An organic compound A reacts with NH_3 on hea | vedclass

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(D) The reaction of $B$ with $Br_2$ and $KOH$ is the Hofmann bromamide degradation reaction,which converts an amide $(RCONH_2)$ into a primary amine $

Vedclass · Accepted Answer

$CH_3CH_2COOH$

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(D) The reaction of $B$ with $Br_2$ and $KOH$ is the Hofmann bromamide degradation reaction,which converts an amide $(RCONH_2)$ into a primary amine $(RNH_2)$ with one carbon atom less.Given that the product is $CH_3CH_2NH_2$ (ethanamine),the amide $B$ must be $CH_3CH_2CONH_2$ (propanamide).Since $A$ reacts with $NH_3$ on heating to form propanamide $(B)$,$A$ must be propanoic acid $(CH_3CH_2COOH)$.The reaction sequence is: $CH_3CH_2COOH + NH_3$ $\xrightarrow{\Delta} CH_3CH_2CONH_2 (B)$ $\xrightarrow{Br_2/KOH} CH_3CH_2NH_2$.

An organic compound $A$ reacts with $NH_3$ on heating to give $B$. Compound $B$ reacts with $Br_2$ in the presence of $KOH$ to give $CH_3CH_2NH_2$. What is $A$?

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