A) I > II > m > IV
B) I > III > II > IV
C) II > III > IV > I
D) III > I > II > IV
Correct Answer: B
Solution :
Monobromination of the aromatic compound is a electrophilic substitution reaction. The electron donating group \[(like-C{{H}_{3}})\] will increase the reactivity of ring while electron withdrawing groups decreases the reactivity of the ring. Hence, among the given, the reactivity of \[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{C}{{\text{H}}_{\text{3}}}\]is highest. Between \[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{COOH}\]and \[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{N}{{\text{O}}_{\text{2}}}\text{,}\]electron withdrawing power of \[-\text{N}{{\text{O}}_{\text{2}}}\] group is higher than ?COOH so the reactivity of \[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{COOH}\]is higher than \[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{N}{{\text{O}}_{\text{2}}}\text{.}\] The decreasing reactivity order of ring towards monobromination is \[\underset{(I)}{\mathop{{{C}_{6}}{{H}_{5}}C{{H}_{3}}}}\,>\underset{(III)}{\mathop{{{C}_{6}}{{H}_{6}}}}\,>\underset{(II)}{\mathop{{{C}_{6}}{{H}_{5}}COOH}}\,>\underset{(IV)}{\mathop{{{C}_{6}}{{H}_{5}}N{{O}_{2}}}}\,\]
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