See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: The reaction of a tosylate (OTs) with a carboxylate nucleophile (RCO2H / RCOO-) proceeds via an SN2 mechanism, which involves backside attack and results in inversion of configuration at the carbon bearing the leaving group (OTs). The question uses stereochemical labeling (filled wedge = above plane, open circle/dash = below plane) and isotopic labeling to track the stereochemical outcome. Step 1: Identify the starting material stereochemistry. The carbon bearing OTs has: phenyl on a wedge (coming toward viewer, labeled carbon) and OTs on a dash/open circle (going away from viewer, no label). The cyclohexane ring provides the other two substituents. Step 2: SN2 mechanism. The carboxylate (RCOO-) attacks the back face of the C-OTs bond. Since OTs is on a dash (going away from viewer), backside attack occurs from the front (wedge side). This results in inversion: the new OCOR group ends up on the dash side, and the phenyl group ends up on the wedge side — wait, let me reconsider the geometry. Step 3: More carefully — OTs is on an open circle (dash, behind plane). Nucleophile attacks from the front (wedge direction). After inversion, the OCOR attaches from the front and ends up behind (dash), while Ph which was on wedge remains... Actually inversion means if OTs was behind (open circle/dash), after SN2 the nucleophile comes from in front, and the configuration inverts: Ph moves from wedge to dash, OCOR ends up on wedge. This matches option (b). Step 4: But the question shows BOTH (a) and (b) are correct (answer = C). This is because the molecule has TWO stereocenters involved or because there is an internal nucleophile / neighboring group participation. Given the labeled carbon (isotopic label on the Ph-bearing carbon) and the 'no label' on the OTs carbon, it appears these are TWO DIFFERENT carbons — one carbon bears Ph (labeled) and an adjacent carbon bears OTs (no label). Step 5: With neighboring group participation — the labeled carbon bearing Ph can assist via an episulfonium-like or phenonium ion intermediate (phenyl participation). The phenyl group migrates from the labeled carbon to the adjacent carbon bearing OTs, forming a bridged phenonium ion intermediate. The carboxylate then opens this intermediate from either face. Step 6: Opening at the original OTs carbon (path 1) gives product (a): Ph ends up on the no-label carbon (inverted), OCOR on the labeled carbon — this gives retention of overall configuration relative to original Ph position but with label scrambling. Opening at the labeled carbon (path 2) gives product (b): OCOR on the no-label carbon, Ph on labeled carbon with inverted configuration. Both products form in equal amounts due to the symmetric phenonium intermediate. Step 7: Therefore both (a) and (b) are formed, making option (c) correct. Options (a) alone or (b) alone would be incorrect because the phenonium intermediate is symmetrically opened. Option (d) is incorrect because definite products are formed. Therefore, the correct answer is C.