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NBS vs radical HBr in the bromination of olefins

  1. Nov 2, 2015 #1
    I am having some trouble understanding why different sources of bromine radicals supposedly brominate an alkene at different positions. What I mean by this:
    m1mFwbV.png
    In the first example, the Bromine radical attacks a hydrogen at the allylic position and then a termination reaction results in a Bromine at that position. In the second reaction, the Bromine radical from a split HBr molecule attacks the π bond and adds in an anti-Markovnikov way, because that allows the radical intermediate to be at the more stable tertiary carbon. The final reaction is the non-radical addition of HBr.

    My question is about the first two reactions here. What favors one over the other. In my mind, both HBr + peroxides and NBS (N-Bromosuccinimide) are a source of Bromine radicals, and the reaction should proceed in much the same way. I know that in the HBr reaction, there are hydrogen radicals floating around also, but I don't think that that is the answer. The mechanisms are different, but I cannot pinpoint what would make one product favored over the others.

    I realize that radical reactions are extremely messy, with lots of side products, but the way my textbook (Brown, Foote, Iverson, Anslyn) teaches it it seems like the NBS reaction primarily targets the allylic position and the HBr + peroxides reaction primarily adds to the π bond. Any ideas on what about the mechanism leads to this result would be hugely helpful, I am struggling to find anything useful online (too many competing search results for my search terms).

    Thanks!
     
  2. jcsd
  3. Nov 3, 2015 #2

    DrDu

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  4. Nov 3, 2015 #3
    Thank you, I found it after some searching.

    The answer is that the above mechanisms do compete with each other in all cases, however, when NBS is used, allylic hydrogen substitution wins.

    In the halogenation of a pi bond, either by ionic or radical mechanisms, the first step of the reaction is the reversible addition of the halogen, in this case bromine. In order for the addition to be permanent, the resultant cation or radical needs an anion or radical to complete the reaction. If the concentration of Br2, HBr, radicals, and nucleophiles is sufficiently low, the reaction will proceed very slowly.

    N-bromosuccinimide provides a low concentration source of radicals and Br2, and scavenges any created HBr:
    aLJjIJ4.png
    r3dvywT.png
    2A9FcEE.png
    The result is a low concentration of Br2 and radical sites on the carbon species being brominated, resulting in the following reaction:

    1fjR5G1.png

    As the radical is most stable in an allylic position, and additions of bromine to the pi bond are reversible and short lived, the radical bromination at the allylic position outcompetes the other reactions, resulting in almost 100% allylic position hydrogen substitution. If the concentrations of bromine or hydrogen bromide were higher, this would not happen.

    From March:
    1McGrath, B. P. & Tedder, J. M. MECHANISM OF ALLYLIC BROMINATION BY N-BROMOSUCCINIMIDE. PROCEEDINGS OF THE CHEMICAL SOCIETY OF LONDON p.199 (1961).
     
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