What happens when tran-2-butene reacts with aqueous bromine in the presence of chloride?

$$\ce{Br2-H2O}$$, containing some $$\ce{NaCl,AlCl3}$$ ?

I initially though that there would be anti addition of Br and OH to give 3-Bromobutan-2-ol but then I also got the intuition that NaCl might ionize in water and the $$\ce{Cl-}$$ might be involved here as well (obviously, with different stereochemical products)

But when I referred my book, it showed the following products only:

Why does this happen?

• First $\ce{Br2}$ is forming first a positive ion with one $\ce{Br}$ atom fixed on the double bond. It makes a positively charged triangular structure with the two central carbon atoms and one bromine atom. Then a negative ion may react with this positive ion. And here there are two possibilities. Either the bromide ion remaining after this first reaction, or a chloride ion from the solution. The probability of having chloride to react is greater. So $\ce{Cl-}$ attacks the triangular cation, and gets fixed from the opposite side of the triangle. – Maurice Mar 23 at 16:51
• @Maurice Yes, I know that the bromonium ion is formed as a transition state. What I'm asking is: Why does $\ce{OH-}$ also participate and attack the cation? – Prajwal Tiwari Mar 23 at 17:00
• @PrajwalTiwari OH- is present in very low concentrations in water. Usually, its H2O which attacks the bromonium intermediate, however, in this case there are lots of Cl- around, and Cl- is a better nucleophile than water, so Cl- is the one that attacks (mostly) – Shoubhik R Maiti Mar 23 at 19:40