Question

I almost finished my pre-lab... but I am stuck! What are the energies of (-)-menthone and (+)-isomenthone? How can I find this (how to do it)?

Thank you for the help!

Answer 1

EXPERIMENTAL RESULTS

We will NOT be doing the laboratory portion of this experiment but we will use results that have been obtained previously. Under acidic conditions menthone was allowed to isomerize to isomenthone and an equilibrium mixture established. For purposes of our calculations we will consider the equilibirium to be written as

isomenthone <------> menthone

with menthone considered to be on the product side. The corresponding equilibrium constant K is given by

K = [menthone] / [isomenthone] .........1

or

K = X_m / X_i ..........2

where X_m and X_i are the mole fractions of menthone and isomenthone, respectively. And if we consider the special case of a binary mixture, since the sum of the mole fractions is equal to 1, one can write

X_m + X_i = 1. ................3

The experimental optical rotation was measured and found to be -3.0 degrees. The specific rotation of the isomenthone is A_i = +92 and menthone is A_m = –30 degrees, and the measured angle of rotation, A, for a mixture is given by

A = (A_m ) ( X_m ) + ( A_i ) ( X_i ) .................4

or combining with Eq. (3), the above can be written as

A = (A_m ) X_m + ( A_i ) (1- X_m ) ..........5

Use the information above to calculate the experimental mole fraction of menthone, X_m, and isomenthone, X_i . Next calculate the experimental equilibrium constant K.

COMPUTATIONAL THEORY

Consider again the following equilibrium

isomenthone <-----> menthone

and corresponding equilbirium constant

K= [menthone] / [isomenthone]

The Gibbs free energy, ∆G, for this reaction is related to the equilibrium constant, K, by

∆G = – RT ln(K) .......................6

Recall that changes in Gibbs free energy, enthalpy, and entropy are related by

∆G = ∆H – T∆S .........................7

and that if the change in entropy is small during the isomerization equilibration then the Gibbs free energy change is dominated by the change in enthalpy which can be estimated from the difference in steric energy in the two isomers

∆G ~ ∆H ~ ∆E_{steric ......................8}

Therefore combining the equations above, we may write the approximation

∆Esteric = – R T ln(K) ..............9

K = e ^{– (∆E_steric/R T) .................10}

The difference in steric energy is given by

∆E_steric = (Em – Ei ) ......................11

where E_m is the calculated steric energy of menthone and E_i is the calcualterd steric energy of isomenthone. Recall that any thermodynamic difference or change is always expressed as (final – initial) or (product – reactant) and that an equilibrium constant is always (“products”/ “reactants”).

For the purpose of using the equations above, we assume the equilibrium is at the boiling temperature of glacial acetic acid 118^o C or 391K and use a value of R= 1.987 (cal/mol K) and appropriate conversion factor since the CAChe calculations of energy are expressed in kcal/mol. After completing CAChe modeling and calculations you will report E_i , E_m , ∆E_steric, K, and X_m values and compute the percent error comparing computed and experimental values of X_m. Assume the experimental value is the “accepted” value and determine the percent error of the molecular modeling Xm relative to this “accepted” value. Answer sheets are provided to guide your calculations and indicate what to report.