Cannizzaro Reaction: Mechanism, Crossed Cannizzaro and Exam Traps

ChemistryClass 12, Aldehydes, Ketones and Carboxylic AcidsFor JEE and NEET aspirants

The Cannizzaro reaction is the self oxidation-reduction of an aldehyde that has no $α$ -hydrogen. Heated with concentrated NaOH or KOH, two molecules of the aldehyde react: one is oxidised to the carboxylate salt and the other is reduced to the primary alcohol. Methanal, for example, gives sodium methanoate and methanol. The Cannizzaro reaction is in the Class 12 NCERT chapter on aldehydes and ketones; JEE Main and JEE Advanced test it most, often together with aldol condensation.

At a glance

Needs: an aldehyde with no $α$ -hydrogen, so it cannot form an enolate: methanal, benzaldehyde and most ring-substituted benzaldehydes, 2,2-dimethylpropanal, ethanedial (glyoxal).
Reagent: concentrated (about 50%) NaOH or KOH, warmed. Acidify afterwards to get the free acid.
Products: one carboxylate salt and one primary alcohol for every two aldehyde molecules.
Methanal:
$2HCHO + NaOH conc. HCOONa + C H_{3} OH$
Benzaldehyde:
$2 C_{6} H_{5} CHO + KOH conc. C_{6} H_{5} COOK + C_{6} H_{5} C H_{2} OH$
Key step: a hydride ion ( $H^{-}$ ) moves from one aldehyde carbon to the other.
Type: disproportionation. One carbonyl carbon rises by 2 in oxidation number (aldehyde to acid), the other falls by 2 (aldehyde to alcohol).

1. Mechanism

Without an $α$ -hydrogen the base cannot make an enolate, so it does the only other thing it can: it adds to the C=O. The anion formed carries a C–H bond that can be pushed out as a hydride ion, and a second aldehyde molecule is right there to accept it.

Figure 1: Cannizzaro reaction of benzaldehyde. The new C–H bond of benzyl alcohol comes from the other aldehyde as a hydride ion, not from water. Step 2 is the slow step; step 3 cannot be reversed, which drives the reaction to completion.

Addition. $H O^{-}$ adds to the carbonyl carbon of one aldehyde, giving a tetrahedral anion. This step is reversible.
Hydride transfer (slow). The $O^{-}$ pushes back to re-form C=O. Its C–H electrons leave as a hydride ion and go straight to the carbonyl carbon of a second aldehyde. The first molecule becomes a carboxylic acid, the second an alkoxide ion.
Proton transfer (fast). The alkoxide, a strong base, takes the acid's proton: carboxylate plus alcohol. This step cannot be reversed, so it pulls the whole reaction forward.

Exam Trick

Read two things: the $α$ -H and the concentration. No $α$ -H with concentrated alkali: Cannizzaro. An $α$ -H with dilute alkali: aldol. In a crossed Cannizzaro with methanal, methanal is the one that gets oxidised, to methanoate.

2. Conditions and scope

An aldehyde with $α$ -hydrogens normally does not give the Cannizzaro reaction: the base removes an $α$ -hydrogen first and aldol condensation takes over. A simple ketone never gives it, because its carbonyl carbon has no H to pass on as a hydride.

Compound	$α$ -H	With concentrated NaOH
Methanal, $HCHO$	None	Cannizzaro: $HCO O^{-}$ + $C H_{3} OH$
Benzaldehyde, $C_{6} H_{5} CHO$	None	Cannizzaro: $C_{6} H_{5} CO O^{-}$ + $C_{6} H_{5} C H_{2} OH$
2,2-Dimethylpropanal, $(C H_{3})_{3} CCHO$	None	Cannizzaro: $(C H_{3})_{3} CCO O^{-}$ + $(C H_{3})_{3} CC H_{2} OH$
Ethanal, $C H_{3} CHO$	3	No Cannizzaro: aldol (and resin with strong alkali)
Phenylethanal, $C_{6} H_{5} C H_{2} CHO$	2	No Cannizzaro: the $C H_{2}$ next to CHO carries $α$ -H
Trichloroethanal (chloral), $CC l_{3} CHO$	None	Not Cannizzaro: the C–C bond breaks, giving $CHC l_{3}$ + $HCO O^{-}$
Propanone, $C H_{3} COC H_{3}$	6	Never: a ketone has no H on its carbonyl carbon

Chloral is the classic trap: it has no $α$ -H, yet it does not disproportionate. $H O^{-}$ adds to its C=O and the $CC l_{3}^{-}$ group leaves instead of a hydride, which is how pure chloroform is made:

CC l_{3} CHO + NaOH CHC l_{3} + HCOONa

3. Variants

3.1 Crossed Cannizzaro

Two different aldehydes without $α$ -hydrogens give four products in principle. With methanal as one partner the outcome is almost one-sided: $H O^{-}$ adds fastest to methanal, the least crowded and most reactive aldehyde, so methanal becomes the hydride donor and is oxidised to methanoate, while the other aldehyde is reduced to its alcohol. Methanal is cheap, so it is spent to turn a valuable aldehyde completely into its alcohol:

C_{6} H_{5} CHO + HCHO + NaOH conc. C_{6} H_{5} C H_{2} OH + HCOONa

3.2 Intramolecular Cannizzaro

When two carbonyl groups without $α$ -hydrogens sit close together in one molecule (a 1,2-dicarbonyl, or two CHO groups ortho on a benzene ring), the hydride moves inside the molecule and one molecule gives one hydroxy-acid salt. Ethanedial (glyoxal) gives the salt of glycolic acid (hydroxyethanoic acid):

OHCCHO + NaOH conc. HOC H_{2} COONa

Phenylglyoxal, $C_{6} H_{5} COCHO$ , gives the salt of mandelic acid: $H O^{-}$ adds to the more reactive aldehyde carbon and the hydride moves to the neighbouring ketone carbon.

C_{6} H_{5} COCHO + NaOH conc. C_{6} H_{5} CH (OH) COONa

Benzene-1,2-dicarbaldehyde (phthalaldehyde) gives 2-(hydroxymethyl)benzoate, which closes to a lactone (phthalide) on acidification.

3.3 After aldol steps

With excess methanal, an aldehyde that has $α$ -hydrogens first undergoes crossed aldol reactions until no $α$ -H is left, then a crossed Cannizzaro with one more methanal. Ethanal gives pentaerythritol, $C (C H_{2} OH)_{4}$ , this way (see the aldol condensation page).

JEE Advanced

Where the new hydrogen comes from. In $D_{2} O$ , benzaldehyde gives $C_{6} H_{5} C H_{2} OD$ : the solvent supplies only the O–D, never a C–D. With $C_{6} H_{5} CDO$ in $H_{2} O$ , the alcohol is $C_{6} H_{5} C D_{2} OH$ and the acid salt carries no deuterium. Both results show that the C–H of the alcohol arrives as a hydride from the other aldehyde.

Rate law. Rate $= k [RCHO]^{2} [O H^{-}]$ , so two aldehyde molecules and one hydroxide are involved up to the slow step. In very concentrated alkali a second term in $[O H^{-}]^{2}$ appears: the tetrahedral anion loses its O–H proton and the dianion is an even better hydride donor.

Tishchenko reaction. With aluminium ethoxide instead of NaOH, the same hydride transfer joins the acid and alcohol parts as an ester, and it works even for aldehydes with $α$ -H: $2C H_{3} CHO$ gives $C H_{3} COOC H_{2} C H_{3}$ .

4. Worked examples

Solved Example 1

Benzaldehyde is warmed with 50% NaOH and the mixture is then acidified. Name the products and say which aldehyde molecule is oxidised.

Solution:

Benzaldehyde has no $α$ -hydrogen, so it gives the Cannizzaro reaction: benzyl alcohol, $C_{6} H_{5} C H_{2} OH$ , and sodium benzoate, which acidification turns into benzoic acid, $C_{6} H_{5} COOH$ . They form in equal amounts. The molecule that took the $H O^{-}$ is oxidised (it gives up the hydride); the one that accepted the hydride is reduced.

Solved Example 2

4-Chlorobenzaldehyde is warmed with methanal and concentrated KOH. What are the main products?

Solution:

A crossed Cannizzaro with methanal: methanal is oxidised to potassium methanoate, $HCOOK$ , and 4-chlorobenzaldehyde is reduced to 4-chlorobenzyl alcohol, $Cl C_{6} H_{4} C H_{2} OH$ . The C–Cl bond on the ring is not touched under these conditions.

Solved Example 3

C_{6} H_{5} CDO

(deuterium on the aldehyde carbon) is warmed with concentrated NaOH in ordinary water. Where does the deuterium end up?

Solution:

The molecule that gives the hydride loses its C–D as $D^{-}$ , so its product, the benzoate, has no deuterium: $C_{6} H_{5} CO O^{-}$ . The molecule that receives it keeps its own C–D and gains the transferred one: the alcohol is $C_{6} H_{5} C D_{2} OH$ , with ordinary H on the oxygen from the water.

5. Exam traps

Watch out

Being an aldehyde is not enough: it must have no $α$ -H. Ethanal and phenylethanal do not give the Cannizzaro reaction.
Concentrated alkali, not dilute. Dilute alkali with an $α$ -H gives aldol condensation.
A simple ketone never gives the Cannizzaro reaction (no H on the carbonyl carbon). A statement such as "the ketone undergoes Cannizzaro reaction" is false.
In a crossed Cannizzaro with methanal, methanal is always the one oxidised (to methanoate); the other aldehyde becomes the alcohol.
The products are a salt and an alcohol in a 1:1 ratio, and the free acid appears only after acidification. In a self Cannizzaro, at most half of the aldehyde can become each product.
The H that reduces the second aldehyde comes from the first aldehyde as a hydride ion, not from water.
Chloral, $CC l_{3} CHO$ , has no $α$ -H but gives chloroform and methanoate, not a Cannizzaro reaction.
Ethanedial and phenylglyoxal react inside one molecule: one molecule gives one hydroxy-acid salt.

Past-paper questions on Cannizzaro reaction

Our past papers: JEE Main 2025-2026, JEE Advanced 2022-2026, NEET 2018-2019, 2022-2026. We found 10 questions in them that test this:

JEE Main 5 (2025-2026)
JEE Advanced 5 (2023-2025)

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Properties and Important Reactions of Aldehydes & Ketones Notes, formulas and examples →

Related named reactions

Aldol condensationAn aldehyde or ketone with an α-hydrogen forms an enolate that adds to a second C=O; heating removes water to give an α,β-unsaturated carbonyl.

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