Question

Methanol containing the oxygen isotope 18O (shown in green below) is allowed to react with the sulfonyl chloride. The intermediate is then treated with NaOH to give unlabeled MeOH. Identify the neutral intermediate and charged product and highlight the atom in both to identify the position of the 18O label.

Answer 1

Answer 2

The methanolic OH (Labeled 18 isotope- purple O) attacks S in sulfonyl chloride, displacing Cl in it. Formation of Methyl sulfonyl chloride takes place.

The labeled O is in C-O-S bond.

Methyl sulfonyl chloride then treated with NaOH solution. Hydroxyl ion OH^- is strongly nucleophilic and hence attacks C-O-S bond following SN2 mechanism and displaces O-S- group and itself get attached to C. Thus the methanol generates back. The labeled O is now in sulfur trioxide molecule. This is clear from the reaction mechanism.