Reaction Sequence of Minor Product from Aniline Nitration

To determine the major product(s) of the given reaction sequence, let us analyze the steps systematically:

Step 1: Identification of Compound R

The nitration of aniline using mixed acid ($\text{conc. HNO}_3 + \text{conc. H}_2\text{SO}_4$) at $288\text{ K}$ yields three isomers:

• $p$-nitroaniline ($51\%$) $\rightarrow \mathbf{P}$
• $m$-nitroaniline ($47\%$) $\rightarrow \mathbf{Q}$
• $o$-nitroaniline ($2\%$) $\rightarrow \mathbf{R}$

Thus, compound $\mathbf{R}$ is $o$-nitroaniline ($2$-nitroaniline).

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Step 2: Conversion of R to S

1. Protection of $-\text{NH}_2$ group: $\text{o-nitroaniline} \xrightarrow{\text{Ac}_2\text{O, pyridine}} \text{o-nitroacetanilide}$

2. Bromination: $\text{o-nitroacetanilide} \xrightarrow{\text{Br}_2, \text{CH}_3\text{COOH}} \text{4-bromo-2-nitroacetanilide}$ The strongly activating $-\text{NHCOCH}_3$ group directs the incoming bromine to its para-position (C-4).

3. Deprotection (Hydrolysis): $\text{4-bromo-2-nitroacetanilide} \xrightarrow{\text{H}_3\text{O}^+} \text{4-bromo-2-nitroaniline}$

4. Diazotization: $\text{4-bromo-2-nitroaniline} \xrightarrow[\text{273--278 K}]{\text{NaNO}_2, \text{HCl}} \text{4-bromo-2-nitrobenzenediazonium chloride}$

5. Deamination: $\text{4-bromo-2-nitrobenzenediazonium chloride} \xrightarrow{\text{EtOH, } \Delta} \text{1-bromo-3-nitrobenzene } (\mathbf{S})$

Thus, compound $\mathbf{S}$ is $1$-bromo-$3$-nitrobenzene ($m$-bromonitrobenzene).

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Step 3: Conversion of S to the Final Product

1. Reduction of $-\text{NO}_2$ group: $\text{1-bromo-3-nitrobenzene } (\mathbf{S}) \xrightarrow{\text{Sn/HCl}} \text{3-bromoaniline}$

2. Poly-bromination: $\text{3-bromoaniline} \xrightarrow{\text{Br}_2/\text{H}_2\text{O (excess)}} \text{2,3,4,6-tetrabromoaniline}$ The $-\text{NH}_2$ group at position $1$ is strongly activating and directs bromine to all open ortho and para positions (C-2, C-4, and C-6), while C-3 already contains a bromine atom.

3. Diazotization: $\text{2,3,4,6-tetrabromoaniline} \xrightarrow[\text{273--278 K}]{\text{NaNO}_2, \text{HCl}} \text{2,3,4,6-tetrabromobenzenediazonium chloride}$

4. Deamination: $\text{2,3,4,6-tetrabromobenzenediazonium chloride} \xrightarrow{\text{H}_3\text{PO}_2} \text{1,2,3,5-tetrabromobenzene}$ The $-\text{N}_2^+$ group at position $1$ is replaced by hydrogen ($-\text{H}$), leaving bromine atoms at positions $1, 2, 3,$ and $5$ relative to the ring IUPAC numbering.

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Conclusion

Comparing the structure of $1,2,3,5$-tetrabromobenzene with the given options:

• (A) represents $1,2,4,5$-tetrabromobenzene
• (B) represents $1,2,3,4$-tetrabromobenzene
• (C) represents $1,3,5$-tribromobenzene
• (D) represents $1,2,3,5$-tetrabromobenzene

Thus, the correct option is (D).