Key Fact
Ozonolysis
Alkene + O₃ → ozonide → cleaves to two carbonyl compounds (with Zn/H₂O reductive workup). Used to identify position of double bond.
-- NCERT Class 11 Chemistry, Ch. 9, p. 313Ozonolysis
8 MCQs9-step worked example
Source: NCERT HydrocarbonsPYQ coverage: NEET 2020, 2022Official key: NTA-verifiedLast updated: 24 Sep 2026
Ozonolysis, explained for NEET
Ozonolysis is a cleavage reaction: ozone (O₃) breaks the carbon–carbon double bond (or triple bond) of an alkene (or alkyne) and replaces it with carbonyl groups. The reaction proceeds in two stages — ozonide formation followed by reductive workup — and the product pattern directly reveals the position and substitution of the original double bond.
The reaction sequence. When an alkene is treated with O₃ (typically in an inert solvent such as CH₂Cl₂ at low temperature), an unstable molozonide forms across the C=C bond. This rearranges to a more stable ozonide. The ozonide is then cleaved reductively — Zn/H₂O or (CH₃)₂S are common reducing agents — to give aldehydes, ketones, or both, depending on the substitution at each end of the original double bond.
Product-prediction logic. Each carbon of the former C=C bond becomes a carbonyl carbon. A terminal =CH₂ gives formaldehyde (HCHO). A =CHR gives an aldehyde (RCHO). A =CR₂ gives a ketone (R₂C=O). This mapping is the single skill NEET tests: given the alkene, predict the carbonyl fragments, or given the fragments, reconstruct the alkene.
Why the reducing agent matters. If H₂O₂ or an oxidative workup is used instead of Zn/H₂O, aldehydes are further oxidised to carboxylic acids. NEET questions that specify "ozonolysis followed by Zn/H₂O" expect aldehyde products to stay as aldehydes — not acids. Watch for stems that change the workup agent.
NCERT anchor. The ozonolysis of alkenes is discussed as a method of locating the double bond (NCERT Class 11 Chemistry, Chapter 9, page 313). The text emphasises that the products are used to deduce the structure of the original alkene — a retrosynthetic logic that appears directly in NEET stems.
Watch-out. Ozonolysis of alkynes with reductive workup gives dicarbonyl products (two C=O groups per original C≡C). Internal alkynes give two molecules of carboxylic acid under oxidative conditions, or two aldehydes/ketones under reductive conditions. Don't confuse alkyne ozonolysis products with alkene ones — count the bonds being broken.
Can you answer these Ozonolysis 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
Ozonolysis of an alkene followed by treatment with Zn/H₂O gives:
Show answer and why every option is right or wrong
Answer: C. Reductive ozonolysis (O₃ then Zn/H₂O) cleaves the C=C bond to give carbonyl compounds — aldehydes, ketones, or both depending on the substitution pattern of the original alkene (NCERT Class 11 Chemistry, Chapter 9, page 313).
Why A is wrong: A is wrong. Ozonolysis cleaves the double bond into carbonyl (C=O) groups, not hydroxyl (–OH) groups. Alcohols are products of hydration or reduction, not ozonolysis.
Why B is wrong: B is wrong. Epoxides result from peroxide-mediated oxidation (e.g., mCPBA or peracid treatment), not ozonolysis. Ozonolysis fully cleaves the C=C bond rather than adding an oxygen bridge across it.
Why D is wrong: D is wrong. Carboxylic acids form only if an oxidative workup (e.g., H₂O₂) is used instead of the reductive Zn/H₂O. Under reductive conditions, aldehyde products remain as aldehydes, not oxidised further to acids.
MCQ 2Easy RecallPractice
Which reagent is used for reductive workup in ozonolysis to prevent further oxidation of aldehyde products?
Show answer and why every option is right or wrong
Answer: A. Zn/H₂O is the standard reductive workup that cleaves the ozonide to carbonyl products without oxidising aldehydes further. (CH₃)₂S is another common reductive agent for the same purpose.
Why B is wrong: B is wrong. KMnO₄ is a strong oxidising agent used for oxidative cleavage of alkenes, not for reductive ozonolysis workup. It would oxidise aldehyde products to carboxylic acids.
Why C is wrong: C is wrong. H₂O₂ is an oxidative workup agent. Using it after ozonide formation would oxidise any aldehyde fragments to carboxylic acids — the opposite of what reductive ozonolysis intends.
Why D is wrong: D is wrong. K₂Cr₂O₇ (acidified dichromate) is a strong oxidant. It would oxidise aldehyde products to carboxylic acids, defeating the purpose of reductive workup.
MCQ 3Easy RecallPractice
Ozonolysis is primarily used to determine:
Show answer and why every option is right or wrong
Answer: B. The carbonyl fragments produced by ozonolysis directly reveal where the C=C bond was located in the original molecule. By identifying the products, the position and substitution of the double bond can be deduced (NCERT Class 11 Chemistry, Chapter 9, page 313).
Why A is wrong: A is wrong. Molecular formula is determined by elemental analysis and molecular mass measurements, not by ozonolysis. Ozonolysis reveals structural information (bond position), not composition.
Why C is wrong: C is wrong. Hybridisation state is inferred from bonding geometry and spectroscopic data. Ozonolysis destroys the double bond entirely, so it reveals bond position, not hybridisation.
Why D is wrong: D is wrong. Degree of unsaturation is calculated from the molecular formula (Index of Hydrogen Deficiency). Ozonolysis reveals which specific bonds are double bonds, not the total count of unsaturations.
MCQ 4Direct ApplicationPractice
Ozonolysis of 2-butene (CH₃–CH=CH–CH₃) followed by Zn/H₂O gives:
Show answer and why every option is right or wrong
Answer: B. 2-Butene has the structure CH₃–CH=CH–CH₃. Ozonolysis cleaves the C=C bond. Each doubly-bonded carbon carries one H and one CH₃ group, so each fragment gives an aldehyde: CH₃CHO. The product is two molecules of acetaldehyde.
Why A is wrong: A is wrong. HCHO (formaldehyde) forms only from a terminal =CH₂ group. In 2-butene, neither carbon of the double bond is terminal — both carry a methyl substituent, so each gives CH₃CHO, not HCHO.
Why C is wrong: C is wrong. HCHO would require a =CH₂ terminus. 2-Butene is an internal alkene (CH₃ on each side of the double bond), so both fragments are CH₃CHO. Only a terminal alkene like propene would give one HCHO and one RCHO.
Why D is wrong: D is wrong. A ketone (CH₃COCH₃) forms when the C=C carbon carries two alkyl substituents and no hydrogen. In 2-butene, each C=C carbon has one CH₃ and one H, giving aldehydes, not ketones.
MCQ 5Direct ApplicationPractice
Ozonolysis of 2-methylpropene ((CH₃)₂C=CH₂) followed by reductive workup gives:
Show answer and why every option is right or wrong
Answer: D. In 2-methylpropene, one doubly-bonded carbon carries two methyl groups and no H → it gives a ketone (acetone, CH₃COCH₃). The other carries two H atoms (=CH₂) → it gives formaldehyde (HCHO).
Why A is wrong: A is wrong. Acetaldehyde (CH₃CHO) would require each C=C carbon to carry one alkyl group and one H. In 2-methylpropene, one carbon has two methyls (→ ketone) and the other has two H's (→ HCHO). Neither end gives CH₃CHO.
Why B is wrong: B is wrong. Acetaldehyde (CH₃CHO) requires =CHR (one alkyl, one H on the C=C carbon). The terminal carbon in 2-methylpropene is =CH₂ (no alkyl group), so it gives formaldehyde, not acetaldehyde.
Why C is wrong: C is wrong. Two molecules of HCHO would require both carbons of the C=C to be =CH₂ (terminal with no substituents). Only ethene (CH₂=CH₂) gives that. Here one carbon is disubstituted, giving acetone.
MCQ 6Direct ApplicationPractice
A compound C₅H₁₀ on ozonolysis followed by Zn/H₂O gives acetone (CH₃COCH₃) and acetaldehyde (CH₃CHO). The compound is:
Show answer and why every option is right or wrong
Answer: C. Reconstructing the alkene: acetone means one carbon of the C=C had two methyl groups (=C(CH₃)₂), and acetaldehyde means the other carbon had one methyl and one hydrogen (=CHCH₃). Joining these gives (CH₃)₂C=CHCH₃, which is 2-methyl-2-butene (C₅H₁₀).
Why A is wrong: A is wrong. 1-Pentene (CH₂=CHCH₂CH₂CH₃) would give formaldehyde (HCHO) from the terminal =CH₂ and butanal (CH₃CH₂CH₂CHO) from the other end — not acetone and acetaldehyde.
Why B is wrong: B is wrong. 2-Pentene (CH₃CH=CHCH₂CH₃) would give acetaldehyde (CH₃CHO) from one end and propanal (CH₃CH₂CHO) from the other — not acetone. For acetone, you need two methyls on the same C=C carbon.
Why D is wrong: D is wrong. Cyclopentane has no C=C double bond and does not undergo ozonolysis. Even cyclopentene (C₅H₈) would give a single dicarbonyl product (a dialdehyde), not two separate fragments.
MCQ 7Concept TrapPractice
An unknown alkene gives ONLY formaldehyde (HCHO) as the ozonolysis product. The alkene is:
Show answer and why every option is right or wrong
Answer: A. If the only product is HCHO, both carbons of the C=C bond must be =CH₂. The only alkene where both doubly-bonded carbons carry only hydrogen atoms is ethene (CH₂=CH₂). Each carbon gives one HCHO, so the sole product is formaldehyde.
Why B is wrong: B is wrong. 1-Butene (CH₃CH₂CH=CH₂) would give propanal (CH₃CH₂CHO) and HCHO. The propanal comes from the monosubstituted end, so HCHO is not the only product.
Why C is wrong: C is wrong. Propene (CH₃CH=CH₂) would give one molecule of acetaldehyde (CH₃CHO) from the CH₃CH= end and one molecule of HCHO from the =CH₂ end. The product is not exclusively HCHO.
Why D is wrong: D is wrong. 2-Butene (CH₃CH=CHCH₃) gives two molecules of acetaldehyde (CH₃CHO), not formaldehyde. Each C=C carbon carries one methyl group, so neither end can produce HCHO.
MCQ 8CalculationPractice
Ozonolysis of an alkene C₆H₁₂ gives two moles of propanal (CH₃CH₂CHO) as the only product. The structure of the alkene and the type of isomerism it can exhibit about the double bond are:
Show answer and why every option is right or wrong
Answer: D. Two moles of propanal (CH₃CH₂CHO) means each C=C carbon carries one ethyl group (–CH₂CH₃) and one H. Joining the two =CHCH₂CH₃ fragments gives CH₃CH₂CH=CHCH₂CH₃ — 3-hexene. Since each doubly-bonded carbon has two different substituents (H and C₂H₅), geometric (cis/trans) isomerism is possible.
Why A is wrong: A is wrong. 1-Hexene (CH₂=CHCH₂CH₂CH₂CH₃) would give HCHO from the terminal =CH₂ and pentanal (CH₃(CH₂)₃CHO) from the other end — not two identical propanal molecules. Also, 1-hexene cannot show cis/trans isomerism as one C=C carbon carries two H atoms.
Why B is wrong: B is wrong. 2-Methyl-2-pentene ((CH₃)₂C=CHCH₂CH₃) would give acetone (CH₃COCH₃) from the disubstituted carbon and propanal from the other — a ketone and an aldehyde, not two moles of propanal. It also lacks geometric isomerism because one C=C carbon bears two identical methyl groups.
Why C is wrong: C is wrong. 2-Hexene (CH₃CH=CHCH₂CH₂CH₃) would give acetaldehyde (CH₃CHO) from one end and butanal (CH₃CH₂CH₂CHO) from the other — two different aldehydes, not two identical propanal molecules.
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How do you solve a Ozonolysis question? A worked example
- 1
Given
• Alkene A: C₇H₁₄• Ozonolysis products: acetone + butanal (reductive workup with Zn/H₂O)
- 2
Required
• Structure of alkene A• Whether A can exhibit geometric (cis/trans) isomerism
- 3
Concept
Ozonolysis cleaves the C=C bond. Each fragment's carbonyl carbon was originally part of the double bond. A ketone product means two alkyl groups on that C=C carbon; an aldehyde means one alkyl group and one H.
- 4
Formula
No numerical formula. The logic is retrosynthetic: remove the C=O from each product and join the two carbons to reconstruct the C=C bond.
- 5
Substitution (reconstruct)
• From acetone (CH₃COCH₃): the C=C carbon carried two methyl groups → fragment is =C(CH₃)₂• From butanal (CH₃CH₂CH₂CHO): the C=C carbon carried one propyl group and one H → fragment is =CHCH₂CH₂CH₃• Join: (CH₃)₂C=CHCH₂CH₂CH₃
- 6
Identification
The structure (CH₃)₂C=CHCH₂CH₂CH₃ is 2-methyl-2-hexene. Carbon count: C₁(CH₃) + C₂(=C) + C₃(H) + C₄(H₂) + C₅(H₂) + C₆(H₃) + branch CH₃ = C₇H₁₄. ✓
- 7
Geometric isomerism check
For cis/trans isomerism, each doubly-bonded carbon must carry two different groups.• C-2: carries –CH₃ and –CH₃ → two identical groups → no geometric isomerism.
- 8
Common trap
A common confusion: assuming that any internal alkene shows geometric isomerism. It only occurs when each C=C carbon has two different substituents. Here, C-2 has two identical methyl groups, so no cis/trans isomers exist despite the double bond being internal.
- 9
Similar NEET-style question
An alkene C₈H₁₆ on ozonolysis with Zn/H₂O gives only 2-methylpropanal [(CH₃)₂CHCHO]. Identify the alkene. *(Answer: 2,5-dimethyl-3-hexene — two identical fragments joined at the C=C.)*
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What to remember before solving Ozonolysis questions
More in Hydrocarbons: 4 exam traps and mistakes · 1 formula · 2 question patterns from its other lessons.
Ozonolysis questions from past NEET papers
2 questions from NEET 2020, 2022. Answers verified against NTA official keys.
NEET 2022
1Pent-2-ene
23-Methylbut-1-ene
32-Methylbut-1-ene
42-Methylbut-2-ene
NTA Answer: Option 2(final)
NEET 2020
An alkene on ozonolysis gives methanal as one of the product. Its structure is :
1cyclohexene ring carrying –CH2CH2CH3 (ring C=C, drawn two bonds away from the substituent)
2cyclohexane ring carrying –CH=CH–CH3
3cyclohexene ring carrying –CH2–CH2–CH3 (ring C=C, drawn next to the substituent carbon)
4cyclohexane ring carrying –CH2–CH=CH2
NTA Answer: Option 4(final)
Sources
NCERT refs: Class 11 Chemistry Chapter 9, p.313
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.
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