Question

Which of the following reagents will best convert nitrobenzene into $2,4,6-$trichlorobromobenzene?

• A. $SnC{l}_2/HCl,C{l}_2\left(aq\right),NaN{O}_2/HCl-0^{o}C;CuBr$
• B. $C{l}_2/AlC{l}_3,SnC{l}_2/HCl,NaN{O}_2/HCl-0^{o}C;CuBr$
• C. $SnC{l}_2/HCl,B{r}_2-H_{2}O,NaN{O}_2/HCl-0^{o}C;CuCl$
• D. $SnC{l}_2/HCl,NaN{O}_2/HCl-0^{o}C,H_{2}O,C{l}_2\left(aq\right);HBr$

Answer 1

The correct option is A $SnC{l}_2/HCl,C{l}_2\left(aq\right),NaN{O}_2/HCl-0^{o}C;CuBr$

Nitrobenzene is first reduced with tin(II) chloride and $HCl$ to form aniline. Then, aniline is chlorinated with aqueous chlorine to form $2,4,6$-trichloroaniline. It is then diazotized with sodium nitrite and $HCl$. Finally, cuprous bromide is used to replace diazonium group with bromo group to obtain $1$-bromo-$2,4,6$-trichlorobenzene.