1. Tardigrade
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  3. Chemistry
  4. Reaction CH 3 CONH 2 stackrel NaOBr longrightarrow gives:

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Q. Reaction $CH _{3} CONH _{2} \stackrel{ NaOBr }{\longrightarrow }$ gives:

Amines

Solution:

This is a Hofmann bromamide reaction. The product is a primary amine with one carbon less.

$CH _{3} CONH _{2} \stackrel{ NaOBr }{\longrightarrow } CH _{3} NH _{2}$