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JEE MAIN 2026 · Organic Compounds Containing Nitrogen

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 68

Updated
Single correct+4 / -1Answer matches NTA’s final key

Which statements are True?

A. In Hoffmann bromamide degradation, 4 moles of NaOH and 2 moles of Br2_2 are consumed per mole of an amide

B. Hoffmann bromamide reaction is not given by alkyl amides.

C. Primary amines can be synthesized by Hoffmann bromamide degradation.

D. Secondary amide on reaction with Br2_2 and NaOH will give secondary amine.

E. The by-products of Hoffmann degradation are Na2_2CO3_3, NaBr and H2_2O.

Choose the correct answer from the options given below:

Choose one

Solution

Solution

The overall reaction is

RCONH2+Br2+4 NaOH⟶RNH2+Na2CO3+2 NaBr+2 H2O.\text{RCONH}_2 + \text{Br}_2 + 4\,\text{NaOH} \longrightarrow \text{RNH}_2 + \text{Na}_2\text{CO}_3 + 2\,\text{NaBr} + 2\,\text{H}_2\text{O}.

  • A, false. One mole of Br2_2, not two.
  • B, false. Alkyl amides (e.g. acetamide → methylamine) react normally.
  • C, true. It gives a primary amine with one carbon fewer.
  • D, false. The reaction needs two hydrogens on the amide nitrogen (an N-bromo amide that can lose HBr to form the nitrene/isocyanate); a secondary amide cannot do this.
  • E, true. By-products are Na2_2CO3_3, NaBr and water.

Answer: C and E only.

Answer checked against the official answer key. The solution is our own working.

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