Question
Which aldehydes undergo the Cannizzaro reaction? Explain the mechanism and why other aldehydes do NOT undergo it.
Solution — Step by Step
The Cannizzaro reaction is a self-disproportionation reaction in which an aldehyde without an alpha-hydrogen (α-H) undergoes simultaneous oxidation and reduction when treated with a concentrated base (NaOH or KOH).
One molecule is oxidised to a carboxylic acid (as its sodium salt), while another molecule is reduced to an alcohol.
Example with formaldehyde (HCHO):
One HCHO is reduced to methanol (CH₃OH), the other is oxidised to sodium formate (HCOONa).
Critical condition: The aldehyde must have no alpha-hydrogen (α-H).
Alpha-hydrogen refers to hydrogen atoms on the carbon adjacent to the carbonyl group ().
Aldehydes that undergo Cannizzaro reaction:
- Formaldehyde (HCHO) — no carbon at all besides carbonyl carbon
- Benzaldehyde (C₆H₅CHO) — aromatic ring has no α-H
- 2,2-Dimethyl propanal (trimethylacetaldehyde, ) — alpha carbon is fully substituted with methyl groups
Aldehydes that do NOT undergo Cannizzaro reaction:
- Acetaldehyde (CH₃CHO) — has 3 α-H atoms on the methyl group
- Propanal (CH₃CH₂CHO) — has α-H atoms
These aldehydes instead undergo Aldol condensation with base.
Step 1 — Nucleophilic attack: The hydroxide ion (OH⁻) from NaOH attacks the carbonyl carbon of one aldehyde molecule, forming a tetrahedral intermediate with a hydride (H⁻) group.
Step 2 — Hydride transfer: The hydride (H⁻) from the tetrahedral intermediate transfers to the carbonyl carbon of a second aldehyde molecule (intermolecular hydride shift). This is the key step — it’s a 1,2-hydride shift between two aldehyde molecules.
Step 3 — Proton transfer: The alkoxide ion formed accepts a proton from the carboxylate species, giving the final products: alcohol + carboxylate salt.
The reaction is a disproportionation — no external oxidant or reductant is needed; one molecule oxidises itself by oxidising the other.
Aldehydes with α-H have acidic alpha protons. Under basic conditions (NaOH), these protons are removed to form enolate ions. The enolate then attacks another aldehyde molecule — this is Aldol condensation, which is much faster and more favourable.
Since Aldol condensation dominates when α-H is present, Cannizzaro reaction only proceeds when there’s no α-H (no competition from Aldol pathway).
Why This Works
The driving force of the Cannizzaro reaction is the thermodynamic stability of the products (alcohol + salt). The mechanism proceeds via a concerted or stepwise hydride transfer — a genuinely unusual step in organic chemistry, as hydrides normally transfer in carbonyl reductions using reagents like NaBH₄ or LiAlH₄.
The absence of α-H is not just a structural requirement — it’s a kinetic requirement. With no enolisable protons, the Aldol pathway is blocked, forcing the reaction to proceed via the Cannizzaro mechanism.
Common Mistake
A very common error: students write that all aldehydes undergo Cannizzaro reaction. Only α-H-free aldehydes do. In JEE Advanced and NEET questions, the examiner tests exactly this distinction. If you see benzaldehyde + NaOH: Cannizzaro. If you see acetaldehyde + NaOH: Aldol condensation. These are opposite reactions — choosing the wrong one is a full-question error.
For JEE: The crossed Cannizzaro reaction is an important variant — mixing formaldehyde with another α-H-free aldehyde. Since formaldehyde is a stronger reducing agent (easier to oxidise), it preferentially gets oxidised to formate, while the other aldehyde gets reduced to its alcohol. This is used in synthetic chemistry to convert benzaldehyde to benzyl alcohol using HCHO as the sacrificial reductant.