How does charge delocalization affect phenoxide anion stability?

[ephemeral1]

Why does delocalization of the negative charge over the ortho and para positions of aromatic rings increase the stability of the phenoxide anion? Why do the ortho and para positions increase the stability but not the meta position? Please explain. Thank you.

 

[sjb-2812]

Can you draw resonance forms of the phenoxide ion?

 

[Char. Limit]

You see, the ortho and para forms allow the resonance to "spread out" the charge to the O- ion at the end, while the meta position is placed just so that it cannot achieve this.

 

[chemisttree]

Of course noting that the negative charge is more delocalized doesn't really explain why it is more stable!

Read about the "particle in a box" for an explanation. Consider a particle like an electron confined in a region of dimension L. The energy of that electron is expressed as a function of dimension L, or length, that the particle can move. If the electron is allowed to move more freely (L increases), it has a lower energy since http://en.wikipedia.org/wiki/Particle_in_a_box#cite_note-Davies5-3" The particle in a box is a one dimensional treatment but this is used as an approximation of the delocation length of an individual electron.

 

[clustro]

http://chemistry.boisestate.edu/people/richardbanks/organic/phenol.gif

Google is just lovely at times.

In general, the as the number of feasible resonance structures increases, the more stable the molecule is, for the reasons chemisttree stated.

 

[Char. Limit]

Nice, clustro.