Oxidation of aromatic alkanes with KMnO4 to give carboxylic acids
Description: Treatment of an alkylbenzene with potassium permanganate results in oxidation to give the benzoic acid.
Notes: The position directly adjacent to an aromatic group is called the “benzylic” position.
The reaction only works if there is a hydrogen attached to the carbon.
Examples:
Notes: Note that in example 2 the extra carbons are cleaved to give the same product as in example 1. And in example 3, two benzoic acids are formed. Finally, when no hydrogens are present on the benzylic carbon, no reaction occurs (example 4).
Mechanism: For the purposes of Org 1/ Org 2 the mechanism isn’t considered that important. Manganese acts in mysterious ways. It’s thought that the first step is removal of a hydrogen by one of the oxygens on MnO4(–) in a free-radical reaction. Beyond that, it gets complicated.
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In the second example, there is a three-carbon chain on the benzene, KMnO4 cleaves the chain leaving just one C (carboxylic acid). What is the other product formed then? Does the 2 carbon-chain cleaved off get oxidized too, so you’d have H3C-COOH? Or does it just stay as ethane or what? Essentially, does KMnO4 oxidize both sides of the benzylic hydrogens so it cleaves both sides into having a -COOH group on the cleaved side, or does KMnO4 only oxidize the side with the benzene to have a carboxylic acid?
Thanks for this great and very helpful website!
Yes, the other product would cleave as a carboxylic acid to give H3C-COOH although it’s quite possible that under these reaction conditions the alpha-position of the carboxylic acid would undergo further oxidation to HOOC-COOH, (oxalic acid) which would then probably oxidize to CO2.
In short, yes – KMnO4 should oxidize both sides.
You said the mechanism for the KMnO4 causing a carboxylic acid formation to the CH3 of the benzene is too complex for this site. Do you know where I can find the mechanism?
You might start with looking through March’s Advanced Organic Chemistry for this reaction and follow the reference from there. Essentially the first step is that KMnO4 removes a hydrogen from the benzylic position, forming a benzyl radical, and the oxygen then “rebounds” back to the carbon to form C-O. This repeats several times; the overall mechanism can go through several different pathways, but this is the essentials of it.
Does H2CrO4 have a similar mechanism/yield the same product?
Yes, I believe that H2CrO4 can also be used for this purpose. Not sure about the mechanism; in both cases, it’s fairly obscure.