See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: Sulfonation as a blocking group and electrophilic aromatic substitution (EAS) directing effects. Step 2 - Reaction of Phenol with conc. H2SO4 at 100°C: Phenol undergoes sulfonation. The OH group is a strong ortho/para director. At 100°C, sulfonation preferentially gives the para product (4-hydroxybenzene sulfonic acid, i.e., phenol-4-sulfonic acid) as the major product. Thus, intermediate A = 4-hydroxybenzene sulfonic acid (para-phenolsulfonic acid), with SO3H at C4 and OH at C1. Step 3 - Reaction of A with excess Br2/H2O: The OH group directs bromine to ortho and para positions. In A, the para position (C4) is already occupied by SO3H. Therefore, bromination occurs at both ortho positions (C2 and C6). With excess Br2/H2O, both ortho positions are brominated, giving 2,6-dibromo-4-hydroxybenzene sulfonic acid initially. Step 4 - Desulfonation: The sulfonic acid group (SO3H) at C4 acts as a blocking/protecting group. Under the aqueous conditions with excess Br2/H2O, the SO3H group is displaced (desulfonation occurs, as sulfonic acid groups are reversible under acidic aqueous conditions). After desulfonation, the C4 position becomes available and gets brominated by the excess Br2, yielding 2,4,6-tribromophenol. Step 5 - Final product B: 2,4,6-tribromophenol — benzene ring with OH at C1, Br at C2, Br at C4, and Br at C6. Step 6 - Why other options fail: - (a) retains SO3H and has only two bromines — sulfonation is not retained after aqueous workup with excess Br2 - (b) shows Br at C2, C3, C4 which is an unusual substitution pattern not consistent with OH directing - (c) shows only one Br at C4 — insufficient bromination given excess Br2 - (d) correctly shows 2,4,6-tribromophenol, which is the expected product after sulfonation-blocking strategy followed by desulfonation and bromination Therefore, the correct answer is D.