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Question

Addition of bromine on propene in presence of Brine yields a mixture of


Solution

Consider the following reaction
CH3−CH=CH2−→−−−−Br2/NaCl?
First of all, the ππ-electrons will attack an electrophile which in this case is a Br+BrX+ which is obtained form polarisation of the Br−Br and it is added to the less substituted carbon, which complies with Markovnikov's rule, forming the carbocation below:
CH3−CH+−CH2−Br



*The first step is Br−BrBr−Br approaching the pi-cloud, being polarized, loosing a Br−BrX−to form a 3-cyclic bromonium cation. Given solubility and reactively inert solvent, Br−BrX− (large poor nucleophile) competes with Cl−ClX− (smaller, better nucleophile) for ring-opening capture. Now factor in relative concentrations. Now factor in thermodynamics (reversibility, Cl−ClX− wins for the stronger C−ClC−Cl bond) versus kinetics (smaller Cl−ClX− 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 ClCl comes in at the 2-position by default. If my cyclic ion intermediate is controlling, the ClCl 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.

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