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AJKOER
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Here is an interesting gas phase path good for demonstration purposes as only small quantities of liberated bromine I would expect forming (due to
scaling limitations), but using only solar light, water, SO2, N2O, air/O2 and a bromide.
First, the action of solar light on aqueous N2O (source, see pages 4 to 5 at https://tspace.library.utoronto.ca/bitstream/1807/14631/1/NQ... ).
N2O + H2O + e-(aq) ---> N2 + •OH + OH-
Followed by (see, for example, reactions (7) to (9) at http://onlinelibrary.wiley.com/doi/10.1029/2009JE003425/full ):
SO2 + •OH + M ---> HOSO2 + M (gas phase reaction where M is another molecule)
HOSO2 + O2 ---> HO2 + SO3
SO3 + H2O + 2 Br- (moist solid) ---> 2 HBr + SO4(2-)
•OH + HBr = H2O + •Br
Next more radical reactions I have cited above with sources:
A44 •Br + •Br --> Br2
A41 •Br + Br- --> •Br2-
A58 •Br2- + •HO2 --> Br2 + HO2- (see "Hydroperoxyl radical (HO2•) oxidizes dibromide radical anion (•Br2−) to bromine (Br2) in aqueous
solution: Implications for the formation of Br2 in the marine boundary layer" by Matthew, M., at http://onlinelibrary.wiley.com/doi/10.1029/2003GL018572/full )
A45 •Br + •Br2- --> Br2 + Br-
A07 Br2 + Br- = Br3-
A43 •Br2- + •Br2- --> Br3- + Br-
A59 Br3- + •HO2 --> •Br2- + Br- + O2 + H+
My take is to employ lots of sunlight with an excess of N2O (needs lots of hydroxyl radicals), followed by SO2 and O2 (need acid and sufficient •HO2
to break up the bromine complexes •Br2- and Br3-).
Note, excess water recycles the Br2 to HBr in the presence of SO2 via the reaction:
SO2 + Br2 + 2 H2O = 2 HBr + H2SO4
See http://www.bromine.chem.yamaguchi-u.ac.jp/library/L02_Global... .
[Edited on 11-3-2017 by AJKOER]
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unionised
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A handy method if you happen to have a linear accelerator.
Sunlight has practically no light with a wavelength below 200nm. Unfortunately, for your idea, Nitrous oxide does not absorb light with a wavelength
above 200 nm or so.
So sunlight will have precisely no effect on NO2
Also, any SO2 present will reduce Br2 back to bromide.
Like I said, why do you keep posting this dross?
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AJKOER
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Quote: Originally posted by unionised  | A handy method if you happen to have a linear accelerator.
Sunlight has practically no light with a wavelength below 200nm. Unfortunately, for your idea, Nitrous oxide does not absorb light with a wavelength
above 200 nm or so.
So sunlight will have precisely no effect on NO2
Also, any SO2 present will reduce Br2 back to bromide.
Like I said, why do you keep posting this dross? |
In the current context of the action of light on a mix containing moist bromide, please remainder that Br- (and I-) are a source of solvated electrons
with light. See http://www.jstor.org/stable/97085?seq=1#page_scan_tab_conten... where the author mentions the creation of a photoelectric current upon
illumination.
My preferred path working with N2O is a micro wave pulse. However, here with solar light:
Br- + hv --> .Br + e- (see, for iodine, page 4 at https://pdfs.semanticscholar.org/d696/b35956e38351dd2eae6706... )
N2O (aq) + e- --> N2 + .O- (aq) (see page 10 at https://pdfs.semanticscholar.org/d696/b35956e38351dd2eae6706... )
.O- + H2O = .OH + OH- (far to the right except in highly alkaline conditions)
Net:
Br- + N2O (aq) + hv ---> N2 + .Br + .OH + OH-
Interestingly also is a possible added path to HO2 via:
e-(aq) + H+ --> .H
.H + O2 --> .HO2 (see Table 1 at http://onlinelibrary.wiley.com/doi/10.1002/bbpc.19870911203/... )
e-(aq) + O2 --> .O2- (Table 1 reference)
.O2- + H+ = .HO2 (pKa = 4.88)
There are also other paths as in the gas phase:
N2O + hv --> N2 + O(1D) (see Table 2 at http://onlinelibrary.wiley.com/doi/10.1029/2009JE003425/full )
O(1D) + H2O --> H2O2
H2O2 + hv --> .OH + .OH
Also, N2O has primary sensitivity to photolysis, as observed in the stratosphere, between 195nm and 215nm (see http://www.atmos-chem-phys.net/11/8965/2011/acp-11-8965-2011... ), which is in the UV spectrum (requiring UV lamp), see http://naturalfrequency.com/wiki/solar-radiation. Here is an interesting source paper on N2O itself in the atmosphere, noting sensitivity not only
to UV in the upper atmosphere but, in the lower atmosphere, even infrared radiation, see http://web.gps.caltech.edu/~wennberg/n2o.pdf .
---------------------------------------
On your second point, my comment is that water is a problem, and much more so in the presence of SO2. Note, I started out by stating that I was
presenting a gas phase demonstration.
[Edited on 11-3-2017 by AJKOER]
[Edited on 12-3-2017 by AJKOER]
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PirateDocBrown
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My experience is that to make bromine from bromide, the cleanest and best oxidizer is hydrogen peroxide.
Sulfuric acid is combined with an alkali metal bromide salt dissolved in water. This makes an aqueous solution of HBr with the metal bisulfate.
Stoichiometry is straightforward.
30% H2O2 is added dropwise, with simple distillation to remove the bromine as it forms. Fumes are neutralized in a sulfite, hyposulfite, or
metabisulfite solution.
My yields have always been well in excess of 80%, and usually over 90%.
If bromine for a reactive solution is desired, collecting the element in a suitable solvent (such as methylene chloride) can make handling
considerably easier, by reducing the vapor pressure. Ensure such collection occurs at a sufficiently low temperature.
Drying bromine is done in a separatory funnel, using concentrated sulfuric acid. Especially if you are using a bromine solution, again make sure the
temperature remains low.
Bromine solutions can be stored for the short term in a bottle, inside a plastic bag, in the freezer. I prefer to ampoule the pure element, but it can
be stored frozen also.
All that said, producing the stuff is actually quite simple. It's the safety precautions that take so much time and effort. The fumes are much more
concentrated than those of chlorine, (which is deadly enough, thank you very much) and do not disperse as readily, either, due to their high density.
Make sure you have very strong ventilation, and your skin well protected also. Bromine burns are said to be very painful.
Working in a professional lab, of course, the element can be ordered as needed. But the added time and expense of safely shipping so hazardous a
material makes in-house production of small quantities reasonable.
Of course, for hobbyists, there is little choice.
[Edited on 3/12/17 by PirateDocBrown]
[Edited on 3/12/17 by PirateDocBrown]
[Edited on 3/12/17 by PirateDocBrown]
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unionised
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| Quote: Originally posted by AJKOER |
. Note, I started out by stating that I was presenting a gas phase demonstration.
[Edited on 11-3-2017 by AJKOER]
[Edited on 12-3-2017 by AJKOER] |
Next time try presenting something that might be useful.
Also, the moist solid you mention in that scheme makes it clear that you have no idea what you are talking about.
[Edited on 14-3-17 by unionised]
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AJKOER
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I have been exploring a new method centering on the irradiation of a mixture of Br-/H2O/N2O/O2/HCO3- upon exiting a gas liquid scrubber (producing an
aersole, in essence).
The processing reagents are interesting being available and low cost, but the implementation employing, for example, a UV lamp and a gas liquid
scrubber like device, would require some investment.
I won't bore anyone with the chemistry quite yet, and perhaps not exactly "Making bromine (super cheap and easy, no distillation, quantitative
yield)".
[Edited on 15-3-2017 by AJKOER]
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Dan Vizine
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Really, it's one of the easiest reactions to perform that I can think of. The heat of solution and heat of reaction cause the Br2 to be distilled out
as the reaction proceeds. The only effort needed after this is washing and drying, and all of the preps. should include this anyway. If your goal is
just to make good Br2 for subsequent reactions, this is a good way to go. If your interests lie with the thought of going your own way, just ignore
this.
[Edited on 3/20/2017 by Dan Vizine]
"All Your Children Are Poor Unfortunate Victims of Lies You Believe, a Plague Upon Your Ignorance that Keeps the Youth from the Truth They
Deserve"...F. Zappa
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theAngryLittleBunny
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| Quote: Originally posted by Dan Vizine |
Really, it's one of the easiest reactions to perform that I can think of. The heat of solution and heat of reaction cause the Br2 to be distilled out
as the reaction proceeds. The only effort needed after this is washing and drying, and all of the preps. should include this anyway. If your goal is
just to make good Br2 for subsequent reactions, this is a good way to go. If your interests lie with the thought of going your own way, just ignore
this.
[Edited on 3/20/2017 by Dan Vizine] |
Hmm, you have a point, since made bromine yesterday through oxidizing NaBr with H2SO4 and KClO3, and I got a 92% yield. I used it to make acetyl
bromide, and the annoying part actually wasn't making the bromine, it was reacting it with phosphorus afterwards, which was the most time consuming
part.
Anyway, nothing wrong with trying something new, I just git a little bit too hyped about it .-.
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clearly_not_atara
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CBDMH + sodium metabisulfite is frankly genius :p the issue is controlling the quantity of bisulfite so you don't end up with HBr or Cl2. If you can
find DBDMH that's much better ofc
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theAngryLittleBunny
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| Quote: Originally posted by clearly_not_atara | | CBDMH + sodium metabisulfite is frankly genius :p the issue is controlling the quantity of bisulfite so you don't end up with HBr or Cl2. If you can
find DBDMH that's much better ofc |
Well, I see how that works out, but I can't imagine you'll get a high yield from that, if you add too much metabisulfite, you will get chloride in the
solution which would from the chlorodibromde ion with the bromine to dissolve it. Wouldn't it also be possible to just ad an equal molar amount of
HBr, so that you'll get the double amount of bromine? Because you get bromine from the reduction of the CBDMH, it seems like the Br+ is a stronger
oxidizer then the Cl+, so it should also react with the HBr first. But this wouldn't be a viable route for me anyway, since I can't get CBDMH .-.
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clearly_not_atara
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The person who reported it initially claimed large yields. Essentially bisulfite won't react with the N-chloro at room temperature so you want to make
sure it's all used up before you start distillation (use a slight excess of CBDMH) and you'll get practically no chloride in solution. It happens
because the N-Br bond is weaker than the N-Cl bond, so the reaction has a lower activation energy.
If you're not consuming the bromine immediately it might be better to store it as pyridinium tribromide or the tribromide salt derived from methyl
nicotinate. The tribromide is an odorless solid and much easier to store. Additional bromide must be added or some bromine must be reduced for this to
work of course.
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AJKOER
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A similar idea I have was to first convert say NaBr into CuBr2 (by, for example, treating the NaBr with CuSO4 and freezing out the Na2SO4).
Then, add metal copper (including nano Cu) to a warm solution of CuBr2 (forming some cuprous) to ones normally acidified H2O2 (with H3PO4 or perhaps
CO2, or employ Na2CO3/H2O2 or NaHCO3/H2O2) all in the presence of a small amount of catalytic humic rich burned wood soot (see http://onlinelibrary.wiley.com/doi/10.1029/2008GL035285/pdf ).
The fenton-like reaction (likely extended into a wider pH range by the presence of humic) may form the hydroxyl (see, for example, http://www.atmos-chem-phys.net/16/1761/2016/acp-16-1761-2016... ) or the carbonate radical to liberate .Br together with .HO2 (by action of .OH on
H2O2). Aiding in liberating Br2 would be using an acidified H2O2 and a separate Cu(nano)/CuBr2/Humic reaction mixture combined together as aerosols
(see, for example, discussion at http://m.pnas.org/content/111/2/623.full), thereby effectively changing the medium, the dielectric constant and lowering the pKa of the mixture
(see comments at https://en.wikipedia.org/wiki/Hydroperoxyl ). Optionally, a Photo-fenton system would likely give a better yield.
Some of the supporting pertinent research, not sourced above, for this proposed idea, I may be able to supply for anyone interested.
[Edited on 1-4-2017 by AJKOER]
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Boffis
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I have used several methods to prepare bromine over the years but my favoured method is the bromide-bromate-sulphuric acid route. All of the reagents
are available from ebay and it is very simple to separate the moist bromine in a separatory funnel. I don't even bother to dry it if I am going to use
it in an aqueous medium otherwise I treat it with some conc sulphuric acid and separate again. The use of bromate overcomes the uncertainty of the
presence of chlorine and potassium bromate is now easier to get than the chlorate too. I use a slight excess of sulphuric acid over this equation to
ensure acid conditions:
5NaBr + KBrO<sub>3</sub> + 6H<sub>2</sub>SO<sub>4</sub> → 3Br<sub>2</sub> +
5NaHSO<sub>4</sub> + KHSO<sub>4</sub> + 3H<sub>2</sub>O
Recently this raised a thought; do you need so much acid? Could you use less acid and obtain the normal sulphate in the residue and if so can you use
sodium hydrogen sulphate as the acid? ie:
5NaBr + KBrO<sub>3</sub> + 7NaHSO<sub>4</sub> → 3Br<sub>2</sub> +
6Na<sub>2</sub>SO<sub>4</sub> + KHSO<sub>4</sub> + 3H<sub>2</sub>O
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Boffis
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| Quote: Originally posted by theAngryLittleBunny | | Quote: Originally posted by clearly_not_atara | | CBDMH + sodium metabisulfite is frankly genius :p the issue is controlling the quantity of bisulfite so you don't end up with HBr or Cl2. If you can
find DBDMH that's much better ofc |
Well, I see how that works out, but I can't imagine you'll get a high yield from that, if you add too much metabisulfite, you will get chloride in the
solution which would from the chlorodibromde ion with the bromine to dissolve it. Wouldn't it also be possible to just ad an equal molar amount of
HBr, so that you'll get the double amount of bromine? Because you get bromine from the reduction of the CBDMH, it seems like the Br+ is a stronger
oxidizer then the Cl+, so it should also react with the HBr first. But this wouldn't be a viable route for me anyway, since I can't get CBDMH .-.
|
This is an interesting bit of chemistry. I can see how the reaction works but the main problem that I see is how to separate the sparingly soluble
dimethylhydantoin from the two phase liquid. You could use a glass frit but mmm.. vacuum filtration of volatile liquid phase, sounds nice.
@Clearly_not_atara where did this method come from? or is the idea homebrewed too.
@Angrylittlebunny, I don't see any advantage in using DBDMH since you will loose half your bromine to HBr assuming the reaction runs something like:
4C<sub>5</sub>H<sub>6</sub>N<sub>2</sub>O<sub>2</sub>ClBr +
3Na<sub>2</sub>S<sub>2</sub>O<sub>5</sub> + 9H<sub>2</sub>O →
4C<sub>5</sub>H<sub>8</sub>N<sub>2</sub>O<sub>2</sub> + 6NaHSO<sub>4</sub> + 4HCl + 4Br
Or is there a different reaction?
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Elemental Phosphorus
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Clearly not atara's idea is actually detailed in Len1's book:
https://books.google.com/books?id=VqosZeMjNjEC&lpg=PP1&a...
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woelen
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| Quote: Originally posted by Boffis | Recently this raised a thought; do you need so much acid? Could you use less acid and obtain the normal sulphate in the residue and if so can you use
sodium hydrogen sulphate as the acid? ie:
5NaBr + KBrO<sub>3</sub> + 7NaHSO<sub>4</sub> → 3Br<sub>2</sub> +
6Na<sub>2</sub>SO<sub>4</sub> + KHSO<sub>4</sub> + 3H<sub>2</sub>O |
Yes, using NaHSO4 as acid works very well. I have written a web page about making bromine, using OTC chemicals only. But with KBrO3 easily available,
things become really easy. Use NaBr, KBrO3, and NaHSO4 to make bromine.
The web page I wrote describes a process using NaBr and electrolysis before adding NaHSO4:
http://woelen.homescience.net/science/chem/exps/OTC_bromine/...
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theAngryLittleBunny
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| Quote: Originally posted by Boffis | I have used several methods to prepare bromine over the years but my favoured method is the bromide-bromate-sulphuric acid route. All of the reagents
are available from ebay and it is very simple to separate the moist bromine in a separatory funnel. I don't even bother to dry it if I am going to use
it in an aqueous medium otherwise I treat it with some conc sulphuric acid and separate again. The use of bromate overcomes the uncertainty of the
presence of chlorine and potassium bromate is now easier to get than the chlorate too. I use a slight excess of sulphuric acid over this equation to
ensure acid conditions:
5NaBr + KBrO<sub>3</sub> + 6H<sub>2</sub>SO<sub>4</sub> → 3Br<sub>2</sub> +
5NaHSO<sub>4</sub> + KHSO<sub>4</sub> + 3H<sub>2</sub>O
Recently this raised a thought; do you need so much acid? Could you use less acid and obtain the normal sulphate in the residue and if so can you use
sodium hydrogen sulphate as the acid? ie:
5NaBr + KBrO<sub>3</sub> + 7NaHSO<sub>4</sub> → 3Br<sub>2</sub> +
6Na<sub>2</sub>SO<sub>4</sub> + KHSO<sub>4</sub> + 3H<sub>2</sub>O |
Bromate salts being easely avaliable would be a dream come true for me *-*, I have several kilos of KClO3, but KBrO3 is like ten times as expensive
.-. I once just bought a little KBrO3 because I read that it's a much more aggressive oxidizer then KClO3 and got excited about it. And holy shit, it
is way more aggressive! O.o. I also did a lot of experiments with iodates, but they're kinda lame honestly. So making bromine with bromate kinda makes
my heart hurt, because KBrO3 is just a really lovely substance on it's own, and honestly way more exciting then bromine.
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clearly_not_atara
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If barium is available, barium bromate is poorly soluble in water and may be precipitated from alkaline mixtures containing Br in oxidized states.
Calcium bromate unfortunately lacks this useful property.
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theAngryLittleBunny
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| Quote: Originally posted by clearly_not_atara | | If barium is available, barium bromate is poorly soluble in water and may be precipitated from alkaline mixtures containing Br in oxidized states.
Calcium bromate unfortunately lacks this useful property. |
I actually did something like that once, I boiled some NaBr dissolved in 100mL of a 14% NaOCl solution, and put some BaCl2 into it. I think I got 7g
of a Ba(BrO3)2 precipitate, which is an about 45% yield based on the NaOCl. I also tried another way to make KBrO3, I just made a super concentrated
KOH solution with KBr dissolved in it, the molar ration of KOH to KBr was 6:1. And then I bubbled chlorine gas into this solution. The chlorine should
oxidize the bromide from the KBr, and then the Br2 should react with the KOH to form KBr and KOBr, and then 3 KOBr make a disproportionation to make 2
KBr and 1 KBrO3. The end result should have been a solution of KCl, with a KBrO3 precipitate. It.....kinda worked, I think I got an incredible 15%
yield or so.
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