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Q.
Nitration of benzene, produce nitrobenzene. During nitration the role of conc. $HNO _{3}$ in nitrating mixture is as a :
BHUBHU 2001
Solution:
Nitrating mixture $= HNO _{3}+ H _{2} SO _{4}$
$HONO _{2} \underset{\text { Nitrating mixture }}{+ H _{2} SO _{4}} \longrightarrow HSO _{4}^{-}+ H\overset{\oplus}{\underset{\overset{|}{H}}{O}}- N\overset{\oplus}{O _{2}}$
$H\overset{\oplus}{\underset{\overset{|}{H}}{O}}- NO _{2} \longrightarrow H _{2} O + N _{2}\overset{\oplus}{O}$
Here, $H _{2} SO _{4}$ protonates $HNO _{3}$ and causes the split of $HNO _{3}$ in $H _{2} O$ and $N\overset{\oplus}{O _{2}}$
$\therefore HNO _{3}$ behaves as a base and $H _{2} SO _{4}$ behaves as acid.
