Question

Chlorobenzene can be prepared by reacting aniline with

• A. hydrochloric acid
• B. cuprous chloride
• C. chlorine in the presence of anhyd $AlCl_3$
• D. nitrous acid followed by heating with cuprous chloride

Answer 1

The Correct Option is D

$C_6H_5NH_2 + HNO_2\rightarrow C_6H_5N_2OH\xrightarrow{CuCl}C_6H_5-Cl$

Answer 2

Let's start by drawing the structure of the reactant. Think about what kind of reactions aniline undergoes with the given reagents in the options. The product formed should not contain any other substituent on the ring except chlorine.

Complete step-by-step solution:

• Aniline is an aromatic amine. It is a base.
• Let’s have a look at the question. We need to prepare chlorobenzene from aniline. That means the amine group needs to be replaced by a chlorine group.
• We have studied that the amine group can only be replaced after it has been converted to diazonium salt form, which is unstable. Direct halogenation of aniline will give rise to halogen derivatives of aniline.
• If HCl is added to aniline, then aniline being a base will abstract the proton and form anilinium chloride salt.
• When aniline is treated with nitrous acid, which is produced in situ by sodium nitrite and hydrochloric acid, at low temperature (0°C−5°C), then aniline is converted to a benzene diazonium salt. This reaction is known as diazotization.
• When a freshly prepared solution of benzene diazonium salt is treated with cuprous chloride (CuCl), it gives chlorobenzene. This reaction is known as Sandmeyer’s reaction.
• Therefore, chlorobenzene can be prepared by reacting aniline with nitrous acid followed by heating with cuprous chloride (CuCl).

Therefore, the correct option is (D).

Note: Chlorobenzene can also be prepared by treating benzene diazonium salt with copper powder and hydrochloric acid. This reaction is known as the Gattermann reaction and has a better yield than the Sandmeyer reaction. Remember in aromatic compounds electrophilic substitution reactions occur readily and therefore, direct halogenation will yield halo-substituted compounds.

Answer 3

Let's start the solution by drawing the structure of the reactant. Recall what type of reactions aniline undergoes with the given reagents in the options. Except for the chlorine atom, the product formed must not contain any substituent on the benzene ring.

Complete answer:

• Aniline is a base that is an aromatic amine.
• The question says that aniline is reacted with a reagent and gives chlorobenzene. This means that the amine group of the aniline is to be replaced on reaction with the reagent with a chlorine atom.
• Recall the knowledge that amine group can only be replaced after it has been converted to diazonium salt form which is an unstable compound and direct halogenation of aniline gives halogen derivatives of aniline.
• When hydrochloric acid is added to aniline, the aniline being a base will abstract the proton forming anilinium chloride salt.
• Nitrous acid produced in situ by sodium nitrite and hydrochloric acid on reaction with aniline at low temperature (0°C−5°C) converts aniline to a benzene diazonium salt and this reaction is known as diazotization.
• When a freshly prepared solution of benzene diazonium salt is treated with cuprous oxide, chlorobenzene is obtained. This reaction is known as Sandmeyer’s reaction.
• Therefore, in the conclusion, we can say that chlorobenzene can be prepared by reacting aniline with nitrous acid followed by heating with cuprous chloride (CuCl).

Hence, the correct answer is option D.

Note: Chlorobenzene can also be prepared by treating diazonium salt with hydrochloric acid and copper powder. This reaction is known as the Gattermann reaction and gives a better yield than the Sandmeyer reaction. Always remember that aromatic compounds readily undergo electrophilic substitution reactions, therefore direct halogenations will yield halo-substituted compounds.