Reaction of alkenes with bromine in presence of sodium chloride
The first step is $\ce{Br-Br}$ approaching the pi-cloud, being polarized, loosing a $\ce{Br^{-}}$ to form a 3-cyclic bromonium cation.  Given solubility and reactively inert solvent, $\ce{Br^{-}}$ (large poor nucleophile) competes with $\ce{Cl^{-}}$ (smaller, better nucleophile) for ring-opening capture.  Now factor in relative concentrations.  Now factor in thermodynamics (reversibility, $\ce{Cl^{-}}$ wins for the stronger $\ce{C-Cl}$ bond) versus kinetics (smaller $\ce{Cl^{-}}$ vs. initial tight ion pair rapidly collapsing to product).  Temperature probably makes a product ratio difference, too.

If your linear ion intermediate is controlling, the $\ce{Cl}$ comes in at the 2-position by default.  If my cyclic ion intermediate is controlling, the $\ce{Cl}$ comes in at the 1-position by steric hindrance. And if you do it in water, you get halohydrin byproducts.  

Keep the reaction way from light or you will get radical abstraction and allylic substitution.