Hydrohalogenation of Butene with Hydrogen Bromide
 

The answer to the following reaction is given as choice E. I thought it should be D because, if the two carbon atoms at the double bond are linked to a different number of hydrogen atoms, the halogen is found preferentially at the carbon with less hydrogen substituents (Markovnikov's rule). Would this not be the second carbon of the butene? Please advise.

It'll be the second carbon of the butene because the major product (2-bromobutane) is always formed through the most stable carbocation.

I too think you are correct. I too reviewed the Markovnikov's rule. The structure should be CH3---CHBr---CH2---CH3

the addition has not an ionic mechanism
 

There is a little detail (UV light) than means we cannot use the Markovnikov rule. The rule is OK for ionic addition, but here we have a radicalic addition. So, the most probably shall be 1-Bromoderivative (the intermediate is a free radical)

I don't think it'll be different, because HBr is in the liquid form and will be dissosiated into H+ and Br-. So, Markovnikov will run the reaction.. :lol:

Markovnikov's Rule will not apply!!
The use of HBr and UV light will not give H+ and Br-.
The UV light photochemically intitiates a radical reaction. Through this reaction 1 electron from bromine forms a bond with 1 electron from the double bond with the Br going to the less substituted side of the double bond and leaving 1 free electron radical on the more substituted carbon because a secondary radical is more stable than a primary radical. This secondary radical then reacts with the hydrogen, giving the anti-Markovnikov addition of HBr.

Hope this helps,

bw800402

So, what exactly about this addition identifies it as a free radical addition. Is it only the UV light? Nothing about regioselectivity, is this structure cis or trans? If it is just the UV light, then what about the UV light causes free radical addition? I have searched the internet high and low, no luck on any solid info about UV addition.

Yes, it's the UV light. I guess the photons of UV light are just the right energy to break chemical bonds, making radicals. It's a method for making haloalkanes from alkanes; the UV light makes radicals, allowing the reaction.