$\ce{Br2-H2O}$, 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?

  • $\begingroup$ 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. $\endgroup$ – Maurice Mar 23 at 16:51
  • $\begingroup$ @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? $\endgroup$ – Prajwal Tiwari Mar 23 at 17:00
  • 2
    $\begingroup$ @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) $\endgroup$ – Shoubhik R Maiti Mar 23 at 19:40

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