Question

Please explain and rank the order of oxidation of methanol, ethanol, 2-propanol, 2-methyl-2propanol reacting with potassium permanganate.

Answer 1

Answer:Methanol>Ethanol>2-Propanol>2-methyl-2Propanol

Explanation:The mechanism of oxidation using Potassium permanganate

involves two steps   :

In the first step permanganate ion abstracts a alpha-hydrogen as hydride ion available at alcohol.

Alpha hydrogen is the hydrogen attached to the carbon bearing functional group.

In the second step the permanganate ion subsequently is reduced from +7 oxidation state of Manganese to +5 oxidation state of Manganese.

So for the oxidation of alcohol using potassium permanganate , there must be availability of alpha hydrogens.

we can also relate the order of oxidation with reference to number of alpha hydrogens present on the substrate.

so

Methanol has 3 available alpha hydrogen

Ethanol has 2 available alpha hydrogen  

2-propanol has 1 available alpha hydrogen as it is a secondary alcohol

2-methyl-2propanol has 0 available alpha hydrogens as it is a tertiary alcohol.

Greater the number of available hydrogens easier would be oxidation of alcohols so the order of oxidation would be the following:

Methanol>Ethanol>2-Propanol>2-methyl-2Propanol

please refer the attachment for the structures of following compounds.