Achiever Test 2 OC Question

[capnamerica]

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Why is phenol a stronger acid than simpler alcohols? I understand that the charge on the conjugate base gets delocalized through resonance, but wouldn't this disrupt the existing aromatic ring by violating the requirement of an available p orbital on the C bonded to the O or Huckel's rule?

 

[DrNutmeg]

You're right. The charge can be more delocalized via resonance (there are a ton of resonance structures!).

Here is where you are confused (I think): The electrons forming the negative charge on the resulting conjugate base (after the hydrogen is removed) aren't in a p orbital! Therefore it won't disrupt the aromaticity.

Where are the electrons then? They are in an sp3 orbital, because the Oxygen atom was sp3 hybridized before the hydrogen was removed. The electrons actually don't move to a p orbital for exactly the reason you mentioned! it will disrupt aromaticity!

 

[capnamerica]

You're right. The charge can be more delocalized via resonance (there are a ton of resonance structures!).

Here is where you are confused (I think): The electrons forming the negative charge on the resulting conjugate base (after the hydrogen is removed) aren't in a p orbital! Therefore it won't disrupt the aromaticity.

Where are the electrons then? They are in an sp3 orbital, because the Oxygen atom was sp3 hybridized before the hydrogen was removed. The electrons actually don't move to a p orbital for exactly the reason you mentioned! it will disrupt aromaticity!

Hmm, so then how do resonance structures with a C=O bond form? I was under the assumption that the electrons forming the negative charge form a pi bond (p orbital) with the C. I think I'm missing something here.

 

[DrNutmeg]

Hmm, so then how do resonance structures with a C=O bond form? I was under the assumption that the electrons forming the negative charge form a pi bond (p orbital) with the C. I think I'm missing something here.

 

[capnamerica]

Yupp. I understand this part as well. I guess my question is whether or not the resonance structures with the lone pair being on a C in the ring are aromatic/anti-aromatic/non-aromatic. In my mind, Huckel's rule is being violated in such resonance forms.

 

[DrNutmeg]

Here is a good read:
http://chemwiki.ucdavis.edu/Organic..._on_resonance_effects_on_acidity_and_basicity

Your question still stands though. I am more unsure now than I was when I first answered lol. Here are two of my theories, though:

1) Huckel's rule is not violated because the ring is no longer planar in the resonance structures (they can make chair formations to avoid anti-aromaticity)

2) It doesn't matter how unstable the resonance structures are; it only matters that there are simply more locations that the negative charge can be moved to.

Hope this helps.