CH_{3}CH_{2}Br xrightarrow[Delta]{aq. KOH} A | vedclass

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Question

(C) Step $1$: $CH_{3}CH_{2}Br$ reacts with $aq. KOH$ (nucleophilic substitution) to form ethanol $(A = CH_{3}CH_{2}OH)$.Step $2$: Ethanol $(CH_{3}CH_{

Vedclass · Accepted Answer

$CH_{3}NH_{2}$

Vedclass · Answer

(C) Step $1$: $CH_{3}CH_{2}Br$ reacts with $aq. KOH$ (nucleophilic substitution) to form ethanol $(A = CH_{3}CH_{2}OH)$.Step $2$: Ethanol $(CH_{3}CH_{2}OH)$ is oxidized by $KMnO_{4} / H^{+}$ to form acetic acid $(B = CH_{3}COOH)$.Step $3$: Acetic acid reacts with $NH_{3}$ followed by heating to form acetamide $(C = CH_{3}CONH_{2})$.Step $4$: Acetamide reacts with $Br_{2}$ in the presence of alkali (Hofmann bromamide degradation reaction) to form methylamine $(D = CH_{3}NH_{2})$.Thus,the final product '$D$' is $CH_{3}NH_{2}$.

$CH_{3}CH_{2}Br$ $\xrightarrow[\Delta]{aq. KOH} A$ $\xrightarrow[\Delta]{KMnO_{4} / H^{+}} B$ $\xrightarrow[\Delta]{NH_{3}} C$ $\xrightarrow[alkali]{Br_{2}} D$. '$D$' is

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