Explain the bromination of aniline by an electrophilic substitution reaction.

Aniline reacts with bromine water at room temperature to give a white precipitate of $2,4,6$-tribromoaniline.
Reaction: $C_6H_5NH_2 + 3Br_2(aq) \rightarrow C_6H_2(NH_2)Br_3(s) + 3HBr$
If we need to prepare a monosubstituted aniline derivative,this can be done by protecting the $-NH_2$ group by acetylation with acetic anhydride,then carrying out the desired substitution followed by hydrolysis of the substituted amide to the substituted amine.
The lone pair of electrons on the nitrogen of acetanilide interacts with the oxygen atom due to resonance as shown below:
$CH_3-CO-NH-C_6H_5 \leftrightarrow CH_3-C(O^-)=N^+-H-C_6H_5$
Hence,the lone pair of electrons on nitrogen is less available for donation to the benzene ring by resonance. Therefore,the activating effect of the $-NHCOCH_3$ group is less than that of the amino group,which prevents polysubstitution and allows for the formation of $p$-bromoacetanilide,which upon hydrolysis yields $4$-bromoaniline.