My book says the following statement:

Due to the higher electronegativity of $sp^{2}$ hybridized carbon of phenol , the electron density in oxygen decreases. This increases polarity of O-H bond and results in ionization of phenol

My question is , since the electron density in oxygen atom decreases , the overall partial negative charge that oxygen has should decrease which makes the O-H bond less polar.

What Have I misunderstood here ?


1 Answer 1


Generally, higher electronegativity of $\ce{X}$ in $\ce{H-O-X}$ leads to higher acidity of the hydrogen atom.

If the oxygen atom in a phenol molecule can take less electron density from the bond with $\ce{C_{\mathrm{sp^2}}}$ (which is more electronegative than $\ce{C_{\mathrm{sp^3}}}$), it takes more electron density for $\ce{O-H}$ bond, polarizing it more, making the hydrogen more acidic.

Another factor, decreasing phenolate basicity and therefore increasing phenol acidity, is delocalization of the charge on the oxygen atom, involving the aromatic ring.

Both effects can be seen in perchloric acid $\ce{H-O-ClO3}$ too, where electronegativity of $\mathrm{Cl}$ is boosted by three oxygen atoms and the charge of perchlorate anion is delocalized among all four oxygen atoms.

  • $\begingroup$ So the electron density belonging to oxygen we are speaking is of two types in the example , oxygen with carbon since carbon has relatively higher electronegativity it allows the second electron density o-h to be more polar(since csp^2 pulls majority of electron density towards itself , this allows more attraction and charge seperation) $\endgroup$
    – Razz
    Sep 8, 2023 at 12:05
  • $\begingroup$ Please correct me if I am wrong $\endgroup$
    – Razz
    Sep 8, 2023 at 12:05
  • $\begingroup$ Not two types, rather two contribution sources. Consider a family analogy: A boy (O) is asking for his monthly allowance at his parents. He is not greedy, he wants just the amount that was promised. If the parent A (C) gives less money than expected, the more money the boy (O) asks at the parent B (H). // Rather, O pulls less of the electron density from C_sp2 than from C_sp3, as the electronegativity order is O >> C_sp2 > C_sp3 $\endgroup$
    – Poutnik
    Sep 8, 2023 at 12:12
  • $\begingroup$ Ah ! That is a very good analogy thank you for clearing this for me $\endgroup$
    – Razz
    Sep 8, 2023 at 15:12

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