Sciencemadness Discussion Board - Substitute for dioxane

Substitute for dioxane

Baphomet - 8-1-2011 at 14:46

I'm about to run out of 1,4-dioxane which is a wonderfully toxic and versatile polar aprotic solvent that is now hard to find outside of chemical suppliers.

Are there any good substitutes? Note that I'm not just after dissolution ability but a low level of chemical reactivity. I'm here scratching my head but all the other solvents that fit the bill are even less 'OTC' than dioxane.

Ozone - 8-1-2011 at 16:56

How about glyme (1,2-dimethoxyethane)?

BromicAcid - 8-1-2011 at 20:17

How about making it? Frogfot had a good procedure for it when his site was up, still available by the wayback archive:

http://web.archive.org/web/20071011001804/www.frogfot.com/sy...

Dioxane is pretty toxic but it's all relative compared to what you're using it for obviously. When my I told my doctor I was a chemist he was worried about exposure to solvents, little did he know what nasties I use the solvents with.

garage chemist - 9-1-2011 at 08:13

There is a german synthesis of dioxane on Lambdasyn which is similar to frogfots method, but uses a mix of 225ml ethylene glycol and 6ml sulfuric acid. Quite surprising how little acid is needed to make this ether.

Whether you can find a substitute depends on the application.
Glyme and Diglyme sound good, but they're hard to get (for me at least) and can't be made at home (well, unless you're prepared to methylate glycol or diglycol with dimethyl sulfate).
Normally I would recommend dioxane as a substitute for glymes.

Baphomet - 10-1-2011 at 01:28

Thanks guys! I like that synth.. will try it out and report the results later.

maxidastier - 14-7-2011 at 09:42

Does it make sense to carry out the synthesis under lower pressure to minimize the formation of those nasty by-products?

blogfast25 - 14-7-2011 at 13:13

Quote: Originally posted by garage chemist

There is a german synthesis of dioxane on Lambdasyn [...]

Where precisely? I looked for it sometime ago and couldn't find it!

garage chemist - 14-7-2011 at 15:07

It's here:
http://www.lambdasyn.org/synfiles/dioxan.htm

It's a lot of work even if it seems easy. Several treatments with KOH are necessary to seperate most of the water. If there's a better way to remove large amounts of water from dioxane, I'd like to know it. CaCl2 forms a complex with dioxane.

maxidastier - 15-7-2011 at 01:39

Quote:

If there's a better way to remove large amounts of water from dioxane, I'd like to know it.

.

Molecular Sieve 4A... for 1 liter around 100g

But still:

Quote:

Does it make sense to carry out the synthesis under lower pressure to minimize the formation of those nasty by-products?

[Edited on 15-7-2011 by maxidastier]

plastics - 15-7-2011 at 01:41

I have done the frogfot method with 1000ml MEG and got approx 400ml of very dry dioxane

Don't overdo the sulphuric acid as mentioned otherwise you end up with black goo very quickly

As garage chemist says it is quite an effort to get all the water out and you get a fair amount of sticky orange polymerised ethanal as well

As the frogfot method states use a Vigreux column for the initial separation and distil the final nearly dry product off Na metal or KOH. Judging by the BP and smell (!?) of the final product it seems quite pure

[Edited on 15-7-2011 by plastics]

DJF90 - 15-7-2011 at 07:03

I suspect p-TsOH can be used to advantage in place of sulfuric acid. You should get less charring and higher yields, as the acid is not dehydrating or oxidising. MsOH may also find use in this synthesis.

maxidastier - 15-7-2011 at 09:38

I know that it works better with p-TsOH, but I simply cannot get that stuff...so
Does it make sense to carry out the synthesis under lower pressure to minimize the formation of those nasty by-products?

blogfast25 - 15-7-2011 at 09:47

Quote: Originally posted by garage chemist

It's here:
http://www.lambdasyn.org/synfiles/dioxan.htm

It's a lot of work even if it seems easy. Several treatments with KOH are necessary to seperate most of the water. If there's a better way to remove large amounts of water from dioxane, I'd like to know it. CaCl2 forms a complex with dioxane.

The frogfot procedure uses K2CO3, IIRW...

not_important - 15-7-2011 at 11:17

Quote: Originally posted by maxidastier

I know that it works better with p-TsOH, but I simply cannot get that stuff...so
Does it make sense to carry out the synthesis under lower pressure to minimize the formation of those nasty by-products?

Can you get toluene?

http://www.sciencemadness.org/talk/viewthread.php?tid=11878

Even crude TsOH, containing some o- and m- isomers, generally is OK for use as a catalyst.

Higher yields of the para isomer are obtained by passing toluene vapour through H2SO4 at 150-160 C, with excess toluene removing the water formed as the azeotrope, taking care to stop at about 95% conversion of the H2SO4 (measure the amount of water distilled off) to minimise poly-sulfonation.

maxidastier - 19-7-2011 at 07:52

No, I can not get toluene. It's restricted in Germany for "common people". Fuck it!

blogfast25 - 19-7-2011 at 08:37

If it 'works better' with TsOH, what precisely does that mean? In what respect is it preferable to conc. H2SO4 that justifies making the TsOH in the first place? In what way would the synth. of dioxane be modified, using TsOH instead of conc. H2SO4?

[Edited on 19-7-2011 by blogfast25]

DJF90 - 19-7-2011 at 12:24

If you read my post you'll notice I mention the criteria on which TsOH makes a better catalyst. Its a non-oxidising, non-dehydrating acid with a pKa similar to that of sulfuric acid. This means you'd expect a cleaner reaction and as such a higher yield of the deisred product.

tmb - 19-7-2011 at 13:09

What is the yield with TsOH?
Is a dehydrating acid not preferred? Since the aim is to remove water?!

maxidastier - 23-7-2011 at 06:26

Quote:

1,4-dioxane was synthesis by dehydration of ethanediol using p-toluene sulphonic acid as the catalyst.Influence of catalyst dosage,reaction temperature and reaction time on the yield was investigated.The optimum reaction condition was as follows: catalyst dosage=4.3% of the amount of ethanediol,reaction time 50 min.1,4-dioxane yield of 84.82% was obtained under the optimum condition.

http://61.185.219.126:88/qikan/epaper/zhaiyao.asp?bsid=14647

Can someone get that paper or tell me how to do it?

maxidastier - 29-7-2011 at 08:29

ok, then. How many grams of TsOH would you use for dehydrating 200ml ethanediol?

Nicodem - 29-7-2011 at 11:55

Quote: Originally posted by maxidastier

ok, then. How many grams of TsOH would you use for dehydrating 200ml ethanediol?

Hm, have you bothered reading that abstract you quote? It says it all right there!

Quote: Originally posted by maxidastier

http://61.185.219.126:88/qikan/epaper/zhaiyao.asp?bsid=14647

Can someone get that paper or tell me how to do it?

You click on the link for the full text and it downloads. You need a PDF viewer installed to view the article.

PS: Are you sure you are able to conduct chemical experiments safely? I would suggest you to first work on your general skills.

DJF90 - 29-7-2011 at 13:00

I've just checked myself what the issue is and it won't allow you to download. Anyone have access?

Nicodem - 29-7-2011 at 13:17

Quote: Originally posted by DJF90

I've just checked myself what the issue is and it won't allow you to download. Anyone have access?

Could be IP country dependent. Looks like the Chinese like my IP:

Attachment: gych02006.pdf (349kB)
This file has been downloaded 1230 times

DJF90 - 29-7-2011 at 15:48

Thanks - no good to me as its only the abstract in English. Should be sufficient to give it a go if necessary though.

maxidastier - 3-8-2011 at 08:18

Thank you anyway Nicodem. I'm not that stupid as it seems. I just don't like making experiments with only an abstract

maxidastier - 8-8-2011 at 10:15

Quote:

Even crude TsOH, containing some o- and m- isomers, generally is OK for use as a catalyst.

1. I've made TsOH by heating 75ml of 96% H2SO4 and 200ml toluene, removing water with a dean stark trap.
I used conc. hydrochloric acid to let the para isomer crystallize, vacuum filtered it and dried it.

2. I've distilled 345g of water-free Ethylene Glycole from anti-freeze agent.

3. Then I added 4,3% (14,8g) of the TsOH as catalyst to the flask, added a short Vigreux column and distilled at the temperature given in the pdf, 110-130°C.

4. I added some NaOH to the distillate and fillled it into a seperating funnel. The next morning the liquid was brown (from polymerised acetaldehyde) and a layer of polymerised acetaldehyde was at the bottom. That would have been ok, but there was not one upper layer, but two.
There should have only been one containing the crude Dioxane.

5. I decided to follow the instructions, separated the acetaldeyde layer, filled the rest into a round-bottom flask together with a few grams of NaOH and distilled again.
But there doesn't seem to be any dioxane, because the steam temperature was 82°C and not 101°C as it should have been for dioxane.
I thought it might be the extra layer formed in the seperatory funnel and decided to continue. However, only at the very end of the distillation the temperature went up to around 100°C, but there was only a small residue left, which began to get black, fuming and stinking.

Can you tell me what I've made wrong. Is it true, that

Quote:

Even crude TsOH, containing some o- and m- isomers, generally is OK for use as a catalyst.

Why didn't I get dioxane then or did I?

Nicodem - 8-8-2011 at 12:22

Quote: Originally posted by maxidastier

4. I added some NaOH to the distillate and fillled it into a seperating funnel. The next morning the liquid was brown (from polymerised acetaldehyde) and a layer of polymerised acetaldehyde was at the bottom. That would have been ok, but there was not one upper layer, but two.

Sounds like you did not use enough NaOH to thoroughly dry the product. Dioxane and water are freely miscible, but dioxane with NaOH(aq) should not be. Still, due to the high higroscopicity of dioxane, water will still efficiently partition between both phases (as will dioxane in the NaOH(aq)). This means the upper layer will not be dry dioxane (assuming it is mainly dioxane at all).

Quote:

5. I decided to follow the instructions, separated the acetaldeyde layer, filled the rest into a round-bottom flask together with a few grams of NaOH and distilled again.
But there doesn't seem to be any dioxane, because the steam temperature was 82°C and not 101°C as it should have been for dioxane.

If you used no distillation column, or the column you used is not efficient, you will not be able to separate the initial azeotrope dioxane/water fraction from the pure dioxane, thus it is quite normal that you got a wide boiling point fraction. Without fractionation, you distill all the intermediate compositions from the azeotrope onward (thus the bp follows the changing composition). This is no evidence that your synthesis gave no dioxane. It only means that, if it does, your work up and equipment are not up to the task and should be further optimized.

Quote:

Can you tell me what I've made wrong. Is it true, that

Quote:

Even crude TsOH, containing some o- and m- isomers, generally is OK for use as a catalyst.

Yes, it should be true.

Quote:

Why didn't I get dioxane then or did I?

You'll only be sure once you prove it, but you need evidence, at least a sharp boiling point, unless you have some spectroscopic or chromatographic mean of analysis. Which again brings up the isolation issue. I would say you are probably only having troubles in properly drying the presumed dioxane layer (the top one, of course). Try drying your final product over NaOH pellets, this time making sure the NaOH does not completely dissolve (the lower layer must become saturated!). Separate the upper layer and distill again, preferably using at least a simple distillation column.

blogfast25 - 8-8-2011 at 12:49

I'm guessing you probably can't really overdo it with the alkali, right? A bit like 'salting', really...

[Edited on 8-8-2011 by blogfast25]

maxidastier - 9-8-2011 at 01:25

Actually, I used quite a lot NaOH and not everything dissolved...
Furthermore the azaeotrope b.p. with water should be 87°C, not 82°C.

Why did the residue become black, smoking and stinking? Is it because the concentration of the NaOH increased?

I'm going to dry the Dioxane with some molecular sieve 3A.
After that distill again and test the b.p.

It's really frustrating, since this is my second attempt now working properly.

Edit: The mid layer would be water you say? Then it would be quite a lot water!
You should get 100ml water during the synthesis.
I've got at least 50ml polymerised Acetaledhyde.
So I should have left roughly 200g pure Dioxane. That would be 100% yield.
Using TsOH as catalyst they got 85% yield= 210g Dioxane.
Now these are optimum conditions, so let's say I would get 75% yield which would be 184 g = 180 ml Dioxane.
After distilling and so on I've left about 150ml of a mixture of water and dioxane. Now, if you get 100ml water during the synthesis, there would be left 50ml Dioxane, a 21% yield.
That's fucking annoying :mad:

And I used a vacuum-jacketed Vigreux column with effective length of 200ml.

If crude TsOH really isn't the problem what else?

[Edited on 9-8-2011 by maxidastier]

blogfast25 - 9-8-2011 at 07:46

2 mol ethylene glycol === > 1 mol dioxane + 2 mol water!

A vacuum jacketed vigreux would provide insulation, not reflux. It would act like a well insulated steam pipe, unless you had a refluxing valve at the top...

[Edited on 9-8-2011 by blogfast25]

I meant that the synthesis inevitably generates more water than dioxane.

Explain what you mean by a 'jacketed vigreux'? Unless you have some coolant running through the jacket (even air would be fine) the jacket acts like an insulator, that's not really what you want from a vigreux...

[Edited on 9-8-2011 by blogfast25]

maxidastier - 9-8-2011 at 10:51

Sry, but what are we talking about right now?
About the synthesis or removing water or...?

As for the synthesis the pdf recommends using a short vigreux. Or do you mean I need a column head?

blogfast25 - 9-8-2011 at 11:36

Synth.: inevitably you get lots of water as a byproduct.

Vigruex: for any fractionation to occur there has to be some refluxing going on. Liquid needs to be dripping down the column and back into the reboiler flask, otherwise you only have a one distillation separation. With a Vigruex reflux is provided by natural heat losses of the column and resulting condensation of rising vapours. Is this the case in your set up?

maxidastier - 9-8-2011 at 12:03

I've got a vacuum jacketed vigreux column. That's what German-English google translates.
The vacuum jacket provides better fractioning and prevents too much heat loss, but there still is heat loss for refluxing!

I also know that the synth produces lots of water, but as I said in my previous post, the yield of pure Dioxane should be around 80% with p-TsOH. I've only got 35% yield, because after the second distillation there were 160mls of a Dioxane-water-mixture. After drying with molecular sieve I've left roughly 90ml!
From 347g dry Ethylen Glycole I should have got around 200ml Dioxane (84% yield)

I'd be very happy if someone could give it a try with p-TsOh.
Also I'd like to know how the other isomeres could harm the formation of dioxane.

[Edited on 9-8-2011 by maxidastier]

blogfast25 - 9-8-2011 at 12:31

Quote: Originally posted by maxidastier

I've got a vacuum jacketed vigreux column. That's what German-English google translates.
[Edited on 9-8-2011 by maxidastier]

That's probably a poor translation because there's no vacuum involved. I understand German fairly well: what was the German term? In English 'jacketed Vigreux' would probably be what you refer to.

A small but steady airflow through the mantle would probably boost fractionation: the more reflux, the more theoretical plates the instrument delivers...

maxidastier - 9-8-2011 at 12:43

Well, I can't let air flow through the mantle, because the mantle is evacuated-->vacuum mantle (german: Vakuummantel) -->better isolation since no air not much transport of heat

blogfast25 - 10-8-2011 at 05:22

Hmmm, in my opinion the vacuum mantle is a bit of an unnecessary option and is best used with a refluxer valve at the still head to control reflux ratio. That allows to carry out fractionations with accurately defined reflux ratio R = L/D. For most purposes a naked or very lightly insulated vigreux is more than adequate and provides good reflux (and it's a lot cheaper)...

maxidastier - 10-8-2011 at 05:36

I always thought an vacuum mantle would be better fro fractionating. Maybe it is, but only in combinations with a still head, which I don't have.
Nevertheless, that can't be the problem. I still have only 35% yield instead of 85%....
So, what could be the problem? Can the other isomeres of TsOH have any influence?

Nicodem - 10-8-2011 at 05:42

Quote: Originally posted by blogfast25

For most purposes a naked or very lightly insulated vigreux is more than adequate and provides good reflux (and it's a lot cheaper)...

If the distillation column is not isolated, its resolution power drops sharply. The more isolated it is, less heat loss, less temperature/composition gradient deterioration, the better the efficiency. It appears to me that you are unaware on how the distillation columns work. Their internal reflux is gradient driven (the less volatile composition goes downstream as the liquid phase, the more volatile goes upstream as the gas phase). Obviously, any heat loss deteriorates the equilibrium, hence all better (expensive) distillation columns, especially those filled with Raschig rings, are always isolated by a vacuum jacket and sometimes even by a reflective layer to reduce radiation heat loss. This topic has been discussed ad nauseam a couple of years ago. You can find the thread if you UTFSE.

blogfast25 - 10-8-2011 at 07:04

Quote: Originally posted by Nicodem

[It appears to me that you are unaware on how the distillation columns work.

[snip]

You can find the thread if you UTFSE.

No, I can assure you I know perfectly well how distillation works and it seems we’re suffering from the same confusion that occurred on the thread you mention (in which I took part).

I worked for months on end with a 10 m (about 15 cm ID) semi-industrial fractionation column for the rectification of technical methanol to UV-grade methanol (very pure, UV transparent and almost completely water free). It was vacuum jacketed and glass ring packed and fitted with a condenser and flow splitting timed reflux valve at the top, to provide accurately controlled reflux.

Most of us haven’t got the luxury of the latter device and do rely on natural cooling to get some condensate flowing downwards and get some fractionation going. I use it all the time with a home made acetone still, packed with good grade steel wool. It refluxes alright and acetone comes over at about 1 C above theor. BP, until near the end. No insulation worth mentioning...

A Vigreux column, perfectly insulated and without a condenser/reflux valve at the top acts as a steam pipe and provides a single stage distillation, no more.

maxidastier - 10-8-2011 at 07:38

I understand what you mean, but did this lead to my bad 35% yield?

blogfast25 - 10-8-2011 at 08:22

Quote: Originally posted by maxidastier

I understand what you mean, but did this lead to my bad 35% yield?

No, I don't think that's the cause. Maybe you should try once with straight conc. H2SO4? I've been meaning to do this synth for ages but haven't got round to it yet...

sternman318 - 10-8-2011 at 08:36

Quote: Originally posted by Nicodem

This topic has been discussed ad nauseam a couple of years ago. You can find the thread if you UTFSE.

Is this the thread?http://www.sciencemadness.org/talk/viewthread.php?tid=12409

maxidastier - 22-8-2011 at 10:41

What is the better way to precipitate the p-TsOH: Boiling and cooling in conc. HCl or bubbling HCl gas through it? Do the two methods differ in consumption of HCl?

blogfast25 - 22-8-2011 at 12:26

If your gas absorption is 100 % (with HCl that shouldn't be a problem) then there should be no significant difference between both approaches.

maxidastier - 23-8-2011 at 00:36

So, it's easier to dissolve the crude TsOH in conc. HCl, boil and cool down?
Because bubbling HCl gas through the solution is more time consuming.
And the p-TsOH will precipitate in both ways in the same amount or do I get a better yield when I bubble HCl through it?

blogfast25 - 23-8-2011 at 04:16

As far as I know that solubility is a state function and only depends on actual, final HCl concentration (and of course temperature), not on how you arrive at that concentration.

Recrystallisations from concentrated HCl are common with a few substances, of the top of my head: hydrated AlCl3, ZrOCl2.8H2O, (NH4)2SnCl6.

maxidastier - 23-8-2011 at 12:30

Ok, thanks for your reply. But can I reach a higher concentration of HCl by bubbling it through water? I've only 30-33% HCl aq. Maybe you could reach 37% with the bubbling method? Or do you reach less then 30%?

blogfast25 - 24-8-2011 at 04:50

37 % is more or less the theoretical limit that can be achieved and it's slightly higher than the 30 - 33 % you've altready got. 37 % is made by absorbing HCl in water, to saturation point but for the hobbyist that's not so easy to achieve: 37 % means just over 1 mol of HCl gas absorbed into 100 ml of solution! If you've got a decent sized HCl generator that's maybe not too difficult (ahem!) And saturating a small amout of solution is of course easier.

Why don't we stay down to earth and you try and precipiatate that stuff from 30 - 33 %, huh? If failed, you have plan B!

By the way: there's no need to bubble the HCl gas through the water: it's so soluble that the 'inverted funnel method' (which prevents suck back) works just as well...

maxidastier - 24-8-2011 at 06:51

Sry, but I didn't now the "inverted funnel method". My literature tells me to bubble through water....
Could you explain this method, please?

blogfast25 - 24-8-2011 at 07:37

HCl is so extremely soluble in water that bubbling isn't even really possible from what I've seen with my own eyes. As the gas flows through the tube, when it hits the water it kind of forms one stationary bubble, with the gas being constantly absorbed at the interface between bubble and solution. The heavier than water solution then tends to sink to the bottom.

In the inverted funnel method, an inverted funnel (spout pointing upwards) is connected to the HCl source and the circumference of the top of the funnel (now bottom) is hovered a few mm above the solution level (by adequately clamping the spout). The gas flows onto the solution surface and is immediately absorbed. It's always best to stir the solution constantly. Inverted funnel avoids suck back.

But suckback can also be prevented with the 'bubbling method' by having an empty vessel between the HCl source and the bubbler: any suck back then just sucks some of the solution into the empty vessel.

Clear?

[Edited on 24-8-2011 by blogfast25]

maxidastier - 25-8-2011 at 02:30

Thank you!

blogfast25 - 25-8-2011 at 08:28

No probs. Let us know how you get on. I'm really interested in making some dioxane but had never heard of the toluene sulphate method...

maxidastier - 25-8-2011 at 09:36

AFAIK you don't use the sulphate but the p-TsOH monohydrate.
I've one last question before I try it again: Since the p-TsOH is a catalyst, it doesn't get consumed right? So can I use the same amount of TsOH for even more ethylene glycol if I continuously add fresh ethylene glycol?
By this way I also don't get that much waste at the end.

blogfast25 - 25-8-2011 at 10:21

Quote: Originally posted by maxidastier

Since the p-TsOH is a catalyst, it doesn't get consumed right? So can I use the same amount of TsOH for even more ethylene glycol if I continuously add fresh ethylene glycol?

Catalysts, chemically speaking, don't get consumed, that's correct. But catalyst recovery can be quite a kerfuffle, as I suspect will be the case here. Just adding more and more glycol won't cut it, I'm fairly sure.

maxidastier - 26-8-2011 at 00:28

I didn't want to recover the catalyst, but if you say a continuous process isn't possible, ok then.

[Edited on 26-8-2011 by maxidastier]

maxidastier - 8-9-2011 at 01:27

Can HCl destroy the p-TsOH if it isn't removed immediately?

blogfast25 - 8-9-2011 at 03:19

No.

maxidastier - 9-9-2011 at 08:38

This is now my second attempt to make p-TsOH from H2SO4 and Toluene.
I've with the help of two literature instructions. One is from Gattermann-Wieland and Experiments in Organic Chemistry by Fieser.

So, then: I've used 40ml 96% H2SO4 and 200ml Toluene and a Dean-Stark-Trap as Gatterman-Wieland suggests.
Everything went fine. I've got around 13ml of H2O after 4 hours, so should it be complete reaction, which means there still were aroung 115ml unreacted Toluene left. I decided to collect this with the same apparatus, and collected aroung 100ml through the Dean-Stark-Trap.
Then I did something, which MIGHT be the reason for the failed attempt: I went upstairs to go to the toilet and drink something.
When I came back, the residue in the flask had become dark brown- I immediately removed the flask from the heating mantle.
I transferred the mixture in a beaker glass to cool down. I added around 1 ml of water to test, if there was some TsOH. Where the drops touched the water, instantly became white and solid.
So I let it cool down and added 100ml conc. HCl. I heated until boiling and cool again to recrystallize and get mainly the para isomere.
But I didn't get white crystals. I only got a brown slurry, when I vacuum-filtrated it.
I thought okay, then another recrystallization, added another 50ml HCl, heated, cooled and still having brow-white slurry.

Is the dark brown colour after the reaction normal? Here, I can't see any brown: http://www.sciencemadness.org/talk/viewthread.php?tid=15522

Does that mean, the TsOH became completely desulfonated while I was upstairs and the heat was to high?

When I cleaned my flask and other stuff with water, there were white solid pieces swimming in it. This can't be TsOH, since it is soluble in water and I couldn't dissolve by shaking or anything. What is it?

[Edited on 9-9-2011 by maxidastier]

Nicodem - 10-9-2011 at 00:52

Quote: Originally posted by garage chemist

When no more water is collecting in the trap (all H2SO4 has reacted), one adds a small, carefully measured amount of water to the mix after cooling which makes the p-TsOH crystallize from the excess toluene as the monohydrate. The byproduct o-TsOH does not form such a hydrate and stays in solution.
The filtered crude p-TsOH*H2O is then purified by dissolving it in half its own weight of water and saturating the solution with HCl gas.

For your needs you can skip the tedious recrystallization step. Impurities don't make much of a difference in your application.

Given you appear to be doing this out of practice/fun rather than the need for dioxane (or else you would just use the H2SO4 based procedures), you might as well try using NaHSO4 as the catalyst (instead TsOH). It is available in any shop selling pool chemicals.
The idea of using a continuous process is a good one (reactive distillation of products as they form and feeding in fresh ethylene glycol), but I think you should optimize the batch process first to get an understanding of the critical reaction parameters. Though with a continuous process you would still end up doing a batch isolation process.

maxidastier - 10-9-2011 at 01:17

Thanks for your input!
But I still don't know what has happened to my TsOH...

maxidastier - 11-9-2011 at 08:23

It's really a mess. The problem is, after adding the water to make the p-TsOH crystallize as the monohydrate, it does not crystallize as long as the solution is to hot, but if it's cool enough, after adding water, everything solidifies, the o-TsOH does not stay in solution. How can I get rid of it then and moreover, how can I get rid of the toluene?
I didn't use HCl this time as you suggested. Now I have some clumpy, pink, wet crystals which dissolve in the air and there are little droplets where the solid used to be.
What am I doing wrong?

maxidastier - 19-9-2011 at 01:57

Is there another way aside from potassium carbonate to create an organic layer containing mainly Dioxane and another layer containing the water, since after the synthesis there would be a 1:1 mix of Dioxane/water?

blogfast25 - 19-9-2011 at 05:09

I'm not sure why specifically it calls for (the more expensive) K2CO3, when I believe Na2CO3 should do just as well (my guess). But the principle of 'salting' cannot be avoided, AFAIK...

maxidastier - 19-9-2011 at 11:32

Yes, that is exactly my question: Could I use Na2CO3 instead?

Nicodem - 19-9-2011 at 11:51

Quote: Originally posted by maxidastier

It's really a mess. The problem is, after adding the water to make the p-TsOH crystallize as the monohydrate, it does not crystallize as long as the solution is to hot, but if it's cool enough, after adding water, everything solidifies, the o-TsOH does not stay in solution. How can I get rid of it then and moreover, how can I get rid of the toluene?

What kind of analysis gave you the confirmation that ortho-toluenesulfonic acid also precipitated? Why don't you nevertheless follow up with the isolation, do the vacuum filtration, wash the product, etc.? Why do you want to remove all the ortho isomer when it is obvious you need the product as an acid catalyst?

blogfast25 - 19-9-2011 at 12:34

Quote: Originally posted by maxidastier

Yes, that is exactly my question: Could I use Na2CO3 instead?

I see no good reason why not.

Takron - 29-10-2011 at 22:51

Quote: Originally posted by blogfast25

Quote: Originally posted by maxidastier

Yes, that is exactly my question: Could I use Na2CO3 instead?

I see no good reason why not.

Could you use sodium bicarbonate instead of the carbonate?

blogfast25 - 30-10-2011 at 03:50

Quote: Originally posted by Takron

Quote: Originally posted by blogfast25

Quote: Originally posted by maxidastier

Yes, that is exactly my question: Could I use Na2CO3 instead?

I see no good reason why not.

Could you use sodium bicarbonate instead of the carbonate?

No. It doesn't have any dehydrating power whatsoever.

bbartlog - 30-10-2011 at 05:56

And Na2CO3 is a pretty poor dehydrating agent compared to K2CO3.

blogfast25 - 30-10-2011 at 12:50

I've never understood whether that is true or not: it seems counter-intuitive that a substance that froms no hydrates would be a better dessicant than a substance that forms a relativly stable decahydrate.