Question
When formaldehyde (HCHO) is treated with concentrated NaOH, it undergoes the Cannizzaro reaction. Write the balanced equation and identify the oxidation and reduction products.
Also explain: why does formaldehyde undergo Cannizzaro reaction but acetaldehyde (CH₃CHO) does not?
Solution — Step by Step
Cannizzaro reaction only happens with aldehydes that have no alpha-hydrogen (no H on the carbon adjacent to the carbonyl). Formaldehyde (HCHO) has no carbon chain at all — so it qualifies. Acetaldehyde has the CH₃ group with alpha-H, so it goes for aldol condensation instead.
In Cannizzaro, two identical aldehyde molecules react. One gets oxidised (loses H, gains O) and one gets reduced (gains H). This is called disproportionation — a single compound acts as both oxidising and reducing agent simultaneously.
In basic medium (conc. NaOH), the formic acid immediately forms sodium formate. The balanced equation is:
Left molecule: oxidised to sodium formate (HCOONa)
Right molecule: reduced to methanol (CH₃OH)
In HCHO, the carbon oxidation state is 0.
In HCOONa (formate), carbon is +2 → oxidation.
In CH₃OH, carbon is −2 → reduction.
The shift is symmetric: +2 and −2 from 0. This confirms disproportionation.
Products: Sodium formate (HCOONa) + Methanol (CH₃OH)
If the question asks for acidification of the mixture, HCOONa gives formic acid (HCOOH). Remember to mention this when asked to identify “all products on workup.”
Why This Works
The mechanism runs through a hydride transfer. The OH⁻ from NaOH attacks one HCHO molecule, forming a tetrahedral alkoxide intermediate. This intermediate then transfers a hydride ion (H⁻) to a second HCHO molecule — the hydride acts as the reducing agent. The first molecule, having lost H, ends up as formate; the second, having received H, becomes methanol.
The reason alpha-hydrogen matters: if the aldehyde has alpha-H, the OH⁻ preferentially abstracts that proton (it’s acidic due to the adjacent carbonyl) and triggers an enolate → aldol condensation pathway. That pathway is faster and completely dominates. With no alpha-H available, the only reaction left is this intermolecular hydride transfer.
Where R = H (formaldehyde), aryl group (benzaldehyde), or any group with no alpha-H.
Alternative Method — Cross Cannizzaro
When formaldehyde reacts with another non-alpha-H aldehyde (like benzaldehyde), a crossed Cannizzaro occurs. Here, formaldehyde is always the one that gets oxidised (it’s the stronger reducing agent), and the other aldehyde gets reduced to its alcohol.
This is heavily tested in JEE Main. Formaldehyde preferentially oxidises because the resulting formate is more stable — remember: HCHO is always oxidised in cross Cannizzaro.
Cross Cannizzaro with benzaldehyde is the industrial route to benzyl alcohol. If a JEE problem says “formaldehyde + benzaldehyde + NaOH,” immediately write HCHO → HCOONa and PhCHO → PhCH₂OH. No need to think twice.
Common Mistake
Students often write the oxidised product as formic acid (HCOOH) instead of sodium formate (HCOONa) when the reaction is done in NaOH medium. Since NaOH is in excess and basic conditions are maintained throughout, the acid is immediately neutralised to its sodium salt. Writing HCOOH as a product in basic medium will cost you the mark — this was specifically penalised in the 2023 JEE Main marking scheme. Always write HCOONa unless the question says the mixture is acidified afterwards.