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⁻ ions. OH⁻ ion is a strong nucleophile, which leads the alkyl chloride to undergo a substitution reaction to form alcohol. R-Cl(Alkyl chloride)+KOH(aq)→R-OH(Alcohol)+KCl On the other hand, an alcoholic solution of KOH contains an alkoxide(RO⁻)ion, which is a strong base. Thus, it can abstract hydrogen from the β^-carbon of the alkyl chloride and form an alkene by eliminating a molecule of HCl.R-C(β)H₂-C(α)H₂-Cl+KOH(alc)→R-CH(Alkene)=CH₂+KCl+H₂O OH⁻ ion is a much weaker base than RO− ion. Also, OH⁻ ion is highly solvated in an aqueous solution and as a result, the basic character of OH⁻ ion decreases. Therefore, it cannot abstract hydrogen from the β-carbon.