Master Organic Chemistry Reaction Guide

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.


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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Further Reading

For some examples “in the wild”, see the following.


Comment section

148 thoughts on “Oxidation of aromatic alkanes with KMnO4 to give carboxylic acids

  1. 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!

    1. 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.

      1. If one carbon is getting cleaved from the molecule then I guess it should form HCOOH instead of CH3COOH (as mentioned by you guys above). And further if KMnO4 is in excess amount then HCOOH should probably get oxidised to CO2.

        Correct me if I am wrong.

        1. IAmAGeek, TA asked about the second example. The second example has a three carbon chain attached to the benzene ring. Therefore, two of those three carbons would be cleaved as an ethane chain which would then be oxidized to form H3C-COOH.

        2. if you start with “propanyl-benzene”? you add the carboxylic acid group on the carbon closest to the ring.. once leaved of, youll have a piece if CH3-CH2 which will be oxidized to CH3COOH

    2. Yes James is halfway correct only thing he missed is is a tertbutyl carboxylic acid
      So we have tbutyly carboxylic acid ,oxalic acid (2 moles to be specific ) and CO2

      The reactions skeletal mechanism can be seen by action of KMnO4 on each of benzenes double bond converting the ends to carboxylic acids , the upper part with the tertbutyl group is oxidized twice

  2. 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?

    1. 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.

    1. Yes, I believe that H2CrO4 can also be used for this purpose. Not sure about the mechanism; in both cases, it’s fairly obscure.

    1. For methylidene cyclopentane, it would cleave the C=C bond, to give cyclopentanone and formic acid. Actually the formic acid would probably become CO2.

  3. what is the difference in the oxidation products of hexane,cyclohexane,xylene,benzene,benzylchloride,and methyl benzene by KMnO4? i cant understand the reactions

    1. It’s a long sequence beginning with hydrogen abstraction and “rebound” of oxygen to carbon. There’s some uncertainty about the exact mechanism, so it hasn’t been shown.

    1. In this example both CH2 groups are cleaved to give carboxylic acids. The alkyl C-C bonds are cleaved in the process, resulting in breakage of the cyclohexane ring.

  4. What if potassium permanganate reacts with a Benzene (none substituted) will there be any product since KMnO4 usually racts with the Benzylic Hydrogen

  5. If there is a halogen let’s say attached to our alkyl group will it form carboxylic acid ? and if there is an alkene not directly attached to the benzene but in the side chain will it form carboxylic acid?

    1. Yes, as long as the carbon has at least one H in addition to the halogen, it will still form a caroboxylic acid – the chlorine will eventually get hydrolyzed. As far as the side chain is concerned, it will be completely cleaved until only the carboxylic acid attached to the ring remains.

  6. Are there any other ways to remove the methyl group where the mechanism is known? i.e. substitute it for a Br or F? and then add a carboxylic acid afterwards?


  7. Since in the second example the second product is acetic acid, would the products of reaction n°3 be phtalic acid and oxalic acid?

  8. Hello James,

    Is it only KMnO4 has the ability to oxidise benzylic carbon to carboxylic acid, is there other oxidant capable of doing this?

    Will this oxidation reaction also possible for allylic carbon?

    1. Great question! Yes, in theory, KMnO4 is useful for such a task. However in practice, other reagents such as SeO2 (among many others) are typically used for this reaction. Google “allylic oxidation”.

    1. That’s an alkene. Ocidation of alkanes gives carboxylic acid except oxidation with CrO2Cl2 wich gives an aldehyde. Oxidation of alkenes gives diols and oxidation of alkynes gives a dioc acid

  9. Do we use acidified or alkaline KMnO4? Or it doesn’t really matter?
    Great website though! Helped me so much! People like you deserve happiness :’)

    1. Hi Ashley.. I believe if you use alkaline KMnO4, then you will not get the observation “purple solution is decolourized”.. instead you might get “purple solution produces brown ppt”. Acidified KMnO4 is needed to give the “decolourized” observation.

      BTW.. a BUG *THUMBS-UP* to those of you who manage this website. It is an EXCELLENT resource for Organic Chem, right till A Level !!! :)

  10. Say there is a halogen attached to the benzene ring, will it remain while the alkyl chain is oxidized? Thank you for this helpful page!

  11. I thought KMnO4 could be used to create double bonds? For example if I used NBS to attach a Bromine to the benzylic carbon, could I then used KMnO4 to create a double bond and remove the Bromine? Do I have my reactants mixed up?

  12. thanks for the great info!
    you mentioned that a benzylic H is necessary for this reaction to occur. However i read in Solomons that even PhCOCh2Ch3 reacts and gives Ph COOH .how can this happen?
    thanks again.

    1. That’s an ester hydrolysis, not an oxidation. Sorry, saw an extra oxygen where there wasn’t one.
      Yes, hydrogens on carbons adjacent to carbonyls contain weaker C-H bonds than do normal alkanes, so this reaction is possible. They are weaker because of an effect known as “captodative stabilization”, where the free radical initially generated is stabilized by interaction with the orbitals of the carbonyl. I know this is a very technical sounding answer, but it’s not dealt with very much in introductory organic chemistry.

      1. I believe also in this particular case, the enol double bond can be broken by KMnO4 as alkenes do, forming the same product.

  13. hi. i need to know the mechanism of the reaction of ethyl benzene with KMnO4…
    actually i want to know why both methyl benzene and ethyl benzene give the same product…
    thank you in advanced

  14. Hi James if the benzene ring has a carbon which has 3 OH groups attached to it will it get oxidised to benzoic acid?
    I the beryllium carbon odes not have a hydrogen?

  15. What about if you had (ethoxymethyl)benzene and attempted this oxidation? Would you oxidize the benzylic carbon to a carbonyl and keep the ether linkage, or would you get a carboxylic acid as is outlined above?

    1. Great question. You’d likely oxidize the two C-H bonds on the benzylic carbon to C-O, which would give you an ester, and under the typical reaction conditions (aqueous acid, heat) , you’d probably hydrolyze it. My guess is that the CH2 group on the ethoxy linkage would oxidize as well, since the reaction proceeds through a free radical mechanism and radicals on carbons directly attached to O are stabilized.

  16. HI James
    Do you know how the oxidizing power (or ability) of KMnO4 compares to microbial enzymes for degrading biomolecules found in plant material or soil organic matter?

  17. Hey,

    If the benzene is attached to a cyclohexene and you add this molecule to kmno4 for 2 hours at room temperature, what is the product? Do you get a ketone at the benzyllic position with a carboxylic acid?


    1. Oh, absolutely. Oxidation of aldehydes to carboxylic acids is extremely easy. If you leave benzaldehyde out long enough, air will oxidize it to the carboxylic acid. Using KMnO4 is like using a thermonuclear bomb to kill a moth.

  18. For the first example, is there a way to add the carboxylic acid group to the structure so that there is a carbon between the benzene ring and the carbonyl carbon? Thanks!

  19. What about activated benzene ring with KMnO4 and heat?
    In my notes o cresol decomposes
    So will every activated ring be decomposed ?

  20. I believe if you use alkaline KMnO4, then you will not get the observation “purple solution is decolourized”.. instead you might get “purple solution produces brown ppt”.

    Can any one have any idaa what is brown ppt??

  21. If the compound we treat with KMnO4 is a group like -CH2COCH3 Would that oxidize by the reagent ? Since this substitute has alpha carbon but also this is a ketone group

    1. If the compound we treat with KMnO4 is a group like R’-CH2COCH3 (R’=benzene) Would that oxidize by the reagent ? Since this substitute has alpha carbon but also this is a ketone group

  22. I have two methyl groups at pyrazole ring at 1,3 position. I want to convert both groups to CA. I tried so many times with KMNO4 +water or KMNO4+ pyridine or KMNO4+NaOH. but I could not get acid groups. Please hewlp me or suggest me some other route.

    1. It’s probably hitting the pyrazole nitrogens, forming an N-oxide. I would suggest doing benzylic bromination or similar reaction first, followed by conversion to alcohol and then mild oxidation to the carboxylic acd.

    1. The double bond will be dihydroxylated by KMnO4. At low temperatures with base this can be isolated as a diol. With heating or acid oxidative cleavage will occur to give an aldehyde which can be further oxidized to give a carboxylic acid.

  23. Hi there… How about this compound.. (C6H5)-CH2C(CH3)2OH?.. Could it be oxidised to benzoic acid as well? Even though it is a tertiary alcohols?

  24. What happens if diphenyl methane is used as reactant? Do 2 molecules of benzoic acid form, or 1 benzoic acid and 1 benzene. Do products differ if oxidant is L.R first and then in excess?

  25. would it still oxidize to benzoic acid if the benzylic position is quartairy but one of the substitutes of the benzylic position is an amine?

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