Sciencemadness Discussion Board - Amine Alkylation Question

Amine Alkylation Question

(Brain)2NH - 21-7-2013 at 04:24

Hi all.

I'm not very experienced with practical organic synthesis.
I know :
R-X + Amine ---------------> RNH₂ and/or R₂NH and/or R₃N and/or R₄N⁺X^-

I want to know if the number of carbons of R in R-X would affect the products proportions ?

Example :
how do you predict whether the salt is formed or not in the following reaction :
C₂₀H_?Br + R₂NH ----------> ???????

If there are any review papers on the subject of "Amine alkylation", I would get my answer by comparing the products of the reactions.

Thanks in advance

(Brain)₂NH

bfesser - 21-7-2013 at 05:39

The short answer? No. The long answer? Yes. It gets complicated when you start mucking about with the carbon chains, turning them into bulkier groups and whatnot. <a href="https://en.wikipedia.org/wiki/Steric_effects" target="_blank">Steric effects</a> <img src="../scipics/_wiki.png" /> will come into play, and the <a href="https://en.wikipedia.org/wiki/Chemical_kinetics" target="_blank">kinetics</a> <img src="../scipics/_wiki.png" /> of the reaction will be affected. The more substituted the <a href="http://en.wikipedia.org/wiki/Amine#Classes_of_amines" target="_blank">amine</a> <img src="../scipics/_wiki.png" />, the slower the <a href="http://en.wikipedia.org/wiki/Nucleophilic_substitution" target="_blank">reaction</a> <img src="../scipics/_wiki.png" /> is expected to progress (rxn. rate <a href="http://dl.clackamas.cc.or.us/ch106-03/clasific.htm" target="_blank">1°>2°>3°</a> <img src="../scipics/_ext.png" />

. Just imagine the difference in effort between swinging one bat (fig. 1), versus swinging 3 bats (fig. 2). Also, the product is not a <a href="https://en.wikipedia.org/wiki/Salt_(chemistry)" target="_blank">salt</a> <img src="../scipics/_wiki.png" />.

See: <a href="http://en.wikipedia.org/wiki/Amine_alkylation" target="_blank">Amine alkylation</a> <img src="../scipics/_wiki.png" />

C20H?Br would be C20H41Br, by CnH2n+2, assuming a <a href="https://en.wikipedia.org/wiki/Alkane#Structure_classification" target="_blank">linear alkane</a> <img src="../scipics/_wiki.png" />.

<a href="http://www.homestarrunner.com/sbemail64.html" target="_blank"><img src="http://secxtanx.com/dump/cantmoveface/EatingBatteries.jpg" width="200" valign="top" /></a> <img src="../scipics/_ext.png" />

Hope this helps. By the way, welcome to ScienceMadness! Not a bad first post. I appreciate the use of subscript.

[Edited on 7/21/13 by bfesser]

sonogashira - 21-7-2013 at 05:54

Recently I tried to alkylate a tertiary amine (BnNR2) with Ethyl Iodide. After several weeks at room temperature, and occasional-periods of refluxing, the reaction had barely progressed. I then gassed it with MeBr and it was all converted after leaving at room temperature overnight. So even with something as "small" as ethyl iodide, the rate of the reaction giving the ammonium salt can be negligible.
When you have large alkyl groups as part of your alkylating agent, you can be reasonably sure that there will be no ammonium salt produced, even under forcing conditions. Of course this depends upon the nature of your amine and alkylating agent, but if you make a model of the reactants, it should become obvious as to whether or not the spacial arrangement of the atoms will hinder their combination.
Unless you are using a tertiary amine, and attempting to form the salt of that, you will need a base to aid the alkylation, since it the reacting amine may function as a base. For practical purposes, triethylamine and tripropylamine are often used as non nucleophillic bases, which should give you an idea of the reasonably small tolerance of alkyl-substituted amines towards being shielded from reactive alkylating amines due to the bulk of the appended alkyl groups - three ethyl groups usually being sufficient to prevent further alkyation; at least withing a reasonable time-scale, by which time the desired alkylation will have proceeded, and the amine usually precipitated as the hydrohalide.

(Brain)2NH - 21-7-2013 at 07:00

I want to be sure that no salt is formed in this reaction :

bulkyR-halide + Small 2^o amine -> bulkyR,R',R"N

Small 2^oAmine : Dimethyl, Diethyl, EthylMethyl
My bulky-R : a hydrocarbon of 10 carbons.

it would be no problem if NH₄Halide salt is formed.
but i don't want any R containing salt to be produced. (like NR₄Halide)

any guidance or tips to achieve this goal ?

thanks to all who care.

sonogashira - 21-7-2013 at 07:52

Your alkyl bromide may be totally unreactive. Why won't say what it is? No one can help you with generalities because not all alkyl bromides will act as alkylating agents. Does it contain 10 or 20 carbons? You have stated each.

bfesser - 21-7-2013 at 07:54

So you're trying to synthesize N,N-dimethyldecylamine (or analogues)? Are you planning on preparing the N-oxide? Unfortunately, I've been unable to find a literature procedure for the synthesis of your target compound(s). Here's what I've found:

<a href="http://pubchem.ncbi.nlm.nih.gov/summary/summary.cgi?cid=8168" target="_blank">N,N-dimethyl-1-dodecanamine</a> <img src="../scipics/_ext.png" /> (PubChem)
C12H27N
CAS № 1120-24-7

<a href="http://orgsyn.org/orgsyn/default.asp?formgroup=basenpe_form_group&dataaction=db&dbname=orgsyn" target="_blank">N,N-dimethyldodecylamine oxide</a> <img src="../scipics/_ext.png" /> Organic Syntheses, Vol. 50, p. 56 (1970); Coll. Vol. 6, p.501 (1988).
<a href="http://webbook.nist.gov/cgi/cbook.cgi?ID=1120-24-7&Units=SI" target="_blank">1-Decanamine, N,N-dimethyl-</a> <img src="../scipics/_ext.png" /> (NIST)
<a href="http://scholar.google.com/scholar?hl=en&q=dimethyldecylamine" target="_blank">Google Scholar Search for "dimethyldecylamine"…</a> <img src="../scipics/_ext.png" />
<a href="http://www.chemicalland21.com/specialtychem/perchem/DIMETHYLDECYLAMINE.htm" target="_blank">

Quote:

<a href="http://www.chemicalland21.com/specialtychem/perchem/DIMETHYLDECYLAMINE.htm" target="_blank">DIMETHYLDECYLAMINE</a> <img src="../scipics/_ext.png" />
This compound (tertiary Amine) is used as an intermediate for the manufacture of quaternary ammonium compounds, amine oxide and betaine surfactants for personal care, and institutional use. It is used as a corrosion inhibitor and acid-stable emulsifier.

</a>[edit] sonogashira is correct. Check out Table 1 on the <a href="http://en.wikipedia.org/wiki/Nucleophilic_aliphatic_substitution" target="_blank">Nucleophilic substitution</a> <img src="../scipics/_wiki.png" /> article I provided earlier.

[Edited on 7/21/13 by bfesser]

(Brain)2NH - 21-7-2013 at 09:16

Quote: Originally posted by sonogashira

Your alkyl bromide may be totally unreactive. Why won't say what it is? No one can help you with generalities because not all alkyl bromides will act as alkylating agents.

R is an indole.
my RX is 3-(2-BROMOETHYL)INDOLE
CAS# : 3389-21-7

2^oamine : Dimethylamine or Diethylamine

Thanks

bfesser - 21-7-2013 at 09:38

Your proposed reaction:

<a href="http://www.evershinechem.com/3389-21-7-p-161.html" target="_blank"><img src="http://www.evershinechem.com/uploadfile/product/3389-21-7.gif" valign="middle" /></a> <img src="../scipics/_ext.png" valign="top" /> + <a href="http://en.wikipedia.org/wiki/Dimethylamine" target="_blank">(CH3

2NH</a> <img src="../scipics/_wiki.png" /> → <a href="http://en.wikipedia.org/wiki/Dimethyltryptamine" target="_blank"><img src="http://upload.wikimedia.org/wikipedia/commons/thumb/8/88/DMT.svg/200px-DMT.svg.png" valign="middle" width="120" /></a> <img src="../scipics/_wiki.png" valign="top" />

(Ambiguity will only land this thread in Detritus.)

solo - 21-7-2013 at 10:44

Reference Information

.....find this book and learn the answers.....solo

Concerning Amines
(Their Properties, Preparations and Reactions)
David Gingsburg
1967

.....read, alkylation of amines Page 69

[Edited on 21-7-2013 by solo]

sonogashira - 21-7-2013 at 13:08

It will work, but I doubt that ammonium salt could be avoided with dimethylamine. With diethylamine one could certainly avoid it, but it will depend upon the CAS number of your intended solvent.

Attachment: Liebigs Ann. Chem., 1935, 520, (01), 019-030.pdf (544kB)
This file has been downloaded 389 times

[Edited on 21-7-2013 by sonogashira]

(Brain)2NH - 21-7-2013 at 13:19

Quote: Originally posted by sonogashira

I doubt that ammonium salt could be avoided with dimethylamine. With diethylamine one could certainly avoid it
[Edited on 21-7-2013 by sonogashira]

So the formation of salt depends more on the secondary amine than Alkyl-Halide ?
Is it what you meant or I'm misunderstanding you purpose ?

thanks again

sonogashira - 21-7-2013 at 13:27

I am saying only that diethyltryptamine will react with your alkylating agent reasonably slowly, and dimethytryptamine will react with it quite rapidly, as you can see in that paper.

Why talk of generalities of reactivities of secondary amines with alkyl halides?

(Brain)2NH - 21-7-2013 at 13:40

The paper is in german but based on pics I think it said the attached image is produced in my proposed reaction, Right ?

sonogashira - 21-7-2013 at 13:52

Yes, it is produced along with the tertiary amine.

bfesser - 21-7-2013 at 13:59

(Brain)2NH, your reaction would be difficult to control, would result in a mixture of products, and uses an expensive substrate. This is neither an efficient nor a cost effective route to the target molecule. If you're genuinely interested in the theory, I recommend that you study the literature to get a better handle on the chemistry involved. If you just want someone to tell you how to synthesize your target, try searching Erowid, Hive, Rhodium, etc.

[Edited on 7/21/13 by bfesser]

(Brain)2NH - 21-7-2013 at 14:12

Sorry, I'm asking too many questions on this thread.

But only this last unresolved question:

If tertiary amine reacts with a released HBr molecule, then this one is made.
how to predict whether this salt (attached image) is more prone to be formed or the previous image ?
my guess (raw guess) is this one is more possible.

sonogashira - 22-7-2013 at 01:16

If you ignore the excess of dimethylamine that you have in solution, then certainly it will be formed in preference. What does it matter anyway? Are you interested in studying the mechanism? The reaction will work, and quaternary ammonium bromides are easy to remove.

[Edited on 22-7-2013 by sonogashira]

Rich_Insane - 25-7-2013 at 13:06

Would addition of a strong amine base, like diisopropylethylamine mitigate the formation of the quaternary amine in this particular reaction?

Hunig's base is the only non-nucleophilic amine that I can think of for this purpose... Sterically hindered pyridines like 2,6-lutidine are probably too acidic to work for this purpose, am I right?

Prometheus23 - 25-7-2013 at 16:15

I know this probably isn't a very practical alternative given that your tertiary amine would be easily separated from the quaternary ammonium salt by-product, but I've been working with dodecanethiol lately so it came to mind.

If you are concerned about overalkylation to the quaternary ammonium salt then one possibility would be to use trimethylamine (or N,N-diethylmethylamine) instead of dimethylamine (or diethylamine). This would of course lead to the quaternary ammonium salt exclusively. However heating this product with sodium dodecanethiolate (formed in situ from dodecanethiol and a strong base) in DMSO will produce the N-demethylated product. If trimethylamine is used then the final product would be N,N-dimethyltryptamine, and if N,N-diethylmethylamine is used then it would be N,N-diethyltryptamine.

Again, there are obviously simpler methods if you already have dimethylamine or diethylamine. But I've been experimenting with dodecanethiol for N-demethylations as well as O-demethylations of aryl methyl ethers recently.