Reaction of alkenes with bromine in presence of sodium chloride

Question

Consider the following reaction
$$\ce{CH3\bond{-}CH\bond{=}CH2 ->T[$\ce{Br2}$/$\ce{NaCl}$]?}$$

What would be the product(s) for this reaction? I am confused because there are two nucleophiles here, $\ce{Br-}$ and $\ce{Cl-}$. So, there are two alternatives for the second step. Let me explain what I am saying.

First of all, the $\pi$-electrons will attack an electrophile which in this case is a $\ce{Br+}$ which is obtained form polarisation of the $\ce{Br\bond{-}Br}$ and it is added to the less substituted carbon, which complies with Markovnikov's rule, forming the carbocation below: $$\ce{CH3\bond{-}CH+\bond{-}CH2\bond{-}Br}$$

Now the carbocation will be attacked by a nucleophile. There are 2 nucleophiles in this problem, as I have stated earlier.
So, my question is which one of them will attack the carbocation and subsequently what will be the product? Will it be only 1,2-dibromopropane or only 1-bromo-2-chloropropane or a mixture of both? Please give the reasons.

Answer 1

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.