Anti-Markovnikov rule will be used in C,D. Bromination will take place at the last carbon, since the last carbon will be achiral, you will get 2 enantiomers They'll have only one chirality centre with R/S configuration. In option A we will get a new center since markovnikov rule will be used but it will again form enantiomers but it has only one chiral center(other one is not! 2 methyl groups are there on same carbon). In option B same type of addition will happens and you will get 2 different products (with 2 chiral centers) which I'll draw and send.