Part III: Record the heats of formation of both compounds. A. Heats of formation: equatorial 2-methylcyclohexanone axial 2-methylcyclohexanone -210-17aKJ/mol -270-501Ks/mol B. which conformer is more stable, and by how much? Calculate the % of each conformier at rorn temperature (25°C). (Hint: we learned how to conver△G to Kol in CHEM 3010). △G:-2.303 RT logKeg R= 8.314 J/K-mol eg
Expert Answer
Given that;
Equatorial 2-methyl cyclohexanone < = > axial 2-methyl cyclohexanone
K = [ axial 2-methyl cyclohexanone ]/ Equatorial 2-methyl cyclohexanone
Now calculate the ΔH fro this conversation as follows
ΔH= ΔH product – ΔH reactant
= -270.567 KJ/ mole – (- 270.172 KJ/ mole)
= -0.395 KJ/ moles
= – 395 J / mole
In this case
Heats of formation
ΔH ≈ ΔG
ΔG =-RTlnK
Here R = 8.3145 J / mole K
T= 25C= 298K
K = e-ΔG/RT
K = e -(-395 J / moles )/ (8.3145 )(298)
K = e ^0.159
= 1.44
K = [ axial 2-methyl cyclohexanone ]/ Equatorial 2-methyl cyclohexanone
Thus axial 2-methyl cyclohexanone / Equatorial 2-methyl cyclohexanone = 1.44/1
axial 2-methyl cyclohexanone = 1.44 * Equatorial 2-methyl cyclohexanone
assume that Equatorial 2-methyl cyclohexanone is 1
% of axial 2-methyl cyclohexanone = 1.44/2.44 ]*100
= 59.0%
And
% of Equatorial 2-methyl cyclohexanone = 1.00/2.44 ]*100
= 41.0%