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

Solution 1:

why a protic solvent?

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.

Solution 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.

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Solution 3:

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.

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