[g-bones]

I am looking at the oxidation of a primary alcohol to an aldehyde in a solvent free reaction.  2 equiv. of CrO2 is needed but I am having trouble understanding the balance of the oxidation.  I attacked the Cr followed by a proton transfer to form the mixed anhydride and deprotonated to yield the final aldehyde and the first equivalent of Cr(III)  but deprotonation would not give the second equiv of Cr(III) because quenching the positive charge as I have done removes electrons making it actually Cr(V).  Anybody have any suggestions on how the second equiv of Cr(III) is formed? thank you

[Dan]

What about a mechanism analogous to a CrO₃ oxidation? This generates a Cr(II) species which can then reduce a Cr(IV) species (electron transfer reaction?), which generates two Cr(III) species. See attached scheme. If you want to generate 2Cr(III) straight off the bat you will need to proceed via radicals as any fully ionic mechanism (ie. full curly arrows rather than fish-hook type) will always do redox in multiples of two

I would have thought it's a reasonable suggestion. If you consider the generally accepted mechanism for a Cr(VI) oxidation, a Cr(IV) species is initially generated which presumably then disproportionates - since it is well known that CrO₃ oxidations ultimately leave you with Cr(III), which is why they go green.

What do you think?

[g-bones]

Seems like a valid argument.  Thank you for your input. I dont know why it would simply stay in the (II) oxidation state. is this unreasonable because there would be so much Cr(IV) around to act as an oxidizing agent with the Cr (II) species as much as with the alcohol?

[Dan]

I dont know why it would simply stay in the (II) oxidation state.

I don't think the Cr(II) will survive, if you look up the reduction potentials of Cr(IV) and Cr(III) I'm 99% sure you'll find that Cr(IV)-->Cr(III) is positive and Cr(III)-->Cr(II) is negative - or at least more negative than IV->III, meaning that Cr(II)+Cr(IV)-->2Cr(III) is energetically favourable.