Sciencemadness Discussion Board

Acid-catalyzed addition to double bonds (Nitroaldol-Reaction)

VitaminX - 3-4-2014 at 07:04

Hello,

I'm wondering about the mechanism of the nitroaldol reaction. I've read various protocols using BuNH2, MeNH2, TEA or ammonium acetate as base to catalyze this reaction between an nitroalkane and an aldehyde/ketone.
First I thought, alright as long as you acidify your reaction mixture while workup you get the nitroalkene and if you add a non-nucleophilic acid you get the nitroalcohol. Then I came across this workup on rhodium:

"1-(4-Fluorophenyl)-2-nitropropene:

6.05 mL of 4-fluorobenzaldehyde (56.4 mmol) was added to a 100 mL round bottomed flask containing 4.26 mL of nitroethane (59.2 mmol). 0.73 mL of 40% methylamine (8.46 mmol) was added to the stirring mixture, and the flask was heated at ~60°C. After about an hour, the flask was cooled to room temperature and 0.81 mL of glacial acetic acid (14.1 mmol) with 50 mL deionized water as added. The flask was then chilled in a freezer, but the light-yellow solution did not crystallize. TLC showed a decent amount of 4-fluorobenzaldehyde remaining, so additional water was added, the flask was swirled vigorously, and once the layers separated the light yellow organic layer was isolated and placed in a round bottom flask fitted with a reflux condensor. Another 4.26 mL of nitroethane (8.46 mmol) was added, followed by 0.25 mL of n-butylamine. The solution was held at reflux for 3 hours forming a dark reddish solution, which was cooled to room temperature and then placed in the freezer. After an hour the mixture was filled with yellow-orange solid. After filtering and recrystallizing the solid from boiling isopropyl alcohol, there was 5.63 g of 4-fluorophenyl-2-nitropropene (31.1 mmol, 55% yield) as shiny yellow spikes, some of which were over an inch long."

How the hell does he get the nitropropene even though there are loads of water in his reaction mixture at times and he adds an excess of GAA? Does he form the alcohol halfway through and dehydrates it again in the seconds step with n-butylamine? I would find that quite odd.

For example in another write-up from rhodium:

"10 g (80.6 mmol) of para-fluorobenzaldehyde (99%+ purity, 124.11 g/mole) was mixed with 15 mL’s of methanol in a 100 mL Erlenmeyer flask, and put in the fridge and refrigerated till the temperature was -10°C. Fortunately the benzaldehyde was soluble at these low temperatures, which means that the reaction can be conducted as originally intended. In a separate beaker, 7.3 g (96.7 mmol) of nitroethane (75 g/mole) was mixed with 8.14 g (80.6 mmol) of triethylamine (99%+ purity, 101 g/mol). This was also cooled to –10°C. Once the mixture was cooled satisfyingly, the contents from the beaker with nitroethane/triethylamine was poured into the Erlenmeyer flask with the fluoro-benzaldehyde. This was swirled a few times, and put back in the freezer. Once every 30 min’s it was taken out and swirled. The color of the mixture went from totally clear to yellowish over the course of the 2½ hours, which the reaction was allowed to run. Once the reaction was complete (2½ hours), the amine was quenched with an 1.1 equimolar amount of acetic acid, while the reaction mixture was still cool. It is very important to quench the reaction while it is still very cold, as the isomers will reach an equilibrium once the temperature rises, if there still is active catalyst present. Most of the solvent was stripped under vacuum, and the remains where dissolved in DCM and washed two times with water and once with brine. The DCM was stripped, leaving behind about 15 g of the crude nitroalcohol. "

With the same aldehyde and nitroalkene he gets the nitroalcohol apparently. Is there something essential I'm missing or does this seem odd to anyone else. Is it the heat/cold? Personally I'm interested in getting a nitroalcohol so I'm looking to prevent ending up with some nitroalkene.
Preferably I'm looking for a protocol that let's me approach both substances so I can check TLC of them and distinguish what has been formed in future experiments.

Thanks guys.

Source:
http://www.erowid.org/archive/rhodium/chemistry/para-fluoro-...
http://www.erowid.org/archive/rhodium/chemistry/pfa.spicybro...

[Edited on 3-4-2014 by VitaminX]

[Edited on 3-4-2014 by VitaminX]

leu - 3-4-2014 at 07:32

Your examples aren't acid catalyzed, they're base catalyzed :( One recent thread with a good reply that could be found using:

http://www.sciencemadness.org/talk/search.php?fid=12

is:

http://www.sciencemadness.org/talk/viewthread.php?tid=29379

Thus it's clear you need to do a lot more studying before another such post :D The end results from the effort applied :cool:

VitaminX - 3-4-2014 at 07:52

I meant acid-catalyzed addition of water to a double bond as you would get from a nitroalkene -> nitroalcohol.
I know that the nitroaldol reaction is base-catalyzed and I also stated that in the text. Do not blame your misenterpretation on my lack of knowledge.
The link you provided revolves around the same issue but doesn't give any answers beside: It is possible to get both and somehow phenyl-substituted derivatives dehydrate easier than other systems. This is irrelevant since both my examples are done with the same phenyl-substituted aldehyde.

The more appropriate question seems to be: How to avoid dehydration after the nitroalcohol is formed so you don't end up with the nitroolefine?
Maybe this is not possible for phenyl-substituted aldehydes so that the newly formed double bond has to be attacked once more by a proton in order to make water attack nucleophilicly (this is the acid-catalysis I was talking about all along). But I'm still not sure if the different conditions (solvent/temperature) could play a role too. It just seems odd that a reaction which undergoes dehydration would be done in aqueous methylamine solution (example #1).

Edit: Maybe the nitroalcohol was formed after the first step in example #1 and then after refluxing with BuNH2 dehydration took place so that would've been an error in the procedure.

[Edited on 3-4-2014 by VitaminX]

zed - 9-4-2014 at 23:55

I have always assumed in the heated butylamine catalyzed version, that the reaction proceeds via an imine intermediate. The imine then reacts with nitroethane. In some versions, the reaction is driven by removing water by distillation, as it is formed. A Dean-Stark trap may be used. See Henzelmann for details. If the imine is the intermediate, it suggests that the water molecule has already been eliminated from the structure, when the nitroethane becomes involved, and there is actually no nitroalcohol formed.

It would appear, that in the low temperature version you have attached.....No imine could actually be formed. Triethylamine doesn't have the required hydrogen. Thus, water would not be instantly eliminated from the molecule.
And, it would appear that low temperatures are employed to preserve the nitroalcohol by preventing dehydration.

At any rate, the Nitroolefin in question would most likely be reduced to a ketone, which would then be processed into an phenylisopropylamine. I have the impression that the ketone may be more easily manufactured by the diazotization of P-Flouroaniline, followed by reaction with Isopropenylacetate.

As a side note. A few years ago I was surprised to discover that the Nitro-Aldol condensation is reversible. Shulgin once used it to produce a rare aromatic aldehyde, from a nitro-propenyl-benzene. Refluxed the Nitropropenylbenzene with a benzylamine, to form a high boiling imine......distilled off nitroethane as it formed.

[Edited on 10-4-2014 by zed]

[Edited on 10-4-2014 by zed]

Chemosynthesis - 10-4-2014 at 14:38

Since you already have procedures, I don't see any reason not to post. The Henry nitroaldol reaction and DOI: 10.1039/A706353I should clear up any mechanistic questions pertinent to your question.

VitaminX - 16-4-2014 at 03:39

Quote: Originally posted by zed  
I have always assumed in the heated butylamine catalyzed version, that the reaction proceeds via an imine intermediate. The imine then reacts with nitroethane. In some versions, the reaction is driven by removing water by distillation, as it is formed. A Dean-Stark trap may be used. See Henzelmann for details. If the imine is the intermediate, it suggests that the water molecule has already been eliminated from the structure, when the nitroethane becomes involved, and there is actually no nitroalcohol formed.

It would appear, that in the low temperature version you have attached.....No imine could actually be formed. Triethylamine doesn't have the required hydrogen. Thus, water would not be instantly eliminated from the molecule.
And, it would appear that low temperatures are employed to preserve the nitroalcohol by preventing dehydration.

At any rate, the Nitroolefin in question would most likely be reduced to a ketone, which would then be processed into an phenylisopropylamine. I have the impression that the ketone may be more easily manufactured by the diazotization of P-Flouroaniline, followed by reaction with Isopropenylacetate.

As a side note. A few years ago I was surprised to discover that the Nitro-Aldol condensation is reversible. Shulgin once used it to produce a rare aromatic aldehyde, from a nitro-propenyl-benzene. Refluxed the Nitropropenylbenzene with a benzylamine, to form a high boiling imine......distilled off nitroethane as it formed.

[Edited on 10-4-2014 by zed]

[Edited on 10-4-2014 by zed]


Thank you, the imine part makes a lot of sense now, I can see why they would use TEA as base at low temperatures now (E1 mechanism favored by high temperatures).