When propene is treated with hot alkaline Potassium permangante (KMnO4), it forms Ethanoic acid (CH3COOH) and Formic acid or simply the Methanoic acid (HCOOH)is formed in the product side.

CH2​=CH−CH3​ ==>CH3COOH + HCOOH.
As you can see the above equation CH2=CH–CH3 (C3H6) Propane is treated with Hot alkaline Potassium permanganate (KMnO4) to form Ethanoic acid and Formic acid.

The formic acid decomposes to water and carbon dioxide.

HCOOH→H2​O+CO2​
This is oxidative cleavage reaction.

Product obtained:
The product obtained when propene is treated with hot alkaline KMnO4 is propan-1,2-diol or 1,2-propylene glycol.

Explanation:
The reaction between propene and hot alkaline KMnO4 involves the addition of two hydroxyl groups to the double bond in propene. The reaction proceeds through the following steps:

1. Addition of hydroxyl groups: In the first step, the double bond in propene attacks the MnO4- ion, resulting in the formation of a cyclic intermediate. This intermediate then reacts with water to form two hydroxyl groups attached to the carbon atoms of the double bond.

2. Oxidation: The two hydroxyl groups are then oxidized by the MnO4- ion to form two aldehyde groups.

3. Acid-catalysed hydration: The aldehyde groups are then hydrated to form 1,2-propylene glycol in the presence of acid catalyst.

Overall reaction:
Propene + KMnO4 + OH- → 1,2-propylene glycol + MnO2 + K+ + H2O

Significance:
Propan-1,2-diol or 1,2-propylene glycol is an important chemical used in the production of various products such as antifreeze, solvents, and cosmetics. The reaction between propene and hot alkaline KMnO4 is a useful method for the production of this compound.