What happens when a ketone is treated with amalgamated zinc (Hg-Zn) and concentrated hydrochloric acid?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Alkane is formed
Zinc amalgam with concentrated HCl reduces the C=O group of a ketone to a \( \ce{CH2} \) group; this is the Clemmensen reduction. The product contains only carbon and hydrogen, so a ketone gives an alkane. For example, propanone gives propane:
\( \displaystyle \ce{CH3COCH3 + 4[H]} \xrightarrow[\text{conc. HCl}]{\ce{Zn-Hg}} \ce{CH3CH2CH3 + H2O} \)
Why the other options are incorrect
Option (2) Alcohol is formed — Incorrect. Reduction of a ketone only as far as a secondary alcohol is done with NaBH4, LiAlH4 or catalytic hydrogenation; Zn-Hg/HCl removes the oxygen completely.
Option (3) Alkylhalide is formed — Incorrect. The C=O becomes \( \ce{CH2} \); no halogen enters the product. HCl only provides the acid medium.
Option (4) Acid is formed — Incorrect. Forming an acid would be an oxidation; Zn-Hg/HCl is a reducing agent, and ketones are oxidised only by strong oxidising agents under vigorous conditions.
NCERT (2026-27 edition):
8.4 Chemical Reactions – “(ii) Reduction to hydrocarbons: The carbonyl group of aldehydes and ketones is reduced to CH2 group on treatment with zinc-amalgam and concentrated hydrochloric acid [Clemmensen reduction] or with hydrazine followed by heating with sodium or potassium hydroxide in high boiling solvent such as ethylene glycol (Wolff-Kishner reduction).”