Oxidative Cleavage of Alkynes [KMnO₄]
Oxidative cleavage with KMnO₄ completely cuts the C≡C triple bond apart. Step-by-step mechanism, reagents and practice.
What happens
Oxidative cleavage with KMnO₄ completely cuts the C≡C triple bond apart. The two carbons of the triple bond become carboxylic acids. For a terminal alkyne, the end carbon is oxidized all the way to CO₂. For an internal alkyne, both sides become carboxylic acids. For a terminal alkyne, one side becomes a carboxylic acid, while the terminal carbon is oxidized to CO₂.
Reaction summary
Internal: R–C≡C–R′ → R–COOH + R′–COOH. Terminal: R–C≡C–H → R–COOH + CO₂
Reagents and conditions
- Starting material: 2-Pentyne (internal alkyne)
- Reagents: KMnO₄ (strong oxidative conditions)
- Conditions: KMnO₄ under strong oxidative conditions (hot/strong KMnO₄)
- Products: Propanoic acid (carboxylic acid); Acetic acid (carboxylic acid)
Mechanism
- Cut Through the C≡C: Oxidative cleavage with KMnO₄ completely cuts the C≡C triple bond apart: the molecule is cut straight through the triple bond, not turned into a double or single bond.
- Product: The two carbons of the triple bond become carboxylic acids. For an internal alkyne, both sides survive as carboxylic acids (R–COOH + R′–COOH). For a terminal alkyne, one side becomes a carboxylic acid and the terminal carbon is oxidized to CO₂.
Common mistakes
- Don't simply turn the triple bond into a double or single bond. KMnO₄ oxidative cleavage CUTS the molecule at the triple bond.
- Also remember the special rule for terminal alkynes: ≡C–H → CO₂.
Memory trick
Hot/strong KMnO₄ = molecular scissors ✂️. Find the C≡C and cut straight through it. Then: internal alkyne → acid + acid; terminal alkyne → acid + CO₂.