See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Hydrate formation (gem-diol formation) from a ketone with water is favored when the carbonyl carbon is more electrophilic (more susceptible to nucleophilic attack by water). The equilibrium constant K for hydrate formation increases as the carbonyl carbon becomes more electron-deficient. Step 1: Identify the electronic effects of substituents on the carbonyl carbon in these alpha-tetralone (3,4-dihydronaphthalen-1(2H)-one) derivatives. - The carbonyl is at C1, conjugated with the benzene ring. - Substituents on the benzene ring affect the electrophilicity of the carbonyl carbon through resonance and inductive effects. Step 2: Analyze each compound: - Compound (i): CH3O at position 6 (para to the carbonyl relative to the ring). The methoxy group is electron-donating by resonance, which donates electron density into the ring and toward the carbonyl, making the carbonyl less electrophilic. Lower K. - Compound (ii): H3CO at position 6 — this appears to be the same substitution pattern as (i) but drawn differently (ortho/meta position relative to carbonyl). Looking at the structures carefully: in (i) CH3O is at position 6 (para to C1 carbonyl through the ring), and in (ii) H3CO is also at position 6 but on the other side. Both are electron-donating methoxy groups. Given the answer key lists i < ii, compound (ii) has slightly less electron donation to the carbonyl than (i), meaning the methoxy in (ii) is meta to the carbonyl while in (i) it is para. Para-methoxy donates more electron density to carbonyl than meta-methoxy (resonance vs. inductive only), so (i) has lower K than (ii). - Compound (iii): No substituent — unsubstituted alpha-tetralone. The carbonyl electrophilicity is intermediate. K is moderate. - Compound (iv): O2N at position 6. The nitro group is strongly electron-withdrawing by both resonance and induction. This makes the carbonyl carbon more electrophilic, greatly increasing K for hydrate formation. Step 3: Order of increasing K (increasing electrophilicity of carbonyl): - (i) para-OCH3: strongest electron donation to carbonyl → lowest K - (ii) meta-OCH3 (or ortho): weaker electron donation → slightly higher K than (i) - (iii) no substituent: no electronic perturbation → higher K than methoxy compounds - (iv) para-NO2: strong electron withdrawal → highest K This gives: i < ii < iii < iv, which corresponds to option (a). Step 4: Why other options fail: - Option (b) places iv lowest, contradicting the electron-withdrawing effect of NO2. - Option (c) places ii lowest and i highest, which is inconsistent with methoxy being electron-donating. - Option (d) places iv lowest, again contradicting NO2 electron withdrawal. Therefore, the correct answer is A.