Exams › JEE Advanced › Chemistry
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?
- five-membered cyclic amide structure shown in option (A)
- six-membered cyclic amide structure shown in option (B)
- cyclopentylamine structure shown in option (C)
- 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).
Related JEE Advanced Chemistry questions
⚔️ Practice JEE Advanced Chemistry free + battle 1v1 →