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.
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.
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.
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.
If I have a molecule like methylidene cyclopentane..does KMnO4 react to give a CA??
If I have a molecule like methylidene cyclopentane..does KMnO4 react to give a CA??
For methylidene cyclopentane, it would cleave the C=C bond, to give cyclopentanone and formic acid. Actually the formic acid would probably become CO2.
what is the difference in the oxidation products of hexane,cyclohexane,xylene,benzene,benzylchloride,and methyl benzene by KMnO4? i cant understand the reactions
Can Kmno4 oxidise methane as well?
If not, why?
Given enough heat, sure. It would give CO2. But so would plain old combustion.
is it possible to get the actual mechanism involed, using example 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.
should the KMnO4 be acididic or alkaline?
what other chemical that can be used to replace KMNO4??
CrO3 is also occasionally used for this reaction.
Explain example 3 again,please?
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.
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?
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.
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
I want to know the mechanism of this reaction!
Since in the second example the second product is acetic acid, would the products of reaction n°3 be phtalic acid and oxalic acid?
Assuming excess KMnO4, it would probably eventually convert to oxalic acid and then 2 equivalents of CO2.
Wouldn’t the methyl group be oxidised to methanoic acid as well?
Oh sorry I forgot that methanoic acid is further oxidised to carbon dioxide and water.
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?
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”.
Why is it preferred to use alkali when KMnO4 uses acid in the oxidation of alcohols?
my textbook says that reacting cyclopentene with KMnO4 will give you a di-ol…
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
Thanks for this great and very helpful website!!
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 :’)
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 !!! :)
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!
yep! it will stay.
Why catalysts work only at side chain carbon atom? Why not carbon-H bond of benzene ring?
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.
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.
Isn’t PhCOCh2Ch3 an aromatic ketone?
Thanks for pointing out my error. Corrected above.
I believe also in this particular case, the enol double bond can be broken by KMnO4 as alkenes do, forming the same product.
Would this also work with chromium trioxide as the oxidising agent ?
Thanks,
Parakh
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?
Sorry. The bezylic position has 3 OH GROUPS
A carbon with three OH groups? How is that stable?
I was under the impression that heat was a necessary catalyst.
I wouldn’t say that heat is a “catalyst” but the reaction is definitely helped by heat, yes.
What is the effect of substituents attached to benzene ring is dere any ?
Should be pretty minimal, although you’d want to avoid any substituents that are easily oxidized, like NH2 or SR.
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?
Thanks.
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.
Does KMn04 has an effect on Ester bond?
Generally, no, although it certainly could oxidize the carbon adjacent to the oxygen.
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?
Thanks
Hey, Would this reaction be given by cumene?will it get oxidised to benzoic acid?
Yes, it absolutely would.
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
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
Yes, it would definitely be oxidized to benzoic acid.
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.
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.
if (C6H5)CH2CH2OH reacted with H+/KMnO4 —>>
what product formed???
benzoic acid or 2-phenylethoic acid ???
Benzoic acid
What about oxidation of secondary alkyl benzene?
Still works.