Cannizzaro Reaction: Simple Mechanism and Examples

The Cannizzaro reaction is the base-induced disproportionation of aldehydes that have no α-hydrogen: two molecules of the aldehyde react with concentrated alkali so that one molecule is oxidized to the carboxylate salt while the other is reduced to the primary alcohol.

It matters because it is the complement of the aldol condensation — aldehydes without α-hydrogens cannot form enolates, so instead of condensing they disproportionate. It is also the classic industrial route to pentaerythritol and the reason formaldehyde solutions slowly generate methanol and formate on standing with base.

Key Concepts

• Disproportionation: one substance is simultaneously oxidized and reduced; here, two aldehyde molecules split roles.
• No α-hydrogen requirement: formaldehyde, benzaldehyde, trimethylacetaldehyde and furfural undergo Cannizzaro; ethanal and propanal (which have α-H) undergo aldol instead.
• Hydride transfer: the key step — a hydride ion (H−) moves from one molecule to another; this is the slow, rate-determining step.
• Crossed Cannizzaro: between formaldehyde and another α-H-free aldehyde; formaldehyde is preferentially oxidized to formate while the other aldehyde is reduced to its alcohol.
• Intramolecular Cannizzaro: dialdehydes like glyoxal (OHC–CHO) disproportionate within the same molecule to give the hydroxy acid salt.

How It Works

Step 1 — hydroxide attack. OH− adds to the carbonyl carbon of one aldehyde molecule, giving a tetrahedral intermediate with two oxygens bearing negative character.

Step 2 — hydride transfer (slow). The intermediate collapses, expelling a hydride ion (H−) to the carbonyl carbon of a second aldehyde molecule. The donor becomes the carboxylate; the acceptor becomes an alkoxide.

Step 3 — protonation. The alkoxide picks up a proton from water, giving the primary alcohol and regenerating OH−. Net: 2 RCHO + OH− → RCOO− + RCH2OH.

Concentrated alkali (about 50% NaOH or KOH) is needed because the hydride transfer is difficult; dilute base gives almost no reaction.

Worked Example

Formaldehyde undergoes the Cannizzaro reaction with concentrated NaOH. Balance the equation and verify masses for 60 g of formaldehyde.

1. Overall: 2 HCHO + NaOH → HCOONa + CH3OH.
2. Molar mass of formaldehyde (CH2O) = 12 + 2 + 16 = 30 g/mol, so 60 g = 2.00 mol.
3. Stoichiometry needs 1 mol NaOH (40 g) for 2 mol HCHO.
4. Products: sodium formate (HCOONa) = 1 + 12 + 32 + 23 = 68 g/mol → 68 g; methanol (CH4O) = 12 + 4 + 16 = 32 g/mol → 32 g.
5. Mass check: 60 + 40 = 100 = 68 + 32 — balanced.

Common Misconceptions

• "Acetaldehyde gives the Cannizzaro reaction." Wrong. Any aldehyde with an α-hydrogen undergoes aldol condensation instead; Cannizzaro is strictly for α-H-free aldehydes like formaldehyde and benzaldehyde.
• "In crossed Cannizzaro, either aldehyde may be oxidized." Wrong. Formaldehyde is always the one oxidized (to formate); the other aldehyde is reduced to its alcohol — this selectivity is synthetically useful.
• "Dilute NaOH is enough." Wrong. The reaction needs concentrated alkali (around 50%); with dilute base the hydride transfer is negligibly slow.

Real-World Applications

• Pentaerythritol: made industrially by crossed Cannizzaro-type chemistry of formaldehyde and acetaldehyde — used in alkyd resins and explosives.
• Benzyl alcohol: prepared from benzaldehyde by Cannizzaro; used in perfumes and as a solvent.
• Formaldehyde storage: explains why old formalin solutions contain formic acid and methanol — slow Cannizzaro on standing.

In short, aldehydes lacking α-hydrogens cannot do the aldol reaction, so with concentrated base they disproportionate instead: hydroxide attack, then rate-determining hydride transfer, gives one molecule of carboxylate and one of primary alcohol. Remember the entry ticket — no α-hydrogen — and the Cannizzaro versus aldol choice becomes automatic.