Benzoic acid on treatment with hydrazoic acid | vedclass

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Question

(C) Benzoic acid $(C_6H_5COOH)$ reacts with hydrazoic acid $(HN_3)$ in the presence of concentrated $H_2SO_4$ to form aniline $(C_6H_5NH_2)$.This reac

Vedclass · Accepted Answer

Aniline

Vedclass · Answer

(C) Benzoic acid $(C_6H_5COOH)$ reacts with hydrazoic acid $(HN_3)$ in the presence of concentrated $H_2SO_4$ to form aniline $(C_6H_5NH_2)$.This reaction is known as the Schmidt reaction.The mechanism involves the formation of an acyl azide intermediate,which loses nitrogen $(N_2)$ to form an isocyanate $(C_6H_5N=C=O)$.The isocyanate then undergoes hydrolysis to yield aniline and carbon dioxide $(CO_2)$.The overall reaction is: $C_6H_5COOH + HN_3 \xrightarrow{conc. H_2SO_4} C_6H_5NH_2 + N_2 + CO_2$.Therefore,option $C$ is correct.

Benzoic acid on treatment with hydrazoic acid in the presence of concentrated sulphuric acid gives:

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