I have a radical reaction which I got from an organic chemistry lecture and was drawing the possible mechanism for it.

Fig 2. shows my thought process. The I is removed making a radical and ISnBu$_3$. The radical opens the ring and the new radical is formed which then reacts with the hydrogen radical. However Fig 1. shows the reactants and products from the lecture which don't match with what I thought.

I don't see how the carbon on the "top" of the propane ring becomes part of the chain.

Could the lecture slides have been wrong or am I missing somethenter image description hereing?

  • $\begingroup$ Without consulting the literature, there is the possibility that fig. 2 is a product of kinetic control whereas fig. 1 is the product of thermodynamic control. The second radical in fig. 2 would recyclize under dilute Bu3SnH concentration. This is certainly the case for the bromo epoxide analog of the cyclopropyl iodide. $\endgroup$
    – user55119
    Sep 5 at 17:57
  • $\begingroup$ @user55119 is there be a reaction mechanism for the thermodynamically controlled product? I just don't see how the carbon can become part of the chain $\endgroup$
    – bobsburger
    Sep 5 at 18:13
  • 3
    $\begingroup$ After you form the radical, break the "bottom" bond in the cyclopropane ring. That produces your terminal double bond and a resonance stabilized benzylic radical. Breaking the other cyclopropane bonds does not produce a stabilized radical. $\endgroup$
    – ron
    Sep 5 at 18:47
  • 1
    $\begingroup$ @Ron just beat me to it. $\endgroup$
    – user55119
    Sep 5 at 18:51

On OP's comment:

I just don't see how the carbon can become part of the chain!

shows confusion on the mechanism. However, the comments by user55119 and Ron correctly suggested the path of the mechanism. I thought it would be helpful to put the matter as a solution. Following scheme shows the possible mechanism to achieve the sought product:

Radical opening of cyclopropyl ring

Note that, there are two bonds that can open the cyclopropyl ring. The bond the OP has shown opening gives the primary radical, which is the less stable. The bond opening shown in my scheme gives the benzyl radical, which is stabilized by resonance. Thus, the given path is the thermodynamically stable path.


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