See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: In electrophilic aromatic substitution (EAS), the rate of reaction depends on the electron density of the aromatic ring. Electron-donating groups (EDGs) activate the ring (increase reactivity), while electron-withdrawing groups (EWGs) deactivate the ring (decrease reactivity). The stronger the EDG, the more reactive the compound. Step 2 - Analyze each substituent: Compound II - N(CH3)2 (dimethylamino group): This is a very strong electron-donating group via lone pair donation (resonance). Nitrogen directly donates electrons into the ring through resonance, making this the most activated ring. Highest reactivity. Compound III - OCOCH3 (acetoxy/ester oxygen group): The oxygen atom donates electron density to the ring via resonance (lone pairs on O), but this donation is partially offset by the adjacent carbonyl (C=O) which withdraws electrons from oxygen. Net effect: weak activating group (weaker than a simple -OR or -OH). Still activating overall due to oxygen's lone pair donation. Second highest reactivity. Compound I - CH2CO2H (phenylacetic acid / carboxymethyl group): The CH2 group insulates the ring from the carboxylic acid. The -CH2- group breaks direct conjugation, so the carboxylic acid EWG effect is attenuated. However, -CH2CO2H is still a weak deactivator compared to direct EWGs. The -CO2H is weakly deactivating through induction via the CH2 spacer. This gives moderate deactivation, less than CCl3. Compound IV - CCl3 (trichloromethyl group): This is a strongly electron-withdrawing group through both inductive and field effects (three electronegative Cl atoms). Strongly deactivates the ring. Lowest reactivity. Step 3 - Order of reactivity: II (strong EDG, N lone pair resonance) > III (weak EDG, O lone pair resonance partially attenuated by carbonyl) > I (weak EWG, inductive through CH2) > IV (strong EWG, three Cl atoms inductively withdrawing) This gives: II > III > I > IV, which matches option (d). Step 4 - Why other options fail: (a) I > II > III > IV: Incorrect; places the weakly deactivated I above the strongly activated II and III. (b) IV > III > II > I: Incorrect; reverses the correct order entirely. (c) III > I > IV > II: Incorrect; places the strongest activator (II) last. Therefore, the correct answer is D.