See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1: Identify the electronic nature of substituents for each compound. - G: C6H5NMe2 — NMe2 is a powerful electron-donating group (EDG) via resonance (+M effect), strongly activates the ring toward electrophilic aromatic substitution (EAS). - A: C6H5CH3 (toluene) — CH3 is a weak EDG via hyperconjugation/induction, activates the ring moderately. - B: C6H6 (benzene) — unsubstituted reference compound. - C: C6D6 (deuterobenzene) — D is slightly more electron-donating than H due to a smaller primary kinetic isotope effect and slightly stronger C–D bond (inductive), but the ring electronic activation is essentially the same as benzene. However, because the C–D bond is stronger than C–H, the rate-determining step (proton/deuteron loss) is slightly slower for C6D6 than C6H6, making C slightly slower than B. - D: C6T6 (tritio-benzene) — T (tritium) has an even stronger kinetic isotope effect than D, making C–T bond even stronger, so rate of C6T6 < rate of C6D6. Thus D < C. - E: C6H5Br3 — Br is a weak deactivator (–I inductive effect dominates over weak +M), deactivates the ring, so rate < benzene. - F: C6H5N+R3 — The quaternary ammonium group (N+R3) is a strong electron-withdrawing group (EWG) via both –I and –M effects, strongly deactivates the ring. This gives the lowest rate among all. Step 2: Establish the order. - G (strong EDG, NMe2) > A (weak EDG, CH3) > B (benzene, reference) > C (C6D6, kinetic isotope slows slightly) > D (C6T6, kinetic isotope slows more) > E (weak EWG, Br) > F (strong EWG, N+R3) Step 3: Match with options. - Option (a): G > A > B > C > D > E > F — this matches the reasoning above perfectly. - Option (c): G > A > B = C = D > E > F — this would mean benzene, deuterobenzene, and tritio-benzene all react at the same rate, ignoring the kinetic isotope effect. - Option (d): G > A > B > C = D > E > F — this would mean C6D6 and C6T6 react at the same rate, ignoring the difference in isotope effect between D and T. - Option (b): G > B > C > D > A > F — places toluene after isotope-substituted benzenes, which is incorrect. However, the given answer is C: G > A > B = C = D > E > F. The reasoning for option C being correct is that for EAS, the rate-determining step is the attack of the electrophile (formation of the arenium ion/sigma complex), NOT the deprotonation step. Since the kinetic isotope effect only matters if C–H/D/T bond breaking is rate-determining, and in EAS the rate-determining step is electrophilic attack (not the elimination of H+/D+/T+), benzene, deuterobenzene, and tritio-benzene all have essentially the same rate of nitration. The electronic effect of D vs H on ring activation is negligible. Therefore B = C = D in rate of nitration. Step 4: Final order: G > A > B = C = D > E > F, which is option (c). Therefore, the correct answer is C.