JEE Advanced Chemistry: Nitrogen-Containing Compounds questions with solutions

Q1. 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

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

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).