[xstetsonx]

why would radical bromination of 2 methylbutane gives 1-bromo-3-methylbutane, 2-bromo-3-methylbutane, 3-bromo-3-methylbutane, and 1-bromo-2-methylbutane? i don't see any methyl shife in this reaction. Some one plz help

[nameless]

free radicals are very reactive so u have to look at the possible places that it can attack during the propagation step of the free radical reaction. It could potentially pull off a hydrogen atom off of each carbon resulting in the products that you mention. There isnt a methyl shift. i hope this somewhat clear.

[orgopete]

I like this question a lot. I hadn't thought of it before. I just assumed the reaction occurred to give the most stable radical upon abstraction. Does any know of any isotope effects or labeling studies that prove an electron + proton is not removed from one of the primary carbons and there is an electron + proton shift?

[xstetsonx]

okay i am totally confuse. so there is a H shift? i though shift can only happen when you have a cabocation.... this one is a radical.

[xstetsonx]

i get 1 bromo 2 methly butane....how come the "answer" i found it is 1 bromo 3 methly butane?

[stewie griffin]

I agree with nameless. Usually the idea is that you form the most stable radical (as orgopete said) which would be the tertiary radical which would give you the 3-bromo-3-methylbutane. That's usually the point of these types of questions. However, if the question is to list all possible products, then you have to go nameless' route and look at each C-H bond that can be broken to give a unique product. Some of the C-H bonds will lead to the same product (the three C-H's of a methyl group on the end of the chain for example). If you work through each C-H cleavage, you find out that you can have four unique constitutional isomers.
BTW, of those four possible constitutional isomers, the 1-bromo-2-methylbutane and the 1-bromo-3-methylbutane are products. So I don't think that you're wrong... you just haven't listed all of the products.