Cannizzaro reaction: practice and mechanism
Aldehydes without α-H cannot form an enolate. In concentrated base one molecule transfers a hydride to a second: an alcohol and a carboxylic acid form at the same time (disproportionation).
Example
Aldehydes without an α hydrogen cannot form an enolate. In concentrated base they disproportionate: one molecule is oxidised to the carboxylic acid, a second one reduced to the alcohol (hydride transfer).
Mechanism step by step
- 1
Hydroxide attacks the carbonyl carbon of one aldehyde, giving a tetrahedral anion.
- 2
The anion transfers a hydride to a second aldehyde molecule; the first becomes the carboxylic acid.
- 3
Proton exchange: carboxylate and alcohol (after acidification, carboxylic acid and alcohol).
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
Often confused with: Aldol addition · Oxidation of aldehydes (Tollens test)
Frequently asked questions
Why only aldehydes without α-H?
With an α-H the base would form an enolate and an aldol reaction would be much faster.
What is the crossed Cannizzaro reaction?
With formaldehyde as a sacrificial aldehyde: formaldehyde is oxidised to formic acid and the more valuable aldehyde is reduced to the alcohol.