1: pyridin‐4‐amine; 2: pyridin‐3‐amine; 3: aniline; 4: 1,4,5,6‐tetrahydropyrimidine

In cases 1, 2, 3, the compounds will resonate, thus giving them more stability.

Firstly, I want to know whether nitrogen has a +M effect? (due to it having a lone pair of electrons?)

In case 1, resonance occurs and the electron density above the $\ce{NH2}$ group is delocalised, even in the cases 2, and 3, this is true. But in 1 the $\ce{N}$ is at the para-position, while in 2 the $\ce{N}$ is at the meta-position. How would that affect the order?

It would help if someone could hint at the factors coming into play here.

Case 4 is non-resonating, so it would be the most basic, right? How do I arrange the rest in order?

  • $\begingroup$ The pyridine nitrogen does not involve in resonance but is boosted by para activation by the amine group. So, 1 is more basic than 2 since amine boosts ortho and para positions more than meta. We can put 3 to the end of the list. You are right about 4 being the most basic because there are just two conjugation structures which imply the lone pair is available more easily than 1 or 2 or 3. Hence I would go with 4>1>2>3. $\endgroup$
    – Suraj S
    Oct 22 '19 at 6:56
  • $\begingroup$ @SurajS Could you please show the 2 conjugation structures of 4? I thought that there would be no resonance at all in 4. $\endgroup$
    – Techie5879
    Oct 22 '19 at 7:00

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