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.
Quiz Yourself!
Q1
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Q2
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Q3
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Further Reading
For some examples “in the wild”, see the following.
- Mechanistic studies supporting an initial C-H abstraction. J. Am. Chem. Soc. 1963, 85, 3487.
- Another mechanistic study with tertiary benzylic carbon. J. Am. Chem. Soc. 1970 92 329
- Oxidation of various hydrocarbons with KMnO4 using trifluoroacetic acid: Can. J. Chem. 1978, 56, 1273
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.
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
Sure, but that will probably eventually oxidize to CO2 unless you’re careful.
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
show the mechansim for oxidation of toluene with kmno4
The first step is C-H abstraction by the Mn=O oxo bond, followed by internal return making a manganate ester. After that, it gets complicated.
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.
can u tell me the whole mechanism
I suggest you read the references at the bottom of the page.
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.
where can i find the reaction that you are talking about, above?
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.
can i have the potassium dichromate oxidation mechanism of toluene followed by heating?
I suggest you check out the references at the bottom in “further reading”.
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.
What if potassium permanganate reacts with a Benzene (none substituted) will there be any product since KMnO4 usually racts with the Benzylic Hydrogen
Under normal conditions, no, there should be no reaction.
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”.
K2Cr2O7 will do the job!!
With acid, yes.
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.
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?
Yeah, KMnO4’s not the one. You want KOtBu or some other good base.
How do we add COOH group while replacing benzylic hydrogen of say toluene?
Lots of ways. Benzylic bromination followed by NaCN followed by aqueous hydrolysis… benzylic bromination, metal halogen exchange, CO2 then acid…
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. 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
See references at bottom of the page.
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?
Is the oxidation of side chain directly attached to cyclo hexane possible ?
Not using these conditions, because the cyclohexane does not afford resonance stabilization.
What will be the result if any organic compound undergo complete oxidation?
COMPLETE oxidation?? CO2 !
Hw about there is a alcohol group attached to the alkyl side chain ? The reaction still same?
It would get oxidized to a ketone.
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.
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?
My understanding is that KMnO4 is less “hot” than some of the iron oxo compounds that are found in liver enzymes, although their chemistry is quite similar.
If there is 10 carbon in second example what would be the reaction and product ? Dose that reaction is too hard to do?
Still benzoic acid. Not significantly harder than ethylbenzene. Still goes through a benzylic radical.
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
Can benzoic acid be further oxidized into an alcohol c6h5ch2OH
That’s not an oxidation, that’s a reduction.
Can xylene isomers be converted to toluene with potassium permanganate?
No, you’d get dicarboxylic acids.
Hey, Would this reaction be given by cumene?will it get oxidised to benzoic acid?
Yes, it absolutely would.
Will a similar reaction occur if the ring is partly saturated (e.g. cyclohexene instead of benzene)?
In that case, the double bond of cyclohexene will be attacked. No reaction on the sidechain will occur because no resonance-stabilized benzylic radical can be formed.
Does C6H5CHO (BENZALDYHYDE) reacts with H+/KMnO4 giving Benzoic acid?
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.
Why doesnt the KMnO4 attack the pi bonds on the aromatic
Aromatic rings are generally quite resistant to electrophilic attack.
does this only happen in acidic and high temperature condition?
Yes, it’s generally a reaction that requires heat. Although… there are some variants of organic-soluble permanganate salts that can oxidize once to give alcohols and not give further reactions. See for example, https://science.sciencemag.org/content/324/5924/238 (full disclosure: I am an author).
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!
Not using this reaction, no.
What about activated benzene ring with KMnO4 and heat?
In my notes o cresol decomposes
So will every activated ring be decomposed ?
Anything with an OH directly attached to the ring will be oxidized to a quinone and from there it will produce a polymerized mess.
Can methanoic acid be oxidized to carbondioxide using acidified KMnO4??
You mean formic acid? Sure.
Where can you draw these chemical equations?
ChemDraw.
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??
MnO2 and other manganese salt crap. Can’t be more precise than that.
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.
Why does styrene react with KMnO4 to give benzaldehyde? Why not benzoic acid?
Benzaldehyde would be the first product formed after oxidative cleavage of the glycol. It wouldn’t be difficult for benzaldehyde to be further oxidized to benzoic acid.
if (C6H5)CH2CH2OH reacted with H+/KMnO4 —>>
what product formed???
benzoic acid or 2-phenylethoic acid ???
Benzoic acid
Why not K2Cr2O7 not used in oxidation of toluene?
With acid, as chromic acid H2CrO4, it can be used in the oxidation of toluene and other aliphatic compounds. See here. Tetrahedron 1941, 20, 1151 . https://www.sciencedirect.com/science/article/pii/S0040402001989823
Hi James, what if side chain have one double bond in example 2. Will the main product be same or change??
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.
What about oxidation of secondary alkyl benzene?
Above given examples are only primary and tertiary alkyl benzene
Still works.
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?
It will oxidize to benzoic acid because it has a benzylic C-H bond.
Does it work if instead of a carbon chain, there is an alkyl halide on the benzene ring?
Such as CH2Cl ? That would be fine, it should still oxidize as it will still form a resonance stabilized free radical.
Can kmno4 oxidize acetophenone to benzoic acid?
The alpha position of acetophenone is reasonably susceptible to radical C-H abstraction. Given enough heating I would not be surprised to see oxidation.
will it work if there’s a ketone attached to the benzylic carbon followed by an R group? So C6H5COCH3 for example
Not sure. If the enol is formed, then that will certainly react with KMnO4.
If benzene side chain alkene present instead of alkane what will happen
Styrene? Likely dihydroxylation first, and then oxidative cleavage, and you’ll get benzaldehyde, which should oxidize further to benzoic acid.
What is the colour change for lets say methylbenzene to benzoic acid?
Color change is a part-per-million phenomenon that is often not diagnostic. Methylbenzene (toluene) is colorless, and benzoic acid is also colorless. But trace impurities can lead to color.
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?
I don’t know. You’d get benzophenone for sure. After that, I’d think that given mild enough conditions you’d be able to stop it there.
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?
I don’t know what it would make if there was an amine; it’s likely going to be a mess.
Can KMnO4 oxidize CARDANOL to Hydroxyl benzoic acid?
I had to look up the structure of cardanol. With a phenol on the ring, you’re likely going to get quinone formation and a total mess. Find another way to do it.
Could the KMnO4 oxidize the benzene ring itself?
Generally not under normal conditions, though anything is possible with enough KMnO4 and heat.