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I know the mechanism for Jones' oxidation is (picture from Organic Chemistry by Clayden):

enter image description here

and it proceeds through a chromate ester. And in general we understand that the Jones' reagent $\ce{H2CrO4}$ cannot oxidize tertiary alcohols.

However, in my book, a question asked to oxidize the following tertiary allylic alcohol:

enter image description here

The solution proceeded via a carbocation intermediate to eventually form 3-methylcyclohex-2-en-1-one.

I wish to ask if this a correct solution. Does this reaction really occur? That too in the way it is described? If so, why isn't it described in any other book, even as an exceptional case? (Most other books simply accept that tertiary alcohols cannot be oxidised) Or is this a by-product or a low-yield reaction?


Source: MS Chouhan; Advanced Problems In Organic Chemistry; 11th ed; Q19; Alcohols, Ethers, Epoxides

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It makes sense. Jones's oxidation occurs in presence of a strong acid. Hence, the oxygen can get protonated and leave as a water molecule, giving rise to a tertiary carbocation. The charge can then delocalize, the secondary alcohol get formed, and this alcohol then undergo oxidation. The oxidation will probably drive the equilibrium towards the product.

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  • $\begingroup$ It does make sense. I was kind of surprised that this hasn't been mentioned before in any other book. Like, I had certainly rote learned the fact that "three degree alcohols cannot be oxidised by Jones the boy" like a song in my brain :P $\endgroup$ Commented Mar 15, 2018 at 7:17
  • $\begingroup$ Well, in principle they can't, this is a secondary alcohol getting oxidized. Another thing that sometimes surprises people is that "ketones can actually be oxidized", via their enol tautomers. $\endgroup$
    – ralk912
    Commented Mar 15, 2018 at 7:19
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    $\begingroup$ Here is a study in the kinetics with $\ce{KMnO4}$. I can't think of a practical application off top of my head, I'd expect the reaction not to be very clean (the carboxylic acids/ketones formed can form enols themselves!). Also, the Baeyer-Villiger reaction is also an oxidation, but of course that's a different mechanism and off-topic (and well combustion is too, but even more off-topic). $\endgroup$
    – ralk912
    Commented Mar 15, 2018 at 7:44
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    $\begingroup$ The oxidation may not go through a cation. The tertiary chromate ester may undergo a [3,3]-sigmatropic rearrangement to the secondary chromate ester and thence to the enone. $\endgroup$
    – user55119
    Commented Mar 15, 2018 at 19:29
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    $\begingroup$ @ralk912: You should! Read this, Dauben and Michno, J. Org. Chem., 42, 682 (1977). $\endgroup$
    – user55119
    Commented Mar 15, 2018 at 19:59

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