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mackolol
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[*] posted on 26-10-2017 at 11:44
Separating phthalic acid isomers


Hi, i was wondered how can i separate phthalic acid isomers. I want to oxidise xylene but its sold as a mix of 3 isomers and without doing vacuum distillation its rather not possible to separate it. I was thinking about freezing different isomers but it rather wouldnt work since its all mixed together. So how can i separate these 3 isomers of phthalic acid. I was thinking about sublimation of ortho phthalic acid to anhydride because thats only thing i need from this but i dont know sublimating point of other isomers. Im planning to do luminol the cheapest way. Can somebody help?
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[*] posted on 26-10-2017 at 11:58


I have been looking at that as well. It's actually possible to separate xylene isomers by a combination of freezing and careful fractional distillation without a vacuum, but I think it would take quite a bit of time and effort to obtain highly pure o-xylene that way. My plan is to eventually make phthalic anhydride to make N-phenylanthranilic acid (which is a redox indicator that can be used to titrate chromates).



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SWIM
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[*] posted on 26-10-2017 at 16:36


The Phthalic acid has a low water solubility, but many times higher than its isomers.

Perhaps you could just extract the ortho isomer, especially if you have a Soxhlet.

I get the feeling that calcium salts of these isomers would be easy to separate too, but I haven't found any data on that yet. But just think about the geometry. The other isomers would have calcium salts that ought to form very different crystal lattices and fit together poorly with the Phthalate.

[Edited on 27-10-2017 by SWIM]
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Boffis
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[*] posted on 27-10-2017 at 22:50


Hi mackolol, if phthalic acid/anhydride are your target then you can enrich the starting xylenes in o and p isomes by boiling with excess 80-85% sulphuric acid, I have a paper that describes this process somewhere, basically the m-isomer forms a water soluble sulphonic acid much more readily than the o and p isomers and can be removed as the aqueous acid phase; diluting with an equal volume of water and refluxing hydrolyses the sulphonic acid and liberates the free m-xylene. In the mean time I have attached an old paper, the first part of which describes the oxidation of xylene and the separation of the isomeric phthalic acids, they use fractional distillation to up grade the starting material with respect of p and m isomers and an o isomer enriched residue but they used a 150cm column! With a bit of ingenuity I think you could modify the procedure to use the direct oxidation product though it might require a cascade of crystallisation steps. The removal of the m-isomer by the sulphonation route would simplify matters since phthalic acid is fairly soluble in hot water but terephthalic acid is almost insoluble under any conditions.

Good luck and let us know how you get on.

I'll post the sulphonation paper when I find it.

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[*] posted on 28-10-2017 at 03:20


If you are trying to get phthalic acid it might be easier to start from naphthalene.

http://www.lookchem.com/Chempedia/Chemical-Technology/Organi...
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Boffis
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[*] posted on 28-10-2017 at 04:43


I think the original poster chose xylene because that's what he has access to. Besides it is easier to oxidise xylenes in a "wet" reaction than naphthalene in an amateur setting at least. Vapour phase oxidation is not very practically in a home environment and wet oxidation of naphthalene is slower.

Attached are a couple more papers about the separation of Xylenes, one as promised is the sulphonation route but I notice they are using partially separated xylenes running>95% of one isomer and a second paper on the freezing method of separation o and p isomers.

My original plan was to use the sulphonation method to separate out most of the m-isomer and then the freezing method to separate the o and p isomers; clean up the p-isomer by further low temperature sulphonation and the o-isomer by fractional distillation.

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Attachment: Separation of Xylenes JOC Clarke & Taylor 1923.pdf (104kB)
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mackolol
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[*] posted on 29-10-2017 at 14:06


Ok thanks very much for Your advices now i have two ideas how to make it. I will write how it goes but for now i forgot that i dont have cooling column to reflux the solution in oxidising xylene reaction i must buy it.
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[*] posted on 29-10-2017 at 19:50


mp for o-phthalic acid = 207°C
mp for m-phthalic acid = 347°C
mp for p-phthalic acid =300-402°C

I would think you could drive off the o-phthalic acid by heating the mix, keeping the temperature below 230°C or so. The purity of the phthalic anhydride could then be determined by taking its mp (131.6°C).

See: https://www.sciencemadness.org/whisper/viewthread.php?tid=25...


[Edited on 30-10-2017 by Magpie]




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[*] posted on 30-10-2017 at 01:29


Looking at Magpie's last post has given me an idea. Since o-phthalic acid forms an anhydrous compound simply on heating to its melting point you could either sublime it off as Magpie suggests or maybe remove it with dihloromethane as phthalic anhydride is supposed to be fairly soluble in this solvent but the acids are all very sparingly soluble/ insoluble in this solvent (and most similar non and weakly polar solvents). The two remaining acids could then be separated as described in the papers posted above.
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mackolol
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[*] posted on 30-10-2017 at 14:17


I thought about melting it but o phthalic acid sublimes far below this temperature. But thanks for idea with DMC because while sublimating phthalic anhydride forms structure like cotton and it would be good to recrystalize it.
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[*] posted on 30-10-2017 at 18:21


If you use my method in Prepublication you form a melted phthalic anhydride that is clear. When this melt is solidified it can be easily ground up and is very pure.

hope this helps




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[*] posted on 27-11-2017 at 07:03


A series of U2U's with mackolol and this recent thread from myristicinaldehyde:

http://www.sciencemadness.org/talk/viewthread.php?tid=78508

have prompted me to continue to develop some ideas about utilizing the xylene isomers to produce various derivatives without necessarily isolating the pure isomers as the derivatives re often easier to purify. Of particular interest in this respect is the way the different xylene isomers interact with sulphuric acid. All three isomers readily sulphonate in concentrated sulphuric (Crafts; Comptes Rendus, 114, p1110 [1892]) yielding various sulphonic acids and water. On the other hand 50% sulphuric acid hydrolyses the sulphonic acids to xylene (Clarke and Tayler above), which may be steam distilled out of the mixture and regenerating the sulphuric acid.

It is clear therefore that for each isomer there must be an equilibrium point. m-Xylene is frequently stated to be the most readily sulphonated and even reacts with 80% sulphuric acid so its equilibrium point must be at <80% sulphuric acid but since it is hydrolysed by 50% acid the equilibrium point must be >50% acid.

So my idea is to reflux a mixture of conc sulphuric acid with a quantity of dry xylene such that the sulphuric is diluted to 70% strength when about 60% of the xylene has reacted but maintain the reflux for many hour to ensure that equilibrium has been reached. Hopefully this will yield a xylene residue enriched in o-p isomers and a sulphonic acid solution highly enriched in m-xylene-4-sulphonic acid. I would then see if I can get a good crystalline sulphonic acid from the aqueous phase, if not I would dilute it a little, reflux for a long period (I'm thinking 8-12 hours), steam distil and attempt to crystallise the m-sulphonic acid again. I would then try to freeze the o-p mixture to remove part of the p-isomer. If this doesn't work I would try removing more m-isomer by the boiling dilute nitric acid method (Annalen, v148 p10(1868)) or by further sulphuric acid treatment.

The above is similar to the method proposed by Jacobsen (Berichte, v10 p1009 [1877]) and criticized by others without them fully appreciating the equilibrium nature of the reaction; nor did Clarke (JACS v45 p830 1923). I have yet to check out Crafts paper in Comptes Rendus (v114 p1110 [1892]) in detail but this paper also looked at the separation of xylenes in part by sulphonation. The point is that some at least of these writers have simply used an excess of cold sulphuric acid and have no idea whether any equilibrium is being established in which case little or no preferential reactions take place.

In myristicinaldehyde's thread there is an illustration of the sulphonation of p-xylene to completion by reacting 5.2g (6ml) of p-xylene with 18.4g (10ml) of conc sulphuric acid. If the initial acid strength was 96% with these weights the final acid strength would 89%. It would be interesting to try this reaction with say equal weights and see if there is some residual xylene, this would allow you to estimate the equilibrium position for p-xylene. I hope myristicinaldehyde is going to keep us posted about his experiments.

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mackolol
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[*] posted on 7-12-2017 at 05:30


I have done this reaction yesterday but i used 80% sulphuric acid. While dilution of h2so4 i didnt want to cool it so it heated a lot this may caused little water evaporation. However i set it refluxing with xylene and after about one hour xylene layer disappeared. I waited half an hour but nothing changed. It seems like all xylene isomers reacted. Now im on crystallisation step. I think i used too strong sulphuric acid and in further reaction i will cool it down while dilution and use ~70% h2so4.


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[*] posted on 8-12-2017 at 08:46


In that case you used too much acid. It is not the strength of the acid but the amount you used that is important. You need to experiment to find out what the ratio of your xylene to your acid is that results in the consumption of just two third of the xylene. This should remove most of the m- isomer.
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mackolol
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[*] posted on 8-12-2017 at 13:06


Ok now i see i used 60ml of conc h2so4 what makes 80 ml of 80% sulph acid and only 36ml of xylene according to myristicinaldehydes post with sulphonation of xylene. I will repeat this experiment with 27ml h2so4 diluted to 80% because it is about 2times more densier than xylene and 36ml of xylene. And if i dilute it more would reaction take place or acid will be too weak?
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Boffis
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[*] posted on 9-12-2017 at 11:54


@mackolol, yes that sounds about right. I did some calculations for my acid (96% sulphuric) with xylene and I got the following results

Assuming 10ml of xylene and 2/3 of this reacts with sulphuric acid then if I had 4ml of 96% acid to start with the concentration of the acid at the end will be about 66%. I think this is too weak and the sulphonation reaction will stop before the 2/3 point is reached.

Making the same assumptions but using 6ml of sulphuric acid the final acid concentration will be about 81.5%. I think this is a little too strong so I think I would try about 5 to 5.5ml of 96% sulphuric acid per 10ml of mixed xylenes and reflux until equilibrium is reached (say 2-3 hrs), cool and pour into a measuring cylinder to see how much acid and xylene I now have. You are looking to get about 3.5ml of residual xylene which, according to my theory :), should consist mainly of o and p-xylenes. There is no need to dilute the acid before mixing with the xylene, this is done by the water liberated from the reaction.
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mackolol
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[*] posted on 9-12-2017 at 12:30


Today i did the reaction and used as I said 36ml of mixed xylenes and 27ml h2so4 diluted to 80%. After about 1,5 - 2h i think, the layer of xylene was about 1/3 of starting and it seemed that reaction has stopped (because before when i shaked flask instead of rapid foaming and lot of xylene vapour filling my reflux condenser nothing happened). The yield was about 14ml (so 1/3 of starting) of yellow - crude xylene. I plan to distill the resulting xylene and then put it in the freezer to separate o isomer. I dont remember now why I used this ratio of h2so4/xylene, I thought amount of acid will be exactly to sulphonate 2/3 of xylene but I was wrong. Anyways the reaction gone good . I will check if meta xylene is pure by oxidising it with kmno4 and checking if any part doesnt dissolve in water (which would mean that there was also o-isomer) and i will do the same with purified o-isomer.
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[*] posted on 15-12-2017 at 06:36


@mackolol, your results are interesting. If I assume that your 80% acid is 80%w/w and density 1.732 then I calculated that the final sulphuric acid strength is 59% H2SO4 w/w when 24ml (20.76g) of xylene have reacted. This is much lower than I had expected and it mean that in my experimental idea described above it would be better with only 4ml of conc. 96% sulphuric acid per 10ml of mixed isomer xylene to remove most of the m-xylene. I'll be home in a few days so I have a go at this and then recover the xylene fraction and oxidise them with potassium permanganate ad see what acids are produced.

By the way how did you prepare your 80% acid or what is the 80% (weight to weight: weight to volume (ie. molarity) or volume to volume). Its important because if you mixed 80ml of 96%w/w H2SO4 with 20ml of water the final strength would be about 84.5%w/w, in which case the final acid strength would be about 66.1%w/w.

By the way have translated the Crafts paper from Comptes rendus v114; if you or anybody else wants a copy I'll post it here.
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mackolol
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[*] posted on 15-12-2017 at 06:58


Youre right i took somewhere around 5ml of watet and about 20ml sulphuric acid (i dont really remember how much exactly) so the concentration was way higher.
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[*] posted on 21-12-2017 at 12:34


I have done a little research and I found this https://www.google.com/patents/US2348329. According to this after separating m isomer as the yielding mixture of xylenes isn't able to separate by cooling it to -20C ( lowest temperature i have access to) is further sulphonated and p xylene sulph. acid has higher solubility. But i dont know exactly how and how to do this as i dont really understand whats written here. I also dont know whats the goal of hydrochloric acid in this reaction. Please help me with planning this reaction. Thanks
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[*] posted on 21-12-2017 at 12:53


No idea if this will help, but ...

I tried a natural extraction which yielded something i could not identify, stated in a paper as an isomer of the target compound.

The process required adjusting the pH to get it to precipitate.

Maybe your isomers will behave similarly - at a low pH they both are soluble, at neutral the first one drops out, at a high pH the first one is again soluble but the second one drops out.




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mackolol
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[*] posted on 26-12-2017 at 13:31


Today i did this reaction one more time and took 40ml of mixed xylenes and 20ml of 98% h2so4 run in reflux separated remaining layer of unreacted xylene in separatory funnel. Then i poured 40ml of water to the mixture and again in reflux. It ran few hours but i didnt observe formation of any xylene layer. Was my acid too diluted or i dont know? Any hints? I can say also that temperature in both reactions was about 140C.
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[*] posted on 30-12-2017 at 08:38


@mackolol; yesterday I had a go at my ideas laid out above. I refluxed 40ml of mixed isomer xylene with 16ml of conc 96-97% sulphuric acid for 1 hour and got a god-awful black gunk. I diluted it with 60ml of water and placed it in a separating funnel. Even the organic layer was very dark and the boundary between the layers difficult to see. I ran off most of the rather viscous black aqueous layer and diluted the contents of the funnel with some fresh water. The aqueous layer was now quite pale and the boundary easily discernable. I let it stand until a sharp separation had occurred and ran the remaining aqueous layer into the first cut; total about 130ml. The xylene was run into a measuring cylinder and gave 23ml.

The black aqueous material was partly neutralised with 50ml of 40% (10 M) NaOH solution and 3g of activated charcoal added to the hot solution, heated it to about 80 C for 20 minutes on a hotplate and then filtered it though a Buchner funnel. The filtrate still looks like creosote :). I left it over night in an outbuilding at about 3C and this morning the solution was full of large black monoclinic prisms, these proved to be stained sodium sulphate, the shape of the crystals indicates the decahydrate. Now I am going to treat the liquor with charcoal again and evaporate it down to see if I can get the sodium xylene sulphonate.

Did you get the same black oxidation product when you used 80-85% acid? Its oxidation because I can smell the SO2.

By the way in the Crafts paper (translation attached) they steam distill the sulphonic acids after adding hydrochloric acid. They also admit to losses due to carbonization.

Attachment: A methode of separating Xylenes Comptes Rendus Crafts 1892 v114 p1110 English.docx (19kB)
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[*] posted on 30-12-2017 at 09:49


xylene is available at the hardware store. Can m-xylene be converted to resorcinol with facility? If so, how?



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mackolol
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[*] posted on 30-12-2017 at 13:46


@Boffis; Yes i got same black product in all my reactions (w using 98 and 85% h2so4) and it smells little weird something between so2 and very mild xylene smell. So what about hydrolysing my m xylene sulphonic acid. Won't it work under reflux? I will try to add sulphuric acid to make it 50% and then try again. I dont know how to steam distill and i guess i dont have needed aparature.
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