= 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. \b[<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.
\b[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.
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