Ozonolysis of Alkenes: How to Find the Alkene Structure from the Products
Chemistry · Hydrocarbons · NEET
Ozonolysis cuts the alkene right at the double bond (C=C) and turns each carbon into a carbonyl group (C=O). So you get two carbonyl products (aldehydes or ketones). To find the original alkene, do the reverse: take the two products, remove both oxygen atoms, and join the two carbonyl carbons with a double bond. Memory hook: "Cut at the double bond, put O on each side. To go back, remove both O and rejoin with =".
Ozonolysis breaks the C=C double bond and puts an oxygen on each carbon, giving a ketone and/or aldehyde. To find the starting alkene, face the two carbonyl products, remove both oxygen atoms, and rejoin the carbons with a double bond.
Your doubts, answered
How do I find the original alkene when the question gives me the ozonolysis products?
Do the reverse of the reaction. Write the two carbonyl products facing each other at their C=O carbons. Erase both oxygen atoms. Then join those two carbons with a double bond (C=C). That rejoined molecule is your alkene. Example: propanone (CH3-CO-CH3) + ethanal (CH3-CHO) rejoin to (CH3)2C=CH-CH3, which is 2-methylbut-2-ene.
Does ozonolysis give an aldehyde or a ketone? How do I know which?
It depends on the carbon of the double bond. If a double-bond carbon carries at least one H atom, that side becomes an aldehyde (R-CHO). If a double-bond carbon carries two carbon groups and no H, that side becomes a ketone (R-CO-R'). A terminal =CH2 carbon has two H atoms, so it always gives formaldehyde HCHO. This single rule solves most NEET ozonolysis problems.
What is the difference between reductive and oxidative ozonolysis?
Reductive ozonolysis uses Zn/H2O (a mild reducing agent) after O3, and it stops at aldehydes and ketones. Oxidative ozonolysis uses H2O2 (or no Zn), which further oxidises the aldehydes into carboxylic acids (ketones stay as ketones). For NEET, if you see Zn/H2O the aldehyde stays an aldehyde. If you see H2O2, any aldehyde becomes -COOH.
Why do we add Zn/H2O in ozonolysis?
Ozone first makes an unstable ozonide. Zn (with water) reduces this ozonide and, importantly, destroys the hydrogen peroxide that forms. If that peroxide is left behind, it would oxidise the aldehydes into acids. So Zn/H2O protects the aldehydes and gives clean aldehyde + ketone products.
How does ozonolysis tell me the position of the double bond?
The break always happens exactly at the C=C. So the two carbonyl carbons in the products were the two double-bond carbons. Count the carbons in each product, join them at those carbons, and the double bond sits right where you rejoined them. This is why ozonolysis is used to locate a double bond in an unknown alkene.
What happens when ozonolysis is done on a cyclic alkene like cyclohexene?
A ring has the C=C inside it, so cutting it does not give two separate molecules. Instead you open the ring into ONE open-chain molecule that has a carbonyl group at each end (a dicarbonyl compound). Cyclohexene gives hexanedial (OHC-(CH2)4-CHO). NEET 2024 tested exactly this idea.
⚠️ The NEET trap ✗ Rejoining the two carbonyl products but leaving the oxygen atoms in, or adding the double bond in the wrong place. ✓ Remove BOTH oxygen atoms from the two carbonyl carbons, then join those two exact carbons with the C=C double bond. The double bond must sit between the two former carbonyl carbons, not anywhere else. 🧠 Two O's out, one double bond in — always between the carbons that carried the O.
Real NEET questions
NEET 2019
An alkene 'A' on reaction with O3 and Zn-H2O gives propanone and ethanal in equimolar ratio. Addition of HCl to alkene 'A' gives 'B' as the major product. The structure of product 'B' is:
A · Cl-CH2-CH2-CH(CH3)-CH3
B · H3C-CH2-CH(CH2Cl)-CH3
C · H3C-CH2-C(CH3)(Cl)-CH3 ✓
D · H3C-CH(Cl)-CH(CH3)-CH3
Solution: Step 1 - find alkene A. Products are propanone (CH3-CO-CH3) and ethanal (CH3-CHO). Remove both oxygen atoms and join the two carbonyl carbons with a double bond: (CH3)2C=CH-CH3. This is 2-methylbut-2-ene. Step 2 - add HCl (Markovnikov). H adds to the C with more H (the =CH- carbon), Cl adds to the more substituted C (the C with two CH3 groups) because it forms the more stable 3-degree carbocation. Product: H3C-CH2-C(CH3)(Cl)-CH3, i.e. 2-chloro-2-methylbutane. Answer: C.
NEET 2022
Compound X on reaction with O3 followed by Zn/H2O gives formaldehyde and 2-methylpropanal as products. The compound X is:
A · 3-Methylbut-1-ene ✓
B · 2-Methylbut-1-ene
C · 2-Methylbut-2-ene
D · Pent-2-ene
Solution: Work backwards. Formaldehyde is HCHO (from a terminal =CH2). 2-Methylpropanal is (CH3)2CH-CHO. Remove both oxygens and join the two carbonyl carbons with a double bond: (CH3)2CH-CH=CH2. Number the chain: CH2=CH-CH(CH3)-CH3 is but-1-ene with a methyl on carbon 3 = 3-methylbut-1-ene. Answer: A.
Solution: Cyclohexene to an open-chain dicarbonyl is exactly ozonolysis: reagent IV, (i) O3, (ii) Zn-H2O. So A-IV. Benzene to benzophenone is Friedel-Crafts (anhyd. AlCl3), B-I. Cyclohexanol to cyclohexanone is mild oxidation with CrO3, C-II. Ethylbenzene to benzoic acid is strong side-chain oxidation with KMnO4/KOH, D-III. Matches option B (A-IV, B-I, C-II, D-III). Answer: B.
Solved Hydrocarbons NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
