Question

Explain why: Aniline with $Br_{2}/ H_{2}O$ gives 2,4,6, - tribromo aniline, while $PhNMe_{2}$ gives meta nitro derivative when mono nitrated with $HNO_{3}/H_{2}SO_{4}$?

Answer 1

Hint: We will write about the formation of 2,4,6- tribromo aniline and meta nitro derivatives when $PhNMe_{2}$ is treated with strong acids. Aniline is an organic compound with the formula $C_6H_5NH_2$. Consisting of a phenyl group attached to an amino group, aniline is the simplest aromatic amine.

Complete step by step answer:

As is a strongly activating group and ortho para directing in nature, the lone pair on the nitrogen gets attached to the delocalized ring electrons and becomes delocalized with them. If the electron density around the central ring has an extra electron from the group, it becomes more and more attractive towards the coming electrophiles. Hence has a greater number of activating positions around the ring than others. So, if bromine water is added to phenylamine and the bromine water gets decolorized and a white precipitate is formed. The precipitate formed is 2,4,6- tribromo aniline.
In the presence of the strong acids like the group is highly protonated and the protonated form dominates the electron-withdrawing group via inductive effect. This reduces the attract at ortho position and hence meta directing product is formed.

Additional information:

In acidic condition, the lone pair of the nitrogen atom takes a proton of the acid and the protonated amine will then be considered as a meta directing group. So, it will act as a deactivator that will draw the electron density away from the ring.

Note:
Halogenation is the fastest electrophilic aromatic substitution reaction and the amine is the most activating group. Hence, the overall reaction is uncontrollable and all activated positions are used up by giving tribromo product.
In acidic medium, aniline converts into anilinium ion having a group which is a strongly deactivating group and as a result meta directing product is formed.