Question

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

Answer 1

in first rxn Sn1 mechanism is used while in second rxn E2 mechanism takes place....thus alcohol is formed as product first due to subsitution while alkene is formed in second due to elemination...
plz mark it as brainliest

Answer 2

Hey !!

In aqueous solution, KOH is almost completely ionised to give OH⁻ ions which being a strong nucleophile brings about a substitution reaction on alkyl halides to form alcohols. In the aqueous solution, OH⁻ ions are highly hydrated. This reduces the basic character of OH⁻ ions which fail to abstract a hydrogen from the β-carbon of the alkyl chloride to form an alkene.

         On the other hand, an alcoholic solution of KOH contains alkoxide (OR⁻) ions which being a much stronger base than OH⁻ ions preferentially eliminates a molecule of HCl from an alkyl chloride to form alkenes.

Hope it helps you !!