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Is H2O2 present? If so, HBr and H2O2 will form Br2. That then reacts to add Br across the double bond, forming 1,2-dibromo-1,2-diphenylethane.
If H2O2 is not used, the HBr is added across the double bond. The result of this is a single C-C bond, with a new H-atom bonded to one side, and a new Br-atom bonded to the other side. Since the molecule is perfectly symmetrical, it doesn't matter which side the H and the Br are added to (although if it wasn't symmetrical, there are rules to determine which atoms goes on which side). I'm not sure about the name but I think it'd be something like this: 1-bromo-1,2-diphenylethane. Note that the carbon bonded to the Br is a stereocenter (it is bonded to a phenyl, H, Br, and C, so it has 4 unique substituents), but the product will be a racemic mixture of the two stereoisomers (I'm pretty sure anyway).

Take a look at the Web Links to the left, but note that most of these pages discuss the addition of Br2 rather than HBr (although they use HBr to prepare the Br2 with H2O2).

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