Reaction of 1,3-butadiene with HBr in the presence of peroxides

Question

I have learnt that the reaction of 1,3-butadiene with HBr depends on the temperature as the temperature determines whether the reaction is under kinetic or thermodynamic control. I believe that this is due to the hybrid nature of the carbocation and relative stabilities of the products. Since the allylic radical is also resonance-stabilized, will the same effects take place in the presence of peroxides?

Answer 1

The reaction of 1,3-butadiene (1) with HBr/peroxide (ascaridole, 5) in the absence of solvent was investigated by Kharasch^[1] in 1936. With a reaction time of 2 hours at -78^o, the kinetic bromide 3 dominated over the thermodynamic bromide 2^[2] (44/56). Under the same conditions at -12^o, the order was reversed (78/22). The kinetic mixture (-78^o) was unchanged when "warmed" to -12^o in the presence of the peroxide 5. However, the thermodynamic mixture was obtained in the presence of HBr and peroxide 5 upon warming. Radical 6 is the likely intermediate in the equilibration.

1. M. S. Kharasch, E. T. Margolis, and F. R. Mayo, J. Org. Chem., 1936, 1, 393-404.
   
2. No assignment of the stereochemistry of 2 was made although it is like of the (E)-configuration given the equilibration through radical 6.

Answer 2

The thing about HBr + Peroxide is that it is a free radical addition reaction. Not Electrophilic as

H-Br + O'H gives Br'

(HO-OH gives 2O'H)

Consider (') as a free radical electron. So technically The product formed would be 1-Bromo 3-Butene

Keep in mind Free radical addition doesn't undergo rearrangement. Since 2nd position would have a free radical, the pi bond won't rearrange.(Resonance won't happen)

With HBr pi bond will rearrange forming 1-Bromo 2-Butene