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).
The rest of this page is available to MOC Members only.
To get access to this page, plus over 2500 quizzes, the Reaction Encyclopedia, Org 1 / Org 2 summary sheets, and flashcards, sign up here for only 30 cents/ day!
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.