See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: Leaving group ability in nucleophilic substitution (SN1/SN2) depends on the stability of the departing anion. A more stable (weaker base) anion is a better leaving group. Stability is governed by: (1) inductive/field effects of electron-withdrawing groups, (2) resonance delocalization of the negative charge, and (3) the pKa of the conjugate acid (lower pKa of conjugate acid = more stable anion = better leaving group). Step 1 – Analyze anion I (CF3SO3^-, triflate): The CF3 group has three highly electronegative fluorine atoms exerting a strong electron-withdrawing inductive effect, combined with the three oxygen atoms of the sulfonate group delocalizing the negative charge over three oxygens. Triflate is one of the most stable anions known. Conjugate acid (triflic acid) has pKa ≈ -14. Triflate is the best leaving group. Step 2 – Analyze anion II (C6H5SO3^-, benzenesulfonate): Benzenesulfonate also has three oxygens delocalizing the negative charge, similar to triflate, but the phenyl group is electron-withdrawing by resonance yet not as strongly withdrawing as CF3 by induction. Conjugate acid (benzenesulfonic acid) has pKa ≈ -1 to -2. Benzenesulfonate is an excellent leaving group, second only to triflate among those listed. Step 3 – Analyze anion IV (CH3COO^-, acetate): Acetate has resonance delocalization of the negative charge over two oxygen atoms (carboxylate resonance). Conjugate acid (acetic acid) has pKa ≈ 4.76. Acetate is a relatively poor leaving group compared to sulfonates but still better than phenoxide because carboxylate resonance provides more delocalization than phenoxide. Step 4 – Analyze anion III (C6H5O^-, phenoxide): Phenoxide has resonance delocalization into the aromatic ring, but the negative charge is primarily on oxygen with partial delocalization into the ring. Conjugate acid (phenol) has pKa ≈ 10. Phenoxide is a stronger base than acetate and therefore a worse leaving group. Phenoxide is essentially not a leaving group in SN2 reactions on sp3 carbons. Step 5 – Rank leaving group ability (best to worst): I (triflate, pKa of conjugate acid ≈ -14) > II (benzenesulfonate, pKa ≈ -1) > IV (acetate, pKa ≈ 4.76) > III (phenoxide, pKa ≈ 10) This gives the order I > II > IV > III. Why other options fail: (a) I > II > III > IV places phenoxide above acetate, which is wrong because acetate (pKa ~4.76) is a weaker base and therefore a better leaving group than phenoxide (pKa ~10). (c) IV > I > II > III incorrectly places acetate as the best leaving group. (d) IV > III > II > I is entirely reversed from the correct reasoning. Therefore, the correct answer is B.