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.

Explanation

`underset("Ethylchloride")(CH_(3)-CH_(2)-Cl)+KOH_((aq))rarr underset(("Substitution reaction"))underset("Ethyl alcohol")(CH_(3)-CH_(2)-OH+H_(2)+KCl)`
`underset("Ethylchloride")(CH_(3)-CH_(2)-Cl)+KOH overset(" Ethanol ")rarr underset(("Elimination"))underset("Ethene")(CH_(2)=CH_(2)+H_(2)O`
This is can be explained as follows : In presence of ethanol as a solvent, the KOH abstracts a proton from ethanol to form potassium ethoxide i.e., `C_(2)H_(5)O^(Θ)K^(o+)` Potassium ethoxide is bulkier than `OH^(Θ)` and also it is stronger base than `""^(Θ)OH`. Thus, It will easily abstract the proton than to attack the tetravalent carbon.