Question

Give reasons of the following:
Although $-N{{H}_{2}}$ is an o/p directing group, yet aniline on nitration gives a significant amount of m-nitroaniline.

Answer 1

Hint: Nitration of aromatic compounds is generally performed using a nitrating mixture, which consists of nitric and sulphuric acid at low temperature. The $-N{{H}_{2}}$ group present in aniline is a strong activating group and is ortho/para directing.

Complete step by step solution:
First of all, let us know about the nature of the aniline in terms of its chemical reaction.
Aniline is an aromatic amine with the chemical formula, ${{C}_{6}}{{H}_{5}}N{{H}_{2}}$. So, the amino group i.e. $-N{{H}_{2}}$ present in aniline is a strong activating group and is ortho and para directing due to its strong $+R$ effect. Nitric acid is a strong oxidizing agent. As a result, when the nitration of aniline is carried out, it not only gives nitration products but also some oxidation products.
Even though the amino group is ortho and para directing, there is a significant production of m-nitroaniline. Because here, very strong acidic conditions are being used due to which some of the aniline molecules get protonated to anilinium ion. The $-NH_{3}^{+}$ group is electron withdrawing and is m-directing due to which we get a significant amount of the meta product and the major product will be a para directing one.

Note: The ortho-nitroaniline is always the minor product due to presence of steric hindrance at the ortho position offered by the group due to which the nucleophilic substitution becomes difficult at the ortho position.