[B.W. McCann]

Hello Everyone.  

An organic chemistry question has been on my mind.  


Whenever an ion is capable of resonance, the resonance structures are drawn with the positive or negative charge be distributed to *every other* atom.  I'm told that the actual structure isn't represented by the resonance structures that are depicted in the texts, but that the actual structure is a resonance hybrid of them all.  

So, my question is this.  If the actual structure is a resonace of all the possible formations, then why do ortho~para directors direct reactions to take place on the ortho and para atoms?  If the actual structure is a resonance hybrid, then wouldn't the probability of reaction be equal for all the atoms?  

I hope I'm making sense.  Please ask for clarification if necessary.


Thanks y'all.


Billy Wayne  

[Yggdrasil]

You're on to something when you see that the positive/negative charges are distributed to every other atom.  For an electron-donating substituent, the delocalized electrons add electron density to every other atom.  Therefore, the ortho and para positions gain electron density and are better nucleophiles, while the meta positions do not gain significant electron density.

[mir]

Other thingsaffects the propotions of para and ortho substituent products too:
Field effects and Inductive forces

Another question about the (Valence Bond) resonance model:

Why are benzene more stable than 1,3-cyclobutadiene, when they have the same amount of resonance structures? :-)

[Yggdrasil]

Explaining why cyclobutadiene is very unstable requires a more complex model than resonance structures.  The effect is one which can be best explained using molecular orbital theory.  Assuming the cyclobutadiene is planar, it should form an aromatic system because the four pi orbitals on each carbon will be conjugated.  By symmetry, these four pi orbitals will form four molecular orbitals (a bonding orbital, two degenerate non-bonding orbitals, and an anti-bonding orbital).  Since each pi orbital contributes one electron, the aromatic system will have four electrons total.  This means two will go into the bonding orbital and the other two will each occupy one of the degenerate nonbonding orbitals.  However, this results in a diradical species (i.e. it has two unpaired electrons).  This configuration is very unstable and will result in cyclobutadiene adopting a distorted geometry so that the pi orbitals are no longer planar and cannot form an anti-aromatic system.

When one carries out a similar analysis with benzene, one finds that there are no non-bonding orbitals and that all electrons are paired and occupy bonding orbitals.  Therefore, benzene is aromatic and stable.

[B.W. McCann]

You're on to something when you see that the positive/negative charges are distributed to every other atom.  For an electron-donating substituent, the delocalized electrons add electron density to every other atom.  Therefore, the ortho and para positions gain electron density and are better nucleophiles, while the meta positions do not gain significant electron density.

Yggdrasil,


Thank you for that clear and cogent response.  

If I could impose upon you just a little farther, why is it that *only* the ortho~para positions gain electron density?  

In other words, why is electron density increased at *every other* atom and not *every* atom?  

Is an understanding of electronic orbitals necessary to understand this?


Thanks for your (astute) attention,

Billy Wayne

[movies]

If I recall correctly, all the positions are activated relative to benzene when an inductive donating group is involved, however, the ortho and para positions are much more activated.  It all has to do with how the orbitals align themselves for optimal overlap.

[Yggdrasil]

See Pic

[mir]

For this image of ressonance to work, the p orbital of oxygen must be in the plane of the rest of the benzen orbitals, right? That means, phenol is planar molecule...

[Yggdrasil]

Yes, phenol is a planar molecule.