Q22 · CSIR-NET Chemistry, June 2015

Paper: CSIR-NET June 2015 · Subject: Organic Chemistry · Chapter: Reagents, Reaction Mechanisms & Named Reactions · Topic: Reduction Reagents · Marks: 2 · Difficulty: Medium

The structures of A and B in the following reaction sequence are Question structure 1 of 9, ABC26OR0262 Question structure 2 of 9, ABC26OR0262 Question structure 3 of 9, ABC26OR0262 Question structure 4 of 9, ABC26OR0262 Question structure 5 of 9, ABC26OR0262 Question structure 6 of 9, ABC26OR0262 Question structure 7 of 9, ABC26OR0262 Question structure 8 of 9, ABC26OR0262 Question structure 9 of 9, ABC26OR0262
(a)Option (a) structure 1 of 3, ABC26OR0262 Option (a) structure 2 of 3, ABC26OR0262 B = Option (a) structure 3 of 3, ABC26OR0262
(b)Option (b) structure 1 of 3, ABC26OR0262 Option (b) structure 2 of 3, ABC26OR0262 B = Option (b) structure 3 of 3, ABC26OR0262
(c)Option (c) structure 1 of 3, ABC26OR0262 Option (c) structure 2 of 3, ABC26OR0262 B = Option (c) structure 3 of 3, ABC26OR0262
(d)A = Option (d) structure 1 of 2, ABC26OR0262 B = Option (d) structure 2 of 2, ABC26OR0262
Answer
Answer: B ✓ checked by 4AB · confidence medium

Key B kept. The explanation was rewritten because the N-acylimine is the dienophile; it does not form a 2-azadiene. The text also now says why (d) is excluded.

Explanation
Ph₃P=CHCO₂Me, a stabilised ylide, converts the aldehyde into the (E,E)-dienoate. On heating, the N-(acetoxymethyl)amide loses AcOH to give an N-acylimine (CH₂=N–C=O). This acts as the dienophile in an intramolecular aza-Diels–Alder reaction with the dienoate, giving indolizidinone A, a lactam that carries the CO₂Me group and the ring C=C. LiAlH₄ then reduces both the lactam C=O, giving the tertiary amine, and the ester, giving CH₂OH, which yields amino alcohol B. Option (d) is excluded because its B still contains the lactam C=O, which LiAlH₄ would reduce. Options (a) and (c) do not form the bicyclic ring. Answer: (b).

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