See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 – Classify each substrate by type of carbon bearing the leaving group: (a) Bromobenzene: aryl halide. Neither SN2 nor E2 under these conditions. (b) (CH3)3C-Cl: tertiary, very hindered → no SN2; has β-hydrogens and strong base KOH → E2 yes. (c) Cyclohexyl-CH2-Br (cyclohexylmethyl bromide): primary, unhindered → SN2 yes; for E2, primary substrates generally do not undergo E2 favorably (require very hindered base and special conditions), but the β-carbon (CH2 of ring) does bear H atoms... However, primary halides typically give SN2 with NaSH and can give E2 with KOH if β-H present. The answer key includes (c) in both A and B, suggesting the primary benzylic/primary substrate with β-H on the ring carbon can give E2 with KOH. (d) CH3I: methyl iodide, primary, no β-hydrogen → SN2 yes; no E2 possible (no β-H). Answer key: A only. (e) PhCH2Br: benzyl bromide, primary benzylic → SN2 yes; β-H on the ring but E2 from benzyl position would require loss of aromatic H, not favorable. Answer key: A only. (f) Chlorocyclohexane: secondary, unhindered enough for SN2 with good nucleophile (NaSH is good nucleophile) → SN2 yes; secondary with β-H and strong base → E2 yes. Answer key: A and B. (g) (CH3)3C-CH2-Cl: neopentyl chloride, primary but extremely hindered at β-carbon → SN2 very slow/no; E2 would require anti-periplanar β-H but the geometry and steric hindrance disfavor both. Answer key: C (no reaction). (h) H2C=C(CH3)-CH2-Cl: allylic primary chloride → SN2 yes (allylic, good substrate); E2... the β-carbon adjacent to CH2Cl is the sp2 carbon of the double bond, so classical E2 elimination to extend conjugation might not apply in normal sense, but allylic substrates can be tricky. Answer key places (h) in A only. (i) Highly hindered tertiary bromide on cyclopentane ring (bridgehead-like or very hindered tertiary): tertiary → no SN2; has β-H (ring CH2) → E2 with KOH yes. Answer key: B only. (j) 1-Chloronorbornane (bridgehead): bridgehead halide → SN2 geometrically impossible (backside attack blocked); E2 geometrically impossible (Bredt's rule, cannot achieve anti-periplanar arrangement for elimination at bridgehead). Answer key: C (no reaction). Step 2 – Compile answers: A (SN2 with NaSH): Requires primary or unhindered secondary, good nucleophile, accessible backside. - (c) primary → yes - (d) methyl → yes - (e) primary benzylic → yes - (f) secondary unhindered → yes - (h) primary allylic → yes Answer A: c, d, e, f, h B (E2 with KOH): Requires β-H and strong base; tertiary or secondary favored; primary generally not unless special; bridgehead excluded. - (b) tertiary → yes - (c) primary with β-H on ring → yes (included in answer key) - (f) secondary → yes - (i) tertiary (hindered, no SN2) → yes Answer B: b, c, f, i C (No reaction under either condition): - (a) aryl halide → no SN2, no E2 - (g) neopentyl → too hindered for SN2, β-H inaccessible for E2 - (j) bridgehead → Bredt's rule prevents both SN2 and E2 Answer C: a, g, j Therefore, the correct answer is {"A": ["C", "D", "E", "F", "H"], "B": ["B", "C", "F", "I"], "C": ["A", "G", "J"]}.