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We use an alcoholic medium for reactions in which we want to prevent the characteristics of the anionic species and an aqueous medium for the reactions in which we want to control the activity of the anionic species (depends on how strong is our solvent).

The principle behind this is that alcohol solvates the cationic species only while water in an aqueous medium solvates both cationic as well as anionic species. The solvation results in neutralizing and reducing the mobility of the solvate and thus reduces the reactivity proportional to the degree of solvation.

Now back to my question Why does alcohol only solvate the cationic species despite being a polar protic solvent just like $\ce{H2O}$?

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  • $\begingroup$ The permittivity epsilon is much greater for water (80) than for ethanol (24.3). $\endgroup$
    – Maurice
    Commented Sep 2, 2023 at 13:00
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    $\begingroup$ "alcohol solvates the cationic species only"? I wouldn't say it's true, $\endgroup$
    – Mithoron
    Commented Sep 2, 2023 at 22:07
  • $\begingroup$ @Mithoron But then what's the reason for Why does alcoholic KOH prefer elimination whereas aqueous KOH prefers substitution? $\endgroup$
    – D13G
    Commented Sep 3, 2023 at 5:49
  • $\begingroup$ @Maurice What does it interpret? $\endgroup$
    – D13G
    Commented Sep 3, 2023 at 5:51
  • $\begingroup$ @D13G. Coulomb's law states that the force between two charges is inversely proportional to the factor epsilon. So in water it is easier than in ethanol to separate electric charges. $\endgroup$
    – Maurice
    Commented Sep 7, 2023 at 9:03

1 Answer 1

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I will come to an answer to this question by inductive reasoning.

If we compare acidity of $\ce{H2O}$ and $\ce{ROH}$:

$\ce{H2O ->[-H+] OH-}$ and $\ce{ROH ->[-H+] OR-}$
$\ce{OH- > RO-}$ [Stability]
$\implies$ $\ce{H2O}$ is more acidic than $\ce{ROH}$

As $\ce{pH}$ of $\ce{H2O}$ is 7 so it implies $\ce{ROH}$ is a base and as the $\ce{ö}$ has lone pairs so it is a lewis base and donating ability of lone pair is higher than that of $\ce{H}$, similar to polar aprotic solvents.
Therefore despite being a polar protic solvent, $\ce{ROH}$ acts as a polar aprotic solvent due to its Hydrogen atom is not acidic enough.

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