I came across this concerning phenols:

The hydroxyl group, in phenol is directly attached to the $\ce{sp^2}$ hybridised carbon of benzene ring which acts as an electron withdrawing group. Due to the higher electronegativity of $\ce{sp^2}$ hybridised carbon of phenol to which $\ce{–OH}$ is attached, electron density decreases on oxygen. This increases the polarity of $\ce{O–H}$ bond and results in an increase in ionisation of phenols than that of alcohols.

My concepts seem to be flawed, and hence I have few questions on the same.

When the $\ce{sp^2}$ carbon atom withdraws electrons from the oxygen atom, the oxygen's electron density decreases. How does this increase the polarity in the $\ce{O-H}$ bond? Shouldn't it be the opposite as oxygen becomes more electropositive due to the carbon atom? I'm missing something here.

Also, does the carbon atom in this case act as electron withdrawing because it is less electronegative than the oxygen atom?

Note that this question isn't about the comparison of acidic character between alcohols and phenols, and the relative stability of alkoxides and phenoxides.

  • $\begingroup$ Related: Inductive vs resonance effects and the acidity of phenol $\endgroup$
    – user7951
    Aug 8, 2015 at 11:50
  • $\begingroup$ possible duplicate of reason for the stronger acidic property of phenol than alcohol $\endgroup$
    – Mithoron
    Aug 8, 2015 at 12:57
  • $\begingroup$ @Mithoron I've mentioned in the end note that this question is not asking for " the comparison of acidic character between alcohols and phenols". I want to know why the sp2 carbon increases the polarity of the OH bond. $\endgroup$
    – Indo Ubt
    Aug 8, 2015 at 12:59
  • $\begingroup$ Explanation you're citing is rather bad one in these links you have better ones $\endgroup$
    – Mithoron
    Aug 8, 2015 at 13:06
  • $\begingroup$ I get the resonance part, which is what the link explains. I need the inductive explanation, which @Loong pointed to. But the way its explained in that link isnt so clear $\endgroup$
    – Indo Ubt
    Aug 8, 2015 at 13:24

1 Answer 1


You are confusing the two terms electronegativity/electropositivity and higher/lower electron density.

The source you read states that the $\ce{sp^2}$ hybridised carbon in phenol is more electronegative than the $\ce{sp^3}$ hybridised one in (say) ethanol. This means, that this carbon has a greater desire to draw electrons towards it.

Therefore, even though any carbon is still more electropositive than oxygen, the phenolic oxygen will be able to draw the electrons towards it less effectively and therefore has a lower electron density than in ethanol.

But the oxygen atom in itself hasn’t changed; specifically, it is still hybridised the same way (assuming that one can speak of hybridisation for oxygens). Thus, oxygen’s electronegativity is still the same.

The combination of still having the same electronegativity but having less electron density means that any other electrons will be drawn towards it more strongly — specifically, this concerns the electrons in the $\ce{O-H}$ bond.

In a nutshell: The oxygen’s electronegativity didn’t change (it sounds weird to talk of electropositivity for oxygen if there is no fluorine around) it only has less electron density on one side due to more pulling. It combats that by pulling more from the other side.

This can be called an inductive effect.

  • $\begingroup$ One thing I don't understand is why does decreased electron density on oxygen increases lit's acidity.... $\endgroup$
    – Scáthach
    Aug 1, 2018 at 22:13
  • $\begingroup$ Or more properly affect it's acidity.Is there a logic behind this thought $\endgroup$
    – Scáthach
    Aug 1, 2018 at 22:21
  • $\begingroup$ @harambe Acidity or loss of a proton leaving a negative charge corresponds rather well with how stabilised the resulting anion is. And anions are typically most stable if there is a low electron density surplus. So if there are other groups drawing electrons away from the oxygen, this reduces the effective negative charge of an anion and thus makes the compound more acidic. $\endgroup$
    – Jan
    Oct 16, 2018 at 13:36

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