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Question: Starting from propanoic acid, the following reactions were carried out \[\text{Propanoic acid }\ove...

Starting from propanoic acid, the following reactions were carried out

Propanoic acid →SOCl2X→NH3Y→Br2+KOHZ\text{Propanoic acid }\overset{\quad SOCl_{2}\quad}{\rightarrow}X\overset{\quad NH_{3}\quad}{\rightarrow}Y\overset{\quad Br_{2} + KOH\quad}{\rightarrow}Z

What is the compound Z

A

CH3−CH2−BrCH_{3} - CH_{2} - Br

B

CH3−CH2−NH2CH_{3} - CH_{2} - NH_{2}

C
D

CH3−CH2−CH2−NH2CH_{3} - CH_{2} - CH_{2} - NH_{2}

Answer

CH3−CH2−NH2CH_{3} - CH_{2} - NH_{2}

Explanation

Solution

CH3CH2COOH→SOCl2CH3CH2COCl+SO2+HClCH_{3}CH_{2}COOH\overset{\quad SOCl_{2}\quad}{\rightarrow}CH_{3}CH_{2}COCl + SO_{2} + HCl

CH3CH2COCl+NH3→CH3CH2CONH2+HClCH_{3}CH_{2}COCl + NH_{3} \rightarrow CH_{3}CH_{2}CONH_{2} + HCl

CH3CH2CONH2+Br2/NaOH→CH3CH2NH2Ethylamine+CO2CH_{3}CH_{2}CONH_{2} + Br_{2}/NaOH \rightarrow \underset{\text{Ethylamine}}{CH_{3}CH_{2}NH_{2}} + CO_{2}