See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: KSH is a nucleophilic thiol reagent (HS⁻ as nucleophile). The starting material appears to be a secondary alkyl halide (or equivalent leaving group) — in this context the structure drawn represents a carbon bearing a leaving group (implied halide, likely the 'D' position represents a leaving group or more likely the molecule is an alkyl halide where the leaving group must be identified). Re-examining: the starting material has Me, H, Et, D, and H on a single carbon — this is a 5-substituent depiction which indicates a two-carbon or the structure shows two adjacent carbons in a wedge-dash notation. Actually, the structure shown is a two-carbon wedge-dash (Newman-like or sawhorse): top carbon bears Me and H, bottom carbon bears Et, D, and H. This is a vicinal substitution scenario, or more likely the molecule is a haloalkane where one substituent is a halogen (not shown explicitly but implied). Re-reading: The structure is a single carbon with Me (back-left wedge), H (back-right), Et (left horizontal), D (right horizontal), H (bottom) — this gives 5 groups on one carbon which is impossible. Therefore this must be a two-carbon structure drawn as a cross: upper carbon has Me (top-left) and H (top-right), lower portion has Et (left), D (right), H (bottom), sharing the central point. This represents a molecule like CH3-CHX-CH(Et)D or similar, where the leaving group is implicit. Step 1: The most reasonable interpretation is that this is a secondary alkyl halide (the halide being implicit, perhaps at the position marked or the structure is an epoxide/halide). Given KSH reacts via SN2, it inverts configuration at the carbon bearing the leaving group. Step 2: SN2 reaction proceeds with inversion of configuration (Walden inversion). The nucleophile HS⁻ attacks the back face of the carbon bearing the leaving group, replacing it with inversion. Step 3: Looking at the starting material — the central carbon (with Et, D, H substituents on the lower part) bears the leaving group. The upper carbon (Me, H) is not the reaction site. The SH replaces the leaving group (implied halide) with inversion. Step 4: In the starting material, the arrangement at the reacting carbon is Et (left), D (right), H (bottom), leaving group (top, toward upper carbon). After SN2 inversion, SH comes from the top and the remaining groups invert: what was Et on left stays relatively but the configuration inverts — D and H swap relative positions. Step 5: Comparing answer choices: Option (d) shows SH on top, H on left, Me on right, Et on lower-left, D on lower-right, H on bottom — this represents the inverted product where SH has replaced the leaving group with inversion at the stereocenter, placing SH in the position opposite to where the leaving group was, with H and Me swapping to reflect inversion. Step 6: Options (a), (b), (c) fail because: (a) shows retention of configuration (SH in same position as leaving group without inversion); (b) places SH at the bottom suggesting a different mechanism; (c) places HS on the left with Me on right which doesn't correspond to correct inversion product. Therefore, the correct answer is D.