I want to produce ethylamine from bromoethane and ammonia, but I'm not sure about the best way to go about capturing the ethylamine after the reaction
has taken place and was hoping someone here might be willing to help me out.
As I understand it, the reaction of bromoethane with excess ammonia should be favorable and require no heat input. From what I've read, ammonia boils
at -33C and ethylamine boils at around 18C. I was thinking I could use a distillation apparatus in order to attempt to evaporate off the ammonia into
a separate container leaving the ethylamine in the original flask.
>Would this require any heat input or would it work at room temperature?
>Will I be able to feasibly condense the ammonia into a second flask or will it try to remain a gas?
I'm ok with losing the ammonia, but I want to capture as much of the ethylamine as possible.
>Will I require a fractional distillation condenser or would this be feasible with a simple condenser?
Another concern I have is the impact of water on the system. I know anhydrous ammonia would be ideal for this job, but if there were a significant
amount of water present with my ammonia, how badly might that affect my yield? As I understand it, the basicity of the ammonia would lead to a lot of
free -OH in my solution which I suspect might compete with the ammonia in bumping off the bromine atom on the bromoethane molecules creating ethanol
rather than ethylamine. Will I need anhydrous ammonia for this to be feasible or would an ammonia-water solution still do the job well enough?karlos³ - 19-3-2017 at 13:56
You will never get pure ethylamine using ammonia, rather the product will be mostly triethylamine or even quarternary ammonium salts, depending on the
excess if ammonia used.
For primary amines, usually synthesis methods like the delepine or gabriel synthesis are employed.
I have even seen a very good, but now lost write-up for making ethylamine using the delepine reaction, on versuchschemie.de back when it existed.
So, reaction of bromoethane with hexamine in chloroform(or EtOH), the hexamine adduct precipitates, and can be cleaved using HCl, so you get
hydrochloride salt of ethylamine then.synchronize - 19-3-2017 at 14:14
It's not that I don't believe you, but I'm just interested in understanding why:
If I were to add the bromoethane dropwise to a solution of excess ammonia, why would the bromoethane molecules be preferentially attacked by ethylated
amines rather than the much more widely available ammonia molecules?
The reaction you're describing between hexamine (hexamethylenetetramine?) and bromoethane followed by HCl is the one from versuch?
I think the gabriel synthesis is probably too cost-ineffective, but I'll definitely look into the delepine synthesis with hexamine. From what I
understand reading your post, you're saying the hexamine attacks the bromoethane resulting in ethyl groups attached to the amines in the hexamine, and
the addition of HCL breaks them off into individual ethylamine molecules right? If I got that right, do you mean the whole hexamine molecule breaks up
into ethylamine molecules (like a polymer breaking into monomers), or there's a 1:1 conversion of bromoethane into ethylbromine?karlos³ - 19-3-2017 at 14:47
Yes, hexamine is hexamethylentetramine.
And the write-up on versuchschemie was using, IIRC, ethyl chloride instead of bromide and hexamine, in chloroform.
I am not sure if I can explain it right:
I think the reason is, that a secondary amine is a stronger nucleophile than a primary amine, and thus rather prone for further alkylation than the
primary one, and the so obtained tertiary amine even more, that guarantees a mixture of products.
This is usually a problem when primary amines are the goal and ammonia is used to achieve this, being a strong nucleophile,
The delepine reaction(or gabriel, or any similar, like using succinimide/saccharin etc as synthon) does only produce the primary amines.
But only one nitrogen is alkylated on each hexamine, the halogen from the alkylhalide ends up producing the salt of the quarternary hexamine adduct.
The acidic hydrolysis using HCl, with EtOH present as well, produces ammonium chloride from the remaining three amino groups in hexamine, and the
methylene groups end up producing formaldehyde, which reacts with the ethanol present to form formaldehyde diethylacetal. Assured Fish - 19-3-2017 at 16:55
Another way you could get to ethylamine would be to react silver nitrite with your halo ethane in the Viktor mayer test, followed by reduction of the
nitro ethane to ethylamine.
This would unfortunately have to be done in ether i think with a ice bath to keep the temp below 15*C, a refux condenser might also be a good idea in
case you add the ethyl bromide too fast but if your careful you could achieve this in a beaker and you shouldn't have to take any drying precautions,
just don't add more water the the reaction.
If you read up on the Viktor mayer test then most examples will use iodide but this paper describes using brimides instead so im certain it should
work. http://onlinelibrary.wiley.com/doi/10.1002/9780470638859.con...
If your still intent on reacting ammonium with your bromo ethane then i would advise using a rather large excess of ammonium and then adding the bromo
ethane to it slowly. You would then need to carefully add a base to convert any hydrogen bromide amine salts to there freebases and metal bromide
salts followed by low temperature fractional distillation, a 200ml vigreux culumn will do.
Ethylamine boiling point 20*C
Diethylamine boiling point 55*C
Triethylamine biling point 88*C
When collecting the ethylamine just drip it into a solution of HCl to get ethylamine hydrochloride which can be reacted with a base to get freebase
ethylamine at will.
No discussion of the delepine in the ethylamine synthesis thread? I am very surprised!
Person on versuchschemie had made it using ideal amateur conditions, using OTC stuff and such(Ethyl chloride is OTC as well as hexamine there).CuReUS - 19-3-2017 at 21:41
If I were to add the bromoethane dropwise to a solution of excess ammonia, why would the bromoethane molecules be preferentially attacked by ethylated
amines rather than the much more widely available ammonia molecules?
that method will give you ethylamine
Quote:
Make sure you have a 10x molar excess of NH3 to EtBr else you will get di and tri ethylamines as the main products. Where as in industry this is a
problem on a small scale home setting this should not be an issue as long as the EtBr is slowly added to NH3 solution.
