ssdd
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Biphenyl, would this method work?
I have seen several references to the following reaction as impurities to a grignard reaction:
Ph-MgBr + PhBr --> Ph-Ph + MgBr2
I noticed that this reaction is known to proceed faster at high temperatures, but what if I wanted to use this type of reaction to form a product.
Under THF as the solvent is it possible to react a grignard reagent with a brominated organic to yield product? And would the yield be worth it?
-ssdd
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Nicodem
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| Quote: | Originally posted by ssdd
I have seen several references to the following reaction as impurities to a grignard reaction:
Ph-MgBr + PhBr --> Ph-Ph + MgBr2 |
Where have you seen that?
It can be achived under Kumada coupling conditions. See these posts:
http://sciencemadness.org/talk/viewthread.php?goto=lastpost&...
http://sciencemadness.org/talk/viewthread.php?tid=8158&p...
http://sciencemadness.org/talk/viewthread.php?tid=8444&p...
...and especially:
http://www.organic-chemistry.org/namedreactions/kumada-coupl...
http://www.acros.com/_Rainbow/pdf/Kumada.pdf
http://en.wikipedia.org/wiki/Kumada_coupling
(search your self for more)
If you need biphenyl there are much simpler ways for making it (starting a Grignard with PhBr is not easy and the catalysts for the Kumada coupling
are often expensive). There is even one OTC method for making biphenyl: http://sciencemadness.org/talk/viewthread.php?goto=lastpost&... Also, if you have cyclohexene and benzene you can make phenylcyclohexane via
Friedel-Crafts reaction and dehydrogenate phenylcyclohexane to bipheny. Actually, most methods are more practical than the Kumada coupling.
[Edited on 4/3/2008 by Nicodem]
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Sandmeyer
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I agree, those transition-metal catalysed coupling reactions look simple on paper but in practice can be very time-consuming and boring (IMO) since so
many parameters are involved. It is robot-chemistry IMO, so it is no wander Japanese are inventing those reactions. I always see them as last resort
options despite their obvious popularity in the literature. This pupularity can in part be explained by wannabe chemists who publish to get more
papers and not for the sace of advancing the synthesis. They put an old coupling in microwawe, get a paper, change base another paper, change ligand -
paper, use co-solvent solvent yet another, so in a sense they are spaming the comunity with junk chemistry (expensive too!).
[Edited on 4-3-2008 by Sandmeyer]
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Nerro
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Actually this is a pretty straight forward reaction. You add the two reagents in equimolar amounts in THF with 0.05%
Ni(ligand)<sub>2</sub>Cl<sub>2</sub> and heat and wait. I'm doing an internship right now which involves making new ligands
for this stuff.
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chemrox
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seems like we got biphenyl when adding water to ph-mg-br
"When you let the dumbasses vote you end up with populism followed by autocracy and getting back is a bitch." Plato (sort of)
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Nicodem
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Nope, that would give you benzene.
Nerro, the Kumada coupling sure can be easy when you are versed and have the know-how, but still, would you really consider using it for preparing
biphenyl itself? I would consider the Kumada coupling only when there would be a need to prepare some precious biaryls that have limited synthetic
alternatives. But biphenyl can even be made by a reductive Ullmann reaction from iodobenzene (of course, for a moment ignoring the fact that biphenyl is a couple of times cheaper than either bromo- or
iodobenzene).
I tend to firmly advocate using simple solutions to simple problems (a rare point of view among academic colleagues).
…there is a human touch of the cultist “believer” in every theorist that he must struggle against as being
unworthy of the scientist. Some of the greatest men of science have publicly repudiated a theory which earlier they hotly defended. In this lies their
scientific temper, not in the scientific defense of the theory. - Weston La Barre (Ghost Dance, 1972)
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Nerro
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true dat.
still though, Kumada works with Chlorobenzene which is pretty cheap, just add Mg, a tiny pinch of iodine and an equally tiny pinch of nickelcomplex
and off you go. Quantitative yields in a matter of minutes. On a large scale Ullmann looks to me like a pretty expensive reaction seing as how it
gobbles up copper and iodine. In the Kumada coupling you use up magnesium and chlorine/bromine. Much easier and cheaper.
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vulture
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You don't explicitely need Mg or grignardreagents. Zinc metal, NiL2Cl2 and PhCl refluxation in a polar solvent will get you biphenyl in high yield. In
fact, this reaction proceeds so well that you'll have a hard time isolating any intermediates or to couple it with anything else.
EDIT: Note that the above is only applicable to symmetrical aryl endproducts which are not sterically hindered.
[Edited on 5-3-2008 by vulture]
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Jor
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EDIT : nevermind
[Edited on 5-3-2008 by Jor]
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Sandmeyer
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| Quote: | Originally posted by vulture
You don't explicitely need Mg or grignardreagents. Zinc metal, NiL2Cl2 and PhCl refluxation in a polar solvent will get you biphenyl in high yield. In
fact, this reaction proceeds so well that you'll have a hard time isolating any intermediates or to couple it with anything else.
EDIT: Note that the above is only applicable to symmetrical aryl endproducts which are not sterically hindered.
[Edited on 5-3-2008 by vulture] |
That sounds very interesting, can you provide reference? To me it seems that the role of Zinc metal is to in-situ provide an arylzinc reagent before
the nickel catalysis starts. It sounds extraordinary that aryl zinc can be formed combining elemental zinc and aryl halide, esp chloride. But I might
be completely wrong regarding the role of zinc. I have tried to make aryl zinc reagent from an aryl iodide, using elemental zinc dust but it didn't
work.
[Edited on 5-3-2008 by Sandmeyer]
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vulture
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The zinc is just there to reduce the nickel(II). Then you have oxidative addition of the aryl halide and two equivalents of the complex combine to
NiX2 and to provide the biaryl. There are some references by Kochi et al in the bulletin of the japanese chemical society.
I "discovered" this to my displeasure when I was trying to prepare an unsymmetrical biaryl using a similar method. Any other reducing agent will do,
but zinc is ofcourse the most economical.
[Edited on 5-3-2008 by vulture]
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kmno4
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BTW, references (for example):
Modern Organonickel Chemistry
Modern Organocopper Chemistry
The Chemistry of Organozinc Compounds
Organomagnesium Methods in Organic Chemistry[BSM]
.......
There is interesting reference (-> Synthesis, 147, 1990) in the last ebook, about reactions RMgX+R'Y(in THF+NiCl2) ->R-R'.
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Nerro
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@Vulture
So the method using Zn would only work for symmetrical biaryls. Since the Zn takes away the halide the outcome would be the same as with an organozinc
reagent. Is there any proof to say which of the two theories is msot likely to be true? Do both occur? (so organozinc ánd
"2NiL<sub>2</sub>XAr --> NiL<sub>2</sub>X<sub>2</sub> + NiL<sub>2</sub>Ar<sub>2</sub>
--> NiL<sub>2</sub>X<sub>2</sub> + NiL<sub>2</sub> + Ar<sub>2</sub>")
And another question, what L did you use? Did you prepare it yourself?
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vulture
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No, no. It's not the same as with organozinc reagents. Anyway, my interest in this reaction is limited as for me it's only a sidereaction. The outcome
is highly dependant on the arylhalide (electron rich/poor), temperature, polarity of the solvent and in some cases even the reducing agent (for
example using DIBAL as a reducing agent causes problems with grignard reagents).
One shouldn't accept or resort to the mutilation of science to appease the mentally impaired.
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