Understanding Aniline Nitration and $m$-Nitroaniline Formation

The question asks about the reason behind the significant formation of $m$-nitroaniline when aniline undergoes nitration in a strongly acidic medium (using $HNO_3/H_2SO_4$). This specific outcome is related to the behavior of aniline under acidic conditions.

Role of Aniline in Strong Acidic Medium

Aniline ($C_6H_5NH_2$) is a base due to the lone pair of electrons on the nitrogen atom. In a strongly acidic medium, like the mixture of nitric acid ($HNO_3$) and sulfuric acid ($H_2SO_4$) used for nitration, aniline readily reacts with the acid protons ($H^+$).

This protonation reaction forms the anilinium ion ($C_6H_5NH_3^+$):

In a strongly acidic solution, a large fraction of aniline molecules are converted into these anilinium ions.

Characteristics of the Anilinium Ion ($C_6H_5NH_3^+$)

The anilinium ion plays a crucial role in directing the incoming electrophile (the nitronium ion, $NO_2^+$) during the nitration reaction.

• Positive Charge: The nitrogen atom in the anilinium ion carries a positive charge ($+1$).
• Electron-Withdrawing Effect: This positive charge makes the $-NH_3^+$ group highly electron-withdrawing. It pulls electron density away from the benzene ring through both inductive and resonance effects, although the inductive effect is dominant here.
• Deactivation: Because it withdraws electron density, the anilinium ion is a strongly deactivating group. This means it makes the aromatic ring less reactive towards electrophilic substitution compared to benzene itself.
• Directing Effect: Positively charged groups attached directly to the benzene ring, like $-NH_3^+$, are typically meta-directing. This is because the formation of the intermediate carbocation (arenium ion) during electrophilic attack is more destabilized at the ortho and para positions due to the proximity of the positive charge on the nitrogen atom. The meta position is relatively less destabilized.

Therefore, the anilinium ion acts as a strongly deactivating and meta-directing group primarily due to its strong inductive electron-withdrawing effect.

Analysis of Options

Let's examine the given options based on this understanding:

• Option 1: Weakly activating and $o,p$-directing group due to hyperconjugation. This is incorrect. The anilinium ion is deactivating, not activating, and it directs meta, not ortho/para. Hyperconjugation is associated with alkyl groups, not charged nitrogen groups like $-NH_3^+$.
• Option 2: Strongly activating and $m$-directing group due to its positive charge. This is incorrect. While the positive charge contributes to its nature, it makes the group deactivating, not activating.
• Option 3: Strongly deactivating and $m$-directing group due to its strong inductive electron-withdrawing effect. This option correctly describes the nature of the anilinium ion. The strong electron withdrawal via induction deactivates the ring, and the positive charge causes meta-direction. This explains the formation of $m$-nitroaniline.
• Option 4: Weakly deactivating and $o,p$-directing group through resonance effects. This is incorrect. The anilinium ion is strongly deactivating, not weakly, and it's meta-directing, not ortho/para. While resonance is involved in aniline's reactivity, the protonated form ($C_6H_5NH_3^+$) primarily exhibits deactivation and meta-direction due to the inductive effect stemming from the positive charge.

Conclusion on $m$-Nitroaniline Formation

The predominant formation of $m$-nitroaniline during the nitration of aniline in a strong acid is a direct consequence of aniline being protonated to form the anilinium ion ($C_6H_5NH_3^+$). This ion acts as a strongly deactivating and $m$-directing substituent on the benzene ring, guiding the incoming nitro group ($NO_2^+$) predominantly to the meta position.