Sciencemadness Discussion Board - Glycine to triethylamine?...

Glycine to triethylamine?...

RebeccaOlson - 20-3-2017 at 09:14

Heyy guys, I'm working on a reaction to synthesize triethylamine from glycine. I'm really crappy at chemistry and don't know much about how the reactions work. Can you see any errors in it? Thank you <3

Amos - 20-3-2017 at 09:35

EDIT: Please ignore my inability to discern functional groups from an image.

Do you absolutely have to synthesize triethylamine from glycine? It's going to be terribly inefficient no matter which way you go about it. If you can settle for just obtaining it from a different source, there are weedkillers and herbicides that use triethylammonium salts, such as this one found at Home Depot in the US: http://www.kellysolutions.com/erenewals/documentsubmit/Kelly...

The above product can yield 18-20 grams of triethylamine if boiled down, treated with base, and distilled.

[Edited on 3-20-2017 by Amos]

Texium - 20-3-2017 at 09:48

Quote: Originally posted by Amos

The first step isn't an esterification; the functional group you're making is an ether, not an ester. This will require a methylating agent such as methyl iodide or dimethyl sulfate, which actually can't be used because they'll end up methylating the amine group on glycine and giving you a mixture of products.

Sure about that? Looks like an esterification to me. The ether is then formed from the ester by Clemmensen reduction. The real problem here is that the ester will not survive the conditions of the Clemmensen reduction (or the Wolff-Kishner reduction, for that matter). If this route is to be used, a different reaction will need to be used for that step.

Also, I'd hazard a guess that based on the OP's previous posts, this is a theoretical homework problem, and not something that they intend to actually do.

CuReUS - 20-3-2017 at 10:22

the errors IMO are:
1.The first step might lead to N-methylation,as Amos mentioned
2.clemmenson's can't reduce ester to ether,so it won't work
3.collin's oxidation step might oxidise the amine
4.the last step would give you tetraethylammonium iodide as well
You could get the target compound in 2 steps from glycine:-
1.reduce the COOH in glycine to CH₃ http://www.sciencemadness.org/talk/viewthread.php?tid=73024#...
2.Do an eschenweiler clarke using acetaldehyde on ethylamine to get TEA
You could also do it the other way round.You might even be able to do a one step reductive ethylation of glycine to get TEA

[Edited on 21-3-2017 by CuReUS]

Amos - 20-3-2017 at 11:18

Quote: Originally posted by zts16

Quote: Originally posted by Amos

Sure about that? Looks like an esterification to me. The ether is then formed from the ester by Clemmensen reduction. The real problem here is that the ester will not survive the conditions of the Clemmensen reduction (or the Wolff-Kishner reduction, for that matter). If this route is to be used, a different reaction will need to be used for that step.

Well now I feel dumb; I think I was thrown off by the way the molecule is oriented in the diagram. Either way, glycine won't form an ester becaus

Also, I'd hazard a guess that based on the OP's previous posts, this is a theoretical homework problem, and not something that they intend to actually do.

Well, now I feel dumb. I guess I got confused by the orientation of the molecule in the diagram. Regardless I'm pretty sure Fischer esterification is not a viable method for glycine due to its zwitterionic nature.

Metacelsus - 20-3-2017 at 11:55

No, it's perfectly possible to make glycine methyl ester hydrochloride by treatment of a solution of glycine in methanol with dry HCl gas. I would call that a Fischer esterification.

But this overall route isn't viable anyway, so it's of little matter.

RebeccaOlson - 20-3-2017 at 20:35

What if I reacted the glycine with benzaldehyde to protect the nitrogen then at the end react it with ethyl iodide, then hydrolize it to break off the benzaldehyde, then react it with ethyl iodide again?

RebeccaOlson - 20-3-2017 at 20:40

I found this: (https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/Ima...). It shows the amine as a salt (hydrochloride maybe) and it doesn't get N-methylated by the Fischer esterification.

How would I get the Clemmensen reduction to work though?

[Edited on 21-3-2017 by RebeccaOlson]

JJay - 20-3-2017 at 20:55

Can hydrobromic acid really cleave that ether?? You could skip straight to the amino alcohol by reacting your starting material with lithium aluminum hydride. It might be possible to reduce the hydroxyl with catalytic hydrogenation with the right catalyst, but if you dehydrate it first (with something like sulfuric acid) or oxidize it to a carbonyl, you'll have an easier time. Would Collins oxidation actually oxidize the amine under normal conditions?

[Edited on 21-3-2017 by JJay]

RebeccaOlson - 20-3-2017 at 21:05

Yes, apparently refluxing it with concentrated hydrobromic acid demethylates an ester.

I don't think the Collins oxidation would oxidise the amine. I'm following this: (https://chemistlibrary.files.wordpress.com/2015/05/oxidation...) page 20, though it doesn't really say anyting about amines.

Lithium aluminium hydride is way too expensive so I can't use that. I could put some sort of protecting group on the nitrogen (benzaldehyde maybe?).

Are you saying I can break off the OH (when it's a carboxylic acid) with sulfuric acid??? I thought only LiAlH4 could do that?

[Edited on 21-3-2017 by RebeccaOlson]

JJay - 20-3-2017 at 21:19

I don't think hydrobromic acid would be able to demethylate that particular ether, though it could demethylate the ester... every example I'm seeing of hydrobromic acid demethylating a methyl ether involves a phenol product.

I don't think you can dehydrate a carboxylic acid with sulfuric acid, but I think you could dehydrate the alcohol with it, as I stated previously. I'm not seeing how a protecting group on the nitrogen would be helpful.

RebeccaOlson - 20-3-2017 at 21:27

How would I go from a carboxylic acid/ester to a ketone (so the Clemmensen reduction would work) (with easily availabe chemicals)?

Sorry if I sound really dumb :$...

[Edited on 21-3-2017 by RebeccaOlson]

[Edited on 21-3-2017 by RebeccaOlson]

JJay - 20-3-2017 at 21:37

You could make an aldehyde by dry distilling the calcium salt with calcium formate as discussed here: http://www.sciencemadness.org/talk/viewthread.php?tid=10336

This would not be high yielding but would require cheap and easily available reactants.

You could then reduce the aldehyde with a Wolff-Kishner reduction and proceed with ethyl iodide.

Oh, sorry, did I say ketone? I meant aldehyde.

[Edited on 21-3-2017 by JJay]

RebeccaOlson - 20-3-2017 at 21:42

Hmmm, that sounds pretty good! What's the yield like?

[Edited on 21-3-2017 by RebeccaOlson]

JJay - 20-3-2017 at 22:05

Probably around 5-60%. I think a vapor phase reaction with formic acid and a manganese dioxide catalyst might form the aldehyde, perhaps with better yields.

CuReUS - 20-3-2017 at 22:38

Quote: Originally posted by JJay

Would Collins oxidation actually oxidize the amine under normal conditions?

Quote: Originally posted by RebeccaOlson

I don't think the Collins oxidation would oxidise the amine. I'm following this: (https://chemistlibrary.files.wordpress.com/2015/05/oxidation...) page 20, though it doesn't really say anyting about amines.

see pg 25(43 in pdf) of that article

Quote:

Selenides are oxidized to selenoxides that normally suVer an in situ elimination.111 amines are destroyed,

Quote: Originally posted by RebeccaOlson

I found this: (https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/Ima...). It shows the amine as a salt (hydrochloride maybe) and it doesn't get N-methylated by the Fischer esterification.

why is the glycine methyl ester unstable ? no zwitter ion to waltz with ?

[Edited on 21-3-2017 by CuReUS]

RebeccaOlson - 20-3-2017 at 23:51

Sorry what's a zwitter ion?

JJay - 21-3-2017 at 00:00

Quote: Originally posted by CuReUS

see pg 25(43 in pdf) of that article

Quote:

Selenides are oxidized to selenoxides that normally suVer an in situ elimination.111 amines are destroyed,

[Edited on 21-3-2017 by CuReUS]

The cited reference is about cyclic nitrones, not amines: http://www.sciencedirect.com/science/article/pii/S0040402001...

I remain skeptical.

RebeccaOlson - 21-3-2017 at 00:16

JJay, wouldn't dry distilling the calcium salt with calcium formate make a ketone? How would I get rid of the methyl group? Like get rid of R2...

[Edited on 21-3-2017 by RebeccaOlson]

JJay - 21-3-2017 at 00:19

Dry distilling with formic acid supposedly makes an aldehyde rather than a ketone. I have not tried this but believe that it actually works.

[Edited on 21-3-2017 by JJay]

RebeccaOlson - 21-3-2017 at 00:22

Oh okey. Thank you so much. You guys are so helpful <3

RebeccaOlson - 21-3-2017 at 00:35

So would this be the final reaction?

[Edited on 21-3-2017 by RebeccaOlson]

RebeccaOlson - 21-3-2017 at 00:49

Would the dry distillation work with two molecules of glycine per atom of calcium?

[Edited on 21-3-2017 by RebeccaOlson]

RebeccaOlson - 21-3-2017 at 00:59

Ummm but wouldnt calcium formate turn the OH into a CH3. How do I get rid of that? I want it to be a H.

[Edited on 21-3-2017 by RebeccaOlson]

RebeccaOlson - 21-3-2017 at 02:24

Would the Wolff kiscner reaction work with esters?

RebeccaOlson - 21-3-2017 at 04:56

Bouveault–Blanc reduction

Look what I found!

Would that work?

1. Fischer esterification
2. Bouveault-Blanc reduction
3. Collins oxidation
4. Clemmensen reduction
5. Ethyl iodide

[Edited on 21-3-2017 by RebeccaOlson]

Texium - 21-3-2017 at 05:46

Yep, that looks like it would work. Although JJay's proposed route also likely works, you'd probably get a funny look from your professor for it being as it's rather unconventional by academic standards. Better to go with the Bouveault–Blanc route that you propose.

And no, Wolff-Kishner reduction doesn't work on esters.

CuReUS - 21-3-2017 at 09:35

Quote: Originally posted by RebeccaOlson

Would that work?

1. Fischer esterification
2. Bouveault-Blanc reduction
3. Collins oxidation
4. Clemmensen reduction
5. Ethyl iodide

why do you want to do a 5 step reaction(3rd and 5th step problematic,4th step might cause 2 molecules of amino acetaldehyde to react with each other to give piperazine) when you can get the target compound in 2 steps(possibly 1) ?

[Edited on 22-3-2017 by CuReUS]

RebeccaOlson - 21-3-2017 at 13:49

Thank you so much zts16! Gosh I feel so bad asking you guys so many questions... But I like being bad

1 step from glycine to triethylamine? Whaaat??

CuReUS - 21-3-2017 at 16:49

Quote: Originally posted by RebeccaOlson

1 step from glycine to triethylamine? Whaaat??

http://www.sciencemadness.org/talk/viewthread.php?tid=73046#...