Question:medium

What is the loss in molar mass when a primary amine is obtained by Hofmann degradation of amide?

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Hofmann degradation "degrades" the chain by one carbon atom ($C=O$).
Updated On: May 14, 2026
  • $32\text{ g mol}^{-1}$
  • $14\text{ g mol}^{-1}$
  • $28\text{ g mol}^{-1}$
  • $30\text{ g mol}^{-1}$
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
Hofmann degradation (bromamide reaction) converts a primary amide into a primary amine with one fewer carbon atom.
Step 2: Key Formula or Approach:
$\text{R-CO-NH}_2 + \text{Br}_2 + 4\text{NaOH} \rightarrow \text{R-NH}_2 + \text{Na}_2\text{CO}_3 + 2\text{NaBr} + 2\text{H}_2\text{O}$
Step 3: Detailed Explanation:
The starting material is an amide: $\text{R-CO-NH}_2$.
The product is an amine: $\text{R-NH}_2$.
The group lost in the process is the carbonyl group ($\text{C}=\text{O}$).
Molar mass of Carbon ($\text{C}$) $= 12\text{ g/mol}$
Molar mass of Oxygen ($\text{O}$) $= 16\text{ g/mol}$
Total mass lost $= 12 + 16 = 28\text{ g/mol}$.
Step 4: Final Answer:
The loss in molar mass is $28\text{ g mol}^{-1}$.
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