In the conversion of $(C)$ to $(E)$ , $(E)$ is $B u - {N H}_{2}$ . This reaction is called:

haloform reaction

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hofmann reaction

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hofmann bromamide reaction

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hofmann bromamide rearrangement reaction

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Solution

The correct option is D hofmann bromamide rearrangement reaction
The reaction is Hofmann bromamide rearrangement reaction. It is also called Hofmann degradation or Hoffmann bromamide reaction. The cyanide ion attacks compound $A$ to from a nitrile which on partial hydrolysis forms amide (compound C). Compound $C$ reacts with $N a O B r$ to from butyl amine (compound E)
$C H_{3} C H_{2} C H_{2} C H_{2} C l C N^{-} - -- \to - C l^{-} C H_{3} C H_{2} C H_{2} C H_{2} C N partial hydrolysis - --------- \to$
$C H_{3} C H_{2} C H_{2} C H_{2} C O N H_{2} N a O B r - --- \to C H_{3} C H_{2} C H_{2} C H_{2} N H_{2}$
In this reaction, the product amine contains one $C$ atom less than the substrate amide.

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Similar questions

R - O | | C - N H_{2} + x N a O H + B r_{2} ⟶ R - N H_{2} + 2 N a B r + N a_{2} C O_{3} + H_{2} O

N a O H

We can obtain ethylamine by Hoffmann bromamide reaction. The amide used in this reaction is:

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