= Solution
The <alkene> attacks <electrophilic> <mercury (element)>(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>. \b[<water> attacks the ring <carbon>], opening the <mercury (element)> 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 \b[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 (element)>-stage facial selection or a detailed <demercuration> mechanism.
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