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Melgar
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That only works at low concentrations though. At high concentrations, you'd get a lot of stuff like tetrachloroethylene.
The first step in the process of learning something is admitting that you don't know it already.
I'm givin' the spam shields max power at full warp, but they just dinna have the power! We're gonna have to evacuate to new forum software!
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dan.vlad
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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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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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clearly_not_atara
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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.
Attachment: adimurthy2006.pdf (309kB)
This file has been downloaded 495 times
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dan.vlad
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"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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ninhydric1
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Here it is @dan_vlad.
Attachment: methanoltoDCM.pdf (119kB)
This file has been downloaded 546 times
The philosophy of one century is the common sense of the next.
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dan.vlad
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Thank you.
But it's too complicated..
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SWIM
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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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NEMO-Chemistry
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| 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.
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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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SWIM
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| 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.
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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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clearly_not_atara
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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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NEMO-Chemistry
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| 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 
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SWIM
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| 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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clearly_not_atara
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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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Assured Fish
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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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AJKOER
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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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madcedar
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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).
Attachment: Preparation of Methylene Chloride (Greene 1879).pdf (213kB)
This file has been downloaded 613 times
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dan.vlad
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Seems interesting, I have to try this reaction
Thank you
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