Author Topic: Formylation w/ Mg(MeO)2 -- translation  (Read 2838 times)

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dioulasso

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Formylation w/ Mg(MeO)2 -- translation
« on: December 22, 2003, 03:25:00 PM »
This is a translation of a great writeup by DTT from the Russian HyperLab:

Post 472860 (missing)

(DDT: "Ôîðìèëèðîâàíèå...", Russian HyperLab)


This formylation method can be a superior alternative to the much discussed Reimer-Tiemann formylation (a selective ortho formylation of phenols).

The translation couldn't have been done wo/ the much appriciated aid of Antoncho!


A 2l RBF equipped w/ a reflux conenser, a mechanical stirrer, and an addition funnel is charged w/ 32g of magnesium followed by 200ml dry Methanol. It is possible to ad more MeOH so Mg dissolves faster. The reaction is vigorus. Sometimes the flask has to be cooled to prevent the condenser from spitting around. When the mixture is beginnig to get thick, 250g of p-MeO-phenol in 300-500ml of hot toluene is added in a slow stream from the addition funnel. The mixture is heated up w/ good stirring. (Whatch out - the precipitated phenolate can stick to the bottom).
 
When all the magnesium has dissolved, the MeOH is evaporated (it is also possible not to evaporate all [Note 1]).  A suspension of 180g of Paraformaldehyde [Note 2] in Toulene is added in small portions every 10 minutes (exotermic!), w/ good stirring of the thick mixture. This takes about an hour. The mixture turns yellow and gets less thick. The mixture is than stirred for another 0,5h and acidified w/ dilute Sulfuric Acid[Note 3]. The toluene layer is separated and the toluene is driven off. The rest is 2-OH-5-MeO-Benzaldehyde.
The thus obtained product is not contaminated w/ p-MeO-Phenol, but containes the dimer as an inpurity. It may be cleaned via the bisulphite adduct or it can be distilled under reduced pressure. Yield is around 80%.

This aldehyde stains hands yellow.

[Note 1] IMHO it is not necesseary…Once I have tried not to evaporate all of it and all seemed to work fine.

[Note 2] Dry paraformaldehyde was used.

[Note 3] In the patent the acidified mixture is stirred for 5h. I'm not sure about that half an hour figure - just act by the situation. You can see the decomposition of phenolate visually as it dissolves in toluene. One has to scratch it dilligently off the flask's wall, though  :)


Notes were added based on the answers of DTT to some of Antonchos questions.

------------------------------------------------

Some final notes/questions by Swim:

- It might be necesseary to efficently stir the acidified mixture in order to achive full conversion of the phenolate. Swim experienced a lot of tar like residue after steam distilling his R-T product. Is it possible that unconverted phenolate was extracted from the not fully acidified reaction mixture, and partially that was the residue left behind? This would explain his low yields (less than 50%)...

- It might be advantageous to initiate the Mg(MeO)2 formation w/ small crystals of Iodine (at least for smaller batches).

---------------------------------------------------

References of similar formylations:

Post 290438

(Lilienthal: "2-hydroxy-5-MeO-BA: The easy way", Chemistry Discourse)

the last section in:

https://www.thevespiary.org/rhodium/Rhodium/chemistry/formylations.html




El_Zorro

  • Guest
So is this reaction selective to phenols, or...
« Reply #1 on: December 23, 2003, 01:22:00 AM »
So is this reaction selective to phenols, or could it be used on benzodioxole?


Saddam_Hussein

  • Guest
formylation
« Reply #2 on: December 25, 2003, 11:05:00 AM »
So is this reaction selective to phenols, or could it be used on benzodioxole?

Only selective ortho-formylation.


El_Zorro

  • Guest
hmm, I guess I should have read it better.
« Reply #3 on: December 29, 2003, 08:25:00 AM »
hmm, I guess I should have read it better.  So forgive a guy with no formal training and who's been out of the organic chemistry game for while, so this means that this reaction is selective to an ortho- formylation of phenols, which means that it couldn't be used to produce piperonal, because first one would have have to use catechol, but even then the formylation would be next to the phenol group, and not in the position needed, correct?


starlight

  • Guest
yes
« Reply #4 on: December 29, 2003, 12:48:00 PM »
and the 2,3 isomer does not work like the 3,4 does

dioulasso

  • Guest
Yeah...
« Reply #5 on: December 29, 2003, 01:22:00 PM »
The value of the above described formylation method is that it provides a high yealding alternative to the often messy and low yealding Reimer-Tiemann formylation.

See threads:

Post 289973

(Antoncho: "2-hydroxy-5-MeO-BA: the continuation of the saga.", Chemistry Discourse)


Post 421783

(Vitus_Verdegast: "Reimer-Tiemann", Methods Discourse)


The precursor in the abowe example is p-Methoxy phenol, wich is easily obtained or synthetized. And the product, 2-Hydroxy-5-Methoxy benzaldehyde can be methylated to the much desired 2,5-Dimethoxy benzaldehyde , the key precursor to 2,5-Dimethoxy substituted PEAs. ;D


As for Benzodioxole formylations see:

Post 382264

(Antoncho: "Zealot: azomethinic (Haack) formylation", Novel Discourse)

might be a bit tricky method

OR

Post 391475

(pHarmacist: "One More piperonal patent, High-yielding", Novel Discourse)

through the mandelic acid intermediate (could be also interresting for the formylation of 1,4-Dimethoxybenzene, or the synthesis of warious substituted pemolines 8) ).

OR
Some forms of the Vilmeier haack could also be used for Benzodioxole.


El_Zorro

  • Guest
Yeah, I recognized what the end target was for
« Reply #6 on: December 30, 2003, 12:44:00 AM »
Yeah, I recognized what the end target was for formylating p-MeO-phenol, but I was just hoping that such an easy and OTC(relatively, anyway) synth could be used on benzodioxole or catechol, but in my excitement, I missed the part about selective ortho formylation.  Dammit, I was so hopeful, too.


Antoncho

  • Guest
Questions, questions...
« Reply #7 on: December 30, 2003, 05:05:00 PM »
As (i think it was) Azole once noted, phenols in alkaline medium form phenolates, which are more susceptible to electrophylic substitution than phenols themselves by order a of thousand. That's why phenol ethers usually don't work in alkaline phenol formylations.

Now - i want to draw everyone's attention to the fact which i have myself posted a long while ago, but completely forgot until now. I'm talking about

Post 300201

(Antoncho: "3-methyl-2,5-diMeO-benzaldehyde", Novel Discourse)
, namely about this Example 1:
 

As you can see, what we have here is a much easier (non-unhydrous, no time-consuming distillations etc) reaction, which uses essentially the same components (a base and a HCHO source) but leads to a completely different product - the benzyl alcohol instead of the benzaldehyde.

Additionally to that, this variation of hydroxymethylation

Post 235445 (missing)

(Antoncho: "Formylations", Chemistry Discourse)
also uses paraform in boiling toluene - only with an acidic catalyst (does that mean it can bee used on phenol ethers as well?).

Now, with this reaction the total equation is pretty clear:

Ph-OH + HCHO ---> 2Ph(-OH)CH2OH ,

but with the original formylation in question - i'm totally lost... Where do the 'extra' 2 hydrogens go? Reduce Mg++? Highly unlikely, to say the least...

Does anyone have ideas on that?


Puzzled,
Antoncho

GC_MS

  • Guest
Benzaldehydes
« Reply #8 on: December 30, 2003, 06:48:00 PM »
Hey Antoncho,


I'm not completely sure about the actual reaction mechanism, but there are a couple of characteristics all paraformaldehyde mediated formylation reaction have in common: presence of a (very) strong base. In case of magnesium methylate, the strong base is MeO-. There are other examples to be on both Rh's site and The Hive that apply NEt3. I think there the key of the reaction mechanism is to be searched for:

Ar-CH2OH ----> Ar-CH2O- + H+ (i)

Ar-CH2O- ----> Ar-CH2=O (ii)

Ar-CH2=O ----> Ar-CH=O + H+ (iii)

I started from the benzyl alcohol. Its formation in the reaction mixture from paraformaldehyde is rather straightforward. The benzyl alcohol will get rid of its hydroxyl hydrogen in the presence of e.g. the methylate ion. The resulting ion will be converted to a carbonyl functional group after the expulsion of another H+.
Sounds OK, no?  ;)  The problem is: where do the H+ go to? It might be that methylate will be the hydrogen acceptor (being converted to methanol), but there are also reactions in which the methylate is not applied (but substituted by e.g. EtN3). They are not always present in excess. Something that is always present in excess however, is paraformaldehyde. It is known that paraformaldehyde is readily converted to formaldehyde in a variety of organic solvents. Now, suggest that paraformaldehyde degrades to formaldehyde, which reacts with a phenolate to form a benzylalcohol. However, it is also known that formaldehyde polymerizes in acidic environments. Could it be that formaldehyde is used for both the formation of the benzyl alcohol as the "hydrogen catcher" necessary for the formation of the benzaldehyde?
Concerning magnesium: some metal ions can be used to catalyze the reaction between phenolates and formaldehyde. anhydrous tin(II) and tin(IV) chloride are amongst them. FeCl3 can cause polymerization, so I doubt it can be applied (although suggested to work on Rh's site:

https://www.thevespiary.org/rhodium/Rhodium/chemistry/formylations.html

).
Let me add that this is purely theoretical  ::) .


Nicodem

  • Guest
The reaction i) is OK, but ii) does not equal...
« Reply #9 on: January 02, 2004, 12:20:00 PM »
The reaction (i) is OK, but (ii) does not equal by charge, so either you put a minus sign on the carbon (making it have 5 bonds - no good) or rearrange it to:

Ar-CH2O- -> Ar-CH=O + H-

Obviously you need a hydride acceptor if this option is to work. Formaldehide might be hydride acceptor in strongly basic conditions so if used in excess I would expect this to be the mechanism (methanol forms).
A similar mechanism goes on in the Canizzaro reaction but reversed as you get the benzyl alcohols as products with the aldehide group being the hydride donor in one molecule and an acceptor in the other.
Off course, a hydride ion per se never forms but it is rather present just in the transition state complexes.
The reaction (iii) is also wrong by charge.


GC_MS

  • Guest
Sommelet
« Reply #10 on: January 07, 2004, 03:39:00 PM »
The interaction between a carbonyl and an alkaline alcohol is believed to involve the transfer of a hydride ion:

R1-C=O + R2-CH2-O- <--> R1-CHO- + R2-CH=O

In case of formaldehyde, R1 = 2H. The methoxide is necessary to create an alkaline reaction medium. I don't know if Mg plays an important role here. Maybe NaOMe works as well. Or maybe the magnesium ion has a catalytic function in the addition of formaldehyde to the substrate (like tin(II) or tin(IV)).


Nicodem

  • Guest
Damn speculative theories
« Reply #11 on: January 07, 2004, 08:39:00 PM »
I don't think the Mg2+ has any real catalytic role here. It seams to be there for two reasons. The most obvious is because you need something to form a strong base. The other (I speculate!) is to direct the hydroxymethylation to the ortho position by forming a hexagonal intermediate in between the phenolate and the formaldehide. I draw this in its ring-nucleophile/HCHO-electrophile resonance structure - a moment before new bond formation (I just love to use this SMILES :)  but they unfortunately don’t remain as I draw them - the first structure should be planar and have the formaldehide part turned inward to form a hexagon) :












Molecule:

proposed mechanism trough a hexagonal intermediate ("[CH-]/1\C=C(/C=CC\1=[O+]/[Mg][O+]=C)OC>>c12cc(ccc1O[Mg]OC2)OC")


Now, who needs ortho selectivity anyway? With the usual NaOH catalyzed hydroxymethylation you would get the same product. The ortho position in relation to the –OH group is the most activated and even if there is some meta product formed it doesn’t matter (after the O-methylation you get the same thing: 2,5-MeO-BA).
But from what I know the usual hydroxymethylation of phenoles yield mostly double- or multi-hydroxymethylated products or bakelite-goo if the formaldehide is used in excess. It could be that the toluene solvent somehow prevents this. Don’t know.
Though, if someone is to confirm it works with MeONa it should work with NaOH as well since phenolate is still less basic than hydroxide, but its harder to evaporate the residual water than methanol.
Anyhow, the formation of the benzaldehide is still a mystery to me. If I remember well, formaldehide can be used as a hydride donor in the Canizzaro reaction to get a better yield of the benzylalcohol on the account of formic instead of the benzoic acid. By the same assumed mechanism we have a contrary equilibrium push here!? I can’t efficiently explain this neither on the excess formaldehide basis neither by some ring substituent or solvent effect. Sure, in the case of Al(i-PrO)3/i-PrOH reduction of ketones the very large excess of isopropanol favors ketone reduction, but thermodynamically the benzylalcoholes are nearly not the same as i-PrOH.
I would also like to get an acceptable explanation so I could better believe in such a simple preparation.


Rhodium

  • Guest
Magnesium ortho-Formylation Mechanism
« Reply #12 on: January 08, 2004, 04:04:00 AM »
According to the article below, this is the actual reaction mechanism:




Convenient Method for the ortho-Formylation of Phenols
Nini U. Hofslokken and Lars Skattebol

Acta Chemica Scandinavica 53, 258-262 (1999)

(https://www.thevespiary.org/rhodium/Rhodium/pdf/mg-ortho-formylation.pdf)

Abstract
Phenolic derivatives are formylated selectively ortho to the hydroxy group by paraformaldehyde with magnesium dichloride-triethylamine as base. With alkyl-substituted phenols excellent yields of the corresponding salicylaldehyde derivatives were obtained. Similar results were obtained with chloro-substituted phenols and with 3-and 4-methoxyphenol, while 2-methoxyphenol was unreactive. A good yield of methyl 3-formyl-4-hydroxybenzoate was obtained by this method as well, but generally phenols with electron-attracting groups reacted sluggishly; the long reaction times required caused the formation of by-products, particularly MOM-derivatives of the phenols.


Nicodem

  • Guest
That's a much nicer haxagon
« Reply #13 on: January 08, 2004, 11:33:00 AM »
Thanks, Rhodium, that is kind of what I miserably failed to draw with those SMILES.
That paper made my finaly belive that it is really possible. Have to read it more carefuly but it looks like the ring substituents are the answer. And it made me think if Mg(OH)2 would also work? They use an amine as base, so there is no real need for a strong base!?
I never ever saw anything like that mentioned anywhere else.
Looks worth trying.


GC_MS

  • Guest
Doubtful
« Reply #14 on: January 08, 2004, 06:20:00 PM »
I don't think Mg(OH)2 is a valid substitute. The base functions as hydrogen acceptor (Et3N -> Et3NH+ or MeO- -> MeOH). In your case, the hydroxide is converted to water, while the reaction requires an anhydrous reaction environment.


Nicodem

  • Guest
Actually, I was thinking more like this: ...
« Reply #15 on: January 09, 2004, 11:01:00 AM »
Actually, I was thinking more like this:
1-melt Mg(OH)2 and p-MeO-phenol together to form the phenolate
2-add toluene
3-disstill of the toluene/water azeotrope
4-add acetonitrile+paraformaldehide, reflux and so on.

Or instead of distiling off the toluene just using a Dean-Stark trap to get the water out and proced like DDT.

Why do you think it needs anhydrous conditions?
The normal hydroxymethylation are mainly done in aqueus conditions, but I do agree that water probably inhibits the hydride transfer in the proposed mechanism. I also think traces of water would not harm that much as DDT found that leaving some methanol does not harm neither.

Besides if triethylamine is enough basic to promote the formylation, so should the phenolate anion (they are both bases of similar strentgh): p-MeO-C6H4-O- + H+->p-MeO-C6H4-OH
That is if the triethylamine functions only as a base.


Nicodem

  • Guest
HCHO/Mg(II) formylation + methylation patent
« Reply #16 on: March 06, 2004, 04:20:00 PM »
This patent is related to the Mg2+ mediated formylations with formaldehyde already discussed in this thread.

Patent US6670510

: Process for preparing 2,5-dimethoxy benzaldehyde

Abstract
The present invention recites a process for the of 2,5-dimethoxy-benzaldehyde by reacting a 2-hydroxy-5 methoxy-benzaldehyde with a suitable metal hydroxide in the presence of a suitable solvent to make a metal salt of 2-hydroxy-5-methoxy benzaldehyde comprising. The invention further provides a method for alkylating said metal salts with dimethylsulfate in the presence of a suitable solvent so as to provide a 2,5-dimethoxy benzaldehyde.

Example 1: Preparation of Crude 2-Hydroxy-5-Methoxybenzaldehyde
Post 0033 (not existing) A mixture of 600 g of acetonitrile, 186.2 g of 4-methoxyphenol (1.5 moles) and 214 g (2.25 moles) of anhydrous magnesium chloride was warmed to 45°C. and 227.4 g (2.25 moles) of triethylamine was added dropwise at that temperature. Paraformaldehyde (150 g, 4.75 moles) was then added and the reaction was heated to reflux. Methanol was removed over 3 hours by distillation, using a 10-inch column packed with Penn-State packing. The weight of the methanol removed (vapor temperature 60-65°C.) was 130 g. The solution was cooled to 60[deg.] C. and added to 900 mL of water. The solution was acidified with 240 mL of conc. HCl and extracted with 150 g of ethyl acetate. The ethyl acetate solution was washed once with 300 mL of tap water, clarified through diatomaceous earth, and stripped on a rotary evaporator at 10 mm pressure and 85°C. bath temperature. The weight of the product was 206.5 g, the liquid chromatograph area percent was 86%, and liquid chromatograph weight percent was 82%. The overall assay yield was 74%.

Comparative Example 1(a): Use of Crude 2-Hydroxy-5-Methoxybenzaldehyde
The above crude oil (6.1 g, 81% assay, 0.33 moles), 25 mL of acetone, 6.8 g of potassium carbonate, and 5 g of dimethylsulfate was refluxed for 3 hours and drowned into 200 mL of water. The dark oil solidified overnight, and the product was collected by filtration and dried to yield 5.6 g of black solid. The weight percent assay was 83%, resulting in an assay yield of 77%. The liquid chromatograph area percentage was only 62%.