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pyrogallol 1,3-dimethyl ether aka syringol

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phenethyl_man:
The preparation of both the 1-methyl ether [1] and 1,2-dimethyl ether [2] by the action of dimethyl sulfate on pyrogallol have been well described here.  The former being prepared by blockage of two of the hydroxyl groups by boric acid using excess DMS, and the second being prepared by carboxylation to pyrogallolcarboxylic acid, methylation and subsequent decarboxylation.

However, the only preparation I could find of the 1,3-dimethyl ether involves the use of gaseous methyl bromide and NaOMe [3].  Can anyone explain the mechanism behind this reaction so perhaps we can come up with a workable synthesis using DMS?

Obviously my interest in this compound is that just about any formylation procedure will prepare syringaldehyde from this compound which can lead 3,4,5-substituted phenethylamines; whereas any other pyrogallic ether will inevitably result in a useless 2,3,4-substitution pattern.

Rhodium posted what is seemingly a preparation of this ether (also known as 2,6-dimethoxyphenol) in Post 477454 (Rhodium: "2,6-Dimethoxyphenol & 2,6-Dimethoxybenzoquinone", Novel Discourse).  Translation anyone?  8)

[1] Synth. Commun. 20(8), 1213-1221 (1990) https://www.thevespiary.org/rhodium/Rhodium/chemistry/croweacinaldehyde.pyrogallol.html
[2] J. Chem Soc. 2542-2549 (1931) https://www.thevespiary.org/rhodium/Rhodium/pdf/tetrahydroxybenzenes.baker-1.pdf
[3] JACS 39, 1433 (1917) https://www.thevespiary.org/rhodium/Rhodium/chemistry/syringylpropenyl.html

moo:
However, the only preparation I could find of the 1,3-dimethyl ether involves the use of gaseous methyl bromide and NaOMe [3].  Can anyone explain the mechanism behind this reaction so perhaps we can come up with a workable synthesis using DMS?

NaOMe is the base used to deprotonate the phenolic hydroxyls, making them good nucleophiles. The reaction is an SN2 type nucleophilic substitution reaction, bromide being the leaving group. Reaction with DMS is goes through the same mechanism.

phenethyl_man:
Thanks for the chemistry lesson.  :P

I know how nucleophilic substitution works, I was just questioning the use of such a strong base in this particular reaction when alkali carbonates and hydroxides are strong enough to deprotonate phenols and if the rapid gassing w/MeBr has any effect on what gets alkylated.

The problem is trying to obtain a good yield of the 1,3-dimethyl ether.  Does this substitution follow any rules to predict which hydroxyl will be methylated first, second?  Will methylation of pyrogallol w/2 mol eq. DMS produce primarily syringol, or should I assume both methyl groups are displaced and use 1 mol eq DMS?  Will the outer phenolic groups be alkylated first or does it just proceed at random?

Any 2,3,4-trimethoxybenzene can be easily removed from the alkaline soln containing the phenolates, but how is one to seperate any unreacted pyrogallol, 1-methyl ether, or 1,2-dimethyl ether that is formed from the desired product.  And what is the optimal way to run the rxn to optimize the yield of the desired phenol.

I suppose it is because 2,6-dimethoxyphenol is such a readily available material commercially that no one cares to synth it themselves but it is strange I can't find any information relating to the production of it; most likely doesn't even use pyrogallol as a precursor.

moo:
You're welcome. :P  I'll give you a good answer then.

To my knowledge the 2-hydroxyl is least acidic of the three, so the hydroxyls at 1- and 3-position are the ones deprotonated first, enabling them to be methylated. Theoretically one needs two equivalents of a base for this, but the authors use 2,5 equivalents of sodium methoxide instead. There is also the possibility of Williamson ether synthesis side reaction, MeBr + NaOMe --> Me2O + NaBr, so I quess the extra half equivalent is used to compensate for that side reaction.

Also note that no yield is given and that the product is purified by vacuum distillation. What else is there produced? Some 1,2,3-trimethoxybenzene prehaps?

I don't actually know if using DMS is viable in practice, theoretically it should be. Sodium hydroxide should be enough to deprotonate the hydroxyls. I reckon there are other methods out there, waiting to be digged up from the library.

Edit: I noticed one more thing so there is one correction to what I said. The methoxide is added simultaneously with the methyl bromide. This is most likely an attempt to not deprotonate two phenolic hydroxyls at once, but to  keep to situation such that only phenolate monoanions are present in the solution, so that pyrogallol first gets methylated at the 1-hydroxyl, and the 1-methoxy-2,3-dihydroxybenzene (whose 3-hydroxyl is the most acidic one) formed is then methylated to 2,6-dimethoxyphenol. Stepwise methylation.

phenethyl_man:
Also note that no yield is given and that the product is purified by vacuum distillation. What else is there produced? Some 1,2,3-trimethoxybenzene prehaps?
--- End quote ---

yeah, it states that the 1,2,3-trimethoxybenzene is steam distilled off before the oil is fractionated.  I assume the trimethyl ether is volatile with steam while the others are not.  The fact that it is an oil implies there is some impurity besides a pyrogallic ether since all possible ethers of pyrogallol are solids.  His b.p is also 11 degC off from most of the literature I have, but since this was almost a century ago I guess I'll give him a break  ;)

Here is the exact wording from the document, perhaps it is a little more clear than the trunicated rendition on rhodium, however, still no sign of a yield:


Pyrogallol Dimethyl Ether Sulfonic Acid.-

The preparation of pyrogallol 1,3-dimethyl ether involves considerable difficulty if attempted by the usual methods. The use of an autoclave and the oxidizing influence of the air can be avoided by the following method:

One mol. pyrogallol is placed in a flask connected with a reflux condenser. Methyl bromide is passed into the flask by a glass tube ending in a capillary and passing through the condenser to near the bottom of the flask until the air is displaced.

Two and a half mols. metallic sodium are dissolved in twenty mols. methyl alcohol absolute.  This sodium alcoholate is run into the flask together with a continuous stream of methyl bromide. The flask is heated on the water bath and the current of alkyl halide continued until the reaction is nearly neutral. The gas is rapidly absorbed and sodium bromide settles out.

The methyl ether is isolated as follows: Water is added until the sodium bromide dissolves. The methyl alcohol is then evaporated and the residue distilled in steam. Any trimethyl ether passes over in the steam. The dimethyl ether is extracted with ether after acidifying.  The ether solution is evaporated and the resulting oil fractionated in vacuo. A Colorless liquid boiling at 250 degC was obtained. To check the purity the methyl-ether content was determined and 40.42% : 40.31% found while the theoretical figure is 40.26%. Further methylation by the above method yields the trimethyl ether.


Not quite sure what they mean by "methyl-ether content."

Edit: I noticed one more thing so there is one correction to what I said. The methoxide is added simultaneously with the methyl bromide. This is most likely an attempt to not deprotonate two phenolic hydroxyls at once, but to  keep to situation such that only phenolate monoanions are present in the solution, so that pyrogallol first gets methylated at the 1-hydroxyl, and the 1-methoxy-2,3-dihydroxybenzene (whose 3-hydroxyl is the most acidic one) formed is then methylated to 2,6-dimethoxyphenol. Stepwise methylation.
--- End quote ---

So I guess the key is to try to keep the reaction mixture as close to neutral as possible through out the reaction which will cause only one hydroxyl to be deprotonated at any given time?

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