Take the 2-minute tour ×
Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers and students. It's 100% free, no registration required.

Let's consider the five-atom ring part of the histidine molecule.

enter image description here

I was wondering why only one of the lone pairs on nitrogen is involved in the delocalized pi system. Specifically, I was wondering why it is that lone pair in particular instead of the other lone pair (there are two lone pairs on nitrogen here if we draw out the Lewis structure before looking for resonance). Picture below.

I've taken the liberty to label the two lone pairs on nitrogen as A and B.

Here is my conjecture about why B's lone pair is delocalized while A's lone pair cannot be.

1) If we delocalize A, we'd get extreme ring strain; we'd make the geometry around that nitrogen $\ce{sp}$. This implies a $120^o$ bond angle. That's problematic.

2) Effective pi-orbital overlap (conjugation) only occurs when the p-lobes are planar - i.e. parallel with each other. If nitrogen with the B lone pair kept its B lone pair, then that nitrogen would be tetrahedral and that would distort the planarity of the molecule; everything else in the molecule appears to be $\ce{sp^2}$ hybridized.

3) We also run into formal charge issues; we get a positive formal charge on the "A" nitrogen and a negative formal charge on the B nitrogen. While this isn't a dealbreaker, it's not preferable, as there is separation of like formal charges.

4) Anything I miss? Any points I mangle?

share|improve this question
add comment

2 Answers 2

up vote 6 down vote accepted

Obs: since you deleted your picture, I'll add this structural diagram to help point out to which nitrogen atoms I am referring to.

enter image description here

The correct answer to this question is more direct, and not listed in your items. The non-bonding electron pair in nitrogen $3$ is in an orbital perpendicular to the $\pi$-bonding $p$ orbitals of all other atoms in the imidazole ring. Thus, it does not have the appropriate geometry to overlap with other orbitals forming $\pi$-bonds, and does not participate in any resonance effects.

This effect of orbital geometry is often invoked to explain the difference in basicity between pyridine and pyrrole. In the former, the nitrogen lone pair is of type $3$ and does not participate in resonance, and so it has a basicity comparable to regular amines and imines. However, in the latter, the lone pair is of type $1$ and does participate in the aromaticity of the ring. Protonation of the nitrogen atom would require localization of the electron pair on the nitrogen, breaking most of the resonance effect. This is energetically unfavourable, so pyrrole displays a much lower basicity than expected for amines or imines.

This section on pyridine has a little more information on why the type A lone pair does not delocalize, including a helpful picture. Notice that the lone pair is in an $sp^2$ orbital parallel to the plane of the ring, while the $\pi$-bonded system is perpendicular.

Edit: In response to the comments, I can make the following observations on your conjectures:

1) I don't think this applies, actually. The fact that orbital overlap is disallowed in the first place precludes the possibility of even thinking what would happen to the geometry of the molecule if the electron pair in nitrogen $3$ formed a $\pi$-bond; it just doesn't.

However, if the molecule had a boron atom instead of carbon in position $2$ (with an empty $sp^2$ orbital pointing in the same direction as the full $sp^2$ orbital in nitrogen $3$), then I'm not sure what would happen. A more simple example of this type of structure would be 1,2-azaborine (of which only the 1,2-dihydro derivative seems to have been isolated as of yet).

I could imagine canonical structures in which the electron pair on nitrogen links with the boron atom, forming an additional bond (analogous to benzyne, which is isoelectronic with 1,2-azaborine). However, to form an extra N-B $\pi$-bond, then either the bond will be a $\pi\ (sp^2-sp^2)$ bond (which is unusual as far as I know), or both the nitrogen and boron atoms would have to partially rehybridize into a mix of $sp^2$ and $sp$ configurations to allow a $\pi\ (p-p)$ bond, which would likely imply an increase in bond angle and a huge increase in ring strain. Neither option seems particularly good, so I suspect such canonical structures would contribute little to the actual bonding picture in the boron-substituted ring.

2) The nature of the nitrogen atom which participates in resonance is actually something between $sp^2$ and $sp^3$. The concept of hybridization isn't very adept at explaining resonance naturally.

I'm also not happy with how much my arguments rely on hybridization, but I don't know other effective means to explain what's going on.

share|improve this answer
    
Thank you for reading over my items. I assume that you have at least found my items to be factually accurate if only tangentially related to the reason one lone pair is delocalized while the other is not. –  Dissenter May 18 at 15:18
    
Also is my formal charge analysis correct? I appear to have a residual positive formal charge in my second picture for whatever reason. –  Dissenter May 18 at 15:19
    
Edit nvm I had nitrogen forming five bonds (not good!). Removed Texas nitrogen picture. –  Dissenter May 18 at 15:32
    
Also I consulted two organic chemistry textbooks and they agree with you. On one nitrogen, the lone pair is in an sp2 orbital pointing away from the ring. It's not possible for this lone pair to participate in the orthrogonal p-orbitals. However, regarding the other nitrogen, that nitrogen 1) cannot have sp3 hybridization because that would screw around with the planarity. Also, none of the single bonds on that nitrogen can be part of any pi system; single bonds are by definition sigma bonds. So the lone pair has to be a pi lone pair. 4n+2 and we get 6; Huckel's rule; we have resonance. –  Dissenter May 18 at 15:41
    
And you're right; protonation can destroy the conjugated pi system. Protonate one nitrogen and you make it sp3, destroying the planarity of the molecule. Protonate the other nitrogen, no worries, because that nitrogen stays sp2 hybridized and thus planar. –  Dissenter May 18 at 15:43
show 1 more comment

You have a problem with:

1) If we conjugate A, we'd get extreme ring strain; we'd make the geometry around that nitrogen $sp^2$. This implies a 180° bond angle.

If the nitrogen is $sp^2$, you get 120° bond angles, not 180°. For 180° bond angles, you would need $sp$ hybridization.

Actually, both A and B are conjugated. Both atoms are $sp^2$ hybridized and both have a p orbital used for pi bonding in the ring.

Added:
Here is the bonding in pyrole. As you will notice, the nitrogen is $sp^2$ hybridized and the lone pair occupies the p orbital for pi bonding in the ring. In pyridine, the nitrogen is again $sp^2$ hybridized but the lone pair sticks out from the ring. These are the 2 nitrogen configurations in imidazole.

share|improve this answer
    
Edited. I meant sp and that's what the nitrogen would be. Also looks like I confused delocalization of electrons with conjugation. –  Dissenter May 18 at 6:07
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.