I have to identify delocalized electron pairs in Isoniazid (pairs not shown in the image below):

I know the nitrogen in the ring has a localized electron pair, since it already forms a pi bond. I'm confused about the other two nitrogen atoms. When we draw one of the resonance structures, we can see the amide nitrogen has a delocalized electron pair:

enter image description here

And if we push the lone pair from the amine nitrogen, we get another resonance structure (even if it is insignificant, due to all of the formal charges, it still seems to be a valid resonance structure):

enter image description here

Because these two lone pairs engage in resonance, I thought that they were both delocalized. But my textbook (Klein's Organic Chemistry, section 2.12, 2nd Ed.) says only one pair is delocalized:enter image description here

Therefore, if the electron pairs on both the amine and amide nitrogens in Isoniazid seem to engage in resonance, why is only one pair delocalized?


1 Answer 1


The reason why the amine lone pair is delocalized is that to make the last conformer you proposed, the nitrogen closest to the oxygen would have to form five bonds. Nitrogen is a second-row element that cannot readily possess 10 electrons from those 5 bonds, even if nitrogen has to possess those 10 electrons for the very short time required for the electrons to switch conformations between different resonance conformers.

  • 3
    $\begingroup$ Molecules do not switch between resonance structures. All resonance structures exist simultaneously. $\endgroup$
    – Andrew
    Commented Feb 16, 2020 at 0:52
  • $\begingroup$ @Andrew can you write a answer for this query by OP , i think i am having same doubt as OP $\endgroup$
    – Orion_Pax
    Commented Mar 28, 2022 at 10:29

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