Question

Post-lab questions (1) In a Michael reaction, an α carbon attacks the β carbon of an α,β-unsaturated carbonyl com pound to afford a 1,5-dicarbonyl compound. Ethyl acetoacetate has two a carbons: why does the a carbon sandwiched between the two carbonyls conduct the attack? Ph Ph 0 Ph Michael OCH2CH3 5 OCH2CH3 OCH2CH 0 This forms... ...not this: Why? (2) In a Robinson annulation, an α,β-unsaturated carbonyl compound reacts with a carbonyl compound to afford a (new) cyclic α,β-unsaturated ketone by a Michael-Aldol-dehydration se- quence. Draw the structure of the reactants that would combine to form the A ring of the steroid desoxy-19-nortestosterone (below) by a Robinson annulation. Preparation of an α,β-unsaturated ketone via Michael and Aldol reactions CH desoxv-19-nortestosterone

Answer 1

11 Ethyl acetoacetate has two α-carbons. one is sandwiched between two carbonyls and the other is terminal-CH3 proton. The central CH2 proton has pKa value of - 11 and the terminal CH3 proton has pKa value of19. Hence the central proton gets deprotonated easily. Besides, the resulting carbanion is well resonance stabillized than the terminal carbanion pKa 19 H3C НН pKa 11 O CH3 H3C O CH3 H3C H2C НН Resonance stabillization is more here 2] First let me show the retro synthesis of the particular reaction sequence Now let me show the forward synthesis. I have not shown complete mechanism but shown how it would have happened in the reaction itself EtONa EtONa EtOH EtOH он EtONa EtOH