C_4H_7OCl (x) xrightarrow{NH_3} C_4H_9NO xrig | vedclass

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Question

(A) The reaction sequence is: $C_4H_7OCl$ $(x)$ $\xrightarrow{NH_3} C_4H_9NO$ (amide) $\xrightarrow{Br_2 + KOH} CH_3-CH_2-CH_2-NH_2$ (propylamine).Thi

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Butanoyl chloride $(CH_3CH_2CH_2COCl)$

Vedclass · Answer

(A) The reaction sequence is: $C_4H_7OCl$ $(x)$ $\xrightarrow{NH_3} C_4H_9NO$ (amide) $\xrightarrow{Br_2 + KOH} CH_3-CH_2-CH_2-NH_2$ (propylamine).This is a Hofmann bromamide degradation reaction,which converts an amide $(R-CONH_2)$ into a primary amine $(R-NH_2)$ with one carbon atom less.Since the product is propylamine $(CH_3CH_2CH_2NH_2)$,the amide must be butanamide $(CH_3CH_2CH_2CONH_2)$.Butanamide is formed by the reaction of butanoyl chloride $(CH_3CH_2CH_2COCl)$ with ammonia $(NH_3)$.Therefore,compound $x$ is butanoyl chloride.

$C_4H_7OCl$ $(x)$ $\xrightarrow{NH_3} C_4H_9NO$ $\xrightarrow{Br_2 + KOH} CH_3-CH_2-CH_2-NH_2$. Compound $x$ is:

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