Mechanism of radical cyclisation to form hydroazulene core

Question

Recent studies on radical-induced cyclisation reactions have led to a simple, one-step method for the preparation of hydroazulenes from appropriately substituted cyclopentanone precursors. Treatment of 1, for example, with triphenyltin hydride and AIBN in refluxing toluene gave 2 in 79% yield. Give a mechanism for this transformation. (TES = triethylsilyl, SiEt₃)

(taken from: McKillop, A. Advanced Problems in Organic Reaction Mechanisms; Tetrahedron Organic Chemistry Series, Vol. 16; Pergamon Press: Oxford, U.K., 1997; p 1.)

I know that triphenyltin hydride, Ph₃SnH, can act as a reducing agent but I have no idea of its reduction mechanism. And I know that AIBN is a radical initiator:

The given conditions strongly suggest a radical pathway. However, looking at the product I also feel like there is a carbocation intermediate, as there is a ring expansion, but I don't know whether or how that happens.

Answer 1

This reaction is taken from Tetrahedron Lett. 1995, 36 (17), 3015. (DOI: 10.1016/0040-4039(95)00502-4)

The heat (provided by refluxing) is required to initiate the reaction by the decomposition pathway you have drawn. The cyanopropyl radical thus formed abstracts a hydrogen atom from the tin hydride, a consequence of the weak Sn–H bond:

The triphenylstannyl radical is the one that actually does interesting chemistry. Generally, there are a few options for radicals at this stage. One common pathway is to abstract a halide or chalcogenide from the starting material, but there is no such atom here (C–F and C–O bonds are stronger and rarely broken, so the OTES group is out of bounds). The other common pathway is to add to a π-bond. In this case the best place to add is the terminal carbon of the alkyne, which is least sterically hindered.

After this, the remainder is a series of addition-elimination steps. Note that in each case the ring size formed is 3, 5, or 6: these are all kinetically fast cyclisations.

In particular, note that:

• Intermediate 4 is stabilised by electron donation from adjacent oxygen (similar to carbocations, radicals - which are also electron-deficient species - can be stabilised by adjacent lone pairs).
• You are correct that a ring expansion is needed – in fact, two are needed – but this does not require a Wagner–Meerwein shift in a carbocation. Radicals are perfectly capable of doing this job, usually by a sequence of cyclisation (5 to 6) and retro-cyclisation (6 to 7). See also J. Org. Chem. 2012, 77 (19), 8634–8647 for another example of this motif, which is very common in radical chemistry.
• There isn't tin in the final product, so we will eventually have to kick it back out again: this can be achieved via a β-scission mechanism, so we have to work towards an intermediate which has a radical β to SnPh₃, i.e. 7.

The original paper supports this mechanism (although they skipped some steps, as is usual in journals):