See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

We address each part using core concepts of nucleophilicity, basicity, and leaving group ability. (a) Stronger nucleophile in APROTIC solvent — F- or I-: In aprotic solvents (e.g., DMSO, DMF, acetone), ions are not solvated by hydrogen bonding. Nucleophilicity therefore follows intrinsic electron density / basicity. F- is a harder, more electron-dense anion with higher charge density, making it a stronger nucleophile than I- in aprotic media. Therefore: F-. (b) Stronger nucleophile in PROTIC solvent — F- or I-: In protic solvents (e.g., water, alcohols), smaller, more charge-dense anions are more heavily solvated by hydrogen bonding and thus less available to attack an electrophile. F- is heavily solvated and effectively 'caged,' while I- is large and polarizable with low charge density, so it is poorly solvated and free to act as a nucleophile. Nucleophilicity in protic solvents follows polarizability: I- > F-. Therefore: I-. (c) Stronger base — F- or I-: Base strength is determined by the pKa of the conjugate acid. HF has pKa ≈ 3.2, while HI has pKa ≈ -10. A stronger acid has a weaker conjugate base. Since HF is a much weaker acid than HI, F- is a much stronger base than I-. Therefore: F-. (d) Stronger nucleophile in PROTIC solvent — NH3 or NH2NH2 (hydrazine): Both are neutral nitrogen nucleophiles. NH2NH2 (hydrazine) is a stronger nucleophile than NH3 because the lone pair on the attacking nitrogen is made more electron-rich (more available) by the alpha-effect: the adjacent nitrogen's lone pair raises the energy of the HOMO on the attacking nitrogen, enhancing its nucleophilicity. This alpha-effect is independent of solvation considerations. Therefore: NH2NH2. (e) Better leaving group — CH3COO- or CH3SO3- (methyl sulfonate): Leaving group ability correlates with the stability of the departing anion (i.e., ability to accommodate the negative charge). CH3SO3- (methanesulfonate, mesylate) is derived from methanesulfonic acid (pKa ≈ -1.9), a very strong acid, meaning the anion is extremely stable. CH3COO- (acetate) is derived from acetic acid (pKa ≈ 4.75), a much weaker acid, so acetate is a poorer leaving group. The more stable (weaker base) anion is the better leaving group: CH3SO3- >> CH3COO-. Therefore: CH3SO3-. Therefore, the correct answer is {"a": "F-", "b": "I-", "c": "F-", "d": "NH2-NH2", "e": "CH3SO3-"}.