JEE Mains Chemistry · Aldehydes, Ketones and Carboxylic Acids

Crossed Aldol and Cannizzaro Reactions

Two different carbonyl compounds give a mixture of aldol products unless one of them has no α-hydrogen; an aldehyde with no α-hydrogen in concentrated alkali disproportionates instead, in the Cannizzaro reaction.

All Aldehydes, Ketones and Carboxylic Acids notes JEE Mains Chemistry strategy

Why this matters

Seventeen PYQs, three numerical, five from 2026. Nine mix two carbonyl partners and ask which products form, or how many; eight use the Cannizzaro reaction of an aldehyde with no α-hydrogen, alone or after an aldol step.

Concept 1 of 2: Crossed aldol: counting and naming the products

Every partner that has an α-hydrogen can act as the enolate, and every partner can be attacked. So count the possible pairs: each enolate with each carbonyl group. A partner with no α-hydrogen can only be attacked, which cuts the mixture down.

Definition

Two different aldehydes that both have α-hydrogens give four aldol products: two self and two crossed. The count assumes each partner has only one kind of α-carbon.
If one partner has no α-hydrogen (an aromatic aldehyde, methanal, 2,2-dimethylpropanal), there is only one enolate, and it gives two products. Using the non-enolisable partner in excess makes the crossed product the main one.
Claisen–Schmidt condensation: an aromatic aldehyde with an aldehyde or ketone that has α-hydrogens, in NaOH. Benzaldehyde and acetophenone give chalcone, $\mathrm{C_6H_5CH{=}CHCOC_6H_5}$ ; two benzaldehydes and one propanone give dibenzalacetone, $\mathrm{C_6H_5CH{=}CHCOCH{=}CHC_6H_5}$ .
Methanal is the best acceptor. With $\mathrm{R_2CHCHO}$ it adds once and stops at $\mathrm{HOCH_2CR_2CHO}$ , which has no α-hydrogen left. With excess methanal, every α-hydrogen is replaced by $\mathrm{CH_2OH}$ and a crossed Cannizzaro step then reduces the CHO: ethanal and four methanal give pentaerythritol, $\mathrm{C(CH_2OH)_4}$ .
An aldehyde with no α-hydrogen, such as vanillin, cannot undergo self-aldol condensation.

Number of aldol products (ignoring stereoisomers)

N = (\text{partners with an }\alpha\text{-H}) \times (\text{all partners})

Worked example

Name every aldol condensation product, ignoring stereoisomers, from a mixture of ethanal and butanal with dilute NaOH and heat.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 1 · Q31Moderate

Example 1 · Aldehydes, Ketones and Carboxylic Acids · Crossed Aldol and Cannizzaro Reactions

x

' is the product which is obtained from propanenitrile and stannous chloride in the presence of hydrochloric acid followed by hydrolysis. '

y

' is the product which is obtained from the but-2-ene by the ozonolysis followed by hydrolysis. From the following, which product is not obtained when one mole of '

x

' and one mole of ' y ' react with each other in the presence of alkali followed by heating?

Name each crossed product by which partner is the enolate

The enolate supplies the α-carbon, and the acceptor supplies the carbon that ends up doubly bonded to it. Swapping roles gives a different product, so write both crossed pairs separately.

A partner with no α-hydrogen is never the enolate

Aromatic aldehydes and methanal can only be attacked. Any option that needs them as the enolate is not formed.

Drill 8 more on crossed aldol: counting and naming the products

Concept 2 of 2: Cannizzaro reaction

An aldehyde with no α-hydrogen cannot form an enolate. In concentrated alkali, hydroxide adds to one molecule, and that adduct hands a hydride from its carbon to a second molecule. One aldehyde is oxidised to the carboxylate, the other reduced to the alcohol.

Definition

Needs an aldehyde with no α-hydrogen (methanal, aromatic aldehydes, $\mathrm{R_3CCHO}$ ) and concentrated alkali, such as 50% KOH.
It is a disproportionation: $\mathrm{2\,ArCHO \to ArCOO^- + ArCH_2OH}$ .
Crossed Cannizzaro with methanal: methanal is attacked first and is oxidised to methanoate; the other aldehyde is reduced to its alcohol.
The hydride moves from carbon to carbon, not from the solvent. In NaOD and $\mathrm{D_2O}$ , the $\mathrm{CH_2}$ of the alcohol carries no deuterium; D appears only on oxygen.
A dialdehyde can react within one molecule: glyoxal, OHC–CHO, gives glycolate, $\mathrm{HOCH_2COO^-}$ .

Cannizzaro and crossed Cannizzaro

\mathrm{2\,ArCHO \xrightarrow{conc.\ OH^-} ArCOO^- + ArCH_2OH} \qquad \mathrm{HCHO + ArCHO \xrightarrow{conc.\ OH^-} HCOO^- + ArCH_2OH}

Worked example

4-Chlorobenzaldehyde is heated with methanal and concentrated NaOH. Give the products.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 2 · Q43Moderate

Example 2 · Aldehydes, Ketones and Carboxylic Acids · Crossed Aldol and Cannizzaro Reactions

The compound A,

C_{8}H_{8}O_{2}

reacts with acetophenone to form a single product via crossAldol condensation. The compound A on reaction with conc. NaOH forms a substituted benzyl alcohol as one of the products. Compound A is:

Concentrated alkali, not dilute

Dilute base with an aldehyde that has α-hydrogens gives an aldol. The Cannizzaro reaction needs concentrated alkali and an aldehyde with no α-hydrogen.

The transferred hydrogen comes from carbon

Run in

\mathrm{D_2O}

with NaOD, the alcohol is

\mathrm{ArCH_2OD}

: the

\mathrm{CH_2}

keeps both hydrogens, because the hydride came from the other aldehyde, not from the solvent.

Drill 7 more on cannizzaro reaction

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Crossed aldol: counting and naming the products
Number of aldol products (ignoring stereoisomers)
$N = (\text{partners with an }\alpha\text{-H}) \times (\text{all partners})$
Cannizzaro reaction
Cannizzaro and crossed Cannizzaro
$\mathrm{2\,ArCHO \xrightarrow{conc.\ OH^-} ArCOO^- + ArCH_2OH} \qquad \mathrm{HCHO + ArCHO \xrightarrow{conc.\ OH^-} HCOO^- + ArCH_2OH}$

Watch out for (4)

Name each crossed product by which partner is the enolate
→ Crossed aldol: counting and naming the products
A partner with no α-hydrogen is never the enolate
→ Crossed aldol: counting and naming the products
Concentrated alkali, not dilute
→ Cannizzaro reaction
The transferred hydrogen comes from carbon
→ Cannizzaro reaction

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