From the following data, calculate the resonance energy (in kcal/mol, with sign, to two decimals) of the cyclopentadienyl cation.
Allyl cation + $H_{2}$ $\to$ propyl cation : ΔH = -14.58 kcal/mol
Cyclopentadienyl cation + 2 $H_{2}$ $\to$ cyclopentyl cation : ΔH = -72.91 kcal/mol

Question

From the following data, calculate the resonance energy (in kcal/mol, with sign, to two decimals) of the cyclopentadienyl cation.

Allyl cation + $H_{2}$ $\to$ propyl cation : ΔH = -14.58 kcal/mol

Cyclopentadienyl cation + 2 $H_{2}$ $\to$ cyclopentyl cation : ΔH = -72.91 kcal/mol

TestHub · Accepted Answer

Two isolated C=C hydrogenations of the allyl type would liberate 2 × (-14.58) = -29.16 kcal/mol. The actual value is -72.91 kcal/mol, so the cation is higher in energy by 72.91 - 29.16 = 43.75 kcal/mol. The resonance energy is negative : -43.75 kcal/mol.

Chemistry - GOC Question with Solution | TestHub

Doubts & Discussion