See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step-by-step solutions for each part: (a) SN2 reaction rate concept: SN2 reactions proceed via backside attack by the nucleophile on the carbon bearing the leaving group. The rate is highly sensitive to steric hindrance. A primary alkyl halide (less hindered) reacts faster than a secondary alkyl halide (more hindered). CH3-Br is a primary (methyl) halide with no steric bulk around the carbon, while CH3-CH(Br)-CH3 (isopropyl bromide) is secondary with two methyl groups creating steric hindrance. Therefore CH3-Br undergoes SN2 more rapidly. Why the other fails: isopropyl bromide has two flanking methyl groups that block backside attack, significantly slowing SN2. (b) SN1 reaction rate concept: SN1 reactions proceed via carbocation intermediate formation. More substituted carbocations are more stable (hyperconjugation, inductive donation). A secondary carbocation (from isopropyl bromide) is more stable than a primary/methyl carbocation (from CH3-Br). Therefore isopropyl bromide, CH3-CH(Br)-CH3, undergoes SN1 more rapidly. Why the other fails: CH3-Br would form a methyl carbocation, which is extremely unstable and essentially does not form under normal SN1 conditions. (c) E2 reaction to give (Z)-1,2-diphenylpropene concept: E2 requires anti-periplanar geometry between H and the leaving group (Br). The product (Z)-1,2-diphenylpropene has Ph and H on the same side (cis) at C1, and Ph and CH3 on C2. Working backward using the Bredt/stereochemical analysis: in the Fischer projection with Ph(top)/H-Br(C1)/H-CH3(C2)/Ph(bottom), the anti relationship between H and Br requires them to be anti-periplanar. In a Fischer projection, groups on the same side are syn (eclipsed conformation) and groups on opposite sides are anti. For E2 elimination to give the Z alkene, the correct stereoisomer is the one where H at C2 and Br at C1 are anti in the relevant conformation. Analysis of the first Fischer projection (Ph top, H on left at C1 with Br on right, H on left at C2 with CH3 on right, Ph bottom): the H (C2, left) and Br (C1, right) are on opposite sides of the Fischer projection, meaning they are anti in the extended chain conformation, satisfying E2 anti-periplanar requirement and giving the Z product. The second Fischer projection gives a different stereochemical relationship leading to the E product. Therefore the first Fischer projection is correct. (d) Reaction with NaI to give (Z)-1,2-diphenylpropene concept: NaI in acetone effects an SN2 reaction (Finkelstein-type), replacing one Br with I, generating a vinyl bromide or an iodo compound that then undergoes elimination. More specifically, NaI acts as a nucleophile in SN2 to give an iodide, which then undergoes E2-like elimination. The dibromide that reacts with NaI (one equivalent) to give (Z)-1,2-diphenylpropene must have the correct stereochemistry for anti elimination. For the first Fischer projection (Ph top, H-Br at C1, Br-CH3 at C2, Ph bottom): SN2 at C2 by I- inverts configuration at C2, then E2 anti elimination of HBr gives Z alkene. This stereochemical pathway is consistent with the first dibromide giving the Z product. The second dibromide (both Br on same side in Fischer) would give the E product after the same sequence. Therefore the first Fischer projection (Ph top / H-Br / Br-CH3 / Ph bottom) is correct. (e) SN1 reaction rate for naphthalene derivatives concept: SN1 rate depends on carbocation stability. The left structure has Br at C1 of a dihydronaphthalene where C1 is allylic to an endocyclic double bond but also part of a vinyl/allylic system - however the carbon bearing Br is sp2-like or has limited stabilization. The right structure has Br at C1 of 1,2,3,4-tetrahydronaphthalene (tetralin) where C1 is a benzylic sp3 carbon adjacent to the aromatic ring. A benzylic carbocation is highly stabilized by resonance with the aromatic ring. The left structure's carbocation at a vinyl/allylic position adjacent to a non-fully aromatic system is less stable than a fully benzylic carbocation. Therefore 1-bromo-1,2,3,4-tetrahydronaphthalene (right structure, Br on benzylic sp3 carbon) undergoes SN1 more rapidly. Why the other fails: the left structure's C1 bearing Br is at a vinylic or allylic position in a system that does not generate as stable a carbocation as a pure benzylic position adjacent to a fully aromatic benzene ring. Therefore, the correct answer is {"a": "CH3-Br", "b": "(CH3)2CHBr (isopropyl bromide)", "c": "Fischer projection: Ph (top) / H-Br / H-CH3 / Ph (bottom)", "d": "Fischer projection: Ph (top) / H-Br / Br-CH3 / Ph (bottom)", "e": "1-bromo-1,2,3,4-tetrahydronaphthalene (Br on the benzylic sp3 ring carbon)"}.