Concept understanding — Oxidative Cleavage by KMnO4
Oxidative Cleavage by KMnO₄
Imagine an alkene as a bridge connecting two carbon atoms. Normally, you can break that bridge and put something new on each end — that's addition. But oxidative cleavage is more violent: it completely destroys the bridge and oxidises each carbon to its highest possible oxidation state.
The Intuition
Hot, acidic KMnO₄ is an extremely strong oxidising agent. When it attacks an alkene, it doesn't just break the π bond — it breaks the σ bond too. Each carbon that was part of the double bond gets ripped apart and turned into a carbonyl group (C=O). But the story doesn't end there. If that carbonyl carbon still has a hydrogen attached (like in an aldehyde), the KMnO₄ keeps oxidising it all the way to a carboxylic acid. If the carbon has no hydrogen (like in a ketone), it survives as a ketone. And if the carbon has no carbon neighbour at all (like in terminal alkenes), it gets oxidised to CO₂.
The Precise Rule
R1R2C=CR3R4KMnO4,H+,ΔProducts
The product at each double-bond carbon depends on how many alkyl (or aryl) groups are attached to it:
Substitution at the carbon
Product
CR2 (two alkyl groups)
Ketone (R2C=O)
CHR (one alkyl, one H)
Carboxylic acid (RCOOH)
CH2 (two H's)
CO2 + H2O
Watch out
A common mistake: students think terminal alkenes give formic acid (HCOOH). They don't. Formic acid itself is oxidised further by hot KMnO₄ to CO₂ and water. So CH2= always gives CO2.
Worked Examples
Example 1: 2-methyl-2-butene
CH3C(CH3)=CHCH3
Left carbon: two alkyl groups (CH₃ and CH₃) → ketone: CH3COCH3 (acetone)
Right carbon: one alkyl (CH₃) and one H → carboxylic acid: CH3COOH (acetic acid)
Example 2: 1-hexene
CH2=CH(CH2)3CH3
Left carbon: two H's → CO2+H2O
Right carbon: one alkyl (pentyl) and one H → CH3(CH2)3COOH (pentanoic acid)
Baeyer's reagent test is a classic test for unsaturation (C=C or C≡C bonds). When ethylene (CH2=CH2) is treated with cold, dilute, alkaline potassium permanganate solution, the purple colour of KMnO4 fades (decolourises) as the alkene is oxidised. Mechanistically, the reagent adds two -OH groups across the double bond (syn-addition, via a cyclic manganate ester intermediate), converting the C=C into a vicinal diol:
Q.The compound which decolorizes alkaline KMnO4 is
(a) C3H8
(b) C2H4
(c) CH4
(d) CCl4
›Reveal solutionSolution
Ethylene (C2H4), having a C=C double bond, decolorizes alkaline KMnO4 (Baeyer's test for unsaturation); the saturated compounds do not.
Baeyer's reagent (cold, dilute, alkaline KMnO4) oxidizes carbon–carbon double/triple bonds (e.g., converting an alkene to a diol), and in doing so the purple permanganate ion is reduced and decolorized (to brown MnO2). C3H8 (propane), CH4 (methane), and CCl4 (carbon tetrachloride) are all saturated (or fully substi …
Alkaline KMnO4 (Baeyer's reagent) turns ethylene into ethylene glycol.
Cold, dilute alkaline potassium permanganate is Baeyer's reagent, used as a test for unsaturation. It syn-adds two hydroxyl groups across the double bond of an alkene. With ethylene (CH2=CH2):
Hot alkaline KMnO4 cleaves ethene's double bond completely, oxidising both terminal =CH2 carbons all the way to CO2.
Cold, dilute alkaline KMnO4 (Baeyer's reagent) adds across a C=C double bond to give a 1,2-diol (glycol) without breaking the carbon skeleton. But under HOT, stronger alkaline KMnO4 conditions (as indicated here by the elevated temperature, 373-383 K, and 4 atoms of oxygen supplied), the oxidation goes further and actually cleaves the C=C bond:
If a doubly-bonded carbon carries one H and one alkyl group (=CHR), it is oxidised to a carboxylic acid (RCOOH).
If a doubly-bonded carbon carries two alkyl groups (=CR2), it is oxidised to a ketone (R2C=O).
If a doubly-bonded carbon is a TERMINAL =CH2 group (two H atoms, as in ethene), it is oxidised all the way to CO2 (with H2O).
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