Following are the alternative chair conformations for trans-4-methylcyclohexanol
ID: 986106 • Letter: F
Question
Following are the alternative chair conformations for trans-4-methylcyclohexanol Using the data for Delta G for mono substituted cyclohexanes at room temperature (25 degree C) and the representative value* for the gauche interaction of two equatorially positioned substituents in the 1.2-position: Calculate the difference in the Gibbs free energy between the second and first conformation including the algebraic sign. Given your value in (a), calculate the percent of the chair, indicated as B. presented in an equilibrium mixture of the confonners at 25 degree C. The actual value will depend on the substituents. The values of the gas constant and the temperature for 25 degree C are 8.314 JK^-1 mol^-1 and 298 K respectively.Explanation / Answer
In the B conformation both the methyl and hydroxyl groups are in equatorial position hence G is zero whereas in the A conformation (methyl group axial and hydroxyl group axial hence) G will be sum of
-3.9 and -7.3 which is equal to -11.2.
So difference in gibbs free energy of two alternative conformations is equal to 0-(-11.2)
Hence G0 =11.2
We know that G0 = -RT * lnKeq
11.2 = - 8.314 * 298 lnKeq
Find Keq from the above calculation
After determining Keq, the percentage of the conformations can be determined by using the below mentined formula
Substitute Keq value in below equation
Keq= B/A
A + B = 100%
Hope you have done the calculation part and got the answer. All the best