ChemistryISC Class 12 · Aldehydes, Ketones and Carboxylic Acids

Aldol and Cannizzaro Reactions: Mechanisms for ISC Class 12

Updated 15 October 2026 · Fact-checked

Aldehydes and ketones with an α-hydrogen undergo aldol condensation with dilute alkali, giving a β-hydroxy carbonyl compound that loses water on heating. Aldehydes without an α-hydrogen (methanal, benzaldehyde) undergo the Cannizzaro reaction with concentrated alkali: one molecule is oxidised to a carboxylate and another reduced to an alcohol.

Understand Aldol and Cannizzaro Reactions

The α-hydrogen decides. The carbon next to a carbonyl group is the α-carbon; hydrogens on it are α-hydrogens. They are acidic, because the anion left after removing one (the enolate or carbanion) is stabilised by resonance with the C=O group. Whether a carbonyl compound has α-hydrogens decides which base-catalysed reaction it undergoes.

Aldol condensation (with α-H, dilute alkali). Two molecules of ethanal with dilute NaOH give 3-hydroxybutanal (aldol, from ALDehyde + alcohOL): $2 C H_{3} C H O \to C H_{3} C H (O H) C H_{2} C H O$ . On warming, the β-hydroxy aldehyde loses water to give an α,β-unsaturated aldehyde, but-2-enal: $C H_{3} C H (O H) C H_{2} C H O \to C H_{3} C H = C H C H O + H_{2} O$ . Ketones react too: propanone with Ba(OH)₂ gives 4-hydroxy-4-methylpentan-2-one (a ketol), $2 C H_{3} C O C H_{3} ⇌ (C H_{3})_{2} C (O H) C H_{2} C O C H_{3}$ , which dehydrates to 4-methylpent-3-en-2-one.

Mechanism of aldol condensation (ethanal). (1) OH⁻ removes an α-hydrogen to give the enolate ion: $C H_{3} C H O + O H^{-} ⇌^{-} C H_{2} C H O + H_{2} O$ . (2) The carbanion, a nucleophile, attacks the carbonyl carbon of a second ethanal molecule, forming a new C–C bond and an alkoxide ion, $C H_{3} C H (O^{-}) C H_{2} C H O$ . (3) The alkoxide takes H⁺ from water, giving the aldol and regenerating OH⁻ (the base is a catalyst). On heating, the α-H and β-OH are lost as water.

Cross aldol condensation. Two different carbonyl compounds, each with α-H, give four products (two self- and two crossed). It is useful when one partner has no α-H: then it can only accept, so fewer products form. Benzaldehyde with ethanal and dilute NaOH gives cinnamaldehyde (3-phenylprop-2-enal) after dehydration: $C_{6} H_{5} C H O + C H_{3} C H O \to C_{6} H_{5} C H = C H C H O + H_{2} O$ (the Claisen–Schmidt reaction).

Cannizzaro reaction (no α-H, concentrated alkali). Aldehydes with no α-hydrogen, such as methanal, benzaldehyde and 2,2-dimethylpropanal, undergo self oxidation-reduction (disproportionation) when heated with concentrated NaOH or KOH: one molecule becomes the alcohol and the other the salt of the carboxylic acid. $2 H C H O + N a O H \to C H_{3} O H + H C O O N a$ ; $2 C_{6} H_{5} C H O + N a O H \to C_{6} H_{5} C H_{2} O H + C_{6} H_{5} C O O N a$ .

Mechanism of Cannizzaro reaction. (1) OH⁻ adds to the carbonyl carbon of one aldehyde molecule, giving a tetrahedral anion. (2) This anion transfers a hydride ion (H⁻) to the carbonyl carbon of a second molecule: the first becomes a carboxylic acid, the second an alkoxide. (3) A proton passes from the acid to the alkoxide, giving the carboxylate ion and the alcohol. The hydride transfer is the key step.

Crossed Cannizzaro. With methanal and another aldehyde without α-H, methanal is always the one oxidised (to formate), because it is the most reactive towards OH⁻ attack; the other aldehyde is reduced: $H C H O + C_{6} H_{5} C H O + N a O H \to C_{6} H_{5} C H_{2} O H + H C O O N a$ . This is a good way to make benzyl alcohol.

Other condensations of benzaldehyde (ISC list). Benzoin condensation: with alcoholic KCN, two benzaldehyde molecules give benzoin, $2 C_{6} H_{5} C H O \to C_{6} H_{5} C H (O H) C O C_{6} H_{5}$ . Perkin reaction: benzaldehyde heated with acetic anhydride and sodium acetate gives cinnamic acid, $C_{6} H_{5} C H O + (C H_{3} C O)_{2} O \to C_{6} H_{5} C H = C H C O O H + C H_{3} C O O H$ .

Aldol and Cannizzaro Reactions: formulas and key facts

Aldol of ethanal: $2 C H_{3} C H O dil. NaOH C H_{3} C H (O H) C H_{2} C H O Δ C H_{3} C H = C H C H O + H_{2} O$; 3-Hydroxybutanal, then but-2-enal.
Aldol of propanone: $2 C H_{3} C O C H_{3} B a (O H)_{2} (C H_{3})_{2} C (O H) C H_{2} C O C H_{3}$; 4-Hydroxy-4-methylpentan-2-one; dehydrates to 4-methylpent-3-en-2-one.
Cannizzaro reaction: $2 C_{6} H_{5} C H O + N a O H conc., heat C_{6} H_{5} C H_{2} O H + C_{6} H_{5} C O O N a$; Aldehydes without α-H; disproportionation.
Crossed Cannizzaro: $H C H O + C_{6} H_{5} C H O + N a O H \to C_{6} H_{5} C H_{2} O H + H C O O N a$; Methanal is oxidised; the other aldehyde reduced.
Claisen–Schmidt (cross aldol): $C_{6} H_{5} C H O + C H_{3} C H O dil. NaOH C_{6} H_{5} C H = C H C H O + H_{2} O$; Cinnamaldehyde; benzaldehyde has no α-H.
Rule of thumb: $α -H present \Rightarrow aldol (dil. alkali); no α -H \Rightarrow Cannizzaro (conc. alkali)$; Check the carbon next to C=O first.

How to solve Aldol and Cannizzaro Reactions questions

Every question here starts with one check: does the carbonyl compound have an α-hydrogen? That decides aldol or Cannizzaro; the rest is writing the right product and mechanism.

1Mark the α-carbon(s) and count α-hydrogens. Methanal, benzaldehyde and 2,2-dimethylpropanal have none.
2With α-H and dilute alkali: join the α-carbon of one molecule to the carbonyl carbon of another; the carbonyl O becomes OH (β-hydroxy compound). With heat, remove H₂O to give C=C next to C=O.
3Without α-H and with concentrated alkali: write the alcohol (–CH₂OH) and the sodium carboxylate (–COONa) from two molecules.
4For cross reactions, list all possible products, then cut down using 'no α-H means acceptor only' and 'methanal is always oxidised'.
5For mechanisms: aldol = enolate formation, nucleophilic attack, protonation; Cannizzaro = OH⁻ addition, hydride transfer, proton exchange.
6Name products with IUPAC names (3-hydroxybutanal, but-2-enal, phenylmethanol, sodium benzoate).

Aldol and Cannizzaro Reactions in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Cross aldol makes the flavour of cinnamon

Suppose a flavour company makes cinnamaldehyde (M = 132), the main flavour compound of cinnamon, by stirring 10.6 g of benzaldehyde (M = 106) with ethanal in dilute NaOH and then warming. It isolates 9.9 g of cinnamaldehyde.

Write the equation, calculate the percentage yield, and explain why benzaldehyde is used with ethanal rather than two different aldehydes that both have α-hydrogens.

Try it first, then show the solutionSolution

$C_{6} H_{5} C H O + C H_{3} C H O \to C_{6} H_{5} C H = C H C H O + H_{2} O$ (cross aldol, then dehydration).
Moles of benzaldehyde $= \frac{10.6}{106} = 0.10$ mol; theoretical cinnamaldehyde $= 0.10 \times 132 = 13.2$ g.
Percentage yield $= \frac{9.9}{13.2} \times 100 = 75%$ .
Benzaldehyde has no α-H, so it cannot form an enolate and only accepts attack; the main crossed product is cinnamaldehyde. Two partners with α-H would give four aldol products.

Answer: Yield 75%. Benzaldehyde can act only as the acceptor, so cross aldol gives mainly one crossed product.

Memory hook: No α-H, no enolate: the aldehyde can only receive.

Tricks to remember Aldol and Cannizzaro Reactions

Short memory aids. Say each one aloud once, then use it on a question the same day.

Alpha decides the fate
α-H present → Aldol (both start with A). No α-H → Cannizzaro.
Use it when: First step in every question.
Dilute for aldol, concentrated for Cannizzaro
Dilute alkali makes a little enolate (aldol); concentrated alkali attacks the C=O directly (Cannizzaro).
Use it when: Reading reaction conditions.
Methanal always gets oxidised
In a crossed Cannizzaro, methanal is the 'giver' of the hydride and ends up as formate.
Use it when: Predicting crossed Cannizzaro products.

Aldol and Cannizzaro Reactions: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: No aldol here · 1 mark · MCQBoard level

Which of the following does NOT undergo aldol condensation?
(a) ethanal
(b) propanal
(c) propanone
(d) benzaldehyde

Try it first, then show the solutionSolution

Aldol condensation needs an α-hydrogen.
Benzaldehyde's carbonyl carbon is attached directly to the ring; it has no α-hydrogen.

Answer: (d) benzaldehyde.

Example 2: Aldol of ethanal · 2 marks · Short answerBoard level

Write the equations for the aldol condensation of ethanal with dilute NaOH and for the dehydration of the product on heating. Name both products.

Try it first, then show the solutionSolution

$2 C H_{3} C H O dil. NaOH C H_{3} C H (O H) C H_{2} C H O$ : 3-hydroxybutanal (aldol) (1 mark).
$C H_{3} C H (O H) C H_{2} C H O Δ C H_{3} C H = C H C H O + H_{2} O$ : but-2-enal (1 mark).

Answer: 3-Hydroxybutanal, then but-2-enal.

Example 3: Mechanism of aldol condensation of propanal · 3 marks · Short answerHarder board level

Write the mechanism of the base-catalysed aldol condensation of propanal, and give the product formed when the aldol is heated.

Try it first, then show the solutionSolution

Step 1: OH⁻ removes an α-H from C-2: $C H_{3} C H_{2} C H O + O H^{-} ⇌ C H_{3} C - H C H O + H_{2} O$ ; the carbanion is resonance-stabilised as an enolate (1 mark).
Step 2: the carbanion attacks the carbonyl carbon of a second propanal molecule, forming a C–C bond and the alkoxide CH₃CH₂CH(O⁻)CH(CH₃)CHO; Step 3: the alkoxide takes H⁺ from water, giving 3-hydroxy-2-methylpentanal and regenerating OH⁻ (1 mark).
On heating, the α-H and β-OH are lost as water: 2-methylpent-2-enal, CH₃CH₂CH=C(CH₃)CHO (1 mark).

Answer: Enolate formation, nucleophilic addition, protonation give 3-hydroxy-2-methylpentanal; heating gives 2-methylpent-2-enal.

Example 4: A crossed Cannizzaro with two aldehydes lacking α-H · 3 marks · Short answerHarder board level

A mixture of methanal and 2,2-dimethylpropanal is heated with concentrated NaOH. Write the equation for the main reaction, name the products, and explain which aldehyde is oxidised.

Try it first, then show the solutionSolution

Neither aldehyde has an α-H, so both can take part in a Cannizzaro reaction (½ mark).
$H C H O + (C H_{3})_{3} C C H O + N a O H \to (C H_{3})_{3} C C H_{2} O H + H C O O N a$ (1 mark).
Products: 2,2-dimethylpropan-1-ol and sodium methanoate (½ mark).
Methanal is oxidised: its carbonyl carbon is the least hindered and most electrophilic, so OH⁻ adds to it fastest and its adduct transfers the hydride (1 mark).

Answer: 2,2-Dimethylpropan-1-ol and sodium formate; methanal is oxidised.

Example 5: Four aldols and a dehydration · 5 marks · Long answerHOTS / competency

A mixture of ethanal and propanal is treated with dilute NaOH.
(a) Write the structures and IUPAC names of the four β-hydroxy aldehydes that can form. [2]
(b) Give the product formed when the self-aldol of propanal is heated, with its IUPAC name. [1]
(c) Explain why a mixture of benzaldehyde and ethanal gives far fewer products under the same conditions. [2]

Try it first, then show the solutionSolution

(a) Ethanal + ethanal: CH₃CH(OH)CH₂CHO, 3-hydroxybutanal. Propanal + propanal: CH₃CH₂CH(OH)CH(CH₃)CHO, 3-hydroxy-2-methylpentanal (1 mark).
Ethanal enolate + propanal: CH₃CH₂CH(OH)CH₂CHO, 3-hydroxypentanal. Propanal enolate + ethanal: CH₃CH(OH)CH(CH₃)CHO, 3-hydroxy-2-methylbutanal (1 mark).
(b) $C H_{3} C H_{2} C H (O H) C H (C H_{3}) C H O \to C H_{3} C H_{2} C H = C (C H_{3}) C H O + H_{2} O$ : 2-methylpent-2-enal (1 mark).
(c) Benzaldehyde has no α-H and cannot form an enolate, so it can only be attacked; the ethanal enolate attacks it (crossed product, cinnamaldehyde after dehydration) (1 mark), and only the self-aldol of ethanal competes, so two products instead of four (1 mark).

Answer: (a) 3-Hydroxybutanal, 3-hydroxy-2-methylpentanal, 3-hydroxypentanal, 3-hydroxy-2-methylbutanal. (b) 2-Methylpent-2-enal. (c) Benzaldehyde can only be an acceptor.

Where students lose marks in Aldol and Cannizzaro Reactions

The slips examiners see most often, and what to write instead.

Mistake: Writing the free acid instead of its salt
ISC examiners reported C₆H₅COOH written for the Cannizzaro product instead of sodium benzoate, C₆H₅COONa (2024).
Wrong working
$2 C_{6} H_{5} C H O \to C_{6} H_{5} C H_{2} O H + C_{6} H_{5} C O O H$
Right working
$2 C_{6} H_{5} C H O + N a O H \to C_{6} H_{5} C H_{2} O H + C_{6} H_{5} C O O N a$
Do this: In alkali the acid is present as its sodium salt; include NaOH in the equation.
Usually costs: 1 mark.
Mistake: Wrong Cannizzaro or decarboxylation equations
Examiners noted wrong Cannizzaro equations and decarboxylation of sodium benzoate written without heat or soda lime (2024).
Do this: Cannizzaro: concentrated alkali, heat. Decarboxylation: $C_{6} H_{5} C O O N a + N a O H CaO, heat C_{6} H_{6} + N a_{2} C O_{3}$ .
Usually costs: 1 mark each.
Mistake: Confusing dilute and concentrated alkali
Examiners stressed the difference between dilute and concentrated NaOH (2024).
Do this: Dilute alkali: aldol (α-H compounds). Concentrated alkali: Cannizzaro (no α-H).
Usually costs: 1 mark.
Mistake: Wrong aldol structure
Students join the two carbonyl carbons directly.
Do this: The new C–C bond is between the α-carbon of one molecule and the carbonyl carbon of the other; the OH ends up on the β-carbon.
Usually costs: 1 mark.

Exam tips

Before writing any product, circle the α-carbon and count its hydrogens; write this check in your answer.
Name the reactions: aldol condensation, Claisen–Schmidt, Cannizzaro, crossed Cannizzaro, benzoin condensation, Perkin reaction.
Use structural formulae, not molecular formulae, for products; examiners asked for this in 2024.
In mechanism answers, show curved arrows for each step and label the enolate or hydride transfer.
When dehydration is asked, show the C=C between the α- and β-carbons, conjugated with C=O.

Practice questions on Aldol and Cannizzaro Reactions

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The Cannizzaro reaction is given by aldehydes that:
(a) have at least one α-hydrogen, with dilute NaOH
(b) have no α-hydrogen, with concentrated NaOH
(c) are aromatic only, with acid
(d) have two α-hydrogens, with Tollens' reagent
Hint
Think of methanal and benzaldehyde.
Answer
(b).
Q2 · 2 marks · Short answerBoard level
Write the equation for the Cannizzaro reaction of methanal and name the products.
Hint
One molecule oxidised, one reduced.
Answer
$2 H C H O + N a O H conc. C H_{3} O H + H C O O N a$ ; methanol and sodium methanoate (sodium formate).
Q3 · 2 marks · Short answerHarder board level
Write the aldol product of butanal with dilute NaOH and the product formed when it is heated. Give IUPAC names.
Hint
The α-carbon is C-2, which carries an ethyl group after joining.
Answer
$2 C H_{3} C H_{2} C H_{2} C H O \to C H_{3} C H_{2} C H_{2} C H (O H) C H (C_{2} H_{5}) C H O$ : 2-ethyl-3-hydroxyhexanal. On heating it loses water to give CH₃CH₂CH₂CH=C(C₂H₅)CHO, 2-ethylhex-2-enal.
Q4 · 3 marks · Short answerHarder board level
Write the mechanism of the Cannizzaro reaction of methanal with concentrated NaOH.
Hint
OH⁻ addition, hydride transfer, proton exchange.
Answer
(1) OH⁻ adds to the carbonyl carbon of one HCHO: H₂C(OH)O⁻. (2) This anion transfers a hydride ion to a second HCHO, giving HCOOH and CH₃O⁻. (3) Proton exchange gives HCOO⁻ (as sodium methanoate) and CH₃OH (methanol).
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): 2,2-Dimethylpropanal undergoes the Cannizzaro reaction with concentrated NaOH.
Reason (R): 2,2-Dimethylpropanal has no α-hydrogen atom.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Look at the carbon next to CHO in (CH₃)₃CCHO.
Answer
(a).
Q6 · 3 marks · Short answerHarder board level
Predict the main organic products and name the reaction in each case: (a) 4-chlorobenzaldehyde heated with concentrated NaOH; (b) ethanal with dilute NaOH at room temperature; (c) methanal and 4-methylbenzaldehyde heated with concentrated NaOH.
Hint
Check for α-hydrogens first.
Answer
(a) Cannizzaro: 4-chlorobenzyl alcohol and sodium 4-chlorobenzoate. (b) Aldol condensation: 3-hydroxybutanal. (c) Crossed Cannizzaro: 4-methylbenzyl alcohol and sodium methanoate (methanal is oxidised).
Q7 · 3 marks · Short answerHOTS / competency
21.2 g of benzaldehyde (M = 106) is heated with excess concentrated NaOH until the Cannizzaro reaction is complete. Calculate the masses of benzyl alcohol (M = 108) and sodium benzoate (M = 144) formed.
Hint
2 mol aldehyde give 1 mol of each product.
Answer
Benzaldehyde $= \frac{21.2}{106} = 0.20$ mol; each product $= 0.10$ mol. Benzyl alcohol $= 10.8$ g; sodium benzoate $= 14.4$ g.
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a student is given an aldehyde X (C₅H₁₀O). With dilute NaOH it shows no aldol reaction, but on heating with concentrated NaOH it gives an alcohol Y and the sodium salt Z of a carboxylic acid, both with five carbon atoms.
(i) Identify X, and explain why it gives no aldol product. [1]
(ii) Write the equation for the reaction with concentrated NaOH. [1]
(iii) Name Y and Z. [1]
Hint
Which C₅ aldehyde has a quaternary carbon next to CHO?
Answer
(i) X is 2,2-dimethylpropanal, (CH₃)₃CCHO; the carbon next to CHO carries no hydrogen, so no enolate can form. (ii) $2 (C H_{3})_{3} C C H O + N a O H \to (C H_{3})_{3} C C H_{2} O H + (C H_{3})_{3} C C O O N a$ . (iii) Y: 2,2-dimethylpropan-1-ol; Z: sodium 2,2-dimethylpropanoate.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is an aldol?Show answer
A β-hydroxy aldehyde (or ketone) formed by aldol condensation, e.g. 3-hydroxybutanal.
2.Key step in the Cannizzaro mechanism?Show answer
Transfer of a hydride ion from the OH⁻ adduct of one aldehyde to a second aldehyde molecule.
3.Product of benzaldehyde with alcoholic KCN?Show answer
Benzoin.
4.Which aldehyde is always oxidised in a crossed Cannizzaro?Show answer
Methanal.
5.Name an aldehyde with α-H that gives aldol condensation.Show answer
Ethanal (or propanal).
6.Product of the Perkin reaction of benzaldehyde?Show answer
Cinnamic acid.

Aldol and Cannizzaro Reactions: frequently asked questions

What is the difference between aldol and Cannizzaro reactions?

Aldol condensation needs an α-hydrogen and dilute alkali and joins two molecules by a C–C bond. Cannizzaro needs an aldehyde without α-hydrogen and concentrated alkali, and converts two molecules into an alcohol and a carboxylate.

Why do aldehydes without α-hydrogen not give aldol condensation?

They cannot form an enolate (carbanion), which is the nucleophile in the aldol reaction.

What is cross aldol condensation?

Aldol condensation between two different carbonyl compounds; if both have α-H, four products form.

Is the Cannizzaro reaction a redox reaction?

Yes. It is a disproportionation: one aldehyde molecule is oxidised to a carboxylate and another reduced to an alcohol.

Why do the products of aldol condensation lose water easily?

Loss of the α-H and β-OH gives a C=C bond conjugated with the C=O group, which is stabilised.