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.

$R-Cl+KO{H}_{\left(aq\right)} \to R-OH+ KCl$
$Alkyl Alcohol$
$chloride$

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.

$R-\underset{\beta }{C}{H}_2-\underset{\alpha }{C}{H}_2-Cl+KOH\left(alc\right) \to R-CH=C{H}_2+ KCl+ H_{2}O$
$Alkyl chloride Alkene$

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.