When propanamide reacts with $B r_{2}$ and NaOH then which of the following compound is formed

Ethyl alcohol

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Propyl alcohol

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Propyl amine

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Ethylamine

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Solution

The correct option is D Ethylamine
$C H_{3} C H_{2} C O N H_{2} B r_{2} / K O H - ------------------ \to H o f m a n n b r o m a r r i d e r e a c t i o n C H_{3} C H_{2} N H_{2} E t h y l a m i n e$

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