Question

Both cyclopentadiene and cyclopentadienyl nion have same valence electrons and same 5-membered Carbon ring structure. However, cyclopentadiene adopts "envelope" shape with C on right tilted out of the plane of 4 Carbon atoms. The anion adopts flat structure. Further, the anion is far more stable chemically than cyclopentadiene. Speculate why this is so. Answer should consider both hybridization and resonance.

Answer 1

Aromatic compounds are those that meet the following criteria

The structure must be cyclic and planar, containing conjugated π-bonds.

Each atom in the ring must have an unhybridized p-orbital.

The unhybridized p-orbitals must overlap to form a continuous ring of parallel orbitals.

Delocalization of the π-electrons over the ring must lower the electronic energy.

It should follow the Huckel’s rule. The rule states that aromatic compounds must contain (4n+2) π-electrons, where n is any whole number. If it contains (4n) π-electrons, the compounds are anti-aromatic compound.

Aromatic systems have 2, 6, or 10 π-electrons, for n = 0, 1, or 2 and antiaromatic systems have 4, 8, or 12 π-electrons, for n = 1, 2, or 3.

Cyclopentadiene is not aromatic because of the presence of sp³ hybridized carbon atom. The cyclopentadienyl anion is aromatic because it has an uninterrupted ring of p-orbital and (4n+2) π-system.

All non-aromatic compouns will adopt non-planar form and all aromatic compouns will adopt cyclic planar form with conjugation of double bonds.

The double bonds of cyclopentadiene are constrained in a planar cisoid form, to overcome this constraint it adopts non-panar form. Generally it exist as dimers in exo- and endo-forms (endo- is much more stable than exo-isomer)