Question

Acetamide reacts with $Br_{2}$ and aqueous KOH to form :

• A. Ethanamine
• B. Ammonia
• C. Methanamine
• D. Ethanenitrile

Hint:

Hoffmann bromamide degradation is a "step-down" reaction: The final Amine always has 1 carbon less than the original Amide.

Answer 1

The Correct Option is C

Step 1: Name the reaction.
The treatment of a primary amide with $Br_2$ and aqueous KOH is the Hofmann bromamide degradation. It converts a primary amide into a primary amine with one fewer carbon atom than the original amide.
Step 2: Identify the starting material and carbon count.
Acetamide is $CH_3CONH_2$, containing 2 carbon atoms. One carbon is in the methyl group ($CH_3$-) and one is in the carbonyl group ($-CO$-).
Step 3: Trace the mechanism.
Bromine reacts with the amide nitrogen to form an N-bromoamide. Base then forms an anion that rearranges, migrating the $-CH_3$ group from C to N via an isocyanate intermediate ($CH_3-N=C=O$). Aqueous base immediately hydrolyses the isocyanate, releasing the carbonyl carbon as $CO_3^{2-}$.
Step 4: Identify the product.
After loss of the carbonyl carbon, the remaining fragment is $CH_3-NH_2$ (one carbon), which is methanamine (methylamine).
\[ \boxed{\text{Methanamine}} \]