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Discussion

The major product formed in the following reaction is

Methyl acetoacetate plus methyl acrylate with MeONa in MeOH

  • A.Linear keto diester, MeO2C-CH2CH2CH2-CO-CH2-CO2Me
  • B.Enol ether diester: MeO2C-CH2CH2-O-C(CH3)=CH-CO2Me
  • C.Methyl 2-acetyl-3-oxopent-4-enoate, a carbon bearing acetyl, acryloyl and methyl ester groups
  • D.Methyl 2-acetylpentanedioate, a diester with an acetyl group on the carbon next to one ester

Answer: D

The major product is (D), the Michael adduct.

Sodium methoxide removes a proton from the active methylene group of methyl acetoacetate, flanked by the ketone and ester carbonyls. The stabilised enolate adds to the β-carbon of methyl acrylate, a Michael acceptor, and protonation gives dimethyl 2-acetylpentanedioate: the former CH2 now carries the acetyl group, the ester and a CH2CH2CO2Me chain, which is structure (D).

(A) would need the attack to come from the terminal methyl group, the less acidic site. (B) is an enol ether, the product of attack through oxygen, which is not the course of a Michael addition with a soft enolate. (C) would need acylation of the enolate by the acrylate ester, giving a vinyl ketone, not conjugate addition.

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