Starting from propanoic acid,the following re | vedclass

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Question

(B) The reaction sequence is as follows:$1.$ Propanoic acid $(CH_3-CH_2-COOH)$ reacts with $SOCl_2$ to form propanoyl chloride $(X = CH_3-CH_2-COCl)$.

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$CH_3-CH_2-NH_2$

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(B) The reaction sequence is as follows:$1.$ Propanoic acid $(CH_3-CH_2-COOH)$ reacts with $SOCl_2$ to form propanoyl chloride $(X = CH_3-CH_2-COCl)$.$2.$ Propanoyl chloride reacts with $NH_3$ to form propanamide $(Y = CH_3-CH_2-CONH_2)$.$3.$ Propanamide undergoes Hoffmann bromamide degradation with $Br_2 + KOH$ to form ethylamine $(Z = CH_3-CH_2-NH_2)$.The overall reaction is: $CH_3-CH_2-COOH$ $\xrightarrow{SOCl_2} CH_3-CH_2-COCl$ $\xrightarrow{NH_3} CH_3-CH_2-CONH_2$ $\xrightarrow{Br_2 + KOH} CH_3-CH_2-NH_2$.

Starting from propanoic acid,the following reactions were carried out: $CH_3-CH_2-COOH$ $\xrightarrow{SOCl_2} X$ $\xrightarrow{NH_3} Y$ $\xrightarrow{Br_2 + KOH} Z$. What is the compound $Z$?

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