Sciencemadness Discussion Board - making a quaternary salt from secondary amine?

making a quaternary salt from secondary amine?

chemchemical - 7-11-2008 at 12:09

Well I'm trying to make a quaternary salt of a piperidine-like compound to use as a leaving group in a subsequent reaction.

I've looking up a lot of journal articles and have found the use of MeI in turning tertiary amines (I think it was tropane) into quaternary salts.

Could I just use excess MeI for making my secondary amine into a tertiary salt? Or I would I have to somehow deprotonate it before doing so? I'm not exacty sure how that hydrogen will effect the methylation.

Also, would iodoethane give similar results? (in making a quat. salt that is a good leaving group?)

smuv - 7-11-2008 at 13:21

Both would work, you need to add a base as well (preferably non-nucleophilic).

UnintentionalChaos - 7-11-2008 at 13:56

This largely depends on what you are trying to do with this reaction. If you are trying to get N-methyl piperine or N-methyltropane to leave during a hoffman elimination, you will probably end up cleaving the piperine/tropane system instead. I don't think iodoethane is a good choice since it may eliminate preferentially (depending on what else is attached to the nitrogen) and give ethylene. I believe that iodomethane is used intentionally because the methyl cannot eliminate to give an unsaturated compound.

More specifics would be extremely helpful.

leu - 8-11-2008 at 16:46

It takes two moles of methyl iodide to form the quaternary salt in the case of a heterocyclic amine

Water is the only product of the Hoffman elimination in this case, unless it's repeated to obtain the olefin and trimethylamine

UnintentionalChaos - 8-11-2008 at 20:29

Hoffman elimination procedes with the formation of the least substituted alkene. If you were to use bromoethane, as shown in the last reaction, elimination of one of the ethyl groups to ethylene is an unhindered path that the reaction may take. I don't know if the abstraction of a proton from a methyl carbon is any more favorable than the cleavage of the 6-membered ring (and abstraction of the proton from a methylene group). If it is more favorable, the reaction will proceed mainly via elimination of ethylene. If not, you will primarily cleave the ring to give N,N-diethyl-pent-4-en-1-amine. If methyl groups are used, there is no beta-carbon to abstract the hydrogen from except for the ones on the ring and it must proceed with ring cleavage.

[Edited on 11-9-08 by UnintentionalChaos]

hoffmanelimination.bmp - 204kB

chemrox - 9-11-2008 at 18:12

I'm troubled. You plan to use a piperidine-like compound and the N center is part of a leaving group? This sounds like a really wasteful thing to do. Please enlighten us further or at least u2u me with some details.

[Edited on 9-11-2008 by chemrox]

chemchemical - 9-11-2008 at 18:25

I'm not trying to do a hoffman elimination.

chemrox: this is what confused me about the reaction too at first but there is literature out there using tropinone compounds, thats where my main concern came from since the tropinones already have a methylated N, so adding MeI just adds another making the quat. I'm curious if my secondary, with the hydrogen, will work the same or if I need to abstract the hydrogen first with like LAH(or milder)?

The rxn I ultimately want to do is to take this quat. salt and react it with a primary amine, my understanding is that the charged nitrogen is somehow eliminated and the primary amine takes its place.

I have 3-chloropiperdine, I want to methylate it to the quat. salt and react it with 3-phenylpropylamine. The product should be the addition of the phenylpropylamine adding to where the quat salt was.

I can pull up the original tropinone paper if you are still having questions

Edit: I'm really just interested in how to make the quat salt since I know the other chemistry works, but I'm worried about that hydrogen.

[Edited on 9-11-2008 by chemchemical]

chemrox - 9-11-2008 at 19:44

Please post that paper!

Nicodem - 10-11-2008 at 01:04

Quote:

Originally posted by chemchemical
I'm curious if my secondary, with the hydrogen, will work the same or if I need to abstract the hydrogen first with like LAH(or milder)?

The acronym "LAH" is generally used for LiAlH4. May I ask where it came into being mentioned? What does it have to do with such a simple N-alkylation?
You can make a quat out of a secondary amine by simply refluxing the amine, 2eq alkyl halide and NaHCO3 in ethanol, filtering off the insolubles and rotavaping (with further purification if necessary). There are numerous other examples in the literature.
However, your more than obvious lack of basic understanding of concepts like acid/base chemistry, nucleophilic substitution and other such, makes it highly doubtful that you actually know what you want to do and what you are talking about.

Quote:

The rxn I ultimately want to do is to take this quat. salt and react it with a primary amine, my understanding is that the charged nitrogen is somehow eliminated and the primary amine takes its place.

It appears to me that you completely misunderstood some particular reaction. What you describe sounds like a Hoffman elimination/Michael addition tandem reaction, but since you did not provide a single reference, this thread is quite unreadable (and that is the reason why I'm moving it to Beginnings section).

Quote:

I have 3-chloropiperdine, I want to methylate it to the quat. salt and react it with 3-phenylpropylamine. The product should be the addition of the phenylpropylamine adding to where the quat salt was.

You can not make N,N-dimethyl-3-chloropiperidinium salts for the more than obvious reasons.

Quote:

Edit: I'm really just interested in how to make the quat salt since I know the other chemistry works, but I'm worried about that hydrogen.

Huh!?
You don't understand basic proton transfer chemistry, but you know all the rest about organic chemistry?

side notes

chemrox - 10-11-2008 at 18:27

Here are a couple of articles you might get something out of. I do rather agree with Nicodem that you have some of the concepts and terms mixed up and also suggest reading a book on mechanisms and reactions. I think his reply was a bit harsher than it had to have been but his points were correct.

Attachment: eq-Hutchinson & Tarbell.pdf (600kB)
This file has been downloaded 746 times

and here's the other one

chemrox - 10-11-2008 at 18:29

Attachment: Analgesics- Patchett & Giarrusso.pdf (360kB)
This file has been downloaded 781 times

chemchemical - 14-11-2008 at 16:00

Quote:

Originally posted by Nicodem
Huh!?

Alright, well a little searching and I was able to find out that I can methylate the piperdine twice (one pot) to make it a quaternary salt. They go from 22 (piperdine derivative) to 23 (a piperidinium iodide with excess MeI. My molecules are simpler too.

Attachment: dimethylation piperine.pdf (874kB)
This file has been downloaded 773 times

chemchemical - 14-11-2008 at 16:03

Quote:

Originally posted by chemrox
Please post that paper!

Here is the paper on the tropinone derivatives, the reason I wanted to know if it was possible to make the quaternary form. Even though piperidine is lacking that carbon bridge, it should still work, probably even better since there is less hindrance.

Attachment: Tropinone.pdf (207kB)
This file has been downloaded 1462 times

chemchemical - 14-11-2008 at 16:15

Quote:

Originally posted by chemchemical
Well I'm trying to make a quaternary salt of a piperidine-like compound to use as a leaving group in a subsequent reaction.

I've looking up a lot of journal articles and have found the use of MeI in turning tertiary amines (I think it was tropane) into quaternary salts.

Could I just use excess MeI for making my secondary amine into a tertiary salt? Or I would I have to somehow deprotonate it before doing so? I'm not exacty sure how that hydrogen will effect the methylation.

Also, would iodoethane give similar results? (in making a quat. salt that is a good leaving group?)

Well, I guess thats solved. MeI will make the iodide salt, I still do not know if iodoethane would work, I'm pretty sure I saw a tropinone with one ethyl and one methyl.

As the author states: "The supposed mechanism consists in the elimination
of dimethylamine yielding cycloheptadien-2,6-one
in situ, followed by addition of primary amine to the
diene to afford the desired tropinone derivative."

Quote:

Originally posted by Nicodem
The acronym "LAH" is generally used for LiAlH4. May I ask where it came into being mentioned? What does it have to do with such a simple N-alkylation?
You can make a quat out of a secondary amine by simply refluxing the amine, 2eq alkyl halide and NaHCO3 in ethanol, filtering off the insolubles and rotavaping (with further purification if necessary). There are numerous other examples in the literature.
However, your more than obvious lack of basic understanding of concepts like acid/base chemistry, nucleophilic substitution and other such, makes it highly doubtful that you actually know what you want to do and what you are talking about.

It appears that you misunderstood me. When I said LAH I meant lithium aluminum hydride. I incorrectly thought you might need something to abstract the proton, something like a base, such as the NaHCO3, or baking soda, for it to work. I was getting the methylation of nitrogen confused with oxygen for some reason as the methylations I have done all have been with oxygen.

I really dont understand why chemists appear to be so arrogant, I mean is my lack of basic understanding really that obvious?
I was talking about abstracting a proton with LAH.
I know nitrogen is acting as a nucleophile to kick out the dimethylamine, I wasnt sure exactly how it did that or could do that with a cyclic though. As for the other such I dont know... I'll work on it.

I said what I wanted to do, make the quat salt and react it with a primary amine like they do in the tropinone paper I posted. My question has been answered by the other paper I posted. Sorry if you didnt understand me.

And yes, I do not know proton chemistry but I am a specialist in anything organic

I guess I have a lot to learn in grad school

[Edited on 14-11-2008 by chemchemical]

Klute - 15-11-2008 at 06:56

LiAlH4 is a strong reducer, so rarely used to deprotonate a compound. Strong bases like NaH or t-BuOK are rather used. Most N-H protons require such a strong base except if they are particularily acid (sulfonamides for example) to be completly deprotonated, but with alkylating the very nucleophilic amines with a free doubet you do not need to deprotonate like you would with a alcohol, the base is just there to capture the HX formed, so things like K2CO3 can be used. Without any base, the amine would be alkylated anyhow, but the more basic unreacted amine would form the HX salt, which cannot be alkylated as it doesn't have a free doublet anymore. If a mild base is added, the freebase is liberated can can be alkylated again.

So most amines only require a alkylating agent and a mild base in a suitable solvent. Hindered amines can require long reflux, especially with larger electrophile.

The doublet in amides in much less "avaible" though, becasue of the mesomerism with the C=O, so it cannot attack an electrophile as readily. The N-H in a primary or secondary amide will have to be deprotonated for the alkylation to happen. The formed anion is also stabilized by mesomery, and can react very easily with an electrophile. Strong bases will be used to entirely deprotonate the amide, or in certain case a milder base like KOH can be used to partially deprotonate the amide.

Phenols can be alkylated with light base like K2CO3, as they are particularily acid for R-OH compounds. So only a small amount is deprotonated (equilibrium), but get immediatly alkylated. The phenolate is consumed, so more phenol gets deprotonated, shifting the equilibrium until all has reacted. Usually excess base is used.
On the other hand, if KOH or another strong base is used, all the phenol is deprotonated, and the very nucleophilic phenolate anion is quickly alkylated.

Hope this helps.

Nicodem - 17-11-2008 at 01:43

Quote:

Originally posted by chemchemical
Well, I guess thats solved. MeI will make the iodide salt, I still do not know if iodoethane would work, I'm pretty sure I saw a tropinone with one ethyl and one methyl.

As the author states: "The supposed mechanism consists in the elimination
of dimethylamine yielding cycloheptadien-2,6-one
in situ, followed by addition of primary amine to the
diene to afford the desired tropinone derivative."

N,N-Dimethyl-3-chloropiperidinium salts, regardless of the impossibility of making them from 3-chloropiperidine, can not go trough such an elimination/Michael addition tandem reaction for the simple reason because they don't have a beta-carbonyl group. Therefore, there can be no easy E1cB double elimination step and since no enone forms there can be no Michael addition of any kind either.

Quote:

LiAlH4 is lithium aluminum hydride!

R3N + RX => R4N+ + X-
As you see, no proton transfer is involved in the quarternisation of tertiary amines with alkyl halides - so no base is needed.
The base is only required to deprotonate the intermediate tertiary amine salt when starting from secondary amines (like in what you want). Their pKa is about 9 so just about any base of comparable strength and low nucleophilicity is enough.
R2NH + RX => R3NH+ + X- ={base]=> R3N
Besides, if hypothetically it was possible to quarternize 3-chloropiperidine this way (which is not), you would need to use an additional equivalent of the base since 3-chloropiperidine can only be stored as its salt (like its hydrochloride; but it is not even commercially available so what do I know what form do you have since you arrogantly don't say).

Quote:

I really dont understand why chemists appear to be so arrogant, I mean is my lack of basic understanding really that obvious?

It was you who started arrogantly by not even bothering telling what you want and even so much arrogant as not to provide the reference. Not to mention your not only arrogant but also dumb claim you know not about the proton transfers which are the base of basis of most polar mechanism based organic chemistry, but knowing all the rest of organic chemistry involved (which you now showed you don't have the slightest clue either).
Now you act surprised if you get an equally arrogant reply? Just read again what you write here:

Quote:

And yes, I do not know proton chemistry but I am a specialist in anything organic

WTF?

Quote:

I guess I have a lot to learn in grad school

You have to learn to learn first.

chemchemical - 19-11-2008 at 17:15

Quote:

Originally posted by Nicodem
N,N-Dimethyl-3-chloropiperidinium salts, regardless of the impossibility of making them from 3-chloropiperidine, can not go trough such an elimination/Michael addition tandem reaction for the simple reason because they don't have a beta-carbonyl group. Therefore, there can be no easy E1cB double elimination step and since no enone forms there can be no Michael addition of any kind either.

Does the impossibility lie in the halide? I should not have used that as an example I suppose, I have other piperidine analogues with non-halide substituents. The paper I posted has a piperidine analogue being turned into the N,N-dimethyl-piperidinium salt with MeI. So it is possible to turn a piperidine into the dimethyl salt. Its in the paper.

I said before that I was never trying to do a Micheal addition. I'm going to have to assume that the reaction is not going through a Micheal addition then, since none of the components of a Michael addition are present (double bonds, carbonyls).
Yes, I did want an addition but someone else thought it was Michael.

Please elaborate on your claim that chloropiperidine will not work. I was using it as an example analogue.

Quote:

Originally posted by Nicodem
It was you who started arrogantly by not even bothering telling what you want and even so much arrogant as not to provide the reference. Not to mention your not only arrogant but also dumb claim you know not about the proton transfers which are the base of basis of most polar mechanism based organic chemistry, but knowing all the rest of organic chemistry involved (which you now showed you don't have the slightest clue either).

It appears we have had a miscommunication, I forget english isnt everyones first language. I have provided the references now so thats not an issue.
Your replies do tend to have a bite to them though, I felt it demeaning which explains the adjective I used.
My very first post I said what I wanted, other people added their own opinions. All I wanted to know was if I could methylate piperidine twice without a base... sorry if the following quote sounded arrogant. I didn't realize people would have so much trouble with this question:

Quote:

Originally posted by chemchemical
Could I just use excess MeI for making my secondary amine into a tertiary salt(I meant quaternary)? Or I would I have to somehow deprotonate it before doing so? I'm not exacty sure how that hydrogen will effect the methylation.

Quote:

Originally posted by Nicodem
Now you act surprised if you get an equally arrogant reply? Just read again what you write here:

Quote:

And yes, I do not know proton chemistry but I am a specialist in anything organic

Sorry, but that was my attempt at using sarcasm against your original claim that I knew everything organic but nothing protonic. I was trying to get some basic answers when you posted that "harsh" (another posters words) reply.
I really do not know that much ochem besides undergrad level. I will be taking graduate level ochem though next year at a top ten research Uni in the USA, so I should learn a lot that year.

....

So I provided the references, my question is answered and we now know it is possible to dimethylate piperidine and then react it with a primary amine, which goes through some unknown addition/elimination reaction

I didn't mean to offend anyone, but I didn't think I deserved such a harsh reply. There were obviously some miscommunications in this post, probably from my poor use of the english language and others to comprehend it.

[Edited on 19-11-2008 by chemchemical]

Nicodem - 20-11-2008 at 01:12

Quote:

Originally posted by chemchemical

Quote:

3-Chloropiperidine is a beta-haloamine and as such is unstable as a free base where it is in equilibrium with the corresponding aziridine hydrochloride which is also unstable and will tend to form polymerization products. In any case you can not selectively N-methylate it before an intramolecular reaction occurs.
Obviously you can turn other piperidines without such sensitive groups to their quaternary salts, that is no problem and there are hundreds of papers with hundreds of examples in the literature. What you fail to understand is the mechanism of the reaction of the next step. You don't get it that for an E1cB elimination you need a carbonyl group attached at the beta position from the amine. Without this carbonyl there can be no Michael addition and therefore your desired reactions does not work.
Why don't you read your referenced paper again? Also, read some others on the same tandem reaction - there are several since it is a relatively well known reaction. Perhaps all this would make more sense to you if you would not limit yourself in your learning process.

Quote:

I said before that I was never trying to do a Micheal addition. I'm going to have to assume that the reaction is not going through a Micheal addition then, since none of the components of a Michael addition are present (double bonds, carbonyls).
Yes, I did want an addition but someone else thought it was Michael.

You are a bit ambivalent here. You say you are not attempting a Michael addition yet the paper you posted is about the double Michael addition on the dienone intermediate.

Quote:

All I wanted to know was if I could methylate piperidine twice without a base...

Of course you can, after all piperidine is a base itself, however the yields will suffer. It is unlikely you can get more than 30-50% yield this way.

making a quaternary salt from secondary amine?

side notes

and here's the other one

Related theoretical question