Question

product of aldol condensation is 1,3-diphenyl pro-3-ene-1-one, then its reactants are

Hint:

To quickly find aldol reactants from an enone, simply cut the $C=C$ bond. Give an 'O' to the carbon further from the carbonyl, and give 'H2' to the carbon next to the carbonyl.

Answer 1

Step 1: Understanding the Concept:
An Aldol condensation reaction between two carbonyl compounds (at least one having an $\alpha$-hydrogen) yields an $\alpha,\beta$-unsaturated carbonyl compound (an enone) after dehydration. To find the reactants, we perform a retrosynthetic analysis on the product, breaking it apart at the double bond.

Step 2: Key Formula or Approach:
Identify the $\alpha,\beta$-double bond relative to the carbonyl group. Cleave this double bond. Add two hydrogen atoms to the $\alpha$-carbon and an oxygen atom to the $\beta$-carbon to regenerate the original carbonyl reactants.

Step 3: Detailed Explanation:
First, let's correct the chemical name provided in the question. "1,3-diphenyl pro-3-ene-1-one" is structurally impossible as a propene chain only has 3 carbons. The correct name for the common aldol product Chalcone is 1,3-diphenylprop-2-en-1-one. Let's proceed with this correct structure.
Structure of the product:
$\text{C}_6\text{H}_5 - \text{CH} = \text{CH} - \text{C}(=\text{O}) - \text{C}_6\text{H}_5$
$\beta \quad \quad \alpha \quad \quad \text{carbonyl}$
This is a cross-aldol condensation product.
Applying the retrosynthetic cleavage at the $\alpha,\beta$ double bond:
1. Break the $\text{C=C}$ bond.
2. The carbon adjacent to the carbonyl ($\alpha$-carbon) originated from the enolate and must get 2 hydrogen atoms back.
$-\text{CH} \rightarrow -\text{CH}_3$
So, the right fragment becomes $\text{CH}_3 - \text{C}(=\text{O}) - \text{C}_6\text{H}_5$, which is Acetophenone.
3. The other carbon of the double bond ($\beta$-carbon) was the electrophilic carbonyl carbon and gets an oxygen atom back to form a $C=O$ group.
$\text{C}_6\text{H}_5 - \text{CH}= \rightarrow \text{C}_6\text{H}_5 - \text{CHO}$
So, the left fragment becomes Benzaldehyde.
Reactants: Benzaldehyde (which lacks $\alpha$-hydrogens) and Acetophenone (which provides the $\alpha$-hydrogens to form the enolate).

Step 4: Final Answer:
The reactants are Benzaldehyde and Acetophenone.