Elimination of water from alcohols to form alkenes using acid

by James

Description: Addition of acid to alcohols can result in elimination to form the corresponding alkene along with a molecule of water.

Notes: The most common acid used for this purpose is H2SO4, since the conjugate base HSO4(–) is a poor nucleophile and unlikely to attack the intermediate carbocation.

Note also that the more substituted alkene is formed preferentially (Zaitsev product)


Notes: Heat generally favors elimination reactions. This works best for secondary and tertiary alkyl halides.

Mechanism: Protonation of the alcohol oxygen (Step 1, arrows A and B) converts it to OH2(+) which is a much better leaving group. Loss of water (Step 2, arrow C) leads to formation of the carbocation. A proton from the most substituted adjacent carbon (β carbon) is then removed, either by water (if available) or HSO4(–) to give the alkene (Step 3, arrows D and E).

Notes: Steps 2 and 3 constitute an E1 mechanism. Note that removal of proton from C6 would give the same product, but removal of proton from C-7 is not as favored because it would lead to a less substituted alkene.

It’s less reasonable to use HSO4(–) as a base, since HSO4(–) is not as strong a base as H2O. Note how acid is regenerated in the last step, so only a catalytic amount of acid is required.

Since this goes through a carbocation, rearrangements can be possible when we have a secondary carbocation adjacent to a tertiary or quaternary carbon.


{ 2 comments… read them below or add one }

Bianca Pries

James, in the note above you state “this works best for secondary and tertiary Alkyl Halides”… I know you mean alcohols, yes?



Yes! Thanks, I’ll fix this.


Leave a Comment