See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Reaction of aniline with Br2/H2O: Aniline is a strongly activating, ortho/para director. When treated with Br2 in water (aqueous bromine), electrophilic aromatic substitution occurs readily at all three activated positions (both ortho positions and the para position), giving 2,4,6-tribromoaniline as product (A). This tribromination occurs because the NH2 group strongly activates the ring, and aqueous bromine is sufficient without a Lewis acid catalyst. Step 2 - Diazotization and reduction: Product (A), 2,4,6-tribromoaniline, is treated with NaNO2/HCl (diazotization) to form the diazonium salt 2,4,6-tribromobenzenediazonium chloride. Then treatment with H3PO2 (hypophosphorous acid) reduces the diazonium group, replacing N2+ with H (reductive deamination). This converts the -N2+Cl- group to -H. Step 3 - Identify product (B): Replacing the diazonium (originally the NH2 at position 1) with H in 2,4,6-tribromoaniline gives 1,3,5-tribromobenzene, since the bromine substituents were at positions 2, 4, and 6 relative to the amino group; after removal of the amino group, renumbering gives 1,3,5-tribromobenzene. Step 4 - Why other options fail: - Option (a) 1,2,4-tribromobenzene: Would require unsymmetrical bromination not consistent with the symmetric trisubstitution by Br2/H2O on aniline. - Option (c) 1,2,3-tribromobenzene: Would require ortho/ortho/meta substitution, not consistent with NH2 directing. - Option (d) bromobenzene: Would require only monobromination, but aniline with Br2/H2O undergoes tribromination. Therefore, the correct answer is B.