Why would a polar protic solvent favor the addition of HBr to an alkene?

I understand that a polar solvent will help dissociate the $\ce{HBr}$ into ions.


But why a protic solvent?


Why would we want a solvent that would encage the $\ce{Br}$ and $\ce{H}$ ions?


Why is the aprotic solvent not favored? My textbook does not discuss this and I searched online, with no result.





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> why a protic solvent?
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> 


Protic solvents tend to have higher dielectric constants than typical, **unreactive**, non-protic solvents. Look at the following table.


\begin{array}{|c|c|c|}\hline
\text{Solvent} & \text{Dielectric constant} \\ \hline
\text{methanol} & 32.6 \\ \hline
\text{ethanol} & 24.6 \\ \hline
\text{methylene chloride} & 9.08 \\ \hline
\text{chlorobenzene} & 5.69\\ \hline
\text{chloroform} & 4.81 \\ \hline
\text{diethyl ether} & 4.27 \\ \hline
\end{array}


The higher dielectric constant stabilizes ions which means that 1) $\ce{HBr}$ will tend to be more dissociated and 2) (as Dissenter noted) the resultant carbocation intermediate will be more stabilized thereby lowering the energy of the transition state leading to it.


 5

High-dielectric solvent assist dissociation because they shield charge. Since the proton's electric field is so high, it always finds a base to bind to, which will in your case be $\ce{Br-}$. So you also have to consider the solvent's basicity. Read [this article](http://pubs.acs.org/doi/abs/10.1021/j100139a045).


The assertion that you need a protic solvent is wrong, you need a sufficiently basic solvent.


 2

A polar protic solvent will help stabilize the carbocation that is formed in the process of electrophilic addition. When the nucleophilic $\pi$ bond grabs the hydrogen from the $\ce{HBr}$, one of the carbons that was part of the $\pi$ bond is left deficient of an octet. 


 2