Please explain and rank the order of oxidation of methanol, ethanol, 2-propanol, 2-methyl-2propanol reacting with potassium perm
anganate.
1 answer:
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.
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