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Aldehydes in concentrated alkali can do two quite different things: an aldol condensation, or a Cannizzaro reaction. Picking the wrong one is the most common error in this part of the chapter, and it usually comes from not checking a single carbon atom. This post shows how to decide in one step, what the products are, how to handle the crossed reaction with formaldehyde, and how the stoichiometry limits the yield.
Where is this in the syllabus?
Both the JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus name the "Cannizzaro reaction" in Unit 17, Organic Compounds Containing Oxygen, next to aldol condensation and the haloform reaction. The JEE (Advanced) 2026 brochure lists it too. The NCERT Class 12 chapter Aldehydes, Ketones and Carboxylic Acids (2026-27 reprint) covers it under "Other reactions". The crossed reaction and the hydride-transfer step go a little beyond the NCERT text but appear in JEE questions. Check your own year's syllabus.
What does the reaction do?
NCERT's definition: aldehydes which do not have an α-hydrogen atom "undergo self oxidation and reduction (disproportionation) reaction on heating with concentrated alkali. In this reaction, one molecule of the aldehyde is reduced to alcohol while another is oxidised to carboxylic acid salt."
The two standard examples:
2HCHO + concentrated KOH → CH₃OH + HCOOK
2C₆H₅CHO + concentrated NaOH → C₆H₅CH₂OH + C₆H₅COONa
The reaction is named after Stanislao Cannizzaro, whose 1853 paper in Liebigs Annalen was titled "Ueber den der Benzoësäure entsprechenden Alkohol", on the alcohol corresponding to benzoic acid, that is, benzyl alcohol. Our profile of Cannizzaro covers the rest of his work.
Which aldehydes undergo it?
Look at the carbon next to the C=O, the α-carbon. If it carries a hydrogen, dilute base pulls that hydrogen off and you get an aldol reaction. If there is no α-hydrogen, no enolate can form, and concentrated base gives Cannizzaro instead. Ketones never give a Cannizzaro reaction: it needs the hydrogen on the aldehyde carbon, which is the one passed on as hydride.
Worked example 1: sort these six.
Methanal, HCHO. No α-carbon at all. Cannizzaro.
Ethanal, CH₃CHO. Three α-hydrogens. Aldol.
Benzaldehyde, C₆H₅CHO. The α-carbon is a ring carbon with no hydrogen. Cannizzaro.
2,2-Dimethylpropanal, (CH₃)₃CCHO. The α-carbon carries three methyl groups and no hydrogen. Cannizzaro, even though the molecule has nine hydrogens elsewhere.
2-Methylpropanal, (CH₃)₂CHCHO. One α-hydrogen is enough. Aldol.
4-Methoxybenzaldehyde. Ring carbon again. Cannizzaro.
How does one molecule reduce another?
Hydroxide adds to the carbonyl carbon of one aldehyde molecule. That intermediate then passes a hydride ion (H⁻) from the same carbon to the carbonyl carbon of a second molecule. The first becomes a carboxylic acid, which the base turns into its salt; the second becomes an alkoxide and then the alcohol. A review entry in Comprehensive Organic Name Reactions and Reagents (2010) states that the hydride transfer takes place directly between aldehydes, and that the reaction is second order in aldehyde.
How much product can you get?
The same review describes the products as an equimolar mixture of alcohol and carboxylic salt, "with optimal yield of 50%". The 50% follows from the equation: two molecules in, one alcohol out.
Worked example 2. You treat 10.6 g of benzaldehyde (molar mass 106.12 g/mol, PubChem) with concentrated NaOH.
Moles of benzaldehyde: 10.6/106.12 = 0.0999 mol.
Maximum benzyl alcohol: 0.0999/2 = 0.0499 mol. With molar mass 108.14 g/mol (PubChem), that is 5.40 g.
Maximum sodium benzoate: also 0.0499 mol. With molar mass 144.10 g/mol (PubChem), that is 7.20 g.
Acidifying the mixture turns the benzoate into benzoic acid.
What is the crossed Cannizzaro reaction?
When two different aldehydes without α-hydrogens react together, the one that is oxidised can differ from the one that is reduced. The 2010 review says this crossed version "is quite useful for synthesizing a high yield of alcohol". With formaldehyde as the partner, formaldehyde is oxidised and the other aldehyde is reduced. Tiwari and Srivastava (1956) describe the crossed reaction as "a very convenient and useful procedure for the reduction of aromatic and heterocyclic aldehydes to the corresponding alcohols", and a 1988 study of two substituted benzaldehydes with formaldehyde reports benzylic alcohols "together with formic acid" as the products.
Worked example 3. Benzaldehyde with excess HCHO and concentrated NaOH:
C₆H₅CHO + HCHO + NaOH → C₆H₅CH₂OH + HCOONa
Now, in principle, every benzaldehyde molecule can become benzyl alcohol, so 0.0999 mol could give up to 0.0999 mol, about 10.8 g. Formaldehyde is the sacrificial reagent.
Which mistakes cost marks?
Writing an aldol product for methanal or benzaldehyde on their own.
Writing Cannizzaro for 2-methylpropanal because it "looks crowded". It has an α-hydrogen.
Using dilute alkali. NCERT specifies concentrated alkali for Cannizzaro; dilute alkali is the aldol condition.
Giving the acid instead of the salt. In alkali the product is the carboxylate; the acid appears only after acidification.
Assuming 100% conversion to alcohol in a simple Cannizzaro reaction. The ceiling is half.
A practice routine
Draw ten aldehydes, circle the α-carbon in each, and sort them into aldol, Cannizzaro or neither.
Do NCERT Exercise 8.7, which asks exactly this for nine compounds.
Redo example 2 for 13.6 g of 4-methoxybenzaldehyde.
Write the crossed products for furfural with formaldehyde.
Revisit the full chapter in our guide to aldehyde and ketone reactions.
If you want to revise the α-hydrogen idea before more reactions, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, each with its own lesson.
Today, sort the six aldehydes in example 1 again from memory, giving the reason for each.
