Aqueous KOHKOH is alkaline in nature i.e. it dissociates to produce a hydroxide ion. These hydroxide ions act as a strong nucleophile and replace the halogen atom in an alkyl halide.

RCl+KOH(aq)⟶ROH+KClRCl+KOH(aq)⟶ROH+KCl

This results in the formation of alcohol molecules and the reaction is known as nucleophilic substitution reaction.

Alcoholic, KOHKOH, specially in ethanol, produces C2H5O−CX2HX5OX− ions. The C2H5O−CX2HX5OX− ion is a stronger base than the OH−OHX− ion. Thus,the former abstracts the ß-hydrogen of an alkyl halide to produce alkenes. This reaction is known as elimination reaction.

CH3CH2Br+KOH(alc)⟶H2C=CH2+KBr+H2O

Alc KOH dissociates in water to give RO- ions which is a strong base and abstracts hydrogen giving rise to elimination where Aq KOH dissociates in water giving - OH  ions which is a good nucleophile and prefers substitution.

Alkyl chloride in the presence of aqueous KOH gives substitution product .In an aqueous solution, KOH completely ionizes to give hydroxide ions. OH− ion is a strong nucleophile, which can easily substitute the strong leaving chloride ion and leads to the formation of alcohol.

R-Cl + Aq. KOH → R – OH + KCl

On the other hand, alkyl chloride containing β-hydrogen atom with alcoholic KOH solution undergo elimination reaction to give alkene as major product. An alcoholic solution of KOH produce alkoxide (RO−) ion, which is a strong base. it can easily abstract a β – hydrogen atom the from β – carbon of alkyl chloride and form an alkene as major product.

Alcoholic KOH is used to form Alkene from Alkyl Halides.where as aqueous KOH is used to form alcohols from Alkyl Halides.