The solubility of phenol is 83 g/L, while that of cyclohexanol is 36 g/L. Both of them have a hydroxy group, and can form hydrogen bonds with water. Also, the hydrophobic part of phenol, as well as cyclohexanol is almost the same. Then what is the reason for phenol's better water solubility in water compared to cyclohexanol?

  • $\begingroup$ Phenol's $\ce{OH}$ is more polar (which is also manifested in its acidity). $\endgroup$ – Ivan Neretin May 11 '16 at 9:53
  • $\begingroup$ The aromatic component draws significant electron density away from the hydroxy group. It is an example of conjugation. One of the Lewis structures involve a double-bonded oxygen. $\endgroup$ – Yunfei Ma May 11 '16 at 11:45
  • $\begingroup$ @IvanNeretin Hmm, could you elaborate a little more on the reason for the higher polarity of Phenol's $\ce{O-H}$ bond you suggested. I don't see a particular link between bond polarity and acidity as my view of the higher acidity of Phenol is that the higher (relative) stability of the phenoxide ion due to resonance stabilisation of the negative charge (which is not present in the cyclohexoxide ion) leads to a an equilibrium that is shifted a bit more to the side of dissociated species compared to the dissociation equilibrium of cyclohexanol and thus to a higher acidity. $\endgroup$ – Philipp May 11 '16 at 11:47
  • $\begingroup$ @Philipp It is all the same; if you have resonance structures that stabilize phenoxide ion by taking its negative charge and distributing it someplace else, then you also have the same structures in protonated form with (+) on oxygen. Or, as YunfeiMa put it, the aromatic ring draws electron density away from oxygen, thus effectively making it more electronegative. $\endgroup$ – Ivan Neretin May 11 '16 at 11:58
  • $\begingroup$ And how 'exactly' does higher bond polarity make up to a better hydrogen bonding? $\endgroup$ – Bhargav Varshney May 12 '16 at 9:01

In addition to the acidity difference described in the comments, there is a more subtle effect involving conjugation in the undissociated phenol. When the phenyl group withdraws some electron density from the π-donating hydroxyl group, it gains some negative charge and so is better able to attract the hydrogen ends of the water dipoles. Ultimately this is a manifestation of the polarizability of the delocalized π electrons.


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