The major product 'A' in the given reaction is
Benzaldehyde + Acetophenone \(\xrightarrow{\text{OH}^- / 293\text{K}}\) 'A' (Major product)
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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.
Step 1: Understanding the Reaction:
The reaction is between benzaldehyde (C\(_6\)H\(_5\)CHO) and acetophenone (C\(_6\)H\(_5\)COCH\(_3\)) in the presence of a base (OH\(^-\)). This is a base-catalyzed condensation reaction.
- Benzaldehyde has no \(\alpha\)-hydrogens.
- Acetophenone has acidic \(\alpha\)-hydrogens on its methyl group.
This type of reaction between an aldehyde with no \(\alpha\)-hydrogens and a ketone with \(\alpha\)-hydrogens is called a Claisen-Schmidt condensation, which is a type of crossed aldol condensation. Step 2: The Reaction Mechanism:
1. Enolate formation: The base (OH\(^-\)) abstracts an acidic \(\alpha\)-hydrogen from acetophenone to form a resonance-stabilized enolate ion.
\[ \text{C}_6\text{H}_5\text{COCH}_3 + \text{OH}^- \rightleftharpoons [ \text{C}_6\text{H}_5\text{COCH}_2 ]^- + \text{H}_2\text{O} \]
2. Nucleophilic attack: The enolate ion acts as a nucleophile and attacks the electrophilic carbonyl carbon of benzaldehyde.
\[ \text{C}_6\text{H}_5\text{CHO} + [ \text{C}_6\text{H}_5\text{COCH}_2 ]^- \rightarrow \text{C}_6\text{H}_5\text{-CH(O}^-\text{)-CH}_2\text{COC}_6\text{H}_5 \]
3. Protonation: The resulting alkoxide ion is protonated by water to give the \(\beta\)-hydroxy ketone (aldol addition product).
\[ \text{C}_6\text{H}_5\text{-CH(O}^-\text{)-CH}_2\text{COC}_6\text{H}_5 + \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_5\text{-CH(OH)-CH}_2\text{COC}_6\text{H}_5 + \text{OH}^- \]
4. Dehydration: The aldol product readily dehydrates (loses a molecule of water) upon gentle warming (293K is room temperature, but dehydration is rapid here) to form an \(\alpha\),\(\beta\)-unsaturated ketone. The dehydration is favorable because the resulting double bond is in conjugation with both the benzene ring and the carbonyl group, leading to a very stable system.
\[ \text{C}_6\text{H}_5\text{-CH(OH)-CH}_2\text{COC}_6\text{H}_5 \xrightarrow{-\text{H}_2\text{O}} \text{C}_6\text{H}_5\text{-CH=CH-COC}_6\text{H}_5 \]
Step 3: Identifying the Major Product:
The final major product 'A' is the dehydrated \(\alpha\),\(\beta\)-unsaturated ketone, which is 1,3-diphenylprop-2-en-1-one, commonly known as chalcone. Its structure is C\(_6\)H\(_5\)-CH=CH-C(=O)-C\(_6\)H\(_5\). This matches the structure in option (4). Step 4: Final Answer:
The major product of the reaction is benzalacetophenone (chalcone), which is represented by structure (4).
