[ubcdumbass]

I've been researching and learning and really struggle to understand...
Why is the lone pair on N from NH in imidazole (pyrrole like N) able to contribute pi electrons but the other lone pair on N (pyridine-like) cannot?
The pyrrole-like N is sp2 hybridized while the pyridine-like N is also sp2 hybridized.
It says that pyridine-like N already has contributed to pi electrons with a Carbon, but doesn't it have another pair of lone electrons it can contribute?
Additionally, why does the O in furan only contribute one pair towards the conjugated pi system?

Thank you so much.

[mjc123]

Try drawing out the structure including the lone pair orbitals. Can you see which orbitals can participate in pi-bonding, and which can't?

[Enthalpy]

Why is the lone pair on N from NH in imidazole (pyrrole like N) able to contribute pi electrons but the other lone pair on N (pyridine-like) cannot?

Neither do I understand. I expect the opposite: -N= to contribute pi electrons and -NH- not.

[mjc123]

See diagram. The pyridine-like N contributes 1 electron to the aromatic sextet in a p_π orbital. The lone pair is in an sp² orbital orthogonal to the pi orbitals, and cannot contribute. The pyrrole-like N contributes 2 electrons to the sextet; its "lone pair" is in the p_π orbital, while the N-H bond pair is in the orthogonal sp² orbital.

[kriggy]

Why is the lone pair on N from NH in imidazole (pyrrole like N) able to contribute pi electrons but the other lone pair on N (pyridine-like) cannot?

Neither do I understand. I expect the opposite: -N= to contribute pi electrons and -NH- not.

Think about this way

[mjc123]

That is an interesting way of looking at it, but it doesn't bring out the key point that the pyridine lone pair has the wrong symmetry to contribute to the pi system. In the right-hand structure for pyridine, if it were possible, the double bond to the left of N must lie in the plane of the ring, and cannot contribute to the aromaticity. (Eight pi electrons would not be aromatic anyway.)

Also, the left-hand structure for pyrrole is not sp³ hybridised - pyrrole is known to be planar about the N. It is sp² with the lone pair in a p_z orbital. In fact, I think it is a general rule for drawing resonance structures, that you are not allowed to change the hybridisation between contributing structures. For the neutral structure, there may be an energy cost for changing tetrahedral sp³ N to planar sp² N, but it is compensated for by the fact that the latter can be stabilised by resonance; hence the molecule is planar.

[kriggy]

That is an interesting way of looking at it, but it doesn't bring out the key point that the pyridine lone pair has the wrong symmetry to contribute to the pi system. In the right-hand structure for pyridine, if it were possible, the double bond to the left of N must lie in the plane of the ring, and cannot contribute to the aromaticity. (Eight pi electrons would not be aromatic anyway.)

Also, the left-hand structure for pyrrole is not sp³ hybridised - pyrrole is known to be planar about the N. It is sp² with the lone pair in a p_z orbital. In fact, I think it is a general rule for drawing resonance structures, that you are not allowed to change the hybridisation between contributing structures. For the neutral structure, there may be an energy cost for changing tetrahedral sp³ N to planar sp² N, but it is compensated for by the fact that the latter can be stabilised by resonance; hence the molecule is planar.

Ok, the pyrrole explanation is not good thx for pointing it out clearly for me. 

Ive seen the pyridine explanation in "Art of writing reasonable reaction mechanism" by Grossman, page 14. Might not be perfect but easily explains why the lone pair doesnt contribute.

[Enthalpy]

Right, the strong dipole moment tells that the shared electrons come from -NH- rather than -N=. I had been fooled by oversimplistic ideas of valence and bond counts. Thanks!