[Violagirl]

I am currently taking O Chem II and am reviewing for my final. In going over a sample exam that my professor had up, I am unable to figure out this question:

The PKa of both diphenylmethane and triphenylmethane are significantly higher than cyclopentadiene. Explain this difference using molecular orbitals. (Hint: You only need to draw the molecular orbitals of the appropriate ionic species generated from cyclopentadiene.)

I'm also told that the pKa of diphenylmethane is 33.5, for triphenylmethane it is 31.5, and for cyclopentadiene, it is 15. I know that cyclopentadiene has 4 pi electrons and so 4 molecular orbitals are drawn where the first and second orbitals are completely filled. However, I also know that an sp3 orbital is present in cyclopentadiene and so there are no empty p orbitals for it to be filled. I was not sure of the ionic species mentioned in this problem or what it meant by that or this shows why it has a lower pKa then that of the two compounds previously mentioned. Any insight into this would be appreciated!

[Dan]

Draw the conjute bases of the three compounds (ie. the anions formed on deprotonation), and consider their relative stability. A more stable conjugate base corresponds to a stonger acid.

Hint: Consider pyrrole