Question

The major product 'A' in the given reaction is
Benzaldehyde + Acetophenone \(\xrightarrow{\text{OH}^- / 293\text{K}}\) 'A' (Major product)

• (1)
• (2)
• (3)
• (4)

Hint:

In crossed aldol condensations, always identify the enolizable component (the one with \(\alpha\)-hydrogens) and the non-enolizable component. The enolate of the first attacks the carbonyl of the second. The final product is almost always the dehydrated \(\alpha\),\(\beta\)-unsaturated compound, especially when conjugation provides extra stability.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This reaction between an aldehyde with no $\alpha$-hydrogens and a ketone with $\alpha$-hydrogens in the presence of a base is a Cross-Aldol condensation, specifically known as the Claisen-Schmidt condensation.
Step 2: Key Formula or Approach:
Identify the acidic $\alpha$-hydrogens on the ketone, form the enolate ion, and perform a nucleophilic attack on the carbonyl carbon of the aldehyde, followed by dehydration to yield an $\alpha,\beta$-unsaturated ketone.
Step 3: Detailed Explanation:
1. The base ($OH^{-}$) abstracts an $\alpha$-hydrogen from acetophenone ($C_{6}H_{5}COCH_{3}$) to form a stable enolate ion: $C_{6}H_{5}COCH_{2}^{-}$.

2. The enolate ion attacks the electrophilic carbonyl carbon of benzaldehyde ($C_{6}H_{5}CHO$), forming a $\beta$-hydroxy ketone intermediate.

3. Protonation and subsequent dehydration (loss of a water molecule) occur easily at 293K (or upon mild heating) due to the extended conjugation of the newly formed double bond with the two benzene rings.
Reaction: $C_{6}H_{5}CHO + CH_{3}COC_{6}H_{5} \xrightarrow{OH^{-}} C_{6}H_{5}-CH=CH-CO-C_{6}H_{5}$ (Chalcone).
Step 4: Final Answer:
Structure (4) representing 1,3-diphenylprop-2-en-1-one (Chalcone) is the correct major product.