Deprotonation of alcohols to give alkoxides

by James

Description: Treatment of alcohols with bases give their conjugate bases, called alkoxy ions (alkoxides)

Notes: In order to be effective the base must be as strong as (or stronger) than HO(–) or RO(–). The alkoxy ions formed are excellent nucleophiles and can be used in substitution reactions such as the Williamson ether synthesis (and many others).


Notes: In cases where the pKa of the acid formed is similar to that of the alcohol, the reaction will be an equilibrium (example 1, ROH and H2O).

Mechanism: The base removes a proton from the alcohol (Step 1, arrows A and B) forming the alkoxide.



{ 3 comments… read them below or add one }

Sean Bannier

In Example 2, you have a potassium ion above your reaction arrow, yet you have a sodium ion in your products. It is completely irrelevant for the reaction, just different spectator, but just thought you should know.


Sarah Shtargot

There was a question on one of my practice exams that asked “which base will not fully deprotonate an alkyl alcohol?”

So I take it is NaOH because of the pKa similarity, but I don’t really understand what it means to “not fully deprotonate”



Hey Sarah – thanks for writing. I think a better way of writing it is “to not irreversibly deprotonate”. In other words, once the acid base reaction occurs, there is no equilibrium between the new acid [the conjugate acid of the base] and the new base [conjugate base of the acid]

For example in the reaction
CH3OH + NaOH —> CH3O- + H2O
there can still be an acid-base reaction between the base [CH3O-] and acid [H2O] formed in this reaction [in other words it is in equilibrium]

but in the reaction
CH3OH + CH3Li —> CH3O- + CH3-CH3
there is no acid base eaction between the base [CH3O-] and acid [CH3CH3] formed here – it is an irreversible acid-base reaction.

I agree that “completely deprotonated” is confusing because of course in any individual reaction the molecule is completely deprotonated. The issue is wherther ALL of the molecules are irreversibly deprotonated.

Does that make sense?



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