Aarti Gupta
m-bromoiodobenzene can be prepared from benzene in the following way-
1.Treating benzene with nitrating mixture which gives nitrobenzene-
C6H6 + HNO3/H2SO4 ----------> C6H5NO2
2.Bromination of nitrobenzene,giving m-bromonitrobenzene-
C6H5NO2 + Br2/Fe --------------> 3-Br-C6H4NO2
3.Reducing the above compound with Sn/HCl-
3-Br-C6H4NO2 + Sn/HCl ------------> 3-Br-C6H4NH2
4.Treating 3-bromoaniline with NaNO2/HCl-
3-Br-C6H4NH2 + NaNO2/HCl --------> 3-Br-C6H4N2Cl
5.Finally treating the above diazonium compound with KI ,yields m-bromoiodobenzene
3-Br-C6H4N2Cl + KI -------------> 3-Br-C6H4I