I saw the following reaction mechanism in paper Tetrahedron Lett. 1976, 17 (33), 2869–2872.:

enter image description here

I'm not able to understand the following parts:

  1. How did the conversion of 27 to 29 take place? I've never seen that kind of a ring expansion, and a detailed explanation would help - maybe including the necessary transition state(s) and why it happens.

  2. 27 to 28 and 30 to 31 aren't clear as well; why should such rearrangements happen and how do they happen? What is the mechanism?

  • $\begingroup$ A key thing to remember is that 4-membered rings are highly strained and not very stable. $\endgroup$
    – Waylander
    Mar 21, 2018 at 14:51
  • $\begingroup$ Alright, I was aware of that. What's next? $\endgroup$ Mar 21, 2018 at 14:52
  • 2
    $\begingroup$ @schrodinger_16 The bridge C-C bond of 27 has just shifted to the right and created that big ring. It is interesting to note though, that there are two ways for it to shift: one creates this 8-member ring, the other creates two five member rings; I'm curious how we can even predict theoretically which combination will have lesser strain. $\endgroup$ Mar 21, 2018 at 15:10

1 Answer 1


I think the possible mechanism for the intermidiate conversions can be as following.Here thr $\ce{R}$ is actually $\ce{-OCHO}$. enter image description here

  • $\begingroup$ What's the reason for that hydride shift? Clearly, it decreases the carbocation's stability by decreasing the number of hyperconjugative structures. $\endgroup$ Mar 22, 2018 at 4:34

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