Sciencemadness Discussion Board

Vanillin to Piperonal

Korialstrasz69 - 16-3-2016 at 12:44

I found another way that makes perfect sense to me at least,but i know it's wrong and i want to know why,i am still a beginer in organic chemistry
Most reactions involve demethylating of Vanillin followed by remythelation by Williamson Ether Synthesis i think.
Point is,why the fuck demethylate and then methylate again ? Just free radical halogenate !
Of what i know is, the alcohol group wont be effected byfree radical bromine ir by hydrobromic acid, And the ring is resistant to halogination without a catalyst i suppose,which leaves us with the methyl and the carbonyl groups to be halogenated.By slowly adding bromine to vanillin while hv ight is directed through the apparatus bromine will homolise brominating the methyl group's first hydrogen,which should rect immediatly with the alcohol group through a Williamson Ether Synthesis reaction,closing a ring.
Further bromination will occure to both the methyl and carbonyl group (I suppose).
So after that the product is dehalogenated,i know things can be dehaligenated but i don't exactly know how really
I also know know that c-Br bond is the weakest bond possible since iodine wont work here
So if the bone Br-c is weak and C-H is strong shouldn't it be fairly easy to debrominate the product back into piperonal ?
That's pretty much it.
And hey sorry about a few spelling errors and whatnot english isn't my first language
And no am making no drugs i don't even have a lab i am just discovering the beauty of orgnic synthesis and thanks ahead ya'll

chemplayer.. - 16-3-2016 at 15:19

Having done lots of experiments with vanillin, I can tell you that the ring is extremely reactive as it's a phenol and very succeptible to electrophilic substitution. Iodine in KI solution will substitute, as will dilute nitric acid. What you are thinking of will work for relatively unreactive rings such as toluene but in this case you're going to run into difficulties.

Due to it being a p-phenol the ring is also fairly easily oxidised and so anything like chlorine stands a reasonable chance of oxidising this as well I would guess, as well as potentially oxidising the aldehyde group to the carboxylic acid.

Demethylation is a pain. We've tried about 6 methods so far, all of which need quite difficult reagents to obtain and none have given us more than about a 10% yield, even with really stringent conditions and drying. There may be a sneaky way but we haven't found it.

That said, don't be discouraged as vanillin itself is easily available, very interesting and you can do a lot of great experiments with it (condensations, benzylidene compounds, nucleophilic substitution on the carbonyl, electrophilic substitution etc.).

zed - 16-3-2016 at 16:46

Aw. Vanillin may be difficult to de-methylate. Ethyl Vanillin....not so much.

Probe via the search engine. Most problems have been broached, and solved, previously.

Also of note....There may be difficulties de-methylating vanillin via a strong base, as aldehydes don't like that. Eugenol however, might be cleaved in such a manner. Of course, it may isomerize to the Iso-propenyl form, during de-methylation, but this might be of little consequence.

As allyl and isopropenyl isomers, may react similarly in some popular reaction sequences.

chemrox - 16-3-2016 at 17:33

Quote: Originally posted by zed  
Eugenol however, might be cleaved in such a manner. Of course, it may isomerize to the Iso-propenyl form, during de-methylation, but this might be of little consequence.

As allyl and isopropenyl isomers, may react similarly in some popular reaction sequences.

Indeed so! Or get an essential oil and isomerize it with base. If you need a small q. of piperonal u2u me. I have rather more than I would need for the next 10 years.

DraconicAcid - 16-3-2016 at 20:08

Quote: Originally posted by chemplayer..  
Having done lots of experiments with vanillin, I can tell you that the ring is extremely reactive as it's a phenol and very succeptible to electrophilic substitution. Iodine in KI solution will substitute, as will dilute nitric acid.


Hmmm...I wonder if you could do a one-pot transformation of vanillic to nitrovanillic acid?

chemplayer.. - 16-3-2016 at 21:29

Yes, heating vanillin with KOH in the molten state gives protocatechuic acid (demethylated, but converted into the carboxylic acid). So you can demethylate using strong base but the aldehyde will certainly react.

And nitration of vanillin to 5-nitrovanillin is easy but does need the right conditions for a quality product to be obtained. (We just did a video on this last week in fact!) I'm guessing it would work likewise for vanillic acid too.

UC235 - 16-3-2016 at 21:53

Quote: Originally posted by chemplayer..  
Yes, heating vanillin with KOH in the molten state gives protocatechuic acid (demethylated, but converted into the carboxylic acid). So you can demethylate using strong base but the aldehyde will certainly react.


A procedure for any interested parties: http://www.orgsyn.org/Content/pdfs/procedures/CV3P0745.pdf

Korialstrasz69 - 16-3-2016 at 23:58

First thanks a whole lot
And about demythelation you're the pro and you seemed to try so many way but just to make sure have you tried AlBr3/nitrobenzene method ? I don't have a lab and never tried it before but it seems like the reagents are easy to aquire for chemists like you,although hazardous.
Looking forward for your video about it lads and thanks again !

chemplayer.. - 17-3-2016 at 06:20

Definitely not pros! But we haven't tried that method no ; nitrobenzene in large quantities (as a solvent) is pretty toxic and although we have access to AlCl3 and AlI3, AlBr3 is going to be a bit more difficult to prepare in practise due to the extremely exothermic reaction between iodine and aluminium. It could be one for us to try at a future date on a small scale perhaps.

Korialstrasz69 - 17-3-2016 at 06:38

Whoa thanks god I don't do chemistry I would've died long ago

solo - 17-3-2016 at 09:28

"Quote: Originally posted by chemplayer..
Yes, heating vanillin with KOH in the molten state gives protocatechuic acid (demethylated, but converted into the carboxylic acid). So you can demethylate using strong base but the aldehyde will certainly react.


A procedure for any interested parties: http://www.orgsyn.org/Content/pdfs/procedures/CV3P0745.pdf"

.....in that procedure they employ the sulfur dioxide gas, i wonder if that 's really needed , hence conc. Hcl would be enough to knock off the K ......since the note says, "5. The sulfur dioxide treatment prevents the formation of a very dark-colored product when the reaction mixture is acidified with a strong acid."........solo


Reference Information


Fussion of vanillin
U. S. pat. 2,547,920

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

Reaction of Vanillin and its Derived Compounds IV The Caustic Fusion of Vanillin
Irwin Pearl
J. Am. Chem. Soc.
68, 2180 (1946).


Summary
Caustic fusion of vanillin below 240-245˚C results in very high yields of vanillic acid free from protocatechuic acid. Fusion of vanillin above 240-245˚C yields protocatechuic acid free from vanillic acid. The critical demethylation temperature varies somewht with the alkali-vanillin ratio. Reactions of vanillin with strong alkali solution at elevated temperatures does not yield either of the acids.

Attachment: Fusion of vanillin-pat2547920.pdf (143kB)
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Attachment: Reactions of Vanillin and its Derived Compounds. IV.1 The Caustic Fusion of Vanillin.pdf (269kB)
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[Edited on 17-3-2016 by solo]

zed - 17-3-2016 at 14:43

Arggg. What I was suggesting was....Vanillin to Piperonal, might be difficult. But....

Eugenol to Iso-Safrole, could be more do-able....Via de-methylation of Eugenol.....followed by Methylenation.

solo - 17-3-2016 at 18:26

.....but how to convert the protocatechuic acid to the desired protocatechuic aldehyde so the OH are not touched or changed.....however there is the conversion of the acid to an acid chloride, then an ester reduced to the desired aldehyde......solo

[Edited on 18-3-2016 by solo]

byko3y - 17-3-2016 at 22:05

Korialstrasz69, aldehyde is the first to react with free radicals. Also, the ring is not resistant to halogenation, as you might know by 5-bromovaniliin procedure. Free radical reaction is not the way Williamson ether synthesis is done.
chemplayer.., I'm sure either you did not try AlCl3+pyridine or you did it wrong. It definitely works giving high yields. Already mentioned AlBr3/ArNO2 is also 100% working, but requires the toxic solvent. AlBr3 and AlI3 can be prepared in a cold solvent (dichloromethane is known to work, but it reacts with aluminium halides, leading to AlCl3 in the end).
I was able find your pyr+hcl and alcl+thiourea videos, I tried AlCl3-thiourea route long time ago and I did not manage to extract any decent amount of protocatechualdehyde(PA) in the end. It's really hard to say anything about products without chromatography, and sadly I was not used to it those days, and I haven't repeat the experiment.
FeCl3 can be used for detecting vanillin/PA: purple for vanillin, green for PA. NaNO2, then heating, then NaOH is another simple one, however, the colors do not depend on the structure of the compound and are rather random (but still the same for the same compound). For PA colors are NaNO2 - yellow-orange, heating - yellow-red, NaOH - yellow-red (same).
Can it be that your vanillin is not a 100% vanillin? Does it have a correct melting point?
zed, do you think isosafrole is more usefull than piperonal? Also, you have much less choice of reagents/conditions with eugenol - you can't use no acidic reagents until you get safrole/isosafrole. Does the eugenol-PTC-AlI3 route even work? I'm pretty sure nobody have ever suceeded doing it.

Loptr - 17-3-2016 at 22:21

What about protecting the carbonyl group first by forming its thioacetal? I recall the thioacetal being more resistant to basic and acidic solutions, which a strong acid is needed for the ether cleavage.

The protocatechuic acid can be turned into the alkylketone using lithium hydride followed by alkyllithium, as seen in the cyclohexyl ketone preparation on OrgSyn (http://www.orgsyn.org/demo.aspx?prep=CV5P0775). This would be a pain though because of the strict requirement of air and moisture exclusion, not something I am equipped for at this time. The reduction of the -OH to -H sounds like an LAH or catalytic H reduction, but something would likely have to be done with the hydroxy groups previously regarding the alkylation and reduction reactions.

If I were to do this, my first attempt would be to follow the procedure outlined in this paper (http://www.ncbi.nlm.nih.gov/pubmed/23127660) where vanillin is taken all the way to MDMA, and the demethylation is accomplished with pyridine and AlCl3.

(clarification: follow the demethylation procedure in the paper, not take it all the way to MDMA. that would conflict with my desire to not be incarcerated.)

[Edited on 18-3-2016 by Loptr]

Loptr - 18-3-2016 at 03:44

It seems that link doesn't provide a download of the PDF.

Attachment: Synthesis and impurity profiling of MDMA prepared from commonly available.pdf (1MB)
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Here is another paper that mentions dealkylation of the ether using sulfuric acid, and the already mentioned lewis acid method.

[Edited on 18-3-2016 by Loptr]

Attachment: Bjorsvik.Liguori.Minisci.Synthesis.Vanillin.Isovanillin.Heliotropin.pdf (183kB)
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[Edited on 19-3-2016 by Loptr]

Loptr - 4-5-2016 at 20:47

Does anyone have more info on the sulfuric acid demethylation?

chemplayer... - 5-5-2016 at 16:55

byko3y - We've tried this many times now and with some variation in conditions and reactants, and we always get the same result which is ~10% yield.

Our vanillin is Rhodia food grade and then dessicated for a week to ensure it's water-free. The DCM is reagent AR grade and we've even tried drying it with P2O5 to be sure.

The only thing we can think of is that the process we use to make the AlCl3 (zinc chloride and Al powder) is resulting in some zinc chloride impurity sublimed into the AlCl3 product. The AlCl3 thus formed works brilliantly for Friedel-Crafts reactions, but perhaps the zinc interferes in this particular case.

For example, we once tried reacting vanillyl alcohol with Lucas' reagent (ZnCl2 in HCl) to see if this would form the chloride - but the result was a dark blue complex. Clearly some vanillin-like compounds and zinc are reactive together.

Loptr - 5-5-2016 at 18:43

ChemPlayer, have you attempted using magnesium iodide for demethylation?

Selective demethylation and debenzylation of aryl ethers by magnesium iodide under solvent-free conditions and its application to the total synthesis of natural products
DOI: 10.1039/b916969e
Kai Bao,a,b Aixue Fan,a Yi Dai,b Liang Zhang,a Weige Zhang,*a Maosheng Chenga and Xinsheng Yao*b

chemplayer... - 6-5-2016 at 03:40

Interesting. Thanks for the reference.

solo - 6-5-2016 at 04:12

Reference Information



Selective demethylation and debenzylation of aryl ethers by magnesium iodide under solvent-free conditions and its application to the total synthesis of natural products
Kai Bao,a,b Aixue Fan
Organic & Biomolecular Chemistry
2009,Issue 24,pp. 5084-5090
DOI: 10.1039/b916969e


dehalogenation_good.jpg - 7kB



Abstract
An efficient selective demethylation and debenzylation method for aryl methyl/benzyl ethers using magnesium iodide under solvent-free conditions has been developed and applied to the synthesis of natural flavone and biphenyl glycosides. A variety of functional groups including glycoside were tolerated under the reaction conditions. Experimental results indicated that the removal of an O-benzyl group was easier than that of an O-methyl group, regardless of wherever they were meta or para to the carbonyl. Thus selective debenzylation can be achieved for substrates bearing both benzyloxy and methoxy groups.


Attachment: Selective demethylation and debenzylation of aryl ethers by magnesium iodide under solvent-free conditions and its appli (145kB)
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[Edited on 6-5-2016 by solo]

careysub - 6-5-2016 at 12:24

Quote: Originally posted by Loptr  
ChemPlayer, have you attempted using magnesium iodide for demethylation?

Selective demethylation and debenzylation of aryl ethers by magnesium iodide under solvent-free conditions and its application to the total synthesis of natural products
DOI: 10.1039/b916969e
Kai Bao,a,b Aixue Fan,a Yi Dai,b Liang Zhang,a Weige Zhang,*a Maosheng Chenga and Xinsheng Yao*b


Unfortunately it does not look like this method will demethylate vanillin. They make vanillin in two different ways (p-demethylation and p-debenzylation) but with the p-hydroxy the m-methoxy seems protected (the do demethylate m-anisaldehyde).

[Edited on 7-5-2016 by careysub]

Loptr - 6-5-2016 at 13:06

Quote: Originally posted by careysub  
Quote: Originally posted by Loptr  
ChemPlayer, have you attempted using magnesium iodide for demethylation?

Selective demethylation and debenzylation of aryl ethers by magnesium iodide under solvent-free conditions and its application to the total synthesis of natural products
DOI: 10.1039/b916969e
Kai Bao,a,b Aixue Fan,a Yi Dai,b Liang Zhang,a Weige Zhang,*a Maosheng Chenga and Xinsheng Yao*b


Unfortunately it does not look like this method will demethylate vanillin. They make vanillin in two different ways (o-demethylation and o-debenzylation) but with the o-hydroxy the m-methoxy seems protected (the do demethylate m-anisaldehyde).


I did notice this but I hadn't evestigated it myself.

clearly_not_atara - 6-5-2016 at 17:51

There's not really good reason to bother with vanillin and eugenol when catechol is so much easier. If you're not trying to make drugs, benzodioxole is about as useful as piperonal -- and if you are, it still is, but the routes are less well-known.

Interestingly, oxidizing salicylic acid with Fenton's reagent (Fe2+/H2O2) gives catechol-3-carboxylate.

Benzodioxole is most easily functionalized by chloromethylation:

https://www.erowid.org/archive/rhodium/chemistry/piperonal.n...

[Edited on 7-5-2016 by clearly_not_atara]

Loptr - 7-5-2016 at 04:49

Just out of curiousity, what if you protected that p-hydroxyl group with O-alkylation? Would this be a non-issue then? I have read how notoriously difficult this demethylation is, but what if both hydroxyl group were ethers.

Chemi Pharma - 8-5-2016 at 12:25

I think the attachment will be very useful for those who intends to demethylate vanilin or iso-eugenol and want to close the benzodioxile ring later to produce piperonal and safrole.

Two methods to demethylation and two methods to close the benzodioxile ring are given.



Attachment: amphetamine.rar (2.4MB)
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RosarioHeis - 26-4-2017 at 13:08

Quote: Originally posted by chemplayer...  
Interesting. Thanks for the reference.


Something you may want to try as far as you can get some aromatic solvent

http://chemistry.mdma.ch/hiveboard/novel/000291912.html

Rhodanide - 27-4-2017 at 05:06

Quote: Originally posted by chemplayer..  

That said, don't be discouraged as vanillin itself is easily available, very interesting and you can do a lot of great experiments with it (condensations, benzylidene compounds, nucleophilic substitution on the carbonyl, electrophilic substitution etc.).


Hey, did you ever get to successfully make Vanillin nitrile/cyanide?

chemplayer... - 3-5-2017 at 08:22

No unfortunately not - all attempts failed but we think that nickel boride may be a workable path to vanillylamine from the oxime.

byko3y - 3-5-2017 at 10:25

Vanillin nitrile would be a relatively unstable compound dure to the fact it's slightly acidic by itself, and reacts with itself in acidic conditions ( https://en.wikipedia.org/wiki/Hoesch_reaction ). I'd suggest to protect hydroxy group first.

clearly_not_atara - 3-5-2017 at 13:57

Doesn't Zn/NiCl2 reduce oximes? There is a patent saying so:

http://www.sciencemadness.org/talk/viewthread.php?tid=20153#...

Although in this case it seems that Zn/NiCl2 reduction of the imine would suffice.

However I think the most efficient route would be the Mannich reaction of formaldehyde and ammonia with guaiacol, would it not?

Cryolite. - 3-5-2017 at 14:02

I always thought the Mannich was a little dodgy with primary amines and ammonia, since the resulting primary/secondary amine forms a more stable imine leading to overalkylation.

byko3y - 4-5-2017 at 02:02

clearly_not_atara, it's called a Duff formylation. I'm not sure whether it's possible to stop the reaction at amine stage. Also ortho-to-hydroxy substitution product might be a major one instead of para.