Question

Nitration of benzene, produce nitrobenzene. During nitration the role of conc. $HN{O}_3$ in nitrating mixture is as a :

• A. catalyst
• B. reducing agent
• C. acid
• D. base

Answer 1

Answer: (D)

Nitrating mixture $=HN{O}_3+H_{2}S{O}_4$
$HON{O}_2\underset{\text{ Nitrating mixture }}{+H_{2}S{O}_4}⟶HS{O}_4^{-}+H\stackrel{\oplus }{\underset{\stackrel{|}{H}}{O}}-N\stackrel{\oplus }{O_2}$
$H\stackrel{\oplus }{\underset{\stackrel{|}{H}}{O}}-N{O}_2⟶H_{2}O+N_2\stackrel{\oplus }{O}$
Here, $H_{2}S{O}_4$ protonates $HN{O}_3$ and causes the split of $HN{O}_3$ in $H_{2}O$ and $N\stackrel{\oplus }{O_2}$
$\therefore HN{O}_3$ behaves as a base and $H_{2}S{O}_4$ behaves as acid.