question_answer

\[{{C}_{6}}{{H}_{6}}\xrightarrow[{{H}_{2}}S{{O}_{4}}]{HN{{O}_{3}}}X\xrightarrow[FeC{{l}_{3}}]{C{{l}_{2}}}Y.\] In the above sequence Y can be

A)  3-nitrochlorobenzene

B)  1 -nitrochlorobenzene

C)  4-nitrochlorobenzene

D)  none of these.

Correct Answer: A

Solution :

\[(HN{{O}_{3}}+{{H}_{2}}S{{O}_{4}})\] reagent is the agent for nitration in aromatic rings. \[HN{{O}_{3}}+2{{H}_{2}}S{{O}_{4}}\to NO_{2}^{+}+HSO_{4}^{-}+{{H}_{3}}{{O}^{+}}\] \[N{{O}_{2}}\] group is a meta directing group so chlorine atom goes to meta position of the ring.