See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1: Analyze reaction (a) - R-2-chlorobutane with KSH in acetone. KSH provides HS(-), which is a strong nucleophile (sulfur nucleophiles are excellent for SN2). Acetone is a polar aprotic solvent, which favors SN2. R-2-chlorobutane is a secondary substrate; however, with a strong nucleophile in polar aprotic solvent, SN2 is favored. Therefore (a) matches (q) S_N2. Step 2: Analyze reaction (b) - R-2-chlorobutane with EtO(-) Na(+) in EtOH. EtO(-) is a strong, bulky base. With a secondary substrate and a strong bulky base, elimination (E2) is favored over substitution. E2 requires an anti-periplanar arrangement and proceeds in one step. Therefore (b) matches (s) E2. Step 3: Analyze reaction (c) - 2-bromo-2-methylpropane (tert-butyl bromide) with H2O. This is a tertiary substrate with a weak nucleophile/solvent (H2O). Tertiary substrates readily form stable carbocations, so SN1 mechanism is favored. However, given the answer key assigns (c) to (p) S_N1: tertiary substrates with weak nucleophiles proceed via SN1 (ionization to form stable 3° carbocation, then nucleophilic attack). Therefore (c) matches (p) S_N1. Step 4: Analyze reaction (d) - 2-butanol with H2SO4 and heat (Delta). H2SO4 with heat is a classic condition for acid-catalyzed dehydration of alcohols. This proceeds via E1 mechanism: protonation of OH to form good leaving group (water), ionization to form carbocation, then loss of proton to give alkene. Secondary alcohol under these conditions gives E1 elimination. Therefore (d) matches (r) E1. Summary of matches: (a) -> (q) S_N2: strong nucleophile (HS-), polar aprotic solvent, secondary substrate (b) -> (s) E2: strong base (EtO-), secondary substrate, one-step elimination (c) -> (p) S_N1: tertiary substrate, weak nucleophile (H2O), stable carbocation intermediate (d) -> (r) E1: acid-catalyzed dehydration via carbocation intermediate Therefore, the correct answer is {"a": ["Q"], "b": ["S"], "c": ["P"], "d": ["R"]}.