This reaction is taken from Tetrahedron Lett. 1995, 36 (17), 3015. (DOI:
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):