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Author Topic: Bromine from 1-Bromo-3-Chloro-5,5-Dimethylhydantoin  (Read 59 times)

no1uno

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Bromine from 1-Bromo-3-Chloro-5,5-Dimethylhydantoin
« on: December 10, 2009, 01:01:03 PM »
1-Bromo-3-Chloro-5,5-dimethylhydantoin, the only feasible route to Bromine in my area, weighs 241.47g/mol., while 5,5-dimethylhydantoin weighs 128.13g/mol., we also know that chlorine weighs ~35.45g/mol and bromine weighs in at 79.90g/mol.

Thus over half the starting BCDMH (241.47g/53.06%, nb there are two additional Hydrogens attached to the ring nitrogens before anyone says anything) consists of the hydantoin itself, Chlorine accounts for 35.45g = 14.68% & Bromine accounts for (79.90g/33.09%).

Of course, that would leave us with a problem, because in order to remove the Bromine, we'll have to use an acid (HCl presents itself - wonder if H3PO4 would work too?, it would certainly be neater). Muriatic Acid (HCl) will liberate both Br2 and Cl2 (0.5mol Br2 and 1.5mol Cl2), so how to separate them? Generating them both is easy enough, so how about running the formed halides into an aqueous solution of sodium sulfite, giving the HCl & HBr respectively (in solution). To prepare the sodium halide salts, simply add a strong solution of calcium hydroxide (Ca(OH)2) to the resultant solution until no more [edit]CALCIUM[How fucking dumb a fuckup was that? Just not looking at what I wrote, sorry] sulfate precipitates. Filter off the solid (Calcium Sulfate), concentrate the NaBr:NaCl solution.

Gaseous Chlorine  will displace Bromide salts and should have NO effect on the equivalent Chloride salts - therefore careful addition of Cl2 until no more bromine is formed would be called for (need an easily discerned endpoint, but someone will figure it out, maybe when the chlorinated non-polar layer (DCM/Chloroform) stops darkening?).

Anyway, using the procedure above, we should end up with bromine solution in DCM/Chloroform (which can be used, amongst other purposes, for the direct bromination of propiophenone (in ~5min according to a JCE paper).

I suspect it would (will have to try it once I manage to get back to paid employment)

Anyone see any traps? Fuckups? ??? Speak up... ::)
« Last Edit: December 11, 2009, 02:20:35 AM by no1uno »
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heisenberg

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Re: Bromine from 1-Bromo-3-Chloro-5,5-Dimethylhydantoin
« Reply #1 on: December 10, 2009, 03:11:30 PM »
I've only ever tried this with the dibromo version. DBDMH gives pretty good yields, so I would imagine that BCDMH would also have pretty good yields.

Distilling with HCl will yield Br2 along with a significant of BrCl contamination. So, it will need to be cleaned.

Woelen over at SM had a pretty good method for cleaning the impure product:

Quote from: woelen
I would not simply leave the bromine in an open bottle at 5 C. You will loose way too much bromine as well. BrCl dissolves in bromine exceptionally well and is not easily removed by non-chemicals means.

There is a good way to get rid of the BrCl, but this step requires a second distillation of all the bromine. The BrCl is destroyed chemically instead of physically removed.

Prepare a solution of NaBr or KBr. In your situation with appr. 200 grams of bromine I would take 25 ml of water, dissolve 2 gram of KBr or NaBr in the water and add this solution to your bromine. Shake well and then let the bromine settle at the bottom. If you do not have NaBr or KBr, then you can sarifice some of your current bromine to make a solution of a bromide (see end of post).

Separate the bromine from the aqueous layer as good as possible. If you have a separating funnel then that is most conveniently used, otherwise pipette away the bromine from under the aqueous layer.
Next, take 10 ml of concentrated sulphuric acid and add the bromine to this. Shake well again. Any water, left over in the bromine, now is bound to the sulphuric acid.

Distill this liquid and what goes over is very pure and dry bromine.

The treatment with KBr/NaBr destroys any BrCl: BrCl + Br(-) --> Br2 + Cl(-)
The Cl(-) goes in the aqueous layer.

The treatment in turn with conc. H2SO4 is done in order to easily separate from the water. Without the acid, you cannot get your Br2 free of water in a single distill. At 50 C or so, also quite some water vapor comes over as well. But if you separate as much as possible of the water using a funnel, then the rest can be tightly bound by the sulphuric acid. Your H2SO4/Br2 mix now can be separated very easily, no water will come over below 150 C, while the Br2 goes over at 60 C or so. You can distill over all bromine and you stop when only sulphuric acid remains left. Do not continue the distillation to also get the bromine, dissolved in the H2SO4. That last little bit must be regarded as lost. It will only be a small amount, I think less than 1 gram if you use 10 ml of concentrated acid.

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In case you have no bromide: Take 10 ml of household ammonia (5% NH3) with NO additives. Mix with 15 ml of water. This makes 25 ml of 2% ammonia, To this, slowly add 1 ml of bromine while swirling (which is excess of bromine, but you mix with your bromine anyway, so that is OK).
This liquid gives a very clean quantitative reaction with bromine:

8NH3 + 3Br2 --> 6NH4Br + N2

You really have no side reactions. Any BrCl in your bromine immediately reacts to chloride and bromine, so your solution will contain ammonium ions and bromide ions and a small amount of chloride ions. This can be usded as your solution of NaBr or KBr.
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