Regarding allylic free radicals and anti-Markovnikov product

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:

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: because here an allylic free radical is formed.

So which is correct?

• 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. – iad22agp Nov 28 '16 at 13:59
• @iad22agp Thank you very much for clarifying. – cumulonimbus Nov 29 '16 at 13:08
• 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? – orthocresol Dec 2 '16 at 18:01
• @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! – cumulonimbus Dec 4 '16 at 6:40