Compound A from the following reaction sequence is:

Solution

Chemistry ma’am · 4 steps

NaNO₂/HCl means an -NH₂ group must be there, and H₃PO₂ then removes it as N₂.

1. Work backwards. NaNO₂/HCl followed by H₃PO₂ is the classic way to remove an -NH₂ group. The amine becomes a diazonium salt, and H₃PO₂ replaces -N₂⁺ with H. So B must be an aromatic primary amine, and so must A.
2. Bromination of aniline. -NH₂ is a very strong ortho, para director. With Br₂, aniline is brominated at both ortho positions and the para position, giving B = 2,4,6-tribromoaniline.
3. Diazotise, then remove. At 0–5 °C, NaNO₂/HCl turns B into C = 2,4,6-tribromobenzenediazonium chloride. Heating with H₃PO₂ replaces -N₂⁺ by H.

   C₆H₂Br₃N₂⁺Cl⁻ + H₃PO₂ + H₂O → C₆H₃Br₃ + N₂ + H₃PO₃ + HCl

4. So. With the -NH₂ gone, Br atoms that sat at 2, 4 and 6 are now at 1, 3 and 5. That is 1,3,5-tribromobenzene, so A is aniline, option (d).

The idea to keep. Need a 1,3,5 pattern that no single group can direct? Let -NH₂ place the groups first, then diazotise and remove it with H₃PO₂. The amino group acts as a temporary director.

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