[edwinksl]

Taken from the Clayden's Organic Chemistry book:

pK_aH is the pKa of the conjugate acid.

So, the pK_aH of imidazole is 7.1 and that of pyridine is 5.2.

The book goes to explain that "imidazole, with its two nitrogen atoms, is more basic than pyridine because pyridine only has one nitrogen on which to stabilize the positive charge."

My question is: doesn't placing a positive charge on an electronegative atom like nitrogen energetically unfavorable?

I got this idea from the following website http://www.cem.msu.edu/~reusch/VirtTxtJml/intro3.htm#strc7 which says the following:

Electronegativity of charge bearing atoms and charge density. (High charge density is destabilizing. Positive charge is best accommodated on atoms of low electronegativity, and negative charge on high electronegative atoms.)

Thanks.

[Yggdrasil]

With regard to having a proton on a nitrogen, acidity/basicity is all relative.  For example, it's more favorable to have the positive charge on a nitrogen in a protonated amine or a protonated imidazolium ring than on an oxygen in hydronium.

Also, I don't like the explanation for why imidazole is more basic.  Imidazole has two nitrogens.  One nitrogen has its lone pair in a p-orbital which is delocalized in the aromatic system.  The other nitrogen has its lone pair in an sp² orbital which lies orthogonal to the conjugated pi system.  This nitrogen, with the lone pair in the sp² orbital is what makes imidazole much more basic than pyridine.  Protonation of this nitrogen does not break up the aromatic ring.

In contrast, the lone pair on pyridine's nitrogen is in a delocalized p-orbital.  Protonation of the nitrogen eliminates the aromaticity of pyridine, which is energetically unfavorable.

[edwinksl]

Yggdrasil,

You are right about imidazole, but wrong about pyridine. The lone pair on N in pyridine is in a sp² orbital which lies orthogonal to the pi system. so protonation at this N atom does not destroy the aromaticity of the resultant pyrindinium ion.

[Yggdrasil]

Ah, you are right.  I was thinking about pyrrole.  Thanks.

[edwinksl]

Ah, you are right.  I was thinking about pyrrole.  Thanks.

Heh, no problem man. But my question remains unresolved.

I should make myself clearer at the same time. The book is claiming that imidazole is more basic than pyridine, and you can deduce so by looking at the protonated ions, i.e. the imidazolium ion and the pyridinium ion respectively.

In the imidazolium ion, you have 2 nitrogen atoms to spread the positive charge over, but the pyridinium ion only has one nitrogen on which to stabilize the positive charge. So, the imidazolium ion is more stable, making imidazole more basic.

My question is, doesn't spreading a positive charge onto an electronegative atom like N actually makes imidazolium ion more unstable?

P.S. Are there only 2 resonance structures for an imidazolium ion?

Look at the ring in B. So the C=C bond stays at the same place in both resonance structures?

[movies]

Spreading out the positive charge has a huge effect on stabilizing the protonated intermediate.  As you pointed out, having a positive charge on an electronegative atom seems like a bad thing (however, it is favorable due to electrostatic attraction between a + charge and a - charge).  Since the two resonance structures of protonated imidazole are identical, the two contribute equally to the real structure.  In essence, you get half a + charge on each nitrogen and therefore the electronegative atom has less positive charge.  Since pyridine can't do that, the lone nitrogen bears almost all of the + charge.

Also, there are other resonance structures for both imidazole and pyridine which have a + charge on a carbon atom, but these are not significant contributors to the overall structure.

[sjb]

...Since the two resonance structures of protonated imidazole are identical, the two contribute equally to the real structure.  In essence, you get half a + charge on each nitrogen and therefore the electronegative atom has less positive charge.  Since pyridine can't do that, the lone nitrogen bears almost all of the + charge.

Also, there are other resonance structures for both imidazole and pyridine which have a + charge on a carbon atom, but these are not significant contributors to the overall structure.

Be careful you don't equate that to the picture drawn however. In the system mentioned the two resonance forms are not the same.

S

[movies]

Be careful you don't equate that to the picture drawn however. In the system mentioned the two resonance forms are not the same.

Very true.  I was addressing the original question concerning the parent compounds.

[english]

P.S. Are there only 2 resonance structures for an imidazolium ion?

Far more than 2.  Most of them are insignificant.   

My question is: doesn't placing a positive charge on an electronegative atom like nitrogen energetically unfavorable?

Nitrogen can deal with a +1 charge, more so than oxygen, because it's more electropositive than oxygen.


Placing a +1 charge on N or O does not necessarily lead to enough instability to give really good reasoning behind lower basicity.

In these cyclic pi systems, aromaticity rules.

[edwinksl]

Spreading out the positive charge has a huge effect on stabilizing the protonated intermediate.  As you pointed out, having a positive charge on an electronegative atom seems like a bad thing (however, it is favorable due to electrostatic attraction between a + charge and a - charge).  Since the two resonance structures of protonated imidazole are identical, the two contribute equally to the real structure.  In essence, you get half a + charge on each nitrogen and therefore the electronegative atom has less positive charge.  Since pyridine can't do that, the lone nitrogen bears almost all of the + charge.

Also, there are other resonance structures for both imidazole and pyridine which have a + charge on a carbon atom, but these are not significant contributors to the overall structure.

I believe you can say in an imidazolium ion, each nitrogen atom effectively has a +1/2 charge.

However, in a pyridinium ion, I don't think most of the positive charge resides on nitrogen. You have 5 other carbon atoms, which are more electropositive than nitrogen. So, placing a positive charge on carbon is more energetically favored than placing on nitrogen. So, I am inclined to think that most of the positive charge is shared by the 5 carbon atoms, and there is only little positive charge on the nitrogen atom which is potentially lesser than +1/2.

Therefore, isn't a pyridinium ion more stable than an imidazolium ion, thus making pyridine more basic than imidazole? I am sure I am missing something here, since literature says imidazole is more basic than pyridine.

[edwinksl]

Spreading out the positive charge has a huge effect on stabilizing the protonated intermediate.  As you pointed out, having a positive charge on an electronegative atom seems like a bad thing (however, it is favorable due to electrostatic attraction between a + charge and a - charge).  Since the two resonance structures of protonated imidazole are identical, the two contribute equally to the real structure.  In essence, you get half a + charge on each nitrogen and therefore the electronegative atom has less positive charge.  Since pyridine can't do that, the lone nitrogen bears almost all of the + charge.

Also, there are other resonance structures for both imidazole and pyridine which have a + charge on a carbon atom, but these are not significant contributors to the overall structure.

One more thing to add: It's true an electrostatic attraction between a positive charge and a negative charge is favored, e.g. between protons and bonding electrons in a chemical bond.

However, when you place a positive charge on an electronegative atom, we are not talking about any attraction between a positive charge and negative charge. By giving an electronegative atom a formal positive charge, you are only simply saying there are more protons than electrons (so the atom clearly has lost some valence electrons). A negative charge is more stable on a more electropositive atom because such an atom has a higher effective nuclear charge to attract and stabilize the incoming additional electron that gives the atom a negative charge. So, I fail to see how placing a positive charge on an electronegative atom is favored because of electrostatic attraction between a positive charge and a negative charge: there is no such electrostatic attraction.

[PRIYA1022]

Since the two resonance structures of protonated imidazole are identical, the two contribute equally to the real structure.  In essence, you get half a + charge on each nitrogen and therefore the electronegative atom has less positive charge.  Since pyridine can't do that, the lone nitrogen bears almost all of the + charge.
 Movies.....This is absolutely correct with respect to the parent imidazole and pyridine..but  in substituted imidazoles( as posted by edwinks) and pyridines, it would still be the IMIDAZOLE that is more basic, and this is because of the 2 nitrogen atoms accommodating the positive charge.. Am I right?
Edwinks post So, I fail to see how placing a positive charge on an electronegative atom is favored kept me thinking about why it is easy for a nitrogen to acquire a positive charge easily.Is it because of the ONE lone pair of electrons that it has?...Just a guess..

[edwinksl]

I think I have a plausible answer.

In the imidazolium ion, the 2 resonance structures are actually stable, despite the positive charges on the 2 N atoms, because all the C and N atoms have an octet structure (the octet rule generally applies only to elements in the 2nd period).

On the other hand, in the pyrindium ion, although the ion has its positive charge spread over the 5 relatively more electropositive C atoms, each of its resonance structure has the positively charged carbon lacking an octet structure.

So, an imidazolium ion is more stable as the additional bonding you get from obeying the octet rule outweighs the destabilization caused by putting the positive charge on a relatively electronegative N.

Case in point: Look at the resonance structures for carbon monoxide at http://www.cem.msu.edu/~reusch/VirtTxtJml/intro3.htm#strc7

The octet rule in this case outweighs the charge separation for carbon monoxide.

[PRIYA1022]

Yes... Sounds reasonable!!

[movies]

Ha!  You answered your own question before I got to respond!  Yes, you are absolutely correct regarding the violation of the octet rule with the carbocation resonance structures.

On to the electrostatic interaction.  What I was referring to by a favorable electrostatic interaction is that between the lone pair of a nitrogen and the proton that will stick to it.  Once you have protonated the nitrogen, the electrostatic interaction doesn't go away, it's still there holding the proton on the nitrogen.  The formal positive charge that you get is really a consequence of the electrostatic interaction, not the reason for the electrostatic interaction.  Does that make sense?