Statement I: Propanone gives propan-2-ol when it is treated with hydrazine and then heated with potassium hydroxide in ethylene glycol.
Statement II: Benzaldehyde gives toluene when it is treated with zinc amalgam and concentrated hydrochloric acid.
In the light of the above statements, choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Only Statement II is correct.
Both reagent sets in the statements reduce the carbonyl group of aldehydes and ketones to a CH2 group, giving a hydrocarbon, not an alcohol: hydrazine followed by KOH in ethylene glycol is the Wolff–Kishner reduction, and zinc amalgam with concentrated HCl is the Clemmensen reduction.
Statement I: Propanone gives propan-2-ol when it is treated with hydrazine and then heated with potassium hydroxide in ethylene glycol. — False. Propanone first forms its hydrazone, which on heating with KOH in the high boiling solvent loses nitrogen and gives propane:
\( \displaystyle \ce{(CH3)2C=O + NH2NH2 -> (CH3)2C=NNH2 + H2O} \)
\( \displaystyle \ce{(CH3)2C=NNH2} \xrightarrow[\text{ethylene glycol, heat}]{\ce{KOH}} \ce{CH3CH2CH3 + N2} \)
Propan-2-ol would be formed by reduction to an alcohol, for example with NaBH4 or LiAlH4.
Statement II: Benzaldehyde gives toluene when it is treated with zinc amalgam and concentrated hydrochloric acid. — True. The –CHO group becomes –CH3:
\( \displaystyle \ce{C6H5CHO + 4[H]} \xrightarrow{\ce{Zn-Hg},\ \text{conc. }\ce{HCl}} \ce{C6H5CH3 + H2O} \)
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).”