Explain Wolf-Kishner reduction reaction. Write preparation of propanone by using ethanoyl chloride and dimethyl cadmium.

Answer:

Wolff–Kishner reduction: The carbonyl group of an aldehyde or ketone is reduced to a CH₂ group by first converting it into the hydrazone with hydrazine and then heating the hydrazone with KOH (or sodium ethoxide) in a high-boiling solvent such as ethylene glycol; N₂ is evolved.

>C=O + H₂N–NH₂ → >C=N–NH₂ + H₂O
>C=N–NH₂ → (KOH, ethylene glycol, Δ) >CH₂ + N₂

e.g. CH₃–CO–CH₃ → (i) NH₂NH₂; (ii) KOH/ethylene glycol, Δ → CH₃–CH₂–CH₃ (propane) + N₂

Propanone from ethanoyl chloride and dimethyl cadmium:

2CH₃COCl + (CH₃)₂Cd → 2CH₃–CO–CH₃ + CdCl₂