Part 8: MECHANISMS

Note that no other reagents are needed in order to complete any of these sequences, you should only be using what is there.

A
The reaction is a radical addition of HBr to an alkene which is a radical chain process.  Since it is a radical process, there are no carbocations formed nor are there any bromide ions formed. Review formal charges ? The regioselectivity is accounted for by the addition of Br first and the preferential formation of a more stable radical as a result of that. 


Notes : 
1. Peroxide is the radical initiator here - homolytic cleavage of the R-O-O-R bond (the weakest bond in the system) gives a peroxy radical that then reacts with the HBr.
2. The Br radical adds to the alkene first giving the radical that is more highly alkyl substituted because that is the more stable radical. The accounts for the observed anti-Markownikov regioselectivity.
3.  This alkyl radical then reacts with another molecule of HBr to give the product. This keeps the process going as it regenerates another Br radical.
4.  Why doesn't it react with the H radical ? The concentration of the H radical is very low and the likelihood of the alkyl radical and H radical colliding is very very low.
5.  The alkene pi bond does not dissociate homolytically to give a diradical (that would be a very unstable species)

B
Application of another mechanism....this is very similar to alkene halogenation via a bridged halonium ion then attack of a nucleophile from the opposite side to give the trans-stereochemistry.

C
Ring opening of an epoxide with a strong nucleophile is SN2 like (i.e. in terms of regiochemistry it attacks the least hindered end). The product is (S)-2-butanol.

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