Cannizzaro Reaction: Mechanism & Examples | Chemca

Organic Chemistry

Cannizzaro Reaction: Mechanism & Examples

By Chemca Editorial Team • Last Updated: January 2026 • 9 min read

The Cannizzaro Reaction is a classic disproportionation (redox) reaction where aldehydes that do not contain any $\alpha$-hydrogen atoms undergo self-oxidation and reduction in the presence of a concentrated alkali.

1. General Reaction

One molecule of the aldehyde is reduced to the corresponding alcohol, while the other molecule is oxidized to the salt of the corresponding carboxylic acid.

$$ 2HCHO \xrightarrow{\text{Conc. NaOH}} \underbrace{CH_3OH}_{\text{Alcohol (Red)}} + \underbrace{HCOONa}_{\text{Salt of Acid (Ox)}} $$

Conditions:

Substrate: Aldehydes with NO $\alpha$-hydrogen (e.g., $HCHO$, $C_6H_5CHO$, $(CH_3)_3CCHO$).
Reagent: Concentrated Base (50% NaOH or KOH).
Type: Disproportionation (Self-Redox).

2. Detailed Mechanism

The reaction involves nucleophilic attack followed by a rate-determining hydride transfer.

Step 1: Nucleophilic Attack

The hydroxide ion ($OH^-$) attacks the carbonyl carbon to form a tetrahedral alkoxide intermediate.

$$ OH^- + R-CH=O \rightleftharpoons R-CH(O^-)(OH) $$

Step 2: Hydride Transfer (RDS)

This is the crucial step. The tetrahedral intermediate reforms the carbonyl double bond, expelling a Hydride ion ($H^-$). This hydride acts as a nucleophile and attacks a second molecule of aldehyde.

$$ R-CH(O^-)(OH) + R-CHO \xrightarrow{\text{Slow}} R-COOH + R-CH_2O^- $$

Step 3: Acid-Base Reaction

The carboxylic acid formed transfers a proton to the alkoxide ion (which is a strong base) to form the alcohol and the carboxylate ion.

$$ R-COOH + R-CH_2O^- \rightarrow R-COO^- + R-CH_2OH $$

3. Cross-Cannizzaro Reaction

When two different aldehydes (both without $\alpha$-H) react, it is called a Cross-Cannizzaro reaction.

Rule of Thumb

If Formaldehyde ($HCHO$) is one of the reactants, it is always oxidized to Formate ($HCOO^-$) because it is sterically less hindered, allowing easier nucleophilic attack by $OH^-$. The aromatic/bulkier aldehyde is reduced to the alcohol.

Example: Formaldehyde + Benzaldehyde

$$ HCHO + C_6H_5CHO \xrightarrow{\text{Conc. NaOH}} HCOONa + C_6H_5CH_2OH $$

Products: Sodium Formate + Benzyl Alcohol

4. Intramolecular Cannizzaro

Occurs in dialdehydes like Glyoxal ($OHC-CHO$). One -CHO group oxidizes while the other reduces within the same molecule.

$$ OHC-CHO \xrightarrow{OH^-} HOCH_2-COO^- \text{ (Glycolate ion)} $$

5. Comparison with Aldol

Feature	Aldol Condensation	Cannizzaro Reaction
Requirement	Requires $\alpha$-Hydrogen	Requires NO $\alpha$-Hydrogen
Reagent	Dilute Base (10%)	Concentrated Base (50%)
Intermediate	Enolate Ion (Carbanion)	Hydride Transfer ($H^-$)

Cannizzaro Quiz

Test your concepts on Disproportionation. 10 MCQs with explanations.

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