a. The ?Go for conversion of “axial” fluorocyclohexane to

Solution 22P

Step 1 of 3:

a.

Here, we are asked to calculate the percentage of fluorocyclohexane molecules that have the fluoro substituent in an equatorial position at equilibrium.

Given:

ΔGo  = -0.25kcal/mol  

T= 25oC = 298.15 K

In fluorocyclohexane,  axial ⇋ equatorial, which means the fluorocyclohexane ring flips are interchangeable.

This can be calculated using the relation between equilibrium constant (Keq) and the change in free energy (G). The equation is as follows:

Go = _ RT ln Keq

Where, R = gas constant = 1.98610-3 kcal/mol/k

          T = Temperature in Kelvin

Now, let’s find the Keq by using the above equation:

         - 0.25 kcal/mol  =  -1.98610-3 kcal/mol/k  298.15 K   ln Keq

        ln Keq = 0.422

        Keq = e0.422 

        Keq= 1.52

Keq= =

both the axial and equatorial are in equilibrium. Hence, [axial]= 1

Now that we have Keq, let’s find the percentage of total equatorial conformer:

= 100

= 100

= 60

Hence, the percentage of fluorocyclohexane molecules that have the fluoro substituent in an equatorial position at equilibrium is 60 %.