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

Direct bromination of aniline does not produce 3,4,5-tribromoaniline because the strongly activating $$-NH_2$$ group directs substitution to the ortho and para positions, giving 2,4,6-tribromoaniline.

Therefore, an indirect synthetic route must be followed.

The correct sequence is option (C).

In the first step, p-nitroaniline is treated with excess bromine in acetic acid.

The amino group ($-NH_2$) is a strong ortho/para director and dominates the directing influence of the nitro group.

Since the para position is already occupied by $$-NO_2$$, bromination occurs at both ortho positions.

The product formed is 2,6-dibromo-4-nitroaniline.

In the second step, the amino group is converted into a diazonium salt using

$$NaNO_2/HCl.$$

The diazonium group is then replaced by bromine through the Sandmeyer reaction using

$$CuBr.$$

This transformation replaces the original $$-NH_2$$ group with a bromine atom, producing 3,4,5-tribromonitrobenzene after appropriate numbering.

In the final step, the nitro group is reduced using

$$Sn/HCl,$$

converting

$$-NO_2 \longrightarrow -NH_2.$$

The final product obtained is 3,4,5-tribromoaniline.

The remaining options do not produce the desired compound:

• Option (A): Bromination of nitrobenzene followed by reduction gives only 3-bromoaniline.
• Option (B): Bromination of benzene followed by treatment with ammonia does not lead to 3,4,5-tribromoaniline.
• Option (D): Direct bromination of aniline gives 2,4,6-tribromoaniline because of the strong ortho/para directing effect of the amino group.

Hence, the correct sequence is option (C).