Question

The treatment of alkyl chlorides with aqueous ${\text{KOH}}$leads to formation of alcohols but in the presence of alcoholic${\text{KOH}}$, alkenes are major products. Explain.

Answer 1

Hint: ${\text{KOH}}$ is an ionic compound. In aqueous medium, it breaks into its constituent ions and hydroxide ions are produced. While in the alcoholic medium, the attacking species formed is different.

Complete step by step answer:
When aqueous ${\text{KOH}}$is used, the attacking species in the reaction is${\text{O}}{{\text{H}}^ - }$. The hydroxide group is a strong nucleophile and thus, when it reacts with alkyl chlorides substitution reaction takes place. Thus, the chloride ion is lost and hydroxide ion replaces it resulting in the formation of alcohol.
${\text{O}}{{\text{H}}^ - }$ion is both a strong base as well as a strong nucleophile. But in an aqueous solution, the hydroxide ions are highly hydrated. This reduces the basic character of ${\text{O}}{{\text{H}}^ - }$ions and thus it fails to abstract a hydrogen from the $\beta $- carbon of the alkyl chloride to form a double bond.
Meanwhile, the alcoholic solution of ${\text{KOH}}$ contains alkoxide ion, which is a strong base. It extracts a proton from the $\beta $- carbon atom of the alkyl chloride and the chlorine substituent leaves as chloride ion. In total, a molecule of hydrogen chloride is lost resulting in the formation of an alkene.

Note:
The dehydrohalogenation of alkyl halides is a good laboratory method for the synthesis of alkenes as alkyl halides are readily available for reactions with various starting materials. Considered by itself, the hydrohalogenation reaction has a very unfavourable equilibrium constant. However, if we use a strong base such as alkoxide anion, as in alcoholic ${\text{KOH}}$ to extract a proton on the carbon adjacent to the bromine, thus making the reaction highly favourable.