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Answer 1 · 2017-10-26T14:04:25

The reason for this is that aqueous $K O H$ $KOH$ being alkaline in
nature gives $O H^{-}$ $OH^-$ ions. These $O H^{-}$ $OH^-$ ions then act as strong
nucleophile and attack on electrophilic carbon of alkyl halide
molecule. Thus the leaving group leaves and $O H^{-}$ $OH^-$ substitute the halogen atom of alkyl halide molecule, forming alcohol as a product. Consequently, aqueous $K O H$ $KOH$ gives substitution reactions.

On the other hand, alcoholic $K O H$ $KOH$ (mostly having ethanol) produces $C_{2} H_{5} O^{-}$ $C_2H_5O^-$ ions which are stronger base than $O H^{-}$ $OH^-$ ions. So these abstract the $β$ $beta$ - hydrogen of alkyl halide molecule, forming an alkene. That's the reason behind the statement; Alcoholic $K O H$ $KOH$ give elimination reactions.

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