CH_3-C(=O)-NH_2 xrightarrow[Delta]{Br_2 + KOH | vedclass

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Question

(C) The reaction is the Hofmann bromamide degradation reaction.In the first step,the amide reacts with $Br_2$ and $KOH$ to form an $N$-bromamide inter

Vedclass · Accepted Answer

$CH_3-C(=O)-NH^{-}Br$ and $CH_3-N=C=O$

Vedclass · Answer

(C) The reaction is the Hofmann bromamide degradation reaction.In the first step,the amide reacts with $Br_2$ and $KOH$ to form an $N$-bromamide intermediate,$CH_3-C(=O)-NHBr$,which then loses a proton to form $CH_3-C(=O)-NH^{-}Br$.Subsequently,this undergoes rearrangement to form methyl isocyanate $(CH_3-N=C=O)$ as a key intermediate.Therefore,the correct intermediates are $CH_3-C(=O)-NH^{-}Br$ and $CH_3-N=C=O$.

$CH_3-C(=O)-NH_2 \xrightarrow[\Delta]{Br_2 + KOH} CH_3-NH_2$. Intermediates of the reaction is/are:

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