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.
When alkyl chlorides are treated with aqueous KOH, a substitution reaction occurs. The hydroxide ion (OH⁻) acts as a nucleophile and replaces the chloride ion, resulting in the formation of alcohols as the major product.
When alkyl chlorides are treated with alcoholic KOH, an elimination reaction takes place. When a haloalkane with β-hydrogen atom is heated with alcoholic solution of potassium hydroxide, there is elimination of hydrogen atom from β-carbon and a halogen atom from the α-carbon atom. This results in the formation of alkenes as the major products.
Explanation
The question tests understanding of how the nature of the solvent affects the reaction pathway. Aqueous KOH provides hydroxide ions in a polar medium that favours nucleophilic substitution, leading to alcohol formation. Alcoholic KOH, being a strong base in a less polar medium, favours elimination reactions where the β-hydrogen is removed along with the halogen to form a double bond (alkene).
Solution Steps
Step 1: Identify that aqueous medium favours substitution reaction where OH⁻ replaces Cl⁻
Step 2: Identify that alcoholic medium favours elimination reaction where H from β-carbon and X from α-carbon are removed