I was searching a question "What is the primary product when cyclohexene reacts with bromine in the presence of UV light?" and in a website they said:

3-Bromocyclohexene will be the product because when an alkene reacts with $\ce{Br2}$ in presence of UV light, free radical substitution reaction takes place. In this case, a stable cyclohexene allylic free radical will form which will further react with bromine free radical to form 3-bromocyclohexene.

A reaction that I already studied was propene + $\ce{HBr}$ in presence of $\ce{H2O2}$ and the answer is:enter image description here

The reason is that carbon radical is present on secondary carbon where the neighboring atoms can donate electron density to stabilize the radical.

But based on the first reaction I speculate that the answer should be:enter image description here because here an allylic free radical is formed.

So which is correct?

  • 1
    $\begingroup$ Both allylic substitution and anti-Markovnikov addition are possible reactions under free radical conditions. What exact conditions favor which reaction is something I expect most chemists would have to look up. Your web source may be correct (although the specific conditions are not mentioned), but I would recommend established references or journal articles as more reliable sources of information. $\endgroup$ – iad22agp Nov 28 '16 at 13:59
  • $\begingroup$ @iad22agp Thank you very much for clarifying. $\endgroup$ – cumulonimbus Nov 29 '16 at 13:08
  • $\begingroup$ Do you notice that in your proposed allylic bromination with HBr, you are consuming $\ce{Br^.}$ and generating $\ce{H^.}$ radicals that don't go anywhere? $\endgroup$ – orthocresol Dec 2 '16 at 18:01
  • $\begingroup$ @orthocresol Homolytic bond cleavage of peroxide gives 2OH free radical.One of the OH free radical react with HBr and forms H2O and Br free radical so H free radical is not generated right! $\endgroup$ – cumulonimbus Dec 4 '16 at 6:40

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