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.

1-conversion of alkylbenzenese to benzoic acids using kmno4.gif

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:

examples of kmno4 oxidation of benzylic carbons

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.

benzylic-oxidation-with-kmno4-proposed-mechanistic-process-using-homolytic-dissocation-of-c-h-bond-followed-by-internal-return

Quiz Yourself!

Click to Flip

Click to Flip

Click to Flip


(Advanced) References and Further Reading

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


Real-Life Examples:

Org. Synth. 1927, 7, 18

DOI Link: 10.15227/orgsyn.007.0018

Click to Flip

Org. Synth. 1922, 2, 53

DOI Link: 10.15227/orgsyn.002.0053

Click to Flip

Org. Synth. 1930, 10, 20

DOI Link: 10.15227/orgsyn.010.0020

Click to Flip

Org. Synth. 1940, 20, 79

DOI Link: 10.15227/orgsyn.020.0079

Click to Flip

Comments

Comment section

175 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?

    Thanks

  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?

    1. Assuming excess KMnO4, it would probably eventually convert to oxalic acid and then 2 equivalents of CO2.

  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. Can you please tell me if KMnO4 will react with one of these, if not which one and the name please?
    CH3-CH-OH-CH2-CH3
    Cyclobutyl alcohol and methyl cycloprophy alcohol?

    1. Under mild conditions, each of these would be oxidized to ketones, to start with. If any enolization starts to occur (acid would help with that) then the enol would be dihydroxylated and you’d end up with oxidative cleavage products.

  10. 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 !!! :)

  11. 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!

  12. 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?

  13. Hi there.. just to confirm. so does that mean if the side chain is a ketone it wont get oxidised? because the adjacent carbon to the aromatic ring has no H. is that a fact?

  14. 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.

  15. 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

  16. 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?