An organic compound A on treatment with NH_3 | vedclass

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Question

(D) The reaction sequence is as follows: $CH_3CH_2COOH (A)$ $\xrightarrow{NH_3} CH_3CH_2COONH_4 (B)$ $\xrightarrow{\Delta} CH_3CH_2CONH_2 (C)$ $\xrigh

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$CH_3CH_2COOH$

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(D) The reaction sequence is as follows: $CH_3CH_2COOH (A)$ $\xrightarrow{NH_3} CH_3CH_2COONH_4 (B)$ $\xrightarrow{\Delta} CH_3CH_2CONH_2 (C)$ $\xrightarrow{Br_2/KOH} CH_3CH_2NH_2$ (Ethylamine). The last step is the Hoffmann Bromamide degradation reaction,where an amide is converted into an amine with one less carbon atom. Since the product is ethylamine $(CH_3CH_2NH_2)$,which has $2$ carbon atoms,the amide $(C)$ must have $3$ carbon atoms. Therefore,compound $A$ is propanoic acid $(CH_3CH_2COOH)$.

An organic compound $A$ on treatment with $NH_3$ gives $B$,which on heating gives $C$. $C$ when treated with $Br_2$ in the presence of $KOH$ produces ethyl amine. Compound $A$ is:

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