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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.
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Solution
R−Cl+KOH(aq)→R−OH+KCl
The ionization of aqueous KOH produces hydroxide ions which are strong nucleophiles. Hence, alkyl chlorides undergo substitution to form alcohol.
R−CH2−CH2−Cl+KOH(alc)→R−CH=CH2+KCl+H2O
Alcoholic KOH solution gives alkoxide ion which is a strong base. It abstracts β hydrogen atom of alkyl chloride. A molecule of HCl is eliminated and an alkene is formed.
Note: The basicity of hydroxide ion is much lower than the basicity of alkoxide ion as hydroxide ion is significantly hydrated in aqueous solution.
Hence, hydroxide ion cannot abstract β hydrogen atom of alkyl chloride.
R−Cl+KOH(aq)→R−OH+KCl
The ionization of aqueous KOH produces hydroxide ions which are strong nucleophiles. Hence, alkyl chlorides undergo substitution to form alcohol.
R−CH2−CH2−Cl+KOH(alc)→R−CH=CH2+KCl+H2O
Note: The basicity of hydroxide ion is much lower than the basicity of alkoxide ion as hydroxide ion is significantly hydrated in aqueous solution.
Hence, hydroxide ion cannot abstract β hydrogen atom of alkyl chloride.
Hence, hydroxide ion cannot abstract β hydrogen atom of alkyl chloride.
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