Author Topic: Insitu methyl bromide questions  (Read 195 times)

Goldmember

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Insitu methyl bromide questions
« on: July 15, 2010, 05:20:41 AM »
Ive been mucking around with some Bromo-chloro-dimethylhydantoin (BCDMH) and metabisulfite ala Lens writeup so as to source some bromine/ide to go on and try some methyl bromide preps.No worries works as written. Then it occured to me that maybe I could skip having to isolate any bromine or bromide and use the same reagents to hopefully produce the methyl bromide in methanol.
Thinking along the lines that yeah the bisulfite will liberate bromine from BCDMH, but shouldnt a 2 fold excess also reduce the bromine to HBr and Sulfuric/sulfurous acid which should then form the alky bromide in situ???

So without any decent set of scales and using ethanol instead of methanol I gave it slap-dash whirl. Chilled all the well ground reactants, @1:2.5 BCDMH to NaHSO3 and healthy splash of ethanol(methylated spirits) to minus zero and threw em all together in a test tube. Initially the solids turned orange as if to indicate bromine but after about a minute as the tube was allowed to warm a touch, the bromine colour completely faded and a mildly exothermic reaction kicked with vigorous gas evolution(HBr I might stupidly assume). The reaction subsided and a slightly ethereal/alcoholic smell was noted permeating from the tube.No halogen odour was noted.(yes I realise you shouldnt go huffing unkown compounds from test-tubes)
Anyway  I was left with a slightly cloudy white  sollution with a white precipitate( likely simply an excess of either reagent).
After one hour sitting at room temperature the sollution has cleared and turned slightly yellow.I havent let my finely tuned analytical nostrils loose yet again.

My questions are,
1) Does this make any sense to anyone?
2) Does the theory back my hypothesis? If so is the stoichiometry close to being correct?
3) Is there any quick and easy way you can suggest I determine if methyl bromide is formed.
4) Any better conditions to try it under?
5) If Im not completely off base and there is a chance of what Im saying does have some merit, would someone a bit better equipped be prepared to give it a propper shot.

 ;D



atara

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Re: Insitu methyl bromide questions
« Reply #1 on: September 26, 2010, 06:20:11 PM »
I'm thinking a better thing than bisulfite would be dithionite, the product of reacting sodium bisulfite with zinc (which can be obtained from pennies). Dithionite is significantly more reactive than bisulfite (it can also be used for reductive aminations).

Anywho the only way you're going to see methyl bromide is by piping the gas into something very cold; it's likely that an hour after sitting at room temperature, all of the MeBr has long since left the building. I would presume the gas produced by the reaction was partly HBr and partly EtBr.

I guess one way to tell if you have an alkyl bromide is to put it into a solution of potassium iodide in acetone. If potassium bromide precipitates, you did something right; that's the Finkelstein reaction (which, unfortunately, is quite slow at room temperature, so this isn't a fast test!). The resulting alkyl iodide will be easier to work with and more reactive, too. Removing any HBr from the solution might be a little harder...
« Last Edit: September 26, 2010, 06:49:02 PM by atara »

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Re: Insitu methyl bromide questions
« Reply #2 on: September 26, 2010, 10:55:21 PM »
"I guess one way to tell if you have an alkyl bromide is to put it into a solution of potassium iodide in acetone. If potassium bromide precipitates, you did something right;"

Good idea,Thanks Atara!

Sedit

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Re: Insitu methyl bromide questions
« Reply #3 on: September 26, 2010, 11:51:18 PM »
If your fear is working with MeBr which I fully understand you can always generate MeBr Via traditional methods such as (H+) + NaBr + (-OH)  to Brominate the alcohol and feed the off gas (MeBr is a volitile substance) in to MeOH which it is very stable and soluble in. Doing this you can weigh your MeOH before and after to know the concentration you are dealing with.
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Re: Insitu methyl bromide questions
« Reply #4 on: September 27, 2010, 08:09:05 AM »
Under certain conditions,and after determining the concentration of Methyl bromide resulting from X+Y reactants,shouldnt using an excess of alcohol to begin with hold  a lot  of the formed MeBr anyway?
Unfortunately  OTC NaBr isnt readily available around here, but Im sure I could use the set-up you suggest with the chemicals  I have  as a means to determine how much is made from X+y.
« Last Edit: September 27, 2010, 08:16:30 AM by Goldmember »

atara

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Re: Insitu methyl bromide questions
« Reply #5 on: September 27, 2010, 07:38:08 PM »
I've heard that peroxide is capable of oxidising bromide to hypobromite, which converts to bromine upon acidification. Might be able to recover some of that HBr.

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Re: Insitu methyl bromide questions
« Reply #6 on: September 28, 2010, 01:56:15 AM »
True that excess MeOH would hold the formed MeBr but in doing so you would deny yourself the opertunity to measure the concentration of the MeOH/MeBr mixture. Also there would be a purity issue if all is formed insitu. I think all chemist could agree that if a reagent can be synthesized easy and not insitu it is always the best way to go unless purity is a non issue.
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zzhuchila_clocker

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Re: Insitu methyl bromide questions
« Reply #7 on: September 28, 2010, 06:50:48 AM »
In alcohol solution there is not sufficient water to form HOHal, so only MeOHal could form in equilibrium reaction with BCDMH(or the latter is not alcoholized at all). Alkyl hypochlorite, can not oxidize anything directly, due to high enerdy required for C-O bond cleavage,  as well as BCDMH which is not having active oxygen. So obviously the main process at first step should be halogenation of sulfur
The equation should be then:
SO3H- + ClBr(DMH) => ClSO3H + [Br(DMH)]- (lets say it is chlorine leaving - it is more powerful oxidizer and the resulting bromohydanthoin anion is more stable, then the one than could be formed if bromine would leave first, but it doesnot make big difference anyway i suppose)
Chlorosulfuric acid and hydanthoin anion in the presence of methanol undergo one or several hydrogen transfers, and finally an alcoholysis to yield MeOSO3Na, HCl and bromohydanthoin.
HCl in low concentrations is not capable for formation of any alkylhalide, so it would remain as HCl, acidifying the solution
The next thing that can happen is that it would react with another molecule of BCDMH to give Cl2, or with NaHSO3 to give SO2, H2O and NaCl, depending on what is more soluble, and the concentration of HCl (chlorine formation requires both Cl- and H+, while SO2 formation requires only H+). Cl2 would react with either NaHSO3 to yield HCl, MeOSO3Na and 2HCl (in the presence of methanol), and reaction with SO2 to give Me2SO4 would much less possible since both gases are at low concentrations and both can form equilibrium mixtures with mehtanolysis products, so the main products should be MeOSO3Na and HCl.
Same would happen with bromohydanthoin.
The more acid there is in reaction, the more fast it should be, but the products would remain same, unless concentration of acid is high enough to produce MeHal. Resulting water would easily hydrolyse MeOSO3Na to give NaHSO4 and MeOH
The Hal would be more likely bromine, as it is a more strong nucleophile.
So as a result you should get: MeBr along with some MeCl, NaHSO4 and dimethylhydanthoin.
It is good that there is no base, otherwise hydanthoin could be easily methylated.

Making in situ MeOH solution is also possible, but the reaction should not be too vigourous to make all volatile substances leave, and besides there would be much HCl remaining, and you should quench it with something that would produce not much water, otherwise this solution would be only for immediate use. Also, as we can see there is more and more acid forming while the reaction ana obviously it would react with BCDMH and NaHSO3 at different rates, at least in the begginnig and in the end of process, and one of the gases (Cl2 or SO2) would leave the reaction vessel unreacted(if reaction is vigourous). So you need to find the conditions to keep both reactants reacting with acid at the same rate, probably that can be achieved by dropwise addition of both (of a mixture of both, if they do not react in solid state) substances into methanol, with chilling
Anyway, not a stupid idea at the first sight

ps: everything above is a theory, i'm not sure that it is right
 
« Last Edit: September 28, 2010, 07:03:07 AM by zz-zhuchila »
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POSEIDON

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Re: Insitu methyl bromide questions
« Reply #8 on: October 01, 2010, 12:42:33 AM »
Look this pdf
The chemists are a strange class of mortals, impelled by an almost insane impulse to seek their pleasures amid smoke and vapour, soot and flame, poisons and poverty; yet among all these evils I seem to live so sweetly that may I die if I were to change places with the Persian king.
— Johann Joachim

atara

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Re: Insitu methyl bromide questions
« Reply #9 on: October 04, 2010, 05:54:47 AM »
Look this pdf

Uh, do you have it in English, or at least in some form of text so I can try to translate it? I'm not feeling the transcription right now...