See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: Electrophilic Aromatic Substitution (EAS) rates are governed by the electron density of the aromatic ring. Electron-donating groups (EDG) activate the ring and increase the rate of EAS, while electron-withdrawing groups (EWG) deactivate the ring and decrease the rate. Step 2 - Analyze each substrate: - Reaction A: Toluene with SO3/H2SO4, sulfonation occurring at the meta position. The methyl group is an EDG (activating group) via hyperconjugation and induction. Even though methyl is ortho/para directing, sulfonation can still occur (the product shown is meta, but the rate of the overall sulfonation of toluene is still faster than benzene). Toluene is more reactive than benzene due to the electron-donating methyl group. - Reaction B: Toluene with SO3/H2SO4, sulfonation at the ortho position (2-methylbenzenesulfonic acid). This is also sulfonation of toluene — same substrate as A. The methyl group activates the ring. Ortho/para positions are most activated. The rate of toluene sulfonation is faster than benzene. - Reaction C: Nitrobenzene with SO3/H2SO4. The nitro group (NO2) is a strong EWG (deactivating group) via resonance and induction. It strongly deactivates the ring, making EAS very slow. Meta product is obtained consistent with EWG meta-directing. Step 3 - Compare rates: - Both A and B involve toluene (same substrate, just different regiochemical products shown). Since the methyl group is EDG, toluene reacts faster than benzene and much faster than nitrobenzene. - Reaction B gives the ortho product of toluene sulfonation; Reaction A gives the meta product of toluene sulfonation. Since methyl is an ortho/para director, the ortho/para positions are more electron-rich and more reactive. However, A and B use the same starting material (toluene), so the overall rate of sulfonation for both A and B is the same — the difference is only regioselectivity of the product shown. - Wait — re-examining: A gives meta product from toluene, B gives ortho product from toluene. The rate referred to is the rate of the overall reaction for each substrate. Since A and B are both toluene, their rates are equal in principle, but the question asks about the rate of each specific reaction as written (forming the specific product). The ortho position (B) is more activated than the meta position (A) by the methyl EDG, so B > A in terms of rate of forming that particular product. - C involves nitrobenzene, which is strongly deactivated — slowest. Step 4 - Order of increasing rate: C (slowest, deactivated ring) < A (toluene, meta position, less activated) < B (toluene, ortho position, more activated by methyl EDG). Increasing rate: C < A < B, which is equivalent to B > A > C. Step 5 - Why other options fail: - (b) B > C > A: Wrong, C (nitrobenzene) cannot be faster than A (toluene meta). - (c) A > B > C: Wrong order between A and B; ortho (B) is more activated than meta (A). - (d) A > C > B: Wrong, nitrobenzene (C) is deactivated and cannot be faster than B. Therefore, the correct answer is A.