Animated Solution for Chemistry - Organic Compounds Containing Nitrogen: Consider the given reaction, percentage yield of

When we first look at the nitration of aniline, our chemical intuition immediately points us toward the nature of the amino group.

The $-{\text{NH}}_2$ group is known to be a strongly activating group. Because of the lone pair of electrons on the nitrogen atom, it donates electron density into the benzene ring through resonance.

This resonance effect increases the electron density specifically at the ortho and para positions. Therefore, we naturally expect the incoming electrophile, the nitronium ion (${\text{NO}}_2^{+}$), to attack these electron-rich sites, yielding mainly ortho and para products.

However, chemistry is rarely that straightforward, and this is where many students fall into a classic trap. We must pay close attention to the reaction conditions.

The nitration is carried out using a nitrating mixture, which is a combination of concentrated nitric acid (${\text{HNO}}_3$) and concentrated sulfuric acid (${\text{H}}_2{\text{SO}}_4$). This makes the reaction medium extremely acidic.

Aniline is a base. In the presence of such a strong acidic medium, an acid-base reaction occurs much faster than the electrophilic aromatic substitution.

The basic $-{\text{NH}}_2$ group readily accepts a proton (${\text{H}}^{+}$) from the acidic medium to form the anilinium ion ($-{\text{NH}}_3^{+}$).

This protonation completely changes the game. The nitrogen atom now carries a positive charge and no longer has a lone pair to donate to the ring. Instead, it exerts a strong electron-withdrawing inductive effect ($-I$).

Because of this, the anilinium ion is strongly deactivating and, crucially, meta-directing. The electron density at the ortho and para positions is depleted, making the meta position the relatively most reactive site for the electrophile.

Because the reaction mixture contains an equilibrium between the highly reactive unprotonated aniline and the less reactive protonated anilinium ion, we get a mixture of products.

The unprotonated aniline reacts very rapidly to give the para product, which is sterically less hindered than the ortho product. The anilinium ion reacts to give the meta product.

Experimentally, the percentage yields are: 51% para-nitroaniline (A), 47% meta-nitroaniline (B), and a mere 2% ortho-nitroaniline (C).

Therefore, the correct order of percentage yield is A > B > C.

This brings up an interesting question: what if we want to synthesize only para-nitroaniline without the meta byproduct?

To achieve this, we must prevent the protonation of the amino group. We do this by first reacting aniline with acetic anhydride to form acetanilide.

The acetyl group pulls the nitrogen's lone pair away from the ring slightly, making it less basic so it doesn't get protonated, while still remaining ortho/para directing. After nitration, we simply hydrolyze the amide back to the amine.