What type of reaction results when a haloalkane is combined with alcoholic KOH?

When a haloalkane is combined with alcoholic KOH, the reaction that occurs is an elimination reaction. This is primarily due to the conditions under which the reaction takes place. The presence of alcoholic KOH promotes the removal of a hydrogen atom and the halogen atom from adjacent carbon atoms, leading to the formation of an alkene.

In this process, KOH acts as a strong base that abstracts a proton from the carbon adjacent to the carbon carrying the halogen. As the base pulls away the proton, the leaving group (the halogen) departs, resulting in the formation of a double bond between the two carbon atoms. This characteristic of forming an alkene from a haloalkane, due to the elimination of smaller molecules such as HX (where X is the halogen), distinguishes it as an elimination reaction.

The context of other types of reactions helps clarify the focus on elimination: substitution reactions involve the replacement of one functional group with another; redox reactions involve the transfer of electrons leading to changes in oxidation states; and addition reactions involve the direct addition of atoms or groups across a double bond. In the case of acidic or neutral conditions with haloalkanes, substitution reactions could occur, but under alcoholic conditions, elimination is favored,