See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the reaction type: NaI in DMSO (a polar aprotic solvent) is a classic SN2 reagent system. Iodide is an excellent nucleophile, and DMSO does not solvate the nucleophile, making it highly reactive. The leaving group is the tosylate (OTs), which is an excellent leaving group for SN2 reactions. Step 2 - Identify the electrophilic center: The OTs group is on the middle carbon (C2). This is the site of SN2 attack. The other carbon bearing OBn (PhCH2O) is not an electrophilic center under these conditions (ethers are not displaced by SN2 with NaI under mild conditions). Step 3 - Determine the stereochemical outcome: SN2 reaction proceeds with inversion of configuration (Walden inversion) at the carbon bearing the leaving group. The middle carbon has OTs on a dashed bond (going back) and H on a wedge bond (coming forward) in the starting material. After SN2 inversion, iodide attacks from the front face (opposite to the leaving group), so I will now appear on the wedge side and H will move to the dash side - wait, let me re-examine. In the starting material the middle carbon: H is on wedge (front), OTs is on dash (back). SN2 attack by I- occurs from the front (where H is on wedge, opposite to OTs on dash side). After inversion, I ends up where OTs was (back/dash) and the configuration inverts. Actually, I- attacks from the back side opposite OTs. OTs is on dash (back), so I- attacks from the front, and after inversion, I occupies the position formerly held by OTs but with inverted geometry: I now on wedge? Re-examining answer (c): middle carbon has I on wedge and H on dash - this represents inversion at the middle carbon relative to starting material where H was on wedge and OTs was on dash. This is consistent with SN2 inversion. Step 4 - Confirm the OBn group is retained: In answer (c), the left carbon still bears OCH2Ph (OBn group is intact), which is correct because NaI/DMSO only displaces the tosylate. Step 5 - Confirm the dimethyl acetal is retained: The right carbon C(OCH3)2 is unchanged in answer (c), consistent with no reaction at that center. Step 6 - Eliminate other options: (a) shows I replacing OBn, not OTs - incorrect, OBn is not displaced. (b) shows I on a dash at the middle carbon but the geometry appears to retain rather than invert relative to the starting material, or the left carbon geometry is inconsistent. (d) shows OTs still present, suggesting no reaction or wrong center attacked, and I appears elsewhere inconsistently. Step 7 - Conclusion: The SN2 reaction of NaI/DMSO selectively displaces the tosylate at the middle carbon with inversion of configuration, leaving the OBn and dimethyl acetal groups intact. Answer (c) correctly shows I on wedge at the middle carbon (inversion of the configuration where OTs was on dash), with OCH2Ph retained on the left carbon and C(OCH3)2 retained on the right carbon. Therefore, the correct answer is C.