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For the proposed mechanism, I would like to think the $\ce{O-O}$ bond would be cleaved first, thus creating two radicals (an $\ce{OH}$ radical and the radical situated on the other Oxygen), then radical on the oxygen will go into the ring to form a double bond in which the alkene double bond will donate 1 electron towards that bond and 1 electron to the carbon on the end of the opened ring. My mechanism would be incomplete as I cannot seem to figure out how this reaction occurs. However, I have another suggestion in which there is an epoxide formation. Can anyone elaborate as to which one is correct or can give me some insight?

  • $\begingroup$ Looks like catalytic O-O cleavage, so rather O+ then radical is created. $\endgroup$
    – Mithoron
    Feb 15, 2016 at 22:40
  • $\begingroup$ After the formation of the oxygen radical, it iniciates a C-C bond cleavage, as two possibilities exist, two products are obtained after subsequent addition of the formed C radical to the double bond. $\endgroup$
    – EJC
    Feb 15, 2016 at 22:48

1 Answer 1


At least for the second part of the product formation, you have pretty much given the answer yourself.

The methylcyclopentane moiety of the products results from two different 5-exo-trig radical ring closures, that is from the intramolecular addition of two different radicals to $\ce{R-\color{\red}{C}H=CH2}$.

How are these two different radicals generated?

Have a look at the starting materials! Isn't the combination of the hydroperoxide $\ce{R-OOH}$ and $\ce{FeSO4}$ very similar to the conditions of a Fenton reaction?

Let's therefore postulate the following: $\ce{R-OOH + Fe^2+ -> R-O* +\quad{}OH- + Fe^3+}$

The resulting 1-methylcylohexan-1-ol derived radical can form methyl ketones by cleaving two different $\ce{C-C}$ bonds; the resulting carbon-centered radicals then give rise to the different cyclization products.


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