Question

Complete the following reactions:
 $ {{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{H}}_{\text{2}}} + {\text{B}}{{\text{r}}_2}\left( {{\text{aq}}} \right) \to $

Answer 1

Hint: Aniline is an organic aromatic amine. It has an amine group attached to a benzene ring. Nitrogen carries lone pairs of electrons and thus makes the benzene ring more susceptible to electrophilic attack. In bromine water, aniline undergoes electrophilic substitution reaction and the hydrogen atoms are replaced by bromine atoms. To answer this question, you must recall the result of the resonance in the benzene ring and identify the most susceptible sites for attack.

Complete step by step solution
Aniline has an amine group attached to a phenyl ring and the molecular formula of aniline can be written as $ {{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{H}}_{\text{2}}} $ . The amine group is an activating group towards aromatic electrophilic substitution due to the electron donating nature of the amine group. The lone pair on nitrogen is donated to the phenyl ring and it is then in conjugation with the $ \pi $ bonds of the ring and delocalized all over the ring. In this delocalization, the electron density is highest at the ortho and para positions to the amine group. Thus, the ortho and para positions of aniline are highly susceptible to electrophilic substitution.
On reaction with bromine water, aniline undergoes electrophilic substitution to produce a white colored precipitate of 2, 4, 6- Tribromo aniline. Thus, the given reaction can be completed as:

Note
An important point about aromatic electrophilic substitution is that the aromaticity of the compound is not lost after the reaction. The configuration of the compound however, may be altered. Nitrogen atom is more electronegative than carbon and thus withdraws electron density inductively but due to the lone pair, it acts as an electron donating group metamerically.