$A c e t a m i d e$ is treated with the following reagents separately. Which one of these would yield $m e t h y l a m i n e ?$

N a O H - B r_{2}

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S o d a l i m e

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H o t c o n c .

H_{2} S O_{4}

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P C l_{5}

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Solution

The correct option is C $N a O H - B r_{2}$
Preparation of amines by reduction of nitro compounds can be done either by catalytic hydrogenation or a metal (tin,neon,zinc) and an acid(HCl) can be used to carry out teh reduction and teh product will be acidic, if acidic conditions are employed which can be converted into basic by adding a base.

$+ H_{2} P d / C - ------- \to$

$S n, H C l - --------- \to$ $O^{⊖} H - ------ \to$
$(N H_{4})_{2} S - --------- \to$

Acetamide is treated with $N a O H + {B r}_{2}$ to give methyl amine and reaction is called hoffmann bromamide reaction.

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Similar questions

If acetamide is treated with the following reagents separately, which one of these would yield methylamine?

Acetamide is treated separately with the following reagents. Which would give methyl amine

[IIT 1983; CPMT 1988, 94; MP PET 1993;

MP PMT 1996; AIIMS 1998]

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