Sciencemadness Discussion Board - How to use AlHg Amalgam on Indole?

How to use AlHg Amalgam on Indole?

BobHawson - 11-12-2008 at 12:19

In the procedure below, could 2,5-Dimethoxynitrostyrene be substituted for 2-(indol-3-yl)nitroethene? It would be reduced to tryptamine, I believe.

*
1 g 2,5-Dimethoxynitrostyrene
10 ml Acetic acid
15 ml 96% Ethanol
5 ml Water
50 mg HgCl2
2 g Aluminum foil
The reaction must be carried at about 60øC, otherwise all of the starting nitrostyrene won't be dissolved. Some tips are:

Make a solution of everything but the Al, then add the foil, cut to oneinch squares at a rate enough to keep the temp around 60øC, and add until the orange colour disappear. About the double weight of nitro will be enough, maybe less.
Filter off the Al sludge, wash it with alcohol, then basify the solution with 25% NaOH. Filter again to remove the crystallized sodium acetate, and wash the filter cake with water.
Remove the solvent under vacuum, dissolve the residue in dilute H2SO4, and wash with 3x20 ml CH2Cl2. Basify with 25% NaOH and extract the solution with 3x25 ml CH2Cl2. Dry the pooled extracts over MgSO4, remove the solvent under vacuum, dissolve the residue in anhydrous ether, and saturate the solution with anhydrous HCl gas to precipitate pure 2C-H hydrochloride. To recover the 2C-H freebase, distill the residue above under vacuum instead.

http://www.erowid.org/archive/rhodium/chemistry/2cb.synthesi...
*

I see no reason why using AlHg on 2-(indol-3-yl)nitroethene shouldn't work.

Quote:

Originally posted by Nicodem
My reply in that thread was not to Ephoton, but to Hempshaman who asked if the acetyl group could be replaced with two methyls and I explained him that is not possible without first removing the N-acetyl and gave him a literature example of melatonin hydrolysis to yield 5-methoxytryptamine.

Reductive amination of 5-methoxytryptamine could perhaps be performed with sodium triacetoxyborohydride (STAB) using paraformaldehyde. STAB does reduce aldehydes but it reduces the imines much faster and since HCHO makes pretty stable imines and relatively rapidly it is quite possible such methylation could actually work.
The reductive amination using aluminium amalgam is problematic as it involves a lot of heat and works better with slightly acidic media (heat and acidic media are incompatible for tryptamines in the presence of formaldehyde). If modified to run at room temperature and bellow, with conc. aqueous formaldehyde at neutral media then it could perhaps work.

However, I am not trying to reductively aminate (e.g., the dimethyl), just reduce as in 2,5-Dimethoxynitrostyrene example.

Would 60øC be considered a lot of heat? Would the acidic conditions or heat matter, as it's not in contact with formaldehyde?

Apologies if the title of the thread is confusing, best name I could think of for the question.

[Edited on 12-12-2008 by BobHawson]

Nicodem - 12-12-2008 at 00:13

Quote:

Originally posted by BobHawson
In the procedure below, could 2,5-Dimethoxynitrostyrene be substituted for ß-indolenideniumethyl nitronate? It would be reduced to tryptamine, I believe.
...
I see no reason why using AlHg on ß-indolenideniumethyl nitronate shouldn't work.

What the hell is "ß-indolenideniumethyl nitronate"? Alkyl nitronates can not be isolated since they decompose to aldehydes, so what you are talking about is probably something else. The nitroethene starting material for obtaining tryptamine analogously to the method you posted would be 2-(indol-3-yl)nitroethene. Are you sure that you are not talking about this one?

stoichiometric_steve - 12-12-2008 at 00:34

Quote:

Originally posted by Nicodem
What the hell is "ß-indolenideniumethyl nitronate"?

Shulgin came up with this one. He apparently thought that the Indole nitrogen is basic enough to deprotonate the Nitroalkene. Does it matter?

Nicodem - 12-12-2008 at 00:44

So, is that supposed to be a name for 2-(indol-3-yl)nitroethene? Weird that anybody would have named it as a zwitterion. If it really is a zwitterion then it would reduce to 2-(indolin-3-yl)ethanamine with aluminium amalgam rather than to tryptamine. Luckily most chemist do not consider it as such. Nevertheless, indoles often reduce to indolines particularly in strongly acidic media or with acidic reducents (like boranes). I don't know how they behave under the aluminium amalgam reduction conditions, but I would certainly advise to make a literature search before doing any experiments.

grind - 12-12-2008 at 04:49

As far as I remember correctly STAB reduces indoles to indolines. LAH should work, and I think also modified NaBH4, what amounts to a borane reduction in most cases.

BobHawson - 12-12-2008 at 09:35

So what would be reduced to tryptamine with AlHg? Anything?

Could AlHg be used instead of LAH in the reaction below?

[Edited on 12-12-2008 by BobHawson]

stoichiometric_steve - 13-12-2008 at 02:02

grind, you and Nicodem are contradicting each other.

Nicodem - 13-12-2008 at 09:00

Not exactly. In some cases you can get the indole system untouched even if using BH3*THF. It pretty much depends on various factors like reaction temperature, solvent, time and who knows what. However, if you use borane with, or generate it with, an acid like BF3, Me3SiCl, RCOOH and similar, you will almost certainly obtain a complete reduction to the indolines (in the case of NaBH4/RCOOH you get N-acylindolines and/or N-alkylindolines since the (RCOO)3B intermediate product from the borohydride easily N-acylates the indoline as it forms, and the amide formed often can get reduced). If you take the time to search the literature you will see this topic is quite confusing and no certain outcome can be easily predicted.

zed - 13-12-2008 at 23:57

B-Nitrostyrenes, the benzene analogs to your Indole intermediate, do not reduce well under basic or neutral conditions.

At least, that was the premise that most of the folks that have produced phenethylamines from B-Nitrostyrenes followed.

Such reductions were usually performed under acetic conditions, supposedly to prevent polymerization. Which is probably why this reduction includes acetic acid.

The Indole ring, may or may not, tolerate acetic conditions. Sometimes it does, sometimes it doesn't. The 2-position may open, be oxidized, or be cyclized (via Pictet- Spengler type reactions). Since Indole production via Acetic Acid/ Metal reduction of O-Nitro Nitrostyrenes is a known method, It would appear that at least under mild conditions, the ring is stable in the presence of acetic acid.

It is possible that the reduction might be accomplished by Zinc and Acetic Acid.
Or, by Catalytic Hydrogenation, in glacial acetic acid at 0 C.

Likewise, your Al amalgam method might work.

Higher temps usually aren't your friend in such reductions, but who knows?

Now that being said, there are a lot better ways to make tryptamine. And, if you are hoping to make Alpha-methyl-tryptamine by such a reduction, it is unlikely you will succeed.

grind - 16-12-2008 at 15:24

By the way, is the Al(OH)3 generated from Al/Hg soluble in NaOH-solution?
Al(OH)3 from LiAlH4 reaction workup is not soluble in NaOH-solution. Why? Is there a reason?

Fleaker - 16-12-2008 at 15:38

You must not be using enough NaOH, as excess NaOH will give you the aluminate anion. I cannot see why the aluminum hydroxide from quenching either of these reactions should be different from each other. Both should be removable with strong base.

grind - 16-12-2008 at 23:38

It would be very nice if enough NaOH really dissolves the Al(OH)3 derived from LAH (to give soluble aluminate). But it definitively does not. Really! That means every basic workup (which is for example necessary with amines) is always a mess. There is absolutely no LAH reduction procedure in the chemical literature which ends with a clear aluminate solution as normally expected.
I also can not see why the Al(OH)3 from quenching either of these reactions should be different from each other. But it´s a fact, and this is a pitty.
And I think there is a reason. May be lithium aluminate is formed and this is insoluble?

Fleaker - 17-12-2008 at 08:56

I don't see why lithium aluminate would be so much soluble than sodium aluminate. The only thing I can possibly think of is that you form a different type of Al(OH)3 which is more chemically inert. I know there are people here who have a much better handle on crystal modifications and could better explain why one is more reactive than the other with respect to complexation.

Klute - 17-12-2008 at 09:42

A bit off topic, but I have had good results by neutralising excess LAH with hydrated Na2So4.10H2O until no more hydrogen forms (adding a few drops of water to be sure), warming to 40650°C for 20min, and filtering the granular white solids on a frit. You need to use a little more solvent that with the usual workup, and extract the cake by boiling a little THF, or at least good trituraation under THF, but it really works well and filtration goes like a breeze.

If the reaction medium gets too thick to stir and the LAH is still not completly hydrolyzed, you can finish with THF/H20, most of the water in excess will be moped up by the anhydrous Na2So4 formed, your filtrate will be dry and you just need to remove your solvent to get your product.

I ahve always had trouble getting a good white solid with the IPA/NaOH method.. Slight warming often helps though.

grind - 17-12-2008 at 11:03

@Fleaker:
May be it is dependent on the crystal modification. But even if you bring the precipitate in solution with acid (HCl or H2SO4) and basify then with an excess of NaOH, no clear aluminate solution is obtained. So I think the Li cant be excluded in the explanation.
@Klute:
This is a very interesting fact! I will try this the next time I work with LAH.

There is another method to isolate amines from LAH reductions:
Acidify with HCl, then add an excess of picric acid and filter off the amine picrate. The amine can isolated in 2 ways from the picrate:
1. Add an excess of LiOH solution to the picrate. Soluble Li-picrate is formed (Na and K picrate are far less soluble in water than Li picrate) and the amine can be isolated by solvent extracion.
2. Dissolve the picrate in hot diluted HCl, extract the picric acid with nitrobenzene and the amine can be isolated in the usual way from its HCl-salt (make alkaline, solvent extraction).

stoichiometric_steve - 17-12-2008 at 11:23

Quote:

Originally posted by Klute
warming to 40650°C for 20min

I think that's a bit hot.

chemrox - 17-12-2008 at 11:56

I've known Sasha Shulgin for many years and have a great deal of respect for him but he has confused me with his nomenclature more than once. Not that that's hard to do ;^). @Klute: did you mean warming to 40-65*C ? @Grind: I wonder if this would help with STAB reductions; do amine picrates generally form colored ppts.? Do they usually ppt. from acidic solutions?

Klute - 17-12-2008 at 13:57

Quote:

Originally posted by stoichiometric_steve

Quote:

Originally posted by Klute
warming to 40650°C for 20min

I think that's a bit hot.

Hehe obviously

Well, at least at that temp it will be a granular solid for sure, as your product

I indeed meant 40-50°C (I don't have a number pad on my keyboard, so have to use the shift key..)

@Chemrox: no need of messing with picrates during a STAB work up, NaOH will dissolve all the borate salts easily. If you obtain a amine, you can also directly steam distill it without needing to add excess base.

grind - 17-12-2008 at 21:10

Quote:

Originally posted by chemrox
I wonder if this would help with STAB reductions; do amine picrates generally form colored ppts.? Do they usually ppt. from acidic solutions?

I agree with Klute, there is no need for picric acid in your case. Of course you can try it, but I think the normal workup procedure is sufficient. Or is your filter problem persistent?

As far as I know all amine picrates are coloured (more or less yellow).
I don´t think that these picrates precipitate from very strongly acidic solutions. It would be better to have a pH near 7 or slightly less. In the case of LAH you should use an amount of acid which is just sufficient to give a clear homogeneous solution.