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there is a problem regarding a Diels-Alder reaction that I do not understand (see image). enter image description here

The Diels-Alder reaction of the diene and the dienophile should lead to the product A. I hope I did it correctly, because the molecular formular shall be C9H6O2N2.

Now, the base triethylamine is added and heat is applied. Because it is a very strong base I would deprotonate the proton next to the ester-group -COOR, since it has a -M effect. I think applying heat leads to decarboxylation (B = carbon dioxide). But I don't know what the product C looks like, since there is another reaction to product D.

Can you please help me?

Thank you!

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  • $\begingroup$ The pKa of the protons alpha to the nitriles is very similar to that of the proton alpha to the ester. $\endgroup$ – Waylander Jul 5 at 20:54
  • $\begingroup$ So if I take the proton alpha to the nitrile, then the negative charge can be stabilized by the nitrile. Can I then make a new double bond in the ring, so that a carboxylate forms? The carboxylate can then form carbon dioxide... So in the end another double bond in the ring forms and a CN- leaves. But does that even make sense? $\endgroup$ – QuestionCookie Jul 6 at 15:16
  • $\begingroup$ Why does CN- leave if CO2 leaves? I think you will get a dicyanocyclohexadiene (C) which will spontaneously aromatise to give 1,2 dicyanobenzene $\endgroup$ – Waylander Jul 6 at 21:16
  • $\begingroup$ I am sorry, I still do not completely understand it. If CO2 leaves, then there is still a negative charge in the molecule. So what happens so that dicyanocyclohexadiene is formed? $\endgroup$ – QuestionCookie Jul 7 at 8:32
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    $\begingroup$ There is also Et3NH+ present which can quench the anion. The diene will spontaneously lose H2 as the driving force for aromatisation is very powerful $\endgroup$ – Waylander Jul 7 at 8:35

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