Aniline is converted into 1,3,5-tribromobenzene. Which sequence of reagents achieves this?

A(i) $Br X_{2}$ water; (ii) $NaNO X_{2}$ / dil. $HCl$ , 273-278 K; (iii) $H X_{3} PO X_{2}$
B(i) $NaNO X_{2}$ / dil. $HCl$ , 273-278 K; (ii) $Br X_{2}$ water; (iii) $H X_{3} PO X_{2}$
C(i) $Br X_{2}$ water; (ii) $H X_{3} PO X_{2}$ ; (iii) $NaNO X_{2}$ / dil. $HCl$ , 273-278 K
D(i) $NaNO X_{2}$ / dil. $HCl$ , 273-278 K; (ii) $H X_{3} PO X_{2}$ ; (iii) 3 equiv. $Br X_{2} / FeBr X_{3}$

Answer: A

The strongly ortho/para-directing

NH X_{2}

group is used first:

Br X_{2}

water gives 2,4,6-tribromoaniline, whose bromines are then all meta to each other. Diazotisation followed by

H X_{3} PO X_{2}

removes the nitrogen as

N X_{2}

, leaving 1,3,5-tribromobenzene. Any order that removes

NH X_{2}

before bromination gives ortho/para bromination of benzene instead.