See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: This is a sequence involving oxime formation, Beckmann rearrangement, and LAH reduction. Step 2 - Formation of A (oxime): Cyclopentanone reacts with hydroxylamine (NH2OH) to give cyclopentanone oxime (A). The C=O is converted to C=N-OH. Step 3 - Formation of B (Beckmann rearrangement): Treatment of cyclopentanone oxime with H+ (acid catalyst) causes a Beckmann rearrangement. In a Beckmann rearrangement, the oxime of a cyclic ketone undergoes ring expansion to give a lactam. Cyclopentanone oxime (5-membered ring oxime) rearranges to give a 6-membered lactam, specifically delta-valerolactam (2-piperidinone), which is compound B. Step 4 - Formation of C (LAH reduction): LAH (lithium aluminium hydride) is a strong reducing agent that reduces amides/lactams to amines. Reduction of 2-piperidinone (a cyclic lactam, 6-membered ring) with LAH reduces the C=O of the amide to CH2, giving piperidine (fully saturated six-membered ring with one NH). This is compound C. Step 5 - Why other options fail: - Option (a) 4-hydroxypiperidine: No source of OH at C4; LAH would not introduce a hydroxyl group in this position from a lactam. - Option (c) 2-piperidinone: This is intermediate B, not the final product C after LAH reduction. - Option (d) 3-hydroxypiperidine: No mechanism accounts for OH at C3 in this sequence. Option (b) piperidine is the correct product because LAH fully reduces the lactam (2-piperidinone) to piperidine. Therefore, the correct answer is B.