See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1: Analyze reaction (a) - Ph-C(Et)(Me)-Cl with H2O. This is a tertiary benzylic chloride reacting with water (a weak nucleophile/polar protic solvent). Tertiary substrates with benzylic stabilization strongly favor S_N1 mechanism. In S_N1, the rate-determining step is ionization to form a carbocation intermediate. The benzylic/tertiary carbocation is highly stabilized. Therefore (a) matches S_N1 (p) and carbocation is intermediate (r). Step 2: Analyze reaction (b) - n-butyl chloride (primary alkyl chloride) with PhSNa (sodium thiophenoxide). PhSNa is a strong, excellent nucleophile (sulfur nucleophiles are very good in S_N2). Primary alkyl halides with strong nucleophiles undergo S_N2. There is no significant carbocation or carbanion intermediate in S_N2. Therefore (b) matches S_N2 (q) only. Step 3: Analyze reaction (c) - o-fluoronitrobenzene with KOH/heat. Aryl fluorides normally resist nucleophilic substitution, but with an ortho (or para) nitro group, nucleophilic aromatic substitution (SNAr) is activated. The mechanism of SNAr proceeds through a Meisenheimer complex (carbanion-like anionic intermediate, an addition-elimination mechanism). The nitro group stabilizes the negative charge on the ring in the Meisenheimer complex, which is a carbanion intermediate. Therefore (c) matches carbanion is intermediate (s). Step 4: Analyze reaction (d) - 2-butene (alkene) with Br2/CCl4. This is electrophilic addition of bromine to an alkene. The mechanism involves formation of a bromonium ion (cyclic carbocation-like intermediate) as the key intermediate - specifically a cyclic bromonium ion which is a three-membered ring with a positive charge on bromine, but the carbons bear partial positive character (carbocation character). The bromonium ion is considered a carbocation intermediate in this context. Therefore (d) matches carbocation is intermediate (r). Summary of matches: - (a) → (p) S_N1 and (r) Carbocation is intermediate - (b) → (q) S_N2 - (c) → (s) Carbanion is intermediate - (d) → (r) Carbocation is intermediate Therefore, the correct answer is {"A": ["P", "R"], "B": ["Q"], "C": ["S"], "D": ["R"]}.