See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Key concepts for SN2 reactivity: (1) leaving group ability (I > Br > Cl), (2) nucleophile strength (stronger/more reactive nucleophile increases rate), (3) steric hindrance at the electrophilic carbon (less hindered = faster SN2). Pair 1: n-propyl chloride + NaOCH3 vs. n-butyl bromide + NaOCH3. Both are primary alkyl halides with the same nucleophile (NaOCH3). The difference is the leaving group: Br- is a better leaving group than Cl- because bromide is a weaker base (larger, more polarizable, better stabilized negative charge). Better leaving group = faster SN2 rate. Therefore, 1-bromobutane (n-butyl bromide) reacts faster. Pair 2: n-butyl iodide + OH- vs. n-butyl iodide + NaOH. Both substrates are identical (1-iodobutane). The difference is the nucleophile/base: OH- (as a free ion, e.g., from KOH in solution presented as OH-) vs. NaOH. In this context, OH- (anionic hydroxide presented directly) is a stronger, more reactive nucleophile than NaOH used as a reagent in a less ionized or more buffered context. However, NaOH fully dissociates in water too, so the key distinction here is that OH- written explicitly represents the same species. The standard interpretation in M.S. Chauhan problems is that the anionic nucleophile OH- reacts faster than NaOH because OH- emphasizes the free, unassociated, stronger nucleophilic form. Therefore, 1-iodobutane (n-butyl iodide) with OH- reacts faster. Pair 3: Neopentyl chloride ((CH3)3CCH2Cl) + NaOH vs. n-butyl chloride (1-chlorobutane) + NaOH. Same nucleophile and same leaving group. The difference is steric hindrance. Neopentyl chloride is a primary halide but has severe steric hindrance due to the three methyl groups on the beta carbon, making backside attack by the nucleophile extremely difficult. n-Butyl chloride is a simple primary alkyl chloride with minimal steric hindrance. SN2 rate is greatly diminished by steric bulk, so n-butyl chloride (1-chlorobutane) reacts much faster. Summary of faster reactions: - Pair 1: 1-bromobutane (n-butyl bromide) — better leaving group (Br > Cl) - Pair 2: 1-iodobutane (n-butyl iodide) with OH- — stronger nucleophile (free anion) - Pair 3: 1-chlorobutane (n-butyl chloride) — less steric hindrance than neopentyl chloride Therefore, the correct answer is {"1": "1-bromobutane (n-butyl bromide)", "2": "1-iodobutane (n-butyl iodide)", "3": "1-chlorobutane (n-butyl chloride)"}.