Chapter 5: Q20 P (page 204)

a. Calculate the percentage of isopropylcyclohexane molecules that have the isopropyl substituent in an equatorial position at equilibrium. (Its ∆G° value at 25 °C is -2.1 kcal/mol.)
b. Why is the percentage of molecules with the substituent in an equatorial position greater for isopropylcyclohexane than for fluorocyclohexane?

Short Answer

Expert verified

97%
Because isopropyl substituent is larger than fluoro substituent.

Step by step solution

a) Given quantities -

∆G°=-2.1 kcal/mol

T= 25 °C

Calculation of the percentage of isopropylcyclohexane molecules that have the isopropyl substituent in an equatorial position at equilibrium -

G^o = ^_RT ln K_eq

-2.1 = -1.986× 10^-3× 298× ln K_eq

ln K_eq = 3.56

K_eq= 35

K_eq= $\frac{[i s o p r o p y l c y c l o h e x a n e] e q u a t o r i a l}{[i s o p r o p y l c y c l o h e x a n e] a x i a l} = \frac{35}{1}$

% of equatorial isopropylcyclohexane

$= \frac{[i s o p r o p y l c y c l o h e x a n e] e q u a t o r i a l}{[i s o p r o p y l c y c l o h e x a n e] e q u a t o r i a l + [i s o p r o p y l c y c l o h e x a n e] a x i a l} \times 100 = \frac{35}{35 + 1} \times 100 = \frac{35}{36} \times 100 = 97 %$

b) Because the isopropyl substituent is larger than the fluoro substituent, isopropylcyclohexane has a higher percentage of the conformer with the substituent in the equatorial position.

Because of the 1,3-diaxial interactions, the conformer with the substituent in the axial position becomes less stable as the substituent grows larger.