[Mike Dacre]

Hi All,

I am confused about radical halogenation of alkenes and arenes with an allylic position. If I have an alkene like this:



And I treat it with NBS, my textbook tells me that the bromine group will substitute a hydrogen at the allylic position, giving me this:



However, the same text book tells me that if I add bromine with a peroxide to a alkene like this one:



I will end up with this:



I understand both of these mechanisms, but I am confused by the competition between them in more complex molecules.

For example, in an alkene like this:



There are two potential allylic positions, neither are terminal. If I add molecular bromine without a peroxide, I will get an addition of two bromines across the double bond and an elimination of the double bond. However, from my textbook it seems that if I add HBr with peroxides I will get this:



But if I do the same reaction with NBS, I will get this:



And if I do the same reaction again with molecular bromine in the presence of heat and light I will get this:



I am very confused by this. To me the three mechanisms are similar, with the one major difference that HBr has a hydrogen that can help reduce the pi bond, but a very similar question arises with molecular bromine that has been homolytically cleaved by heat and light. Is there some inherent difference between the bromine radicals from HBr, Br₂, and NBS? Is it possible to predict when a radical addition will happen across the pi bond, and when a substitution will occur at the allylic position?

Thanks!

[Mike Dacre]

I got the answer from March's Advanced Organic Chemistry following a hint on physicsforums.com, I wrote a full summary of the answer here: https://www.physicsforums.com/threads/nbs-vs-radical-hbr-in-the-bromination-of-olefins.841060/#post-5277213

The short answer is that N-Bromosuccinimide generates a continuous low concentration of Br₂ and free radicals, while scavenging all HBr. Because of this, addition to the pi bond is unable to compete with radical substitution at the allylic position. If the concentration of HBr or Br₂ were higher, addition at the pi bond would outcompete allylic addition, but because of the very low concentrations, they cannot, and allylic substitution 'wins'. There is some extra subtlety also, but I have covered that in my response at the forum linked above, I have also included supporting quotes and citations.

[orgopete]

Let me give a slightly shorter version than posted by Mike. If bromine radicals are generated (heat, light, or initiator like H2O2), they can add to an alkene in a kinetically fast reaction. If the new radical is trapped, the product will be an addition product.

That could be the simple answer. If HBr/H2O2 is used, the bromine radicals add, HBr can quench the carbon radical and regenerate a bromine radical in a propagation step. If NBS is used, NBS also traps HBr and competes with the addition reaction. In the absence of a radical trap, the bromine radical may form a more stable allylic radical plus HBr. The HBr can be trapped by NBS to give Br2 (the source of bromine radicals). The Br2 can react with the allylic radical to give an allylic bromide and a bromine radical (propagation step).

This model suggests that using Br2 in an allylic bromination may give both because the concentration of HBr increases as the reaction progresses. At that point, the HBr reaction products should become begin to form. I don't know the results of any bromine reactions with an alkene.