Dronami_inc - 15-4-2017 at 09:42
Is it possible to get 2 useful products from the available (for me) DMF with the reaction of demethylation?
Something like that:
HC(O)N(CH3)2 + 2KBr + H2SO4 -> 2CH3Br + K2SO4 + HCONH2
As you can see, I am interested in Bromomethane and Formamide.
Give advice, please, or maybe somebody have a guide how to do this.
It could be made easier, but the fact is that it is very difficult to buy methanol in our region. Much more difficult than anything...
Crowfjord - 15-4-2017 at 09:58
That would not work. DMF can hydrolyze to formic acid and dimethylamine salt, but there is no good way to remove those methyl groups. Once you had
formic acid or formate you could amidate that with ammonia or an ammonium salt perhaps to get the formamide.
That's too bad about your access to methanol. How about methyl esters or formaldehyde? Hydrolysis or reduction (perhaps with formate from DMF?)
respectively, would get you to what you need.
[Edited on 15-4-2017 by Crowfjord]
Dronami_inc - 15-4-2017 at 11:11
Now it is clear.
1) For the purpose of interest: would it work when using dimethylaniline?
2) I have an access to few 20 l. cans of formaline. But as I know if I react it with NaOH it will result in many reaction products inluding
Formal(Methylal), Methyl formate, etc.
Crowfjord - 15-4-2017 at 11:49
HBr with dimethylaniline or any other methyl amine will not cleave the methyl groups. Von Braun reaction (cyanogen bromide) will cleave one methyl
from a tertiary methyl amine like dimethylaniline to give methyl bromide and the cyano-nor-amine. Cyanogen bromide is really nasty stuff though.
If you wanted to go a roundabout way of getting methanol from foormalin you would of course have to purify somehow. Distillation seems like the first
step.
clearly_not_atara - 15-4-2017 at 17:34
One possibility is the reaction of HBr with e.g. anisole, eugenol, anethole, guaiacol, guaifenisin, mequinol, vanillin, etc. In all cases the aryl
ether will cleave to release MeBr. Unlike amines, ethers can be cleaved by HBr, and methoxyarenes are quite common.
Condensing MeBr may be difficult as its boiling point is quite low. One possibility is to filter the efflux first through water at room temperature to
condense most of the products and then through acetone containing dry ice which will condense MeBr.
[Edited on 16-4-2017 by clearly_not_atara]
Melgar - 15-4-2017 at 17:36
Yeah, forget abstracting methyl groups from nitrogen, unless MAYBE you're dealing with a quaternary salt. you'd have better luck making methyl
bromide by heading up a methoxybenzene compound under pressure in the presence of HBr. You might have to heat to like 400-500C to get a reaction
though.
clearly_not_atara - 16-4-2017 at 19:53
400 C? Heavens no. This paper gives rate constants in glacial acetic acid using hydrobromic acid at 20 C:
http://parazite.pp.fi/hiveboard/picproxie_docs/000537223-The...
For anisole k2 = 9.42...
[Edited on 17-4-2017 by clearly_not_atara]
Melgar - 17-4-2017 at 03:52
Oh right. I was thinking of vanillin pyrolysis in a reducing environment, but forgot to take the catalytic action of the HBr into account. I wonder
if this would work with lignin, incidentally? Take dry sawdust and mix with H2SO4 to get black crap that is high in lignin. Dilute mixture so H2SO4
doesn't oxidize bromide to bromine, add bromide salt, then gradually raise the temperature until bromomethane formation started. There sure are a lot
of methoxy groups in lignin, and the yield would be low, but with the benefit of everything being quite cheap.
[Edited on 4/17/17 by Melgar]