Radical Question

[potbelly]

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I found that Br2 w/ hv and NBS w/ light perform the same way.
They both take out allylic H to form "a stable radical formation." Correct?
(they keep double bond and form bromo alkene compound)

However, HBR w/ hv and HBR w/ HOOR also involve the radical formation.
Anti-M applies here and double bond breaks to form bromo alkane.


Then, what is the difference b/w these two types of rxns?
The way I understood these rxns is right or wrong?
Is top rxn considered as Anti-M as well?


Thx in advance.

 

[Dotoday]

OK, the top reaction where you have Br and Hv or NBS in peroxide is a reaction that is used to Halogenate Alkanes and/or halogenate alkenes by removing the allylic hydrogen. They are not considered an anti-Markovnikov reactions. Now HBr with peroxide is the Anti-M (Anti Markovnikov reaction). If you look at the mechanism of all of these reactions you will see that Br in light goes through 3 stages Initiation, Propagation, Termination. First it makes a Halogen radical, which then reacts with the Alkane to make the most stable radical.
HBr in Peroxide in used primarily to put Br on the double bonded Carbon that has the least number of Hydrogens. Only Br can be placed anti-M, if you use HCl or HI with ROOR they will still go Markovnikov.
Hope this helps, Good luck.