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Author Topic: A question about Grignard reagent and Kumada coupling  (Read 1441 times)

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Sach

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A question about Grignard reagent and Kumada coupling
« on: January 10, 2018, 11:13:13 PM »

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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clarkstill

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Re: A question about Grignard reagent and Kumada coupling
« Reply #1 on: January 11, 2018, 04:20:10 AM »

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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pgk

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Re: A question about Grignard reagent and Kumada coupling
« Reply #2 on: January 11, 2018, 05:22:31 AM »

But even being a halothiophene derivative, the amide hydrogen will immediately deactivate the Grignard reagent.
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Sach

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Re: A question about Grignard reagent and Kumada coupling
« Reply #3 on: January 11, 2018, 05:23:58 AM »

Thnx alot
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Sach

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Re: A question about Grignard reagent and Kumada coupling
« Reply #4 on: January 11, 2018, 05:25:46 AM »

Is it because the amide is acidic and will protonate the Grignard reagent?
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pgk

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Re: A question about Grignard reagent and Kumada coupling
« Reply #5 on: January 11, 2018, 05:27:45 AM »

Yes, Grignards are strong bases.
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Sach

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Re: A question about Grignard reagent and Kumada coupling
« Reply #6 on: January 11, 2018, 05:30:24 AM »

Ok, thank you very much
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rolnor

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Re: A question about Grignard reagent and Kumada coupling
« Reply #7 on: January 11, 2018, 08:10:38 AM »

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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Sach

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Re: A question about Grignard reagent and Kumada coupling
« Reply #9 on: January 11, 2018, 11:41:29 AM »

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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kriggy

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Re: A question about Grignard reagent and Kumada coupling
« Reply #10 on: January 11, 2018, 07:29:42 PM »

You can use the halothiophene and grignard, just use more of the grignard when the first eq. will be consumed by the amidic proton.

rolnor

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Re: A question about Grignard reagent and Kumada coupling
« Reply #11 on: January 11, 2018, 11:10:46 PM »

The Grignard will attack the carbamate-carbonyl?
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pgk

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Re: A question about Grignard reagent and Kumada coupling
« Reply #12 on: January 12, 2018, 06:42:02 AM »

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).
« Last Edit: January 12, 2018, 06:57:04 AM by pgk »
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rolnor

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Re: A question about Grignard reagent and Kumada coupling
« Reply #13 on: January 12, 2018, 12:50:25 PM »

If, for example, methylmagnesiumbromide reacts with a acidic hydrogen, does this not form methane, not hydrogen?
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pgk

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Re: A question about Grignard reagent and Kumada coupling
« Reply #14 on: January 13, 2018, 08:29:20 AM »

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.
« Last Edit: January 13, 2018, 08:59:26 AM by pgk »
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