Answer: A
(i) A free secondary cation forms and undergoes a methyl shift, giving 2,3-dimethylbutan-2-ol. (ii) A mercurinium ion prevents rearrangement, giving the Markovnikov alcohol 3,3-dimethylbutan-2-ol. (iii) Hydroboration is anti-Markovnikov, giving 3,3-dimethylbutan-1-ol.