See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: The reaction involves two steps. First, aniline is treated with concentrated H2SO4, which converts aniline into sulfanilic acid (4-aminobenzenesulfonic acid) via the 'sulfanilic acid trick.' The -NH2 group first forms an anilinium hydrogen sulfate salt, and on heating, sulfonation occurs predominantly at the para position to give 4-aminobenzenesulfonic acid (sulfanilic acid) as intermediate X. However, the key purpose of this step is to protect the para position and also to use the -SO3H group as a blocking/protecting group. Step 2 - Understanding the purpose of sulfonation: When aniline reacts with Conc. H2SO4, it undergoes sulfonation to give sulfanilic acid (X), where the -SO3H group is at the para position relative to -NH2. This blocks the para position. Step 3 - Reaction of X with excess Br2/H2O: Sulfanilic acid has -NH2 (strong ortho/para director) and -SO3H (meta director, but also acts as a blocking group). With excess Br2/H2O, bromination occurs at the ortho positions (positions 2 and 6 relative to NH2) since para is already blocked by SO3H. This gives 2,6-dibromo-4-aminobenzenesulfonic acid... but wait, we need to reconsider. Step 4 - Re-examining: Actually, the purpose of the H2SO4 step is to protonate the amine (forming anilinium salt) to deactivate it, OR to sulfonate. In the context of this classic reaction pathway used to make 2,4,6-tribromoaniline selectively: Aniline + Conc H2SO4 forms anilinium sulfate, which on heating gives sulfanilic acid (para-aminobenzenesulfonic acid). Then with excess Br2/H2O, the -NH2 is a powerful activating group. The -SO3H at para blocks that position. Bromine goes to ortho (positions 2 and 6). After bromination at 2 and 6, the -SO3H group can be removed by hydrolysis (desulfonation occurs in the reaction conditions or upon workup with hot water/steam distillation). This yields 2,4,6-tribromoaniline. Step 5 - Why 2,4,6-tribromoaniline (option c): The sulfonic acid group serves as a temporary blocking group at the para position. After bromination at both ortho positions (2 and 6), when the reaction is worked up, the -SO3H group is hydrolyzed off (desulfonation), and the para position becomes free. However, with excess Br2, bromination also occurs at para after desulfonation, OR the -SO3H leaves and Br takes its place. The net result of this sequence is 2,4,6-tribromoaniline (option c), which has Br at positions 2, 4, and 6 relative to NH2. Step 6 - Why other options fail: Option (a) retains SO3H and has only 2 Br atoms in non-symmetric positions - incorrect. Option (b) has SO3H retained at para with Br at 2,6 - this would be an intermediate but not the final product Y after excess Br2. Option (d) has SO3H and extra Br in wrong positions. The use of excess Br2 and the desulfonation step leads to complete substitution at all three activated positions (2, 4, 6), giving 2,4,6-tribromoaniline. Therefore, the correct answer is C.