See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Concept: When a compound bearing both a thioester (S-C(=O)-CH3) and an acetate ester (O-C(=O)-CH3) on adjacent carbons of a bicyclic framework is treated with hydroxide (HO-), selective hydrolysis and intramolecular cyclization can occur. Step 1 - Identify the stereochemistry of the starting material: The S-acetyl group is on a wedge bond (above the plane) at one carbon, and the O-acetyl group is on a dashed/hashed bond (below the plane) at the adjacent carbon. This means S and O are on opposite faces (trans to each other) on the bicyclic ring system. Step 2 - Reaction with HO-: Hydroxide first hydrolyzes the thioester selectively (thioesters are more reactive toward hydrolysis than oxygen esters under mild basic conditions), generating a thiolate anion (RS-) in situ. Alternatively, the thiolate is liberated and the acetate ester remains momentarily. Step 3 - Intramolecular displacement: The thiolate anion, now on one face of the ring, performs an intramolecular backside (SN2) attack on the carbon bearing the acetate (leaving group = acetate, OAc-). For an intramolecular SN2 to occur, the nucleophile (S-) must approach from the back face relative to the leaving group (OAc). Since S is trans to OAc on the bicyclic system, this backside attack geometry is accessible, leading to inversion at the carbon bearing OAc. Step 4 - Product formation: The intramolecular SN2 by the thiolate on the adjacent carbon bearing the acetate displaces acetate and forms a three-membered sulfur-containing ring (episulfide/thiirane) bridging the two carbons. This gives a bicyclic compound fused with an episulfide ring. Step 5 - Why other options fail: - (a) and (b): These show free SH and OH groups (trans and cis respectively), which would result from simple hydrolysis of both esters without cyclization. Under the reaction conditions with the trans geometry, intramolecular cyclization is favored over simple double hydrolysis. - (b): Wrong stereochemistry even for a diol/dithiol product. - (d): An epoxide would require oxygen to displace, but oxygen is less nucleophilic than sulfur, and the thiolate is a much better nucleophile for SN2 reactions. - The thiolate (S-) is a far superior nucleophile compared to an alkoxide for SN2, making episulfide formation (option c) strongly preferred over epoxide formation (option d). Therefore, the correct answer is C.