Use hot manganate(VII) cleavage to locate an alkene
Interpret oxidative cleavage products to reconstruct the position of a C=C bond in a larger molecule.
Section 14 · Hydrocarbons
AS Level; Cambridge International Chemistry 9701, 2025–2027 syllabus. Apply the stated reagents and conditions, balance equations, and keep mechanistic claims within the named examples. Written practice is self-checked
Before you begin
- Use displayed and structural formulas, covalent bond polarity, balanced equations, and the organic reaction terms from Topic 13.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What happens to C=C in hot concentrated acidified KMnO₄?
What you will learn
- Distinguish hot cleavage from diol formation
- Relate each alkene carbon to its oxidation product
- Infer a double-bond position from products
Hot concentrated oxidation splits the C=C
Hot, concentrated acidified KMnO₄ can break the carbon–carbon double bond. The two alkene carbons become separate carbonyl-derived fragments, with further oxidation depending on the groups attached.
For syllabus-level interpretation, an alkene carbon bearing H can be oxidised to a carboxylic acid; one with no attached H can give a ketone. A terminal =CH₂ end can be oxidised further to CO₂. Read the stated products and conditions carefully.
Work backwards from the fragments
Imagine cutting the C=C and restoring a double bond between the two carbons that formed the carbonyl ends. Then join the fragments without changing their carbon skeletons.
For example, but-2-ene gives two molecules of ethanoic acid under full oxidation: the identical fragments indicate a symmetrical internal alkene. This pattern method is useful for larger molecules but is not a substitute for checking atom counts.
Pause and explain
Two identical ethanoic-acid fragments from butene cleavage support which starting arrangement?
Worked example
A hydrocarbon gives two molecules of ethanoic acid on hot acidified KMnO₄ cleavage. Identify a compatible alkene.
Show the worked solution
- Each ethanoic-acid fragment preserves two carbons from one side of C=C.
- Join the two carbonyl carbons with a double bond.
- The resulting four-carbon alkene is symmetrical.
Answer But-2-ene, CH₃CH=CHCH₃, can form two molecules of ethanoic acid on oxidative cleavage.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Hot oxidation of an alkene gives propanone and CO₂. Reconstruct a compatible terminal alkene.
Give me a hint
A terminal =CH₂ can give CO₂; the other side forms the ketone.
Compare my reasoning
- Propanone has three carbons and forms when the internal C=C carbon has two CH₃ groups.
- CO₂ comes from the terminal =CH₂ carbon after oxidation.
- Join those two alkene carbons and preserve both methyl groups.
2-methylpropene, CH₂=C(CH₃)₂, gives propanone and CO₂ on oxidative cleavage.
Look for these in your work
- Terminal end interpreted
- Carbonyl fragment checked
- Mismatch diagnosed
A symmetrical alkene gives two identical propanoic-acid fragments. Suggest a structure.
Give me a hint
Each acid fragment contains a two-carbon alkyl group attached to a carbonyl carbon.
Compare my reasoning
- Each C=C carbon must carry an ethyl group and one hydrogen.
- Join the two carbonyl carbons by C=C.
- The result is a symmetrical six-carbon alkene.
Hex-3-ene, CH₃CH₂CH=CHCH₂CH₃, can give two molecules of propanoic acid.
Look for these in your work
- Fragments preserved
- Alkene linkage restored
- Structure named
A learner sees one carboxylic acid and one ketone after cleavage and claims the starting alkene was symmetrical. Explain the problem.
Give me a hint
Compare the two different fragments on either side of the original C=C.
Compare my reasoning
- Different products indicate different substituent patterns at the two alkene carbons.
- A symmetrical alkene would give matching fragments under the same cleavage.
- Reconstruct the C=C by joining the carbonyl ends and verify the full carbon count.
The unequal acid and ketone fragments indicate different alkene ends, so the starting alkene is not symmetric; join the carbonyl carbons and check the total atoms.
Look for these in your work
- Unequal fragments noted
- Symmetry inference corrected
- Reconstruction method stated
Common mistakes
- Using the cold diol outcome for hot concentrated reagent
- Joining the fragments at the wrong carbon atoms
- Treating product evidence as unique without checking alternative structures
What can you explain now?
What happens to C=C in hot concentrated acidified KMnO₄?
Compare with the explanation
The double bond is cleaved
The carbon skeleton separates at the alkene linkage.
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Make it stick
How can oxidative cleavage products reveal which carbon atoms were joined by the original C=C?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.