Animated Solution for Chemistry - Organic Chemistry: In the following reaction the reason why meta-nitro product also formed is

Imagine you are a chemist trying to add a nitro group to aniline. You know the fundamental rules of electrophilic aromatic substitution: the amino group ($-{\text{NH}}_2$) is a powerful electron donor due to the lone pair on its nitrogen atom. Therefore, it should strongly activate the benzene ring and direct the incoming electrophile exclusively to the ortho and para positions.

You run the reaction, isolate your products, and boom! You discover that you have obtained a massive $47\%$ yield of the meta-substituted product. What went wrong? Did the rules of chemistry suddenly change?

Well, nothing went wrong, but there is a hidden reaction happening right under our noses that completely alters the playing field.

The secret to this anomaly lies entirely in the reagents we use. To nitrate a benzene ring, we typically employ a "nitrating mixture," which is a potent combination of concentrated nitric acid (${\text{HNO}}_3$) and concentrated sulfuric acid (${\text{H}}_2{\text{SO}}_4$).

The most critical characteristic of this mixture is that it is highly acidic.

Now, let's look back at our starting material, aniline. The nitrogen atom of the $-{\text{NH}}_2$ group possesses a lone pair of electrons, making aniline a very good Lewis base. When you drop a good base into a strongly acidic medium, an acid-base reaction is inevitable.

Before any electrophilic nitration can even begin to occur, the basic aniline rapidly accepts a proton (${\text{H}}^{+}$) from the acidic medium. This protonation transforms the neutral aniline molecule into the positively charged anilinium ion.

This transformation is where the magic happens. Let's analyze the newly formed anilinium ion. The nitrogen atom now carries a formal positive charge ($-{\text{NH}}_3^{\oplus }$).

Because nitrogen is electronegative and now positively charged, it becomes incredibly electron-hungry. The $-{\text{NH}}_3^{\oplus }$ group exerts a very strong $-I$ (inductive electron-withdrawing) effect. It aggressively pulls electron density away from the benzene ring through the sigma bonds.

This massive withdrawal of electron density does two things: 1. It strongly deactivates the benzene ring towards electrophilic attack. 2. It completely flips the directing nature of the substituent, turning it into a meta-directing group.

So, why do we get $47\%$ meta-nitroaniline?

In the strongly acidic nitrating mixture, a very large equilibrium fraction of the starting aniline exists not as the free amine, but as the protonated anilinium ion. While the small amount of unprotonated aniline continues to yield the expected para ($51\%$) and ortho ($2\%$) products, the vast sea of anilinium ions forces the incoming nitronium electrophiles (${\text{NO}}_2^{\oplus }$) to attack the meta position.

This beautiful interplay between acid-base chemistry and electrophilic aromatic substitution perfectly explains the surprisingly high yield of the meta-nitro product.