See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: SN2 reaction proceeds with inversion of configuration (Walden inversion) at the carbon bearing the leaving group. Step 1 – Identify the electrophilic carbon. The leaving group is Cl, which is on C2 (the bottom carbon). The nucleophile PhS(-) will attack C2. Step 2 – Establish the configuration at C2 in the starting material. In the drawn structure, C2 has: Cl on the left (coming toward viewer, wedge), H on the right (coming toward viewer, wedge), Ph pointing downward, and the bond to C1 going upward. This defines a specific spatial arrangement (one particular stereoisomer). Step 3 – Apply SN2 inversion at C2. The nucleophile PhS(-) attacks from the back face (anti to Cl). This inverts the configuration at C2. Whatever groups were on wedge bonds (toward viewer) on C2 now go to dash bonds (away from viewer) relative to the incoming nucleophile, and the SPh group takes the position previously occupied by Cl but from the opposite face. Step 4 – Determine the product stereochemistry at C2. In the starting material, Cl is on the left as a wedge and H is on the right as a wedge. After inversion at C2: the new arrangement has H on the left (wedge) and SPh on the right (wedge) — i.e., both H and SPh are on the same side (both wedge/toward viewer), with Ph at the bottom. This matches option (b). Step 5 – C1 is unaffected. C1 retains its configuration: Ph on top, H on left (wedge), CH3 on right (wedge). This is consistent with option (b). Why other options fail: - Option (a): PhS is on the left as a wedge along with H on the right as wedge — this would correspond to retention of configuration at C2, not inversion. SN2 always inverts. - Option (c): The substituents at C1 are changed (Ph replaces H on left), which is wrong since C1 is not the site of reaction. - Option (d): Ph replaces Cl, which would imply a different nucleophile or no reaction with PhS(-). Therefore, the correct answer is B.