Sciencemadness Discussion Board - Bromine from KBr, MnO2 and hydrochloric acid

Bromine from KBr, MnO2 and hydrochloric acid

metalresearcher - 18-11-2012 at 12:37

My second attempt for making elemental Br2 from these chemicals was rather succesful.
However, according to the stoichiometric ratios

2 KBr + 4 HCl + MnO2 => 2KCl + MnCl2 + 2H2O + Br2

240 146 86 => 160

14.4g 8.8g 5.2g => 9.6g == 3ml Br2

8.8g equals to 24ml 36% conc. HCl acid.

but after some time the Br2 generation stopped, while the liquid in the round flask was still dark brownish red.

The setup for distilling with a mushroom distiller.

The bromine collecting in the water cooled erlenmeyer flask.

Endimion17 - 18-11-2012 at 12:49

Unless you work in a deep freezer or Antarctica, your setup won't give great yields. You simply can't use such primitive cooling methods for bromine. Liebig or vertical spiral cooler with ice water are essential.
At least simple distillation kits aren't that expensive anymore.

blogfast25 - 18-11-2012 at 13:44

It would help a LOT to just put an icepack on your mushroom condenser and its outlet tube: it's clear your bromine vapours (BP = 59 C) have trouble condensing...

And I was fearing to see your camera lens catch fire or possibly melt any time!

[Edited on 18-11-2012 by blogfast25]

mr.crow - 18-11-2012 at 22:00

Haha I love the mushroom condenser!

KMnO4 can also oxidize HCl to Cl2, which works but can form impurities if in excess. You also need good stirring.

Your HCl should really be in a glass bottle, not plastic. It leaks out and can rust everything in your lab.

AJKOER - 19-11-2012 at 09:13

I think this synthesis is a bit too dangerous as you are, in fact, generating Chlorine (via MnO2 + HCl) and inviting the possible formation of Bromine monchloride (see http://www.sciencemadness.org/talk/viewthread.php?tid=1091 ). The creation of any BrCl not only reduces yield, but requires the use of a fume hood if you want to keep your lungs functional.

My suggestion: treat aqueous KBr with an acid (forming HBr) and then react with some chlorate:

KClO3 + 6 HBr --> 3 Br2 + 3 H2O + KCl

For this purpose, a Chlorate/chloride solution can easily be formed by warming a solution of an unstable hypochlorite (includes Magnesium or Copper hypochlorite), or the more stable variety (like KClO or NaClO or Ca(ClO)2).

[EDIT] Note working via an acidified hypochlorite reaction is essentially a chlorine path akin to your original synthesis.

[Edited on 19-11-2012 by AJKOER]

blogfast25 - 19-11-2012 at 10:10

Quote: Originally posted by mr.crow

Your HCl should really be in a glass bottle, not plastic. It leaks out and can rust everything in your lab.

HDPE or PP are perfectly resistant to most mineral acids, even 70 % HNO3 and 95 % H2SO4. Conc. HCl is almost always sold in hydrocarbon polymer bottles.

blogfast25 - 19-11-2012 at 10:13

Quote: Originally posted by AJKOER

My suggestion: treat aqueous KBr with an acid (forming HBr) and then react with some chlorate:

KClO3 + 6 HBr --> 3 Br2 + 3 H2O + KCl

For this purpose, a Chlorate/chloride solution can easily be formed by warming a solution of an unstable hypochlorite (includes Magnesium or Copper hypochlorite), or the more stable variety (like KClO or NaClO or Ca(ClO)2).

[EDIT] Note working via an acidified hypochlorite reaction is essentially a chlorine path akin to your original synthesis.

[Edited on 19-11-2012 by AJKOER]

KClO3 + acid is a recipe for (potentially explosive) ClO2, unless you know EXACTLY what you're doing. And chlorates are fairly expensive anyway.

See for instance:

http://www.sciencemadness.org/talk/viewthread.php?tid=17940

Chilled H2O2 is probably safest, with reported yields of over 90 % and no interhalogens possible.

Perhaps most importantly, the choice of acid (HCl) was unfortunate but as an oxidiser MnO2 would work well with H2SO4 as an acid, then no Cl2 (or BrCl) can form. I assume the poster didn’t have any sulphuric acid or didn’t consider it.

[Edited on 19-11-2012 by blogfast25]

elementcollector1 - 19-11-2012 at 12:07

Wouldn't the chlorine produced react with the bromide first, and then when all bromide was converted to free bromine start making BrCl? I'm planning a similar setup for my eventual bromine distillation, and I'm not using sulfuric acid. (I'm certainly using sulfuric acid to dry the product, however).

AJKOER - 19-11-2012 at 14:57

Quote: Originally posted by blogfast25

Quote: Originally posted by AJKOER

Actually, while I agree with Blogfast25 sentiments, if one mistakenly somehow prepares concentrated HClO3, it is probably a greater risk. The reason I would down play the ClO2 in my current suggested synthesis in the presence of sufficient HBr is based on the reaction quoted in Mellor for Iodide, page 289, http://books.google.com/books?id=AnnVAAAAMAAJ&pg=PA289&a... to quote:

"Iodine separates from an acidified soln. of potassium iodide: 2Cl02+10HI=2HCl+4H20+5I2; in neutral soln.: 6Cl02+10KI=4KI03 +6KCl+3I2; and in the bicarbonate soln.: 2Cl02+2KI=2KCl02+I2, whereby 80 per cent, of the chlorine dioxide is converted into the chlorite."

So for an acidified KBr solution, I would similarly expect:

2 Cl02 +10 HBr --> 2 HCl + 4 H20 + 5 Br2

so any Chlorine dioxide created oxidizes the HBr (or, the HBr presence would reduce any ClO2 formed), so it is important to have sufficient (or an excess) of HBr present to address any ClO2 formation. Also, any Chlorate formed by my suggested disproportionation of a dilute hypochlorite is not likely to result in an unstable concentration of Chloric acid.
----------------------------------------------------------------

Also, one does not need H2O2 if one has access to concentrated H2SO4 as per one source (see Equation #7 at http://wenku.baidu.com/view/bf135124a5e9856a5612602a.html ):

2 NaBr + 2 H2SO4 (Conc) --> Na2SO4 + Br2 + SO2 (g) + 2 H2O

[Edited on 20-11-2012 by AJKOER]

woelen - 19-11-2012 at 23:48

The latter is a very bad way of making Br2. Only a small part of the bromide reacts to bromine, the main part reacts to HBr. Besides this, you get Br2, contaminated with SO2. On contact with water this mix IMMEDIATELY reverts back to HBr and H2SO4.
Mixing H2SO4 with solid bromides is a bad way for making Br2 and it is a bad way for making HBr. This reaction simply is not useful at all for synthetic purposes.

Any reaction in which chlorine is present as well (either as element, or as oxidizer, or as chloride) is suboptimal. Indeed, as AJKOER writes, you can assure that only Br2 is formed if there is enough bromide in the mix. In order to be sure, you need excess bromide. The bad thing, however, of having excess bromide in the mix, is that you loose bromine as bromide, and even worse, the bromide also fairly strongly keeps bromine in solution as Br3(-) ions. This makes the method of simply pipetting away bromine from under the aqueous layer less attractive, because a lot of bromine remains in solution.
If there is insufficient bromide, then you get BrCl and you loose bromine as well, through formation of bromate ions (hydrolysis of BrCl yields Br2, bromate and HCl). If you have a large excess amount of oxidizer, then you'll see that all bromine disappears again. It all will go in solution as HBrO3 and you will not be able to separate this from the liquid.

Best is to use an oxidizer like H2O2 and dilute H2SO4 or dilute HNO3 as acid. MnO2 or KMnO4 also works. I myself make bromine by means of electrolysis of a solution of a bromide, such that 1/6 of the bromide is converted to bromate and then I acidify this solution with dilute H2SO4.

AJKOER - 20-11-2012 at 08:10

OK, a route I would be interesting in testing out is as follows.

Step 1. Prepare HBr from heating NaBr and aqueous NaHSO4 and condensing the vapor.

Step 2. Prepare HOCl. Start by adding a weak acid (Boric, Acetic, Critic or very dilute H2SO4, HNO3,..) to Bleach and distilling half the volume to form HOCl for immediate use. Adding a small amount of alkali metal silicate may increase the stability of the HOCl (see Patent EP 0 892 042 A1 and comment "Silicate is said to prevent decomposition of hypochlorite and corrosion by hypochlorite" at https://data.epo.org/publication-server/html-document?PN=EP0... .

Step 3. Combine HBr and HOCl in an hourglass like container set on ice. The volume of the lower part of the hourglass should correspond closely to the expected yield of Br2. Target reaction:

2 HBr + HOCl --> Br2 + H2O + HCl

Side reactions:

Br2 + H2O <---> H[+] + Br[-] + HOBr

and the presence of H[+] from the HCl should push the equilibrium to the left. Also important for limiting:

3 HOBr --> 2 HBr + HBrO3

Now, hopefully the hourglass vessel shape may limit the surface area contact of any separated Bromine and solution which is important for the undesirable reaction:

Br2 + Br- --> Br3-

Step 4. Now apply the method of pipetting away the bromine from under the aqueous layer.

Now, in comparison to the HBr plus H2O2 method:

2 HBr + H2O2 --> Br2 + 2 H2O

my suggested synthesis produces less water at a better (meaning more acidic) pH to hopefully relatively reduce the formation of HBrO3.
------------------------------------------------------------
[EDIT] After coming across a good source that comments that the reaction between an Iodide and HOCl does form an intermediary of ICl (see "Kinetics of hydrolysis of iodine monochloride measured by the pulsed-accelerated-flow method" by Yi Lai Wang , Julius C. Nagy , Dale W. Margerum, published in J. Am. Chem. Soc., 1989, 111 (20), pp 7838–7844 at http://pubs.acs.org/doi/abs/10.1021/ja00202a026 ), and that the final product in the hydrolysis of ICl in an acid is given by:

5 ICl (aq) + 3 H2O <--> HIO3 + 2 I2 + 5 HCl

Base:
3 ICl(aq) + 6 NaOH <--> NaIO3 + 2 NaI + 3 NaCl + 3 H2O

I would now suspect similarly that the action of HOCl on a Bromide would also produce BrCl and with further hydrolysis a Bromate (see http://cat.inist.fr/?aModele=afficheN&cpsidt=1007184 ). As such, I would not be surprised if this suggested HOCl route does not result in a significantly better yield over the H2O2 approach.

[Edited on 21-11-2012 by AJKOER]

SM2 - 20-11-2012 at 08:40

My suggestion, go to an Ace Hardware store or a small, local hardware store, where everything costs a little more then Home Depot. Get Rooto brand H2S04, and proceed as if you are trying to make hydrobromic acid too quickly. Post reaction chamber, DO HAVE a west condenser circulating a freezing cold, liquid brine solution, stored in a simple polystyrene disposable cooler. Also, insulate condenser, especially at top. Have a short path, yet enough head space so that no solid NaBr or dilute aq HBr comes over. The trick to making bromine it to know what your doing, AND to do it quickly. It doesn't smell deadly, like Cl2, but is has a strange fecal nose, along with the chlorine irritation. Wear double nitrile gloves, I told you so! If you have an organic (activated charcoal) mask, use it, or buy one from the big hardware stores or big auto body paint shops. This is a practical way to bromine for the home chemist.

woelen - 20-11-2012 at 13:02

I strongly advice against both two methods, presented in the last two posts. Maybe of academical interest, but not of any preparative value!

Distilling HOCl, just for making Br2? Why do it the difficult way if it can also be done easily

?

As I wrote before, mixing H2SO4 and a bromide is not useful for preparative purposes. Too many impurities, too many losses, a pain in the ass.

There are quite a few threads on making Br2, please have a look at those, there are quite a few practical ways of making Br2 and to my opinion the subject of making and isolating Br2 is resolved long ago already.

blogfast25 - 20-11-2012 at 14:04

Woelen:

Amen to that.

elementcollector1 - 30-11-2012 at 23:14

I'm trying to decide between a few things:
I don't have the heat setup for distillation of bromine, so could I;
-Pre-produce the bromine in solution (NaOCl, HCl and NaBr) and suck out the water with concentrated (Rooto, 96%) sulfuric acid?
-If I do get the heat setup, should I use the bleach + HCl method to avoid the potential hazards of a chlorine generator (instead forming in-situ chlorine)?

I might borrow a hot plate from a teacher fairly soon, but I wanted to know if there was a bit of a faster way.

woelen - 1-12-2012 at 03:00

Bleach + HCl + bromide makes bromine, but the bromine will be very dilute. Isolating the bromine from all the water and acid will be hard without a proper distillation setup. Adding conc. H2SO4 does not help, because bromine also dissolves to some extent in H2SO4.

If you don't have a distillation setup, then you need a highly concentrated oxidizer. Best is calcium hypochlorite, added in nearly stoichiometric amount to NaBr or KBr. Add the bromide to some conc. HCl and then add the hypochlorite, in small steps and swirl each time. When all is added, quite some liquid bromine will settle at the bottom and you can pipette away the bromine. In this way, yields of appr. 70% are feasible. It is important to have the reagents in stoichiometric ratios, best to have slight excess amount of bromine, otherwise you get impure Br2, which contains BrCl.

elementcollector1 - 1-12-2012 at 17:57

Never mind, it turns out my parents had a hotplate of sorts lying around! It's got a coil for a surface, and only a rudimentary temperature control, but it should work perfectly. Now all I need is a 10/18 glass stopper for my distillation adapter and I'm all set!

metalresearcher - 12-12-2012 at 12:47

Here a new attempt with another distillation bridge and the Br2 received in an ice cold round bottom flask.

<iframe sandbox width="640" height="360" src="http://www.youtube.com/embed/VuMsnKx2ZuE" frameborder="0" allowfullscreen></iframe>

Endimion17 - 12-12-2012 at 13:52

It seems your fume hood is so powerful you don't have to fear any poisoning so the paper plug at the end is not necessary. The fan sound like a jet engine.

To increase yields, use a salt-water-ice bath for the receiver.
If you don't have running water for the cooler, improvise using an elevated bucket filled with water and chunks of ice. Syphon the water into a lower bucket and occasionally transfer it back up. That's what I used before I started using recirculating pumps. It's efficient, but you have to monitor the water level.

Also, avoid using open flame when working with glassware. Yes, it's a small flask and it's round bottomed, and the flame is not "oxidizing blue", but nevertheless, it's a good practice to use a heating bath for a steady supply of evenly distributed heat. Boiling water is probably enough. If you don't want a steady supply, wrap it with two layers of iron net.

sbbspartan - 12-12-2012 at 15:36

Woelen, you mentioned that you make bromine via electrolysis of a solution of potassium bromide. How exactly does this work? Does the potassium bromate react with the sulfuric acid to make the bromine, which then sinks to the bottom and you harvest? On your website, it looked like there was some bromine produced even before you added acid. Did this then dissipear as the bromate was produced? Would there be some excess bromate then so the bromine does not dissolve in the sulfuric acid and you lose your product? How is the yield? Is there any contamination from the potassium dichromate used in the electrolysis, or is it pretty pure? This seems like a pretty easy and safe way to make bromine. Pretty cheap too.

woelen - 13-12-2012 at 00:01

The electrolysis process indeed first makes bromine, which however is not easily separated. I just keep it in the liquid and allow it to react with the hydroxide, formed at the cathode. This process gives bromate ions:

3Br2 + 6OH(-) --> BrO3(-) + 5Br(-) + 3H2O (through intermediate step of BrO(-))

Electrolysis is continued, such that 1/6 of all bromide ions is converted to bromate ion and then acid is added. In the page to which I posted a link it is described how long you need to perform electrolysis. Not too short and not too long, in both cases you loose bromine, but too short is much worse than too long.

The dichromate does not cause any contamination of the bromine.

elementcollector1 - 13-12-2012 at 20:53

How does this method compare to bubbling chlorine in, in terms of percent yield?

woelen - 14-12-2012 at 04:11

The yield depends on how accurately you are able to perform just enough electrolysis. The process, however, is very forgiving when the solution is somewhat electrolysed too much. Even electrolysis of 50% too much leads only to a loss of 10%. The process, however, is extremely sensitive to electrolysing too little. Just a few percent too little of electrolysis leads to huge losses. So, be sure to have sufficiently long electrolysis times. Better 50% too long than 5% too short!

Lambda-Eyde - 14-12-2012 at 04:50

Quote: Originally posted by metalresearcher

Here a new attempt with another distillation bridge and the Br2 received in an ice cold round bottom flask.

You don't have running water in your condenser? What was your yield?

elementcollector1 - 14-12-2012 at 11:52

Checked your website, and I'm liking the setup so far. It doesn't matter if I use sulfuric or even hydrochloric acid, right? I don't have access to bisulfate.
I'm off to do the stoichiometry equations for the amount of NaBr I have.

[Edited on 14-12-2012 by elementcollector1]

[Edited on 14-12-2012 by elementcollector1]

woelen - 15-12-2012 at 06:56

Using HCl is not the best option. As I wrote, it is best to overrun the electrolysis in order to keep losses to a minimum, but overruning the electrolysis also leads to an excess amount of oxidizer. With sulphuric acid or sodium bisulfate this is no problem. The excess oxidizer simply is not used in that case. With hydrochloric acid, however, the excess bromate is capable of oxidizing chloride to BrCl and this compound mixes extremely well with Br2. Actually, expect almost all BrCl to end up, dissolved in the Br2. So, you'll end up with Br2 which is contaminated with BrCl, or you'll have to accept large losses by underrunning the electrolysis (this assures that no Cl2 or BrCl will remain present in the system, because if any were formed, then the excess bromide will react with this to chloride ions and Br2).

elementcollector1 - 15-12-2012 at 10:44

Actually, I ended up using sulfuric acid. This left the solution a deep red color. However, I tried distilling this (my hotplate sucks, and it's freezing outside here!), and only got a weaker solution of bromine. What am I doing wrong?

EDIT: Added some more sulfuric acid, as well as some H2O2. Solution has turned opaque orange (what?) with a red, almost precipitate-like substance on the bottom of the flask. Will try re-distilling tomorrow, with sulfuric acid in the receiver as well as the boiling flask.

[Edited on 16-12-2012 by elementcollector1]

crazyboy - 16-12-2012 at 13:19

http://www.youtube.com/watch?v=OB4MmPTOBxg&feature=related

This uses Cl2 to displace Br in KBr to produce bromine. As previously mentioned however this would contaminate the Br2 with BrCl. Is there a good way of removing the BrCl or is this method simply not viable?

elementcollector1 - 16-12-2012 at 13:21

Well, I got some bromine! It's ampouled, and the ampoule is inside a test tube. The stuff's very black, but does have a red tint (and unfortunately, does wet glass).
I believe plante posted something about keeping the bromine at 10C, and letting the BrCl evaporate off? Read up a few posts.

Endimion17 - 16-12-2012 at 14:45

Quote: Originally posted by crazyboy

http://www.youtube.com/watch?v=OB4MmPTOBxg&feature=related

This uses Cl2 to displace Br in KBr to produce bromine. As previously mentioned however this would contaminate the Br2 with BrCl. Is there a good way of removing the BrCl or is this method simply not viable?

It's one of the worst methods available. I suppose one could get rid of most bromine chloride by refluxing the raw product with finely divided potassium bromide during one day. I'd also employ vigorous stirring. It is possible to purify it just by leaving the bromide inside bromine, but I think that would require at least a week, maybe a lot more, like a month. Savage stirring and reflux could solve the problem in full 24 hours.
BrCl<sub>(l)</sub> + KBr<sub>(s)</sub> -> Br<sub>2(l)</sub> + KCl<sub>(s)</sub>
(actually BrCl is a gas, but it's inside the mixture; I have no idea in what state, solvated or just there, like most of H<sub>2</sub>S inside water is)
You can imagine it's a slow reaction - there are almost no free ions except some small quantities due to moisture involved. Those might even serve as a catalyst, I don't know.

It's much easier simply not to use elemental chlorine or chlorine containing oxidizers for oxidation so all you have to do next is extraction of water and distillation.
Even if you do that, there will always be some bromine chloride inside because the starting material (analytical purity alkali bromide) always contains some chlorine in the form of chloride. If there's 0.5% of chloride in your alkali bromide, you'll get around 0.5% BrCl in your bromine, too.
Also don't forget iodine compounds. Chlorine, bromine and iodine always follow each other, just like there's sodium in potassium samples (which is why its true violet flame can't be visible without cobalt glass).

Leaving BrCl to evaporate can't solve the problem.

[Edited on 16-12-2012 by Endimion17]

elementcollector1 - 16-12-2012 at 17:28

Could sulfuric acid suck the BrCl out?

neptunium - 16-12-2012 at 17:43

trying to seperate out each halogen from any element preparation sounds like an isotopic seperation....it aint gonna happen at home unless you invest in a $90,000 mass spectrometer and run it for a few thousand years..
you ll just have to admit that no matter how pure the sample is, it will always have some Fluoride chloride etc...
the Bromine obtained and dry is usually good enough for element collectors and most reaction at home..

elementcollector1 - 16-12-2012 at 17:54

And that's why I went with the sulfuric acid method. No chlorine = no interhalogens.

woelen - 16-12-2012 at 23:27

@Endimion17: I think that it is not true that you will always have BrCl in your bromine. Using normal equipment at home and wanting to put some effort in it you can remove any BrCl and also you can do quickly. Most people (including me), however, do not take the effort to remove the last few tenths of percent of BrCl.

Indeed, normal commercial grade NaBr or KBr certainly will contain a few tenths of percent of chloride as well. Using my method of electrolysing such that 1/6 of the bromide is converted to bromate and using slight overrunning of the electrolysis step in order to get decent yield results in formation of BrCl in appr. the same percentage as you had chloride in your bromide species.

Removal of this BrCl is not really hard though, but it leads to losses. One way of doing this is dissolving some NaBr in water (make the solution highly concentrated) and add 1 ml of this solution to each 10 ml or so of your Br2. Shake the mix well and after a few minutes all BrCl will be converted to Br2.
After that step, however, you have another impurity (water, NaBr and NaCl). You need to distill the bromine first to get rid of the NaBr and NaCl and you then need to dry the bromine with conc. H2SO4 to get rid of the water (the distillation is not enough to leave all water behind) and then you need to pipette away the bromine from under the H2SO4. All these steps lead to additional losses (mostly mechanical) and especially if you work on a small scale of milliliters then the mechanical losses may be considerable.

Endimion17 - 17-12-2012 at 04:25

Few tenths of percent is still "some".

Of course, it's impossible to remove all BrCl from a macroscopic sample, but 0.5% is not a small amount. Consider 100 g of the sample. It would contain 0.5 g of BrCl. That's 0.23 ml of the BrCl gas at STP stuffed into roughly 32 ml of liquid. Now it looks as a nuisance, at least to me. And if you look at it from an organic chemist's view, it's a pain in the ass.

Shaking with alkali bromide (salt or conc. solution - that works, too, but introduces water) could push that below detectable levels, at least detectable using usual analytical methods.

Mechanical losses are the worst, I agree on that. The stuff just evaporates. If I had lots of crude bromine, I'd try to do this purification, but it's just a stupid thing to do when you've got few mililiters. The losses are incredibly high. For display purposes, it's important to get rid of the water so it doesn't wet the glass and looks nice.

bromobagel - 23-12-2012 at 19:53

I tried and documented a similar prep, and thought I'd post it here.

The yield I obtained was disappointing, but the reason for this is obvious and the problem is easily fixable (see Discussion).

(1) 2KBr + MnO2 + 3H2SO4 ---(heat)---> Br2 + 2KHSO4 + MnSO4 + 2H2O

Source : http://143.239.128.67/academic/chem/dolchem/html/elem035.html

Experimental

In a 500 mL 24/40 RBF was loaded 40.0 g (0,336 mol) KBr and this was dissolved in a minimal amount of distilled water (The dissolution is endothermic, at the end the solution temperature was 5 C). 100 mL dilute H2SO4 [30% or 5,625M] (0,5625 mol, a slight excess) is added to the cold KBr solution with swirling of the flask. Finally 16.3 g (0,182 mol, a slight excess) MnO2 and a few boiling chips were added and the flask was fitted for simple distillation. The receiving flask (150 mL RBF) was submerged in an ice bath, and ice was also added regularly to the condenser water tank. The vac. outlet of the distillation apparatus was fitted with a conc. NaOH bubbler to trap bromine vapours. The reaction flask was then gently heated by means of an oil bath. When the bath temperature reached 44-45C bubbles evolved and the air in the flask took an orange tint. When the bath temp. reached 60C the whole apparatus was orange but no bromine had condensed. When the bath temperature reached 75C bromine started coming over at a rate of one drop about every 4 seconds. After 1h I noticed that the pressure fluctuated wildly, and that a large amount (~60 mL) of NaOH solution had been sucked back into the receiver flask. The distillation was continued anyway over the course of 2,5 additional hrs, watching the pressure closely, until no more bubbles formed in solution and the air inside the apparatus was almost colourless. At this point the bath temperature was 107C. The oil bath was removed and the NaOH bubbler disconnected. The bromine was pipetted into a 50mL Erlenmeyer and 20mL of concentrated H2SO4 were added. This was swirled around a few times and the layers allowed to separate. The bromine was pipetted into a tared graduated cylinder and the volume and weight were recorded. It was the transferred into a 25mL vial, the taps were wrapped in Teflon tape and the lid was tightly closed. The vial itself was put in a Mason canning jar along with a few mL of sodium thiosulphate solution. Final yield : 12,3g and 3,90mL

(46% of theoretical based on KBr, d=3,15 g/mL)

Discussion

It seems obvious that the low yield was a result of the NaOH being sucked back into the collection flask and neutralizing part of the product. Much better yields would be obtained simply by placing a guard flask between the trap and the distillation apparatus. Overall, I think this method is better than the H2O2 one (if it can be repeated with better yields) because MnO2 is much more easily obtainable than concentrated peroxide (it is easily found by opening alkaline batteries and washing the crude MnO2 with water to get rid of the electrolyte). Also, the fact that the reaction is not exothermic and does not start at room temperature avoids the possibility of a thermal runaway, disastrous when working with Br2.

Details about experimental : the original prep mentioned dilute sulphuric acid, for the exact concentration I just took a guess and figured 30% was dilute enough. It seemed to work okay. The temperature at the distilling head was not recorded because my thermometer was broken, and I couldn’t use my bath thermometer because it doesn’t fit in my 24/40 thermometer adaptor.

General notes about working with bromine : Wear full PPE and nitrile gloves throughout. Workup of the product and cleaning of the glassware were extremely unpleasant. I don’t have a sink in my fume hood, so I had to wipe the glassware with thiosulphate-soaked towels before washing them, and some Br2 still escaped, bathing the general area with the irritant and persistent smell of pool water. I don’t think I will be doing this again until I renovate my fume hood with plumbing and better ventilation.

I hope this can help someone...

-- Bromobagel

[Edited on 24-12-2012 by bromobagel]

AJKOER - 4-1-2013 at 17:18

It may be interesting to note without the use of a strong acid, the following quoted reactions from http://infoscience.epfl.ch/record/33222/files/EPFL_TH2746.pd... page 122 :

"Cl2O(g) + H2O(a) → 2HOCl(a) (4.6.6)

HOCl(a) + K-Br(s) → KOCl(s) + HBr(a) (4.6.8)

HOCl(a) + HBr(a) → BrCl(a) + H2O(a) (4.6.9)

BrCl(a) + K-Br(s) → K-Cl(s) + Br2(g) (4.5.4 and 4.5.5)

Net: Cl2O(g) + 2K-Br(s) → K-Cl(s) + KOCl(s) + Br2(g)"

From which I would infer:

2 HOCl + 2 KBr --> KCl + KOCl + H2O + Br2

Note, it is important to avoid a large excess of HOCl as:

Br2 + 5 HOCl + H2O --> 2 HBrO3 + 5 HCl

However, the above process has been described by those performing Bromine extraction as most likely sub optimal

AJKOER - 24-12-2015 at 16:37

Has anyone tried the original path? To quote:

"Bromine was discovered by A J Balardin 1826AD, by the action of Chlorine on the residues (i.e. Bromide salts) after the crystallisation of the salt from the salt-marshes of Montpellier.

NaBr + Cl2 ==> NaCl + Br2 "

Source: http://www.ucc.ie/academic/chem/dolchem/html/elem035.html

However, per a more recent reference given above (link: http://infoscience.epfl.ch/record/33222/files/EPFL_TH2746.pd... ), per page 92, an updated surface chemistry rendition, to quote:

"The fact that the uptake coefficient does not significantly change whereas the yield of Br2 increases after an exposure to H2O(g) may lead to the conclusion that 1) the presence of H2O(a) enhances the conversion of surface adsorbed chlorine to surface-adsorbed bromine or polyhalide species, 2) bromine species are not removed from the surface under ambient conditions, 3) the adsorbed bromine species may be displaced by chlorine adsorbing on the surface. 1) and 2) explain the yield of Br2 of almost 100% of the lost Cl2 and 3) may explain that the uptake coefficient remains almost unchanged after an exposure of the KBr sample to ambient conditions, which means that a “bromine site” is kinetically almost equivalent to a “free site”.

This may be explained by the following reaction scheme:

Cl2 + S = Cl2-S
Cl2-S + KBr → BrCl-S + KCl
BrCl-S + KBr → Br2-S + KCl
Br2-S = Br2 + S

Net: Cl2 + 2KBr → Br2 + 2KCl "

[Edited on 25-12-2015 by AJKOER]

Texium - 24-12-2015 at 16:50

Yes, that method does work fine, but it also leads to the formation of interhalogens, so it isn't great for preparing pure bromine.