Show the mechanism of KOH and alcohol for forming double bond in an alkyl halide?

1 Answer
Mar 3, 2018

Here's what I get.

Explanation:

In a solution of alcoholic #"KOH"#, we have the equilibrium:

#"CH"_3"CH"_2"O-H" +"OH"^"-" ⇌ "CH"_3"CH"_2"O"^"-" + "H"_2"O"#

The position of equilibrium lies to the left, so there is not much ethoxide ion present.

However, ethoxide is a much stronger base than hydroxide, so it is the favoured reactant.

The ethoxide attacks an α-hydrogen atom in an #"E2"# elimination reaction.

chem.libretexts.org

The products are ethanol, an alkene, and a halide ion.