Benzamide xrightarrow{Br_2 / NaOH} X xrightar | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

Carbylamine reaction

Vedclass · Answer

(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

Similar Questions

$R-CONH_2 \xrightarrow[\Delta]{KOBr} R-NH_2$. The correct statement about the product is:

Action of $HNO_2$ on $CH_3NH_2$ gives the following as the major product:

$Isopropyl$ amine with an excess of acetyl chloride will give:

The above $(C-N)$ coupling reaction takes place at

The product $(C)$ of the following reaction is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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