Q28 / 150 · MHT CET 2025 (19 Apr Shift 2) · Chemistry

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

  • A
    $32 \mathrm{~g} \mathrm{~mol}^{-1}$
  • B
    $14 \mathrm{~g} \mathrm{~mol}^{-1}$
  • Correct answer
    C
    $28 \mathrm{~g} \mathrm{~mol}^{-1}$
  • D
    $30 \mathrm{~g} \mathrm{~mol}^{-1}$

Solution

Hofmann degradation converts primary amides to primary amines with one fewer carbon atom:

$\text{R-CO-NH}_2 \xrightarrow{\text{Br}_2, \text{NaOH}} \text{R-NH}_2 + \text{Na}_2\text{CO}_3 + 2\text{NaBr} + 2\text{H}_2\text{O}$

The carbonyl group ($-\text{CO}-$) is lost during this transformation and appears as $\text{Na}_2\text{CO}_3$.

Molar mass loss corresponds to the carbonyl group: $12 + 16 = 28\ \text{g mol}^{-1}$.

Final answer: $\boxed{\text{C}}$

Hence, the correct answer is option: 'C'.

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