For this question, my professor hinted at an intramolecular reaction, and he asked us to recall the bromonium ion as well. As such, I believe the SMe group attacks the open carbocation when OTs leaves. I think I understand how the left-side enantiomer is formed: SMe attacks the open carbocation, then water attacks from the back because the front-side is occupied by the hydrogen (not shown). However, what is the mechanism for the right-side enantiomer? How does the SMe end up in front (on a wedge) along with water after hydrolysis?