What are the products of the following reactions????

Well, you haven't specified the reaction conditions. What is the solvent? Is there light catalysis?

#A)#

The first reactants will not react as-written... I assume room temperature.

At best you might temporarily form #"C"_6"H"_11"OBr"#, but that isn't particularly stable...

#B)#

Nothing will happen as-written... The only thing I can assume is that light is photo-inducing a radical reaction.

The most likely product is then a branched bromide, since more-substituted radical intermediates are stabilized by hyperconjugation ("delocalization" of electron density into partially-filled or empty central #pi# orbitals):

i.e. #3^@ > 2^@ > 1^@ > "methyl"# radical intermediate in terms of thermodynamic stability.

See the mechanism in detail here.

[And in fact, radical bromination shows markedly more drastic favoring of the more-substituted product compared to radical chlorination.]

Only the branched product here is chiral. You can draw the #"Br"# with a wedge to indicate it is the #bbR# stereoisomer (verify this using priority designations), or a dash to indicate it is the #bbS# stereoisomer.

#C)#

Same idea here, except at this point it should be more obvious; there is a branch on the alkane, so that should immediately tell you to look for which carbon is most substituted. That is carbon-2.

You do form quite a few more products though. None of the terminal bromides are chiral, and only the middle-left product is chiral (the middle-right product has two identical methyl groups around the only potentially chiral carbon).

For the middle-left product, the bromine can have a wedge to indicate #bbR# stereoisomerism (verify this using priority designations) or a dash to indicate #S# stereoisomerism.