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) → \({NO_2}^+\)
Nitration of benzene is an electrophilic substitution. The mixture of concentrated nitric and sulphuric acids produces the electrophile. Sulphuric acid, the stronger acid, protonates nitric acid, which then loses water to give the nitronium ion:
\( \displaystyle \ce{HNO3 + 2H2SO4 <=> NO2^+ + H3O^+ + 2HSO4^-} \)
The nitronium ion attacks the benzene ring, and loss of a proton restores the aromatic ring:
\( \displaystyle \ce{C6H6 + HNO3} \xrightarrow[\text{323-333 K}]{\text{conc. }\ce{H2SO4}} \ce{C6H5NO2 + H2O} \)
So the nitrating species is \( \ce{NO2^+} \).
Why the other options are incorrect
Option (1) \({NO_2}^-\) — Incorrect. The nitrite ion is a negative ion, a nucleophile; it cannot attack the electron-rich benzene ring.
Option (3) \(NO^+\) — Incorrect. The nitrosonium ion is formed from nitrous acid, not from the nitrating mixture; it is a much weaker electrophile and nitrosates only strongly activated rings such as phenol.
Option (4) \(NO_2\) and \({NO_2}^+\) — Incorrect. Neutral nitrogen dioxide is not the attacking species in the mixed-acid nitration; \( \ce{NO2^+} \) alone is.