Acetamide is treated separately with the following reagents. Which one of these would give methylamine?

{P C l}_{5}

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Sodalime

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N a O H + {B r}_{2}

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H_{2} {S O}_{4}

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Solution

The correct option is C $N a O H + {B r}_{2}$
We need to convert $C H_{3} C O N H_{2}$ to $C H_{3} N H_{2}$ .

First, we observe that the product contains a CO group less than the reactant.

Now we note that to remove a CO group, Hoffmann bromamide reaction is used. In this reaction, we use bromine and NaOH

$C H_{3} C O N H_{2} + B r_{2} + N a O H \to C H_{3} N H_{2}$

So, C is the correct answer.

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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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