 Solution

Direct iodide attack at the terminal CH2–OMs would replace the leaving group without disturbing the adjacent stereocentre, so it cannot by itself explain racemization. sulfur participation produces an intermediate with two equivalent methylene paths. Label the original S-adjacent methylene A, the stereogenic carbon C, and its CH2–OMs substituent B. The thioether sulfur attacks B intramolecularly and expels mesylate, forming a four-membered thietanium circuit S–A–C–B–S. The rest of the original six-membered ring remains as the other path between S and C.

In this bridged sulfonium ion, A and B are both CH2 groups joining the same S and C centres. Exchanging them no longer changes the labelled molecular connectivity: the distinction that made C stereogenic has been removed. iodide can open the strained intermediate by backside attack at either A or B. Opening one S–CH2 bond gives one configuration of the six-membered product, while opening its equivalent partner exchanges which methylene is in the ring and which is the iodomethyl substituent, giving the opposite configuration.

The carbon iodinated in each opening is a methylene; racemization appears at the adjacent carbon through exchange of equivalent paths. It is not inversion at a CH2 stereocentre, nor does the mesylate carbon have to form a primary free carbocation. An achiral intermediate in an achiral medium gives equal mirror-related outcomes. nitrogen participation alone in a simple terminal displacement would not provide this specific equivalent-path explanation.