Benzamide xrightarrow{Br_2 / NaOH} X xrightar | vedclass

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Question

(D) $1$. The first step,$Benzamide \xrightarrow{Br_2 / NaOH} X$,is the Hoffmann bromamide degradation reaction,where $X$ is $Aniline$ $(C_6H_5NH_2)$.$

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Carbylamine reaction

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(D) $1$. The first step,$Benzamide \xrightarrow{Br_2 / NaOH} X$,is the Hoffmann bromamide degradation reaction,where $X$ is $Aniline$ $(C_6H_5NH_2)$.$2$. The second step,$X \xrightarrow[	ext{alc. } KOH]{CHCl_3} Y$,involves the reaction of a primary amine $(Aniline)$ with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$.$3$. This specific reaction is known as the Carbylamine reaction,which produces an isocyanide ($Y = Phenyl \ isocyanide$ or $C_6H_5NC$).$4$. Therefore,the conversion of $X$ to $Y$ is the Carbylamine reaction.

$Benzamide$ $\xrightarrow{Br_2 / NaOH} X$ $\xrightarrow[\text{alc. } KOH]{CHCl_3} Y$
The conversion of $X$ to $Y$ is

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$Benzamide$ $\xrightarrow{Br_2 / NaOH} X$ $\xrightarrow[\text{alc. } KOH]{CHCl_3} Y$The conversion of $X$ to $Y$ is