How exactly does the oxidation of TEMPO by NaOCl/KBr work?

I would like to understand the exact process by which TEMPO is oxidized to the nitrosonium cation by NaOCl, as in the part of the mechanism shown in the diagram (1) surrounded by the red box:

The confusion for me is that TEMPO clearly needs to donate 1 electron to $\ce{OCl-}$ in order to form the neutral, nonradical nitrosonium cation, but we also get a $\ce{Cl-}$ anion, so there must also be a 2 electron process involving the $\ce{Br-}$ "co-catalyst" coming into play. How do I reconcile these 1 electron and 2 electron processes so as to draw a formally appropriate mechanism for this part of the reaction?

What exactly happens to the $\ce{OCl-}$ oxygen, and how does it happen?

(1) Org. Synth. 2005, 81, 195. DOI:10.15227/orgsyn.081.0195.

• Are you referring to the total mechanism? One specific step within it? Not sure what you're asking here. – hBy2Py Feb 13 '17 at 0:31
• Just the oxidation of TEMPO to the nitrosonium cation. – gannex Feb 13 '17 at 4:40
• Can you show me where you found this mechanism? This might be a case of a poorly drawn mechanism. – Mr Flux Feb 21 '17 at 14:50
• @MrFlux it doesn't matter. I just want to know the details of the oxidation of the TEMPO radical by NaOCl. Here is another picture: pubs.rsc.org/services/images/… some of then say the stoichiometry is 1/2 NaOCl, but I don't really see how this works either. – gannex Feb 21 '17 at 18:46
• A simple google search tells me you found this mechanism off Sigma-Aldrich page on TEMPO oxidation, not exactly a reliable source. Looks like poorly drawn mechanism. – Mr Flux Feb 23 '17 at 1:06

I would propose the following mechanism: Note that the oxygen radical anion isn't that bad as a leaving group as the Cl-O bond is weak. It can participate in an electron transfer with chloride anion, or rather the bromide at the beginning of the reaction as it is more abundant (I have written the protonated form - hydroxyl radical as it is less likely that two anions will come close enough for electron transfer):

The bromine radical can in turn react with another TEMPO molecule. That is why two equivalents of TEMPO react with one equivalent of hypochlorite.

Keep in mind that there are countless possibilities and deviations from the mechanism I proposed.

• This mechanism is good. In the first step, there is no change in enthalpy (if we assume the bonds between oxygen and chlorine are the same). In the second step, since TEMPO has very good electron-donating groups attached to the nitrogen it would have a greater tendency to donate electrons to the oxygen. Also the chloride ion is a good leaving group so it makes forming of the nitrosonium ion easier as well. – Linus Choy Feb 26 '17 at 9:49
• Thanks @Marko Does it make sense yo use two electron arrows for that first step? – gannex Feb 26 '17 at 23:38
• Sorry for that. It is a bimolecular homolytic displacement and fishhook arrows should be used. Fixed. – RBW Feb 26 '17 at 23:45