Why is cyclopentadiene ($\mathrm{p}K_\mathrm{a} = 16$) less acidic than

A) 2,4-pentanedione $(\mathrm{p}K_\mathrm{a} = 9);$
B) phenol $(\mathrm{p}K_\mathrm{a} = 10);$
C) water $(\mathrm{p}K_\mathrm{a} = 15.7)?$

Cyclopentadiene forms an aromatic compound as conjugate base and is quasiaromatic, while phenol has just a negative charge near an aromatic benzene ring, nothing special.

Pentanedione has got active methylene but still how can resonance (even equivalent resonance) can dominate over the aromatic effect?

Finally, why is it even less acidic than $\ce{H2O}?$ $\ce{H2O}$ gives $\ce{OH-}$ which has a localised electron oxygen atom; how is it better than aromatic delocalisation?

It has defied most of my concepts of organic chemistry.

  • 6
    $\begingroup$ Two hints: a) cyclopentadiene is still just a hydrocarbon, b) hydrogen bonds. $\endgroup$
    – Karl
    Feb 2, 2019 at 9:34
  • 4
    $\begingroup$ Oxygen beats aromaticity. $\endgroup$ Feb 2, 2019 at 9:47
  • 2
    $\begingroup$ Because aromaticity isn't the only factor that determines how stable something is, maybe? $\endgroup$ Feb 2, 2019 at 10:15
  • 1
    $\begingroup$ For phenol, think about which bond is easier to break in the solution phase(that is, is more polar),O-H or C-H? $\endgroup$ Feb 2, 2019 at 10:34
  • 2
    $\begingroup$ @orthocresol Yeah,but I think that happens because we are moving down a group, whereas C and O are of the same period. So, I think my statement was valid as we move across a period $\endgroup$ Feb 2, 2019 at 12:06

1 Answer 1


Aromaticity does not rule over everything else. It merely impacts a species relative to others that have formally similar bonding.

To take an example involving a simpler aromatic species: 3-chlorocyclopropene transfers a chloride ion to a Lewis acid, such as $\ce{SbCl5}$, more readily than 3-chloropropene or chlorocyclopropane because the 3-chlorocyclopropene forms an aromatic cation. But sodium chloride, in which the bonding to chlorine is radically different from chloroalkenes or chloroalkanes, still reacts even more readily and with more Lewis acids than 3-chlorocyclopropene.

Likewise, when cyclopentadiene is compared with other candidates for acids, it will be stronger than most hydrocarbons due to the aromatic conjugated base. But it does not achieve equality with bonding the hydrogen to, say, oxygen or chlorine.


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