions. On the other hand haloarenes are comparatively very little reactive.

38

ELECTROPHILIC SUBSTITUTION MECHANISM (ARENIUM ION MECHANISM) The most widely accepted mechanism of aromatic electrophihc substitution is

‘bimolecular and involves arenium ion intermediate. Sometimes this mechanism is

called SE2 mechanism, as it is bimolecular. In this mechanism the electrophile attacks

the substrate in the first step to give a carbocation (known as arenium ion or Wheland

-complex, or benzenonium ion or cyclohexadienyl cation in the case of

benzenoid systems); the leaving group (electrophile) departs in the second step. The

‘mechanism resembles the tetrahedral mechanism but the charges are reversed here.

The first step is rate determining and the reaction follows second order kinetics.

The attacking electrophile may be a positive ion or a dipole. The arenium ion is stabilized

by resonance but due to the loss of aromaticity its resonance energy is much less than

that of the parent aromatic system. Thus, the arenium ion loses a proton and reverts to

the more stable aromatic state. A base present in the reaction mixture helps in the

removal of the proton.

39

Delocalised electrons and resonance

Rate = k [Aromatic compound] [Electrophile

40

Energy profile for a typical aromatic electrophilic substitution reaction: A more

detailed picture of the arenium ion mechanism may be presented by its energy profile

(Fig.)

Reacting coordinate (progress of the reaction) →

Similar energy profiles can be drawn for substituted aromatic rings. The rate of

substitution at ortho, para or meta positions depends on the height of the energy *) between the reactants and the T.S.

Reactions of aromatic compounds are carried out in the dark because in the light they

get evaporated due to their high volatile nature. The aroma in all perfume or any

substance or like in flowers is due to only aromatic compounds.

Aromatic Compounds Never Undergo Addition Reaction When electrophilic addition takes place then one electrophile and one nucleophile is

added to the compound. So, therefore if it happens in benzene ring then one of the pi-

bond will be broken of the benzene and electrophile and nucleophile will be added

which leads to loss of aromaticity according to Huckel's rule i.e. (4n+2)pi electrons and

thus imparts instability to the benzene ring.

When nucleophile is added to the benzene ring then benzene ring is not able to

accommodate the high electron density of the nucleophile which leads to rupture of the

ring.

41

DIRECTIVE INFLUENCE OF FUNTIONAL GROUPS ON THE BENZENE RING

• An electron-withdrawing resonance effect is observed in substituted benzenes

having the general structure C6H5-Y=Z, where Z is more electronegative than Y.

• Seven resonance structures can be drawn for benzaldehyde (C6H5CHO). Because

three of them place a positive charge on a carbon atom of the benzene ring, the

CHO group withdraws electrons from the benzene ring by a resonance effect.

• To predict whether substituted benzene is more or less electron rich than

benzene itself, we must consider the net balance of both the inductive and

resonance effects.

• For example, alkyl groups donate electrons by an inductive effect, but they have

no resonance effect because they lack non-bonded electron pairs or bonds.

• Thus, any alkyl-substituted benzene is more electron rich than benzene itself.

42

43

44

45

46

47

48

49

Ortho, para directors Meta directors

50

The effect of substituents on the reactivity of a benzene ring toward electrophilic substitution Activating substituents Most activating

51

Halogenation Aromatic compounds can be chlorinated or brominated with chlorine or bromine in the

presence of a catalyst often called as a halogen carrier, for example, Lewis acids such as

FeCl3, FeBr3, AlCl3 or AlBr3. Sometimes iron is used but the real catalyst is not the iron

itself but FeCl3 or FeBr3 formed from the reaction between the halogen and iron is the

actual catalyst.

Reagent for the reaction is X2/AlCl3 or FeCl3 or ZnCl2 etc.

X2 + AlX3 X+ + AlX4-

Function of Lewis acid

1. It helps in the generation of electrophile

2. It functions as a catalyst

52

The active electrophile is either the halogen-Lewis acid complex or positive halogen:

The positive end of the Lewis acid complex or the positive halogen itself attacks the

aromatic ring to form the resonance stabilised arenium ion intermediate:

Finally, proton is removed from the arenium by a base to give the halogenated product:

53

A similar mechanism operates when halogenation is carried out with HOCI or HOBr in

the presence of a strong acid. The active electrophile in this case is either XOH2

or

positive, halogen XOH(2

XOH2

), where X = Cl or Br.

Iodination of benzene cannot be done in the same way because HI formed in the

reaction is powerful reducing agent due to low bond dissociation enthalpy (299 kJ mol-

1). It will therefore, make the reaction reversible in nature.

On account of this, HI formed in the reaction is oxidised by carrying the reaction of iodic

acid (HIO3) or conc. HNO3.

222

heat

3IOH2NO2HNO2HI2

22

heat

3I3OH3HIOHI5

Fluorination of benzene is not very successful because the reaction is extremely violent

due to the reactive nature of fluorine and cannot be easily controlled.

Note:

(i) Fluorine is highly reactive. It reacts so rapidly with benzene that aromatic fluorination

requires special conditions and special types of apparatus.

(ii) Iodine is so unreactive that a special technique has to be used to effect direct

iodination. The iodination is carried out in the presence of an oxidising agent such as

HNO3.

54

SULPHONATION

Sulphonation is usually done with fuming sulphuric acid (oleum

) or concentrated sulphuric acid. In these cases the active

electrophile is SO3 which is present as such in oleum, and may be formed as follows in

the case of concentrated sulphuric acid:

S

O

O

HO O

Sulphuric acid

S

O

O

HO OH

Sulphuric acid

S

O

O

H2O OH+

Good Leaving group

S

O

O

OH+

Electrophile

++ H2O + S

O

O

HO O-H

OR

The electrophile SO3 attacks the aromatic ring to give the resonance stabilised arenium

ion:

Finally, proton is removed from the arenium ion by a base to give the substitution

product:

In certain cases the electrophile may be 3OHS

or molecules that can readily transfer

SO3 or 3OHS

.

It should be noted that sulphonation is reversible, especially at high temperature. For

example, when benzenesulphonic acid is heated with an aqueous acid under pressure,

benzene is regenerated:

H2SO2+SO3 H2S2O7

55

The mechanism for this transformation is just reverse of the sulphonation of benzene:

NITRATION

The introduction of a nitro (NO2) group into an aromatic system is called nitration. The

most common reagent for the nitration of aromatic compounds is a mixture of conc.

nitric and cone. Sulphuric acid (mixed acid). It has been established that for nitration

with mixed acid, the active nitrating agent is nitronium ion (2

ON

), which is formed as

follows:

S

O

O

HO O

Sulphuric acid

+H HO N O

O

H2O N O

O

+ ++N O

O

H2O +

Electrophile

Good Leaving group

S

O

O

HO O-

The overall equation:

H2SO4 + HNO3 NO2+ + HSO4- + H2O

This is an acid-base reaction in which nitric acid is the base. Then, attack of nitronium

ion electrophile on the aromatic ring takes place to give the arenium ion intermediate

which is stabilised by resonance:

56

Finally, removal of proton (by a base) from the arenium ion and the formation of the

substitution product (nitro compound) occur:

Besides mixed acid, other reagents have also been used, for nitration, e.g., concentrated

or fuming nitric acid alone, nitric acid in organic solvents (acetic acid, nitromethane,

etc.), acyl nitrates (acetyl or benzoyl) in organic solvents (acetic acid, nitrometlane, etc.),

nitronium salts )OSCFON,EtONO,ON,OClON,FBON 3322524242

in organic

solvents, and nitrosation followed by oxidation.

In the case of concentrated or fuming nitric acid alone the 2

ON

is formed as follows:

OHNO

O

HO N O

O

+ H2O N O

O

+ ++N O

O

H2O ONO

O

+ -

Electrophile

Good Leaving

group

Overall reaction

Here, one molecule of HNO3 is the acid and another the base.

In the case of N2O5 in CCl4, there is spontaneous dissociation to give the nitronium ion:

57

Evidence for the existence of nitronium ion as the attacking electrophile:

1. The salts of nitronium ion, e.g., nitronium perchlorate ( 42OClON

) and nitronium

tetrafluoroborate ( 42FBON

) have been isolated and have been shown to be

nitrating agents.

2. On addition of HNO3, the freezing point of H2SO4 is lowered about four times the

amount expected if no ionisation has taken place. This indicates the production

of four particles which is strong evidence for the ionisation reaction between

HNO3 and H2SO4 to form 2

ON

as given above.

NITROSATION

For Nitrosation HNO2 is prepared in-situ i.e. in the reaction mixture itself.

NaNO2 + HCl

KNO2 + HCl

NaNO2 + H2SO4

KNO2 + H2SO4

HNO2 + NaCl

HNO2 + KCl

HNO2 + NaHSO4

HNO2 + KHSO4

Now from HNO2, NO+ (nitroso) electrophile is formed as follows.

N

HO O

Nitrous acid

H+ (from HCl or H2SO4) +

N

H2O O+

Good Leaving group

N O+

H2O+

58

H2SO4 is generally not used for in-situ preparation of nitrous acid because it is a reagent

of sulphonation also so therefore it forms some amount of sulphonated product also as

side products.

Gattermann Koch Reaction The Gattermann–Koch reaction, named after the German chemists Ludwig Gattermann

and Julius Arnold Koch, in organic chemistry refers to a Friedel–Crafts acylation reaction

in which carbon monoxide and hydrochloric acid are used in situ with Friedel–Crafts

catalyst, namely AlCl3 to produce a benzaldehyde derivative from a benzene derivative

in one step. Benzaldehyde and many aromatic aldehydes are conveniently synthesized

by this reaction. Presence of traces of copper (I) chloride are also needed.

CO, HCl

AlCl3/CuCl

CH

O

benzene or activated benzene needed

59

Mechanism:

CO H C

O

Cl+ ++

HClAlCl3

H C

O

AlCl4-

Benzene

O

Benzaldehyde

H

+

H C

O

+

O

H

+H

GATTERMANN ALDEHYDE SYNTHESIS

This method synthesizes aldehydes by treatment of an aromatic compound with

hydrogen chloride and hydrogen cyanide (or metallic cyanide as zinc cyanide) in the

presence of Lewis acid catalysts.

Mechanism:

+ HClHC N HC NH

Cl

AlCl3HC NH

++ AlCl4

-

Benzene

+ HC NH+

+H

HC NH

NH

Benzyl imine

H2O/H+

O

Benzaldehyde

H

60

Friedal craft alkylation The Friedel–Crafts reactions are a set of reactions developed by Charles Friedel and

James Crafts in 1877. There are two main types of Friedel–Crafts reactions: alkylation

reactions and acylation reactions. This reaction type is a form of electrophilic aromatic

substitution. The general reaction scheme is shown below.

Reagent for the reaction is R-X/AlCl3 or FeCl3 or any other lewis acid.

This Lewis acid-catalyzed electrophilic aromatic substitution allows the synthesis of

alkylated products via the reaction of arenes with alkyl halides or alkenes. Since alkyl

substituents activate the arene substrate, polyalkylation may occur. A valuable, two-

step alternative is Friedel-Crafts Acylation followed by a carbonyl reduction.

Mechanism of the Friedel-Crafts Alkylation

Friedel–Crafts alkylation involves the alkylation of an aromatic ring with an alkyl halide

using a strong Lewis acid catalyst. With anhydrous lewis acid as a catalyst, the alkyl

group attaches at the former site of the chloride ion. The general mechanism is shown

below.

61

Benzene

+

R X AlX3 AlX4-

R+ ++

R+

+H

R R

Limitations of Friedel-Crats alkylation:

1. Rearrangement in the alkylating group: For example, the alkylation of benzene

n-propyl chloride gives a mixture of n- and isopropylbenzenes, in which there is

mostly isopropylbenzene (cumene). The rearrangement is possible even with

Lewis acid complex. Therefore, usually it is not possible to introduce a primary

alkyl group (except Me and Et) into an aromatic ring by Friedel-Crafts alkylation.

Thus, n-alkylbenzenes are often prepared by Friedel-Crafts acylation followed by

reduction.

When d-block element's lewis acid is introduced then R group can be introduced

into the ring without rearrangement.

2. Aromatic compounds containing m-directing groups do not undergo Friedel-Crafts

alkylation. This is because in such cases the ring is deactivated and the attacking

carbocation (electrophile) is less reactive to bring about the reaction. On the other

hand, electrophiles 32

SO,ON

and positive halogen, etc. are reactive enough to bring

about the substitution even on deactivated aromatic rings.

3. Aryl halides cannot be used in the place of alkyl halides in Friedel-Crafts reactions.

There is delocalisation of electrons in haloarenes due to resonance. For example,

chlorobenzene is considered to be a resonance hybrid of the following structures:

62

The contribution of structures III, IV and V imparts a partial double bond character to

the carbon-chlorine bond. The shortening of bond length imparts stability to aryl halides

and as a result, the bond cleavage becomes rather difficult. The aryl halides are,

therefore, less reactive than alkyl halides.

4. The activating groups like OH, NH2, OR, etc. do not facilitate Friedel-Crafts alkylation,

because the catalyst coordinates with these basic groups. Even When excess catalyst is

used, reaction is not facilitated because due to co-ordination of the atom directly

attached to the aromatic ring acquires positive charge resulting in the deactivation of

the ring.

Friedel-Crafts Acylation

The introduction of an acyl group into an aromatic ring is called Friedel-Crafts acylation.

This is the most important method for the preparation of aryl ketones. The reaction is

brought about by acyl halides, carboxylic acids, anhydrides, and ketenes in the presence

of Lewis acid catalyst similar, to that in Friedel-Crafts alkylation. In acylation by

acetylchloride a little more than one mole and acylation by acetic anhydride a little

more than two mole of catalyst is required per mole of reagent because the one and

two mole of catalyst co-ordinates with oxygen of the reagent respectively.

In the case of acyl halides the order of reactivity is usually:

RCOI > RCOBr > RCOCl > RCOF

Similar to Friedel-Crafts alkylation, in most cases the attacking species is a carbocation,

the acylium ion. In certain cases instead an acylium ion, the acyl halide-Lewis acid 1:1

complex is the active electrophile. The acylium ion (or the positive end of the Lewis acid

1:1 complex) attacks the aromatic ring to form the resonance-stabilised arenium ion

intermediate:

63

Finally, proton is removed by a base to give the acylated product:

The major limitations of Friedel-Crafts alkylation are not present in the case of Friedel-

Crafts acylation. For example, rearrangement of R group is never found, because the

carbocation in this case is highly stabilized by the adjacent oxygen. Polysubstitution does

not occur because RCO group is deactivating group. Friedel-Crafts acylation is usually

prevented by meta directing groups. Friedel-Crafts acylation can be carried out with a

trace or even sometimes with no catalyst at all. Thus, compounds containing ortho-

para-directing groups, including alkyl, alkoxy, hydroxy, halogen, and acetamido groups,

are easily acetylated and mainly give para products because of the relatively large size

of the acyl group.

Mechanism:

R C

O

Cl AlCl3+ R C

O

++

Benzene

+ R C

O

+

+H

CO

R

CO

R

AlCl4-

64

Examples:

Benzene

Diphenyl methane

CH2Cl2 + AlCl3 AlCl4-

CH2Cl

+

++

+H

CH2Cl

Cl

Benzyl chloride

Benzene

AlCl3

+

CH2Cl+

+H

Benzene

+ CHCl3AlCl3

Triphenyl methaneTry Yourself

65

Nucieophillc Substitution Reactions

We have studied that haloalkanes are highly reactive and readily take part in the

nucleophilic substitution reactions involving the cleavage of C–X bond. But haloarenes

are comparatively little reactive and the cleavage of the C—X bond is quite difficult.

Explanation for the Low Reactivity of Haloarenes

The low reactivity of haloarenes can be attributed to the following reasons

1. Conjugation effect. In haloarenes, the electron pairs on the halogen atom are in

conjugation with the pi-electron pairs of the ring. Following resonating structures for

chlorobenzene are possible.

The C–Cl bond acquires a partial double bond character (C — Cl) and its bond length

(169 pm) is, therefore, less compared to C–Cl bond length (177 pm) in chloroalkane

molecules. As a result, the bond cleavage in haloarenes is difficult compared to

haloalkanes and these are therefore, less reactive towards nucleophilic substitution (SN)

reactions.

2. Difference in hybridisation states of carbon atom in C—X bond. In haloalkanes, the

carbon atom of the C—X bond is sp3 hybridised while in haloarenes, the carbon atom is

in sp2 hybridisation state.

The sp2 hybridised carbon with a greater s–character is more electronegative and can

hold the electron pair of the C—X bond more tightly than sp3 hybridised carbon atom in

haloalkanes with less s–character. Thus, the bond cleavage in haloarenes is more

difficult and these are therefore, less /reactive compared to haloalkanes.

3. Polarity of C—X bond. We have seen that sp2 hybridised carbon atom in haloarenes is

more electronegative than the sp3 hybridised carbon atom in the molecules of

66

haloalkanes. Therefore, the polarity of the C–X bond in haloarene (based upon the

electronegativity difference of carbon and halogen atoms) is less than the bond polarity

in haloalkanes. This is supported by the fact that the dipole (moment of C–Cl bond in

chlorobenzene is 1.73 D while in chloroethane it is 2.05D. As the reactivity of the bond is

directly proportional to the bond polarity, haloarenes are less reactive than alkanes.

From the above discussion, it is quite obvious that the nucleophilic substitution

reactions involving the cleavage of C–X bond in haloarenes are expected to proceed

with difficulty compared to haloalkanes. Some of these reactions are listed below:

Dow's Process. Chlorobenzene can be converted into phenol by heating with aqueous

sodium hydroxide (6–8%) solution at about 623 K and under a pressure of 300

atmospheres. This is called Dow’s process and is used for the commercial preparation of

phenol from chlorobenzene.

Effect of Substituents on the Reactivity of Haloarenes.

Haloarenes are very little reactive chemically towards nucleophilic reactions because of

the difficulty in the cleavage of C–X bond. However, when the withdrawing groups such

as –NO2, –CN, –CHO, –COOH etc. are present at ortho or para positions in the ring (but

not at meta position), the bond cleavage becomes easy, Greater the number of such

groups present, more will be the reactivity of haloarene. This is illustrated by the

presence of electron withdrawing nitro group on the reactivity of chlorobenzene.

67

It may be noted that the electron withdrawing substituents tend to withdraw the

electron charge from the ring. As a result, the electron density on the carbon atom of C–

X bond decreases and bond polarity therefore, increases. With the increase, in the

number of such groups, the bond cleavage becomes more and more easy. This is quite

evident from the condition of temperature as listed above.

The effect of electron withdrawing nitro group or any other such group is more

pronounced at the ortho or para positions than when, it is present at the meta position

in the ring. This is further supported by the mechanism involving the nucleophile attack

of OH- ion on the nitro substituted chloro benzene.

68

In both ortho and para nitro substituted chlorobenzene, one of the contributing

structures enclosed in bracket has a negative charge on the carbon atom of the ring to

which the electron withdrawing –NO2 group is attached. As a result, the negative charge

on the carbanion is stabilised and, this will facilitate the nucleophilic substitution.

Howeyer, in meta nitro substituted chlorobenzene none of the contributing structures

has negative charge on the carbon atom of the ring bearing the –NO2 group. As a result,

the carbanion is not that much stabilised in this case. Therefore, no effect on reactivity

is observed when –NO2 group is present at the meta position. Cl

Chlorobenzene

OH

Phenol

Steam

700K

69

Replacement by cyano group (Formation of nitriles). Haloarenes when heated with

cuprous cyanide to about 473K in the presence of pyridine catalyst form benzonitrile

also called phenyl cyanide or cyanobenzene.

The C N bond in cyanobenzene is as reactive as in alkyl into useful products in the

same way.

Reaction with copper powder. When iodobenzene is heated with copper powder in a

sealed tube, the product is diphenyl. The reaction is called Ullmann reaction.

Haloarenes can also be converted into aniline by reacting with a mixture of sodamide

and liquid ammonia at about 193K.

Mendius

Reduction

70

The reaction with NH3 described above is nucleophilic substitution. But the reaction

with NaNH2 occurs through the formation of a benzyne intermediate as follows:

Benzyne

Possibility I

Possibility II

+

-

-+ NH2

-H+

NH2-

H+

H

NH2

NH2

H

Direct substitution

Cine substitution

Benzyne is a highly reactive intermediate which contains an additional bond between

carbon atoms of the benzene ring. Out of the two atoms involved, only one halogen

atom while the carbon atom adjacent to it contains hydrogen atom. The extra bond is

formed as a result of the sidewise overlap of the sp2 hybridised orbitals of the two

carbon atoms. There is a very little interaction between the electron cloud of the new

bond and the –electron cloud of the benzene ring, Since the new sidewise overlapping

is very poor, benzynes are highly reactive in nature.

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Reaction with magnesium metal. Bromo and iodoarenes react with a dry eithereal

solution of dry magnesium metal to form the magnesium derivatives also called

Grignard’s reagents.

Preparation of Haloarenes

Preparation from Aromatic Hydrocarbons (Arenes)

Haloarenes (chlorobenzene and bromobenzene) can be prepared by the halogenation of

benzene at low temperature (310 to 320 K) in dark (i.e. absence of sun light) and in the

presence of anhydrous ferric halides or aluminum halides known as halogen carriers.

These are Lewis acids and as the name suggests, they carry halogen to the benzene ring.

In place of ferric halide, even iron can be used which reacts with halogen to form the

desired halogen carrier. For example,

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Iodination of benzene cannot be done in the same way because HI formed in the

reaction is powerful reducing agent due to low bond dissociation enthalpy (299 kJ mol-

1). It will therefore, make the reaction reversible in nature.

On account of this, HI formed in the reaction is oxidised by carrying the reaction of iodic

acid (HIO3) or conc. HNO3.

222

heat

3IOH2NO2HNO2HI2

22

heat

3I3OH3HIOHI5

Fluorination of benzene is not very successful because the reaction is extremely violent

due to the reactive nature of fluorine and cannot be easily controlled.

Preparatón from Dlazonium Salts

Diazonium salts such as. benzene diazonium chloride is quite useful in the synthesis of

all the members of the haloarenes containing one halogen atom.

Chlorobenzene and bromobenzene can be prepared by treating benzene diazonium

chloride with CuX dissolved in HX or CuBr/CuCl dissolved in HBr/HCl. The reaction is

called Sandmeyer’s reaction.

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In the Sandmeyers reaction, the halogen atom of cuprous halide gets attached to the

ring. The yield can be increased by modifying the reaction i.e. cuprous halide is replaced

by a mixture of copper powder and halogen acid (HCl or HBr). The modified form of the

reaction is called Gattermann reaction.

Iodobenzene is prepared by simply warming the diazonium salt with an aqueous

solution of KI.

Fluorobenzene can also be prepared by treating benzene diazonium chloride with

fluoroboric acid (HBF4). The compound formed is dried and then heated to give

fluorobenzene. The reaction is called Balz-Schiemann reaction

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Preparation from Silver salt of Aromatic Acids.

When silver salt of an aromatic acid such as silver beuzoate is refluxed with bromine in

the presence carbon tetrachioride, bromobenzene is formed. The reaction is called

Hundsdiecker reaction.