Why C4 position in pyrimidine ring is more reactive then C2 position in nucleophilic aromatic substitution and Pd chemistry? Should not be the C2 position the most reactive because of the effect of the 2 N atoms?

  • $\begingroup$ The SNAr bit is already asked here: chemistry.stackexchange.com/q/80995/16683 I think the Pd coupling bit would be good as a stand-alone question. $\endgroup$
    – orthocresol
    Jan 9 '19 at 23:43
  • $\begingroup$ That q does not have an answer with a positive score. So it is not "really" answered. $\endgroup$ Jan 10 '19 at 0:53
  • $\begingroup$ @OscarLanzi I never said it was "answered", merely "asked" ;) and arguably, asking both questions together might be too broad. $\endgroup$
    – orthocresol
    Jan 10 '19 at 1:03
  • $\begingroup$ It is true, it might be a lot to ask in one go but, I asked them toghether because on every books or paper they explicitly say that since in the SNAr reaction, C4 is the preferred electrophilic site of reaction, it is the same for Pd chemistry. But I am very confused. $\endgroup$
    – Chem19
    Jan 13 '19 at 11:39
  • $\begingroup$ I thought that the presence of the 2 Nitrogens would have made the C2 position more electrophilic and not the other way around. Again, if you use a negatively charged nucleophile it goes on C4, if you use a neutral one, it goes on the C4 first. So, it does not really depend on the nucleophile. Might it depend on the presence of the orbital on the N preventing some how the reaction t C4? and it is the same for Pd chemistry: if you run a Pd coupling on the C4 it goes in 1 h, on the C2 lack of reactivity or anyway harsh conditions. $\endgroup$
    – Chem19
    Jan 13 '19 at 11:39

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