We are preparing an a,b-unsaturated ketone via Michael and aldol condensation reactions. The reactants are trans-chalcone and ethyl acetoacetate (in ethanol and NaOH). This creates 6-ethoxycarbonyl-3,5-diphenyl-2-cyclohexenone. What I'm wondering is:
Why is it possible to separate the product from sodium hydroxide using acetone?
Could you tell me what is the purpose of this reaction (Michael and aldol condensation reactions), and the stepwise mechanism of the reaction.
Also if we started with 0.24 gr of trans-chalcone , 0.15 gr of ethyl acetoacetate what would be the theoretical yield ?© BrainMass Inc. brainmass.com September 19, 2018, 6:23 pm ad1c9bdddf - https://brainmass.com/chemistry/organic-reactions/the-reactants-are-trans-chalcone-and-ethyl-acetoacetate-in-105561
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 solution provides detailed explanations and answer, including reaction mechanism and drawings, for the problem.