An organic compound A upon reacting with NH_3 | vedclass

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Question

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

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$CH_3CH_2COOH$

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(D) The reaction sequence is as follows:$CH_3CH_2COOH (A)$ $\xrightarrow{NH_3} CH_3CH_2COONH_4 (B)$ $\xrightarrow{\Delta} CH_3CH_2CONH_2 (C)$ $\xrightarrow{Br_2/KOH} CH_3CH_2NH_2$The final step is the Hoffmann Bromamide degradation reaction,where an amide $(C)$ is converted into an amine with one less carbon atom.Since the product is ethylamine $(CH_3CH_2NH_2)$,the amide $C$ must be propanamide $(CH_3CH_2CONH_2)$.This amide is formed by heating the ammonium salt of propanoic acid.Therefore,compound $A$ is propanoic acid $(CH_3CH_2COOH)$.

An organic compound $A$ upon reacting with $NH_3$ gives $B$. On heating $B$ gives $C$. $C$ in presence of $KOH$ reacts with $Br_2$ to give $CH_3CH_2NH_2$. $A$ is:

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