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.
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 + KOH_(aq) -> R - OH + KCl`
Alkyl chloride Alcohol
On the other hand, an alcoholic solution of KOH contains alkoxide (RO−) ion, which is a strong base. Thus, it can abstract a hydrogen from the β-carbon of the alkyl chloride and form an alkene by eliminating a molecule of HCl.
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 a hydrogen from the β-carbon.
If aqueous solution, KOH is almost completely ionized to give OH– ions which being a strong nucleophile brings about a substitution reaction on alkyl halides to form alcohols. Further in the aqueous solution, OH– ions are highly solvated (hydrated). This solvation reduces the basic character of OH– ions which, therefore, fails to abstract a hydrogen from the P-carbon of the alkyl chloride to form alkenes. In contrast, an alcoholic solution of KOH contains alkoxide (RO–) ion which being a much stronger base than OH– ions perferentially eliminates a molecule of HCl from an alkyl chloride to form alkenes.
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