See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: In the Finkelstein reaction, NaI in acetone selectively replaces C-Br bonds over C-F bonds because NaI is a good source of iodide (a soft, polarizable nucleophile), and the driving force is the precipitation of NaBr (insoluble in acetone), while NaF is soluble in acetone. C-F bonds are generally not reactive toward SN2 substitution under these conditions because fluorine is a very poor leaving group (C-F bond is very strong and F- is a poor leaving group in SN2). Step 1: Identify the leaving group selectivity. With NaI in acetone (Finkelstein conditions), only the C-Br bond reacts. The C-F bond does not react because F- is an extremely poor leaving group. Step 2: Determine the stereochemical outcome. The SN2 reaction at the C-Br carbon proceeds with inversion of configuration at that carbon. The starting material shows trans-1-bromo-3-fluorocyclopentane (Br and F are trans to each other). After SN2 inversion at the C-Br carbon, the iodide replaces bromine with inversion. If Br was on the dash (below plane) at C1, after inversion I goes to the wedge (above plane) at C1. F remains unchanged at C3 (on wedge). This gives I and F both on the same face (cis relationship) at C1 and C3. Step 3: Match to options. Option (b) shows the cyclopentane ring with I on wedge at C1 and F on wedge at C3 (both above the plane, cis to each other), which corresponds to inversion at the bromine-bearing carbon with retention of the fluorine-bearing carbon. This is consistent with SN2 inversion at C-Br. Why other options fail: - Option (a): Shows I on dash at C1 and F on wedge at C3, meaning I and F are trans - this would require retention at C-Br, not inversion, so it is incorrect for SN2. - Option (c): Shows Br still present and I at C3, implying substitution at C-F instead of C-Br, which does not occur under Finkelstein conditions. - Option (d): Shows both Br and I present in a manner inconsistent with selective C-Br substitution giving the correct stereochemistry. Therefore, the correct answer is B.