When a radical is alpha to a three-membered ring, it is common for the ring to open via β-scission (to relieve ring strain):

Ring opening of cyclopropylmethyl radical

When the ring contains a heteroatom there are a couple of different possibilities:

Epoxide/aziridine analogues

Ab initio studies of these reactions1 showed that for both X = O and X = NH, the C–C cleavage was thermodynamically favoured (presumably due to stabilisation of the subsequent radical by the adjacent lone pair). For X = O, the C–C cleavage was also found to be kinetically favoured, but for X = NH, the C–N cleavage is kinetically favoured!

This regioselectivity is also backed up by experiment2 and also has potential for use in synthesis.3

C–N bond cleavage, taken from ref 3

Is there an explanation for why the kinetically favoured products are different for X = O and NH?


  1. Pasto, D. J. Ab Initio Theoretical Studies on the Ring-Opening Modes of the Oxiranyl-, Aziridinyl-, Oxaziridinyl-, and Thiaranylmethyl Radical Systems. J. Org. Chem. 1996, 61 (1), 252–256. DOI: 10.1021/jo9505942.

  2. Marples, B. A.; Toon, R. C. An investigation into the β-cleavage of aziridinylcarbinyl radicals. Tetrahedron Lett. 1999, 40 (26), 4873–4876. DOI: 10.1016/S0040-4039(99)00900-4.

  3. Prévost, N.; Shipman, M. Intramolecular Radical Rearrangement Reactions of 2-Methyleneaziridines: Application to the Synthesis of Substituted Piperidines, Decahydroquinolines, and Octahydroindolizines. Org. Lett. 2001, 3 (15), 2383–2385. DOI: 10.1021/ol016194k.


1 Answer 1


The 2-methyl-3-phenyloxiranylcarbinyl radical has been shown EXPERIMENTALLY to cleave the C-O bond kinetically while the C-C bond is cleaved thermodynamically. This means that, in this example, The oxygen-centered radical forms faster than the highly stable benzyl radical. The formation of the oxygen radical is necessarily reversible. See ref. 11 in your ref. 2 (Marples paper)

Ziegler, F. E.; Petersen, A. K. Allyloxy Radicals are Formed Reversibly from Oxiranylcarbinyl Radicals: A Kinetic Study, J. Org. Chem. , 1995, 60, 2666.

  • $\begingroup$ Fair enough, but there's the followup question: why is C–X cleaved kinetically despite forming a less stable radical? Thanks for pointing this out, nonetheless. I'll edit the question when I get a chance to. $\endgroup$ Commented Nov 19, 2017 at 21:19

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