Question

Which is More acidic in Cyclopentadiene and cyclopropane and why?

Answer 1

Hint: First, we are assuming that the central $$C-H$$ bond in cyclopropane ionizes. The $$C\left(s{p}^2\right)-H\left(s\right)$$ bond should ionize because of two factors:
1.The $$s{p}^2$$ carbon would better accommodate the negative charge than the central sp3 carbon.
2.Consequent conjugate base would not contain anti aromaticity.

Complete step by step answer:
Cyclopentadiene is an organic compound with the formula $$C_5{H}_6$$. It is often abbreviated CpH because the cyclopentadienyl anion is abbreviated $$C{p}^{-}$$. Cyclopropane is the cycloalkane with the molecular formula $$C_3{H}_6$$, consisting of three carbon atoms linked to each other to form a ring, with each carbon atom bearing two hydrogen atoms resulting in $$D_{3h}$$ molecular symmetry.
Aromaticity gives stability to the structure, so the formation of the carbanion is favourable and the position of the equilibrium for the reaction is to the right, thus cyclopentadiene is acidic.
Indeed, according to the Evans pKa table the cyclopropane $$C-H$$ bond $$\left(pKa\sim 46\right)$$ is more acidic than the $$C-H$$ bond on the central carbon in propane $$\left(pKa\sim 51\right)$$. We know that, in general, the acidity of $$C-H$$ bonds follow the order $$sp>s{p}^2>s{p}^3$$.
Cyclopentadiene has aromaticity in its conjugate base while conjugate base of cyclopropane is more aromatic.

Therefore, cyclopentadiene is more acidic.
Cyclopentadiene has conjugated double bonds; cyclopentadiene has both $$s{p}^2$$ and $$s{p}^3$$ hybridized carbon atoms.
cyclopentadiene is a strain -free cyclic system cyclopentadiene ion, the conjugate base of cyclopentadiene, is an aromatic species and hence has higher stability.

Note: Hydrocarbons are not usually very acidic $$\left(pKa>50\right)$$. But cyclopentadiene has a much lower pKa due to the aromatic stability of its aromatic conjugate base. Pyridine is like benzene but an N has replaced one CH.