See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: LiAlH4 is a strong reducing agent that reduces carbonyl groups (C=O) to CH2 groups when part of an amide or imide linkage, converting C=O into CH2-OH... wait, more precisely, LiAlH4 reduces amide C=O to C-N (amine), not to alcohol. For an imide (cyclic or not), LiAlH4 reduces each C=O of the imide to CH2, giving a cyclic amine. Step 1: Identify the starting material. The starting material is 1-phenylpiperidine-2,6-dione (N-phenyl glutarimide). It is a six-membered ring containing nitrogen with two flanking carbonyl groups (at C-2 and C-6), and the nitrogen bears a phenyl group. Step 2: Action of excess LiAlH4 on the imide. LiAlH4 reduces amide/imide C=O bonds to C-H2 (methylene), converting the N-C(=O) bonds into N-CH2 bonds. Each carbonyl carbon in the imide ring loses its oxygen and becomes a methylene (CH2) group. Step 3: Result of reduction. Both carbonyl groups (C2=O and C6=O) are reduced to CH2 groups. The ring remains intact. The nitrogen still bears the phenyl group. The product is therefore 1-phenylpiperidine: a piperidine ring (six-membered ring with one nitrogen) with a phenyl group on the nitrogen, i.e., N-phenylpiperidine. Step 4: Why other options fail. - Option (a): An open-chain diol would require ring opening, which does not occur here under these conditions. - Option (b): Aniline would require cleavage of the C-N bonds of the ring and loss of the carbon skeleton, which does not happen with LiAlH4 reduction of an imide. - Option (d): Hemi-reduction to give diol (keeping OH groups) is characteristic of NaBH4 or mild reducing agents acting on ketones, not LiAlH4 acting on amides/imides; LiAlH4 fully reduces amide C=O to give amine (C-N), not C-OH. Step 5: Conclusion. The excess LiAlH4 fully reduces both imide carbonyls to methylenes, giving the cyclic tertiary amine 1-phenylpiperidine. Therefore, the correct answer is C.