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








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[Why does alcoholic KOH prefer elimination whereas aqueous KOH prefers substitution?](/questions/15728/why-does-alcoholic-koh-prefer-elimination-whereas-aqueous-koh-prefers-substituti)

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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.


$$\ce{RCl + KOH (aq) -> ROH + KCl}$$


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.


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


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