...And I'm not sure how you plan to simply kick out the bromine.
What bromine are you referring to? In the reaction bromine is not lost, but gained.
...Carbocations next to phenyl rings are relatively stable ...
I'm aware that iodide is a great leaving group, but I didn't know it would just up and leave if you dropped benzyl iodide in some wet alcohol. I suppose if I knew that benzyl cations were particularly stable, then that would have been a giveaway, now I know
Maybe I am supposed to consider how the intermediate was formed, could somebody guess how I would have formed that carbocation?
I would have thought that the bromide would attack the C+ right way, I just can't see why it wouldn't.
As for my experience, I'd say it is extremely limited. I have a BS chem degree that I may as well have picked up off the corner for $5. Also, it has been 3 years since I was in school, and 6 years since I finished Ochem.
Oh, and I know the answer, that is not the problem. I now I just need to know why.
A seems like a familiar reaction (I'm afraid I must rely on intuition)
B seems like an obvious hydride transfer to stabilize the cation.
C looks exactly like any free radical halogenation, although I couldn't back up why it would attack that site in particular.
D looked perfectly reasonable as the second step of an electrophilic addition (hydrohalogenation), except that it would have started with benzene, which wouldn't happen. If it was the second step of Sn1, then sure bromine would attack it, except what would have left in the first step? I can see bromine leaving in this step in favor of forming an aromatic system, but I'm not sure if that is what I was supposed to see.
E I didn't think this would happen at all, in a neutral alcoholic solvent, but apparently benzylic carbocations are particularly stable.