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As I know, when $\ce{Br2/FeBr3}$ is added to acetophenone ($\ce{C6H5-CO-CH3}$), Br will substitute a H from the meta position which leads to 3-bromoacetophenone. So why doesn't a Br substitute a H from the right side benzene (connected to the carbonyl group)?
Because that is not the most electron rich aromatic. The ring at other end of the molecule is, because it is an aniline. The first, and probably the second, bromination will go there. If you keep throwing in Bromine you will eventually get bromination of the Acetophenone ring.
Aniline is one of the strongest activating groups . It makes the ring more stable by +I and +R effects which are predominant on the Ortho and Para positions. Bromine is slightly deactivating and Ortho-Para directing . So bromine on the right benzene ring will slightly deactivate the benzene ring . Moreover, the electronegative oxygen of CO might exert a deactivating effect on the ring as well. Hence, taking the highly activating nature of aniline into account and other factors I mentioned , bromine will attach at the Para position on the left benzene ( it's Para not Meta according to the picture).
