Oxidation under hot concentrated KMnO4/H+

The C=C double bond in the alkene breaks down entirely. The products that form depend on what is bonded to the carbon atoms in the C=C double bond.
If a carbon atom is bonded to two hydrogen atoms, it oxidises to form a CO2 molecule.
\( \text{H}_2\text{C}=\text{CH}_2 + 6[\text{O}] \rightarrow \text{CO}_2 + \text{CO}_2 + 2\text{H}_2\text{O} \)
If a carbon atom is bonded to one hydrogen atom and one alkyl group, it oxidises to form a CO (aldehyde) group first, which oxidises again to form a COOH (carboxylic acid) group.
\( \text{RHC} = \text{CHR} + 2[O]\rightarrow \text{RCHO} + \text{RCHO} + 2[O]\rightarrow \text{RCOOH} + \text{RCOOH} \)
If a carbon atom is bonded to two alkyl groups, it oxidises to form a C=O (ketone) group.
\( R_1 R_2 C = C R_3 R_4 + 2[O]\rightarrow R_1 (CO) R_2 + R_3 (CO) R_4 \)
To obtain aldehydes, the solution must be heated with a distillation apparatus. This is done to collect the aldehyde vapour before it oxidises again to form carboxylic acids.
To obtain other oxidised substances, the solution must be heated under reflux.

Oxidising under cold dilute KMnO4

Under cold dilute conditions, the C=C bond is not fully broken. Instead, -OH groups are attached to the two carbon atoms, forming diols.
The diols can be oxidised again to form carboxylic acids, ketones or CO2 depending on the chemical structure of the compound.