HBr reacts with alkenes such as propene in the presence of Peroxide as per the Anti-Markovnikov rule. This is an addition reaction, forming 1-bromopropane as the major product.

This happens due to the formation of Br• (Bromine Free radical) formed due to the weak peroxide linkage. The free radical attacks the alkene to form a stable 2-degree carbon free radical, hence forming 1-bromopropane.

But at high temperatures, we can observe yet another reaction... allylic substitution. If we react Br2 with propane at very high temperatures, we get 2-bromopropene (an allylically substituted product) This is because when Br• takes away H• in the form of HBr, it leaves a resonance-stabilised allylic carbon free-radical.

My question is, why doesn't this happen in the case of HBr in the presence of peroxide? It follows the same mechanism of free radicals, why doesn't it have an allylic substituted product?

(I have a few ideas... I guess allylic substitution happens at high temperatures because it forms less stable product... and is hence endothermic. At room temperature however, endothermic reactions like that would not be favoured... that does make sense to me... just need some more clarification)


1 Answer 1


I just realised...

(I would find it helpful if someone confirms that I am on the right track)

HBr while doing allylic substitution would form propene itself 😂

Br radical would take away allylic Hydrogen as HBr, leaving behind the allylic carbon free-radical. This would in turn take up Hydrogen from HBr (since H radical is less stable... so that is to be reacted first) forming propene again, which yet another Br radical...

So for any change to be observed, either both radicals would have to be Br (Br2 for instance) or you would simply do an addition. (or I reckon there is a very minor product of allylic substitution during reaction with HBr if somehow Br radical reacts first.)


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