Step 1: Understanding the Question:
The reaction involves an amide (acetamide) reacting with bromine and aqueous KOH, which is the reagent set for the Hoffmann bromamide degradation.
Step 2: Detailed Explanation:
Hoffmann bromamide degradation converts a primary amide into a primary amine with one fewer carbon atom.
Acetamide (\( \text{CH}_3\text{CONH}_2 \)) contains 2 carbon atoms.
Upon reaction with \( \text{Br}_2/\text{KOH} \), the carbonyl group is removed as carbonate, leaving methylamine (\( \text{CH}_3\text{NH}_2 \)), which contains 1 carbon atom.
\[ \text{CH}_3\text{CONH}_2 + \text{Br}_2 + 4\text{KOH} \xrightarrow{\Delta} \text{CH}_3\text{NH}_2 + 2\text{KBr} + \text{K}_2\text{CO}_3 + 2\text{H}_2\text{O} \]
Step 3: Final Answer:
The product obtained is methylamine (\( \text{CH}_3\text{NH}_2 \)).