[chiddler]

Looking at aniline, its pKb is around 9 and ammonia is around 4. Ammonia is much more basic which means that the aromatic ring is eating up electron density from the ammonia causing the lone pair to be less available. However, it CAN swing both ways. It can donate electron density or take electron density. Which is preferable?

So my question is: generally, is it easier for an aromatic ring to take in electrons or to give them up?

[Arctic-Nation]

If you're talking about simple compounds, your question does not make sense. The aromatic ring just donates electrons to electron-withdrawing substituents, and accepts electrons from electron-donating ones. That's how it is, and there's nothing easy or difficult about either accepting or donating electrons.

It is only in comparing different aromatic rings that you'll find that electron-poor rings are less inclined to give up electrons, while electron-rich rings won't easily accept electrons. Just check up on aromatic substitution reactions to see what happens when you try, for example, to do a nitration of benzene, pyridine and pyrimidine. Conversely, try and figure out why the Chichibabin reaction works on pyridine, but not benzene.

[chiddler]

You're right. I mistakenly thought that the amine group can accept electron density as well as donate. It only donates electrons.

It seems in the chichibabin reaction, an amine anion is formed and it can attack the ring. So it's giving up electrons to the ring. I'd expect it to be more reactive with benzene since, as you wrote, "electron rich rings won't easily accept electrons". Nitrogen heterocyclics are relatively electron rich and should be less reactive. Yet this trend is opposite.

Can you please explain why i'm wrong?

[Arctic-Nation]

You've made a beginner's mistake there: nitrogen atoms are electron-rich when it comes to their free electron pair, but they are electron-poor (or electron-withdrawing) because of their higher electronegativity when that electron pair is not available. In pyridine, the free electron pair is in the plane of the ring and not part of the pi system, making the aromatic ring electron-poor and suited for nucleophilic aromatic substitutions. In the Chichibabin reaction, the nitrogen atom of pyridine acts as an electron sink in the addition step, something that is completely impossible in benzene itself.

If, on the other hand, the free electron pair is part of the pi system, as with pyrrole, then the ring as a whole is more electron-rich and more suited towards electrophilic aromatic substitutions.

[chiddler]

thanks very much for your response. helped a lot.