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Author: Subject: Diclhloromethane synthesis
Melgar
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[*] posted on 9-11-2017 at 11:38


That only works at low concentrations though. At high concentrations, you'd get a lot of stuff like tetrachloroethylene.



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dan.vlad
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[*] posted on 11-11-2017 at 09:06


The teacher,author of many chemistry books, replied to the mail and thanked for that :
"The methods of preparation of compounds given in my book are what I have been reading and teaching for a number of years from differnt books without actually synthesizing them. You may have to go through books on synthetic organic chemistry to know exact synthetic conditions to prepare dichloromethane. "
So..the synthesis exist somewhere..
If somebody know about that..please help
Thank you all for responses and sugestions
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clearly_not_atara
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[*] posted on 11-11-2017 at 09:42


Quote: Originally posted by dan.vlad  
So..the synthesis exist somewhere..
If somebody know about that..please help
Thank you all for responses and sugestions

What's wrong with the route I posted? I don't think you're going to find an "easy" method.
Here's a synthesis of dibromomethane from bromoform, which *may* be an effective replacement -- apparently you're not looking for a solvent:
http://pubs.acs.org/doi/abs/10.1021/es00047a016?journalCode=...

[Edited on 11-11-2017 by clearly_not_atara]




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[*] posted on 12-11-2017 at 17:02


Regarding the preparation of methylene halides, I found a paper which describes the bromination of chloroacetic acid to bromochloroacetic acid using aqueous bromine and sodium hypochlorite in 72% yield. Assuming that this can be decarboxylated by heating similar to trichloroacetic acid (see AssuredFish's post on the last page) this gives a reasonably simple preparation of chlorobromomethane.

"62 g (0.83 moles) of sodium hypochlorite was added into 894 mL of alkaline bromine intermediate, containing a total of 262 g (3.26 moles [of Br]) of dissolved bromine in the form of 4.7:1 ratio of Br-:BrO3-. The contents were mixed thoroughly and allowed to react for 24h in a closed 5L round bottom flask, to obtain the 2:1 Br-:BrO3- reagent having 241g of available bromine."

The reaction is essentially 2 Cl+ + 2 Br2 >> 4 Br+ + 2 Cl-, although Br+ and Cl+ are never free ions in solution. Note that the moles of bromine quoted are moles of atomic bromine, Br, not molecular Br2.
The bromination of chloroacetic acid is performed with hydrochloric acid to activate the bromine, in the ratio 1.5g bromine (~3x molar excess) : 5 mmol substrate : 15 mL of 0.9N HCl, rxn for about 3.5 hours or until complete on TLC and quenched with sodium sulfite. The active brominating reagent may be BrCl or its congener HBrCl2, which is reported to form in solutions of bromine (I) in hydrochloric acid. It may be fruitful to experiment with a smaller excess of bromine or other methods of preparing dichlorobromic acid.
Chlorobromomethane boils at 68 C and probably reacts similar to methylene chloride.

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[*] posted on 19-11-2017 at 05:18


"Direct conversion of Methane to Methanol, Chloromethane and Dichloromethane at room temperature"

https://www.nature.com/articles/319308a0

If somebody can read this article,please help
Thank you ,all
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[*] posted on 19-11-2017 at 10:09


Here it is @dan_vlad.

Attachment: methanoltoDCM.pdf (119kB)
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[*] posted on 19-11-2017 at 22:24


Thank you.
But it's too complicated..
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[*] posted on 20-12-2017 at 15:52


Ha Ha, this bit of info's been knocking around the back of my mind since ephedrine was OTC in bulk quantities:

Freebase Ephedrine, if left dissolved in chloroform for an extended period, slowly deposits crystals of ephedrine hydrochloride.

Sounds like ephedrine is capable of de-halogenating chlorofrom on an extended time-scale. Don't recall if this required exposure to light or not.

Makes me wonder if ammonia or some alkyl amines might do this, and be re-cycleable to just keep making more and more DCM.





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[*] posted on 20-12-2017 at 18:35


Quote: Originally posted by SWIM  
Ha Ha, this bit of info's been knocking around the back of my mind since ephedrine was OTC in bulk quantities:

Freebase Ephedrine, if left dissolved in chloroform for an extended period, slowly deposits crystals of ephedrine hydrochloride.

Sounds like ephedrine is capable of de-halogenating chlorofrom on an extended time-scale. Don't recall if this required exposure to light or not.

Makes me wonder if ammonia or some alkyl amines might do this, and be re-cycleable to just keep making more and more DCM.


With Light/UV wouldnt Phosgene become an issue? Or would the conditions take care of such side reactions?

I ask because the OP has mentioned various routes as being difficult, that would make me nervous suggesting anything that carried a significant risk, unless of course there is a high chance that Phosgene is not an issue in this situation.
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[*] posted on 20-12-2017 at 19:11


Quote: Originally posted by NEMO-Chemistry  
Quote: Originally posted by SWIM  
Ha Ha, this bit of info's been knocking around the back of my mind since ephedrine was OTC in bulk quantities:

Freebase Ephedrine, if left dissolved in chloroform for an extended period, slowly deposits crystals of ephedrine hydrochloride.

Sounds like ephedrine is capable of de-halogenating chlorofrom on an extended time-scale. Don't recall if this required exposure to light or not.

Makes me wonder if ammonia or some alkyl amines might do this, and be re-cycleable to just keep making more and more DCM.


With Light/UV wouldnt Phosgene become an issue? Or would the conditions take care of such side reactions?

I ask because the OP has mentioned various routes as being difficult, that would make me nervous suggesting anything that carried a significant risk, unless of course there is a high chance that Phosgene is not an issue in this situation.


Don't know anything about the reaction mechanism. In fact, now that I think about it I have no idea how this reaction happens or what the products actually are other than ephedrine hydrochloride.

I read about this in a book about the extraction and purification of alkaloids on a commercial scale. I had never given it much thought but methylene chloride as a product of this reaction really doesn't make much sense...

Maybe it does make phosgene, but I'll be damned if I know where the extra oxygen would come from. Maybe it just forms until the air in the bottle's headspace is used up?





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[*] posted on 20-12-2017 at 21:59


Christ with the phosgene. Phosgene is reactive and will never "build up" in any solution that contains any kind of alcohol (incl ephedrine) or which is open to the air. It is hard to make phosgene, you won't do it by accident.

Ephedrine probably slowly reacts with chloroform via Sn2, forming N-dichloromethylephedrine which then cyclizes to 2-chloro-3,4-dimethyl-5-phenyloxazolidine and then maybe eliminate to 3,4-dimethyl-5-phenyl-4H-oxazolinium chloride. I don't expect chloroform to be reduced in this process.




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[*] posted on 21-12-2017 at 04:24


Quote: Originally posted by clearly_not_atara  
Christ with the phosgene. Phosgene is reactive and will never "build up" in any solution that contains any kind of alcohol (incl ephedrine) or which is open to the air. It is hard to make phosgene, you won't do it by accident.

Ephedrine probably slowly reacts with chloroform via Sn2, forming N-dichloromethylephedrine which then cyclizes to 2-chloro-3,4-dimethyl-5-phenyloxazolidine and then maybe eliminate to 3,4-dimethyl-5-phenyl-4H-oxazolinium chloride. I don't expect chloroform to be reduced in this process.

Interesting, i thought there was an inherent danger with Chloroform that didnt have ethanol added. many of the posts on here warn to keep in the dark and add ethanol.

I thought it worth asking, just in case we accidentally 'offed' a member :D
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[*] posted on 21-12-2017 at 16:49


Quote: Originally posted by clearly_not_atara  
Christ with the phosgene. Phosgene is reactive and will never "build up" in any solution that contains any kind of alcohol (incl ephedrine) or which is open to the air. It is hard to make phosgene, you won't do it by accident.

Ephedrine probably slowly reacts with chloroform via Sn2, forming N-dichloromethylephedrine which then cyclizes to 2-chloro-3,4-dimethyl-5-phenyloxazolidine and then maybe eliminate to 3,4-dimethyl-5-phenyl-4H-oxazolinium chloride. I don't expect chloroform to be reduced in this process.


You also don't seem to expect ephedrine hydrochloride to be formed in the reaction you proposed, which makes it hard to see how it could be germane to the deposition of ephedrine hydrochloride in the situation described.
I suspect that even in the early 20th century chemists were capable of telling ephedrine from 3,4-dimethyl-5-phenyl-4H-oxazolinium chloride.




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[*] posted on 21-12-2017 at 17:57


One molecule of ephedrine reacts with chloroform to generate two equivalents of hydrogen chloride which precipitates two equivalents of ephedrine hydrochloride. Not that surprising IMO.

NEMO: if you use stabilized chloroform it shouldn't be an issue

[Edited on 22-12-2017 by clearly_not_atara]




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[*] posted on 21-12-2017 at 22:12


I think i may know what is happening with this ephedrine thing.

The secondary amine of the ephedrine may be sufficiently basic enough to dehydrohalogenate the chloroform producing dichlorocarbene.

This would make sense as secondary amines tend to be the most basic of the amines due to the electron donation from the carbons to the nucleophilic amine and the lack of sterric hindrance.
This would leave the dichlorocarbene in the chloroform which could in theory be oxidized by atmospheric oxygen to phosgene, given enough time.

A way to test for this may be to throw some ethylamine or another primary amine into the solution which would result in the rather unpleasant odor of the isocyanide.

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[*] posted on 24-12-2017 at 17:36


Quote: Originally posted by dan.vlad  
"Direct conversion of Methane to Methanol, Chloromethane and Dichloromethane at room temperature"

https://www.nature.com/articles/319308a0

If somebody can read this article,please help
Thank you ,all


Don't have access to the paper, but suspect this path to CH2Cl2 (and other products) is a variation of the classic photolysis of a mix of Cl2 and CH4, which also uses light to break apart the chlorine gas:

Cl2 + hv --> .Cl + .Cl

However, in the new process, required chlorine gas and perhaps some of the chlorine radical can also be formed by the electrolysis of a chloride, along with the light path above:

Cl- --> .Cl + e- (see https://link.springer.com/article/10.1023/A:1016774717213 )
.Cl + .Cl = Cl2
Cl2 + hv --> .Cl + .Cl

My take on a possible radical reaction chains leading to CH2Cl2:

CH4 + .Cl --> .CH3 + HCl
.CH3 + .Cl --> CH3Cl
CH3Cl + .Cl --> .CH2Cl + HCl
.CH2Cl + .Cl --> CH2Cl2

but CH3Cl and CH2Cl2 are not likely the sole products of the reaction system.

I am basing my comments on the following quote from the abstract of the cited article above:

"The key steps in this process are: (1) electrochemical oxidation of the chloride ion; (2) generation of the chlorine radical under illumination; and (3) formation of the methyl radical by the reaction of methane with the chlorine radical."

[Edited on 25-12-2017 by AJKOER]
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[*] posted on 23-1-2018 at 01:35


Here is the paper that S.C. Wack was linking/referencing earlier in this thread, Preparation of Methylene Chloride from Chloroform, Ethanol and Zinc (Greene 1879).




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[*] posted on 11-2-2018 at 08:27


Seems interesting, I have to try this reaction
Thank you
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