When hex-1-ene is treated with bromotrichloromethane in the presence of a peroxide initiator (e.g. dibenzoyl peroxide), what is the regioselectivity of the addition?

I know that the radical addition of HBr to an alkene is considered anti-Markovnikov, but is this reaction considered a Markovnikov or anti-Markovnikov addition?

Reaction scheme


1 Answer 1


This looks like a typical Kharasch (chain) reaction. The reaction is initiated by the formation of peroxide radicals (by homolytic cleavage of the weak O–O bond). These peroxide radicals then go on to abstract Br• from the starting material:

Initiation of Kharasch reaction

Note that the alternative generation of Br• is unfeasible, as radicals tend to attack at less hindered sites: hence the phenyl radical attacks the bromine which only has one attached atom, and not the carbon which has four flanking halogens. The weaker C–Br bond is also preferentially broken over the other three C–Cl bonds, so there is no chance of generating a CCl2Br• radical.

The trichloromethyl radical then adds to C-1 of the alkene, resulting in the more stable secondary radical. This can now abstract a bromine atom from unreacted bromotrichloromethane to give the product and a new Cl3C• radical.

Propagation of Kharasch reaction

Unfortunately, with additions of X–Y across a C=C double bond (as opposed to additions of H–X), the designations Markovnikov and anti-Markovnikov lose their meaning. A Markovnikov addition of H–X would be one where the hydrogen adds to the less substituted carbon. However, here it is not clear whether to define "Markovnikov" as the regioselectivity where X adds to the less substituted carbon or where Y adds to the less substituted carbon.


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