See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step-by-step analysis of each part: (a) SN2 reaction comparison: Concept: SN2 reactions proceed via a backside attack transition state. The rate depends critically on steric accessibility of the electrophilic carbon. Less hindered (less substituted) carbons react faster in SN2. First structure: cyclopentylethyl bromide — the Br is on a primary carbon at the end of a two-carbon chain attached to cyclopentane (cyclopentyl-CH2-CH2-Br). The carbon bearing Br is primary and relatively unhindered (the cyclopentyl group is two carbons away). Second structure: 1-bromo-1-ethylcyclopentane — the Br is directly on a tertiary carbon (C1 of cyclopentane, which also bears an ethyl group). This is a tertiary alkyl bromide, extremely hindered toward backside attack. Reasoning: Primary substrates react far faster than tertiary substrates in SN2. The cyclopentylethyl bromide (primary) will undergo SN2 much more rapidly than 1-bromo-1-ethylcyclopentane (tertiary, essentially inert to SN2). Why the other fails: Tertiary halides have three substituents blocking the backside of the C–Br bond, making SN2 essentially impossible. (b) SN1 reaction comparison: Concept: SN1 reactions depend on the stability of the carbocation intermediate formed after ionization. The leaving group also affects the rate: better leaving groups (lower C–X bond strength, more stable X⁻) increase SN1 rate. The carbon skeleton is identical in both (tertiary butyl), so the leaving group is decisive. (CH3)3C-Br vs (CH3)3C-I: Both form a tertiary carbocation. The C–I bond is weaker than C–Br (bond dissociation energy: C–I < C–Br), and I⁻ is a better leaving group than Br⁻ (I⁻ is a weaker base, more stable anion). Therefore, (CH3)3C–I ionizes faster. Reasoning: The iodide is a superior leaving group relative to bromide. With identical carbocation stability, (CH3)3C–I undergoes SN1 more rapidly. Why the other fails: (CH3)3C–Br forms the same carbocation but the C–Br bond is stronger and Br⁻ is a poorer leaving group than I⁻, so it ionizes more slowly. (c) SN1 reaction comparison: Concept: SN1 rate depends on carbocation stability. In bicyclic systems, bridgehead carbocations are extremely destabilized because Bredt's rule prevents the needed planar geometry at a bridgehead for sp2 hybridization (the ring constraints prevent flattening). A secondary (non-bridgehead) carbocation in the norbornane system is far more accessible. First structure: 2-bromonorbornane — Br is at C2 (a secondary, non-bridgehead carbon). Ionization gives a secondary norbornyl carbocation, which, while strained, can still form and is even assisted by neighboring group participation (non-classical carbocation stabilization). This is feasible. Second structure: 1-bromonorbornane — Br is at C1, the bridgehead. Ionization would require a bridgehead carbocation, which violates Bredt's rule for small bicyclic systems. This carbocation is essentially inaccessible due to ring geometry preventing sp2 hybridization at the bridgehead. Reasoning: 2-bromonorbornane can ionize to give a (non-classical) norbornyl cation; 1-bromonorbornane cannot ionize to give a Bredt-rule-violating bridgehead cation. Hence 2-bromonorbornane undergoes SN1 far more rapidly. Why the other fails: Bridgehead bromide in norbornane is essentially inert to SN1 because the resulting bridgehead carbocation is geometrically impossible in the small bicyclic framework. Therefore, the correct answer is {"a": "cyclopentylethyl bromide (cyclopentane-CH2-CH2-Br)", "b": "(CH3)3C-I (tert-butyl iodide)", "c": "2-bromonorbornane (Br on a ring carbon adjacent to the bridgehead, C2)"}.