[GoldShadow]

I'm in general chem and a question about cyclopentadiene appeared as a bonus on one of our exams.  We had to explain cyclopentadiene's relatively strong proton donating ability (compared to other hydrocarbons) and draw a lewis structure of the cyclopentadienyl anion.

I think it's because all but one of the carbons in the molecule are sp² hybridized.  The fifth is sp³ hybridized (since it has two C-C bonds and two C-H bonds).  By removing a H⁺, the two unpaired electrons on that carbon are able to participate in delocalized pi-bonding with unhybridized p-orbitals of the other four carbons.  Thus, the cyclopentadienyl anion is a more stable aromatic structure than its protonated counterpart, cyclopentadiene.

Is that the correct explanation or am I way off?

[english]

By removing a H+, the two unpaired electrons on that carbon are able to participate in delocalized pi-bonding with unhybridized p-orbitals of the other four carbons.  Thus, the cyclopentadienyl anion is a more stable aromatic structure than its protonated counterpart, cyclopentadiene.

What you've just described is known as aromaticity.  Aromatic compounds, as you've stated in your description, are unusually stable.

There are some implications to iron out, but overall you got the idea.


Nice work. 

[Yggdrasil]

Your explanation is absolutely correct.  In addition to what you have just said, the cyclopentadienyl anion will have 6 electrons delocalized into the aromatic ring that has been created, which makes its electronic structure similar to that of benzene (which is also very stable).  The number six is important because a theoretical chemist (Huckle) showed that aromatic rings with 4n+2 electrons (e.g. 2, 6, 10, 14, etc.) gain stability from delocalization with 4n electrons (e.g. 2, 8, 12, 16, etc.) lose stability from delocalization.

http://en.wikipedia.org/wiki/H%C3%BCckel%27s_rule