The correct sequence of conversion of 4-Nitrotoluene to 2-Bromobenzoi…

The correct answer is Option (1) → \(C\longrightarrow B \longrightarrow D \longrightarrow E \longrightarrow A\)

In 4-nitrotoluene the –\( \ce{CH3} \) group is ortho/para directing and the –\( \ce{NO2} \) group is meta directing. Both send an incoming bromine to the same position, ortho to \( \ce{CH3} \) (and meta to \( \ce{NO2} \)). The nitrogen is then removed through the diazonium salt, and finally the methyl group is oxidised to –COOH. The route runs:

Compound C, 4-nitrotoluene, is brominated with \( \ce{Br2} \) in the presence of \( \ce{FeBr3} \), which gives 2-bromo-4-nitrotoluene. Its nitro group is then reduced with Sn/HCl to give compound B, 3-bromo-4-methylaniline.

    \( \text{4-nitrotoluene} \xrightarrow{\ce{Br2/FeBr3}} \text{2-bromo-4-nitrotoluene} \xrightarrow{\ce{Sn/HCl}} \text{3-bromo-4-methylaniline} \)

The amine B is diazotised with \( \ce{NaNO2} \) and HCl at 273–278 K, which gives compound D, the diazonium chloride.

    \(\ce{ArNH2 + NaNO2 + 2HCl} \xrightarrow{\text{273-278 K}} \ce{ArN2+Cl- + NaCl + 2H2O}\)

Treating D with hypophosphorous acid replaces –\( \ce{N2^{+}Cl^{-}} \) by H, and nitrogen gas escapes. This gives compound E, 2-bromotoluene.

    \(\ce{ArN2+Cl- + H3PO2 + H2O -> ArH + N2 + H3PO3 + HCl}\)

Finally, oxidation of the side chain with alkaline \( \ce{KMnO4} \), followed by acidification, turns –\( \ce{CH3} \) into –COOH and gives compound A, 2-bromobenzoic acid.

    \( \text{2-bromotoluene} \xrightarrow[\text{(ii) }\ce{H3O+}]{\text{(i) }\ce{KMnO4/KOH}} \text{2-bromobenzoic acid} \)

Shortcut: only Options (1) and (4) start with C and end with A. In Option (4) the diazonium salt D comes before the amine B from which it is made, which is impossible, so Option (1) remains.

NCERT (2026-27 edition):
Importance of Diazonium Salts in Synthesis of Aromatic Compounds – …

Note: The same/similar question has been asked in different years by NTA as per the details given below: