Sciencemadness Discussion Board - Alternate route of AlCl3 synthesis

Alternate route of AlCl3 synthesis

weiming1998 - 8-4-2012 at 03:39

Today, it seems that I might have found an alternate (though more complex) route of anhydrous AlCl3 synthesis that doesn't involve high temperatures (not quite) or corrosive gases.

Firstly, some aluminum foil is boiled in ethanol (methylated spirits dried with MgSO4) with NaOH added (might not be necessary). The Al slowly became a black colour and got disintegrated by the ethanol. The solution was then filtered and evaporated. A white powder remains at the bottom. Some CuCl2 is added after, with some more anhydrous ethanol poured in. (I thought that my white powder was Al(CH3CH2O)3, and the CuCl2 would react with it to form a green precipitate of Cu(CH3CH2O)2, which, according to this:http://www.multivalent.co.uk/chemicals/products.php?item_id=13 doesn't dissolve in any liquids unless reacted, and AlCl3, which dissolves in ethanol.) A green substance indeed precipitated(the green substance also reacts with water to form a blue precipitate, a definite evidence in proving that it was copper ethoxide), leaving behind a colourless ethanol solution. I filtered it off, and poured the now colourless solution in some water. NaOH was then added, and a suspension of powder was observed.

Using the previous two evidence, there is a high possibility that some AlCl3 was produced, anhydrous or not. If not, the anhydrous AlCl3 can probably be produced through this method by more careful drying.

[Edited on 8-4-2012 by weiming1998]

[Edited on 8-4-2012 by weiming1998]

blogfast25 - 8-4-2012 at 05:17

Weiming, this is going to throw up a lot of questions, like:

Was the NaoH really necessary or not? Where did it go?

What kind of CuCl2 did you use? Hydrated or anhydrous? Added as a solution or how else?

You claim also that a ‘powder’ precipitated when adding NaOH to the ‘AlCl3’ water/alcohol mixture. Al(OH)3.nH2O is a very voluminous, floccular precipitate. I wouldn’t describe it as a ‘powder’…

Ultimately only an elemental analysis of the product can tell whether it is anhydrous or hydrated AlCl3. Your description, although somewhat tantalising, lacks facts and rigour to make any definitive claims. For such a product to have real use it really needs to be very dry. Hydrated or 'slightly wet' AlCl3 has little practical uses.

[Edited on 8-4-2012 by blogfast25]

weiming1998 - 8-4-2012 at 05:30

Yes, I believe there is still lots of error in my synthesis. The CuCl2, for example, needed to be anhydrous, and I used the hydrate, which will totally ruin the AlCl3. I will try again tomorrow, with more carefully-planned out steps, and see what happens.

AJKOER - 8-4-2012 at 05:35

Weiming:

I hope you liked the experiment.

I do recall that they may be a preparation via the aqueous AlCl3 using the appropriate solvent. I'll see if I can find the reference. Note, Wiki mentions ethanol, chloroform, carbon tetrachloride and that AlCl3 is slightly soluble in benzene, however, these do not ring the bell (a more esoteric solvent per my vague recollection).

The problem then becomes the synthesis of the organic solvent.

EDIT: I did find a Sciencemadness thread with pictures. DCM may be the solvent name I am searching for. Note, the preparation appears to have failed with a recommendation to just react Al and Cl2.

Link:
http://www.sciencemadness.org/talk/viewthread.php?tid=14111

For those interested, here is the direct synthesis of Al and Chlorine:

"Anhydrous Aluminum Chloride [1,2]

One end of a large diameter Vycor tube (25 to 40 mm) is inserted through one of two holes in a cork closing a wide-neck flask. The second hole holds a tube leading to the hood. The cork can dispensed with if the reaction tube can be made to fit tightly into the neck of a long-neck, round bottom flask. The reaction tube is placed in an electric furnace. The distance between the hot zone and the receiver flask should not be longer than 8 cm, to prevent plugging of the tube by the sublimate. The cork is protected from the heat by an asbestos wrapping. A porcelain boat containing aluminum turnings or powder is inserted into the reaction tube. Gaseous HCl is then passed through the end opposite the receiver. The rubber tubing connections should be as short as possible. When the air has been completely displaced by the HCl, the furnace is slowly heated until a white mist begins to appear. The flow of HCl is then increased and the temperature raised to prevent premature condensation of the sublimate. The reaction is then allowed to continue until the aluminum has been completely consumed."

Link:
http://www.erowid.org/archive/rhodium/chemistry/aluminumchlo...

For less demanding conditions, see US Patent 4,493,784.

[Edited on 8-4-2012 by AJKOER]

blogfast25 - 8-4-2012 at 05:57

Quote: Originally posted by weiming1998

Yes, I believe there is still lots of error in my synthesis. The CuCl2, for example, needed to be anhydrous, and I used the hydrate, which will totally ruin the AlCl3. I will try again tomorrow, with more carefully-planned out steps, and see what happens.

Yes, that would be interesting. Please also include some references, as I assume this idea did not come falling out of the blue sky. Nothing arises out of a vacuum.

blogfast25 - 8-4-2012 at 06:13

Quote: Originally posted by AJKOER

EDIT: I found it and a Sciencemadness thread with pictures.

Link:
http://www.sciencemadness.org/talk/viewthread.php?tid=14111

That thread remains highly disputed. The tentative consensus is that water contaminated the experiment, in which case you’ll always get some AlCl3, albeit wet to some degree (see discussion on page 4).

It remains highly unbelievable to me that anhydrous AlCl3 can be made from Al powder, dry HCl gas and an inert solvent (dichloromethane) AT ROOM TEMPERATURE. AlCl3 can be made by reacting strongly heated Al powder in bone dry HCl, that’s a standard lab preparation for AlCl3 anh., but temperature is the key here.

The requirement for heat strongly suggests that nothing will happen at room temperature (see Boltzmann Maxwell distributions, activation energy and related concepts), ‘solvent’ or not. To run this at RT would require a suitable catalyst to bring down the activation energy. None is used.

Peach also never continued his line of investigation. Presumably he accepted defeat.

bbartlog - 8-4-2012 at 09:04

I'm surprised that ethanol would be a suitable solvent for AlCl3. I would expect a reaction along the lines of AlCl3 + C2H5OH -> AlCl2OH + C2H5Cl, i.e. the AlCl3 would produce ethyl chloride in the process of turning itself into a chlorohydrate.

AndersHoveland - 8-4-2012 at 14:02

One would think the best route would be to pass dry anhydrous HCl gas into ethyl ether that contains some aluminum foil. The resulting AlCl3 is soluble in the ether. AlBr3, however, apparently reacts with ether, evidence that the Al-halogen bonds are covalent.

I really do not think it is practically possible to make anhydrous AlCl3 from aluminum triethanate. The oxygen-aluminum bonds are very strong. The "Al+3 ion" acts as a very strong acid when it is not hydrated. Aluminum triethanate, in the absence of water, is probably not going to be as basic as you think. AlCl3, for example, reacts very violently with water.

weiming1998 - 9-4-2012 at 00:46

I just made another batch of unknown product, dissolved in anhydrous ethanol.
Here are the steps:
Some aluminum foil is placed in a long-necked glass bottle (really just a glass soda bottle) with 100mls of anhydrous ethanol in it (dried methylated spirits first with MgSO4, then With CuSO4 to ensure that it has been dried properly.

According to Wikipedia, http://en.wikipedia.org/wiki/Methanol
short chained alcohols can dissolve the Al2O3, forming water and the corresponding aluminum salt. The reaction probably then snaps back, making a precipitate of Al(OH)3/Al2O3 and regenerating the ethanol. Once the oxide layer has been removed, the alcohol then attacks the aluminum directly, forming hydrogen and the alcohol aluminum salt. So the NaOH wasn't really necessary, and probably formed a byproduct of NaAlO2 from it and the formed H2O attacking aluminum.

The bottle was covered with a piece of aluminum foil at the top, to limit ethanol evaporation and prevent water vapour from entering. After a few hours, another 100mls of ethanol was poured in, to wash the salts formed out. The leftover aluminum was dark and it was weakened, indicating corrosion. The ethanol was then poured out into another container and some CuCl2 (dried on an iron plate) was placed in the solution. Instead of a green precipitate, a dark powdery precipitate appeared, which reacts instantly with water to form a fluffy brown precipitate that is lighter than water. I assume that happened because the wet CuCl2 reacted partially with the iron, generating most of it to FeCl2/FeCl3.

Now the product is slowly evaporating in the same soda bottle, in a water bath, with foil covering the top to prevent water vapour from getting in. The question is, how do I test whether my product is anhydrous AlCl3 or not? Also, any questions, criticisms and suggestions are welcome.

Edit: I don't think that worked well. I have now dissolved my AlCl3(?) in kerosene. It didn't react that much to water poured on it. Also, there were only something like a few milligrams of product. But a precipitate (white) was formed when the kerosene solution was added to a mixture of NaOH and water, so at least there is aluminum in the solution. The kerosene solution didn't even react with the NaOH solid pellets. I might, just might try again (the whole thing took me hours upon hours of work).

Edit again: Considering the previous source from Wikipedia, a bottle of anhydrous ethanol packed with dessicants and reacting with finely powdered aluminum oxide might do the trick to form the aluminum ethoxide. If that doesn't work, smaller aluminum chunks with CuCl2 might work too.

[Edited on 9-4-2012 by weiming1998]

[Edited on 9-4-2012 by weiming1998]

blogfast25 - 9-4-2012 at 05:41

Quote: Originally posted by AndersHoveland

One would think the best route would be to pass dry anhydrous HCl gas into ethyl ether that contains some aluminum foil. The resulting AlCl3 is soluble in the ether. AlBr3, however, apparently reacts with ether, evidence that the Al-halogen bonds are covalent.

And why, oh why, 'One', would one think that, huh? On what basis?

Why does an aqueous HCl solution react with Al at ROOM TEMPERATURE (and vigorously too, depending on concentration)? Because of deprotonation of the HCl and formation of oxonium ions, H3O+. It is these that oxidise the aluminium.

Now HCl is considerably soluble in ether and that may be due in part to protonation of the ether (forming 'diethyl oxonium ions' so to speak). It would remain to be seen how reactive these would be toward solid Al at RT. Not a sure thing at all.

But it would perhaps have been a more logical choice of solvent than CH2Cl2 in Peach's experiment, I'll grant you that...

And in what way does reactivity of AlBr3 to ether indicate covalence? AlX3 go from 'quite ionic' (AlF3) to almost 100 % covalent (AlI3)

Weiming:

Where does the iron pop into all of this?

As to producing Al methoxide, you are referring to this I believe:

"One of the potential drawbacks of using high concentrations of methanol (and other alcohols, such as ethanol) in fuel is its corrosivity to some metals, particularly aluminium. Methanol, although a weak acid, attacks the oxide coating that normally protects the aluminum from corrosion:
6 CH3OH + Al2O3 → 2 Al(OCH3)3 + 3 H2O"

It's very unlikely that significant amounts of Al trimethoxide can be made from reacting alumina with methanol. It's one thing for methanol to burrow itself through the (very thin) passivation layer of Al, quite another to react alumina in appreciable quantities with methanol. Really dry alumina doesn't even dissolve all that easily in real Bronsted acids, acids that are much stronger than methanol...

[Edited on 9-4-2012 by blogfast25]

[Edited on 9-4-2012 by blogfast25]

weiming1998 - 9-4-2012 at 06:56

Quote: Originally posted by blogfast25

Quote: Originally posted by AndersHoveland

Yes, the reaction can be ridiculously slow, but what if we use Al2O3 powder with a high surface area made of, say dehydrated Al(OH)3? The ethanol could be loaded with dessicants that sucks up all the water formed by the reaction: Al2O3+6C2H5OH<===>2Al(C2H5O)3+3H2O. This reaction is probably at an equilibrium and reversible, since aluminum ethoxide probably hydrolyzes upon contact with water. So, if we remove a product of the reaction, which is H2O, the equilibrium will be shifted to the left, speeding up the reaction and allowing to proceed.

Also, concerning synthesis of Al isopropoxide, HgCl2 can be added to the alcohol aluminum to speed up the reaction to favourable rates. It removes the oxide layer and also amalgamates with the aluminum, increasing surface area. This can probably be used with the synthesis of Al ethoxide, since ethanol is the more acidic alcohol in this situation. But if I don't have HgCl2, what about CuCl2? A small amount dissolved in ethanol? That would remove the oxide layer, although it wouldn't amalgamate.

Finally, that iron thing happened probably because this happened: CuCl2+Fe===>FeCl2+Cu or 3CuCl2+2Fe===>2FeCl3+3Cu. As the FeCl3 probably either has decomposed or is still a hydrate, it caused the failure of the experiment.
Note: The iron came from a plate that I was using to heat-dry the CuCl2.

[Edited on 9-4-2012 by weiming1998]

[Edited on 9-4-2012 by weiming1998]

barley81 - 9-4-2012 at 06:59

Quote: Originally posted by weiming1998

dried methylated spirits first with MgSO4, then With CuSO4 to ensure that it has been dried properly.

By the way, wiki says that MgSO4 is slightly soluble in alcohol, so you probably have it in your alcohol.

weiming1998 - 9-4-2012 at 07:06

Quote: Originally posted by barley81

Quote: Originally posted by weiming1998

dried methylated spirits first with MgSO4, then With CuSO4 to ensure that it has been dried properly.

By the way, wiki says that MgSO4 is slightly soluble in alcohol, so you probably have it in your alcohol.

They might have meant methanol instead of ethanol. Surprisingly, lots of inorganic salts are soluble in methanol, but not ethanol.

Even if it is slightly soluble, the amount might be so small it doesn't really matter.

blogfast25 - 9-4-2012 at 12:07

weiming:

"[...] but what if we use Al2O3 powder with a high surface area made of, say dehydrated Al(OH)3? The ethanol could be loaded with dessicants that sucks up all the water formed by the reaction:"

So a lot of something that absorbs water quickly and irreversibly but does not react with Al, Al methoxide or alumina? Not much springs to mind right now...

weiming1998 - 9-4-2012 at 16:06

The only one I could think of is MgSO4. Would that work, or will it react with aluminum ethoxide?

weiming1998 - 10-4-2012 at 01:20

I have completed my third attempt. Here are the process:
A mix of anhydrous CuSO4 is combined with NaCl in dessicant-dried ethanol. The mix at the bottom of the jar slowly turned a green colour with a bit of brown at the bottom. Aluminum is packed in and the jar is placed on a water bath for a few hours, with the anhydrous ethanol being replaced once. After the jar is taken out of the water bath, the dirty, grey mix is filtered. The resulting grey solution is then boiled dry in a long-necked soda bottle in a water bath. A white powder that slowly turned grey as the bottle was left there was the product. The white powder dissolves very rapidly in kerosene, leaving grey gunk that gives the same opaque solution as before when dissolved in water. The kerosene solution is still the original blue and hasn't changed colour/gone opaque. Some of the kerosene solution+water+NaOH produces a white precipitate, indicating the is Al3+ ions in the kerosene(I know other metals make white hydroxide precipitates as well, but out of the variety of metal salts that I used in this reaction( mainly Cu2+, Al3+ and Na+), only the Al3+ can produce a white precipitate when reacted against NaOH.

The setup was meant to generate anhydrous AlCl3 by first making Na2SO4/CuCl2 by a double displacement reaction(CuSO4 and NaCl are both(?) soluble, at least to a tiny extent, in ethanol, while Na2SO4 is not). CuCl2 is the most soluble salt out of the four. Then a reaction between CuCl2 and Al will occur, generating the desired product. A side reaction that will alse generate the product will occur, as CuCl2 removes the oxide layer, exposing the Al metal to attack, the formed ethoxide reaction with more CuCl2 to form copper ethoxide and AlCl3.

The question is, as I said before, how could I prove, using a chemical test, that the kerosene solution contains the anhydrous form of AlCl3? Also, is there any problems about my setup?

blogfast25 - 10-4-2012 at 08:42

Combining Cu2+ with Al metal would in aqueous conditions plate out the copper (3 Cu2+(aq) + 2 Al(s) === > 3 Cu(s) + 2 Al3+(aq)) but that doesn't seem to happen here.

Have you tried sublimating the suspected AlCl3? How does it react with water?

[Edited on 10-4-2012 by blogfast25]

Polverone - 10-4-2012 at 13:58

Try boiling a portion of the kerosene solution to dryness with exclusion of moisture. Anhydrous AlCl3 should sublime, while a hydrated product will hydrolyze and leave behind nonvolatile solids. Also, just to double check, you should run blank tests against your original kerosene, both for the boiling to dryness and the sodium hydroxide solution test.

weiming1998 - 10-4-2012 at 21:40

Unfortunately, my AlCl3 got hydrolyzed by some methylated spirits with water accidentally, so I cannot sublime it right now, but still I made a few observations:
1, The AlCl3 solution bubbled a lot when swirled with water, compared to the control.
2, The original kerosene with nothing added first did make some precipitate, but after careful washing of the beaker, it did not make any. The solution, while it made less precipitate, still made some.
3, The AlCl3 solution discoloured the kerosene (destroyed the colouring somehow, so the blue kerosene turned transparent) when heated in a water bath. Plain kerosene, kerosene with ethanol added in/with hydrated AlCl3 added in did not exhibit the discolouration effect.