Reaction of R-CONH_2 with a mixture of Br_2 a | vedclass

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(C) The reaction described is the $Hofmann$ bromamide degradation reaction. The mechanism involves the following steps: $1$. Formation of $N$-bromamid

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$A, C$

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(C) The reaction described is the $Hofmann$ bromamide degradation reaction. The mechanism involves the following steps: $1$. Formation of $N$-bromamide: $R-CONH_2 + Br_2 + KOH \rightarrow R-CONHBr + KBr + H_2O$. $2$. Formation of an acyl nitrene or direct rearrangement: The $N$-bromamide reacts with base to form an anion,which loses $Br^-$ to form an acyl nitrene,which then rearranges to form an isocyanate intermediate: $R-CONHBr + KOH \rightarrow R-N=C=O + KBr + H_2O$. $3$. Hydrolysis of isocyanate: $R-N=C=O + 2KOH \rightarrow R-NH_2 + K_2CO_3$. Thus,the intermediates involved are $R-CONHBr$ $(A)$ and $R-N=C=O$ $(C)$.

Reaction of $R-CONH_2$ with a mixture of $Br_2$ and $KOH$ gives $R-NH_2$ as the main product. The intermediates involved in this reaction are :-
$(A)$ $R-CONHBr$
$(B)$ $R-NH-Br$
$(C)$ $R-N=C=O$
$(D)$ $R-CONBr_2$

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Reaction of $R-CONH_2$ with a mixture of $Br_2$ and $KOH$ gives $R-NH_2$ as the main product. The intermediates involved in this reaction are :-$(A)$ $R-CONHBr$$(B)$ $R-NH-Br$$(C)$ $R-N=C=O$$(D)$ $R-CONBr_2$