Chapter 8 · Chemistry
The Cannizzaro Reaction
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Two identical aldehyde molecules, two different products. How an aldehyde with no α-hydrogen, in concentrated alkali, is oxidised and reduced in the same reaction.
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Key points
- Condition: an aldehyde with no α-hydrogen (HCHO, C₆H₅CHO) heated with concentrated NaOH or KOH.
- Result: one molecule is oxidised to a carboxylate salt, the other is reduced to an alcohol.
- 2 C₆H₅CHO + NaOH → C₆H₅COONa + C₆H₅CH₂OH
- Oxidation and reduction of the same compound in one reaction is called disproportionation.
- The key step is a hydride (H⁻) transfer from one molecule to the other, passed directly, never free in solution.
The mechanism in three steps
- Step 1, nucleophilic addition: OH⁻ adds to the carbonyl carbon of one molecule, giving a tetrahedral intermediate (O⁻, OH and H on the same carbon).
- Step 2, hydride transfer: the intermediate re-forms C=O and passes its H, with the bonding pair, directly to the carbonyl carbon of a second molecule. The first becomes an acid, the second an alkoxide.
- Step 3, proton transfer: the acid gives its proton to the alkoxide: a carboxylate ion and an alcohol.
Why it is oxidation and reduction
| Carbon | Bonds | Oxidation number |
|---|---|---|
| C of C₆H₅CHO | C=O, C–H, C–C | +1 |
| C of C₆H₅COO⁻ | C=O, C–O⁻, C–C | +3 (oxidised) |
| C of C₆H₅CH₂OH | C–O, 2 C–H, C–C | −1 (reduced) |
With an α-hydrogen the aldehyde gives aldol instead. Benzoic acid itself is obtained only after the salt is acidified.
Try it yourself
Goal: Predict whether an aldehyde gives aldol, Cannizzaro or neither, and see the oxidation number of the carbonyl carbon change.
- Choose an aldehyde and count its α-hydrogens.
- Choose concentrated or dilute NaOH.
- Predict, then run the reaction.
- Read the two products and their oxidation numbers.
- Try HCHO and (CH₃)₃CCHO, then an aldehyde with an α-hydrogen.
Quick quiz
5 questions. Choose an answer for each, then submit to see your score and the explanations.