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The major products X and Y in the following reaction sequence are

N-(2-methoxy-5-methylbenzoyl)pyrrolidine with 1. Na, NH3 (liquid), THF; 2. allyl bromide, giving X; then 1. aqueous HCl; 2. 1,2-dichloro benzene, reflux, giving Y

  • A.X is a methyl methoxy cyclohexadiene with the allyl and the pyrrolidine amide on one quaternary carbon; Y is a 4-allyl-4-methylcyclohexenone bearing the pyrrolidine amide
  • B.X is a cyclohexadiene with a quaternary carbon carrying methyl and allyl and an OCH3 on an sp3 carbon; Y is a 6-allyl methylcyclohexadienone pyrrolidine amide
  • C.X is a methyl methoxy cyclohexadiene with the allyl and the pyrrolidine amide on one quaternary carbon; Y is a 4-allyl-4-methylcyclohexenone carboxylic acid
  • D.X has an O-allyl ketene N,O-acetal on a methoxy cyclohexadiene ring; Y is a cyclohexenone with the allyl and the pyrrolidine amide on the same carbon next to the ketone

Answer: A

X and Y are the pair shown in (A).

Step 1 is a Birch reductive alkylation: sodium in liquid ammonia reduces the benzamide to an enolate at the amide-bearing carbon, and allyl bromide alkylates that carbon. X is a 1-allyl-1-carboxamide cyclohexa-2,5-diene in which the methoxy group sits on one ring double bond as an enol ether.

Aqueous HCl hydrolyses the enol ether to a ketone next to the quaternary carbon. Refluxing in 1,2-dichlorobenzene then drives a [3,3]-sigmatropic (Cope) rearrangement of the 1,5-diene formed by the allyl group and the remaining ring double bond: the allyl group moves to the methyl-bearing carbon and a new C=C forms at the amide carbon, conjugated with both carbonyls. That is Y in (A).

(B) and (D) put the allyl group in the wrong place in X (on carbon or on oxygen), and (C) wrongly converts the amide to an acid, which aqueous HCl at this stage does not do.

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