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Question

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

A
PCl5
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B
Sodalime
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C
NaOH+Br2
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D
Hot concentrated H2SO4
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Solution

The correct option is C NaOH+Br2
We need to convert CH3CONH2 to CH3NH2.

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

CH3CONH2+Br2+NaOHCH3NH2

So, C is the correct answer.

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