(X) C_4H_7OCl xrightarrow{NH_3} C_4H_9ON xrig | vedclass

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Question

(A) The reaction sequence is as follows:$1$. The final product is $CH_3CH_2CH_2NH_2$ (propan$-1-$amine),which is formed by the Hofmann bromamide degra

Vedclass · Accepted Answer

Butanoyl chloride $(CH_3CH_2CH_2COCl)$

Vedclass · Answer

(A) The reaction sequence is as follows:$1$. The final product is $CH_3CH_2CH_2NH_2$ (propan$-1-$amine),which is formed by the Hofmann bromamide degradation reaction.$2$. The Hofmann bromamide degradation reaction converts an amide $(RCONH_2)$ into a primary amine $(RNH_2)$ with one carbon atom less.$3$. Therefore,the intermediate amide must be butanamide $(CH_3CH_2CH_2CONH_2)$.$4$. The amide is formed by the reaction of an acid chloride $(RCOCl)$ with ammonia $(NH_3)$.$5$. Thus,the compound $(X)$ is butanoyl chloride $(CH_3CH_2CH_2COCl)$.

$(X) C_4H_7OCl$ $\xrightarrow{NH_3} C_4H_9ON$ $\xrightarrow[KOH]{Br_2} CH_3CH_2CH_2NH_2;$ Compound $(X)$ is

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