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. \[HN{{O}_{3}}+{{H}_{2}}S{{O}_{4}}\xrightarrow[{}]{{}}{{H}_{2}}NO_{3}^{+}+HSO_{4}^{-}\]\[{{H}_{2}}O+\underset{electrophile}{\mathop{NO_{4}^{-}}}\,\]You need to login to perform this action.
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