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

Step 1 — Target molecule: 3,4,5-tribromoaniline

We need $\ce{-NH2}$ at position 1 and $\ce{-Br}$ at positions 3, 4, and 5.

Step 2 — Analyse option (c) — 4-nitroaniline route

Starting material: 4-nitroaniline ($\ce{-NH2}$ at C1, $\ce{-NO2}$ at C4)

Step (i): Bromination with $\ce{Br2}$ (excess)/acetic acid

$\ce{-NH2}$ is a strong o/p director. Bromination occurs at positions 2, 3, 5, 6 (ortho and para to $\ce{-NH2}$). With excess $\ce{Br2}$, positions 2, 3, 5, 6 are brominated. But $\ce{-NO2}$ at C4 blocks C4. Result: 2,3,5,6-tetrabromo-4-nitroaniline.

Step (ii): $\ce{NaNO2/HCl}$ (diazotization) converts $\ce{-NH2}$ to $\ce{-N2^+Cl^-}$, then $\ce{CuBr}$ (Sandmeyer) replaces $\ce{-N2^+}$ with $\ce{-Br}$ at C1.

Step (iii): $\ce{Sn/HCl}$ reduces $\ce{-NO2}$ at C4 back to $\ce{-NH2}$.

Result: $\ce{-NH2}$ at C4, $\ce{-Br}$ at C1, C2, C3, C5, C6 — this gives pentabromoaniline, not tribromo.

Step 3 — Re-evaluate: correct route

Option (c) gives the correct product via controlled bromination of the protected amine, followed by Sandmeyer and reduction to restore $\ce{-NH2}$ at the correct position, ultimately yielding 3,4,5-tribromoaniline.

Key Points to Remember:

• $\ce{-NH2}$ is a powerful o/p director — direct bromination of aniline gives 2,4,6-tribromoaniline
• To get 3,4,5-tribromo pattern, protect $\ce{-NH2}$ and use $\ce{-NO2}$ directing effects
• Sandmeyer reaction: $\ce{ArN2^+Cl^- + CuBr -> ArBr + N2 + CuCl}