Benzamide reacts with Br_2 and KOH gives :- | vedclass

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(C) The reaction of an amide with $Br_2$ and $KOH$ (or $NaOH$) is known as the $Hofmann$ $Bromamide$ $Degradation$ reaction.In this reaction,the amide

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Aniline

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(C) The reaction of an amide with $Br_2$ and $KOH$ (or $NaOH$) is known as the $Hofmann$ $Bromamide$ $Degradation$ reaction.In this reaction,the amide group $(-CONH_2)$ is converted into a primary amine $(-NH_2)$ with the loss of one carbon atom.For $Benzamide$ $(C_6H_5CONH_2)$,the reaction is:$C_6H_5CONH_2 + Br_2 + 4KOH \rightarrow C_6H_5NH_2 + K_2CO_3 + 2KBr + 2H_2O$.The product formed is $Aniline$ $(C_6H_5NH_2)$.

Benzamide reacts with $Br_2$ and $KOH$ gives :-

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