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For the electrophilic addition of $\ce{X2}$ to alkene in $\ce{CCl4}$ solvent, a dihaloalkane is formed. This is a halogen addition reaction, with the intermediate being a halonium ion.

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However, if instead of $\ce{CCl4}$, $\ce{H2O}$ is used instead, a halohydrin would be formed.

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However, in the second case (halohydrin formation reaction), the $\ce{X-}$ ion is still present in the reaction. Furthermore, $\ce{X-}$ is a stronger nucleophile than $\ce{H2O}$. Why is it then that it is $\ce{H2O}$ that reacts with the halonium rather than $\ce{X-}$, the stronger nucleophile?

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The halohydrin is usually formed by means of a solvolysis reaction where water is used as the solvent. So, it is present in such high amounts that it overpowers the larger nucleophilic strength of the $\ce{X-}$ ion since the latter is only present in relatively minute quantities. Furthermore, water is a polar protic solvent and thus it has the effect of weakening the nucleophilic strength of $\ce{X-}$ in solution by "encasing" it in a large hydration shell.

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  • $\begingroup$ Is this reaction regiospecific or regioselective? (i.e. Will the product be 100% halohydrin or perhaps 99% halohydrin and 1% dibromoalkane)? Does the hydration shell exclude the X- completely from reaction? Thanks! $\endgroup$
    – t.c
    Oct 2 '14 at 9:49
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    $\begingroup$ @t.c Unfortunately, I lack practical experience with this reaction so I can't really say how good the yield will be or whether the dihaloalkane is a noticable by-product. But in theory I'd say that yes, there would be some small quantity of dihaloalkane formed but that might even be less than 1% because water is present in an overwhelming excess. You will have a water concentration of about 55.55 mol/L and if you have, say, a bromine concentration of 0.1 mol/L then the difference is already a factor of about 500. But from the bromine an even much smaller amount of bromide is formed. $\endgroup$
    – Philipp
    Oct 2 '14 at 14:18

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