Aldehydes WITHOUT α-H (HCHO, C₆H₅CHO) in presence of conc. NaOH undergo disproportionation: 2 RCHO → RCOO⁻ + RCH₂OH. One gets oxidised, other reduced.
-- NCERT Class 12 Chemistry, Ch. 8, p. 242Cannizzaro
8 MCQs9-step worked example
Source: NCERT Alcohols, Phenols, Ethers and Carbonyl CompoundsOfficial key: NTA-verifiedLast updated: 27 Sep 2026
Cannizzaro, explained for NEET
The Cannizzaro reaction is a disproportionation — one molecule of aldehyde is oxidised to a carboxylate salt while an identical molecule is reduced to an alcohol. The key structural requirement: the aldehyde must lack an α-hydrogen. If α-H is present, the base extracts it and an aldol condensation occurs instead.
Reaction conditions: concentrated NaOH (or KOH), no catalyst, no external oxidant/reductant. The base is stoichiometric, not catalytic — one equivalent is consumed forming the carboxylate salt.
Mechanism (simplified for NEET recall):
- OH⁻ attacks the carbonyl carbon of one HCHO molecule → alkoxide tetrahedral intermediate.
- Hydride transfer from this intermediate to a second HCHO molecule.
- Products: HCOONa (sodium formate, oxidised) + CH₃OH (methanol, reduced).
Substrates that undergo Cannizzaro:
- Formaldehyde (HCHO) — the classic example.
- Benzaldehyde (C₆H₅CHO) — gives sodium benzoate + benzyl alcohol.
- Trimethylacetaldehyde ((CH₃)₃CCHO) — no α-H due to quaternary carbon.
Substrates that do NOT undergo Cannizzaro:
- Acetaldehyde (CH₃CHO) — has α-H → aldol instead.
- Propionaldehyde (CH₃CH₂CHO) — has α-H → aldol instead.
- Any ketone — Cannizzaro is specific to aldehydes without α-H.
Crossed Cannizzaro: When HCHO is mixed with another non-α-H aldehyde (e.g., benzaldehyde) in conc. NaOH, HCHO preferentially gets oxidised (to formate) and the other aldehyde gets reduced (to alcohol). HCHO is a better hydride donor because its carbonyl is least sterically hindered.
NEET watch-out: The trap is confusing "no α-hydrogen" with "no hydrogen at all." The requirement is specifically no hydrogen on the carbon adjacent to carbonyl. Formaldehyde has no α-carbon at all (carbonyl carbon is bonded only to H atoms), so it qualifies.
(NCERT Class 12 Chemistry Chapter 8, Part 2, page 242)
Can you answer these Cannizzaro MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Which of the following aldehydes will undergo Cannizzaro reaction when treated with concentrated NaOH?
Show answer and why every option is right or wrong
Answer: A. Benzaldehyde (C₆H₅CHO) has no α-hydrogen — the carbon adjacent to the carbonyl is part of the aromatic ring with no H attached. It undergoes Cannizzaro reaction with conc. NaOH to give sodium benzoate and benzyl alcohol. (NCERT Class 12 Chemistry Chapter 8, Part 2, page 242)
Why B is wrong: B is wrong because acetaldehyde has three α-hydrogens on the methyl group adjacent to carbonyl — it undergoes aldol condensation with base, not Cannizzaro.
Why C is wrong: C is wrong because propionaldehyde has two α-hydrogens on the CH₂ adjacent to carbonyl — it undergoes aldol condensation, not Cannizzaro.
Why D is wrong: D is wrong because isobutyraldehyde has one α-hydrogen on the CH adjacent to carbonyl — it undergoes aldol condensation, not Cannizzaro.
MCQ 2Easy RecallPractice
In the Cannizzaro reaction of formaldehyde with concentrated NaOH, the products are:
Show answer and why every option is right or wrong
Answer: D. Two molecules of HCHO undergo disproportionation: one is oxidised to HCOONa (sodium formate) and the other is reduced to CH₃OH (methanol). The base is consumed stoichiometrically. (NCERT Class 12 Chemistry Chapter 8, Part 2, page 242)
Why A is wrong: A is wrong because Cannizzaro is a redox disproportionation producing two different products (one oxidised, one reduced), not a dimerisation.
Why B is wrong: B is wrong because the product is sodium formate (HCOONa), not free formic acid — the concentrated NaOH provides the Na⁺ counterion and the medium is strongly basic.
Why C is wrong: C is wrong because the reduced product is methanol (CH₃OH from HCHO gaining 2H), not ethanol. Ethanol would require a two-carbon aldehyde as substrate.
MCQ 3CalculationPractice
In the Cannizzaro reaction, formaldehyde (HCHO) disproportionates into sodium formate (HCOO⁻Na⁺) and methanol (CH₃OH). Using standard oxidation-number rules for carbon (each C–O bond, including one bond of a double bond, gives carbon +1, since O is more electronegative than C; each C–H bond gives carbon −1, since H is less electronegative than C), what are the oxidation numbers of carbon in HCHO, HCOO⁻, and CH₃OH, and by how many units is carbon oxidised and reduced respectively?
Show answer and why every option is right or wrong
Answer: A. In HCHO, carbon has 2 C–H bonds (2 × −1 = −2) and a C=O double bond, counted as two bonds to O (2 × +1 = +2); total = 0. In HCOO⁻, carbon has 1 C–H bond (−1), a C=O double bond (+2), and a C–O⁻ single bond (+1); total = +2. In CH₃OH, carbon has 3 C–H bonds (3 × −1 = −3) and one C–O single bond (+1); total = −2. Going from HCHO (0) to HCOO⁻ (+2) is oxidation by 2 units, and from HCHO (0) to CH₃OH (−2) is reduction by 2 units — matching the disproportionation NCERT describes: "one molecule of the aldehyde is reduced to alcohol while another is oxidised to carboxylic acid salt." (NCERT Class 12 Chemistry Chapter 8, Part 2, page 242)
Why B is wrong: B assigns HCHO an oxidation number of +2 instead of 0, incorrectly treating each C–H bond as +1 rather than −1 (H is less electronegative than C, so C–H bonds lower carbon's oxidation number, they do not raise it).
Why C is wrong: C treats each C–O bond as contributing only about +0.5 instead of the correct +1 per bond, which understates the swing to ±1 instead of the correct ±2.
Why D is wrong: D reverses the sign of both changes, treating the oxygen-richer carboxylate carbon as more reduced and the hydrogen-richer alcohol carbon as more oxidised — the opposite of the correct assignment.
MCQ 4Direct ApplicationPractice
In a crossed Cannizzaro reaction between formaldehyde and benzaldehyde in concentrated NaOH, which statement is correct?
Show answer and why every option is right or wrong
Answer: A. In crossed Cannizzaro, HCHO preferentially acts as the hydride donor (oxidised to formate) because its carbonyl is least sterically hindered. Benzaldehyde accepts the hydride and is reduced to benzyl alcohol. (NCERT Class 12 Chemistry Chapter 8, Part 2, page 242)
Why B is wrong: B is wrong because benzaldehyde is the hydride ACCEPTOR in crossed Cannizzaro — it gets reduced to benzyl alcohol, not oxidised to benzoate.
Why C is wrong: C is wrong because in crossed Cannizzaro, formaldehyde is the preferred hydride DONOR — it gets oxidised to formate, not reduced to methanol.
Why D is wrong: D is wrong because disproportionation requires one molecule oxidised and another reduced. Both cannot be oxidised — there is no external oxidant. In crossed Cannizzaro, HCHO is selectively oxidised.
MCQ 5Direct ApplicationPractice
Which of the following will NOT undergo Cannizzaro reaction?
Show answer and why every option is right or wrong
Answer: C. Acetone (CH₃COCH₃) is a ketone, not an aldehyde. Cannizzaro reaction is specific to aldehydes without α-hydrogen. Ketones do not undergo this reaction because the mechanism requires hydride transfer from the carbonyl carbon, which in ketones bears R groups instead of H. (NCERT Class 12 Chemistry Chapter 8, Part 2, page 242)
Why A is wrong: A is wrong because formaldehyde (HCHO) is the classic Cannizzaro substrate — it has no α-carbon at all, hence no α-hydrogen.
Why B is wrong: B is wrong because trimethylacetaldehyde has a quaternary carbon adjacent to carbonyl (no α-H), making it a valid Cannizzaro substrate.
Why D is wrong: D is wrong because benzaldehyde has no α-hydrogen (the adjacent carbon is aromatic ring carbon with no H), making it a classic Cannizzaro substrate.
MCQ 6Easy RecallPractice
The essential structural requirement for an aldehyde to undergo Cannizzaro reaction is:
Show answer and why every option is right or wrong
Answer: C. Cannizzaro reaction occurs only with aldehydes that lack α-hydrogen. If α-H is present, the strongly basic conditions favour aldol condensation instead. (NCERT Class 12 Chemistry Chapter 8, Part 2, page 242)
Why A is wrong: A is wrong because presence of α-hydrogen causes aldol condensation to dominate under basic conditions, preventing Cannizzaro reaction.
Why B is wrong: B is wrong because electron-withdrawing groups are not a requirement for Cannizzaro — the only requirement is absence of α-hydrogen. EWGs may affect rate but are neither necessary nor sufficient.
Why D is wrong: D is wrong because a methyl group adjacent to carbonyl provides three α-hydrogens, which would make the substrate undergo aldol condensation rather than Cannizzaro reaction.
MCQ 7Direct ApplicationPractice
When benzaldehyde undergoes Cannizzaro reaction with concentrated KOH, the organic products are:
Show answer and why every option is right or wrong
Answer: B. With KOH as the base, the oxidised product is potassium benzoate (C₆H₅COOK), not sodium benzoate or free acid. The reduced product is benzyl alcohol (C₆H₅CH₂OH). The counterion matches the base used. (NCERT Class 12 Chemistry Chapter 8, Part 2, page 242)
Why A is wrong: A is wrong because the product is the carboxylate salt (potassium benzoate), not free benzoic acid — the reaction medium is strongly basic (conc. KOH), so the acid exists as its potassium salt.
Why C is wrong: C is wrong because the base used is KOH, not NaOH — the counterion in the product salt must be K⁺, giving potassium benzoate, not sodium benzoate.
Why D is wrong: D is wrong because the reduced product is benzyl alcohol (C₆H₅CH₂OH), not toluene. The carbonyl is reduced to -CH₂OH, not to -CH₃. Reduction to toluene would require removal of oxygen entirely.
MCQ 8Concept TrapPractice
Consider the following aldehydes:
(i) CH₃CHO (ii) HCHO (iii) Cl₃CCHO (iv) C₂H₅CHO
How many of these undergo Cannizzaro reaction with concentrated NaOH?
Show answer and why every option is right or wrong
Answer: B. Only HCHO (no α-carbon) and Cl₃CCHO (α-carbon is CCl₃ with no hydrogen) lack α-hydrogens. CH₃CHO has 3 α-H; C₂H₅CHO has 2 α-H. So 2 aldehydes undergo Cannizzaro. (NCERT Class 12 Chemistry Chapter 8, Part 2, page 242)
Why A is wrong: A is wrong because both HCHO and Cl₃CCHO qualify — HCHO has no α-carbon at all, and trichloroacetaldehyde has a CCl₃ group (no α-hydrogen on the carbon adjacent to carbonyl). The count is 2, not 1.
Why C is wrong: C is wrong because it counts 3 aldehydes; only HCHO and Cl₃CCHO lack α-hydrogens, while CH₃CHO and C₂H₅CHO both have them, so the count is 2.
Why D is wrong: D is wrong because CH₃CHO (3 α-H) and C₂H₅CHO (2 α-H) both have α-hydrogens and will undergo aldol condensation, not Cannizzaro. Only 2 of the 4 qualify.
Free NEET study resources
Get a structured 30-day study plan and a complete formula booklet — delivered to your inbox instantly.
How do you solve a Cannizzaro question? A worked example
- 1
Given
• Equimolar HCHO + C₆H₅CHO• Concentrated NaOH (excess)• Both substrates lack α-hydrogen
- 2
Required
Major organic products and reasoning for selectivity.
- 3
Concept
Crossed Cannizzaro reaction. When two different non-α-H aldehydes react with conc. base, the less sterically hindered aldehyde preferentially donates hydride (gets oxidised).
- 4
Principle
HCHO has the least steric hindrance at its carbonyl carbon (only H atoms bonded). The tetrahedral intermediate formed from OH⁻ attack on HCHO more readily transfers hydride to the second aldehyde.
- 5
Substitution
• HCHO → oxidised → HCOO⁻Na⁺ (sodium formate)• C₆H₅CHO → reduced → C₆H₅CH₂OH (benzyl alcohol)
- 6
Calculation
No numerical calculation required. This is a product-identification problem.
- 7
Final answer
Major products: sodium formate (HCOONa) and benzyl alcohol (C₆H₅CH₂OH).
HCHO is preferentially oxidised because its carbonyl carbon is the least sterically hindered, making hydride transfer from its tetrahedral intermediate kinetically favoured. - 8
Common trap
Students sometimes assume the bulkier aldehyde (benzaldehyde) is oxidised because it is "more complex." The selectivity is governed by steric accessibility of the carbonyl for hydride donation, not by molecular size or stability of the carboxylate product.
- 9
Similar NEET-style question
"An equimolar mixture of 2,2-dimethylpropanal and HCHO is treated with excess conc. NaOH. What are the products?" Answer: Sodium formate + 2,2-dimethyl-1-propanol (neopentyl alcohol). HCHO is still preferentially oxidised due to lower steric demand.
What to remember before solving Cannizzaro questions
More in Alcohols, Phenols, Ethers and Carbonyl Compounds: 3 exam traps and mistakes · 2 formulas · 3 question patterns from its other lessons.
Cannizzaro questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
All 23 past-paper questions from Alcohols, Phenols, Ethers and Carbonyl Compounds →
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
Test yourself on this topic with real past-paper questions:
Practice this topic →