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After the formation of the cyclic intermediate, $\mathrm{S}_{N}2$ reaction takes place. This reaction would be favored by an aprotic solvent.

How will the products of the reaction be affected if I use a polar protic solvent?

I think that the formation of the cyclic intermediate is independent of the solvent used but I can't figure out what would happen after that.

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  • $\begingroup$ The $SN_2$ reaction won't take place if you use a polar protic solvent. If you use a polar protic solvent such as water, you would form $\ce{HX}$. This would lead to an electrophilic substitution reaction. $\endgroup$ – Yashas Apr 7 '17 at 14:04
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In the presence of a polar protic solvent, the halogen will react with the solvent to form acids ($\ce{HX}$ and ${HOX}$).

$$\ce{Cl2 + H2O -> HCl + HOCl}$$

The high concentration of the cations formed (the hydrohalic acid and the hypohalous acid will exist as ions in a polar protic solvent) can act as electrophiles and initiate an electrophilic substitution reaction which results in the formation of a range of products.

$$\ce{CH2-CH=CH2 + X2 + H2O -> CH2-CH(Cl)-CH3 + CH2-CH(OH)-CH2(Cl) + CH2-CH(Cl)-CH3(Cl)}$$

This reaction is primarily initiated by the $\ce{H+}$ attacking the electron-rich double bond. This causes an electomeric shift of electrons to a carbon such that the carbocation formed is stable (Markonivkoff's rule). As carbocations are formed, rearrangements are possible. Not only are the mechanisms different, the products will be different too. This reaction is entirely different from a $SN_2$ reaction.

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