Starting from propanoic acid,the following re | vedclass

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Question

(B) The reaction sequence is as follows:$1.$ $CH_3-CH_2-COOH + SOCl_2 \rightarrow CH_3-CH_2-COCl + SO_2 + HCl$ (Formation of acid chloride $[X]$)$2.$

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$CH_3-CH_2-NH_2$

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(B) The reaction sequence is as follows:$1.$ $CH_3-CH_2-COOH + SOCl_2 \rightarrow CH_3-CH_2-COCl + SO_2 + HCl$ (Formation of acid chloride $[X]$)$2.$ $CH_3-CH_2-COCl + NH_3 \xrightarrow{\Delta} CH_3-CH_2-CONH_2 + HCl$ (Formation of amide $[Y]$)$3.$ $CH_3-CH_2-CONH_2 + Br_2 + 4KOH \rightarrow CH_3-CH_2-NH_2 + K_2CO_3 + 2KBr + 2H_2O$ (Hoffmann Bromamide Degradation to form amine $[Z]$)Thus,$[Z]$ is $CH_3-CH_2-NH_2$.

Starting from propanoic acid,the following reactions were carried out: $CH_3-CH_2-COOH$ $\xrightarrow{SOCl_2} [X]$ $\xrightarrow[\Delta]{NH_3} [Y]$ $\xrightarrow[KOH]{Br_2} [Z]$. Product $[Z]$ will be?

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