Please see the attached file.

1. It's possible to separate the product from sodium hydroxide using acetone because the organic product (6-ethoxycarbonyl-3,5-diphenyl-2-cyclohexenone) dissolves in acetone, while NaOH is insoluble in acetone. Thus NaOH will precipitate (cloudy). Upon centrifuging, NaOH is separated from the product.

2. The Michael reaction or Michael addition is the nucleophilic addition of an carbanion to an alpha, beta unsaturated carbonyl compound. It belongs to the larger class of conjugate additions. This is one of the most useful methods for the mild formation of C-C bonds.

The aldol reaction is an important C-C bond forming reaction in organic chemistry involving the addition of an enol or enolate anion to an aldehyde or ketone. In the aldol addition, the reaction results in a β-hydroxy ketone (or aldehyde), also called an "aldol" (aldehyde + alcohol). In the aldol condensation, the initial aldol adduct undergoes dehydration (loss of water) to form an α,β-unsaturated ketone (or aldehyde).

3. Mechanism: First is the sodium hydroxide-catalyzed conjugate addition of ethyl-acetoacetate to trans-chalcone, a Michael rxn; then a base-catalyzed aldo condensation rxn; then the aldol intermediate is dehydrated to form the final product. See full mechanism next page.

4. Calculation of theoretical yield. Write down balanced equation:

C6H10O3 + C15H12O = C21H20O3 + H2O

MW ethyl acetoacetate = 130 g/mol
MW trans-chalcone = 208 g/mol
n ethyl acetoacetate = 0.15/130 = 0.00115 (mol)
n trans-chalcone = 0.24/208 = 0.00115 (mol)
So we should have a theoretical yield of 0.00115 mol of C21H20O3
MW C21H20O3 = 21*12+20+48 = 320 (g/mol)
Theoretical yield of C21H20O3 = 320*0.00115 = 0.368 (g)