# Basicities of nitrogen atoms in purine

Arrange the sites in purine in order of basicity.

I feel that 9 is the least basic as nitrogen's lone pair is delocalised. However, I am unable to rank 1, 3, and 7 in a particular order. Is there a particular theory to decide basicity of sites?

• Why don't you try to draw out all the resonating structures? – Avnish Kabaj Feb 2 '18 at 7:57
• @AvatarShiny What should I look for in those ? – user488460 Feb 2 '18 at 8:03
• Which site has the most number of times a negative charge in the Canonical structures ? – Avnish Kabaj Feb 2 '18 at 8:10
• Possible duplicate of Most basic nitrogen in Adenine – Archer Apr 7 '18 at 9:51
• I'm not convinced that this is a duplicate, but on a conceptual level they are clearly related. – pentavalentcarbon Apr 7 '18 at 14:17

I don't think that question is trivial and except for what you have already deduced, I would not have an idea how to solve that without employing computational chemistry.

The problems already start that I suspect that the proton from position 9 readily exchanges to position 7 in any kind of polar solvent (and in absence why not with itself).

Well from that point on forward, I'd say you need at least a lot of acid to protonate 9, as the π system is aromatic and will most likely not react first.

After that you could probably deduce from the approximate bond angles which of the nitrogen would have the largest s- and p- character, then rank them from highest to lowest p contribution. The one with the most will probably be the HOMO and therefore react first. To be honest though, this is just guessing.
I have absolutely no clue what the lesson with this exercise it, except for: "They're pretty much the same."

Let's have a closer look anyway. I have computed the neutral molecule at the DF-B97D3/def2-TZVPP level of theory, and used NBO6.0 for the partial charges:

As expected, there is not much of a difference; I'd even go as far and say they are the same. Position 9 is a little bit more negative, because it has a hydrogen to draw from.

Next up, let's have a look at the highest molecular orbital, as we would assume to protonate there:

Again, there is only little to no difference. Or in numbers (contributions > 3%):

Alpha occ 31 OE=-0.217 is
N1-p=0.3635 N1-s=0.0382
N3-p=0.2551
N7-p=0.0667
C5-p=0.0563
C4-p=0.0473
C2-p=0.0452
C6-p=0.0382


From this we would conclude that since N1 has the largest contribution, the proton is most likely to go there.

Therefore I also calculated all the protonated species, and their relative energies are: \begin{array}{lr}\hline \text{Position} & \Delta G(\pu{298.15 K}, \pu{1 atm})/(\pu{kJ/mol})\\\hline 1 & 0.0 \\ 3 & 41.2 \\ 7 & 26.1 \\ 9 & 196.8 \\\hline \end{array}

From this we conclude that the most likely position to be protonated is N1, followed closely by N7, and also N3. Off the charts is, as expected, N9 as it breaks the aromaticity.

TL;DR: Order of protonation $1 > 7 > 3 > 9$, calculated at DF-B97D3/def2-TZVPP.

• tl;dr: Do you mean to say this question cannot be solved theoretically on pen and paper and needs special software? – Gaurang Tandon Feb 13 '18 at 5:55
• @GaurangTandon Yes. And even then it is only a very educated guess. – Martin - マーチン Feb 13 '18 at 6:03
• @Martin Why "educated guess"? Aren't computational softwares accurate enough to accurately mimic real molecules? – Apoorv Potnis Feb 13 '18 at 6:50
• @ApoorvPotnis While computational chemistry can certainly do a lot, and explain a lot, the level of theory I used is also an approximation. Without proper calibration and cross-reference with more accurate levels it remains a very good educated guess. Given the small energy differences a quantitative statement is not possible. At least the calculations are good enough to gain the understanding, that there is not much of a difference in this case. – Martin - マーチン Feb 13 '18 at 6:57
• @KingTut that's really more of a philosophical kind of question. I believe that with a better understanding (which may include yet unknown mathematics, physics, or chemistry) we would be able to predict anything about any compound. With the current state of the art, that may already be in reach for simple compounds. However, for complicated molecules, there will be changes which we can't get tackle. In any case, the situation here is by far too complicated to be treated without (rudimentary) quantum chemical models. – Martin - マーチン Mar 16 '18 at 13:07