Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Sach on January 11, 2018, 05:13:13 AM
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I wanted to couple 2 thiophenes (to reactive 2- and 5-positions) to the protected thiophene that you can find in the attachment. First I thought about doing this using Kumada coupling but I in order to perform a Kumada coupling, grignard reagents are required. I could not make Grignard reagent because they react with a carbonyl group and there is a carbonyl group present in my protected thiophene. This was the explanation I had for not using Kumada coupling (because Grignard reagent can't be made). My professor recently told me that there is another reason because of which I can't use Kumada coupling. I did some research but I really can't see another reason (Pd or Ni catalyst is used which is fine I guess).
Thank you in advance
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The Kumada coupling forms a C-C bond from a grignard and a molecule containing a C-X bond (X=I,Br,Cl; normally aryl-X or alkenyl-X) - I'm not sure exactly how you hope to use it here.
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But even being a halothiophene derivative, the amide hydrogen will immediately deactivate the Grignard reagent.
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Thnx alot
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Is it because the amide is acidic and will protonate the Grignard reagent?
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Yes, Grignards are strong bases.
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Ok, thank you very much
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You could try a Stille-coupling where you use a tin-derivative instead of a boronic acid. I have done this reaction and you can use PdCl2 as catalyst, I think this is because the tin-derivative reduces it in situ to Pd(0).
https://en.wikipedia.org/wiki/Stille_reaction
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The stannane is cheap;
https://www.sigmaaldrich.com/catalog/search?term=Tributylstannylthiophene&interface=All&N=0&mode=match%20partialmax&lang=en®ion=SE&focus=product
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Thnx
I can certainly propose this in my thesis as I have no time more left to work in the lab.
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You can use the halothiophene and grignard, just use more of the grignard when the first eq. will be consumed by the amidic proton.
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The Grignard will attack the carbamate-carbonyl?
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Alkylmagnesium halide will give trialkylcarbinol + benzyl alcohol + (N-)alkyl 3-thiophenamide.
But usually, low yields are obtained by using excess of Grignard reagents when active (acidic) hydrogens are present; in addition to the vigorous (and dangerous) formation of hydrogen gas (windows must be open during the reaction, in order to avoid any formation of 2/1 explosive mixture).
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If, for example, methylmagnesiumbromide reacts with a acidic hydrogen, does this not form methane, not hydrogen?
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Yes, but methane is not a big problem, if the fume’s ventilation is “on”.
The problem is the unreacted magnesium, which is in micro/nanosuspension form and activated by the remaining traces of the preliminary added, iodine crystal that will form magnesium hydride with the “active” hydrogen; which by its turn, will react with another “active” hydrogen and will finally lead to the formation of gaseous hydrogen.
Of course, this problem can be avoided by purchasing pre-prepared methylmagnesium bromide from suppliers.
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Please, note that the working up of the old Chugaev-Tserevitinov method for “active hydrogen” determination with methylmagnesium iodide, is as follows:
After the methylmagnesium iodide preparation in methoxybenzene (anisole), unreacted methyl iodide is removed by distillation and the so prepared methylmagnesium iodide is decanted from the precipitated, unreacted magnesium.
This technique leads to lose of significant amount of the so prepared methylmagnesium iodide, which may be tolerated for analytical purposes (for samples of known mass); but it is unacceptable in synthetic chemistry.