Question
Formaldehyde (HCHO) undergoes the Cannizzaro reaction when treated with concentrated NaOH. Write the mechanism and identify the products formed. Why does formaldehyde undergo this reaction while acetaldehyde does not?
Solution — Step by Step
Cannizzaro reaction occurs with aldehydes that have no α-hydrogen — meaning no hydrogen on the carbon adjacent to the carbonyl group. Formaldehyde (HCHO) fits perfectly: the carbon carries only a hydrogen directly on the carbonyl, with no α-carbon at all.
Acetaldehyde (CH₃CHO) has three α-hydrogens on the methyl group, so it undergoes aldol condensation instead.
Concentrated NaOH provides a hydroxide ion, which attacks the electrophilic carbonyl carbon of one HCHO molecule. This forms a tetrahedral intermediate — a geminal diol anion:
The carbonyl carbon becomes hybridised at this point.
This is the heart of the mechanism. The hydride ion () is transferred intramolecularly from the tetrahedral intermediate to a second molecule of HCHO. Think of it as one formaldehyde molecule “donating” its hydrogen to reduce the other.
One molecule gets oxidised to formate, the other gets reduced to methanol.
After the reaction and acidification:
- Formic acid (methanoic acid) — HCOOH — from oxidation
- Methanol — CH₃OH — from reduction
In the presence of NaOH, you get sodium formate (HCOONa) directly. The final products are methanol and sodium formate in a 1:1 molar ratio.
When two different aldehydes (both without α-H) are used, the more easily oxidised one acts as the hydride donor. Formaldehyde is always the reducing agent in a crossed Cannizzaro because it is the most reactive aldehyde — a fact that appeared directly in JEE Main 2021.
Why This Works
The driving force is the stability of the products. Formate ion (HCOO⁻) is a resonance-stabilised carboxylate, and methanol is a neutral, stable alcohol. Both are thermodynamically very favourable compared to formaldehyde.
The reaction is a disproportionation — the same element (carbon) is simultaneously oxidised and reduced. No external oxidising or reducing agent is needed. The hydride transfer step is concerted or near-concerted, which is why conc. NaOH is required: you need a strong, concentrated nucleophile to push the reaction.
The absence of α-hydrogens is non-negotiable. If α-H were present, the base would deprotonate that position instead of acting as a nucleophile, and the reaction would divert to an enolate pathway (aldol).
Alternative Method — Cross-Cannizzaro
When formaldehyde is mixed with another α-hydrogen-free aldehyde (say, benzaldehyde) and treated with conc. NaOH, a crossed Cannizzaro occurs:
Formaldehyde acts as the reducing agent (it gets oxidised to formate), and benzaldehyde gets reduced to benzyl alcohol. This selectivity is because HCHO is more reactive toward nucleophilic addition — its carbonyl is less hindered and more electrophilic.
In any crossed Cannizzaro involving HCHO, formaldehyde is always the one that gets oxidised. This is a standard 1-mark MCQ fact in JEE Main and CBSE 12 practicals viva.
Common Mistake
Students often write that both formaldehyde molecules produce formic acid and methanol simultaneously. That is wrong. One molecule acts as the hydride donor (gets oxidised → formate) and the other is the hydride acceptor (gets reduced → methanol). The ratio is strictly 1:1, and they have opposite fates. Confusing this in a mechanism question costs marks in JEE Main and CBSE long-answer problems.
A related slip: writing the mechanism with a free H⁻ ion floating in solution. Hydride is never released as a free ion — the transfer happens directly from the tetrahedral intermediate to the second aldehyde molecule. Show the arrow pushing from C–H bond of the intermediate to the carbonyl carbon of HCHO.