I'm working with a certain molecule, and could see that it was only on the ether group that it was possible to make these resonance structures. First, I'd like to know if the structures are correct and second how does it effect the molecule that this particular group can make these resonance structure. Does it contribute to its stability?


  • $\begingroup$ Some arrows are missing and others are drawn poorly but otherwise they are correct. $\endgroup$ – bon Dec 18 '16 at 14:23
  • $\begingroup$ It's just a sketch -I'll draw a better one . Can you point out which arrows are missing? $\endgroup$ – sunshine257 Dec 18 '16 at 14:33
  • $\begingroup$ I'm not sure what "It's just a sketch" means as a response here. If you're drawing resonance structures, you should either have no arrows because you think it's obvious, or you should have all arrows. The intermediate case isn't so good. The third structure is missing one arrow. In the second structure, the arrow at the 5 o'clock position should point to the carbon atom, not to the bond. $\endgroup$ – Zhe Dec 18 '16 at 15:05
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    $\begingroup$ You should use mesomery arrows ($\ce{<->}$) and not equilibrium arrows ($\ce{<=>}$) for resonance structures. $\endgroup$ – Jan Dec 20 '16 at 0:57

Personally, I would recommend drawing the resonance process this way (first line):


Please note the difference in arrows between forms; you have used the reversible equilibrium arrows instead of the double-headed resonance arrows. This may seem like merely a grammatical concern, but the processes of equilibrium and resonance are fundamentally different and indicating the wrong process is incorrect.

Assuming the substituent is indeed a methoxy group as you have drawn it, this compound is known as anisole (not phenol as mentioned elsewhere).

These resonance structures have significance in rationalizing and predicting experimental observations. They demonstrate the electron-donating nature of the methoxy group, and the electron-rich nature of anisole's aromatic ring (relative to unsubstituted benzene) (see resonance hybrid).

These structures explain the experimental observation that, when anisole reacts with electrophiles, the electrophiles bond at the ortho (C2,6) and para (C4) positions, and not at the meta positions (C3,5). (See third line.) We say this occurs because the bond to an electrophile will be made by an aromatic ring donating electrons from the carbon(s) with the highest electron density (in the absence of steric factors), which are the three carbons featuring lone pairs in our resonance structures (line 2).

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Well the following resonance structures are of the phenolate ion, which are correct. The existence of this resonance in phenolate ion provides it quite some stability, hence phenol is acidic. Phenolate ion, phenol's conjugate base, being stable by this resonance, is a weak base, so phenol becomes acidic.

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  • $\begingroup$ Phenol is not a strong acid. $\endgroup$ – orthocresol Dec 19 '16 at 13:26
  • $\begingroup$ Depends upon with what it is being compared to. I just added the word strong to make it compatible for the conjugate base argument. Of course it is not a strong acid compared to HCl and other "strong" acids. $\endgroup$ – Mrityunjay Gupta Dec 19 '16 at 13:31
  • $\begingroup$ Phenol (pKa~10) is a stronger acid than most other alcohols, but weaker than a carboxylic acid. But that is NOT phenol, it is anisole. $\endgroup$ – electronpusher Dec 20 '16 at 0:59
  • $\begingroup$ Damn! Didn't look at the methyl attached to the oxygen!! $\endgroup$ – Mrityunjay Gupta Dec 20 '16 at 2:07

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