The conversion of an alkene to an alcohol using acidic potassium permanganate KMnO4 as the oxidizing agent is an example of an oxidative cleavage reaction. The mechanism for this reaction can be broken down into the following steps:1. Formation of the cyclic intermediate: The alkene reacts with the acidic potassium permanganate KMnO4 to form a cyclic intermediate. This occurs through the transfer of one electron from the alkene double bond to the permanganate ion MnO4- and the formation of a bond between the manganese atom and one of the carbon atoms of the alkene. Simultaneously, one of the oxygen atoms from the permanganate ion forms a bond with the other carbon atom of the alkene. This results in the formation of a cyclic intermediate with a five-membered ring containing manganese.2. Cleavage of the cyclic intermediate: The cyclic intermediate is unstable and undergoes cleavage, breaking the carbon-carbon double bond. This is facilitated by the transfer of a proton from the acidic solution to one of the oxygen atoms in the intermediate. The cleavage results in the formation of two carbonyl groups C=O on the two carbon atoms that were initially part of the alkene.3. Reduction of the carbonyl groups: The carbonyl groups formed in the previous step are then reduced to hydroxyl groups OH by the reduced manganese species Mn2+ . This is achieved through the transfer of a hydride ion H- from the manganese species to the carbonyl carbon, followed by the addition of a proton H+ from the acidic solution to the oxygen atom. This results in the formation of two alcohol functional groups.In summary, the functional group of the alkene C=C changes to two alcohol functional groups OH during the reaction. The overall reaction can be represented as:R-CH=CH-R' + KMnO4 + H2O + H+ R-CH OH -CH OH -R' + Mn2+ + K+ + 2OH-