Question

In the reaction of \( NaOBr \) with amide, the carbonyl carbon is lost as :

• A. \( HCO_{3}^{-} \)
• B. \( CO_{3}^{2-} \)
• C. \( CO_{2} \)
• D. \( CO \)

Hint:

Hoffmann "Degradation" literally means stepping down the carbon chain.
If you start with propanamide (\( C_{3} \)), you get ethanamine (\( C_{2} \)). The "lost" carbon is always in the carbonate byproduct.

Answer 1

The Correct Option is B

Step 1: Identify the reaction being described.
Treating an amide $R-CONH_2$ with $Br_2$ and excess $NaOH$, which together generate $NaOBr$ in situ, is the Hoffmann bromamide degradation, and it shortens the amide to a primary amine with one carbon fewer.
Step 2: Write the overall balanced change.
\[ R-CONH_2 + Br_2 + 4NaOH \rightarrow R-NH_2 + Na_2CO_3 + 2NaBr + 2H_2O \]
Step 3: Follow the carbonyl carbon through the mechanism.
The reaction proceeds through an isocyanate intermediate $R-N=C=O$, and in the final step this carbon is expelled from the organic fragment entirely, reacting with the leftover hydroxide ions in solution rather than staying attached to $R$.
Step 4: Identify what that expelled carbon becomes.
The carbon combines with hydroxide to form sodium carbonate, which in solution exists as $Na^+$ and $CO_3^{2-}$ ions. \[ \boxed{CO_3^{2-}} \]