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.