Question

(A) 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 used to reduce a carbonyl group (C=O) to a methylene group (CH\(_2\)) by employing hydrazine (H\(_2\)NNH\(_2\)) in the presence of a strong base such as potassium hydroxide (KOH), with heating. This reaction effectively removes the oxygen atom from the carbonyl group, converting it into a hydrocarbon. Reaction: \[ \text{R-CO-R'} + \text{H}_2\text{NNH}_2 \xrightarrow{\text{KOH, heat}} \text{R-CH}_2\text{R'} \] This process reduces aldehydes or ketones to alkanes. \bigskip (b) Etard Reaction The Etard reaction involves the oxidation of toluene (methylbenzene) to benzaldehyde using chromium-based reagents such as chromium trioxide (CrO\(_3\)) or pyridinium chlorochromate (PCC). In this reaction, an aldehyde group is added to the methyl group of toluene. Reaction: \[ \text{C}_6\text{H}_5\text{CH}_3 + \text{CrO}_3 \rightarrow \text{C}_6\text{H}_5\text{CHO} \] Here, toluene is oxidized to benzaldehyde. \bigskip (c) Cannizzaro Reaction The Cannizzaro reaction is a redox disproportionation reaction that occurs with aldehydes lacking an alpha hydrogen. In the presence of a strong base, one molecule of the aldehyde is reduced to an alcohol, while another 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 participate: one is reduced to alcohol, and the other is oxidized to a carboxylate.