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We were taught in class that when a carbon atom and an oxygen atom both have a negative charge in a resonating structure, then carbon atom attacks the nucleophile rather than oxygen atom because it acts as a better nucleophile.

In the question I have attached, in the first step the hydroxide ion abstracts the hydrogen from the oxygen of o-cresol. The negative charge so formed is delocalised onto ortho and para position.

There are 4 possibilities:

  1. Oxygen atom attacks $\ce{CH3I}$.
  2. The carbon atom attached to $\ce{CH3}$ attacks.
  3. The carbon atom adjacent to oxygen (not attached to $\ce{CH3}$) attacks.
  4. Carbon at para position attacks.

Now, based on the statement that Carbon atom should attack, Option 1 should be ruled out. Option 2 also is unfavourable because the carbon atom attached to $\ce{CH3}$ would have less negative charge due to inductive destabilisation by $\ce{CH3}$. Option 3 is unfavourable because the product formed would be sterically crowded. Option 4 seems correct.

But the answer given to the above question is Option 2.

I confirmed from my teacher and he says that the statement that Carbon atom is a better nucleophile than oxygen is true but here attack from oxygen would take place (he doesn't explain why).

Also, in the well known Riemer Tiemann Reaction and Kolbe Reaction of Phenol, the attack takes place from Carbon side.

Question- where is my reasoning wrong? According to my reasoning, answer should be Option 1. Why isn't it?

  • 2
    $\begingroup$ Phenols are fairly acidic, the -OH is completely deprotonated and the -ve charge stays on the oxygen - forget any potential resonance structures, they do not apply here. $\endgroup$
    – Waylander
    Jan 19, 2019 at 19:21
  • $\begingroup$ Fair explanation is given in: chemistry.stackexchange.com/questions/38006/…. $\endgroup$ Jan 19, 2019 at 20:44

2 Answers 2


You may be seeing the impact of the aromatic ring. Putting a negative charge on carbon in your structure disrupts the aromatic coupling. So carbanion contributions are reduced and the oxyanion contribution, where the oxygen is nucleophilic, is favored.


Well, here's the catch. When C attacks, the reaction proceeds through a carbanion intermediate. This carbanion in case of phenyl systems is stable enough to react only when the reaction takes place in a polar protic solvent which stabilizes the intermediate. But in case of an aprotic solvent such as iodomethane, O- wins the nucleophilicity race and hence it attacks. You must not forget when we're talking about the product, we mean the major product. Experimentally, we're bound to obtain both products in certain quantities.

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  • 2
    $\begingroup$ -1 from me: There's no carbanion intermediate when carbon acts as a nucleophile. Sure, some of the resonance forms have a formal negative charge on carbon, but these are relatively small contributors and it's a serious stretch to call the entire thing a carbanion. Methyl iodide is a reagent not a solvent here. I'm afraid your answer seems like you are applying rules you've read without thinking about the underlying chemistry and context. $\endgroup$
    – orthocresol
    Aug 11 at 11:20

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