The reaction involves Zn-Hg and HCl, which is known as the Clemmensen reduction. Clemmensen reduction is typically used to reduce carbonyl groups (like aldehydes and ketones) to methylene (-CH2-) groups.
Analyzing the Given Compound:
The starting compound likely has a carbonyl (C=O) group. In the presence of Zn-Hg and HCl, the carbonyl group will be reduced to a -CH2- group.
Identifying the Product:
Based on the Clemmensen reduction, the product will have the carbonyl group replaced by a methylene (-CH2-) group. Among the given options, Option (4) represents the structure where the carbonyl group has been reduced to a -CH2- group.
Conclusion:
The correct product of the reaction is represented by Option (4).
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Approach Solution -2
Step 1: Given reaction.
The given compound is an aromatic aldehyde having a methyl ketone group in the meta position.
It is treated with Zn–Hg/HCl, which indicates the Clemmensen reduction reaction.
Step 2: Concept of Clemmensen reduction.
The Clemmensen reduction reduces the carbonyl group (–CHO or –CO–) to a methylene group (–CH₂) using Zinc amalgam (Zn–Hg) and concentrated HCl under acidic conditions.
Only the carbonyl groups attached directly to the ring (aromatic aldehydes or ketones) are reduced to –CH₂ groups.
Step 3: Reaction pathway.
Starting compound: m–methylacetophenone (3–acetylbenzaldehyde).
Under Clemmensen reduction, the aldehyde (–CHO) group is reduced to –CH₃, while the methyl ketone (–COCH₃) group remains unaffected because it is less reactive toward Clemmensen conditions compared to the aldehyde.
Hence, the product formed is: m–ethylacetophenone (1–bromo–3–ethylbenzene) (as shown in the correct answer image).
Step 4: Final Answer.
The product is m–ethylacetophenone (structure matches the image).
Final Answer:
\[
\boxed{\text{Product: 1–(3–ethylphenyl)ethanone}}
\]
