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

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Answer 1 · 2018-03-03T16:16:16

Here's what I get.

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.

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