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.
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.
Solution and Explanation
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.
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