[Ranowa]

Reduced form of my problem: "When a Br atom is adjacent to the site of the abstraction by a free radical, retention of stereochemistry is observed... propose a mechanism..." Example given is a butane with Br at one end and a methyl at the second Carbon; after halogenation the methyl is replaced with Br, and stereochemistry at the chiral center is retained.

I get why usually you have a racemic mixture as products, and the prof has given us a clear hint that the intermediate formed is the reason why it differs here. (Duh.) And, I'm completely stuck, and the internet/book/lecture notes have been of no help. Any hints?

[orgopete]

This looks simple enough. Let me change the example slightly so the result will match the premise. (I think the major product of one isomer of 1-bromo-2-methylbutane will be racemic.) Draw the products of bromination of (2R)-2-bromo-3-methylbutane.

[Ranowa]

My professor has said that when there's a Br atom adjacent to the site of H abstraction, though, the product will not be a racemic mixture but will retain stereochemistry. I found an article that's helping me craft my answer a lot; not sure how accurate it is since it's 2001, but here's the link if you/anyone else is interested

http://pubs.acs.org/doi/pdf/10.1021/jp0117056

[orgopete]

I understood that, however in the example given, bromination will destroy the chirality center, so in this case will give a racemic product, 1,2-dibromo-2-methylbutane. I had presumed this fact is what had led to your confusion. If the chirality center is not the reactive center, what would you predict the result to be? In the example I gave bromination would occur adjacent to the bromine. What would you get in my example?