How Does KMnO4 Oxidize Toluene to Benzoic Acid?

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The discussion centers on the oxidation of toluene with KMnO4 to produce benzoic acid, specifically focusing on the mechanism of how KMnO4 interacts with the hydrogen on the aliphatic chain. The initial step involves the formation of either radicalic or cationic species, with the consensus that a benzyl cation or radical is generated. The process begins with hydrogen atom abstraction, leading to the formation of a benzyl radical, which can then react with water to form benzyl alcohol. The subsequent oxidation of benzyl alcohol to benzoic acid is considered more straightforward and likely does not involve radical intermediates.
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Hi,

It's been quite some time now I've spent on figuring out the mecanism of the oxydation of toluene with KMnO4 to give benzoic acid. My problem is i don't know how exactly the MnO4 attacks the hydrogen on the aliphatic chain. So, if anyone can help me figure out the first steps of which O on MnO4 will attack, that'll be great! :confused:
 
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Hello

In this process, either radicalic or cationic species form, as seen in a closely related reaction:

Ar_2CH_2+CrO_3\longrightarrow Ar_2C=O
rate~determining~step:Ar_2CH_2 \longrightarrow Ar_2CH \cdot ~or~Ar_2CH_2 \longrightarrow Ar_2CH^+

I'm certain about a benzyl cation or benzyl radical will be produced in this reaction you're asking about.
 
I think chem_tr is right. The first step is a hydrogen atom abstraction (that is, a proton plus an electron) to give a benzyl radical. That can then combine with a water molecule (water is typically the solvent for these oxidations) to give benzyl alcohol. The oxidation from the alcohol to the acid is much more straight forward and probably doesn't involve radicals.
 
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