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.