# Use of aqueous KOH and alcoholic KOH in dehydrohalogenation reactions [duplicate]

What is the difference between aqueous KOH and alcoholic KOH? How do they react differently in dehydrohalogenation?

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.
$$\ce{RCl + KOH (aq) -> ROH + KCl}$$
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.
$$\ce{CH3CH2Br + KOH (alc) -> H2C=CH2 + KBr + H2O}$$