SolutionsMichael (conjugate) addition
Exercise 1
Write the product of the Michael addition of ethyl acetoacetate to acrylonitrile with a catalytic amount of sodium ethoxide, and explain why a catalytic amount of base is enough.
Ethyl 2-acetyl-4-cyanobutanoate, CH3COCH(CO2Et)CH2CH2CN, is obtained. The intermediate carbanion formed after the addition (α to the nitrile) is much more basic than the acetoacetate anion (pKa ≈ 11), so it removes a proton from another acetoacetate molecule and regenerates the nucleophile: the base is recovered in every cycle. (With excess acrylonitrile a second addition can occur.)
Exercise 2
Propose a synthesis of hexane-2,5-dione from acetone and a nitroalkene.
The lithium enolate of acetone (LDA, −78 °C), or better its enamine or its silyl enol ether with a Lewis acid, adds to 2-nitropropene to give 5-nitrohexan-2-one. The Nef reaction (NaOMe then aqueous H2SO4, or TiCl3) converts the secondary nitro group into a carbonyl and gives the 1,4-diketone.
Exercise 3
Why do organocuprates give conjugate addition to enones while organolithiums attack the carbonyl?
Organolithiums are “hard”, very reactive nucleophiles and attack, under kinetic control, the position with the greatest positive charge, the carbonyl carbon (1,2-addition). Cuprates are “soft” nucleophiles and react by a different mechanism (coordination of copper to the alkene and oxidative addition at the β carbon), which favours 1,4-addition and forms an enolate that can then be alkylated.