See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

SN2 reactivity depends on steric hindrance at the carbon bearing the leaving group, the leaving group ability, and electronic effects such as allylic/benzylic activation. (a) SN2 with NaOCOCH3 in methanol on three bromides: - Compound 1: CH3CH2CH2Br — primary, unactivated alkyl bromide. - Compound 2: (CH3)2CHBr — secondary, more sterically hindered than primary, slower SN2. - Compound 3: CH2=CHCH2Br — allylic primary bromide; the allylic position is both primary AND activated by resonance (allylic system lowers the transition state energy for SN2), making it the most reactive. Order: 3 (allylic primary, most reactive) > 1 (normal primary) > 2 (secondary, least reactive). Answer: 3>1>2. (b) SN2 with NaI in acetone on three chlorides: - Compound 1: C6H5Cl — aryl chloride; SN2 at sp2 carbon is essentially impossible due to geometry and ring electron density. Least reactive. - Compound 2: C6H5CH2Cl — benzylic primary chloride; activated by benzylic resonance and primary carbon, most reactive for SN2. - Compound 3: C6H5CHClCH3 — benzylic secondary chloride; benzylic activation present but secondary carbon adds more steric hindrance compared to compound 2. Order: 2 (benzylic primary) > 3 (benzylic secondary) > 1 (aryl, SN2 inactive). Answer: 2>3>1. (c) SN2 with NaCN in methanol on three ethyl halides (same carbon skeleton, different halogens): - All are primary (ethyl), so steric effects are identical. Reactivity is governed by leaving group ability. - Leaving group order for SN2: I > Cl > F (iodide is the best leaving group; fluoride is the worst due to high C–F bond strength). - Compound 3: CH3CH2I — iodide leaving group, best. Most reactive. - Compound 1: CH3CH2Cl — chloride leaving group, intermediate. - Compound 2: CH3CH2F — fluoride leaving group, worst (C–F bond very strong, poorest leaving group). Order: 3 (iodo) > 1 (chloro) > 2 (fluoro). Answer: 3>1>2. (d) SN2 with NaSCH3 in methanol on three bromides: - Compound 1: (CH3)2CHCH2CH2Br — primary bromide, the reactive carbon (C1) is a normal primary carbon; the branching is on C3 (isobutyl-type chain extended), so steric hindrance at the reaction center is relatively low. - Compound 2: CH3CH2CHBrCH2CH3 — secondary bromide (C3 of pentane); more sterically hindered than primary, slower SN2. - Compound 3: (CH3)3CCH2Br — neopentyl bromide; although the reacting carbon is primary, it has a quaternary carbon directly adjacent (beta-branching), causing extreme steric hindrance (neopentyl system), making it the least reactive in SN2. Order: 1 (primary, least hindered) > 2 (secondary) > 3 (neopentyl, most hindered primary). Answer: 1>2>3. Therefore, the correct answer is {"a": "3>1>2", "b": "2>3>1", "c": "3>1>2", "d": "1>2>3"}.