Reactions of Alkenes with KMnO4
A concise revision guide to the oxidation of alkenes using potassium manganate(VII), including the formation of diols, the purple to colourless observation and how this reaction can be used as a test for the alkene functional group.
Hot, concentrated KMnO4: the cleavage of the double bond and locating a C=C bond from the products (14.2.2(c)) now have their own page: Oxidative Cleavage of Alkenes.
GCSE Recap: Saturated, Unsaturated and a Positive Test
Before you start, check the GCSE ideas this page builds on: what makes a hydrocarbon unsaturated, and what a positive test looks like.
The Overall Reaction
Alkenes react with KMnO4, potassium manganate(VII), in acidified solution to form diols.
A diol is an organic compound containing two alcohol, -OH, groups. In this reaction, the C=C double bond opens and an -OH group is added to each carbon atom from the original double bond.
For example, propene is oxidised to form propane-1,2-diol.
Key idea: KMnO4 oxidises the alkene functional group, converting the C=C double bond into a diol.
The displayed reaction shows the alkene functional group in propene changing into a diol after oxidation with KMnO4.
Quick Check: Name the Diol
Apply the reaction to an alkene that this page does not use.
Conditions and Observation
This reaction is carried out using acidified KMnO4 at room temperature.
The important observation is that the purple colour of the MnO4– ion decolourises to colourless.
This colour change is useful because it provides a simple chemical test for the presence of the alkene functional group.
Exam focus: State the full observation as purple to colourless. Do not write only “goes clear”.
Quick Check: Two Unlabelled Bottles
Drag the words into place to complete the account of the two tests.
Why It Tests for Alkenes
The reaction can be used as a test for alkenes because alkenes contain a C=C double bond that can be oxidised under these conditions.
Alkanes do not contain a carbon-carbon double bond. They are saturated hydrocarbons, so they do not react in the same way with acidified KMnO4 at room temperature.
Therefore, an alkene gives the purple to colourless change, but an alkane would show no colour change.
Diol: An organic compound containing two alcohol, -OH, groups. In this alkene oxidation reaction, the two -OH groups form on the two carbon atoms from the original C=C double bond.
Remember: The test depends on oxidation of the alkene double bond, not substitution or hydrogenation.
Quick Check: Which Samples React?
Click every hydrocarbon that makes the purple colour fade.
Quick Check: The Limits of the Test
Write a short explanation, then compare it with the mark points and the model answer.
Hot Concentrated Acidified KMnO4: Cleaving the Double Bond
Cambridge also requires the oxidation of alkenes by hot concentrated acidified potassium manganate(VII). Under these harsher conditions the diol formed at first is oxidised further and the carbon–carbon double bond is broken completely.
Each carbon atom of the original C=C bond becomes a carbonyl carbon. The products depend on what was attached to that carbon in the alkene:
| Groups on the C=C carbon | Product from that carbon | Reason |
|---|---|---|
| two alkyl groups (R2C=) | a ketone, R2C=O | no C–H bond to oxidise further |
| one alkyl group and one H (RHC=) | a carboxylic acid, RCOOH | the aldehyde formed first is oxidised to the acid |
| two H atoms (H2C=) | carbon dioxide and water | methanal is oxidised all the way to CO2 |
For example, hot concentrated acidified KMnO4 oxidises but-2-ene to two molecules of ethanoic acid, and 2-methylpropene to propanone plus carbon dioxide and water.
Using the products to locate the double bond
Because the products reveal what was attached to each end of the C=C bond, this reaction can be used to determine the position of an alkene linkage in a larger molecule. Working backwards from the ketones and carboxylic acids formed tells you where the double bond was and how the carbon chain was substituted.
Exam focus: Cold dilute acidified KMnO4 gives the diol; hot concentrated acidified KMnO4 cleaves the C=C bond. State the conditions precisely and name the products formed from each end of the double bond.
Quick Check: Cleaving the Double Bond
Work out the products of each reaction before you flip the card.
Common Exam Mistakes
Students often lose marks by forgetting that the product is a diol, not a single alcohol.
Another common error is to describe the reaction as addition only. The expected reaction type here is oxidation.
For observation questions, the safest answer should include both the starting colour and the final colour.
Exam focus: Link the colour change to the reaction of MnO4– with the alkene C=C double bond.
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Some ionic radii are shown.
| Ion | Ionic radius / nm |
|---|---|
| Na+ | 0.102 |
| K+ | 0.138 |
| F− | 0.133 |
| Cl− | 0.180 |
Which compound has the strongest ionic bonding?
Explain why the metallic bonding in magnesium is much stronger than that in sodium.
Copyright notice: This OLS revision content, including the explanations, layout, diagrams, tables and embedded learning structure, is authored for Online Learning System by Dr. Mohammed Al-Fatah. It may not be copied, reproduced, redistributed or adapted without written permission.
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