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Chemistry · JEE Advanced only

Reactions of Alkenes with KMnO4 and Ozone (JEE Advanced)

Not in the JEE Main syllabus, but it is in JEE Advanced. Study time: about 3 to 4 hours for thorough understanding and practice problems.

Last updated: 26 October 2026

Direct answer

Alkenes react with KMnO4 and ozone to break the C=C double bond, producing carbonyl compounds.

Alkenes react with KMnO4 and ozone to break the C=C double bond, producing carbonyl compounds. KMnO4 oxidation can be mild (cold, dilute) giving diols, or vigorous (hot, concentrated) cleaving the bond completely. Ozonolysis involves ozone addition followed by reductive or oxidative workup to give aldehydes, ketones, or carboxylic acids.

Official syllabus

What the JEE Advanced syllabus says

“Reactions of alkenes with KMnO4 and ozone”

Source: JEE Advanced 2026 Information Bulletin, Annexure-I. Closest JEE Main topic: Ozonolysis and polymerization.

JEE Main covers ozonolysis basics and polymerization, but Advanced expects you to distinguish between different KMnO4 conditions and predict products based on alkene substitution patterns. You need to master both reagents and their workup variations to handle Advanced level problems.

Key ideas

What you must understand

  • KMnO4 (cold, dilute, alkaline) adds two OH groups across C=C to form a vicinal diol (syn addition), this is called Baeyer's test
  • KMnO4 (hot, concentrated, acidic) cleaves the double bond completely, terminal CH2 gives CO2, monosubstituted C gives COOH, disubstituted C gives ketone
  • Ozonolysis step 1: O3 adds across C=C to form an unstable molozonide, which rearranges to ozonide
  • Reductive workup (Zn/H2O or Me2S) of ozonide gives aldehydes and ketones without further oxidation
  • Oxidative workup (H2O2) of ozonide converts any aldehyde to carboxylic acid
  • Both reactions are used to determine the position of double bond in unknown alkenes by analyzing the products
Formulas

Formulas and key facts

  • RCH=CHR' + O3 → ozonide; then Zn/H2O → RCHO + R'CHO
  • RCH=CR'R'' + O3 → ozonide; then Zn/H2O → RCHO + R'R''C=O
  • R2C=CR2 + KMnO4 (hot, conc.) → 2 R2C=O (ketones)
  • RCH=CHR' + KMnO4 (hot, conc.) → RCOOH + R'COOH
  • CH2=CHR + KMnO4 (hot, conc.) → CO2 + RCOOH
  • RCH=CHR' + KMnO4 (cold, dilute) → RCH(OH)−CH(OH)R' (diol)
  • Ozonide + H2O2 → carboxylic acids (aldehydes oxidized to acids)
  • Terminal =CH2 with hot KMnO4 → CO2 and H2O
Avoid

Common mistakes

  • Confusing reductive vs oxidative workup in ozonolysis: Zn/H2O preserves aldehydes, H2O2 oxidizes them to acids
  • Forgetting that terminal CH2=CH gives CO2 (not HCOOH) with hot acidic KMnO4 because formic acid is further oxidized
  • Mixing up cold dilute KMnO4 (gives diol, no cleavage) with hot concentrated KMnO4 (cleaves the bond)
  • Not recognizing that a disubstituted carbon (R2C=) gives a ketone, while monosubstituted (RCH=) gives aldehyde or acid depending on conditions
Worked example

A JEE Advanced-level example

Question. An alkene A (C5H10) on ozonolysis followed by reductive workup gives only one product B (C2H4O) and one product C (C3H6O). Identify A, B, and C. What would be the products if A is treated with hot acidic KMnO4?

Show solution
  1. Total carbons in products: 2 + 3 = 5, matches C5H10. Reductive workup gives aldehydes/ketones, so B and C are carbonyl compounds.
  2. C2H4O with carbonyl: CH3CHO (acetaldehyde). C3H6O with carbonyl: CH3COCH3 (acetone) or CH3CH2CHO (propanal).
  3. If B = CH3CHO (from CH3CH=) and C = CH3COCH3 (from (CH3)2C=), then A = CH3CH=C(CH3)2 which is 2-methylbut-2-ene.
  4. Check: 2-methylbut-2-ene is C5H10. Ozonolysis with Zn/H2O cleaves at C=C giving CH3CHO and (CH3)2C=O. This matches.
  5. With hot acidic KMnO4: The CH3CH= part gives CH3COOH (acetic acid), and the (CH3)2C= part gives (CH3)2C=O (acetone).
  6. Final products with KMnO4: CH3COOH (acetic acid) and CH3COCH3 (acetone).

Answer: A = 2-methylbut-2-ene (CH3CH=C(CH3)2), B = acetaldehyde (CH3CHO), C = acetone (CH3COCH3). With hot acidic KMnO4: acetic acid and acetone.

Solved twice independently; both solutions agree.

FAQs

Frequently asked questions

Is Reactions of Alkenes with KMnO4 and Ozone in the JEE Main syllabus?

No. It is in the JEE Advanced 2026 syllabus only. The closest JEE Main topic is Ozonolysis and polymerization.

How long does Reactions of Alkenes with KMnO4 and Ozone take to prepare for JEE Advanced?

About 3 to 4 hours for thorough understanding and practice problems, if you already know the related JEE Main topic.

WA✎
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