Reaction of benzaldehyde with acetophenone in the presence of methanolic NaOH yields a major stable product as

trans- 1,3-diphenylprop-2-en-1-one

The correct answer is Option (3) → trans- 1,3-diphenylprop-2-en-1-one

The reaction between benzaldehyde and acetophenone in the presence of dilute alkali (like methanolic NaOH) is a classic example of a Claisen-Schmidt condensation (a type of crossed aldol condensation).

The correct answer is: trans-1,3-diphenylprop-2-en-1-one.

Reaction Mechanism & Stability

1. Enolate Formation: Acetophenone has a-hydrogens, so it forms an enolate ion in the presence of NaOH. Benzaldehyde has no a-hydrogens, so it acts purely as the electrophile.

2. Nucleophilic Attack: The enolate of acetophenone attacks the carbonyl carbon of benzaldehyde.

3. Dehydration: The resulting $β$-hydroxy ketone (aldol) undergoes dehydration to form an $α, β$-unsaturated ketone, commonly known as chalcone.

4. Stereochemistry: The trans (or E) isomer is the major stable product because it minimizes steric repulsion between the two bulky phenyl groups.

The Product Structure

The IUPAC name for chalcone is 1,3-diphenylprop-2-en-1-one. In the trans-configuration, the two large phenyl groups are on opposite sides of the double bond, making it energetically more favorable than the cis-isomer.