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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 H2O
?

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Final answer:

Alcohols solvate cations effectively due to their polar nature but solvate anions less efficiently because the bulky alkyl group creates steric hindrance, allowing anions to remain more reactive. This contrasts with water, which effectively solvates both cations and anions due to its smaller molecular structure and lack of steric hindrance.

Step-by-step explanation:

The reason alcohol primarily solvates cationic species, even though it is a polar protic solvent like water, is due to its steric hindrance and inability to form strong ion-dipole interactions with anionic species. In alcohol, the bulky alkyl group hinders the approach of anions to the partial positive charge on the solvent molecule.

This is in contrast to water, which, having a smaller molecular structure, can effectively solvate both cations and anions. Consequently, anions in alcoholic solutions have greater freedom of movement and retain more of their reactivity compared to when they are solvated in water.

In a polar aprotic solvent, the solvation of ionic compounds occurs via ion-dipole interactions, but the anion is less tightly solvated due to steric hindrance from the solvent. This effect allows the anion to remain more reactive, which is advantageous for certain types of chemical reactions, such as SN2 reactions, where a strong nucleophile is required.

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