Free Radical Addition of HBr To Alkenes
Description: Treatment of an alkene with HBr in the presence of catalytic amounts of a radical initiator (such as peroxides) in addition to heat or light leads to addition of HBr to the alkene in anti-Markovnikov fashion (note that the Br adds to the less substituted side of the alkene and the H adds to the more substituted side).
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Regarding quiz 1838: Why does C-6 of the monobrominated intermediate bind to C-2? If C-6 instead binds to C-1, a secondary instead of a primary radical would form.
Good question! 5-membered rings form significantly faster (10x) faster than 6 membered rings. Search Wikipedia for “Baldwin’s rules”.
Why doesn’t the Br radical abstract the allylic hydrogen as it does in the allylic bromination mechanism. In both mechanisms, you have a step in which a Br radical reacts with an alkene, but they do different things. Please explain the difference.
It probably happens to some extent, but the reaction is a cul-de-sac that doesn’t result in a brominated product.
The reason is that once you form that allylic radical, what can it do?
– there is no source of Br2 that would result in bromination
– it can react with H-Br (forming C-H and breaking H-Br) to regenerate the allyl group and bromine radical
– reacting with H-Br to form C-Br and the H radical is very unlikely as the H radical is more unstable than the allylic radical.
This is why you need a source of Br2 to do an allylic bromination
Minor note, light is traditionally represented by h*nu not h*gamma. Nu is the frequency of the photon, and h*nu is then the energy of the light being supplied. https://en.wikipedia.org/wiki/Planck%E2%80%93Einstein_relation
thanks – embarrassing mistake.
fixed.