Use of aqueous KOH and alcoholic KOH in dehydrohalogenation reactions
Aqueous $\ce{KOH}$ 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.

$$\ce{RCl + KOH (aq) -&gt; ROH + KCl}$$

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

Alcoholic, $\ce{KOH}$, specially in ethanol, produces $\ce{C2H5O-}$ ions. The $\ce{C2H5O-}$ ion is a stronger base than the $\ce{OH-}$ ion. Thus,the former abstracts the ß-hydrogen of an alkyl halide to produce alkenes. This reaction is known as elimination reaction.

$$\ce{CH3CH2Br + KOH (alc) -&gt; H2C=CH2 + KBr + H2O}$$