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I tried to explain the equilibrium by comparing the basicity of different nitrogen atoms in a purine anion, but I can’t see any obvious difference. Seems like that the 9-N will be the most basic because it can have two adjacent imine structure while keeping the structure of pyrimidine untouched. Computational results will be appreciated.

9H-purine

P.S. I don’t think the question is a duplicate of the “purine nitrogen basicity” one because that one compares the basicity of nitrogens in a neutral purine molecule whereas my question compares the basicity in a purine anion.

andselisk
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HFerKerman
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1 Answers1

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The given purine has two structures as shown below. enter image description here

In $\ce{9H–Purine}$ (figure 2), lone pairs are far apart from each other (3 and 7). Comparatively in $\ce{7H–Purine}$ (figure 1) lone pairs are very near each other (3 and 9) . Nearness of lone pairs in $\ce{7H–Purine}$ could leed to greater repulsions between them. This could make it energetically unfavorable in comparison to $\ce{9H–Purine}$.

This could be the reason why $\ce{9H–Purine}$ is more favored tautomer compared to $\ce{7H–Purine}$.

enter image description here

Chakravarthy Kalyan
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