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In the above question I was asked to compare the basic strength order. So I did it like this:

The Nitrogen in I and II will not participate in resonance so the only effect it could produce is inductive (-I) . Now the compound IV has less resonating structures than the other compounds so the the Nitrogen atom will have larger electron density.

So IV should be most basic then III > II > I because of (-I) effect. But my book mentions (C) as the answer.

How? Any help will be appreciated


1 Answer 1


You don't have problem with assigning compound IV (1, 4, 5, 6- tetrahydropyrimidine) as the most basic compound among given four amines. I agree with you, but not the same reasons. According to Ref.1, the 1,4,5,6-tetrahydropyrimidines (THPs) are extremely basic compounds, as might be expected from their amidine structure (e.g., 1,8-Diazabicyclo[5.4.0]undec-7-ene or DBU). The parent THP, 1,4,5,6-tetrahydropyrimidine (IV) has a $\mathrm{p}K_\mathrm{a}$ of about $13$ at $\pu{20 ^\circ C}$, which is even reacted with water (Ref.1). This highly basic nature of IV can be attributed to its stabilization by resonance between two equivalent structures, V and V', when protonated:

Protonation of THP

Also, keep in mind that it can also get protonated on 3-nitrogen atom (VI; inside the blue box) as well (Ref.1).

Among the rest of other 3 amines, two are aminopyridines (I and II), and the third one is aniline (III). Since the lone pair of nitrogen in pyridines does not contributed to aromatic system (hence it is readily available for proton) while that of aniline is involved in aromatic system by resonance (not readily available), generally pyridine is more basic than aniline. Even though I and II both have extra amino group, they are expected to behave like pyridines. By this analysis, we can conclude that the least basic amine among given four amines is aniline (III).

Now, let's look at what happens when I (4-aminopyridine) is protonated:


From above illustration, you can clearly see conjugate acid of I is well stabilized by resonance including full contribution of the lone pair of 4-amino-nitrogen. If you visualize the same with II (3-aminopyridine), it is not stabilized by the extra amino grop as depicted in following diagram:


Thus, by these facts, we can conclude that II is less basic than I. Therefore, the order of basicity strength of four given amines is: IV > I > II > III.


  1. D. J. Brown, R. F. Evans, “100. Hydropyrimidines. Part I. 1, 4, 5, 6- Tetrahydropyrimidine and its derivatives,” Journal of the Chemical Society 1962, 527–533 (https://doi.org/10.1039/JR9620000527).
  • $\begingroup$ How to decide which is the basic site of a compound? $\endgroup$ Dec 18, 2019 at 1:19
  • $\begingroup$ The lone pair of pyridine $\ce{N}$ is fully available than that on 3- or 4-amino $\ce{N}$. $\endgroup$ Dec 18, 2019 at 1:28
  • 1
    $\begingroup$ @MathewMahindaratne I had some confusion regarding (I) and (IV). Both have lone pair of electrons which are localized hence good basic sites. The (+) charge on both is stabilized by nitrogen. They both suffer (-I) due to nitrogen and the benzene ring respectively. So how do you decide between them? Infact (I) has more resonance structures hence its (+) charge should be better stabilized without compromising the electron density since lp are not in the plane $\endgroup$
    – Ashish
    May 24, 2021 at 20:04
  • $\begingroup$ @Ashish: (IV) is not aromatic amine, and everything else are. Thus, most basic amine is (IV) due to that fact alone. (III) has on ring nitrogen (not a pyridine), hence it is the least basic. Your argument applies between (I) and (II), which I distinguished in my answer. $\endgroup$ May 25, 2021 at 16:12
  • 1
    $\begingroup$ But thats my doubt, why does being an aromatic amine reduces its basicity even though the electrons of the amine are not delocalised in the ring ? $\endgroup$
    – Ashish
    May 25, 2021 at 16:48

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