Hoffmann bromamide degradation of benzamide g | vedclass

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Question

(B) Step $1$: Hoffmann bromamide degradation of benzamide $(C_6H_5CONH_2)$ involves the reaction with $Br_2$ and $NaOH$ to form aniline $(C_6H_5NH_2)$

Vedclass · Accepted Answer

$A = 	ext{aniline}, B = 	ext{phenyl isocyanide}$

Vedclass · Answer

(B) Step $1$: Hoffmann bromamide degradation of benzamide $(C_6H_5CONH_2)$ involves the reaction with $Br_2$ and $NaOH$ to form aniline $(C_6H_5NH_2)$ as product $A$.Step $2$: Aniline $(C_6H_5NH_2)$ is a primary amine. When heated with $CHCl_3$ and $NaOH$ (or $KOH$),it undergoes the carbylamine reaction to form phenyl isocyanide $(C_6H_5NC)$ as product $B$.Thus,$A$ is aniline and $B$ is phenyl isocyanide.

Hoffmann bromamide degradation of benzamide gives product $A$,which upon heating with $CHCl_{3}$ and $NaOH$ gives product $B$. The structures of $A$ and $B$ are

Similar Questions

$C_6H_5NH_2 \xrightarrow{H_2SO_4, 180^{\circ}C} H_2NC_6H_4(SO_3H)$ (para). Which statement is correct for the product?

Explain the chemical reactions of benzene-diazonium chloride.

The increasing order of $pK_b$ for the following compounds will be:
$(A)$ $NH_2-CH=NH$
$(B)$ $1,5,7-Triazabicyclo[4.4.0]dec-5-ene$
$(C)$ $(CH_3)_2NH$

The sequence of reagents required to convert aniline to benzene nitrile is:

What is the product of the following reaction?

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