Solution Found!
a. The ?Go for conversion of “axial” fluorocyclohexane to
Chapter 5, Problem 22P(choose chapter or problem)
Problem 22P
a. The ΔGo for conversion of “axial” fluorocyclohexane to “equatorial” fluorocyclohexane at 25 oC is -0.25kcal/mol. Calculate the percentage of fluorocyclohexane molecules that have the fluoro substituent in an equatorial position at equilibrium.
b. Do the same calculation for isopropylcyclohexane (its ΔGo value at 25 oC is -2.1 kcal/mol).
c. Why is the percentage of molecules with the substituent in an equatorial position greater for isopropylcyclohexane?
Questions & Answers
QUESTION:
Problem 22P
a. The ΔGo for conversion of “axial” fluorocyclohexane to “equatorial” fluorocyclohexane at 25 oC is -0.25kcal/mol. Calculate the percentage of fluorocyclohexane molecules that have the fluoro substituent in an equatorial position at equilibrium.
b. Do the same calculation for isopropylcyclohexane (its ΔGo value at 25 oC is -2.1 kcal/mol).
c. Why is the percentage of molecules with the substituent in an equatorial position greater for isopropylcyclohexane?
ANSWER:
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 %.