Reaction:
C₂H₄ (ethene) + [O] (from KMnO₄) → CH₂(OH)CH₂(OH) (ethane-1,2-diol)
Explanation:
* Acidified potassium permanganate (KMnO₄) acts as an oxidizing agent. In acidic conditions, it provides nascent oxygen ([O]) for the reaction.
* Ethene is an unsaturated hydrocarbon with a double bond.
* The double bond in ethene is broken, and each carbon atom gains an -OH group (hydroxyl group). This results in the formation of ethane-1,2-diol (also known as ethylene glycol).
Important Note:
* The reaction requires an acidic medium. Typically, dilute sulfuric acid is used.
* The reaction is exothermic and often results in a color change. Potassium permanganate is a purple solution, but it gets decolorized as it oxidizes the ethene.
Overall, the product of the reaction between ethene and acidified potassium permanganate is ethane-1,2-diol (ethylene glycol).
