Ozonolysis

8 MCQs9-step worked example

Source: NCERT Aldehydes, Ketones and Carboxylic AcidsPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

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, Part 2, page 20). 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.

Practice 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:

MCQ 2Easy RecallPractice

Which reagent is used for reductive workup in ozonolysis to prevent further oxidation of aldehyde products?

MCQ 3Easy RecallPractice

Ozonolysis is primarily used to determine:

MCQ 4Direct ApplicationPractice

Ozonolysis of 2-butene (CH₃–CH=CH–CH₃) followed by Zn/H₂O gives:

MCQ 5Direct ApplicationPractice

Ozonolysis of 2-methylpropene ((CH₃)₂C=CH₂) followed by reductive workup gives:

MCQ 6Direct ApplicationPractice

A compound C₅H₁₀ on ozonolysis followed by Zn/H₂O gives acetone (CH₃COCH₃) and acetaldehyde (CH₃CHO). The compound is:

MCQ 7Concept TrapPractice

An unknown alkene gives ONLY formaldehyde (HCHO) as the ozonolysis product. The alkene is:

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:

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.)* ---

Before solving, remember these

Key Fact

Ozonolysis

Alkene + O₃ → ozonide → cleaves to two carbonyl compounds (with Zn/H₂O reductive workup). Used to identify position of double bond.

-- NCERT, p. 20

Formulas

Markovnikov's rule (and anti-Markovnikov)

H^{+} to C with more H

Without peroxide: ionic mechanism — H goes to carbon with MORE hydrogens (carbocation stability rule). With peroxide (HBr only, Kharasch): radical mechanism — anti-Markovnikov.

Symbol	Quantity	SI Unit
H,X	added atoms	-

Valid when

Asymmetric alkene
H-X with X = Cl, Br, I
Without peroxide for Markovnikov

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Trap

Markovnikov reverses with peroxide

Category: Organic Reaction Conditions

HBr addition to alkene: WITHOUT peroxide → Markovnikov (H to C with more H). WITH peroxide (Kharasch effect) → anti-Markovnikov (radical mechanism). Specific to HBr only — not HCl, HI.

When it triggers

Question gives HX addition to alkene with explicit peroxide condition or hints (e.g. ROOR, light).

How to avoid

Without peroxide: ionic mechanism, carbocation stability → Markovnikov. With peroxide: radical mechanism, radical stability → anti-Markovnikov. Effect ONLY for HBr (HCl too strong, HI too weak).

Trap

Solvent changes product distribution

Category: Organic Reaction Conditions

Same starting materials give different products depending on solvent. Polar protic (water, alcohols): SN1/E1 favoured. Polar aprotic (DMSO, DMF): SN2 favoured. Affects substitution vs elimination.

When it triggers

Question contrasts product when solvent is changed; or specifies solvent type.

How to avoid

Polar protic stabilises carbocation → SN1/E1 (3° preferred). Polar aprotic doesn't solvate nucleophile → strong SN2 nucleophile (1°/2° preferred). Bulky base (t-BuOK) favours E2 over SN2.

Mistake

Confuses ortho/para directors with meta directors.

Root cause: concept gap

Correction

o,p-directors (activators except halogens): -OH, -OR, -NH₂, -NHR, alkyl. m-directors (deactivators): -NO₂, -CN, -COOH, -CHO. Halogens: o,p-directors but DEACTIVATORS.

Mistake