question_answer

Nitrobenzene on reaction with cone. \[HN{{O}_{3}}/{{H}_{2}}S{{O}_{4}}\]at \[80-100{}^\circ C\] forms which one of the following products?

A)  1, 2-dinitrobenzene

B)  1, 3-dinitrobenzene

C)  1, 4-dinitrobenzene

D)  1, 2, 4-trinitrobenzene

Correct Answer: B

Solution :

\[\text{N}{{\text{O}}_{\text{2}}}\] group being electron withdrawing reduces electron density at output positions. Hence, now the mete-position becomes electron rich on which the electrophone (nitronium ion) attacks during nitration. \[HN{{O}_{3}}+{{H}_{2}}S{{O}_{4}}\xrightarrow[{}]{{}}{{H}_{2}}NO_{3}^{+}+HSO_{4}^{-}\]\[{{H}_{2}}O+\underset{electrophile}{\mathop{NO_{4}^{-}}}\,\]