Confused about identifying delocalized electron pairs in Isoniazid

Question

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:

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):

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:

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?

Answer 1

Technically, the last structure follows all the rules for valence bonds. But it is highly unfavorable because of the extra formal charge separation and especially to the assignment of a negative charge to a carbonyl carbon (which we generally shouldn't see*).

It is true that the structure wuth the $\ce{C=N}$ double bond also forces some charge separation, but that one is more favored (and generally accepted for amides) because the carbonyl carbon remains formally neutral and the greater electronegativity of oxygen versus nitrogen makes this polarization plausible.

There are, in fact, five delocalized pi-electron pairs. When the amido nitrogen couples its available pair with the carbonyl group, the pi bond in the latter is also delocalized into a three-center four-electron bond. The other three pairs in this count of five are, of course, those in the aromatic ring.

*I am not forgetting enolate ions, whose carbanion contribution has the negative charge adjacent to the carbonyl carbon.

Answer 2

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.