Draw the structure of the neutral product formed in the following reaction

1.Explanation | Common mistakes

Deprotonation Step

It is an acid-base reaction. An ethoxide ion is a strong base and deprotonate the acidic alpha hydrogen present in the diketone (Michael donor) forming the carbanion which is stabilized by the resonance, giving the most stable enolate ion because the oxygen atom is more electronegative than the carbon atom.

2.Explanation | Common mistakes | Hint for next step

1,4-Addition Step

The nucleophilic carbanion (enolate) attacks the carbon-carbon double bond present in the via 1,4-conjugated addition, forming an intermediate enolate ion.

3.Explanation | Common mistakes

Protonation Step

Again, it is an acid-base reaction. The intermediate enolate ion is a strong base that can abstract a proton from ethanol, regenerating the ethoxide base back and gives the more stable thermodynamic product as the Michael adduct.

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