The sequence from the following that would result in giving predominantly 3,4,5- Tribromoaniline is :

Solution

Chemistry ma’am · 3 steps

Use the amino group to place the bromines, then turn it into a bromine and make a new amine.

1. Read the target. In 3,4,5-tribromoaniline the three Br atoms sit next to each other, all meta and para to -NH₂. A direct -NH₂ group sends Br to ortho and para, never to two meta positions. So the final -NH₂ must come from somewhere else.
2. Walk through (c). In p-nitroaniline the -NH₂ directs excess Br₂ to both of its ortho positions, giving 2,6-dibromo-4-nitroaniline. Diazotisation with NaNO₂/HCl and then CuBr (Sandmeyer reaction) replaces -NH₂ with a third Br. This gives 1,2,3-tribromo-5-nitrobenzene. Finally, Sn/HCl reduces -NO₂ to -NH₂.

   Option	What happens	Main product
   (a)	-NO₂ sends one Br meta, then it is reduced	3-bromoaniline (only one Br)
   (b)	Bromobenzene forms. NH₃ cannot displace an aryl Br under normal conditions	bromobenzene, no aniline
   (c)	Two Br go ortho to -NH₂, Sandmeyer turns -NH₂ into Br, then -NO₂ is reduced to -NH₂	3,4,5-tribromoaniline
   (d)	-NH₂ activates the ortho and para positions	2,4,6-tribromoaniline

3. So. Counted from the new -NH₂ at C1, the bromines sit at C3, C4 and C5. Option (c) gives 3,4,5-tribromoaniline.

The idea to keep. When a substitution pattern seems impossible, use -NH₂ as a temporary director: place the groups first, then remove it or replace it through the diazonium salt.

Answer from the NTA final answer key. Solution by PrepWiser.