Predict the correct sequence of reagents to carry out the following c…

The correct answer is Option (2) → (i) \(Br_2/FeBr_3\) (ii) \(Sn/HCl\) (iii) \(NaNO_2/HCl\) (iv) \(Cu_2Br_2/HBr\)

4-Nitrotoluene has to become 2,4-dibromotoluene: one bromine goes next to the methyl group and the nitro group is replaced by a second bromine. A nitro group is replaced by a halogen through the amine and its diazonium salt (Sandmeyer reaction). The ring bromine is put in first, while the nitro group is still there, because both groups then direct to the same position: –\( \ce{CH3} \) is an ortho/para director and –\( \ce{NO2} \) a meta director, and C2 is ortho to –\( \ce{CH3} \) and meta to –\( \ce{NO2} \).

First, bromination with iron(III) bromide as the Lewis acid catalyst gives 2-bromo-4-nitrotoluene:

    \( \displaystyle p\text{-}\ce{CH3C6H4NO2 + Br2} \xrightarrow{\ce{FeBr3}} \text{2-bromo-4-nitrotoluene} + \ce{HBr} \)

Next, Sn/HCl reduces the nitro group to an amino group:

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

The amine is then diazotised at 273–278 K:

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

Finally, copper(I) bromide in HBr replaces the diazonium group by bromine (Sandmeyer reaction):

    \( \displaystyle \ce{ArN2^+Cl^- + HBr} \xrightarrow{\ce{Cu2Br2}} \ce{ArBr + N2 + HCl} \)

The product is 2,4-dibromotoluene.

Shortcut: \( \ce{NaNO2} \)/HCl needs an amine and \( \ce{Cu2Br2} \)/HBr needs the diazonium salt, so the last three steps must be Sn/HCl, \( \ce{NaNO2} \)/HCl, \( \ce{Cu2Br2} \)/HBr in that order. Only Option (2) ends this way with bromination first; Option (4) would brominate the strongly activating amine instead.