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Reduction to Amines

Ethanoic acid...

Question

Ethanoic acid on heating with ammonia forms compound $A$ which on treatment with bromine and sodium hydroxide gives compound $B$ . Compound $B$ on treatment with $N a N O_{2} / d i l . H C l$ gives compound $C$ . The compounds $A, B$ and $C$ respectively are:

A

Ethanamide, methanamine, methanol

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B

Propanamide, ethanamine, ethanol

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C

N-ethylpropanamide, methaneisonitrile, methanamine

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D

Ethanamine, bromoethane, ethanediazoniumchloride

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E

Methanamine, ethanamide, methanol

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Solution

The correct option is A Ethanamide, methanamine, methanol
$C H_{3} C O O H + N H_{3} △ - \to C H_{3} C O N H_{2} e t h a n a m i d e^{'} A^{'}$

$C H_{3} C O N H_{2}^{'} A^{'} + B r_{2} + 3 N a O H \to C H_{3} N H_{2} m e t h y l a m i n e (^{'} B^{'})$
$+ N a_{2} C O_{3} + N a B r + H_{2} O$
$C H_{3} N H_{2} H N O_{2} - --------- \to (N a N O_{2} / d i H C l) C H_{3} O H m e t h a n o l^{'} C^{'}$

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