# Aromaticity fails?

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.

• Two hints: a) cyclopentadiene is still just a hydrocarbon, b) hydrogen bonds. – Karl Feb 2 at 9:34
• Oxygen beats aromaticity. – Ivan Neretin Feb 2 at 9:47
• Because aromaticity isn't the only factor that determines how stable something is, maybe? – Oscar Lanzi Feb 2 at 10:15
• For phenol, think about which bond is easier to break in the solution phase(that is, is more polar),O-H or C-H? – Yusuf Hasan Feb 2 at 10:34
• @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 – Yusuf Hasan Feb 2 at 12:06

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.