See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the intermediate. The intermediate shown after tosylation has a cyclopentane ring where at C1: methyl is on a wedge (coming forward) and H is on a dash (going back); at C3: OTs is on a wedge (coming forward) and H is on a dash (going back). Both methyl and OTs are on the same face (cis relationship, both on wedge bonds). Step 2 - Determine Reactant X. Tosylation of an alcohol with TsCl/pyridine proceeds with retention of configuration at the carbon bearing the OH group. Therefore, the OH in reactant X must have the same configuration as OTs in the intermediate. In the intermediate, OTs is on a wedge at C3 with H on a dash. So X must have OH on a wedge at C3 with H on a dash. Also, the CH3 at C1 is on a wedge with H on a dash. This matches option (b): cyclopentane with H3C on wedge at C1, H on dash at C1, OH on wedge at C3, H on dash at C3 — meaning both CH3 and OH are on the same face (cis). Step 3 - Determine Product Y. The reaction of the tosylate with NaN3 in ethanol-water is an SN2 reaction. SN2 proceeds with inversion of configuration at the carbon under attack (C3). In the intermediate, OTs is on a wedge at C3 (coming forward). After SN2 inversion by azide, N3 will replace OTs with inversion, so N3 goes to the dash position at C3, and H remains on the wedge at C3. This gives product Y: cyclopentane ring with H3C on wedge at C1, H on dash at C1, H on wedge at C3, N3 on dash at C3. Step 4 - Match to options. Option (b) shows Reactant X with H3C wedge at C1, OH wedge at C3 (cis), and Product Y with H3C wedge at C1, H wedge at C3, N3 dash at C3 — consistent with SN2 inversion. Step 5 - Eliminate other options. Option (a) product Y retains the configuration (N3 on wedge), which would be retention — not SN2. Options (c) and (d) have the wrong reactant X stereochemistry (OH on dash, not matching the intermediate's OTs on wedge). Therefore, the correct answer is B.