Question

The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.

Answer 1

In an aqueous solution, KOH almost completely ionizes to give OH^– ion, a strong nucleophile and reacts with alkyl halides to form alcohols. In an aqueous solution, OH^– ions are highly hydrated. This reduces the basic character of OH^– ions, due to which they fail to separate hydrogen atoms from the β-carbon of alkyl halide and cannot form an alkene.

On the other hand, alcoholic solution of KOH contains alkoxide (RO^–) ions which, being a stronger base than OH^–, easily remove HCl molecule from alkyl chloride to form alkene.

Answer 2

Simple nucleophilic substitution occurs when alkyl chlorides react with aqueous KOH to form alcohols.

\[\ce{CH3 - CH2 - Cl + KOH ->[H2O]CH3 - CH2 - OH + KCl}\]

When aqueous KOH is substituted with alcoholic KOH, HCI is eliminated from an alkyl halide, resulting in the formation of alkenes instead of alcohols.

\[\ce{CH3 - CH2Cl + KOH ->[EtOH] CH2 = CH2}\]

This can be explained by the size of the nucleophile in both reactions. In an aqueous medium, the \[\ce{N\overset{\ominus}{u}}\] is \[\ce{\overset{\ominus}{O}H}\], which is small, whereas in an alcoholic medium, the \[\ce{N\overset{\ominus}{u}}\] is \[\ce{C2H^-5}\] is bulky.

The bulky \[\ce{N\overset{\ominus}{u}}\] always finds it easier to abstract a proton rather than attack a tetravalent carbon and form a substitution product.

If C₂H₅O^Θ attacks a carbon-carrying halogen, steric repulsions can delay the attack and prevent substitution.