See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: The reactivity of an alkyl-substituted benzene towards EAS depends on the electron-donating ability of the alkyl group via hyperconjugation and induction. The methyl group activates the ring through hyperconjugation, which involves the overlap of C–H (or C–D or C–T) sigma bonds with the aromatic pi system. Step 1: Identify the substituents. - Compound (a): CH3 (methyl with protium, H) - Compound (b): CD3 (methyl with deuterium, D) - Compound (c): CT3 (methyl with tritium, T) Step 2: Apply the hyperconjugation concept. Hyperconjugation involves donation of electron density from the C–H bond into the aromatic ring. The effectiveness of hyperconjugation depends on the ease of C–H bond ionization/overlap, which is related to bond strength and the mass of the isotope attached. Step 3: Consider isotope effect on hyperconjugation. The C–H bond is weaker and more polarizable than C–D, which in turn is weaker and more polarizable than C–T. This is because heavier isotopes form stronger, shorter, less polarizable bonds due to lower zero-point energy. Therefore, C–H undergoes hyperconjugation more readily than C–D, and C–D more readily than C–T. Consequently: - CH3 (a) donates the most electron density → highest EAS reactivity - CD3 (b) donates intermediate electron density → intermediate EAS reactivity - CT3 (c) donates the least electron density → lowest EAS reactivity Step 4: Arrange in decreasing order of reactivity. a > b > c Step 5: Why other options fail. - Option (b) c > b > a: Incorrectly reverses the order; heavier isotopes do NOT enhance hyperconjugation. - Option (c) a > c > b: Incorrectly places CT3 above CD3; tritium forms stronger C–T bonds than deuterium forms C–D bonds, so CT3 hyperconjugates less effectively than CD3. - Option (d) c > a > b: Incorrect ordering with no physical basis. Therefore, the correct answer is A.