Compound (D) is :CH_3CH_2COOH xrightarrow{SOC | vedclass

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(B) Step $1$: Reaction of propanoic acid with $SOCl_2$ gives propanoyl chloride $(B)$.$CH_3CH_2COOH + SOCl_2 \rightarrow CH_3CH_2COCl + SO_2 + HCl$Ste

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$CH_3CH_2NH_2$

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(B) Step $1$: Reaction of propanoic acid with $SOCl_2$ gives propanoyl chloride $(B)$.$CH_3CH_2COOH + SOCl_2 \rightarrow CH_3CH_2COCl + SO_2 + HCl$Step $2$: Reaction of propanoyl chloride with $NH_3$ gives propanamide $(C)$.$CH_3CH_2COCl + NH_3 \rightarrow CH_3CH_2CONH_2 + HCl$Step $3$: Reaction of propanamide with $Br_2$ and $KOH$ is the Hofmann bromamide degradation reaction,which yields ethanamine $(D)$.$CH_3CH_2CONH_2 + Br_2 + 4KOH \rightarrow CH_3CH_2NH_2 + K_2CO_3 + 2KBr + 2H_2O$Thus,compound $(D)$ is $CH_3CH_2NH_2$.

Compound $(D)$ is :
$CH_3CH_2COOH$ $\xrightarrow{SOCl_2} (B)$ $\xrightarrow[\Delta]{NH_3} (C)$ $\xrightarrow[Br_2]{KOH} (D)$

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Compound $(D)$ is :$CH_3CH_2COOH$ $\xrightarrow{SOCl_2} (B)$ $\xrightarrow[\Delta]{NH_3} (C)$ $\xrightarrow[Br_2]{KOH} (D)$