09-26-02 06:53
No 360655
      Products of the oxone reduction?
(Rated as: excellent)

Recently, I read two interesting articles that are available online for free.  Both can probably be located using just about any online search engine using the keywords "bromination, arenes, sodium bromide, oxone."

Here are the citations:

Bull. Korean Chem. Soc. 2002, Vol. 23, No. 5, pp. 773-775
"                     " 2001, Vol. 22, No. 6, pp. 549-550

In former publications wherein arenes were halogenated using the oxone/K-halogen(KCl, KBr, or KI) system, equimolar amounts of arene, K-halogen, and oxone in methanol are used in the reaction.  In the above publications, equimolar amounts of arene, Na-halogen, and oxone in aqueous methanol (1:1) are used in the reaction.

According to Synthetic Communications Vol. 32, no. 15, pp. 2313-2318, 2002, specifically on p. 2316, the reaction between an arene, KI, and oxone occurs in the following manner:

ArH + KBr + 2KHSO5-KHSO4-K2SO4 =
  ArBr + KOH + K2S2O8-KHSO4-K2SO4 + H2O

2KHSO5-KHSO4-K2SO4 + KBr =
  HOBr + KOH + K2S2O8-KHSO4-K2SO4


KBr + 2HO +2OSO3K = KOH +HOBr + K2S2O8

ArH + HOBr = ArBr + H2O

For those a little confused like I was at first, K2S2O8 is potassium persulfate (formed, I would guess from the combination at the oxygen poles of two bisulfate ions; is this right?).  Like oxone, it has the reputation of being a powerful oxidant.  Acidification of potassium persulfate solutions, according to the peracid heading of the Encylopedia of Chemical Technology, yields peroxymonosulfuric acid and hydrogen peroxide.  Other references indicate that through heat and/or moisture, potassium persulfate "decomposes".

According to Synthetic Communications, 31(23), 36-27-3632 (2001), specifically on p. 3628, 8 mmol of acetophenone, 40 mmol of sodium bromide, and 16 mmol of oxone are reacted in 30 mL of aqueous methanol (1:1) under reflux and stirring for 20 hours to provide a-bromoacetophenone in 71% yield.  Under the same conditions, propiophenoe yields 67% of the a-bromoketone.  This is interesting in that "free halogen" under acidic conditions was used to halogenate the ketone.  However, most commercial sources of oxone are composed of "buffer" mixtures (approximately 15% sodium carbonate, according to some information I found online).  Since the above reaction requires acidic conditions (why, I'm not sure) the addition of acid (sulfuric or bisulfate) would probably be necessary to ensure that a low pH is maintained.  In aqueous methanol mixtures, this would prove to be an easy matter; however, performing such an addition to methanol-only solutions might be a bit more complicated.  This would be especially true in the iodination reaction of KI/arene in methanol solvent using, presumably, commercial buffered-oxone as the oxidant.

Unlike the typical oxone/sodium or potassium bromide reactions, free bromine is desired rather than the hypobromite.  So my question is, what are the products of the following reaction and what is the final balanced equation?:

8 mmol acetophenone + 40 mmol of NaBr + 16 mmol of oxone

Some more questions:

1.) Is the NaBr in excess of the oxone?  If so, by how much and why? 
2.)  What is the oxidation potential of oxone? 
3.)  If upon reduction, oxone forms potassium persulfate, then what is its oxidation potential?
4.)  What is the final product of a COMPLETE oxone reduction?  Potassium bisulfate? 
5.)  How is it that in the above reaction using acetophenone as a substrate (is that what it's called?) does free Br2 form?

FYI, methanol is typically used for these reactions because neither oxone (JACS Vol. 56, p. 2198) nor bromine (Brittish Patent 607,538; a-halogenation reaction of arylketones using Br2 in methanol solvent) reacts with it to any appreciable extent.  According to Tetahedron Letters, Vol. 38, No. 16, pp. 2805-2808, 1997, specifically on p. 2806, "Reaction of KBr with KHSO5 gave rise to a bright yellow-orange solution that is decolorized instantly upon addition of any of the three pyrimidine nucleotides."  Just a little info as to what oxone/Na or K halide reactions might look like before and after they are completed.

As for the reputed explosive tendancies of oxone solutions, they seem to be highly overrated.  According to JACS, Vol. 59, pp. 552-555 (1937)--which is a good article detailing the reactions between peroxymonosulfuric acid and lower alcohols, BTW--when concentrated peroxymonosulfuric acid was reacted with the tertiery alcohol ethyldimethylcarbinol, an explosion ensued.  This is quite a unique circumstance and does not seem to apply to typical oxone reactions in which the potassium acid salt is quite stable.

Anyone ever wonder why it is that in JOC Vol. 25, pp. 1901-1906 (1960), sulfuric acid is added slowly drop by drop to alcohol non-aqueous solutions of oxone to form the desired ketone or alkylacetate?  Simply to hydrolize the potassium peroxymonsulfate salt into free peroxymonosulfuric acid, which is soluble in the alcohol solvent.  Interesting. 

BTW, the above JOC article has experimental details on how to synthesize benzylhalides from refluxing mixtures of sodium halide/oxone/and toluene.  Yields are low though.

It is also interesting to note that in a recent Synthetic Communications article, more specifically in a 2002, Vol. 32 issue, the oxone/NaBr system is used to effectively oxidize benzyl alcohols to benzaldehydes in high yield.  I'd give you the exact citation, but the article from which I got the reference only has it listed as "in press".  Sorry.

I hope somebee can answer my questions about the products of the oxone reduction using 40 mmol of NaBr to 16 mmole of oxone.  It's the least somebee can do after my putting so much work into this thread.

(Chief Bee)
09-26-02 08:05
No 360675
      Answer  Bookmark   

First, bromide is oxidized to bromine: 4 HSO5- + 2 Br- -> Br2 + 2 S2O82- + 2 OH-

Second, the bromine reacts, releasing bromide: Acetophenone + Br2 -> Bromoacetophenone + H+ + Br-

Third, the bromide is recycled to bromine again: S2O82- + 2 Br- -> Br2 + 2 SO42-


2 Br- + 4 HSO5- + 2 Acetophenone -> 2 Bromoacetophenone + 2 H2O + 2 SO42-


Br- + 2 HSO5- + Acetophenone -> Bromoacetophenone + H2O + SO42-

If I have calculated correctly, you have 2.5x molar excess of bromide compared to oxone/acetophenone.

The oxidation potential of persulfate at room temp is 2.01 volts (2-electron oxidant), while at 40-100C it is 2.6 volts (thermal radical dissociation).
09-27-02 05:43
No 361076
      So what's left when the reaction is done?  Bookmark   

It would seem to me that since sodium bromide reacts readily with oxone, the oxone would definitely get used up rather quickly as would the potassium persulfate.  At the end of pretty much all of these bromination reactions, the experimental details call for quenching with saturated sodium thiosulfate to reduce excess oxidant.  But what exactly would that excess oxidant be?  Oxone?  Potassium persulfate?  Hypobromite?  Bromine?  What? 

This is what I'm confused about.  By the end of the reaction, aside from the a-bromoacetophenone, what is actually left over that needs to be reduced?  Also, according to a wide variety of articles that I've read, everything from the oxone, to the potassium persulfate, to the hypobromite and the free Br2, can be reduced with ethanol, or more preferably, 2-propanol.  So do you think that substituting 2-propanol for the sodium thiosulfate might work as a substitute reducing agent?  If so, then the final remaining mixture should only be composed of potassium sulfate, potassium bisulfate, acetone, 2-propanol, methanol, possibly some HBr, and the a-bromoacetophenone, correct?

Furthermore, if indeed the NaBr is in excess over just about everything else at the start of the reaction, why do you think this is?  To ensure that all of the oxone readily gets used up so that it no longer plays a role in the subsequent bromination reaction?  Do you think the oxidation of the bromide ion into bromine by oxone and the following re-oxidation of the bromide ion by persulfate occurs faster than the a-bromination of the ketone?  If so, this would explain alot about the reaction mechanism.

Even guessing the answers to the above questions would be greatly appreciated.

(Chief Bee)
09-27-02 06:14
No 361093
      My own guess is that bromide ion is cheap enough ...  Bookmark   

My own guess is that bromide ion is cheap enough for the authors to just use it in a large excess to ensure that all the added oxone (and formed persulfate) gets converted to bromine, and won't be causing  any damage to the molecules on their own, for example by forming radicals.

Likewise, the addition of a reducing agent in the end is just a safety measure IF any kind of oxidant would be left in solution (stray useless hypobromite ions perhaps?). Yes, sure you could add 2-propanol  as the reducing agent instead, I cannot think of any species in solution right now that wouldn't react with that, given a few minutes.

Are you sure all these issues are valid concerns? In my experience a lot of authors are very confused about what they are really doing, and there are often no reason whatsoever for this and that to be added or why reagent A or B is present in excess...