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Author: Subject: Help with 2,2'-bisthiazole
Boffis
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[*] posted on 11-8-2016 at 11:28
Help with 2,2'-bisthiazole


I am trying to prepare 2,2'-bisthiazole or 2,2'-dithiazole and I have a problem.

I am using a modified version of the reactions described in Vogel's Textbook of organic chemistry 3rd Ed p840-841 where he describes the preparation of 2-aminothiazole and 2-amino-4-methylthiazoles by the condensation of 2-chloroacetaldehyde or chloroacetone with thiourea. The chloroacetaldehyde is generated in situ from 1,2-dichloroethyl-ethylether. I do not have any of the latter compound but I do have some chloroacetaldehyde dimethyl acetal which should yield the chloroacetaldehyde in situ in a similar fashion. To test this I ran the 2-aminothiazole synthesis as per Vogel but substituted the 1,2-dichloroethyl ethyl ether with chloroacetaldehyde acetal in a molar equivalence. The reaction was completely successful giving a similar yield to the original procedure. The reaction time was reduced from about 2 hours to about 25 minute.

thiourea and thiooxamide.jpg - 5kB
The comparison between thiourea and dithiooxamide thioamide and enol forms

The reactive species in this reaction is probably the enolized thioamide and several examples of this type of reaction are known. It appears that in thiourea the concentration of the tautomeric amino-formothiolimine is sufficient to bring about the react. In my modification of the reaction I am trying to use dithiooxamide which is the diamide of thiooxalic acid or basically two thioamide groups back to back. It acts as an acid by virtue of the partial enolization of the thioamide groups to a similar formothiolimine group. So dithiooxamide (rubeanic acid so called) should react with two molar equivalences of chloroacetaldehyde to give the required 2,2'-bisthiazole.

C<sub>2</sub>N<sub>2</sub>S<sub>2</sub>H<sub>4</sub> + 2C<sub>4</sub>H<sub>9</sub>O<sub>2</sub>Cl &rarr; C<sub>6</sub>H<sub>4</sub>S<sub>2</sub>N<sub>2</sub> + 2HCl + 4CH<sub>4</sub>O

2-aminothiazole and dithiazole.jpg - 13kB

I tried refluxing 3g of dithiooxamide with 6.3g chloroacetaldehyde dimethyl acetal in 30ml of methanol, most of the dithiooxamide dissolves but even after hours of refluxing simply crystallises out again on cooling. I am wondering if I might be able to add a catalysts. A base might encourage enolization of the dithiooxamide but this compound is not stable towards strong bases such as sodium hydroxide or an acid to accelarate the hydrolysis of the acetal?

Does anyone have any experience of similar reaction or have any ideas/suggestions regarding possible catalysts or other modifications to the conditions?


[Edited on 11-8-2016 by Boffis]
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clearly_not_atara
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[*] posted on 11-8-2016 at 14:41


Chloroacetaldehyde, huh? Nasty looking thing. Anyway TsOH cleaves acetals, giving aldehydes + TsOMe.

You could see success using a base such as DBU to make the chloroacetal react with the deprotonated iminothiol, then use TsOH to cleave the acetal and provoke cyclization.

But recall that haloethers are very reactive compounds. It's possible that 1,2-dichloro-3-oxapentane actually reacts with the amine residue at the 2-position and this hemiacetal eliminates ethanol to the imine. Acetals do not cleave quite as easily, and won't react with most bases. I assume chloroacetaldehyde polymerizes or is otherwise very unpleasant to work with.

Another idea is to use glycolaldehyde phenyl (or mesityl) ether. Phenyl ethers react with thiolates, after all, and this compound should not polymerize. A catalyst is probably required.
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AvBaeyer
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[*] posted on 11-8-2016 at 17:58


Typically, the reaction you are trying to carry out requires the presence of a strong acid. The role of the acid is two-fold: 1.) Hydrolyze the acetal to the aldehyde and 2.) Catalyze the enolization of the rubeanic acid which is less reactive at sulfur than thiourea. My suggestion is to try the reaction in aqueous methanol with an innocuous acid such as phosphoric acid. The acid concentration may have to be fairly high - not catalytic.

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Boffis
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[*] posted on 12-8-2016 at 06:19


Thanks for the replies guys! :)

My intuitive response was to try a base because I have found that dithiooxamide reacts readily with hydroxylamine and hydrazine solutions to give diaminoglyoxime and oxalimidazine (C2H8N6) respectively, however, in both cases hydrogen sulphide is eliminated.

clearly_not_atara, could I replace DBU with DABCO as I have the latter base?

Chloroacetaldehyde dimethyl acetal has a fairly low vapour pressure, is not a lachrymator and is stable so it scores over free chloroacetaldehyde.

I am going to try the acid catalysed process first because in the reaction between chloroacetaldehyde and either thiourea or thioacetamide hydrogen chloride eliminated in the reaction combines with the resulting base, so I am going to try to add a little extra acid. I'll let you know how I get on.
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Boffis
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[*] posted on 12-8-2016 at 15:01


Well I tried adding a little hydrochloric acid and another 20ml of methanol to the reaction mixture and refluxed for another 3 hours. The colour remained the same orange red but copious amounts of orange red crystals formed. I am not sure whether these are the product or just dithiooxamide crystallizing out of the reaction mixture. One promising sign though is they are fairly soluble in water and practically insoluble in methanol which sound good for a dihydrochloride of 2,2-bisthiazole, the problem is they are bright orange and I would have thought they should be colourless. What do people think?

2,2-bisthiazole dihydrochloride.jpg - 735kB 2,2'-bis(thiazole) dihydrochloride ??
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PHILOU Zrealone
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[*] posted on 12-8-2016 at 15:14


"The comparison between thiourea and dithiooxamide thioamide and enol forms"
I would write "enthiol" or "thioenol" for A=C-SH sequence.

Beware that since dithiooxamide is bifunctional just like you chloroacetal that you may have different possible products...
1°) the product you want with two cyclopentarings...thus intramolecular cyclisation via geminal N and S on the same C.
2°) a similar product with two cyclohexarings...thus intramolecular cyclisation via viccinal N and S from different initially neightbourgs C.
3°) a polymer kind of linear with cycli inbetween (I can imagine two types) from the binding of different dithiooxamide molécules via intermolecular bridging.
4°) a mix of intramolecular cyclisation and intermolecular bridging with limited lenght and Molecular Weight.

The color is highly probable because of the hyperconjugaison of the molecule all sp2 are resonating and the free S doublets propagates the "aromaticity".


[Edited on 12-8-2016 by PHILOU Zrealone]




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[*] posted on 13-8-2016 at 12:07


Isn't there some sort of reaction condition that naturally degrades thioamides? If you're trying to differentiate 2,2'-bisthiazole from dithiooxamide, you can try doing something that would hydrolyze the latter and give oxamide; the arene should be significantly more stable.

Reading a few sources it seems like thiazoles form complexes with Cu2+ and this paper mentions something called "thiazole orange":

http://www.ncbi.nlm.nih.gov/pubmed/18066832

So you may be on the right track.

PHILOU's point, about the possibility of forming thiazino[2,3-b]thiazines, is also worth considering, though.

[Edited on 13-8-2016 by clearly_not_atara]
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PHILOU Zrealone
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[*] posted on 14-8-2016 at 05:50


Quote: Originally posted by PHILOU Zrealone  
"The comparison between thiourea and dithiooxamide thioamide and enol forms"
I would write "enthiol" or "thioenol" for A=C-SH sequence.

Beware that since dithiooxamide is bifunctional just like you chloroacetal that you may have different possible products...
1°) the product you want with two cyclopentarings...thus intramolecular cyclisation via geminal N and S on the same C.
2°) a similar product with two cyclohexarings...thus intramolecular cyclisation via viccinal N and S from different initially neightbourgs C.
3°) a polymer kind of linear with cycli inbetween (I can imagine two types) from the binding of different dithiooxamide molécules via intermolecular bridging.
4°) a mix of intramolecular cyclisation and intermolecular bridging with limited lenght and Molecular Weight.

The color is highly probable because of the hyperconjugaison of the molecule all sp2 are resonating and the free S doublets propagates the "aromaticity".


[Edited on 12-8-2016 by PHILOU Zrealone]

A little drawing is better than a long writting...and it can fix the ideas...so here is the general picture.
I have found 3 polymeric types and just thought to another mixed polymer possibility.
The polymer can be finite if the ends close a macrocyclic loop or "infinite" (if the chain propagates to very high MW).

other condensation products.jpg - 72kB




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Boffis
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[*] posted on 15-8-2016 at 13:20


Hi c_n_a and Philou. I have done a good deal of research and reading since my previous post and it appears that the original method of producing 2,2-dithiazole was by heating 2-bromothiazole with copper powder at 150-170 C (Erlenmeyer & Schmid; Helv. Chim. Acta, v22 p698 (1939)). They describe the dithiazole as a white solid melting at 102.5 C.

The strange orange product was obtained in 2.496g recovered and has several interesting properties. Evaporating down the filtrate produced a little more impure material and finally a brownish oily residue that was rapidly decomposed and dissolved by sodium hydroxide solution. The first crop of orange crystals is exceedingly well crystallized and apparently free of tarry or polymeric products. It is insoluble in cold water, sparingly soluble in hot and a little more soluble in hot alcohols. It does not react with sodium carbonate suggesting that it is not a hydrochloride. I will check the Mp tomorrow.

Do you think this sounds like c_n_a's thiazino[2,3-b]thiazine? Its certainly not 2,2'-dithiazole. It is interesting that in spite of the theoretical possibility of forming polymers they are actually a very minor product at most.
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clearly_not_atara
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[*] posted on 15-8-2016 at 14:15


Thiazino[2,3-b]thiazine is an interesting molecule: 1,4-thiazine fails to be aromatic because the 6-carbon is tetrahedral, but this double ring is fully planar, so it should be an aromatic system.

Apparently nitrogen-sulfur bidentate ligands form Pd complexes which are useful in e.g. nucleophilic substitution at allylic esters:

http://www.sciencedirect.com/science/article/pii/S0022328X98...
https://www.researchgate.net/profile/Jacek_Skarzewski/public...

They also form complexes with Zn. So this compound may be quite an interesting ligand (even if it is not the one you expected)... this particular N-S ligand is less bulky and possibly more stable than many others, and might (if you're really lucky) even be usable in hydrogenation, though most other sulfur ligands will be reduced by those conditions so don't hold your breath.

Quote:
It is interesting that in spite of the theoretical possibility of forming polymers they are actually a very minor product at most.


Polymer formation is a decreasing-entropy reaction, so at "high" temperatures (which may be anywhere from 10-1000 K) it should proceed more quickly backwards than forwards.


[Edited on 15-8-2016 by clearly_not_atara]

[Edited on 15-8-2016 by clearly_not_atara]
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[*] posted on 15-8-2016 at 18:20


The old method is vapor volume versus molecular mass. If you can vaporize the stuff...?

That doesn't help with the hexagonal isomer; which was my first thought reading your post of a likely impurity. And - come to think of it, it looks like something that would be colored as well...

So if you have failed to make the desired isomer you do have the chance to diazotize the amino thiazole and follow that with CuI / Cu(metal) - that's, uh, Sandmeyer and Ullman coupling successively - should give you a small amount if all else fails.

One thing you might also try is to react the anions of thiooxamide with your acetal, then hit it with acid.

bis thiazole should form colored complex with metal copper or nickel. Or cobalt. Those complexes with the obvious isomer could be polymeric, so maybe less soluble in things, because it's mainly the nitrogen the metal wants and there's a better geometry too. Could be a way of separating.

ARE you making it for purpose of a ligand? ;)

[Edited on 16-8-2016 by halogen]




F. de Lalande and M. Prud'homme showed that a mixture of boric oxide and sodium chloride is decomposed in a stream of dry air or oxygen at a red heat with the evolution of chlorine.
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Boffis
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[*] posted on 16-8-2016 at 04:30


Thanks Halogen, I think your're right about the 2-aminothiazole route. I have prepared 2-aminothiazole and the route worked well but I think the acetal route I used is faster and the reflux period in Vogel for the chloroether route is too long and you get a bit of brown tar that is difficult to remove.

I have done some more investigation on my orange crystals. They form coloured complexes with Co, Ni and copper that exactly match those of dithiooxamide... suspicious. So then I tried a melting point test, the powder suddenly turns black at 198 C and then begins to decompose with copious evolution of hydrogen sulphide from 200 upwards to about 206 when decomposition apears to be complete. This fits the description of dithiooxamide perfectly.

So it appears that I have recovered most of my dithioxamide and that very little reaction has taken place apart from possibly polymerisation of the chloroacetaldhyde. I have found several more reaction of dithiooxamide in the literature and there is a strong common theme:- basic conditions. So I am going to try this experiment again using a) the addition of a stochiometric amount of aqueous sodium hydroxide and b) a tertiary base, I am going to try DABCO.
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Boffis
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[*] posted on 17-8-2016 at 08:58


@halogen, re "One thing you might also try is to react the anions of thiooxamide with your acetal, then hit it with acid"

I just tried this with the carefully measure out quantity of NaOH in standard 1M solution on 1g of dithiooxamide to give the disodium salt and then added the acetal. The dithiooxamide dissolved fairly quickly as I have observed before so I added a good deal of methanol to try and get the acetal into solution. I then tried refluxing gently for 30 minute.

I got a very dark homogenous solution, curiously no salt precipitate, so I rendered it moderately acid and refluxed again for a brief period and cooled. Filtering removed a very small amount of black water soluble grains. I evaporated off the methanol and got an oil. I tried triturating this with a little ether and this caused the precipitation of a small amount of minute pale hairy ponpoms (filtered off about 40 mg). I then tried partial neutralisation with sodium acetate and I am now waiting to see if anything is going to crystallise from the dark brown solution. If not I am going to try some sodium hydroxide to free base any thiazole present. But it doesn't look promising:(.
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clearly_not_atara
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[*] posted on 17-8-2016 at 12:28


The trouble is that 2-chloroethylidene dimethoxide is not nearly as strong an electrophile as 2-chloroacetaldehyde, which is an alpha-haloketone, nor is it as reactive as 1,2-dichloro-3-oxapentane, which is an alpha-haloether. Both of these increase the reaction rate of nucleophile exchange by a factor around 10^2-10^5 relative to a simple alkyl halide -- hence chloroacetone is more reactive than propyl chloride and methoxymethyl chloride is extremely reactive as well.

However, in order to use the acetal as an electrophile before the acetal is destroyed, you have got yourself a quandary.

In this case you can cut the knot by preparing the alpha,beta-dihaloether. Methyl vinyl ether is the obvious choice, but there is no reason not to use, e.g., benzyl vinyl ether, or isobutyl vinyl ether, but it is the benzyl compound I'll preoccupy the discussion with. Anyway, acetaldehyde reacts with two equivalents of alcohol in the presence of concentrated H2SO4 to form a dibenzyl acetal:

MeCOH + 2 BnOH [H2S2O4] >> MeC(OBn)2H + H2O

This acetal is separated from the reaction mixture and destructively distilled with catalytic (clean) H2SO4 so that it fragments into benzyl vinyl ether and benzyl alcohol:

MeC(OBn)2H (l) [cat H2SO4] >> H2C=C(OBn)H (g) + BnOH (g)

Fractional distillation separates the alcohol from the vinyl ether.

This then reacts with a halogen at low temperature:

H2C=C(OBn)H + Cl2 (CH2Cl2?) [-78 C?] >> H2ClCC(OBn)ClH

to give the alpha,beta-dichloroether. I am familiar with the preparation of vinyl ethers, and I of course expect chlorine to react preferentially with alkenes at least at low enough temperatures, but I haven't actually seen a reference for the last part of this tek, so you may want to look around for one. However, I believe that something like this will achieve the preparation of your haloether, and, therefore, your heterocycle.

[Edited on 17-8-2016 by clearly_not_atara]
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PHILOU Zrealone
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[*] posted on 17-8-2016 at 13:09


I just thought the fact that thiazole and/or thiazine moeties could be involved into Diels-Alder type of cyclisations (cage tricyclo compounds)...Right? So problem might be more complex.

I have rewritten the equations of clearly_not_atara for convenience purpose...I hope he/she doesn't mind. :(
CH3-CH=O + 2 C6H5-CH2OH --H2S2O4--> CH3-CH(-OCH2-C6H5)2 + H2O

CH3-CH(-OCH2-C6H5)2 --cat H2SO4--> CH2=CH-OCH2-C6H5 (g) + HOCH2-C6H5 (g)

CH2=CH-OCH2-C6H5 + Cl2 --CH2Cl2/-78 °C--> Cl-CH2-CHCl-OCH2-C6H5


[Edited on 17-8-2016 by PHILOU Zrealone]




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