In this preparation, http://www.orgsyn.org/Content/pdfs/procedures/CV5P1106.pdf, 2,4,6-trimethylpyrylium perchlorate is prepared from 4 molecules of acetic anhydride and 1 molecule of t-butanol.
However, I cannot find a mechanism of this reaction, nor can I think of one. Could someone please provide the mechanism? Thanks.

  • $\begingroup$ From some quick research it seems the first step involves forming isobutylene from the t-butanol. Both ac. anhydride and t-butanol have basicity potential (oxygen lone pairs), but the t-butanol conjugated acid can undergo [Dn] and [De] steps to form isobutylene, which can then act as a nucleophile on the reaction (the piece that seems to be missing in the reactants). I didn’t go further than that yet, but I’ll try to figure it out later if no one does it by then. $\endgroup$ – IanC Jun 22 '19 at 15:00
  • $\begingroup$ Look at Method II in your link. They use mesityl oxide (4-methylpent-3-en-2-one) and 2 equiv. of Ac2O. Mesityl oxide is formed from isobutylene and Ac2O. Form the extended enol of mesityl oxide, acylate it and cyclize. $\endgroup$ – user55119 Jun 22 '19 at 22:05

I tried to work out a mechanism from the information I could gather about the reaction. If someone notices something odd/wrong please comment and I'll fix it.

The steps are either showing the protonation or skipping directly to the protonated forms of the reactants, them being in acidic medium.

Formation of isobutylene


Not very complicated, in acidic conditions we remove the hydroxyl group from the t-butanol to form isobutylene. The latter behaves as a nucleophile in the next steps of the reaction. This step is acid catalysed, so the acid is not consumed.

Formation of Mesityl Oxide

Mesityl Oxide

Also relatively simple step. Nucleophile addition to the protonated C=O bond [Adn] followed by elimination of Acetic Acid.



Shows the Mesityl Oxide tautomerism, important because the tautomer that will react to form the final product is the one in the right (with the double bond in the terminal carbon).

Enol Formation


Here a Enol is formed by a nucleophilic attack of the Mesityl Oxide over another Acetic Anhydride molecule (protonated by the acidic medium). In the last step I added the deprotonation of the more acidic oxygen to respect Pauling's rule of charges by keeping an overall +1 charge in the molecule. The left oxygen is more acidic since the other oxygen can better stabilize the charge with delocalization. I wrote the molecule with wavy bonds because I'm unsure about the stereospecificity of the step, but maybe the Z conformation is favored due to the charges on the oxygens repelling themselves?



Final step is cyclization of the intermediate, by nucleophilic attack of the carbon by the deprotonated oxygen. The intermediate is dehydrated and rearranged to form the final product (releasing a proton to reach aromatic status). This step had an intermediate with +2 charge, which I'm a little skeptical about due to the Pauling's rule of charges, so I also draw the mechanism as a concerted step:

Cyclization Concerted

If the rule of charges isn't that strict, maybe the non-concerted path is a closer description of what's happening, but I'm keeping both here for now. If anyone thinks one of the two is the right one, comment and I'll update the answer.

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    $\begingroup$ I would enolize mesityl oxide by protonating the oxygen to form the dienol and then protonate acetic anhydride to the nucleophilic addition. You don't want a negative charge on oxygen in acidic medium. The structure above acetic acid needs to be a Z-double bond in order to cyclize. $\endgroup$ – user55119 Jun 23 '19 at 3:21
  • $\begingroup$ @user55119 I omitted the protonation trying to make the mechanism less messy, but I'll try to draw the mechanism without omitting them later. By the way, regarding the Z double-bond, is the reaction of the mesityl oxide with acetic anhydride stereoselective? Or will I have both the Z and E forms? And do you have any idea why the method claims more acetic anhydride equivalents are required? $\endgroup$ – IanC Jun 23 '19 at 22:03
  • $\begingroup$ The yield is only 55% so some alpha acylation of mesityl oxide (MO) may occur. As to E an Z isomers, they interconvert via the enol of the diketone. The stoichiometry is correct. Two equivalents of Ac2O are required to form the diketone. Another Ac2O reacts with water from t-BuOH and the fourth equivalent of Ac2O reacts with the water formed in the last step. If you start from MO, then only two equivalents of Ac2O are required. The idea is to sequester all water as acetic acid. $\endgroup$ – user55119 Jun 23 '19 at 23:38
  • $\begingroup$ One other thing, in your seventh structure change the -OH to =O. – $\endgroup$ – user55119 Jun 24 '19 at 15:41
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    $\begingroup$ @IanC maybe use some enol instead of carbonyl during your mesityl oxide final stages/ ring closing steps. You don't need to have all the protonation you've drawn and it consistently keeps your charges down to +1 $\endgroup$ – Beerhunter Jun 25 '19 at 21:10

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