In the nitration of benzene using a mixture of conc. $H_2SO_4$ and conc. $HNO_3$, the nitrating species is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → ${NO_2}^+$
In the nitration of benzene using a mixture of concentrated $\text{H}_2\text{SO}_4$ and concentrated $\text{HNO}_3$ (known as the nitrating mixture), the active nitrating species is the Nitronium ion, which has the formula:
$\text{NO}_2^+$
Mechanism of Formation
Potassium sulfuric acid ($\text{H}_2\text{SO}_4$) is a stronger acid than nitric acid ($\text{HNO}_3$). In this mixture, $\text{H}_2\text{SO}_4$ acts as an acid and protonates the $\text{HNO}_3$, which acts as a base.
- Protonation: $\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightleftharpoons \text{H}_2\text{NO}_3^+ + \text{HSO}_4^-$
- Dissociation: The protonated nitric acid ($\text{H}_2\text{NO}_3^+$) then loses a molecule of water to form the nitronium ion:
$\text{H}_2\text{NO}_3^+ \to \text{NO}_2^+ (\text{Nitronium ion}) + \text{H}_2\text{O}$
Role in the Reaction
The $\text{NO}_2^+$ ion is a powerful electrophile. Because the benzene ring is electron-rich due to its delocalized $\pi$ electrons, it attacks the positive nitronium ion. This is the first and rate-determining step of the Electrophilic Aromatic Substitution (EAS) mechanism.
Summary of Other Options
- $\text{NO}_2^-$ (Nitrite ion): This is a nucleophile and will not attack the electron-rich benzene ring.
- $\text{NO}^+$ (Nitrosonium ion): This is the active species in nitrosation (e.g., reaction with primary amines), not nitration.
- $\text{NO}_2$ (Nitrogen dioxide): This is a neutral radical and is not the primary species formed in the concentrated acid mixture for this substitution.
The correct answer is: $\text{NO}_2^+$