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"An alkene gives propanone on ozonolysis. Identify it." Questions like this are quick marks once you have a method, and a trap if you guess. This post explains what ozonolysis does, gives a three-step join method for working backwards, and runs it on four cases, including a ring and a diene.
What does ozonolysis do?
The NCERT Class 11 unit on hydrocarbons describes it as the addition of ozone to an alkene to form an ozonide, followed by cleavage of the ozonide with Zn–H₂O into smaller molecules. NCERT adds that the reaction "is highly useful in detecting the position of the double bond". Its examples:
Propene gives ethanal and methanal.
2-Methylpropene gives propan-2-one and methanal.
The mechanism was set out by Rudolf Criegee. His 1975 review of the mechanism describes ozonide formation as a sequence of three cycloaddition and cycloreversion steps through a primary ozonide and a carbonyl oxide. You do not need those steps for JEE Main or NEET, but they explain why the C=C ends up fully cut.
The JEE (Main) 2026 syllabus and NEET (UG) 2026 syllabus both list ozonolysis under alkenes; the JEE Advanced 2026 syllabus has "Reactions of alkenes with KMnO₄ and ozone". Check the current year's syllabus.
The forward rule
Cut every C=C. Put an =O on each carbon that was in the double bond. Everything else stays where it was.
A carbon with one hydrogen becomes an aldehyde (–CHO).
A carbon with no hydrogens becomes a ketone (C=O between two carbons).
A terminal =CH₂ becomes methanal, HCHO.
The join method for working backwards
Write each product with its C=O carbon facing the other product's C=O carbon.
Remove the oxygens.
Join those two carbons with a double bond, then name the alkene.
Two checks then catch most errors. Ozonolysis adds only oxygen, so the carbons and hydrogens of the products add up to the alkene's formula. And if one product contains two C=O groups, the double bond was in a ring or the molecule had more than one double bond.
Our hydrocarbons map works one example, 2-methylbut-2-ene from propanone and ethanal. Here are four more.
Four worked examples
1. One product only. C₆H₁₂ gives only propanone. Two molecules of (CH₃)₂C=O face each other: (CH₃)₂C=C(CH₃)₂. Check: 2 × C₃H₆O minus the oxygens is C₆H₁₂. The alkene is 2,3-dimethylbut-2-ene. A single product means the alkene is symmetrical about its double bond.
2. A terminal alkene. An alkene gives butanone and methanal. Face CH₃CH₂(CH₃)C=O and O=CH₂, remove the oxygens and join: CH₂=C(CH₃)CH₂CH₃. Check: C₄H₈O + CH₂O gives C₅H₁₀. The alkene is 2-methylbut-1-ene. Methanal always signals a =CH₂ end.
3. A ring. A compound C₆H₁₀ gives one product, hexanedial, OHC–CH₂CH₂CH₂CH₂–CHO. Both C=O carbons are in the same molecule, so they were joined to each other: close the chain into a six-membered ring with one double bond. The answer is cyclohexene. Check: C₆H₁₀O₂ minus two oxygens is C₆H₁₀.
4. A diene. Buta-1,3-diene, CH₂=CH–CH=CH₂, has two C=C bonds, so cut both. The ends give two HCHO, and the middle two carbons become ethanedial, OHC–CHO. Working backwards, a product with two C=O groups plus two separate methanal molecules tells you there were two double bonds.
Which mistakes cost marks?
Forgetting methanal. A terminal =CH₂ gives HCHO, and missing it gives a formula that does not add up.
Joining the wrong carbons. Join the carbonyl carbons, not the end of a chain.
Assuming two products. Symmetrical alkenes give one; dienes give three or more.
Confusing ozonolysis with permanganate. NCERT shows acidic KMnO₄ taking 2-methylpropene to propan-2-one and CO₂, and but-2-ene to ethanoic acid. With ozone and Zn–H₂O, but-2-ene gives ethanal instead.
Naming slips. Number from the end nearer the double bond. Our guide to IUPAC naming has the rules.
A practice routine
Write the ozonolysis products of five alkenes forwards, including one terminal alkene and one cyclic alkene.
Cover the alkenes and work back from the products using the join method.
Do the formula check on every answer.
Try NCERT exercises 9.4 to 9.7 at the end of the hydrocarbons unit, which ask for exactly this.
If you keep picking the wrong carbons to join, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Next study session, take any three alkenes from your textbook, write their ozonolysis products, then rebuild the alkenes without looking.
