Question
Nitrobenzene can be prepared from benzene by using a mixture of $$conc.$$ $$HN{O_3}$$ and $$conc.$$ $${H_2}S{O_4}.$$ In the mixture, nitric acid acts as a/an
A.
reducing agent
B.
acid
C.
base
D.
catalyst
Answer :
base
Solution :
$$Conc.$$ $${H_2}S{O_4}$$ and $$conc.$$ $$HN{O_3}$$ react in the following manner
\[\begin{align}
& \underset{\text{Base}}{\mathop{HN{{O}_{3}}}}\,+\underset{\text{Acid}}{\mathop{{{H}_{2}}S{{O}_{4}}}}\,\to {{H}_{2}}NO_{3}^{+}+HSO_{4}^{-} \\
& {{H}_{2}}NO_{3}^{+}\to \underset{\begin{align}
& \,\,\,\,\,\,\,\,\text{Nitronium}\text{.} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ion}\text{.} \\
& \text{Attacking species} \\
& \,\,\,\,\,\,\text{(electrophile)} \\
\end{align}}{\mathop{NO_{2}^{+}}}\,+{{H}_{2}}O \\
\end{align}\]
Hence, in this reaction $$HN{O_3}$$ acts as a base and $${H_2}S{O_4}$$ as an acid.