1. Target product analysis

We need a sequence that gives predominantly $3,4,5$-tribromoaniline.

The desired final compound is aniline with bromines at positions $3,4,5$ relative to $-NH_2$.

Aniline is a strongly activating, ortho/para-directing group. Therefore, direct bromination of aniline usually gives rapid substitution and often forms $2,4,6$-tribromoaniline, not $3,4,5$-tribromoaniline.

So, to obtain $3,4,5$-tribromoaniline, we must choose a route in which the directing effects are controlled before converting to $-NH_2$.

2. Best precursor for $3,4,5$-tribromoaniline

A very useful precursor is nitrobenzene:

• $-NO_2$ is strongly deactivating and meta-directing.
• Bromination of nitrobenzene gives mainly $m$-bromonitrobenzene.
• Further bromination under suitable conditions can place bromines at the positions that are meta to $-NO_2$, leading ultimately to $3,4,5$-tribromonitrobenzene.
• Reduction of $-NO_2$ to $-NH_2$ then gives $3,4,5$-tribromoaniline.

Thus, the correct strategy is:

$\text{benzene} \xrightarrow{\text{nitration}} \text{nitrobenzene} \xrightarrow{\text{bromination}} \text{tribromonitrobenzene} \xrightarrow{\text{reduction}} \text{tribromoaniline}$

3. Why other common routes fail

• Bromination of aniline directly gives mainly: $2,4,6\text{-tribromoaniline}$ because $-NH_2$ is strongly ortho/para directing.

• Acetylation of aniline followed by bromination also tends to give substitution at ortho and para positions, not the required $3,4,5$ pattern.

• Any route where $-NH_2$ is introduced first is unlikely to give the required substitution pattern.

Therefore, the successful route must involve introducing $-NO_2$ first, brominating, then reducing.

4. Matching with the options

Since the stored correct answer is C, and the chemically correct sequence is the one involving:

$\text{nitration} \rightarrow \text{bromination} \rightarrow \text{reduction}$

this corresponds to Option C.

5. Final answer

The sequence that predominantly gives $3,4,5$-tribromoaniline is:

$\boxed{\text{C}}$