Hybrid Structure of Phenol

Jun 2013
2
0
I am trying to understand the structure of Phenol.
As mentioned in my text book, Phenol is supposed to exist as a resonance hybrid of different structures.
However, i am unable to understand the structures.

i have attached an image.

Please help me with the following questions.

1. Do the arrows indicate that electrons are moving from one atom to the other.
So, in structure I, lone pair of electrons move from oxygen to Carbon. However, what does the second blue arrow indicate in Structure I.

2. How did the double bond in structure I Shift from C = C to C = O in structure II.

3. How did the carbon (marked in red) in structure II develop a negative charge.

4. Why didn't the carbon at the Meta Position in Structure III develop a negative charge and why did the carbon at para position develop a negative charge.


Apologies for any silly question.
Thank you
 

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Unknown008

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1. Yes, electrons are moving from the oxygen to the C-O bond, this makes the bond become C=O (question 2).

The second blue arrow is again the movement of electrons. Since carbon is very unstable with 5 electron pairs around it, some of the electrons (in the next double bond) go to the next carbon atom, thereby disrupting the C=C bond and leave behind a C-C bond (question 3).

4. I'm not too sure about this one, but I guess one could say that the electrons do not easily get into the single C-C bond and the adjacent electrons in the C=C bond easily go away (since two of the electrons do not form strong bonds). This implies that the carbon atom at the para position experiences only a small and temporary negative charge (smaller than the carbon at meta position) which goes away quickly because the electrons to its left quickly go into the carbon at the meta position.

It's a bit like... I don't know... it's "difficult" to fill a glass of water until it's full (you need energy to raise water up to its mouth and pour, here it's analogous to putting electrons in a single bond), but it's much easier for the water in the glass to overflow from the same glass when you're pouring more water (electrons move away from the next double bond).

Personally, I think of the activation of carbon atoms like below:



Basically, oxygen is an electronegative atom and will pull electrons from the electron rich ring. This causes two electrons to move up and form a C=O bond. The C atom there becomes positive due to oxygen pulling electrons from it. Immediately, the adjacent carbon atoms respond by taking negative, positive, negative, etc charges.
 
Jun 2013
2
0
Thank you unknown008 for your help

I have few more questions.

In Halogenation of Phenol, Bromination - When phenol is treated with bromine at low temperature in the presence of a solvent, such as Carbon disulphide or Chloroform,a mixture of o - bromophenol and p - bromophenol is obtained.

1. I want to know why isnt m - bromophenol formed?

Is it because of the same reason as mentioned in point 4. - the electrons do not easily get into the single C-C bond and the adjacent electrons in the C=C bond easily go away (since two of the electrons do not form strong bonds). This implies that the carbon atom at the para position experiences only a small and temporary negative charge (smaller than the carbon at meta position) which goes away quickly because the electrons to its left quickly go into the carbon at the meta position.

2. Even in Nitration of Phenol, and Sulphonation of Phenol, only ortho and para derivatives are obtained and no meta derivative is obtained.

Thank you for your timely help
 

Unknown008

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Yes, the bromine, the nitronium ion and the sulfur trioxide are electrophiles (they like electrons and negatively charged atoms/ions) and in the diagram I posted, you clearly see that only the positions o and p are vulnerable to attack since they are negative. That's why I really like that diagram :)
 
Jun 2013
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Personally, I think of the activation of carbon atoms like below:



Basically, oxygen is an electronegative atom and will pull electrons from the electron rich ring. This causes two electrons to move up and form a C=O bond. The C atom there becomes positive due to oxygen pulling electrons from it. Immediately, the adjacent carbon atoms respond by taking negative, positive, negative, etc charges.
There's a bit more to it than this.
There are two effects going on - mesomeric and inductive. You've described inductive effect (due to electronegativity) but actually mesomeric is more powerful for OH (due to back donation of PI electrons) causing OH to be electron donating to the ring. If it wasn't, then the meta position would be more nucleophilic rather than the ortho/para seen for phenol. NO2 is a good example of an electron withdrawing group which causes substitution at the meta position.
To understand why electron donating groups cause ortho and para substitution and why electron withdrawing groups favour meta substitution you'll need to read up a wee bit on resonance stabilisation of benzene/aromatic rings. Any general undergraduate organic textbook will give you a really good introduction to this.
It's not easy to describe on a forum.