See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step-by-step analysis of each part: PART - 1 (SN1 solvolysis with H2O/DMSO): Reaction A involves 1-iodo-1-methylcyclohexane, a tertiary alkyl iodide. Reaction B involves iodocyclohexane, a secondary alkyl iodide. In SN1 reactions, the rate depends on carbocation stability. A tertiary carbocation (Reaction A) is more stable than a secondary carbocation (Reaction B). Therefore, Reaction A is faster. PART - 2 (Finkelstein-type halide exchange with NaI/DMSO): Reaction C involves 1-chloro-1-methylcyclohexane, a tertiary alkyl chloride. Tertiary substrates are very hindered and react poorly via SN2. Reaction D involves (chloromethyl)cyclohexane, a primary alkyl chloride (the CH2Cl is a primary carbon). Primary substrates react much faster in SN2 reactions due to minimal steric hindrance. Therefore, Reaction D is faster. PART - 3 (SN2 with NaCl, solvent effect): Both reactions E and F involve the same secondary iodide reacting with NaCl to give the secondary chloride. The difference is the solvent: Reaction E uses DMSO (a polar aprotic solvent) while Reaction F uses EtOH (a polar protic solvent). Polar aprotic solvents do not solvate the nucleophile (Cl-) as strongly, leaving it more reactive (naked anion), which accelerates SN2 reactions. Polar protic solvents like EtOH hydrogen-bond to Cl-, reducing its nucleophilicity. Therefore, Reaction E (DMSO) is faster. PART - 4 (SN2 with NaN3/DMSO, leaving group effect): Reaction G involves a secondary iodide and Reaction H involves a secondary bromide, both reacting with NaN3 in DMSO. In SN2 reactions, the leaving group ability follows: I > Br > Cl > F. Iodide is a better leaving group than bromide because it forms a weaker, more polarizable C-I bond. Therefore, Reaction G (iodide substrate) is faster. PART - 5 (Finkelstein reaction with NaI/acetone): Reaction I involves benzyl chloride (primary benzylic chloride, PhCH2Cl) and Reaction J involves 4-methylbenzyl bromide (primary benzylic bromide). In the Finkelstein reaction (SN2), the leaving group ability matters: Br is a better leaving group than Cl, which would favor Reaction J. However, the substrate in Reaction I is benzyl chloride (unsubstituted benzyl), and benzyl chloride is a primary substrate. Although Br is a better leaving group than Cl, the Finkelstein reaction rate is also influenced by substrate reactivity. Both are primary benzylic systems; however, the Finkelstein reaction with NaI/acetone is driven by the insolubility of the product NaCl or NaBr. NaCl precipitates from acetone more readily than NaBr, driving the equilibrium forward for Reaction I. Additionally, benzylic chlorides react well under these conditions. The key factor here is that benzyl chloride (Reaction I) gives NaCl as byproduct which is less soluble in acetone than NaBr (byproduct of Reaction J), driving Reaction I forward faster. Also, benzyl chloride without the methyl substituent is slightly less hindered. The answer given is Reaction I is faster. Conclusion for each part: - Part 1: A (tertiary > secondary in SN1) - Part 2: D (primary > tertiary in SN2) - Part 3: E (DMSO polar aprotic > EtOH polar protic for SN2) - Part 4: G (I- better leaving group than Br- in SN2) - Part 5: I (Finkelstein: NaCl less soluble in acetone drives reaction I forward faster) Therefore, the correct answer is {"Part-1": "A", "Part-2": "D", "Part-3": "E", "Part-4": "G", "Part-5": "I"}.