ChemistryCBSE Class 12 · Aldehydes, Ketones and Carboxylic Acids

Aldol and Cannizzaro Reactions: Class 12 Chemistry

Updated 15 October 2026 · Fact-checked

Aldehydes and ketones with at least one α-hydrogen undergo aldol condensation in dilute alkali, forming β-hydroxy aldehydes or ketones that lose water on heating to give α,β-unsaturated compounds. Aldehydes with no α-hydrogen, such as methanal and benzaldehyde, undergo the Cannizzaro reaction in concentrated alkali: one molecule is reduced to an alcohol and another oxidised to a carboxylate salt.

Understand Aldol and Cannizzaro Reactions

Acidity of α-hydrogens. The carbon next to a carbonyl group is the α-carbon, and its hydrogens are α-hydrogens. They are weakly acidic for two reasons: the C=O group withdraws electrons, and when an α-hydrogen is removed by a base, the resulting carbanion (enolate) is stabilised by resonance, with the negative charge shared between the α-carbon and the carbonyl oxygen. This acidity is the basis of aldol reactions; whether α-hydrogens are present decides between aldol and Cannizzaro.

Aldol condensation. In the presence of dilute alkali, an aldehyde or ketone with at least one α-hydrogen forms an enolate, which attacks the carbonyl carbon of a second molecule. The product contains both an aldehyde (ald) and an alcohol (ol) group, hence the name aldol (from ketones, ketol). Two ethanal molecules give 3-hydroxybutanal: $2 C H_{3} C H O dil. NaOH C H_{3} C H (O H) C H_{2} C H O$ . On heating it loses water to give the α,β-unsaturated aldehyde but-2-enal: $C H_{3} C H = C H C H O$ . Propanone with $B a (O H)_{2}$ gives 4-hydroxy-4-methylpentan-2-one, which dehydrates to 4-methylpent-3-en-2-one. The overall process, addition followed by loss of water, is called aldol condensation, though the dehydrated product is the 'condensation product'.

Cross aldol condensation. When two *different* aldehydes and/or ketones, each with α-hydrogens, are used, each can act as the enolate and as the carbonyl partner. A mixture of four products results. Ethanal and propanal give but-2-enal and 2-methylpent-2-enal (self condensation) and pent-2-enal and 2-methylbut-2-enal (cross condensation). Cross aldol is useful when one partner has no α-hydrogen, because then only it can act as the carbonyl partner. Ketones can be one component: benzaldehyde and acetophenone in dilute NaOH give 1,3-diphenylprop-2-en-1-one (benzalacetophenone). Such reactions of aromatic aldehydes with ketones are called Claisen–Schmidt reactions.

Cannizzaro reaction. Aldehydes with no α-hydrogen cannot form an enolate. With concentrated alkali they undergo self oxidation and reduction (disproportionation): one molecule is reduced to an alcohol and another is oxidised to the salt of a carboxylic acid. $2 H C H O + conc. K O H \to C H_{3} O H + H C O O K$ ; $2 C_{6} H_{5} C H O + conc. N a O H \to C_{6} H_{5} C H_{2} O H + C_{6} H_{5} C O O N a$ . Other examples are 2,2-dimethylpropanal and substituted benzaldehydes. In methanal, carbon's oxidation state is 0; it becomes −2 in methanol and +2 in methanoate, a textbook disproportionation.

Ring substitution in aromatic carbonyl compounds. The –CHO and –COR groups withdraw electrons from the ring, so they are deactivating and meta-directing: benzaldehyde with conc. $H N O_{3}$ / $H_{2} S O_{4}$ gives 3-nitrobenzaldehyde.

Quick rule: α-H present + dilute alkali → aldol; no α-H + concentrated alkali → Cannizzaro.

Aldol and Cannizzaro Reactions: formulas and key facts

Aldol of ethanal: $2 C H_{3} C H O dil. NaOH C H_{3} C H (O H) C H_{2} C H O Δ, - H_{2} O C H_{3} C H = C H C H O$; 3-Hydroxybutanal (aldol) → but-2-enal.
Ketol of propanone: $2 C H_{3} C O C H_{3} B a (O H)_{2} (C H_{3})_{2} C (O H) C H_{2} C O C H_{3} Δ, - H_{2} O (C H_{3})_{2} C = C H C O C H_{3}$; 4-Hydroxy-4-methylpentan-2-one → 4-methylpent-3-en-2-one.
Claisen–Schmidt (cross aldol): $C_{6} H_{5} C H O + C H_{3} C O C_{6} H_{5} dil. NaOH C_{6} H_{5} C H = C H C O C_{6} H_{5} + H_{2} O$; Benzaldehyde has no α-H, so it is only the carbonyl partner.
Cannizzaro reaction: $2 H C H O + conc. K O H \to C H_{3} O H + H C O O K; 2 C_{6} H_{5} C H O + conc. N a O H \to C_{6} H_{5} C H_{2} O H + C_{6} H_{5} C O O N a$; Only for aldehydes without α-H.
Products of cross aldol: $two different carbonyls, both with α -H \Rightarrow 4 products$; Two self and two cross products.

How to solve Aldol and Cannizzaro Reactions questions

First decide which reaction a compound can undergo by checking α-hydrogens; then build the product carefully by joining the right atoms.

1Mark the α-carbon(s) next to C=O and count α-hydrogens.
2If there is at least one α-H and the alkali is dilute: aldol. If there is no α-H and the alkali is concentrated: Cannizzaro (aldehydes only).
3For aldol, take the α-carbon of one molecule (minus one H) and bond it to the carbonyl carbon of the other; the old C=O becomes C–OH. Name the β-hydroxy product.
4On heating, remove OH from the β-carbon and H from the α-carbon to form the C=C conjugated with C=O (possible only if the α-carbon still has an H).
5For cross aldol, write all four combinations (A+A, B+B, A+B, B+A) unless one partner lacks α-H.
6For Cannizzaro, write one alcohol and one carboxylate salt from two molecules of aldehyde, and give oxidation states if asked.
7For yield or mass questions, remember that two aldehyde molecules give one alcohol and one salt.

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 to make a flavour compound

Suppose a flavour chemist wants to make cinnamaldehyde, $C_{6} H_{5} C H = C H C H O$ , the compound responsible for the smell of cinnamon, from benzaldehyde and ethanal using dilute NaOH.

Show how the reaction forms cinnamaldehyde and explain why benzaldehyde cannot react with itself in this mixture. Which self-condensation product of ethanal might also form?

Try it first, then show the solutionSolution

Ethanal has α-hydrogens; dilute NaOH converts it into the enolate $^{-} C H_{2} C H O$ .
The enolate attacks the carbonyl carbon of benzaldehyde, giving $C_{6} H_{5} C H (O H) C H_{2} C H O$ ; on heating, loss of water gives $C_{6} H_{5} C H = C H C H O$ (cinnamaldehyde): $C_{6} H_{5} C H O + C H_{3} C H O \to C_{6} H_{5} C H = C H C H O + H_{2} O$ .
Benzaldehyde has no α-hydrogen, so it cannot form an enolate and cannot attack another benzaldehyde in an aldol reaction.
Ethanal can still condense with itself to give but-2-enal, so it is usually added slowly to excess benzaldehyde.

Answer: Ethanal's enolate adds to benzaldehyde and the product dehydrates to cinnamaldehyde; benzaldehyde has no α-H; but-2-enal is the possible side product.

Memory hook: In a cross aldol, the partner without α-H can only accept: it is the target, never the attacker.

Tricks to remember Aldol and Cannizzaro Reactions

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

α-H decides the road
Has α-H → aldol (dilute base). No α-H → Cannizzaro (concentrated base).
Use it when: First step of every question.
Cannizzaro = one up, one down
Two aldehyde molecules: one goes up to acid (salt), one goes down to alcohol.
Use it when: Writing Cannizzaro products.
β-OH, then double bond
Aldol puts OH on the β-carbon; heating removes it with an α-H to make the α,β-double bond.
Use it when: Dehydration products.

Aldol and Cannizzaro Reactions: solved examples

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

Example 1: Which gives Cannizzaro? · 1 mark · MCQBoard level

Which of the following undergoes the Cannizzaro reaction?
(a) Ethanal (b) Propanal (c) Propanone (d) Benzaldehyde

Try it first, then show the solutionSolution

Benzaldehyde has no α-hydrogen (the CHO is attached directly to the ring).

Answer: (d) Benzaldehyde.

Example 2: Aldol condensation of ethanal · 2 marks · Very short answerBoard level

Write the reactions involved in the aldol condensation of ethanal, naming the aldol and the final product.

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: Ketol from propanone and α-H acidity · 3 marks · Short answerHarder board level

(a) Write the aldol reaction of propanone in the presence of $B a (O H)_{2}$ and the product of its dehydration, with IUPAC names. (b) Why are the α-hydrogens of propanone acidic?

Try it first, then show the solutionSolution

(a) $2 C H_{3} C O C H_{3} \to (C H_{3})_{2} C (O H) C H_{2} C O C H_{3}$ : 4-hydroxy-4-methylpentan-2-one (1 mark).
On heating: $(C H_{3})_{2} C = C H C O C H_{3}$ , 4-methylpent-3-en-2-one (1 mark).
(b) The C=O group withdraws electrons from the α-carbon, and the carbanion formed by loss of an α-H is stabilised by resonance (negative charge shared with the carbonyl oxygen) (1 mark).

Answer: (a) 4-Hydroxy-4-methylpentan-2-one → 4-methylpent-3-en-2-one. (b) −I of C=O and resonance-stabilised enolate.

Example 4: Cross aldol of ethanal and propanal · 3 marks · Short answerHarder board level

A mixture of ethanal and propanal is treated with dilute NaOH and heated. Write the structures and names of the four α,β-unsaturated products and classify each as a self or cross product.

Try it first, then show the solutionSolution

Self (ethanal + ethanal): $C H_{3} C H = C H C H O$ , but-2-enal (½ mark).
Self (propanal + propanal): $C H_{3} C H_{2} C H = C (C H_{3}) C H O$ , 2-methylpent-2-enal (1 mark).
Cross (ethanal enolate + propanal): $C H_{3} C H_{2} C H = C H C H O$ , pent-2-enal (½ mark).
Cross (propanal enolate + ethanal): $C H_{3} C H = C (C H_{3}) C H O$ , 2-methylbut-2-enal (1 mark).

Answer: But-2-enal and 2-methylpent-2-enal (self); pent-2-enal and 2-methylbut-2-enal (cross).

Example 5: An aldehyde with no α-hydrogen · 5 marks · Long answerHOTS / competency

An aromatic compound A ( $C_{7} H_{6} O$ ) gives a silver mirror with Tollens' reagent. Heated with concentrated NaOH, it gives B ( $C_{7} H_{8} O$ ) and the sodium salt C. With acetophenone in dilute NaOH, A gives D ( $C_{15} H_{12} O$ ). (a) Identify A, B, C and D. (b) Name the reactions forming B/C and D. (c) If 2.12 g of A undergoes complete Cannizzaro reaction, what masses of B and C form? (A = 106, B = 108, C = 144 g mol⁻¹)

Try it first, then show the solutionSolution

(a) A: benzaldehyde, $C_{6} H_{5} C H O$ . B: benzyl alcohol, $C_{6} H_{5} C H_{2} O H$ . C: sodium benzoate, $C_{6} H_{5} C O O N a$ . D: 1,3-diphenylprop-2-en-1-one, $C_{6} H_{5} C H = C H C O C_{6} H_{5}$ (2 marks).
(b) B and C: Cannizzaro reaction (no α-H, conc. alkali). D: Claisen–Schmidt (cross aldol) condensation (1 mark).
(c) Moles of A $= 2.12/106 = 0.020$ mol; $2 C_{6} H_{5} C H O + N a O H \to C_{6} H_{5} C H_{2} O H + C_{6} H_{5} C O O N a$ , so 0.010 mol each (1 mark).
Mass of B $= 0.010 \times 108 = 1.08$ g; mass of C $= 0.010 \times 144 = 1.44$ g (1 mark).

Answer: (a) A benzaldehyde, B benzyl alcohol, C sodium benzoate, D benzalacetophenone. (b) Cannizzaro; Claisen–Schmidt. (c) 1.08 g B, 1.44 g C.

Where students lose marks in Aldol and Cannizzaro Reactions

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

Mistake: Applying Cannizzaro to aldehydes with α-H
Do this: Cannizzaro needs an aldehyde with no α-H (HCHO, $C_{6} H_{5} C H O$ , $(C H_{3})_{3} C C H O$ ).
Usually costs: 1-2 marks.
Mistake: Putting the OH on the wrong carbon in the aldol
Wrong working
$C H_{3} C H_{2} C H (O H) C H O$ from ethanal
Right working
$C H_{3} C H (O H) C H_{2} C H O$ (3-hydroxybutanal)
Do this: The former carbonyl carbon of the acceptor molecule carries the OH; it is β to the remaining CHO.
Usually costs: 1 mark.
Mistake: Listing only one product for a cross aldol
Do this: If both partners have α-H, give all four products.
Usually costs: 1-2 marks.
Mistake: Using dilute alkali for Cannizzaro or concentrated for aldol
Do this: Aldol: dilute alkali. Cannizzaro: concentrated alkali.
Usually costs: ½ mark.

Exam tips

Start every answer by stating whether the compound has α-hydrogens.
Name aldol products by IUPAC rules: number from the CHO carbon; the OH is on C3.
For Cannizzaro, write both products and name them (alcohol and salt).
Mention 'Claisen–Schmidt' by name for aromatic aldehyde + ketone condensations.
Show the dehydration step separately with 'heat' over the arrow.

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
For an aldehyde to undergo aldol condensation, it must have:
(a) no α-hydrogen (b) at least one α-hydrogen (c) an aromatic ring (d) a –COOH group
Hint
The enolate forms at the α-carbon.
Answer
(b).
Q2 · 2 marks · Very short answerBoard level
Write the Cannizzaro reaction of methanal with concentrated KOH and name the products.
Hint
One alcohol and one salt.
Answer
$2 H C H O + conc. K O H \to C H_{3} O H + H C O O K$ : methanol and potassium methanoate.
Q3 · 3 marks · Short answerHarder board level
Which of the following undergo aldol condensation and which undergo the Cannizzaro reaction? Methanal, 2-methylpropanal, benzaldehyde, propanone, 2,2-dimethylpropanal, ethanal.
Hint
Count α-hydrogens.
Answer
Aldol (have α-H): 2-methylpropanal, propanone, ethanal. Cannizzaro (no α-H): methanal, benzaldehyde, 2,2-dimethylpropanal.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The α-hydrogens of aldehydes and ketones are acidic.
Reason (R): The carbanion formed by loss of an α-hydrogen is stabilised by resonance with the carbonyl group.
(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
Enolate ion.
Answer
(a).
Q5 · 2 marks · Very short answerHarder board level
Write the Claisen–Schmidt reaction of benzaldehyde with acetophenone. Why does benzaldehyde not form a self-aldol product here?
Hint
Which partner has α-H?
Answer
$C_{6} H_{5} C H O + C H_{3} C O C_{6} H_{5} dil. NaOH C_{6} H_{5} C H = C H C O C_{6} H_{5} + H_{2} O$ (benzalacetophenone). Benzaldehyde has no α-hydrogen, so it cannot form an enolate; it acts only as the carbonyl partner.
Q6 · 2 marks · Very short answerHarder board level
What product is formed when benzaldehyde is nitrated with a mixture of conc. $H N O_{3}$ and conc. $H_{2} S O_{4}$ ? Explain the position of substitution.
Hint
–CHO withdraws electrons.
Answer
3-Nitrobenzaldehyde (m-nitrobenzaldehyde). The –CHO group withdraws electrons from the ring by resonance and induction, deactivating ortho and para positions more than meta, so substitution occurs at meta.
Q7 · 3 marks · Short answerHOTS / competency
2-Methylpropanal reacts with dilute NaOH. (a) Write the structure and IUPAC name of the aldol formed. (b) Can this aldol be dehydrated to an α,β-unsaturated aldehyde on heating? Explain. (c) What would 2,2-dimethylpropanal do under the same conditions, and in concentrated NaOH?
Hint
Look at the α-carbon of the product.
Answer
(a) $(C H_{3})_{2} C H - C H (O H) - C (C H_{3})_{2} - C H O$ : 3-hydroxy-2,2,4-trimethylpentanal. (b) No: the α-carbon (C2) now carries two methyl groups and no hydrogen, so water cannot be lost to form the C=C bond conjugated with CHO. (c) 2,2-Dimethylpropanal has no α-H, so it gives no aldol in dilute alkali; in concentrated NaOH it undergoes Cannizzaro reaction to 2,2-dimethylpropan-1-ol and sodium 2,2-dimethylpropanoate.
Q8 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose a student warms methanal with concentrated potassium hydroxide and finds that the solution contains methanol and potassium methanoate in equal amounts.
(a) Write the balanced equation. (1 mark)
(b) Give the oxidation state of carbon in methanal, methanol and the methanoate ion. (1 mark)
(c) Why is this called a disproportionation reaction? (1 mark)
(d) Why does ethanal, under similar alkaline conditions, not give this reaction? (1 mark)
Hint
Methanal has no α-H; ethanal has three.
Answer
(a) $2 H C H O + K O H \to C H_{3} O H + H C O O K$ . (b) 0 in methanal, −2 in methanol, +2 in methanoate. (c) The same substance is simultaneously reduced (to methanol) and oxidised (to methanoate). (d) Ethanal has α-hydrogens, so alkali forms its enolate and it undergoes aldol condensation instead.

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.Aldol of ethanal?Show answer
3-Hydroxybutanal.
2.Cannizzaro products of benzaldehyde?Show answer
Benzyl alcohol and sodium benzoate.
3.Number of products in a cross aldol of two different aldehydes with α-H?Show answer
Four.
4.Alkali strength for aldol?Show answer
Dilute.
5.Why can methanal not give aldol condensation?Show answer
It has no α-hydrogen.
6.Directing effect of –CHO on a benzene ring?Show answer
Meta-directing, deactivating.

Aldol and Cannizzaro Reactions: frequently asked questions

What is aldol condensation in Class 12 Chemistry?

A reaction in which an aldehyde or ketone with α-hydrogen, in dilute alkali, adds to another molecule to give a β-hydroxy aldehyde or ketone, which can lose water on heating to give an α,β-unsaturated carbonyl compound.

What is the Cannizzaro reaction?

A disproportionation of an aldehyde with no α-hydrogen in concentrated alkali, giving an alcohol and the salt of a carboxylic acid.

Which aldehydes give the Cannizzaro reaction?

Aldehydes without α-hydrogens, such as methanal, benzaldehyde and 2,2-dimethylpropanal.

Why are α-hydrogens acidic?

The carbonyl group withdraws electrons and the carbanion formed after losing an α-hydrogen is stabilised by resonance with the C=O group.

What is a Claisen–Schmidt reaction?

A cross aldol condensation between an aromatic aldehyde (no α-H) and an aldehyde or ketone with α-H, such as benzaldehyde with acetophenone.