See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Concept: The starting material is a spiro or fused tricyclic system containing a cyclopropane, an epoxide (with 18O), and an episulfide (thiirane). When treated with a Grignard reagent (RMgX), selective ring opening occurs. When treated with CH3I, alkylation of the resulting anion/nucleophile occurs. Step 1 - Reactivity of the starting material with RMgX: The molecule has both an epoxide (18O) and an episulfide (S) fused to a cyclopropane. Grignard reagents preferentially open the more electrophilic/reactive ring. Epoxides are more reactive toward Grignard reagents than episulfides because oxygen is more electronegative and the C-O bond is more polarized, making the carbon more electrophilic. Therefore, RMgX opens the epoxide (18O-containing ring) selectively. Step 2 - Ring opening of epoxide by RMgX: The Grignard reagent (R-) attacks the less hindered carbon of the epoxide, opening it. This generates an alkoxide (18O-) on one carbon and introduces the R group. The thiirane ring remains intact. The product (I) contains: intact thiirane, a carbon bearing 18O- (alkoxide/after workup 18OH), and the R group attached. Step 3 - Reaction of intermediate (I) with CH3I: Intermediate (I) contains a thiirane (episulfide) ring. Upon treatment with CH3I, the sulfur of the thiirane (which is nucleophilic) gets alkylated by CH3I. However, more precisely, the alkoxide (18O-) formed in step 1 is the nucleophile that reacts with CH3I to give 18OCH3 (methyl ether with labeled oxygen). This methylates the 18O-alkoxide to give an 18OCH3 group. Step 4 - Final product (II): The product retains the intact thiirane ring (episulfide), has an 18OCH3 group (from methylation of the 18O-alkoxide by CH3I), and has the R group from the Grignard. This matches option (c): a thiirane ring with -CH2-(18OCH3) and -CH2-R substituents. Why other options fail: - (a) and (b): These show free -SH groups suggesting episulfide opening, but the episulfide is less reactive than the epoxide toward RMgX, so it remains intact. - (b): Shows an intact epoxide after reaction, but the epoxide was opened by RMgX. - (d): Shows methylation of sulfur (18SCH3) which would require 18S label on sulfur, but the sulfur has no isotope label; the 18O label is on oxygen, and the alkoxide is what gets methylated by CH3I. Therefore, the correct answer is C.