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Question

(D) The reaction sequence is as follows:$1$. $C_6H_5COOH + NH_3 \xrightarrow{\Delta} C_6H_5CONH_2$ ($P$,benzamide).$2$. $C_6H_5CONH_2 + NaOBr \rightar

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sulphanilic acid

Vedclass · Answer

(D) The reaction sequence is as follows:$1$. $C_6H_5COOH + NH_3 \xrightarrow{\Delta} C_6H_5CONH_2$ ($P$,benzamide).$2$. $C_6H_5CONH_2 + NaOBr ightarrow C_6H_5NH_2$ ($Q$,aniline) (Hofmann bromamide degradation).$3$. $C_6H_5NH_2 + 	ext{conc. } H_2SO_4 \xrightarrow{460 \ K} NH_2-C_6H_4-SO_3H$ ($R$,sulphanilic acid).Thus,$R$ is sulphanilic acid.

$R$ is

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