why is furan aromatic??

[Kneecoal]

even though it's cyclic and conjugated, i'm counting 8 pi electrons. am i somehow counting wrong?

for reference, it's 2009 destroyer ochem #87

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[Kneecoal]

oh, i just looked it up on wiki - so you don't count the extra lone pair of O's electrons?

 

[failedmagician]

No, only one of the two electron pairs is counted.

 

[ethanyoo726]

You have to also remember that it is planar. I see why the 2 lone pairs of electrons around the Oxygen is confusing. Only 1 pair out of the 2 contributes to the pi electrons. That one pair is delocalized around the entire structure, while the other pair always stays with the oxygen. So there are only 6 pi electrons. So it is aromatic.

Click on this link to view the resonance. It explains what I'm talking about.
http://upload.wikimedia.org/wikipedia/commons/archive/8/86/20090510170215!Furan_Resonance.png

Hope that helps.

 

[krazchikin]

The way i think about it is that the Pi electrons of the double bonds, which we know participate in the conjugated ring system, are above and below the plane of the bond. Although you have 2 lone pairs of electrons on the hydrogen, only one of them participate in the pi system because the oxygen atom is sp3 hybridized (one pair located directly above the plane of the oxygen and the other on one of the vertices of the tetrahedron). I'm not sure if that's how it is, but it seems to work...lol....

but another tricky one is pyridine, which is a six membered ring system enclosing a nitrogen atom with 3 double bonds (wiki for drawing). The nitrogen has one pair of lone pair electrons and a double bond on one of the carbons it is bonded to. At first look, it seems as though there are 8 pi electrons, which would make it anti-aromatic, but in fact, it actually is aromatic....the lone pair electrons do not participate in the ring system because they are sp2 hybridized, in which the atoms/electrons bonded to nitrogen are roughly 120 degrees (angle is slightly skewed due to lone pair) and planar....remember that the pi electrons of the double bonds are above and below the ring....hope this makes sense and helps!

 

[cyrano]

The way i think about it is that the Pi electrons of the double bonds, which we know participate in the conjugated ring system, are above and below the plane of the bond. Although you have 2 lone pairs of electrons on the hydrogen, only one of them participate in the pi system because the oxygen atom is sp3 hybridized (one pair located directly above the plane of the oxygen and the other on one of the vertices of the tetrahedron). I'm not sure if that's how it is, but it seems to work...lol....

but another tricky one is pyridine, which is a six membered ring system enclosing a nitrogen atom with 3 double bonds (wiki for drawing). The nitrogen has one pair of lone pair electrons and a double bond on one of the carbons it is bonded to. At first look, it seems as though there are 8 pi electrons, which would make it anti-aromatic, but in fact, it actually is aromatic....the lone pair electrons do not participate in the ring system because they are sp2 hybridized, in which the atoms/electrons bonded to nitrogen are roughly 120 degrees (angle is slightly skewed due to lone pair) and planar....remember that the pi electrons of the double bonds are above and below the ring....hope this makes sense and helps!

Aromatic:
1) Planar: To be planar all the sigma bonds and non participating electron pairs must be in sp2 hybridized orbitals with the remaining P orbital (Py) containing one of oxygen's lone pairs.
2) 4n+2 conjugated pi system in the p orbitals. This is done, as you can see, by having the Py orbital inhabited by one of the lone pairs on oxygen. It can now participate in pi bonding along with the other electrons on the carbon atoms in their Py orbitals.

 

[ah7391]

I'm a bit confused because isn't the oxygen sp3 hybridized? Therefore aren't the two lone pairs pointing "inside the board" and "towards us" since the two bonds to carbon are in the "plane of the board?"

 

[Troyvdg]

I'm a bit confused because isn't the oxygen sp3 hybridized? Therefore aren't the two lone pairs pointing "inside the board" and "towards us" since the two bonds to carbon are in the "plane of the board?"

its sp2 hybridized because it's second lone pair is delocalized through an empty p orbital👍