Question

Problem #8. The strain energy of a cyclic hydrocarbon can be estimated using the experimental enthalpy of formation of that compound and the calculated enthalpy of formation for a hypothetical strain-free ring. This would be a ring containing an infinite number of methylene (CH2) groups. Calculated AH for AH per CH2 strain-free (kcal/mol) molecule (kcal/mol) Strain Energy per CH2 (kcal/mol) + 4.2 - 14.7 Strain Energy (kcal/mol) 27.4 26.4 5.8 9.1 + 1.7 - 19.6 - 3.7 - 24.5 Measured AH Molecule (kcal/mol) cyclopropane +12.7 cyclobutane + 6.8 cyclopentane - 18.7 cyclohexane - 29.5 cycloheptane - 28.3 cyclooctane - 29.7 (CH2). n = infinity - 29.4 - 34.3 a) Explain how Strain Energy is calculated in the above table. Use the first entry (cyclopropane) as an example. b) According to this data, cyclopentane and cycloheptane have the same strain energy per methylene. The total strain of the molecule is sum of the angle strain and the torsional strain. Comparing these two molecules, which do you think has more angle strain? Which do you think has more torsional strain? Explain. c) What does it mean for the experimental enthalpy of formation of a compound to be positive? Is the compound more or less stable that its constituent elements in their standard state?

Answer 1

$\Delta$H_f per CH₂ = Measured $\Delta$H_f/n

Where n = Number of CH₂ units in the given cycloalkane

For cyclopropane, n = 3

i.e. $\Delta$H_f per CH₂ = 12.7 kcal/3 = 4.2 kcal/mol

Strain energy = Measured $\Delta$H_f - calculated $\Delta$H_f for strain-free molecule

For cyclopropane, strain energy = 12.7 - (-14.7) = 27.4 kcal/mol

Strain energy per CH₂ = Strain energy/n

For cyclopropane, strain energy per CH₂ = 27.4/3 = 9.1 kcal/mol