What occurs when a haloalkane undergoes elimination with KOH (alc)?

When a haloalkane undergoes elimination with potassium hydroxide (KOH) in an alcoholic solution, the primary reaction that takes place is the formation of an alkene. This elimination reaction, often known as a dehydrohalogenation, involves the removal of a hydrogen atom and a halogen atom from adjacent carbon atoms of the haloalkane.

In the presence of KOH, which serves as a strong base, the hydroxide ion (OH-) abstracts a proton (H+) from one carbon atom of the haloalkane. Simultaneously, the bond between the halogen and its carbon breaks, leading to the release of the halogen as a halide ion (like Cl- or Br-) from the adjacent carbon. This results in the formation of a double bond between the two carbon atoms, yielding an alkene.

The specific alcohol solvent enhances the outcome of elimination over substitution, as it favors the formation of alkenes through this route. The product alkene can vary depending on the structure of the starting haloalkane, but the fundamental transformation from a haloalkane to an alkene through the loss of halogen and hydrogen is consistent in this reaction pathway.

Thus, the correct understanding of the reaction mechanism