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 ?
To understand why it is possible to separate the product from sodium hydroxide using acetone, we need to consider the solubility and polarity properties of the compounds involved.
Sodium hydroxide (NaOH) is highly soluble in water but not in organic solvents like acetone. On the other hand, the product of the condensation reaction, 6-ethoxycarbonyl-3,5-diphenyl-2-cyclohexenone, is generally more soluble in organic solvents than in water. Acetone, being a common organic solvent with moderate polarity, can effectively dissolve the organic product while leaving behind the aqueous sodium hydroxide solution.
The purpose of the Michael and aldol condensation reactions is to synthesize a,b-unsaturated ketones. These reactions are important in organic synthesis as they allow the introduction of functional groups and formation of carbon-carbon bonds. The trans-chalcone and ethyl acetoacetate react to form the product, which contains a conjugated system of double bonds, making it useful as an intermediate in the synthesis of various organic compounds.
Now, let's look at the stepwise mechanism of the Michael and aldol condensation reactions:
1. Michael Addition:
- The trans-chalcone (an α,β-unsaturated ketone) acts as the Michael acceptor, containing a β-carbonyl system that is electron-deficient.
- The ethyl acetoacetate (a nucleophilic compound) acts as the Michael donor, containing an activated α-carbonyl carbon.
- In the presence of a base (sodium hydroxide), the α,β-unsaturated ketone and ethyl acetoacetate undergo a Michael addition reaction.
- The enolate ion of ethyl acetoacetate attacks the β-carbonyl carbon of the trans-chalcone, resulting in the formation of a new carbon-carbon bond.
2. Aldol Condensation:
- The newly formed compound after Michael addition is an aldol, which is a β-hydroxy ketone.
- In the presence of the base, a dehydration process occurs, resulting in the elimination of water, and the formation of a double bond.
- The resulting compound is the desired a,b-unsaturated ketone, 6-ethoxycarbonyl-3,5-diphenyl-2-cyclohexenone.
Now, let's calculate the theoretical yield based on the given amounts of starting materials:
1. Convert the given amounts to moles:
- 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. Identify the limiting reagent:
- Compare the moles of trans-chalcone and ethyl acetoacetate to determine the limiting reagent (the one that generates fewer moles of product).
3. Calculate the theoretical yield:
- Theoretical yield = moles of limiting reagent * molar mass of the product
Remember to use the balanced equation of the reaction to determine the stoichiometry between the reactants and product.
I hope this explanation helps! Let me know if you have any further questions.