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.

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

  • $\begingroup$ A good answer would also mention the order of each reaction (S<sub>N</sub>1, S<sub>N</sub>2, E1, or E2) for completeness. $\endgroup$ Mar 13 '18 at 16:36

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