Ozonolysis of Alkenes: Mechanism, Reading the Products and Exam Traps

ChemistryClass 11-12, Hydrocarbons; Aldehydes and KetonesFor JEE and NEET aspirants

Ozonolysis splits a carbon-carbon double bond with ozone. The alkene adds ozone to give an ozonide, and zinc dust with water then breaks the ozonide into carbonyl compounds (two molecules, or one if the C=C was in a ring): each carbon of the old C=C becomes a C=O carbon. Because the products show where the double bond was, ozonolysis is used to find the position of a double bond and to make aldehydes and ketones. It is in the Class 11 NCERT chapter on hydrocarbons, and again among the ways to prepare aldehydes and ketones in Class 12. JEE Main, JEE Advanced and NEET usually ask it backwards: which alkene gave these products?

At a glance

Propene:
$C H_{3} CH = C H_{2} (i) O_{3} (ii) Zn, H_{2} O C H_{3} CHO + HCHO$
2-Methylpropene:
$(C H_{3})_{2} C = C H_{2} (i) O_{3} (ii) Zn, H_{2} O (C H_{3})_{2} C = O + HCHO$
Rule: each carbon of the C=C becomes a C=O carbon. $= C H_{2}$ gives $HCHO$ , $= CHR$ gives an aldehyde, $= C R_{2}$ gives a ketone.
Intermediate: the ozonide, a five-membered ring with an O–O bond, formed through the primary ozonide.
Zn and water (or $(C H_{3})_{2} S$ ) give aldehydes and ketones; $H_{2} O_{2}$ instead turns the aldehyde fragments into carboxylic acids.
Carbons: none are lost. The products together hold all the carbons of the alkene.

1. Mechanism

Ozone, $O_{3}$ , is a bent molecule; in each of its resonance structures the central oxygen is positive and one end oxygen is negative. It adds across the double bond in one step, like a pair of tongs closing on both carbons at once. The ring it forms is unstable and rearranges to the ozonide; the C=C carbons are no longer joined to each other.

Figure 1: Ozone adds to the C=C in one step, the ring splits, and the pieces join again as the ozonide, in which the two alkene carbons are no longer bonded to each other. Zinc and water then free the two C=O compounds.

Addition. The two end oxygens of ozone bond to the two alkene carbons at the same time, giving the primary ozonide (also called the molozonide), a ring with three oxygens in a row.
Splitting. The primary ozonide breaks into a carbonyl compound and a carbonyl oxide, $R_{2} C = O^{+} - O^{-}$ . The C–C bond of the old double bond breaks here.
Ozonide. The two pieces add together the other way round, giving the ozonide: a ring in which the two carbons are joined through an O and through an O–O bond. This route is the Criegee mechanism, and the carbonyl oxide is called the Criegee intermediate.
Work-up. Zinc dust and water break the ozonide into the two carbonyl compounds.

Water alone would also break the ozonide, but it gives $H_{2} O_{2}$ as well, which oxidises any aldehyde formed to a carboxylic acid. The zinc removes this peroxide, so the aldehydes survive.

Exam Trick

"Cut and cap." Cut the C=C and cap each carbon with =O. Going backwards, "strip and stitch": strip the O from the two C=O carbons that must have been joined, and stitch those carbons together with a double bond.

2. Reading the products

Figure 2: Cut the C=C and put =O on each carbon. A carbon with two carbon groups gives a ketone; a carbon with an H gives an aldehyde. Read backwards, join the two C=O carbons to find the alkene.

Alkene	Products with Zn and water
Ethene	Two $HCHO$ (methanal)
Propene	$C H_{3} CHO$ + $HCHO$
But-1-ene	$C H_{3} C H_{2} CHO$ + $HCHO$
But-2-ene (cis or trans)	Two $C H_{3} CHO$ (ethanal)
2-Methylpropene	$(C H_{3})_{2} CO$ + $HCHO$
2-Methylbut-2-ene	$(C H_{3})_{2} CO$ + $C H_{3} CHO$
Cyclohexene	One molecule, $OHC (C H_{2})_{4} CHO$ (hexanedial)
1-Methylcyclohexene	One molecule, $C H_{3} CO (C H_{2})_{4} CHO$ (6-oxoheptanal)

Three things decide a question of this kind:

What sits on each alkene carbon. An H on the carbon gives an aldehyde (CHO); two carbon groups give a ketone; a terminal $= C H_{2}$ gives methanal.
A C=C inside a ring gives one molecule. The ring opens, but the product is one molecule with a C=O at each end. An exocyclic C=C is different: methylenecyclohexane gives cyclohexanone and $HCHO$ .
Count the double bonds. Each C=C is cut, so an open-chain diene gives three fragments (a ring diene gives two, because the ring holds one of the pieces together). Two C=O groups on one fragment mean that fragment sat between two double bonds, or in a ring.

3. Zinc or hydrogen peroxide: the work-up decides

Group on the alkene carbon	With zinc and water (reductive)	With hydrogen peroxide (oxidative)
$= C H_{2}$	$HCHO$	$HCOOH$ (it may be oxidised further to $C O_{2}$ )
$= CHR$	$RCHO$ , an aldehyde	$RCOOH$ , a carboxylic acid
$= C R_{2}$	$R_{2} C = O$ , a ketone	$R_{2} C = O$ , a ketone (not oxidised)

Hot alkaline or acidic $KMn O_{4}$ also breaks a C=C, but like the $H_{2} O_{2}$ work-up it oxidises the aldehyde fragments to acids, and a terminal $= C H_{2}$ ends as $C O_{2}$ and water. Only a reductive work-up of the ozonide (zinc and water, or $(C H_{3})_{2} S$ ) stops at the aldehydes.

JEE Advanced

Benzene has three C=C bonds for ozone. Benzene forms a triozonide, which gives three molecules of glyoxal, $OHC - CHO$ . 1,2-Dimethylbenzene (o-xylene) gives glyoxal, methylglyoxal and dimethylglyoxal in the ratio 3 : 2 : 1, which is only possible if both Kekulé structures contribute: one Kekulé form gives two methylglyoxal and one glyoxal, the other gives one dimethylglyoxal and two glyoxal.

Alkynes. With zinc and water a C≡C gives a 1,2-dicarbonyl compound, and the two carbons stay joined: ethyne gives glyoxal, $OHC - CHO$ , and but-2-yne gives $C H_{3} COCOC H_{3}$ . With water alone, the $H_{2} O_{2}$ formed splits it into two carboxylic acids (but-2-yne gives two $C H_{3} COOH$ ).

Stereocentres survive, chirality may not. Ozonolysis breaks only the C=C, so a stereocentre elsewhere keeps its spatial arrangement (its R/S label can still change, because the priorities of its groups change). And once two of its groups become identical CHO groups, that carbon is no longer chiral.

Ozonolysis then aldol. A ring alkene gives a dicarbonyl compound that can close again by an intramolecular aldol condensation in base, usually to a five- or six-membered ring.

4. Worked examples

Solved Example 1

An alkene

C_{5} H_{10}

gives propanone and ethanal on ozonolysis (Zn, water). Identify it, and name the main product when it adds HCl.

Solution:

Strip the oxygens and stitch the two carbonyl carbons: $(C H_{3})_{2} C = CHC H_{3}$ , 2-methylbut-2-ene. HCl adds by Markovnikov's rule (H to the CH carbon, Cl to the carbon with two methyls), giving 2-chloro-2-methylbutane, $(C H_{3})_{2} CClC H_{2} C H_{3}$ .

Solved Example 2

Which cyclic alkene gives 6-oxoheptanal,

C H_{3} CO (C H_{2})_{4} CHO

, on ozonolysis? What does that compound give with aqueous NaOH on heating?

Solution:

Both C=O groups are on one molecule, so the C=C was in a six-membered ring, with a methyl on one of its carbons: 1-methylcyclohexene. In base, the $C H_{2}$ next to the ketone C=O attacks the CHO carbon, closing a five-membered ring; loss of water gives 1-acetylcyclopentene.

Solved Example 3

Can ozonolysis distinguish cis-but-2-ene from trans-but-2-ene?

Solution:

No. Both give two molecules of ethanal, $C H_{3} CHO$ , because the arrangement around the C=C is lost when the bond is cut. A physical property such as the dipole moment tells them apart instead.

5. Exam traps

Watch out

A terminal $= C H_{2}$ gives methanal, $HCHO$ . Do not forget it when counting products.
Zinc and water give aldehydes and ketones; $H_{2} O_{2}$ gives carboxylic acids from the aldehyde ends. Ketones are the same either way.
An alkene whose C=C is inside a ring gives one molecule with two C=O groups; an exocyclic $= C H_{2}$ (methylenecyclohexane) gives two: cyclohexanone and $HCHO$ .
cis and trans isomers give the same products.
No carbon is lost with zinc and water: add up the carbons of the products to check the alkene's formula.
The ozonide has an O–O bond and a C–O–C bridge; the two alkene carbons are no longer bonded to each other.
Hot $KMn O_{4}$ also cleaves a C=C, but it does not stop at the aldehydes.

Past-paper questions on Ozonolysis

Our past papers: JEE Main 2025-2026, JEE Advanced 2022-2026, NEET 2018-2019, 2022-2026. We found 16 questions in them that test this:

JEE Main 9 (2025-2026)
JEE Advanced 5 (2022, 2024-2026)
NEET 2 (2019, 2024)

Each question was read and checked by hand (list checked 2026-10-15). A question that shows its structures only as pictures can be missed by our search. Each opens in a new tab, with the question to try and our step-by-step solution.

Learn it with the chapter

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