Major Product C in Reaction Sequence Starting from Aniline

To determine the product $\text{C}$ in the given reaction sequence, let us analyze each step systematically:

1. Step 1: Reaction of Aniline with $\text{Br}_2/\text{H}_2\text{O}$ Aniline contains an amino group ($-\text{NH}_2$) which is a strongly activating, ortho/para-directing group. In aqueous medium ($\text{Br}_2/\text{water}$), polybromination occurs rapidly at all available ortho and para positions relative to the $-\text{NH}_2$ group: $\text{Aniline} + 3\text{Br}_2 \xrightarrow{\text{H}_2\text{O}} \text{2,4,6-Tribromoaniline (A)} + 3\text{HBr}$

2. Step 2: Diazotization of Compound A Reaction of 2,4,6-tribromoaniline with nitrous acid ($\text{NaNO}_2/\text{HCl}$) at $273-278\text{ K}$ converts the primary aromatic amine group ($-\text{NH}_2$) into a diazonium salt: $\text{2,4,6-Tribromoaniline (A)} \xrightarrow[\text{273--278 K}]{\text{NaNO}_2 / \text{HCl}} \text{2,4,6-Tribromobenzenediazonium chloride (B)}$

3. Step 3: Conversion to Nitro Compound Treatment of the diazonium salt $\text{B}$ with fluoroboric acid ($\text{HBF}_4$) converts it into the insoluble diazonium fluoroborate intermediate. Subsequent heating with aqueous sodium nitrite ($\text{NaNO}_2$) in the presence of copper powder ($\text{Cu}$) replaces the diazonium group ($-\text{N}_2^+$) with a nitro group ($-\text{NO}_2$): $\text{2,4,6-Tribromobenzenediazonium chloride (B)} \xrightarrow{\text{(i) HBF}_4} [\text{Ar-N}_2^+\text{BF}_4^-] \xrightarrow[\Delta]{\text{(ii) NaNO}_2, \text{Cu}} \text{1,3,5-Tribromo-2-nitrobenzene (C)}$

According to IUPAC nomenclature rules, assigning the lowest locant set gives the name 1,3,5-Tribromo-2-nitrobenzene.

Correct Answer: B (1, 3, 5-Tribromo-2-nitrobenzene)