Consider the given reaction, percentage yield of :

In electrophilic aromatic substitution, the $$\mathrm{-NH_2}$$ group of aniline is strongly activating and ortho/para directing due to resonance donation of the lone pair on nitrogen.

$$\mathrm{Aniline \longrightarrow o\text{-}Nitroaniline + p\text{-}Nitroaniline}$$

However, nitration is carried out in strongly acidic medium using:

$$\mathrm{Conc.\ HNO_3 + Conc.\ H_2SO_4}$$

In this acidic medium, aniline behaves as a base and gets protonated:

$$\mathrm{C_6H_5NH_2 + H^+ \rightleftharpoons C_6H_5NH_3^+}$$

The anilinium ion $$\mathrm{(-NH_3^+)}$$ no longer possesses a free lone pair for resonance donation.

Hence, it becomes strongly electron-withdrawing due to the $$\mathrm{-I}$$ effect and acts as a meta-directing group.

$$\mathrm{Anilinium\ Ion \longrightarrow m\text{-}Nitroaniline}$$

Therefore, the reaction mixture contains both:

1. Unprotonated aniline → ortho/para directing

2. Protonated anilinium ion → meta directing

As a result, a significant amount of meta product is formed.

Major products:

$$\mathrm{o\text{-}Nitroaniline,\ m\text{-}Nitroaniline,\ p\text{-}Nitroaniline}$$

Meta-nitroaniline is formed in high proportion because a large fraction of aniline exists as the anilinium ion in acidic medium.