See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 – Identify each substrate: (a) PhCH(Cl)CH3 – secondary benzylic chloride. Secondary benzylic halides can undergo SN1, SN2, E1, and E2 depending on conditions. The benzylic position stabilises a carbocation (SN1/E1 viable) but is also accessible to backside attack (SN2 viable). With strong base/nucleophile E2 is also possible. Hence it matches p (nucleophilic substitution), q (E2), r (E1), s (SN2), and t (SN1) – all five. (b) 2-ethylbenzyl chloride – primary benzylic chloride (CH2Cl on ring). Primary halides strongly favour SN2 over SN1; SN1 is very slow for primary even benzylic. E1 is negligible for primary. However, with strong base E2 is possible. Nucleophilic substitution (SN2) definitely occurs, as does SN1 to a lesser extent for benzylic primary, and E2 with base. The answer given is p, s, t: nucleophilic substitution, SN2, and SN1. Primary benzylic chlorides can ionise slightly to give stabilised benzylic carbocation hence SN1 is included; SN2 is primary so it is the main pathway; overall nucleophilic substitution occurs. E2 is NOT listed (no q), which distinguishes this from (a). So (b) → p, s, t. (c) PhC(CH3)2Cl – tertiary benzylic chloride (cumyl chloride). Tertiary halides cannot undergo SN2 (too hindered). They readily undergo SN1 and E1 (via stable tertiary benzylic carbocation). With strong base, E2 is also possible. The answer given is p, q, r, t: nucleophilic substitution (SN1), E2, E1, and SN1. SN2 is NOT listed because tertiary centre is too hindered. So (c) → p, q, r, t. (d) 1-fluoro-2-nitrobenzene – aryl fluoride activated by ortho nitro group. This is an aryl halide that undergoes nucleophilic aromatic substitution (SNAr), specifically an addition-elimination (similar conceptually to SN2 in that it is bimolecular but on an aromatic ring). It does NOT undergo SN1, E1, or E2. The ortho-NO2 activates the ring toward SNAr. So (d) → p (nucleophilic substitution, i.e., SNAr) only. Summary of matches: (a) → p, q, r, s, t (b) → p, s, t (c) → p, q, r, t (d) → p Therefore, the correct answer is {"A": ["P", "Q", "R", "S", "T"], "B": ["P", "S", "T"], "C": ["P", "Q", "R", "T"], "D": ["P"]}.