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ExamsJEE MainChemistry

Cyclopentanone reacts with hydroxylamine (NH2OH) to first form an oxime (A). The oxime (A) is then treated with acid (H+) to give product B via the Beckmann rearrangement. What is B?

  1. five-membered cyclic amide structure shown in option (A)
  2. six-membered cyclic amide structure shown in option (B)
  3. cyclopentylamine structure shown in option (C)
  4. None of the above

Correct answer: six-membered cyclic amide structure shown in option (B)

Solution

Cyclopentanone reacts with NH2OH to give cyclopentanone oxime (A). Beckmann rearrangement under acid inserts N between the carbonyl carbon and the anti-substituent, ring-expanding the 5-membered carbocyclic ring to a 6-membered lactam (delta-valerolactam / piperidin-2-one), which is a six-membered cyclic amide (B).

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