Question 63: Identify product (A) in the following reaction:
An aromatic compound containing two carbonyl groups is treated with \mathrm{Zn\!-\!Hg} and concentrated \mathrm{HCl} to give product (A) along with 2\mathrm{H_2O}.
Answer: Option (3)
Explanation:
The given reaction conditions \mathrm{Zn\!-\!Hg/conc.\ HCl} correspond to the Clemmensen reduction.
Clemmensen reduction converts carbonyl groups \mathrm{(C=O)} of aldehydes
and ketones into methylene groups \mathrm{(-CH_2-)} under strongly acidic conditions.
In the given molecule, both carbonyl groups present on the side chains are
reduced completely to hydrocarbon chains.
Thus, each \mathrm{COCH_3} group is converted into an ethyl group \mathrm{CH_2CH_3}.
No hydroxyl groups remain in the product, as Clemmensen reduction does not produce alcohols.
Therefore, the final product is the compound with ethyl substituents on both sides,
which corresponds to option (3).