Reaction scheme

What is the mechanism for the above reaction?

  • $\begingroup$ 3+2 cycloaddition and retro-cyclo maybe but I'm not sure. $\endgroup$ – Mithoron Aug 7 '15 at 15:06
  • $\begingroup$ (I removed the fluff so that it's easier for future readers.) $\endgroup$ – orthocresol Jun 28 '17 at 7:03

This starting material is not actually an enone; it is a 1,3-dicarbonyl, confusingly drawn as the enol tautomer. 1,3-Dicarbonyls are mildly acidic ($\mathrm pK_\mathrm a \sim 10$) and can be deprotonated by weak bases, such as triethylamine ($\mathrm pK_\mathrm a \sim 9$), which was mentioned in the question. So, now, we have an enolate ion which is nucleophilic.

Tosyl azide, $\ce{TsN3}$, is an odd reagent. It's easy to think that it is acting as a nucleophile of some sort, because the azide ion is nucleophilic, but actually it is an electrophile.

Mechanism - part 1

The next part is the removal of the formyl group, which is a bit tricky. Here I am not 100% sure, so take it with a pinch of salt, but I think Mith was on the right track with the retro-(3+2) cycloaddition. This is my guess:

Mechanism - part 2

In fact, the cyclisation of the terminal nitrogen onto a carbonyl group has been proposed before, in mechanistic studies of electrophilic azidation by D. A. Evans.1 That makes me feel a bit more confident in my guess!

Proposed triazoline formation

Anyway, the product above is simply a resonance form of the diazo product you want.

Mechanism - part 3


  1. Evans, D. A.; Britton, T. C.; Ellman, J. A.; Dorow, R. L. The asymmetric synthesis of α-amino acids. Electrophilic azidation of chiral imide enolates, a practical approach to the synthesis of (R)- and (S)-α-azido carboxylic acids. J. Am. Chem. Soc. 1990, 112 (10), 4011–4030. DOI: 10.1021/ja00166a045. Non-paywall version available from the Evans group website.

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