Ozonolysis of Alkenes – From Intuition to Precision
Imagine an alkene's double bond as a tightly linked pair of carbon atoms, each holding one of the two shared electron pairs. Ozone (O3) is a highly reactive, bent molecule that acts like a chemical pair of scissors. It attacks this electron-rich double bond, inserting itself and breaking the π bond first, then the σ bond. The result? The two carbons that were once joined are now separated, each carrying an oxygen atom.
But the story doesn't end there. The initial product is an unstable, explosive compound called a molozonide. This quickly rearranges into a more stable ozonide — a five-membered ring containing three oxygen atoms. This ozonide is still reactive and must be broken down deliberately.
That's where the work-up step comes in. You have two choices:
Reductive work-up (using zinc dust in water or dimethyl sulfide) gives aldehydes or ketones — clean, predictable products.
Oxidative work-up (using hydrogen peroxide) gives carboxylic acids or ketones — more oxidized products.
For a first meeting, focus on the reductive path. It's the standard way to locate where the double bond was.
R2C=CR21.O3,2.Zn/H2OR2C=O+O=CR2
Each carbon of the original double bond becomes the carbonyl carbon of a separate product.
The Precise Statement
Ozonolysis is a two-step reaction:
Ozone addition: Ozone adds across the C=C bond, forming a molozonide that rearranges to an ozonide.
Reductive cleavage: The ozonide is treated with a reducing agent (Zn/H₂O or Me₂S) to cleave the O–O and C–O bonds, yielding two carbonyl compounds.
The identity of each product tells you exactly what was attached to that carbon in the alkene:
A carbon with two alkyl groups (disubstituted) becomes a ketone.
A carbon with one alkyl group and one hydrogen (monosubstituted) becomes an aldehyde.
A carbon with two hydrogens (terminal =CH₂) becomes formaldehyde (HCHO).
Tip
To predict products, mentally "cut" the double bond and put an oxygen atom on each cut end. The groups that were attached stay attached.
Example
Take 2-methyl-2-butene: CH3−C(CH3)=CH−CH3
The double bond carbons are:
Left carbon: attached to two methyl groups → ketone (acetone, CH3COCH3)
Right carbon: attached to one methyl and one hydrogen → aldehyde (acetaldehyde, CH3CHO)
Ozonolysis cleaves an alkene's C=C double bond, converting each of the two doubly-bonded carbons into a carbonyl (C=O) carbon while leaving the rest of each fragment's chain intact. …
Pent-2-ene, CH3-CH=CH-CH2-CH3, has its double bond between C2 and C3. Ozonolysis cleaves exactly at the C=C bond, turning each doubly-bonded carbon into a C=O: the C1-C2 fragment becomes ethanal (CH3-CHO) and the C3-C4-C5 fragment becomes propanal (CH3-CH2-CHO).
Ozonolysis (O3 then Zn/H2O reductive workup) cleaves the C=C double bond, converting each alkene carbon into a carbonyl carbon (C=O), while the rest of each fragment's chain and hydrogens stay as they were:
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2026Set ANNUAL1 markMCQ
Q.But-2-ene --(i) O3,
(ii) H2O/Zn--> Product. Product is
(a) ethanol
(b) ethanal
(c) acetone
(d) propanal
›Reveal solutionSolution
Ozonolysis cleaves a C=C double bond and replaces it with two C=O groups. For but-2-ene, the double bond is cleaved right at its centre, giving two identical carbonyl fragments.
But-2-ene: CH3-CH=CH-CH3
Step 1: Ozone (O3) adds across the C=C double bond to form an ozonide intermediate.
Step 2: Reductive workup with Zn/H2O cleaves the ozonide, converting each carbon that was part of the double bond into a C=O (carbonyl) group, while any H originally on that carbon is retained.
Each of the two double-bonded carbons in but-2-ene carries one H and one CH3 group. After cleavage, each becomes CH3-CHO (ethanal/acetaldehyde).
Ozonolysis splits a C=C double bond exactly at the bond, converting each sp2 carbon into a C=O carbonyl carbon; propene's two different ends give two different carbonyl products, ethanal and methanal.
Ozonolysis is a reaction where an alkene's C=C double bond is cleaved by ozone (O3) followed by reductive workup (e.g. Zn/H2O), converting each carbon of the double bond into a carbonyl group (C=O), with the identity of the carbonyl (aldehyde vs ketone) depending on how many H atoms/alkyl groups were originally on that carbon.
Propene structure: CH3-CH=CH2
The double bond is between C2 and C3 (using CH3 as C1): CH3-CH(=)-CH2(=)
Ozonolysis cleaves exactly at this C=C bond, converting each carbon into its own carbonyl compound:
Ozonolysis breaks the C=C double bond and replaces it with C=O on each fragment; since ethene's carbons are both CH2 groups, cleavage gives two identical methanal (formaldehyde) molecules.
Ozonolysis is a reaction where an alkene's C=C double bond is cleaved by ozone (followed by reductive workup, typically with Zn/H2O), converting each doubly-bonded carbon into a carbonyl (C=O) carbon.
Q.In the above reaction (ozonolysis of the alkene shown, O3 then H2O), compound B will be ____.
›Reveal solutionSolution
The drawn alkene is pent-2-ene, CH3-CH=CH-CH2-CH3. Ozonolysis (O3, then H2O) cleaves the C=C bond, giving acetaldehyde (CH3CHO) and propanal (CH3CH2CHO) as compound B.
The starting alkene is pent-2-ene, CH3-CH=CH-CH2-CH3 (the same molecule is printed as pent-2-ene in Q18(ii) of this paper). Ozonolysis proceeds by O3 forming the ozonide (A), which on hydrolysis with water splits the former C=C bond so that each carbon becomes a carbonyl:
Q.The IUPAC names of the products obtained by the ozonolysis of Pent-2-ene is
(a) ethanal and propanal
(b) ethanal and methanal
(c) ethanal and propanone
(d) none of these
›Reveal solutionSolution
Pent-2-ene, CH3-CH=CH-CH2-CH3, has its double bond between C2 and C3. Ozonolysis cleaves exactly at the C=C bond, turning each doubly-bonded carbon into a C=O: the C1-C2 fragment becomes ethanal (CH3-CHO) and the C3-C4-C5 fragment becomes propanal (CH3-CH2-CHO).
Ozonolysis (O3 then Zn/H2O reductive workup) cleaves the C=C double bond, converting each alkene carbon into a carbonyl carbon (C=O), while the rest of each fragment's chain and hydrogens stay as they were: