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Author: Subject: deacetylation, please help?
mr423
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smile.gif posted on 7-8-2009 at 21:22
deacetylation, please help?


say you have an acetylated organic compound (R-COCH3), where acetyl is the only side chain. if you add lithium metal, would it work like this?:
R-COCH3 + Li > R-H + LiOH + C2H2

also if instead of adding lithium you added lithium hydride or aluminum hydride, would it work like this?:
R-COCH3 + 2LiH = R-CH2CH3 + LiOH/LiO

[Edited on 8-8-2009 by mr423]
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[*] posted on 7-8-2009 at 21:36


the acetyl group is ~COOCH3. You have a methyl ketone as the formula is written.
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mr423
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[*] posted on 7-8-2009 at 21:43


are you sure? i mean i know wikipedia isn't that reliable but
http://en.wikipedia.org/wiki/Acetyl
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[*] posted on 7-8-2009 at 21:50


oops, sorry thought acetate... or something. I shouldn't post whilst intoxicated! :o
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mr423
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[*] posted on 7-8-2009 at 21:52


no problem man, thanks for the quick responses
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ammonium isocyanate
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[*] posted on 7-8-2009 at 22:03


The wikipedia article is talking about the acetyl group, as in acetaldehyde (where the R is an H), acetic acid (where the R is an OH), or acetamide (where the R is NH2). When one is referring to that specific side chain, then it is generally a methyl ketone, and so a different nomeculture is used, and thus the term acetyl is rarely used.

However, you may be mistakenly using acetyl to describe an acetate amide or ester, which are compounds derived from acetic acid and contain the functional group -COOCH3.

Perhaps you can clear this up for us by telling us what compound you are dealing with, or at least whether or not it truly contains the acetyl group.

As for the reaction you listed above, I'm 98% certain it wouldn't work. I just don't see any plausible intermediate that would result in the formation of lithium hydroxide and acetylene. Additionally, the reaction would be thermodynamically unfavorable, as acetylene has very stressed bonds.

Edit: The second reaction you just posted would work, at least with lithium aluminum hydride as the reducing agent. Sodium borohydride would also do IIRC. Not sure about LiH. Only problem is that such strong reducing agents could damage other parts of the molecule (i.e. say bye-bye to nitro groups, double bonds, etc.).


[Edited on 8-8-2009 by ammonium isocyanate]




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mr423
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[*] posted on 7-8-2009 at 22:10


well in the case of trying to deacetylate melatonin (see link below) at R-N1 to 5-methoxy tryptamine (don't worry, both legal)

http://en.wikipedia.org/wiki/Melatonin

Edit: ok thanks a lot man. would aluminum hydride by itself work?


[Edited on 8-8-2009 by mr423]
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not_important
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[*] posted on 7-8-2009 at 22:28


In that case you're dealing with an amide, and generally hydrolysis is used to remove it; strong reducing agents are more likely to convert the acetyl group to an ethyl group.

Amides can be hydrolysed by strong acids, not applicable in your example because of the sensitivity of the indole nucleus to acids, strong base, or enzymes. The specific amid will determine which general method to use, and the particulars needed to optimise the reaction.

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ammonium isocyanate
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[*] posted on 7-8-2009 at 22:29


The biggest problem I see with removing that acetyl group is the methoxy group on the other side of the molecule. Many deacetylation reactions might cleave it off.

I would try hydrolysis under basic conditions, using either triethylamine or sodium hydroxide as the base. The problem is that this might result in methanol formation from the methoxy group being replaced by a hydroxy group. However, I don't think this would happen to a great extent, as the reverse reaction would be more favorable.

Again, the big problem with the first rxn is the lack of a suitable intermediate. Besides, IMHO lithium would wreak havoc on melatonin.

Edit: I have no experience with aluminum hydride, so I'm not sure if it would work, but I haven't seen it mentioned as a method of choice in any scholarly journals as far as I can recall (although I just had surgery and so the hydrocodone I'm on probably isn't helping my memory any).

[Edited on 8-8-2009 by ammonium isocyanate]




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mr423
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[*] posted on 7-8-2009 at 22:36


ah right that methoxy group. yeah that definitely turns down the idea of using lithium.

i'm going to have to look into hydrolysis, but i guess not_important says it won't work...

interesting. i will have to think about this. thanks a lot!

[Edited on 8-8-2009 by mr423]
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[*] posted on 8-8-2009 at 00:25


It is a very easy transformation using acid or base hydrolysis - Take your pick.:)

Best luck!;)
[i don't know why the files have been renamed but open as pdf manually]
Attachment: phpCNCVpY (190kB)
This file has been downloaded 1433 times

Attachment: phpXjxhJJ (42kB)
This file has been downloaded 1270 times

[Edited on 8-8-2009 by sonogashira]
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not_important
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[*] posted on 8-8-2009 at 00:43


That methoxy is part of an aromatic ether, and isn't too easy to cleave. Basic or enzymatic hydrolysis shouldn't touch it.

Note that both nitrogens are attached to hydrogens that will react with alkali metals and their hydrides. That double bond in the pyrol ring is somewhat reactive too, and may not tolerate some of the more aggressive reducing agents or associated conditions.
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mr423
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[*] posted on 8-8-2009 at 01:16


so how exactly does base hydrolysis work? simply putting the amine in an NaOH solution?

i'm sorry i honestly don't know anything about chemistry

[Edited on 8-8-2009 by mr423]
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[*] posted on 8-8-2009 at 06:36


If you know nothing about chemistry then why are you starting with such a complicated substrate... sounds to me like you're after a quick fix.
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mr423
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[*] posted on 8-8-2009 at 06:51


lol ok was just interested as to how it would work theoretically...

i any case, i think i got it all figured out. thanks!

[Edited on 8-8-2009 by mr423]
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[*] posted on 8-8-2009 at 09:29


That is either an ether (R-CH2-O-CH3) or a ketone (R-C-O-CH3). The acetyl group comes from the ester (R-CH2-O-C-O-CH3) with acet-ic acid (HO-C-O-CH3).

Cleaving an ether is a difficult proposition and usually involves the application of a strong Lewis (e.g. BBr3) or mineral acid (HI'>'HBr'>'HCl) and heat.

The really crummy part is that the products are usually less volatile than the starting material so that work-up is a mess.

If you are not in touch with the nomenclature, you are probably not aware of the intricacies involves with handling organolithiums (the metal itself is easier to handle than Na). This week's C&EN news details the sad and untimely demise, by fire, of a graduate student handling tBu-Li. Take care.

Cheers,

O3




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[*] posted on 8-8-2009 at 11:54


Quote: Originally posted by mr423  
say you have an acetylated organic compound (R-COCH3), where acetyl is the only side chain. if you add lithium metal, would it work like this?:
R-COCH3 + Li > R-H + LiOH + C2H2

also if instead of adding lithium you added lithium hydride or aluminum hydride, would it work like this?:
R-COCH3 + 2LiH = R-CH2CH3 + LiOH/LiO

No and no. Read a book about basic organic chemistry for more details.
Quote:
well in the case of trying to deacetylate melatonin (see link below) at R-N1 to 5-methoxy tryptamine

Please UTFSE before opening new threads! Asking questions that have been already answered and doing it in such a shady way as to confuse others is not particularly welcomed here. Also, opening new threads without providing any reference should be done in the Beginnings section where I'm moving this.
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8-8-2009 at 11:55
meohmix
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[*] posted on 19-9-2018 at 08:55


Thanks to sonogashira for the two deacetylation links given above.

Having tried the sulfuric acid deacetylation given in the first link and being (very) unimpressed with my results, I am going to give the hydroxide method given in the second link a try and was hoping members could help with me with my interpretation of the procedure:

This compound was obtained by utilizing a modification of the procedures of the Hag and Nestle companies, developed in making 5-methoxytryptamine, which was the starting material for this procedure (17). Thus melatonin (2.2 g, 0.019 mole) dissolved in about 50 ml 1:1 aqueous ethanol was treated with 50 ml 1:1 of aqueous ethanolic KOH in excess (8–12 molar) and catalytic amounts (200 mg) of Na2S2O4. The mixture was heated to reflux under N2 for 16 hr. The reaction mixture was cooled and extracted with CHCl3. The organic extract was, washed (H2O), dried (Na2SO4) and evaporated to dryness on a rotary evaporator, to give a colorless syrup of 5-methoxytryptamine; this was dissolved in ether and EtOH and treated with HCl in EtOH/ether to give the hydrochloride, 1.7 g of 1 (70%). Mp 243–245°C (dec). Recrystallization with EtOH/ether gave mp 246–248°C (dec). This compound was converted to the free base with aqueous KOH and extracted into CH2Cl2, dried (Na2SO4) and evaporated.

Please confirm that these are reasonable interpretations:

"about 50 ml 1:1 aqueous ethanol" means 25ml of ethanol in 25ml H20...Correct?

"treated with 50 ml 1:1 of aqueous ethanolic KOH in excess (8-12 molar)" means 25ml of ethanol in 25ml of an 8-12 molar KOH solution...Correct?

I appreciate that this procedure is obviously not an exacting one ("about", "excess", "(8-12)") just want to be assured I am not to tolerably confused before I expend any more time and materials.

Your time is appreciated!
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[*] posted on 19-9-2018 at 09:22


Aqueous ethanolic KOH is confusing, something cannot be aqueous and ethanolic simultaneously. Similar nomenclature problems arise when potassium ethoxide is dissolved in water or potassium hydroxide in ethanol. The hydroxide and the alkoxide are really in equilibrium.

Hydroxide does the hydrolysis, ethanol is added to keep substrate and product in solution. I would simplify the procedure to dissolve the substrate in EtOH, add it to conc. KOH, and add more ethanol if the reaction becomes heterogenous and reflux until completion.

The real question is why sodium dithionite (Na2S2O4) is added, supposedly to prevent oxidative dimerization of the indole and why not another, more readily available, oxygen scavenger such as sulfite or thiosulfate?
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[*] posted on 20-9-2018 at 00:24


The hydrolysis of melatonin to the amine is a difficult reaction to run. I can tell you from experience that the sodium hydroxide route described above does work but not well. It is absolutely imperative to run this reaction under nitrogen or other inert atmosphere else you will end up with a significant purification problem. The hydrolysis under acidic conditions is even worse, claims in papers and patents not withstanding.

There is also another route to removal of the acetyl group which uses ethylenediamine/NH4Br (not hydrazine as originally posted) which is reported to be quite clean. I do not have the reference at hand but it is easily found with a Google scholar search. Input the data for the reference (authors, Journal, etc.) quoted above and look under citing references. I have not tried this reaction yet but plan to do so eventually.

AvB

[Edited on 20-9-2018 by AvBaeyer]
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[*] posted on 20-9-2018 at 18:04


This transamidation method:

https://onlinelibrary.wiley.com/doi/abs/10.1002/anie.2011073...

might work with a suitable cyclic secondary amine (e.g. pyrrolidine, piperidine) as the "acetyl" scavenger. These cyclic amines tended to be the most efficacious transamidation targets, so I infer their amides are more stable. Since the indole NH is neither protonated nor deprotonated it should be less reactive under these conditions. Apparently the reaction of hydroxylamine with amides is fastest near pH 6:

https://pubs.acs.org/doi/abs/10.1021/ja01078a042?journalCode...

According to this second paper, hydroxylaminolysis may be much faster than hydrolysis in the presence of sufficiently high concentrations of hydroxylamine.




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[*] posted on 23-9-2018 at 14:58


There is a recent article on this hydrolysis, previously posted on SM.

Softbeard, at the end of the discussion, provided a PDF link. I can't seem to post the link here, but you can go there.

https://www.sciencemadness.org/whisper/viewthread.php?tid=79...


To demonstrate the advantages of utilizing the N-acetyl
protecting group, we sought to deprotect melatonin (3b) and
N-acetyl-2-methyltryptamine (3h). To our delight, by treating
3b,h with a combination of ammonium bromide and
ethylenediamine under microwave irradiation,20 we were able
to isolate the free 5-methoxytryptamine (7b) and 2-
methyltryptamine (7h)21 in 89% and 78% yields, respectively,
avoiding the use of strong acids or bases (Scheme 2).




[Edited on 23-9-2018 by zed]

[Edited on 23-9-2018 by zed]
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