question_answer

Nitration of benzene, produce nitrobenzene. During nitration the role of cone.\[HN{{O}_{3}}\]in nitrating mixture is as a:

A)  catalyst                               

B)  reducing agent

C)  acid                                      

D)  base

Correct Answer: D

Solution :

Nitrating mixture\[=HN{{O}_{3}}+{{H}_{2}}S{{O}_{4}}\]. \[HON{{O}_{2}}+\underset{Nitrating\text{ }mixture}{\mathop{{{H}_{2}}S{{O}_{4}}}}\,\xrightarrow[{}]{{}}HSO_{4}^{-}+\underset{\begin{smallmatrix}  | \\  H \end{smallmatrix}}{\overset{\oplus }{\mathop{HO}}}\,-N{{O}_{2}}\]                         \[\underset{\begin{smallmatrix}  | \\  H \end{smallmatrix}}{\overset{\oplus }{\mathop{HO}}}\,-N{{O}_{2}}\xrightarrow[{}]{{}}{{H}_{2}}O+\overset{\oplus }{\mathop{N{{O}_{2}}}}\,\] Here,\[{{H}_{2}}S{{O}_{4}}\]protonates\[HN{{O}_{3}}\]and causes the split of\[HN{{O}_{3}}\]in\[{{H}_{2}}O\]and\[\overset{\oplus }{\mathop{N{{O}_{2}}}}\,\] \[\therefore \]\[HN{{O}_{3}}\]behaves as a base and\[{{H}_{2}}S{{O}_{4}}\]behaves as acid.