[chiddler]

I hate these questions so much!

I need help :<

So correct me if i'm wrong:

Pyridine is the most basic.

3-nitropyridine is next because it has EWG = more resonance = more e- delocalization

Then the 5 pyrrole look-alike with 2 nitrogens because one of the nitrogens has non-aromatic lone pair.

Then pyrrole because e- are part of aromaticity.

I am unsure of the 2nd and 3rd place.

Thanks very much!

[orgoclear]

I would say that 2-(N)-pyrrole is most basic followed by pridine, then nitro-pyridine and finally pyrrole..

My reason for the first choice would be because:
If you draw the resonance structures for 2-(N)-pyrrole, then, the second N (whose electrons are not involved in delocalisation) have a partial negative charge hence higher electron density => more basicity

Nitro-pyridine will be less basic than pyridine because the nitro group has (-I,-R) effect leading to less electron density => less basicity than pyridine

Pyrrole will be the least basic as the l.p. of N are delocalised in maintaining aromaticity

I hope this order is correct

[chiddler]

That is a good point. Thank you.

[Doc Oc]

Just FYI, that pyrrole "look-alike" is called imidazole.  Much easier to say than "sort-of-pyrrole"

I would actually say that pyridine is more basic than imidazole because at any given time either nitrogen could be the one participating in the aromaticity of the ring.  Those double bonds aren't static and you have no way of knowing which nitrogen is free.  In pyridine, that lone pair doesn't participate.  Ever.  That means it's always available, whereas in imidazole you don't get that certainty.

I do agree that 3 and 4 are m-nitropyridine and pyrrole.  Remember though, it's not just the electron withdrawing nature of the nitro that's important, but that it's in the meta position which lines it up perfectly to suck the electron density away from the nitrogen.

[AWK]

http://en.wikipedia.org/wiki/Imidazole

[Doc Oc]

Ha ha!  Well, that set me straight there, didn't it?