Choose the correct sequence of reagents to carry out the following co…

The correct answer is Option (1) → (A), (C), (B), (D)

In a multi-step synthesis on benzene, the order of the steps is decided by the directing effect of the group already on the ring. In the product, Br and OH are meta to each other. –OH and –\( \ce{NH2} \) are ortho/para directors, but the –NO₂ group is a meta director. So bromine must enter while the nitrogen is still a nitro group; the nitro group is converted into –OH only afterwards.

First, benzene is nitrated with the nitrating mixture (A):

    \( \displaystyle \ce{C6H6 + HNO3} \xrightarrow{\text{conc. }\ce{H2SO4}} \ce{C6H5NO2 + H2O} \)

Next, nitrobenzene is brominated with \( \ce{Br2}/\ce{FeBr3} \) (C); the nitro group sends bromine to the meta position:

    \( \displaystyle \ce{C6H5NO2 + Br2} \xrightarrow{\ce{FeBr3}} m\text{-}\ce{BrC6H4NO2 + HBr} \)

Then the nitro group is reduced to an amino group with Sn/HCl (B):

    \( \displaystyle m\text{-}\ce{BrC6H4NO2 + 6[H]} \xrightarrow{\ce{Sn/HCl}} m\text{-}\ce{BrC6H4NH2 + 2H2O} \)

Finally, the amine is diazotised with \( \ce{NaNO2}/\ce{HCl} \) at 273-278 K and the diazonium salt is boiled with water (D), which replaces \( \ce{N2+} \) by –OH:

    \( \displaystyle m\text{-}\ce{BrC6H4NH2} \xrightarrow[\text{273-278 K}]{\ce{NaNO2 + HCl}} m\text{-}\ce{BrC6H4N2^+Cl^-} \)

    \( \displaystyle m\text{-}\ce{BrC6H4N2^+Cl^- + H2O} \xrightarrow{\text{boil}} m\text{-}\ce{BrC6H4OH + N2 + HCl} \)

Shortcut: The nitrogen must be put on the ring first, so the sequence starts with (A): Options (1) and (2). Bromination must come before reduction, or the –\( \ce{NH2} \) group would send bromine to the ortho/para positions; that leaves Option (1).