Question

Explain the following reactions and write chemical equations involved:
(a) Wolff-Kishner reduction
(b) Etard reaction
(c) Cannizzaro reaction

Hint:

In the Wolff-Kishner reduction, remember that hydrazine and strong base under heat are used to reduce carbonyl compounds, whereas in the Cannizzaro reaction, aldehydes without alpha hydrogens undergo disproportionation.

Answer 1

(a) Wolff-Kishner Reduction The Wolff-Kishner reduction is a method of reducing a carbonyl group (C=O) to a methylene group (CH\(_2\)) using hydrazine (H\(_2\)NNH\(_2\)) in the presence of a strong base like potassium hydroxide (KOH) and heating the mixture. This reduction removes the oxygen atom from the carbonyl group, yielding a hydrocarbon. Reaction: \[ \text{R-CO-R'} + \text{H}_2\text{NNH}_2 \xrightarrow{\text{KOH, heat}} \text{R-CH}_2\text{R'} \] In this reaction, aldehydes or ketones are reduced to alkanes.
(b) Etard Reaction The Etard reaction is the oxidation of toluene (methylbenzene) to benzaldehyde using chromium-based reagents such as chromium trioxide (CrO\(_3\)) or pyridinium chlorochromate (PCC). This reaction adds an aldehyde group to the methyl group of toluene. Reaction: \[ \text{C}_6\text{H}_5\text{CH}_3 + \text{CrO}_3 \rightarrow \text{C}_6\text{H}_5\text{CHO} \] In this reaction, toluene is oxidized to benzaldehyde.
(c) Cannizzaro Reaction The Cannizzaro reaction is a redox disproportionation reaction that occurs with aldehydes that do not have an alpha hydrogen. In the presence of a strong base, one molecule of the aldehyde is reduced to an alcohol, and the other is oxidized to a carboxylate anion. Reaction: \[ 2 \text{R-CHO} \xrightarrow{\text{strong base}} \text{R-CH}_2\text{OH} + \text{R-COO}^- \] In this reaction, two molecules of aldehyde react, with one being reduced to alcohol and the other oxidized to a carboxylate.