Formation of Acetals from Aldehydes and Ketones

by James

Description: Ketones (or aldehydes) can be converted into acetals when treated with alcohols and acid. The byproduct is water. [private_ReactionGuide]


Notes: Acid is often simply written as “H+”. Any strong acid will do. The reaction is actually an equilibrium – the large excess of alcohol relative to water drives the reaction to completion. For this reason you generally won’t see “2 equiv” as you do here, it’s generally assumed that the alcohol is the solvent. Acid is catalytic.

The reaction can be used to “protect” the carbonyl against attack from nucleophiles, since acetals are unreactive towards everything except aqueous acid.

The acetals of ketones are sometimes called “ketals”.


Notes: Example 3 is an example where both alcohol oxygens come from the same molecule. The result is a cyclic acetal.

Mechanism: The reaction begins by protonation of the carbonyl oxygen (Step 1, arrows A and B) which makes the carbonyl carbon a better electrophile. Then, one equivalent of the alcohol attacks the carbonyl carbon in a 1,2-addition reaction (Step 2, arrows C and D) followed by a migration of a proton to the OH group (Step 3, arrows E and F). This event allows water (H2O) to be a leaving group instead of (HO-), and then 1,2-elimination of water occurs (Step 4, arrows G and H) followed by another 1,2-addition of the second alcohol (Step 5, arrows I and J). All that remains to do is to deprotonate the oxygen (Step 6, arrows K and L) and the acetal is formed.


Notes: Many other strong acids work just as well as TsOH – it’s generally fine to be vague and just write “H+”.  As with any reaction involving lots of proton transfer steps, there are other reasonable ways of drawing the mechanism for step 3. Also, there are several other species that could act as bases in step 6, but TsO(-) is chosen for simplicity. Again, for clarity, single arrows are shown here although the reaction is an equilibrium.

Video Walkthrough:


{ 3 comments… read them below or add one }

Brenda Greene

Thanks for this video!



You’re very welcome, thanks Brenda.



Hi James!
Wow, sincere thanks for the video.
Just to clarify… You add a base to the reaction only when you need to deprotonate a H, like you did in Step 5. Otherwise, only the nucleophile will participate in addition reactions? Since protonating a ketone will make it more electrophilic, the strength of the nucleophile will not matter?


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