Reaction of CH3–CH=CH2 with acidified KMnO4:

- The reaction of CH3–CH=CH2 with acidified KMnO4 results in the oxidation of the alkene to form a diol.
- The double bond in the alkene is broken, and two hydroxyl groups are added to the carbon atoms previously involved in the double bond.

Explanation:

- The acidified KMnO4 acts as an oxidizing agent in this reaction.
- The purple color of the KMnO4 solution fades as it is reduced during the oxidation process.
- The alkene undergoes addition reactions with the KMnO4, resulting in the formation of a diol.
- The mechanism involves the cleavage of the double bond and the addition of hydroxyl groups to the carbon atoms.
- The overall reaction can be represented as follows:
CH3–CH=CH2 + KMnO4 + H2O → CH3–CHOH–CHOH–CH3 + MnO2 + KOH
- The products of the reaction include a diol (CH3–CHOH–CHOH–CH3), manganese dioxide (MnO2), and potassium hydroxide (KOH).

Conclusion:

- The reaction of CH3–CH=CH2 with acidified KMnO4 results in the formation of a diol through the oxidation of the alkene.
- This reaction is an important transformation in organic chemistry, demonstrating the addition of hydroxyl groups to an alkene in the presence of an oxidizing agent.