[davidenarb]

I would like to understand the aromaticity of the below compound: There is two conditions that must be met.
Condition 1: presence of a continuous p orbitals system.
Condition 2: the total number of electrons that are in the p orbitals must be a Huckel number.

The second condition is satisfied (6=4(1)+2).
I have difficulty to understand why does the first condition satisfy the requirement.

To do so, I need to be able to see a continuous p orbital system. The four carbons that are in double bond have p orbitals because they are participating in the 2 pi bonds   

We know that whenever a lone pair participates in resonance, it will occupy a p orbital rather than a hybridized-orbital.

However, only a carbocation possesses a an empty orbital not an anion, so how can this lone pair be in p orbital?
Where is the continuous p orbital system?

[Corribus]

The two pi-bonds are only there because you've drawn them there. In an idealized case, every carbon atom is identical and has an empty p-orbital to participate - until you fill the system's electrons in, of course.

[clarkstill]

The molecule has two choices:

i) sp3 hybridization: in an isolated anion (e.g. CH3-) this is favoured since it allows minimal repulsion between lone pairs and bonding pairs (in simple terms, by VSEPR).

ii) sp2 hybridization with a filled p-orbital: this would never happen in CH3-, since electrostatic repulsion can be reduced by forming tetrahedral bond angles.  However, in this case the energy penalty for forcing bonds into closer proximity is more than offset by the energy gain of the filled lone pair in the p-orbital contributing to an aromatic system.

So overall (ii) is favoured, since aromaticity trumps increased repulsion due to sp2 hybridization.

PS everyone feel free to criticize my abandonment of MO theory and quantum mechanics in this explanation, but life is just too short.

[davidenarb]

The two pi-bonds are only there because you've drawn them there. In an idealized case, every carbon atom is identical and has an empty p-orbital to participate - until you fill the system's electrons in, of course.

Do you mean resonance? because if you mean so, I would like just to say that resonance is not a physical process, so the compound either has the 5 p orbitals or not. (sorry but I still can't see the five p orbitals)

[Corribus]

You can't see that if each carbon has a free p-orbital, and there are five carbons, then there are five p-orbitals? Forget the electrons, for the moment. The concept of aromaticity is founded from Huckel theory, which is an LCAO-MO treatment. In this kind of theoretical model, you build the molecular orbitals from individual atomic orbitals, and then fill all the electrons in at the end.

[spirochete]

Understanding hybridization in resonance stabilized molecules can be tricky at first. The first thing you should do is draw all of the significant resonance contributors, even minor ones. In this case there are no minor structures because all of the resonance structures are perfectly equivalent, but minor structures are important in other molecules.

What's important is that a carbon atom is making a pi bond in at least one of the significant structures. We know that resonance, and especially aromaticity is stabilizing so electrons will be delocalized by resonance if possible. P orbitals and P orbital overlap are required for delocalization. A sigma bond, on the other hand, can be formed by overlap of an type of hybrid orbitals: SP3, SP2 or SP. So if an atom "looks" SP2 hybridized in one resonance structure and SP3 in another structure, it's almost always going to actually choose SP2 hybridization.

Another good example of this is the SP2 hybridization and planar geometry of amides.

[davidenarb]

So if an atom "looks" SP2 hybridized in one resonance structure and SP3 in another structure, it's almost always going to actually choose SP2 hybridization.

Can I understand just why is that?

Let's take my example. We have 3 sigma bonds (2 with C and 1 with H) and one lone pair making Sp3 hybridized carbon! where is my mistake?

In my textbook, the author said "The carbon atom has a delocalized lone pair, and it therefore occupies a p orbital (rather than a hybridized orbital), so that it can overlap with the p orbitals of the pi bonds" 

But where does this p orbital came from?!!!! (if it is from resonance, I would say that resonance is not a physical process like a a propoer chemical reaction; it's different from mechanism)

[spirochete]

So if an atom "looks" SP2 hybridized in one resonance structure and SP3 in another structure, it's almost always going to actually choose SP2 hybridization.

Can I understand just why is that?

Let's take my example. We have 3 sigma bonds (2 with C and 1 with H) and one lone pair making Sp3 hybridized carbon! where is my mistake?

In my textbook, the author said "The carbon atom has a delocalized lone pair, and it therefore occupies a p orbital (rather than a hybridized orbital), so that it can overlap with the p orbitals of the pi bonds" 

But where does this p orbital came from?!!!! (if it is from resonance, I would say that resonance is not a physical process like a a propoer chemical reaction; it's different from mechanism)

It doesn't "come from" resonance, the atoms have a hybridization that allows for a larger delocalized pi system, aka resonance. Yes we all agree that resonance is not the same as a chemical reaction and none of the arguments made by anyone in the thread are inconsistent with what resonance is.

Like I said before you can't just look at one resonance structure you have to look at all the resonance structures.

[davidenarb]

So if an atom "looks" SP2 hybridized in one resonance structure and SP3 in another structure, it's almost always going to actually choose SP2 hybridization.

Can I understand just why is that?

Let's take my example. We have 3 sigma bonds (2 with C and 1 with H) and one lone pair making Sp3 hybridized carbon! where is my mistake?

In my textbook, the author said "The carbon atom has a delocalized lone pair, and it therefore occupies a p orbital (rather than a hybridized orbital), so that it can overlap with the p orbitals of the pi bonds" 

But where does this p orbital came from?!!!! (if it is from resonance, I would say that resonance is not a physical process like a a propoer chemical reaction; it's different from mechanism)

It doesn't "come from" resonance, the atoms have a hybridization that allows for a larger delocalized pi system, aka resonance. Yes we all agree that resonance is not the same as a chemical reaction and none of the arguments made by anyone in the thread are inconsistent with what resonance is.

Like I said before you can't just look at one resonance structure you have to look at all the resonance structures.

I think I understand it please check whether my thinking is logical:

In general situations, we should say that the carbanion atom is sp3 hybridized, but as a result of the presence of pi bond, the lone pair of the carbanion will participate in resonance so that the compound will be more stable. the electrons participating in resonance are in the p orbitals of the pi bonds, and in order for the lone pair to be delocalized, they must also be in the same orbital: the p orbital. This is why the carbanion will have an sp2 orbital producing thereby a p orbital that allow the lone pair to participate in resonance.

[Corribus]

At the risk of confusing you further, it's important to understand that the orbitals involved in pi-delocalization are no longer p-orbitals. They are molecular orbitals derived (to an approximation) from carbon atom p-orbitals. Each carbon has a p-orbital perpendicular to the plan of the molecule. Those p-orbitals all overlap with each other. So what you do is combine them all into large, delocalized pi-orbitals. Then you look at how many electrons you have, and fill them all in just like you would in a solo atom. Whether or not there are two electrons, one electron, or no electrons in the p-orbitals that make up the molecular orbitals makes no difference in this model. The electrons are "removed" from the system and then filled back in only after the molecular orbitals are constructed.

The remaining s and p orbitals on each carbon are then free to hybridize and bond with each other, forming the sigma-bond framework of the molecule.

[davidenarb]

Thank you for your details.

At the risk of confusing you further, it's important to understand that the orbitals involved in pi-delocalization are no longer p-orbitals. They are molecular orbitals derived (to an approximation) from carbon atom p-orbitals. Each carbon has a p-orbital perpendicular to the plan of the molecule. Those p-orbitals all overlap with each other. So what you do is combine them all into large, delocalized pi-orbitals. Then you look at how many electrons you have, and fill them all in just like you would in a solo atom. Whether or not there are two electrons, one electron, or no electrons in the p-orbitals that make up the molecular orbitals makes no difference in this model. The electrons are "removed" from the system and then filled back in only after the molecular orbitals are constructed.

The remaining s and p orbitals on each carbon are then free to hybridize and bond with each other, forming the sigma-bond framework of the molecule.

Can I have a confirmation about my previous response please.

I think I understand it please check whether my thinking is logical:

In general situations, we should say that the carbanion atom is sp3 hybridized, but as a result of the presence of pi bond, the lone pair of the carbanion will participate in resonance so that the compound will be more stable. the electrons participating in resonance are in the p orbitals of the pi bonds, and in order for the lone pair to be delocalized, they must also be in the same orbital: the p orbital. This is why the carbanion will have an sp2 orbital producing thereby a p orbital that allow the lone pair to participate in resonance.

[Corribus]

More or less, but as I said some of the language you are using is not very rigorous.

[zsinger]

Orbitals are NOT REAL THINGS.  They are simply mathematical functions to describe a probability.
            -Zack

[Enthalpy]

Orbitals are NOT REAL THINGS.  They are simply mathematical functions to describe a probability.
            -Zack

Why do you express opinions about topics you ignore?

[zsinger]

Pretty sure thats a fact, but I was only trying to help.  My bad guys.
              -Zack