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Author: Subject: dibromomethane via formaldehyde
Mike
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[*] posted on 16-6-2004 at 15:58
dibromomethane via formaldehyde


Hi,

What do you thing about synthesizing CH2Br2 via aqueous solution of formaldehyde and HBr or Br2/P ? Will the polymerization of formaldehyde occur fast and disturb the reaction?




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Esplosivo
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[*] posted on 16-6-2004 at 21:29


The reaction with the C=O occurs when using PBr3 if you want a subsitution of the double bond across the carbon and the oxygen - although the rxn I am sure of occurs with PCl5 for the chlorination. I am not sure if PBr3 would have the same effect. Neither am I sure if PBr5 exists, but as far as I know it doesn't. Well this would (or should rather) give you dibromomethane. Addition of HBr will result to a small extent to the addition of Br in the C=O which is unsaturated giving a bromoalcohol, in this case bromomethanol (by nucleophilic addition if I remeber correctly).

[Edited on 17-6-2004 by Esplosivo]




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Mike
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[*] posted on 17-6-2004 at 09:14


But remember that formaldehyde reacts with water forming methandiol. 99,9% of the aldehyde converts to CH2(OH)2 in present of water. PBr5 exists for sure because I’m just looking at it :)



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Esplosivo
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[*] posted on 17-6-2004 at 09:56


Always learning some new thing. That's good, thanks.

Well since you have PBr5 you can use that to make your CH2Br2 from methanal. The rxn should proceed as follows:

HCHO + PBr5 ---> CH2Br2 + POBr3

99.9% turns into the diol?!?! I don't think thats right. If I recall correctly the reaction is an equilibrium one. Diols having the -OH groups on the same carbon are very unstable and therefore the equilibrium is shifted towards the methanal. Well this applies for ethanal and higher members of aldehydes, maybe it is different for methanal. But because of the lack of alkyl groups (and their inductive effect) in methanal then the rxn could proceed forming the diol (more than in the higher members) but not to such a large extent!

[Edited on 17-6-2004 by Esplosivo]




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[*] posted on 17-6-2004 at 10:58


The reaction of HCl and formaldehyde generates bis chloromethyl ether, a potent carcinogens.
I know HBr isn't the same, but wouldn't try this at home.:o
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Marvin
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[*] posted on 17-6-2004 at 11:04


HCl and formaldehyde forms bis-(chloromethyl) ether which is toxic and carcinogenic.....

Somehow I think brominating agents or HBr and formaldehyde are a really bad idea.

Just noticed on replying Ive been pipped at the post by unionised.
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[*] posted on 17-6-2004 at 11:30


Its probably worth saying twice.
Benzene, known carcinogen OSHA permitted exposure 1 ppm in air.
Bis CME, real bastard, OSHA PEL 0.001 ppm.
(these data are a bit old (1992), the limits may have changed,but the point still stands.)
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Tungsten
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[*] posted on 17-6-2004 at 13:10


Quote:
99.9% turns into the diol?!?! I don't think thats right. If I recall correctly the reaction is an equilibrium one.


That information is correct . Believe me.
PBr5 is expensive and I don’t want to waste it for this. I found out that bis(chloromethyl)ether is formed by reacting formaldehyde(gas) and HCl(gas).
What’s the mechanism of this reaction? But also find out that bis(chloromethyl)ether is decomposed by water.

So the questions are:
1) Will the 0,1% of HCOH react with HBr(aq) forming bis(bromomethyl)ether?
2) Will and how fast will it decompose in water and what will be the products?

PS
Sorry that I register as a new member but a can’t login as Mike. To moderators –just cancel Mike. Sorry again.




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[*] posted on 2-9-2005 at 17:26


Quote:
Originally posted by Esplosivo
Always learning some new thing. That's good, thanks.

Well since you have PBr5 you can use that to make your CH2Br2 from methanal. The rxn should proceed as follows:

HCHO + PBr5 ---> CH2Br2 + POBr3

99.9% turns into the diol?!?! I don't think thats right. If I recall correctly the reaction is an equilibrium one. Diols having the -OH groups on the same carbon are very unstable and therefore the equilibrium is shifted towards the methanal. Well this applies for ethanal and higher members of aldehydes, maybe it is different for methanal. But because of the lack of alkyl groups (and their inductive effect) in methanal then the rxn could proceed forming the diol (more than in the higher members) but not to such a large extent!

[Edited on 17-6-2004 by Esplosivo]

Procedure
In a 2-l. round-bottomed flask placed on a steam bath and fitted with a stirrer, a separatory funnel, and a reflux condenser is placed 540 g. (1.9 moles) of commercial (88 per cent) bromoform (Note 1). There is then added 10 cc. of a solution of sodium arsenite made by dissolving 230 g. (1.16 moles) of c. p. arsenious oxide and 440 g. (11 moles) of sodium hydroxide in 1.4 l. of water. The mixture is warmed gently to start the reaction, and then the remainder of the sodium arsenite solution is added during about one hour at such a rate that the solution refluxes gently. When the addition is complete, the flask is heated for four hours on the steam bath. The reaction mixture is distilled with steam, the lower layer of methylene bromide separated, and the water layer extracted once with 100 cc. of ether (Note 2). The methylene bromide is dried with 10 g. of calcium chloride and distilled. The yield of slightly yellow liquid boiling at 97-100°C is 290-300 g. (88-90 per cent of the theoretical amount).

Notes
1. The commercial bromoform used contained 12 per cent of alcohol. Its specific gravity was 2.59/25° as compared with 2.88/25° for pure bromoform.

2. The chief function of the extraction is to collect the fine droplets of methylene bromide which remain in the water layer.

[Edited on 3-9-2005 by verode]
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