# Nitration of aniline

(b) Nitration: Direct nitration of aniline yields tarry oxidation products in addition to the nitro derivatives. Moreover, in the strongly acidic medium, aniline is protonated to form the anilinium ion which is meta directing. That is why besides the ortho and para derivatives, significant amount of meta derivative is also formed.

Why there is more para-substituted nitroaniline formed than ortho-substituted one even though it can form hydrogen bonding?

In the case of nitration of aniline intermediate stability determines products.The following figure depicts possible electrophillic attack on aniline. Structures 4 and 12 are additional stable resonance structures obtained by attack of nitronium electrophile at ortho and para positions.

Structure 8 is relatively unstable due to negative effect of Nitrogen.

In nitration of aniline in strong acid (HNO3, H2SO4) aniline changes to anillium ion .Anillium withdraws electron density.Its effect is felt maximum at ortho followed by meta and then para position. Consequently very little of ortho nitrated product is formed.

Since electron withdrawing inductive effects are minimum at para position ,maximun nitration occurs here.

Summarizing ,

Even though there exists a possibility of hydrogen bonding in ortho nitro aniline ,strong deactivating influence of $$\ce{NH3^+}$$ on benzene ensures least electrondensity at ortho position in aniline leading to only 2% of ortho nitro aniline.

Such electronwithdrawing tendencies of $$\ce{NH3^+}$$ are minimum at para position in aniline therefore 51% of para nitro aniline is formed.

In case of nitration in aniline, the Nitric Acid protonates the aniline to form the Anilinium ion.

Now since the nitrogen atom has no lone pair to conjugate, it has no mesomeric effect on the ring, the but as nitrogen is now protonated, it has a high negative inductive effect.

Which simply leads to the fact that the position farthest from the nitrogen substituted carbon will have the highest electron density (as inductive effect decreases as we move away from the atom causing it).

So in this case the best site for attack for the electrophile was para>meta>ortho.

The Hydrogen bonds would stabilise then ortho product once its formed, but the Rate Determining step of an electrophilic substitution on benzene is the sigma complex intermediate wherein the most stable intermediate is formed as major product. The positive charge in sigma complex stays the farthest in para position from the anilinium ion hence(most stable) is the major product.

• The Diagram is wrong. $\ce{NO2^+}$ is an electrophile, not a nuleophile. It's the $\pi$-electron cloud of Benzene ring which attacks, not $\ce{NO2^+}$. So, the arrow should be on the opposite direction. Please change the direction of the arrow or upload any different picture. – Soumik Das May 25 at 9:54

One other aspect is that the oxygen in the nitro group is not a particularly good electron donor for the hydrogen bond. True, it has a negative formal charge; but this is shared between two atoms and the polarity of the bonds in the nitro group is relatively low (compared with, for example, a carbonyl or hydroxyl oxygen). It is possible to eliminate meta attack by modifying the substrate, but this still gives a predominance of para over ortho.