halohydrin formation
When you add a halogen such as Br2 or Cl2 to an alkene but do it in water instead of an inert solvent, you do not get the usual dihalide. Instead the two ends pick up different groups — a halogen on one carbon and a hydroxyl on the other — giving a halohydrin (a molecule with both a halide and an -OH on adjacent carbons). It is the same opening move as halogenation, but water muscles in as the nucleophile.
The mechanism explains both the regiochemistry and the stereochemistry. The halogen first adds to the pi bond and bridges the two carbons as a three-membered halonium ion. Now water, the most abundant nucleophile around, attacks this ring from the back side. It does not attack at random: it goes to the more-substituted carbon, because that carbon carries more of the positive charge (it better stabilizes the partial cation). So the OH ends up on the more-substituted carbon and the halogen on the less-substituted one. Because water attacks opposite the bridging halogen, the OH and X add anti, on opposite faces.
Halohydrin formation matters as a way to install two different functional groups across a double bond in one step, with predictable orientation and anti stereochemistry. It is also the doorway to epoxides: treat a halohydrin with base and the alkoxide displaces the neighbouring halide in an intramolecular substitution, closing a three-membered oxygen ring. So this humble reaction is a quiet workhorse of synthesis.
Propene plus Br2 in water gives 1-bromopropan-2-ol (Br-CH2-CHOH-CH3): the OH goes to the more-substituted middle carbon and the Br to the terminal carbon, with the two added anti across the former double bond.
OH attacks the more-substituted carbon of the halonium ion; X and OH add anti.
The OH lands on the more-substituted carbon here even though the simple Markovnikov phrasing would put a nucleophile there too — but the cause is different: it is the lopsided positive charge on the halonium ion, not a free carbocation.