Michael & Aldol Condensation Reaction

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 1.2 grams of trans-chalcone , 0.75 grams of ethyl acetoacetate what would be the theoretical yield ?

1.845g

In the Michael and aldol condensation reactions, the product can be separated from sodium hydroxide using acetone because acetone is a polar aprotic solvent. Sodium hydroxide is a strong base and can be dissolved in a polar solvent like water, but it is not highly soluble in acetone. Therefore, when acetone is added, the sodium hydroxide will precipitate out, allowing for the separation of the product.

The purpose of the Michael and aldol condensation reactions is to form a carbon-carbon bond, resulting in the synthesis of a,b-unsaturated ketones. These reactions are commonly used in organic chemistry to build complex molecules.

The stepwise mechanism of the Michael and aldol condensation reactions can be described as follows:

1. Michael Reaction:
- The trans-chalcone, which contains an alpha,beta-unsaturated carbonyl group, acts as a nucleophile and attacks the carbon of the Michael acceptor, which is ethyl acetoacetate in this case.
- This leads to the formation of an intermediate called the Michael adduct.

2. Aldol Condensation:
- The Michael adduct undergoes a base-catalyzed dehydration reaction. A hydroxide ion abstracts a proton from the alpha position of the Michael adduct, resulting in the formation of an enolate ion.
- The enolate ion is then able to attack the carbonyl carbon of another molecule of trans-chalcone.
- This leads to the formation of a beta-hydroxyketone intermediate.

3. Elimination:
- The beta-hydroxyketone intermediate undergoes an elimination reaction in the presence of a base, such as sodium hydroxide, to form the final product, which is 6-ethoxycarbonyl-3,5-diphenyl-2-cyclohexenone.

The theoretical yield can be calculated using the molecular weights of the reactants and products. The balanced chemical equation for the reaction is required to determine the stoichiometry. Without this information, it is not possible to provide an accurate estimation of the theoretical yield.

To understand why it is possible to separate the product from sodium hydroxide using acetone, we need to consider the solubility of the different compounds involved. Acetone is a polar solvent and has different solubility properties compared to ethanol, water, and sodium hydroxide.

In the given reaction, trans-chalcone and ethyl acetoacetate react in the presence of ethanol and sodium hydroxide to form the product, 6-ethoxycarbonyl-3,5-diphenyl-2-cyclohexenone. The reaction involves two steps: Michael addition and aldol condensation.

1. Michael Addition: The trans-chalcone reacts with the ethyl acetoacetate through a Michael addition, where the carbon-carbon double bond in the chalcone undergoes nucleophilic addition by the enolate of ethyl acetoacetate. This leads to the formation of a new carbon-carbon bond, resulting in an intermediate compound.

2. Aldol Condensation: The intermediate compound formed in the previous step undergoes a base-catalyzed aldol condensation. The enolate ion of the intermediate compound attacks another molecule of trans-chalcone, leading to the formation of an aldol product. This product undergoes dehydration to give the final a,b-unsaturated ketone, 6-ethoxycarbonyl-3,5-diphenyl-2-cyclohexenone.

Now, coming to the separation of the product from sodium hydroxide using acetone, sodium hydroxide is a strong base and can form salts with acidic compounds like carboxylic acids. In this case, the product, 6-ethoxycarbonyl-3,5-diphenyl-2-cyclohexenone, contains a carbonyl group (C=O), which can act as an acid. Sodium hydroxide can react with this acidic carbonyl group to form a sodium salt.

Acetone, being a less polar solvent compared to ethanol, can dissolve the product but does not favor the dissolution of the sodium salt. Therefore, when the reaction mixture is treated with acetone, the product can be separated as it dissolves in acetone, while the sodium hydroxide salt remains in the aqueous layer.

Now, to calculate the theoretical yield, we need to determine the limiting reagent and use stoichiometry.

1. Convert the masses of trans-chalcone and ethyl acetoacetate to moles by dividing each mass by their respective molar masses.
Moles of trans-chalcone = 1.2 g / molar mass of trans-chalcone
Moles of ethyl acetoacetate = 0.75 g / molar mass of ethyl acetoacetate

2. Compare the molar ratios of trans-chalcone and ethyl acetoacetate in the balanced chemical equation to determine the limiting reagent. The reactant that gives the lesser amount of product is the limiting reagent.

3. Use the stoichiometry to calculate the moles of the product that can be formed based on the limiting reagent.

4. Convert the moles of the product to grams using its molar mass.

The calculated mass obtained in step 4 will be the theoretical yield of 6-ethoxycarbonyl-3,5-diphenyl-2-cyclohexenone.