Aromatic Chemistry_Final

1. IIT-JEE Syllabus

Electrophilic substitution reactions, bromination, nitration, sulfonation, friedel crafts alkylation and acylation. Effect of o, m, p directing groups in mono substituted benzenes.

  1. Aromaticity

The modern definition of aromaticity is the ability to sustain induced ring current by a flat or nearly flat cyclic system with (4n + 2) delocalised π electrons, where n is zero or any whole number.

Benzene is the ideal molecule possessing aromaticity. It has a planar hexagonal ring structure with (4 × 1 + 2) or 6π electrons. There are six p-orbitals, one on each sp2 atom. All the six p electron orbitals being parallel to each other and perpendicular to the plane of the molecule, they form a delocalised electron cloud below and above the plane.

When a magnetic field is applied perpendicular to the plane of the benzene molecule, the circulation of delocalized π electrons takes place in one direction and thereby a ring current, which induces a counteraction magnetic field  inside the ring, is produced (see figure). The benzene has the ability to sustain induced ring current.

Systems possessing aromaticity have some physical and chemical properties which are also collectively known as aromatic character.

  1. i) Chemical stability in excess of analogous acyclic unsaturated system e.g.:
  2. a) Although aromatic compounds have unsaturated ring, electrophilic substitution rather than addition reaction occur and the unsaturated ring does not get ruptured.
  3. b) Resistance to oxidation by aqueous solution of normal oxidising agents except group VI oxidising agents (V2O5) etc.
  4. ii) Unusual thermal stability
  5. a) Low heat of hydrogenation
  6. b) Low heat of combustion

iii) Physical measurement

  1. a) Diamagnetic susceptibility
  2. b) Unique nuclear magnetic resonance spectra

There are three theories regarding the cause of aromaticity:

  1. i) Valence bond theory: This theory tells us that resonance stabilization of a planar cyclic structure is the cause of its aromaticity.
  2. ii) Molecular orbital theory: According to this theory high delocalisation energy is the cause of aromaticity of planar cyclic systems.

Six p-orbitals, one form each sp2C atom remain perpendicular to the plane of the hexagonal benzene ring and parallel to each other. All the six p A.O.s overlap to form six delocalised M.O’s-three bonding and three anti-bonding. The three boding M.O.’s accommodate all six electrons.

iii) The Hückel 4n + 2 rule: One the basis of M.O. theory Hückel stated that the presence of 4n+2 delocalized π electrons in the flat or nearly flat cyclic systems is the cause of aromaticity and is known as Hückel’s rule. Here n may be 0, 1, 2, 3 etc., whole number. According to him 4n+2 delocalized  π-electrons of a cyclic system develop high delocalization energy or high resonance energy in it. Thus a compound to be aromatic it must be cyclic planar structure containing (4n+2) delocalized π electrons in a closed shell. This rule is applicable to monocyclic, fused ring, heterocyclic and cyclic ionic systems.

  1. Non-aromatic and Anti-aromatic Compounds

If a cyclic compound has lower π electron energy than its open chain analogue, the compound is said to be an aromatic compound,. When 1, 3, 5 hexatriene is converted to benzene by the hypothetical abstraction of two H-atoms, the π-electron energy diminishes in the system.

I contains higher π-electron energy than II

On the other hand, if a cyclic compound contains higher π-electron energy than its acyclic counterpart, the compound is called anti-aromatic, in fact planar cyclic conjugated systems with 4nπ-electrons are called anti-aromatic. Hence cyclobutadiene is an anti-aromatic compound.

If in the case of hypothetical conversion of an acyclic conjugated system into cyclic conjugated system the π-electron energy remains unchanged the cyclic system is said to be non-aromatic.

Illustration 1: Classify the following compounds as aromatic, anti-aromatic or non-aromatic compounds.

  1. a) Cyclopropenyl cation b) Napththalene
  2. c) Furan d) Pyrrole
  3. e) Pyridine f) Cyclooctatetraene
  4. g) 1,3,5 cycloheptatriene
Solution: a) is aromatic due to 

ring is planar

4n + 2 = 2, π-electrons (n = 0)

π electrons are delocalized

b) It has planar structure with delocalized (4 × 2 + 2) or 10 π electrons obeys Hückel’s rule. It is aromatic.
c) The structure of Furan is The O atom and each of the four C atoms are in sp2 hybridized state forming a flat pentagonal structure. One of two lone pairs of electrons of the O atom is in sp2 orbital and the other in p orbital. Four p orbitals of four. C atoms each containing are electron and the filled p orbital of O are parallel to each other and perpendicular to the plane of furan. 

Therefore furan has 4p electrons from four C atoms and 2p electrons from the O atom, i.e. it has six delocalized π electrons which is a Hückel number. Hence it is an aromatic compound. It has also resonance stabilization shown below


d) Hückel number of  delocalized
π-electrons are present in a planar cyclic structure Hence it is aromatic.
e) Since there are 6 delocalized
p-electrons in planar hexagonal structure of pyridine, it is aromatic.
f) Cyclooctatetraene has 8 π-electrons. It has a puckered ring structure and proper overlap of π-electrons is not possible. Hence it is not an aromatic compound.
g) It has 4 ×1 + 2= 6 π-electrons but they are not delocalized. Hence it is non-aromatic.


Exercise 1: a) 7-Bromo-cycloheptatriene (tropylium bromide) completely dissociates in water and gives a precipitate of AgBr with AgNO3. Explain.

  1. b) Identify which of the following compounds are aromatic, anti-aromatic or non-aromatic?
  2. Electrophilic Aromatic Substitution Reactions

Aromatic hydrocarbons are known generally as arenes. An aryl group is one derived from an arene by removal of a hydrogen atom and its symbol is Ar –. Thus, arenes are designated ArH just as alkanes are designated RH.

The most characteristic reactions of benzenoid arenes are the substitution reactions that occur when they react with electrophilic reagents. These reactions are of the general type shown below.

ArH + E+  ––→  Ar – E + H+ or

The electrophiles are either a positive ion (E+) or some other electron-deficient species with a large partial positive charge. For example, benzene can be brominated when it reacts with bromine in the presence of FeBr3. Bromine and FeBr3 reacts to produce positive bromine ions, Br+. These positive bromine ions act as electrophiles and attack the benzene ring replacing one of the hydrogen atoms in a reaction that is called an electrophilic aromatic substitution (EAS).

Electrophilic aromatic substitutions allow the direct introduction of a wide variety of groups into an aromatic ring and because of this they provide synthetic routes to many important compounds. The five electrophilic aromatic substitutions that we shall study in this package are outlined in fig. All of these reactions involve the attack on the benzene ring by an electron-deficient species – by an electrophile. Later we shall learn what the electrophile is in each instance.

  1. Mechanism for Electrophilic Aromatic Substitution 

π and σ complexes

It might be expected that the first phase of reaction would be interaction between the approaching electrophile and delocalised π orbitals leading to π complexes (a weakly bonded charge-transfer complex which exists in solution only and is formed by the association of an electrophilic species (E) and on electron-donating species) e.g. toluene form a 1:1 complex with HCl at –78°C, the reaction being readily reversible. DCl also forms π complex with toluene. This complex on decomposition does not form deuterium substituted toluene. Formation of complex leads to a solution that is a non-conductor of electricity.

When an electrophile reacts with an arene usually in presence of a catalyst, a salt is formed. This salt is composed of an anioin and a complex, resonance stabilized carbenium ion (arenium ion) in which only two of the total π electrons of arene are utilized to form a α bond between a particular C atom of the ring and the electrophile, known as α complex (also called Wheland intermediate).

Of course one may assume that an aromatic substitution reaction consists of four stpes involving two π and are α complexes as shown below.

  1. Formation of a π-complex
  2. Conversion of π α complex to a α complex
  3. Conversion of a complex into π complex
  4. Decomposition of π complex to products

However the most widely accepted mechanism for aromatic electrophilic substitution reaction is the two step mechanism.

  1. i) α complex formation
  2. ii) Loss of proton from the α complex 
Step 1: +E+ ⎯→ H 
Step 2:

Of the following two steps, step 1, the formation of the arenium ion, is the rate-determining step in electrophilic aromatic substitution.

Step 1:
Step 2:

Step 2, the loss of a proton, occurs rapidly relative to step 1 and has no effect on the overall rate of reaction.

Energy profile of an aromatic electrophilic substitution without formation of π complex.

The evidence for the two step mechanism are

  1. i) Detection and isolation of σ complex 

A large number of σ complexes as intermediate has been detected, some of them have also been isolated.

  1. ii) Displacement of an H (or D) atom of C6H6 (or C6D6) by a nitronium ion NO2+ (formed by the action of concentrated H2SO4 on concentrated HNO3) is an aromatic electrophile substitution reaction. If the C – H bond breaking is the r/d, step, then the reaction will exhibit a primary kinetic isotope effect. By contrast C6H6 and C6D6 are found to undergo nitration at essentially the same rate, thus C – H bond breaking cannot be involved in the r/d step, that means α  complex formation is the r/d step.

5.1 Nitration of Benzene

Benzene reacts slowly with hot concentrated nitric acid to yield nitrobenzene. The reaction is much faster if it is carried out by heating benzene with a mixture of concentrated nitric acid and concentrated sulfuric acid.

Concentrated sulfuric acid increases the rate of the reaction by increasing the concentration of the electrophile –– the nitronium ion (NO2+).

Step 1:
Step 2:

In step 1 nitric acid acts as a base and accepts a proton from the stronger acid, sulfuric acid. In step 2 the protonated nitric acid dissociates and produces a nitronium ion. The nitronium ion reacts with benzene by attacking the π cloud and forming an arenium ion.

Step 3:

The arenium ion then transfers a proton to some base in the mixture such as HSO4 and becomes nitrobenzene.

Step 3:

5.2 Halogenation of Benzene

Benzene does not react with bromine or chlorine unless a Lewis acid is present in the mixture, (as a consequence, benzene does not decolorize a solution of bromine in carbon tetrachloride). When Lewis acids are present, however, benzene reacts readily with bromine or chlorine, and the reactions give bromobenzene and chlorobenzene in good yields.

The Lewis acids most commonly used to effect chlorination and bromination reactions are FeCl3, FeBr3, and AlCl3 all in the anhydrous form. Ferric chloride and ferric bromide are usually generated in the reaction mixture by adding iron to it. The iron then reacts with halogen to produce the ferric halide :

2Fe + 3X2  ––→  2FeX3

The mechanism for aromatic bromination is as follows :

Step 1:
Step 2:
Step 3:

The function of the Lewis acid can be seen in step 1. The ferric bromide reacts with bromine to produce a positive bromine ion, Br+ (and FeBr4). In step 2 this Br+ ion attacks the benzene ring to produce an arenium ion. Then finally in step 3 the arenium ion transfers a proton to FeBr4. This results in the formation of bromobenzene and hydrogen bromide –– the products of the reaction. At the same time this step regenerates the catalyst FeBr3.

The mechanism of the chlorination of benzene in the presence of ferric chloride is analogous to the one for bromination. Ferric chloride serves the same purpose in aromatic chlorinations as ferric bromide does in aromatic brominations. It assists in the generation and transfer of a positive halogen ion.

The rate reaction found is often of the form. Rate = K [Ar – H] [Hal2] [Lewis acid]

Hypo-halous acids (HO – Hal) in presence of strong acid also become a very powerful halogenating agent.

Fluorine reacts so rapidly with benzene that aromatic fluorination requires special conditions and special types of apparatus. Even then, it is difficult to limit the reaction to monofluorination. Fluorobenzene can be made, however, by an indirect method.

Iodine, on the other hand, is so unreactive that a special technique has to be used to effect direct iodination; the reaction has to be carried out in the presence of an oxidizing agent such as nitric acid.

Kinetic isotope effects  have not been observed for chlorination and rarely for bromination, but iodination shows primary kinetic isotope effect. 

5.3 Sulfonation of Benzene

Benzene reacts with fuming sulfuric acid at room temperature to produce benzenesulfonic acid. Fuming sulfuric acid is sulfuric acid that contains added sulfur trioxide (SO3). Sulfonation also takes place in concentrated sulfuric acid alone, that more slowly.

In either reaction the electrophile appears to be sulfur trioxide. In concentrated sulfuric acid, sulfur trioxide is produced in the following equilibrium in which H2SO4 acts as both an acid and a base.

Step 1 2H2SO4 SO3 + H3O+ + HSO4

When sulfur trioxide reacts with benzene the following steps occur.

Step 2:
Step 3:
Step 4:

All of the steps are equilibria, including step 1 in which sulfur trioxide is formed from sulfuric acid. This means that the overall reaction is an equilibrium as well. In concentrated sulfuric acid, the overall equilibrium is the sum of steps 1 – 4.

In fuming sulfuric acid, step 1 is unimportant because the dissolved sulfur trioxide reacts directly.

Because all of the steps are equilibria, the position of equilibrium can be influenced by the conditions we employ. If we want to sulfonate benzene we use concentrated sulfuric acid or better yet fuming sulfuric acid. Under these conditions the position of equilibrium lies appreciably to the right  and we obtain benzenesulfonic acid in good yield.

On the other hand, we may want to remove a sulfonic acid group from a benzene ring. To do this we employ dilute sulfuric acid and usually pass steam through the mixture. Under these conditions with a high concentration of water the equilibrium lies appreciably to the left and desulfonation occurs. The equilibrium is shifted even further to the left with volatile aromatic compounds because the aromatic compound distills with the steam.

We shall see later that sulfonation and desulfonation reactions are often used in synthetic work. We may, for example, introduce a sulfonic acid group into a benzene ring to influence the course of some further reaction. Later, we may remove the sulfonic acid group by desulfonation. Like iodination, sulfonation  exhibits a kinetic isotopic effect, indicating that 
C – H  bond breaking is involved in the rate determining step.

5.4 Friedel – Crafts  Alkylation

In 1877 a French chemist, Charles Friedel, and his American collaborator James M. Crafts, discovered new methods for the preparation of alkylbenzenes (ArR) and acylbenzenes (ArCOR). These reactions are now called the Friedel – Crafts alkylation and acylation reactions. We shall study the Friedel – Crafts alkylation reaction here and then take up the Friedel – Crafts acylation reaction.

A general equation for a Friedel – Crafts alkylation reaction is the following :

The mechanism for the reaction (shown in following steps with isopropyl chloride as R –– X) starts with the formation of a carbocation (step 1). The carbocation then acts as an electrophile (step 2) and attacks the benzene ring to form an arenium ion. The arenium ion (step 3) then loses a proton to generate isopropylbenzene.

Step 1:
Step 2:
Step 3:

When R –– X is a primary halide, a simple carbocation probably does not form. Rather, the aluminum chloride forms a complex with the alkyl halide and this complex acts as the electrophile. The complex is one in which the carbon – halogen bond in nearly broken –– and one in which the carbon atom has a considerable positive charge.

  δ+           δ–

RCH2  —-  Cl : AlCl3

Even though this complex is not a simple carbocation, it acts as if it were and it transfers a positive alkyl groups to the aromatic ring. These complexes are so carbocation like that they also undergo typical carbocation rearrangements.

Friedel – Crafts alkylations are not restricted to the use of alkyl halides and aluminum chloride. Many other pairs of reagents that form carbocations (or carbocation like species) may be used as well. These possibilities include the use of a mixture of an alkene and an acid.

A mixture of an alcohol and an acid may also be used

There are several important limitations of the Friedel – Crafts reaction. 


5.5 Friedel – Crafts Acylation

The group is called an acyl group, and a reaction whereby an acyl group is introduced into a compound is called an acylation reaction. Two common acyl groups are the acetyl group and the benzoyl group.

The Friedel – Crafts acylation reaction is an effective means of introducing an acyl group into an aromatic ring. The reaction is often carried out by treating the aromatic compound with an acyl halide. Unless the aromatic compound is one that is highly reactive, the reaction requires the addition of at least one equivalent of a Lewis acid (such as AlCl3) as well. The product of the reaction is an aryl ketone.

Acyl chlorides, also called acid chlorides, are easily prepared by treating carboxylic acids with thionyl chloride (SOCl2) or phosphorus pentachloride (PCl5).

Friedel – Crafts acylations can also be carried out using carboxylic acid anhydrides. For example :

In most Friedel – Crafts acylations the electrophile appears to be an acylium ion formed from an acyl halide in the following way :

The remaining steps in the Friedel – Crafts acylation of benzene are the following :

In the last step aluminum chloride (a Lewis acid) forms a complex with the ketone (a Lewis base). 

After the reaction is over, treating the complex with water liberates the ketone.

Several important synthetic applications of the Friedel – Crafts reaction are later in the package.

Limitations of Friedel – Crafts Reactions

Several restrictions limit the usefulness of Friedel – Crafts reactions.

  1. When the carbocation formed from an alkyl halide, alkene, or alcohol can rearrange to a more stable carbocation, it usually does so and the major product obtained from the reaction is usually the one from the more stable carbocation.

When benzene is alkylated with butyl bromide, for example, some of the developing butyl cations rearrange by a hydride shift –– some developing 1o carbocations (see following reactions) become more stable 2o carbocations. Then benzene reacts with both kinds of carbocations to form both butylbenzene and sec-butyl-benzene :

  1. Friedel – Crafts reactions do not occur when powerful electron-withdrawing groups are present on the aromatic ring or when the ring bears an –NH2,  –NHR,  or  –NR2 group. The applies to alkylations and acylations.

We shall learn that groups present on an aromatic ring can have large effect on the reactivity of the ring towards electrophilic aromatic substitution.

Electron – withdrawing groups make the ring less reactive by making it electron deficient. Any substituent more electron withdrawing (or deactivating) than a halogen, that is, any meta-directing group, makes an aromatic ring too electron deficient to undergo a Friedel – Crafts reaction. The amino groups, –NH2,  –NHR, and  –NR2, are changed into powerful electron – withdrawing groups by the Lewis acids used to catalyze Friedel – Crafts reactions. For example :

  1. Aryl and vinylic halides cannot be used as the halide component because they do not form carbocations readily.
  2. Polyalkylations often occur Alkyl groups are electron – releasing groups, and once one is introduced into the benzene ring it activates the ring toward further substitution.

Polyacylations are not a problem in Friedel – Crafts acylations, however the acyl group (RCO–) by itself is an electron – withdrawing group, and when it forms a complex with AlCl3 in the last step of the reaction, it is made even more electron withdrawing. This strongly inhibits further substitution and makes monoacylation easy.

Synthetic Applications of Friedel – Crafts Acylations 

Rearrangements of carbon chain do not occur in Friedel – Crafts acylations. The acylium ion, because it is stabilized by resonance, is more stable than most other carbocations. Thus, there is no driving force for a rearrangement. Because rearrangements do not occur, Friedel – Crafts acylations often give us much better routes to unbranched alkylbenzenes than do Friedel – Crafts alkylations.

As an example, let us consider the problem of synthesizing propylbenzene. If we attempt this synthesis through a Friedel – Crafts alkylation, a rearrangement occurs and the major product is isopropylbenzene.

By contrast, the Friedel – Crafts acylation of benzene with propanoyl chloride produces a ketone with an unrearranged carbon chain in excellent yield.

zinc. [Caution : As we shall discuss later zinc and hydrochloric acid will also reduce nitro groups to amino groups.]

When cyclic anhydrides are used as one component, the Friedel – Crafts acylation provides a means of adding a new ring to an aromatic compound. 

Illustration 2: Convert benzene to α-tetralone


Exercise 2: a) Predict the product

  1. b) What happens when chlorobenzene and benzene is treated together in the presence of AlCl3.
  2. Effect of Substituents on Reactivity & Orientation

When substituted benzenes undergo electrophilic attack, groups already on the ring affect both the rate of the reaction and the site of attack. We say, therefore, that substituent groups affect both reactivity and orientation in electrophilic aromatic substitutions.

We can divide substituent groups into two classes according to their influence on the reactivity of the ring. Those that cause the ring to be more reactive than benzene itself we call activating groups. Those that cause the ring to be less reactive than benzene we call deactivating groups.

We also find that we can divide substituent groups into two classes according to the way they influence the orientation of attack by the incoming electrophile. Substituents in one class tend to bring about electrophilic substitution primarily at the positions ortho and para to themselves. We call these groups ortho – para directors because they tend to direct the incoming group into the ortho and para positions. Substituents in the second category tend to direct the incoming electrophile to the meta position. We call these groups meta directors.

Several examples will illustrate more clearly what we mean by these terms.

6.1 Activating Groups: Ortho – Para Directors

The methyl group in an activating group is an ortho – para director. Toluene reacts considerably faster than benzene in all electrophilic substitutions.

We observe the greater reactivity of toluene in several ways. We find, for example, that with toluene, milder conditions –– lower temperatures and lower concentrations of the electrophile – can be used in electrophilic substitutions than with benzene. We also find that under the same conditions, toluene reacts faster than benzene. In nitration, for example, toluene reacts 25 times as fast as benzene.

We find, moreover, that when toluene undergoes electrophilic substitution, most of the substitution takes place at its ortho and para positions. When we nitrate toluene with nitric and sulfuric acids, we get mononitrotoluenes in the following relative proportions.

Of the mononitrotoluenes obtained from the reaction, 96% (59% + 37%) has the nitro group in an ortho or para position. Only 4% has the nitro group in a meta position.

Predominant substitution at the ortho and para positions of toluene is not restricted to nitration reactions. The same behavior is observed in halogenation, sulfonation, and so forth.

All alkyl groups are activating groups, and they are all also ortho – para directors. The methoxy group, CH3O–, and the acetamido group, CH3CONH–, are strong activating groups and both are ortho – para directors.

The hydroxyl group and the amino group are very powerful activating groups and are also powerful ortho – para directors. Phenol and aniline react with bromine in water (no catalyst is required) to produce products in which both of the ortho positions and the para position are substituted. These tribromo products are obtained in nearly quantitative yield.

6.2 Deactivating Groups: Meta Directors

The nitro group is a very strong deactivating group. Nitrobenzene undergoes nitration at a rate only 10-4 times that of benzene. The nitro group is a meta director. When nitrobenzene is nitrated with nitric acid sulfuric acids, 93% of the substitution occurs at the meta position.

The carboxyl group (–CO2H), the sulfo group (–SO3H), and the trifluoromethyl group (–CF3) are also deactivating groups; they are also meta directors.

6.3 Halo Substituents: Deactivating Ortho – Para Directors

The chloro and bromo groups are weak deactivating groups. Chlorobenzene and bromobenzene undergo nitration at rates that are, respectively, 33 and 30 times slower than for benzene. The chloro and bromo groups are ortho – para directors, however. The relative percentages of monosubstituted products that are obtained when chlorobenzene is chlorinated, brominated, nitrated, and sulfonated are shown in Table 1.

TABLE 1 Electrophilic Substitution Of Chlorobenzene
Reaction Ortho Product (%) Para Product


Total Ortho and Para


Meta Product



















Similar results are obtained from electrophilic substitutions of bromobenzene.

  1. Classification of Substituents

Studies like the ones that we have presented in this section have been done for a number of other substituted benzenes. The effects of these substituents on reactivity and orientation are included in Table 2.

TABLE  2 Effect of substituents on electrophilic aromatic substitution
Strongly Activating

–NH2, –NHR, –NR2

–OH, –O:

Moderately Activating


–OCH3, –OR

Weakly Activating

–CH3, C2H5, –R


Weakly Deactivating

–F:, –Cl:, –Br:, –I:

Moderately Deactivating

–C≡ N


–CO2H, –CO2R


Strongly Deactivating



–CF3, –CCl3


  1. Reactivity: the effect of electron-releasing and electron-withdrawing groups

We have now seen that certain groups activate the benzene ring toward electrophilic substitution, while other groups deactivate the ring. When we say that a group activates the ring, what we mean, of course, is that the group increases the relative rate of the reaction. We mean that an aromatic compound with an activating group reacts faster in electrophilic substitutions than benzene. When we say that a group deactivates the ring, we mean that an aromatic compound with a deactivating group reacts slower than benzene.

We have also seen that we can account for relative reaction rates by examining the transition state for the rate-determining steps. We know that any factor that increases the energy of the transition state relative to that of the reactants decreases the relative rate of the reaction. It does this because it increases the free energy of activation of the reaction. In the same way, any factor that decreases the energy of the transition state relative to that of the reactants lowers the free energy of activation and increases the relative rate of the reaction.

The rate-determining step in electrophilic substitutions of substituted benzenes is the step that results in the formation of arenium ion. We can write the formula for a substituted benzene in a generalized way if we use the letter S to represent any ring substituent including hydrogen. (If S is hydrogen the compound is benzene itself). We can also write the structure for the arenium ion in the way shown here. By this formula we mean that S can be in any position – ortho, meta, or para – relative to the electrophile, E. Using these conventions, then, we are able to write the rate-determing step for electrophilic aromatic substitution in the following general way.

When we examine this step for a large number of reactions, we find that the relative rates of the reactions depend on whether S withdraws or release electrons. If S is an an electron – releasing group (relative to hydrogen), the reaction occurs faster than the corresponding reaction of benzene. If S is an electron  withdrawing group, the reaction occurs  slower than that of benzene.

It appears, then, that the substituent (S) must affect stability of the transition state relative to that of the reactants. Electron – releasing groups apparently make the transition state more stable, while electron withdrawing groups make it less table. That is so is entirely reasonable, because the transition state resembles the arenium ion, and the arenium ion is a delocalized carbocation.

8.1 Inductive and Resonance effect: Theory of orientation

We can account for the electron-withdrawing and electron – releasing properties of a group on the basis of two factors: inductive effect and resonance effects. We shall also see that these two factors determine orientation in aromatic substitution reactions.

The  inductive effect of a substituent S  arises from the electrostatic interaction of the polarized S to ring bond with the developing positive charge in the ring as it is attacked by an electrophile. If, for example, S  is a more electronegative atom (or group) than carbon, then the ring will be at the end  of the dipole:

Attack by an electrophile will be retarded because this will lead to an additional full positive charge on the ring. The halogens are all more electronegative than carbon and exert an electron – withdrawing inductive effect. Other groups have an electron – withdrawing inductive effect because the atom directly attached to the ring bears a full or partial positive charge. Examples are the following:

Then resonance effect of a substituent S  refers to the possibility that the presence of S may increase or decrease the resonance stabilization of the intermediate arenium ion. The S substituent may, for example cause one of the three contributors to the resonance hybrid for the arenium ion to the better or worse than the case when S is hydrogen. Moreover, when S is an atom bearing one or more nonbonding electron pairs, it may lend extra stability to the arenium ion by providing a fourth resonance contributor in which the positive charge resides on S.

This electron-donating resonance effect applies with decreasing strength in the following order:

This is also the order of the activating ability of these groups. Amino groups are highly, activating, hydroxyl and alkoxyl groups are somewhat less activating, and halogen substituents are weakly deactivating. When X = F, this order can be related to the electronegativity of the atoms with the nonbonding pair. The more electronegative the atom is the less able it is to accept the positive charge (fluorine is the most electronegative, nitrogen the least). When X = Cl, Br, or I, the relatively poor electron-donating ability of the halogens by resonance is understandable on a different basis. These atoms (Cl, Br, and I) are all larger than carbon and, therefore, the orbitals that contain the nonbonding pairs are further from the nucleus and do not overlap well with the 2p orbital of carbon. (This is a general phenomenon: resonance effects are not transmitted well between atoms of different rows in the periodic table.)

8.2 Meta-Directing Groups

All meta-directing groups have either a partial positive charge or a full positive charge on the atom directly attached to the ring. As a typical example let us consider the trifluoromethyl group.

The trifluoromethyl group, because of the three highly electronegative fluorine atoms, is strongly electron withdrawing. It is a strong deactivating group and a powerful meta director in electrophilic aromatic substitution reactions. We can account for both of these characteristics of the trifluoromethyl group in the following way.

The trifluoromethyl group affects reactivity by causing the transition state leading to the arenium ion to be highly unstable. It does this by withdrawing electrons from the developing carbocation thus increasing the positive charge in the ring.

We can understand how the trifluoromethyl group affects orientation in electrophilic aromatic substitution if we examine the resonance structures for the arenium ion that would be formed when an electrophile attacks the ortho, meta, and para positions of (trifluoromethyl) benzene.

We see in the resonance structures for the arenium ion arising from ortho and para attack that one contributing structure is highly  unstable relative to all the others because the positive charge is located on the ring carbon that bears the electron-withdrawing group. We see no such highly unstable resonance structure in the arenium ion arising from meta attack. This means that the arenium ion formed by meta attack should be the most stable of the three. By the usual reasoning we would also expect the transition state leading to the meta – substituted arenium ion to be the most stable and, therefore, that meta attack would be favoured. This is exactly what we find experimentally. The trifluoromethyl group is a powerful meta director.

We bear in mind, however, that meta substitution is favoured only in the sense that it is the least unfavorable of three unfavourable pathways. The free energy of activation for substitution at the meta position of (trifluoromethyl) benzene is less than that for attack at an ortho or para position, but it is still far greater than that for an attack on benzene. Substitution occurs at the meta position of (trifluoromethyl) benzene faster than substitution takes place at the ortho and para positions, but it occurs much more slowly than it does with benzene.

The nitro group, the carboxyl group, and other meta-directing groups are all powerful electron-withdrawing groups and all act in similar way.

8.3 Ortho – Para – Directing Groups 

Except for the alkyl and phenyl substituents, all of the ortho-para-directing groups in Table 2 are of the following general type:

All of these ortho-para directors have at least one pair of nonbonding electrons on the atom adjacent to the benzene ring.

This structural feature – an unshared electron pair on the atom adjacent to the ring – determines the orientation and influence reactivity in electrophilic substitution reactions.

The directive effect of these groups with an unshared pair is predominantly caused by an electron – releasing resonance effect. The resonance effect, moreover, operates primarily in the arenium ion and, consequently, in the transition state leading to it.

Except for the halogens, the primary effect on reactivity of these groups is also caused by an electron-releasing resonance effect. And, again, this effect operates primarily in the transition state leading to the arenium ion.

In order to understand these resonance effects let us begin by recalling the effect of the amino group on electrophilic aromatic substitution reactions. The amino group is not only a powerful activating group, it is also a powerful ortho–para director. Aniline reacts with bromine in aqueous solution at room temperature and in the absence of a catalyst to yield a product in with both ortho positions and the para position are substituted.

The inductive effect of the amino group makes it slightly electron withdrawing. Nitrogen, as we know, is more electronegative than carbon. The difference between the electronegativities of nitrogen and carbon in aniline is not large, however, the carbon of the benzene ring is sp2  hybridized and thus is somewhat more electronegative than it would be if it were sp3  hybridized.

The resonance effect of the amino group is far more important than its inductive effect in electrophilic aromatic substitution, and this resonance effect makes the amino group electron releasing. We can understand this effect if we write the resonance structures for the arenium ions that would arise from ortho, meta, and para attack on aniline.

Exercise 3: Arrange the following in increasing order of monobromination


  1. Theory of Substituent Effects on Electrophilic Aromatic Substitution

We see that four reasonable resonance structures can be written for the arenium ions resulting from ortho and para attack, whereas only three can be written for the arenium ion that results from meta attack. This, in itself, suggests that the ortho- and para-substituted arenium ions should be more stable. Of greater importance, however, are the relatively stable structures that contribute to the hybrid for the ortho- and para-substituted arenium ions. In these structures, nonbonding pairs of electron from nitrogen form an extra bond to the carbon of the ring. This extra bond – and the fact that every atom in each of these structures has a complete outer octet of electrons – makes these structures the most stable of all of the contributors. Because these structures are unusually stable, they make a large – and stabilizing– contribution to the hybrid. This means, of course, that the ortho- and para-substituted arenium ions themselves are considerably more stable than the arenium ion, that results from the meta attack.

Phenol reacts with acetic anhydride, in the presence of sodium acetate, to produce the ester, phenyl acetate.

Phenyl acetate, like phenol is an ortho – para director, (a) What structural feature of phenyl acetate explains this ? (b) Phenyl acetate, although an ortho – para director, is less reactive toward electrophilic aromatic substitution than phenol. Use resonance theory to explain why this is so. (c) Aniline is often so highly reactive toward electrophilic substitution that undesirable reactions take place. One way to avoid these undesirable reactions is to convert aniline to acetanilide (above), by treating aniline with acetyl chloride or acetic anhydride.

What kind of directive effect would you expect the acetamido group (CH3CONH–) to have ?  (d)  Explain why it is much less activating than the amino group, –NH2.

The directive and reactivity effects of halo substituents may, at first, seem to be contradictory. The halo groups are the only ortho – para directors (in Table 2) that are deactivating groups. All other deactivating groups are meta directors. We can readily account for the behavior of halo substituents, however, if we assume that their electron-withdrawing inductive effect influences reactivity and their electron-donating resonance effect governs orientation.

Let us apply these assumptions specifically to chlorobenzene. The chloro atom is highly electronegative. Thus, we would expect a chloro atom to withdraw electrons from the benzene ring and thereby deactivate it.

On the other hand, when electrophilic attack does take place, the chloro group stabilizes the arenium ions resulting from ortho and para attack relative to that from meta attack. The chloro group does this in the same way as amino groups and hydroxyl group do – by donating an unshared pair of electron. These electrons give rise to relatively stable resonance structures contributing to the hybrids for the ortho- para-substituted arenium ions.

We see that four reasonable resonance structures can be written for the arenium ions resulting from ortho and para attack, whereas only three can be written for the arenium ion that results from meta attack. This, in itself, suggests that the ortho- and para-substituted are

What we have said about chlorobenzene is, of course, true of bromobenzene. We can summarize the inductive and resonance effects of halo substituents in the following way. Through their electron-withdrawing inductive effect halo groups make the ring more positive than that of benzene. This causes the free energy of activation for any electrophilic aromatic substitution reaction to be greater than that for benzene, and, therefore, halo groups are deactivating. Through their electron donating resonance effect, however, halo substituents cause the free energies of activation leading to ortho and para substitution to be lower than the free energy of activation leading to meta substitution. This makes halo substituents ortho-para directors.


  1. Summary of Substituent Effects on Orientation and Reactivity

We can summarize the effects that groups have on orientation and reactivity in the following way.

Full or partial (+) charge At least one nonbonding pair on directly attached atom Alkyl or aryl
on directly attached atoms. Halogen –NH2,  –OH, etc.
← meta  directing →  ←–––––– ortho – para directing ––––––→

←––––––  deactivating ––––––→  ←––––––  activating––––––→

  1. Introduction of a third group into the benzene ring

The position takes up by a third gruop entering the ring depends on the nature of the two groups already present.

  1. i) If the two substituents belong to the activating group, the stronger o/p director will control the orientation; nevertheless several tri-substituted isomers will form. The order of decreasing directing powers of the activating groups is.

O > NH2 > NR2 > OH > OMe, >  NHAc > Ar > Me > X

  1. ii) When both groups belong to the deactivating groups then its difficult to introduce a third group. However the order of decreasing directing powers of deactivating groups are

iii) If one substituent belongs, to activating group and other belongs to deactivating group the first group determining the orientation. Further, if the orientations reinforce each other, the third group enters almost entirely one position. The above scheme is purely qualitative.

Exercise 4: Predict the major product if the following compounds are undergoing electrophilic substitution reaction.

12. Reactions of Alkyl Benzenes


12.1 Halogenation of the side chain

The side halogenation of alkyl benzenes takes place in the presence of light or high temperatures. Side chain halogenation is many times carried out by using the reagent N-bromosuccinimide (NBS) in the presence of light.

An alkyl benzene with side chain other than methyl may lead to the formation of more than one products.

Product (I) is the only product obtained, and formation of such product can be attributed to mechanism governing this reaction. Side chain halogenation has a similar mechanism as that of alkanes. It involves the formation of free radical intermediates. Now if we observe the free radicals formed by attack of Bromine free radical on Ethyl benzene, we will find that product (I) involves formation of Benzyl free radical and (II) involves formation of Primary free radical.

The benzyl free radical is more stable than primary free radical because its bond dissociation energy is 19 kcal/mole less than that of primary free radical.

Order of stability of free-radicals.

Benzyl > Allyl > 3° > 2° > 1° (from bond dissociation energy data).

12.2 Oxidation of side chain

Although benzene itself is not susceptible to oxidation but side chain attached to the benzene ring undergoes oxidation and converts itself into a –COOH group. Oxidation of side chain takes place after substance is heated for a long time with KMnO4.

The above reaction can be used for identifying substitution pattern in aromatic compound. If any compound gives phthalic acid on heating with KMnO4, then we can infer that it is ortho disubstituted benzene.

There is no reaction because of the absence of a benzylic hydrogen

13. Aryl halides

Aryl halides are the compounds that contain halogen atom directly attached to the benzene ring. They have general formula ArX.

Any halogen compound that contains a benzene ring is not classified as aryl halide. e.g. Benzyl chloride is not an aryl halide, but is a substituted alkyl halide.

13.1 Preparation methods of Aryl halides


For introducing only the halogen at para position, the Lewis acid thallium acetate is used.

From diazonium salts

13.2 Properties of Aryl halides

Reactivity: Unlike alkyl halides, aryl halides are less reactive towards Nucleophilic substitution reactions, this can be attributed to their electron release via resonance.

Structures III, IV and V stabilise chlorobenzene molecule and give a double bond character to the carbon-chlorine bond. Now because of this the carbon-chlorine bond has more strength and hence aryl halides are more stable towards Nucleophilic substitution reactions. In Alkyl halides the carbon atom attached to halogen is sp3 hybridized and in aryl halides it is sp2 hybridized, as sp2 hybridized carbon is more electronegative it does not permit the chlorine atom to get displaced with the bonded pair of electrons.

Nucleophilic Substitution reactions: Aryl halides undergo Nucleophilic substitution reactions when a strong Electron withdrawing group is present on the benzene ring. Electron withdrawing groups activate the benzene ring towards nucleophilic substitution in aryl halides whereas Electron donating groups deactivate the ring.

Mechanism: Bimolecular displacement mechanism

Any factor that stabilizes the carbanion will increase the rate of Nucleophilic substitution reaction by dispersing the charge present on resonating structures. An electron withdrawing group present at meta position does not activate the ring as much as it does from ortho and para position. This can be known by looking at following resonance structures.

Elimination – addition mechanism/Benzyne Mechanism: In the absence of an electron withdrawing group, nucleophilic substitution takes place in presence of very strong bases, but the mechanism is entirely different from what we have seen in bimolecular nucleophilic substitution reactions. These reaction proceed by a mechanism called benzyne mechanism.


Benzyne is a symmetrical intermediate and can be attacked by nucleophile at both the positions.

Isotopic labelling confirmed that there is an equal chance of abstraction from both carbons. An aryl halide which does not contain alpha hydrogen with respect to halogen does not undergo this reaction. In the reactions involving Benzyne intermediates, two factors affect the position of incoming group, the first one is direction of aryne formation. When there are groups ortho or para to the leaving group, then, the following intermediates should be formed.

when a meta group is present, aryne can form in two ways, In such cases 

more acidic hydrogen is removed, i.e., an electron attracting ‘Z’ favours removal of ortho hydrogen while an electron donating ‘Z’ favours removal of para hydrogen.


  1. Solution to Exercises
Exercise 1: a) Cycloheptatrienyl cation formed by a loss of Br is very stable because it become aromatic.
  1. b) I) Aromatic
  2. II) Anti-ariomatic

III) Aromatic

  1. IV) Aromatic
  2. V) Anti-aromatic
  3. VI) Non-aromatic
Exercise 2: a)
b) Since chlorobenzene does not form unstable phenyl cation.
Exercise 3:


Exercise 4:


  1. Solved Problems

Problem 1: During reaction of benzene with neopentyl chloride with benzene in the presence of AlCl3 the major product is 2-methyl-2-phenylbutane and not neopentylbenzene. Explain


Solution: The initially formed primary carbocation rearranges by methyl migration to form a more stable tertiary carbocation which acts as a electrophile and attack the benzene to give the major product.

Problem 2:

Identify A and also the Electrophile involved in the reaction.


(A) is iso propyl benzene or cumene and electrophile involved is CH3 – CH – CH3 (Isopropyl Carbocation). Rearrangements are also possible in Friedel Crafts, reactions when benzene is treated with n-butyl chloride and Lewis acid the product obtained is isobutyl benzene.

Problem 3: How will you synthesize

from benzene?


If we would have tried to synthesize it using CH3CH2CH2Br then we would have got iso-propylbenzene. Try to explain it.

Problem 4: Arrange the following substances in order of activating a benzene ring.

  1. a) Benzamide, Acetanilide, Aniline
  2. b) Phenylacetate, Acetophenone, Phenol
Solution: a)

In aniline, benzene ring is directly attached –NH2 which releases electrons by resonance effect, where as the activation of –NH2 group can be decreased by converting it to anilide, and in amide, ring is attached to electron withdrawing group which deactivates the ring.


Problem 5: Write the mechanism of the following reaction


Tertiary butyl carbocation is Electrophile, which further attacks benzene ring. Many other reagents that can generate carbocations can be used for Friedel Craft’s reactions. Some of the reagents are

  1. Alkenes in presence of acids like BF3 or Mineral acid (HF).
  2. Alcohols in presence of acids like BF3 or Mineral acid (HF)

Problem 6: Explain Friedel Crafts acylation requires on excess of the catalyst but Friedel-Crafts alkylation requires only a catalytic amount.

Solution: The product of acylation co-ordinates with the catalyst and removes the latter from the reactant side and thereby stops further acylation. For this reason, Friedel-Crafts acylation requires an excess of the catalyst. The product of alkylation does not co-ordinate with the catalyst. So, the catalyst can form complex with the alkylating agent and the catalyst propagates the reaction.

Problem 7: Indicate which positions in each of the following compounds will be preferentially deuterated when 1 mole of DCl is employed for the reaction

  1. i) Anisole; (ii) Benzotrifluoride

Solution: i) – o/p; (ii) –m

Problem 8: Complete the equations and comment:

  1. i) ?
  2. ii) PhCH2CN ?
  1. v) ?
  2. vi) ?
Solution: i)

Two molecules of aluminimum chloride must be used, since the acylating species is acetic anhydride. The formation of the p-product is probably due to steric effects.


The cyano-group directly attached to the benzene ring has a very strong –R effect and so is m-orienting. When ‘isolated’ from the benzene ring, it exerts only a –I effect on the ring. This effect must be weak, otherwise m-substitution would have occurred. At the same time, the CH2 group exerts a hyperconjugative, which favours o/p substitution. Presumably the steric effect largely decides p-substitution rather than o.


Both alkyl groups have a +I effect, Me2CH > Me. Both also have a hyperconugative effect, Me> Me2CH. Since these effects are in opposition, it is not easy to predict the orientation. However, steric effects due to Me2CH are far larger than those due to Me, and this decides (presumably) in favour of substitution ortho to Me. 


The explanation is similar to that given for (iii).

  1. v) Ph(CH2)3Me PhCO2H

NO matter how long the side – chain is, oxidation with strong oxidising reagents results in benzoic acid.

  1. vi) PhNO2 + EtCl no reaction

Problem 9: Suggest oxidizing agents for the conversions 


Solution: Here, the student either known the answer or looks up a source of information. Possible oxidising agents are:


The presence of a –I group ortho to the side-chain usually require the use of alkaline permanganate.

Problem 10: Predict the products in the following reactions.


Problem 11: An aromatic compound of molecular weight 92 (A) on oxidation (Etard’s reaction) gives a compound (B). (B) on reaction with concentrated sodium hydroxide yielded an aromatic acid (C) and an alcohol (D). Acid (C) can be obtained by oxidation of (A)?  Identify A to D and write the reactions.

Solution: A = Toluene B = Benzaldehyde

C = Benzoic acid D = Benzyl alcohol

Problem 12: Complete the following equations

ii) ?

Solution: i) Since the final product is an aldehyde, one precursor could be the 1,1-dichloride; these compounds are readily hydrolysed to carbonyl compounds.

There is also an alternative route.

  1. ii)   Ph2C=CPh2

The use of two molecules of a 1,1-dichloride and copper suggests the formation of an alkene. This is an extension of the Wurtz reaction.


The use of the reagents given clearly indicates that a Grignard reaction is involved. However, aryl chlorides do not form Grignard reagents in ether; THF is necessary as solvent. On the other hand, aryl bromides (and iodides) readily react in ether solution.


The use of aqueous ammonium chlorides is a safety precaution to prevent the possibility of dehydration of the alcohol (which might  occur if acid is used),


The use of iron indicates nuclear substitution. Me is o/p – orienting, and since there is always the possibility of a steric effect at the o-position, as we can suggest is that the 4-Br product will perdominate. Experimental work would have to be done to find out what are the actual results. From the literature it appears that the 4-Br compound is formed exclusively.

  1. Assignments (Subjective Problems)



1. Identify which of the following compounds are aromatic, anti aromatic or non aromatic. Reasons also.
  1. Among pyrrole and pyridine which is more basic and why?
  2. Classify the following substituents as activating or deactivating groups and write the structural formula of mono-substitution products.
  3. a) C6H5CF3, Mono bromination 
  4. b) C6H5CH(CH3)CH2CH3, mono sulfonation 
  5. c) Mononitration, C6H5COOCH3
  6. When toluene is treated with DCl at – 78°C no deuterium is exchanged and the solution does not conduct electricity. What is the structure of the complex?
  7. What happens when toluene reacts with Br2
  8. a) in presence of light
  9. b) in presence of FeBr3
  10. Phenyl group is known to exert –I effect. But each phenyl ring in biphenyl
    (C6H5 – C6H5) is more reactive than benzene towards electrophilic substitution. Why?
  11. Nitrobenzene, but not benzene, is used as a solvent for the Friedel-Craft alkylation of bromo-benzene. Explain.
  12. When aniline is subjected to the Friedel – Crafts alkylation in the presence of catalytic amount of AlCl3 alkylation does not occur. While in the presence of large excess of AlCl3 small amount of m – alkyl aniline is obtained.

In the above reaction which product is not likely to be formed and why?

  1. Predict the product of the following reactions



  1. Compound A (C8H9Cl) is hydrolysed by dilute acids to give compound B (C8H10O) compound B can be oxidised under mild conditions to compound C (C8H8O). Compound (C) upon reduction with Zn/Hg and concentrated HCl gives Ethyl benzene. Identify structures from (A) to (C).
  2. Convert

(ii) PhH ⎯⎯→o-bromotoluene 

  1. The alkylation of phenol and aniline with alkyl halide in presence of AlCl3 gives poor yields. Explain.
  2. Predict the product
  3. What happens when benzene is treated with methyl chloride in presence of anhydrous AlCl3 and the product is treated with excess of chlorine in presence of UV light?
  4. What happens when p-xylene is treated with concentrated sulphuric acid and the resultant product is fused with KOH?

In the above reaction which is the major product and why?

  1. An organic compound (A) C9H12 can be oxidised with KMnO4 to (B) C8H6O4. (B) a dicarboxylic acid, does not form inner anhydride onheating, but gives one mono bromo substitution product (C). Identify A, B and C.
  2. The rates of bromination (FeBr3 catalyst) of C6H6 and C6D6 are the  same. Discuss.

Predict the major and minor product with reason.


  1. The reaction of o-bromoanisole gives only meta-aminoonisole on reaction with KNH2 in liquid NH3. Explain this regioselectivity.
  2. Complete the equations
i) ii)
iii) iv)
  1. Cinnamic acid is ortho-para directing. Explain
  2. Complete the series of reaction

O —HOOC—C6H4—CH2—C6H5 ? ?

  1. Complete and comment:
  2. Write the structures of the compounds from the following data.
  1. Identify the compounds from A to D
  2. Predict the products of the following reactions.
  1. Predict the product in the reactions


9. a)


  1. The aromatic compound (A) C8H7Cl2 Br gives a pale yellow precipitate insoluble in dilute nitric acid, when warmed for a few minutes with alcoholic AgNO3 solution. The compound (A) on reaction with aqueous KOH yields (B) having molecular formula C8H8Cl2O. The compound (B) on reaction with I2 and NaOH a yellow precipitate (C) and a compound (D). Acidification of (D) with conc. HCl yields an organic acid (E) having equivalent weight 191. Heating of (E) with soda-lime gives (F) which forms a single mono-nitro derivative (G) on nitration. Give structures of (A) to (G) with reasons.

C = CHI3


  1. Assignments (Objective Problems)



  1. A deactivating group in electrophilic substitution reaction

(A) Deactivates only ortho and para

(B) Deactivates only meta

(C) Deactivates meta more than ortho and para

(D) Deactivates ortho and para more than meta

  1. Among the following which should be used as solvent in Friedel Craft’s reaction

(A) Toluene (B) Benzeen

(C) Nitrobenzene (D) Phenol

  1. Among the following which have aromatic character
(A) (B)
(C) (D)
  1. Bbasicity order of following heterocycles are 

(A) 1 > 2 > 3 > 4 (B) 4 > 1 > 2 > 3

(C) 3 > 2 > 1 > 4 (D) 4 > 1 >3 > 2

  1. When –N+ (CH3)3 group is attached to the benzene ring, it orients the incoming electrophile into

(A) Meta position

(B) Ortho position

(C)Para position

(D) Equally directs group to ortho-para position

  1. When Benzene is treated with isobutyl bromide in the presence of AlCl3 it mainly gives

(A) Tertiary butyl benzene (B) Isobutyl benzene

(C) n-butyl benzene (D) Secondary butyl benzene

  1. Among the three possible isomers of dibromo benzenes, the highest melting point is possessed by

(A) o-dibromobenzene (B) p-dibromobenzene

(C) m-dibromobenzene (D) Both B & C

  1. Which is true about the rate of nitration of the following compounds

(A) I is faster than II (B) II is faster than I

(C) They follow the same rate (D) cannot be determined

  1. Chlorination of toluene in the presence of light and heat followed by treatment with aqueous NaOH

(A) o-cresol (B) p-cresol

(C) 2,4-dihydroxytoluene (D) benzoic acid

  1. β-phenyl ethyl chloride is the minor product obtained by reaction of chlorine with
(A) (B)
(C) (D)
  1. Terephthalic acid is obtained by oxidation of which of following compounds.
(A) (B)
(C) (D)
(A) (B)
(C) (D)
(A) (B)
(C) (D) No reaction
  1. Benzal chloride on hydrolysis gives

(A) Benzyl alcohol (B) Benzoic acid

(C) Benzaldehyde (D) Benzotrialcohol

  1. In the reaction of p-chlorotoluene with KNH2 in liq NH3, the major product is

(A) o–toluidine (B) p–toluidine

(C) p–chloroaniline (D) m-toluidine



  1. When nitrobenzene is treated with Br2 in presence of FeBr3, the major product formed is m-bromonitrobenzene. Statements which are related to obtain the m-isomer are

(A) The electron denstiy on meta carbon is more than that on ortho and para positions.

(B) The intermediate carbonium ion formed after initial attack of Br+ at the meta position is least destabilised.

(C) Loss of aromaticity when Br+ attacks at the ortho and para positions and not at meta position.

(D) Easier loss of H+ to regain aromaticity from the meta position than from ortho and para positions.

  1. In chlorobenzene, the – Cl group

(A) Activates the benzene ring more via resonance effect than deactivating it via inductive effect.

(B) Deactivates the benzene ring more via inductive effect than activating it via resonance effect.

(C) activates the benzene ring via resonance effect and deactivates it via inductive effect. Both these effects are evenly matched.

(D) Is a net deactivating group with meta director characteristics.

  1. Among the following which statement is not correct

(A) Rates of nitration of C6H6 and C6D6 under same experimental condition are equal.

(B) C – D bond stronger than C – H bond

(C) Sulphonation is a reversible reaction

(D) In sulphonation σ  complex formation is the r/d step

  1. Xylene on treatment with Br2/FeBr3 give only one product. Xylene was

(A) Ortho-xylene (B) Para Xylene

(C) Meta-xylene (D) Cannot be determined

  1. The species which is the attacking reagent in Reimer-Tiemann reaction is

(A) Negatively charged nucleophile (B) Positively charged electrophile

(C) Free-radical (D) Neutral electrophile

6. then the major product A is
(A) (B)
(C) (D)

7 –CH2Cl group is an example of

(A) Strongly deactivating group (B) Strongly activating group

(C) Weakly activating group (D) Weakly deactivating group

  1. Which of the following is most reactive towards aqueous HBr

(A) 1–phenyl–1–propanol (B) 1–phenyl–2–propanol

(C) 3–phenyl–1–propanol (D) None

  1. The order of the basicity in the following compounds is

(A) IV > I > III > II (B) III > I > IV > II

(C) I > III > II > IV (D) II > I > III > IV

  1. The correct order of reactivity towards electrophilic substitution is

(A) Benzoic acid > chlorobenzene > benzene > phenol

(B) Benzoic acid > phenol > benzene > chloro benzene

(C) Phenol > benzene > chlorobenzene > benzoic acid

(D) Phenol > chlorobenzene > benzene > benzoic acid

  1. Which of the following represents decreasing order of reactivity of given organic compounds towards nitration. 

Terephthalic acid Toluene p-Toluic acid p-xylene m-xylene

            I     II         III     IV       V

(A) IV > V > II > III > I (B) I > II > III > V > IV

(C) III > I > III > V > IV (D) V > IV > II > III > I

  1. The product obtained by reduction of Benzyl bromide with LiAIH4 is
(A) (B)
(C) (D)
  1. Cinnamic acid towards electrophilic aromatic substitution 

(A) metal directing (B) o-p directing

(C) Ipso substitution takes place (D) Equally give ortho, para and meta substituted product

(A) (B)
(C) No reaction (D)
  1. Arrange the following groups in decreasing activating order

– NR2 – Me NHAc –OH


(A) I > IV > III > II (B) II > IV > III > I

(C) III > II > IV > I (C) IV > I > III > II

  1. Answers to Objective Assignments




  1. D 2. C
  2. C 4. B
  3. A 6. A
  4. B 8. C
  5. D 10. B
  6. C 12. D
  7. A 14. B
  8. A
  1. A 2. B
  2. D 4. B
  3. D 6. A
  4. C 8. A
  5. A 10. C
  6. D 12. C
  7. B 14. B
  8. A