Select the correct name for the following reaction:

The correct answer is Option (2) → Wolff-Kishner Reduction

The carbonyl group of an aldehyde or ketone can be reduced right down to a methylene (\( \ce{CH2} \)) group in two named ways, and the conditions tell them apart:

• Clemmensen reduction – zinc amalgam and concentrated HCl (acidic conditions).

• Wolff-Kishner reduction – hydrazine, then heating with KOH or NaOH in a high-boiling solvent such as ethylene glycol (basic conditions).

In the given scheme the ketone first reacts with hydrazine to give a hydrazone; on heating with KOH in ethylene glycol the hydrazone loses nitrogen gas and the carbonyl carbon ends up as a \( \ce{CH2} \) group:

    \( \displaystyle \ce{R2C=O + NH2NH2 -> R2C=N-NH2 + H2O} \)

    \( \displaystyle \ce{R2C=N-NH2} \xrightarrow[\text{heat}]{\text{KOH/ethylene glycol}} \ce{R2CH2 + N2} \)

Hydrazine followed by KOH and heat is the Wolff-Kishner reduction.

Why the other options are incorrect

• Option (1) Clemmensen Reduction — Incorrect. It gives the same \( \ce{CH2} \) product, but with zinc amalgam and conc. HCl; no hydrazine is used.

      \( \displaystyle \ce{R2C=O} \xrightarrow{\ce{Zn-Hg}/\text{conc. }\ce{HCl}} \ce{R2CH2 + H2O} \)

• Option (3) Etard Reaction — Incorrect. It is the oxidation of toluene to benzaldehyde by chromyl chloride, through a chromium complex that is hydrolysed.

      \( \displaystyle \ce{C6H5CH3} \xrightarrow[\text{(ii) }\ce{H3O+}]{\text{(i) }\ce{CrO2Cl2},\ \ce{CS2}} \ce{C6H5CHO} \)

• Option (4) Stephen Reaction — Incorrect. It reduces a nitrile to an imine with \( \ce{SnCl2} \) and HCl, and the imine is hydrolysed to an aldehyde.

      \( \displaystyle \ce{RCN + SnCl2 + HCl -> RCH=NH} \xrightarrow{\ce{H3O+}} \ce{RCHO} \)

NCERT (2026-27 edition):
Reduction to Hydrocarbons