Past exam of the chemistry B course of the University of Cambridge / 2022 / Paper 1 / B4 / b / Solution

The alkene attacks electrophilic mercury(II), with acetate departure as appropriate, to give a bridged mercurinium ion between the ring carbon and the exocyclic methylene. That bridge is polarized: the more substituted ring carbon can better stabilize developing positive charge than the terminal CH2 carbon. water attacks the ring carbon, opening the mercury bridge there, and deprotonation gives the tertiary alcohol with an exocyclic CH2–HgOAc substituent.

Subsequent sodium-borohydride demercuration replaces the terminal C–Hg bond by C–H. Thus the original exocyclic CH2 becomes Me, and the new OH remains on the substituted ring carbon. The result is Markovnikov hydration of the exocyclic alkene. There is no need for a freely equilibrating tertiary carbocation, and no rearrangement of the carbohydrate carbon skeleton or cleavage of the methyl acetal is required. The question does not require the mercury-stage facial selection or a detailed demercuration mechanism.