Question

${ C }_{ 6 }{ H }_{ 5 }COOH \xrightarrow [ 1. N{ H }_{ 3 } ]{ 2. d \Delta } P \xrightarrow [ NaOBr ]{ } Q \xrightarrow [ 1. Conc. { H }_{ 2 }S{ O }_{ 4 } ]{ 2. Heat to 460K } R$
R is:
A. sulphanilamide
B. p-bromo sulphanilamide
C. o-bromo sulphanilic acid
D. sulphanilic acid

Answer 1

Hint: The reaction in the question involves the conversion of a carboxylic acid into an amide. Following this, Hoffmann bromamide reaction takes place in which the amide converts into an amine.

Complete step by step answer: Let us now look at what is happening in the given reaction step by step.
Firstly, benzoic acid is heated in the presence of ammonia. As a result, it produces the product P. When benzoic acid is reacted with ammonia, it reacts to give ammonium benzoate. Ammonium benzoate is a salt of ammonia with benzoic acid.

$\underset { Benzoic acid }{ { C }_{ 6 }{ H }_{ 5 }COOH } \xrightarrow [ 1. N{ H }_{ 3 } ]{ 2. \Delta } \underset { Ammonium benzoate }{ { C }_{ 6 }{ H }_{ 5 }CO{ O }^{ - }N{ H }_{ 4 }^{ + } }$

On heating ammonium benzoate again, it loses a molecule of water. As a result, it forms benzamide. Therefore, the product P is benzamide.

$\underset { Ammonium benzoate }{ { C }_{ 6 }{ H }_{ 5 }CO{ O }^{ - }N{ H }_{ 4 }^{ + } } \xrightarrow [ -{ H }_{ 2 }O ]{ Heat } \underset { Benzamide }{ { C }_{ 6 }{ H }_{ 5 }CON{ H }_{ 2 } }$

After this, the benzamide is treated with NaOBr. Here, the reaction that will take place is Hoffmann Bromamide reaction. In this, the primary amide converts into a primary amine which has one less carbon atom than the corresponding amide. This takes place when the amide is heated with ${Br}_{2}$ in the existence of NaOH which is NaOBr. Thus, benzamide reacts with NaOBr to give product aniline as the primary amine. Hence, the product Q is aniline.

$\underset { Benzamide }{ { C }_{ 6 }{ H }_{ 5 }CON{ H }_{ 2 } } \xrightarrow [ NaOBr ]{ } \underset { Aniline }{ { C }_{ 6 }{ H }_{ 5 }N{ H }_{ 2 } }$

After this, aniline is treated with conc. ${H}_{2}S{O}_{4}$ to form a salt of aniline and sulphide. This process is known as sulphonation. Here, aniline reacts with conc. ${H}_{2}S{O}_{4}$ to produce aniline hydrogen sulphide.

$ \underset { Aniline }{ { C }_{ 6 }{ H }_{ 5 }N{ H }_{ 2 } } \xrightarrow [ { H }_{ 2 }S{ O }_{ 4 } ]{ } \underset { Aniline hydrogen sulphide }{ { C }_{ 6 }{ H }_{ 5 }\overset { + }{ N } { H }_{ 3 }HS{ O }_{ 4 }^{ - } }$

This aniline hydrogen sulphide formed is heated at 460K to form sulphanilic acid. Therefore, the product R formed is sulphanilic acid.

$ \underset { Aniline hydrogen sulphide }{ { C }_{ 6 }{ H }_{ 5 }\overset { + }{ N } { H }_{ 3 }HS{ O }_{ 4 }^{ - } } \xrightarrow [ 460K ]{ } \underset { Sulphanilic acid }{ \overset { + }{ N } { H }_{ 3 }{ C }_{ 6 }{ H }_{ 4 }S{ O }_{ 3 }^{ - } }$

$ \underset { Sulphanilic acid }{ \overset { + }{ N } { H }_{ 3 }{ C }_{ 6 }{ H }_{ 4 }S{ O }_{ 3 }^{ - } } \xrightarrow [ Rearranging ]{ } \underset { Sulphanilic acid }{ N{ H }_{ 2 }{ C }_{ 6 }{ H }_{ 4 }S{ O }_{ 3 }H }$

Hence, the correct answer is option (D).

Note: After the sulphonation of aniline, the product formed, that is, aniline hydrogen sulphide, the sulphonate group is present at the para position. This is due to the steric hindrance which is caused when the sulphonate group approaches the ortho position.