Chapter 6, Problem 6.37P

Interpretation Introduction

(a)

Interpretation: The conformation which is present in higher concentration when $\text{R}={\text{CH}}_{\text{2}}{\text{CH}}_3$ is to be identified.

Concept introduction: The change in Gibbs free energy is represented by $\Delta G°$. It is a state function. The value of $\Delta G°$ depends on $K_{\text{eq}}$. The free energy change is calculated as,

$\text{Δ}G\text{°}=-2.303\text{R}T\log K_{\text{eq}}$

If the $\Delta G°$ is greater than zero and $K_{\text{eq}}$ is smaller than one, then the formation of starting material is favored at equilibrium. However, if the $\Delta G°$ is smaller than zero and $K_{\text{eq}}$ is greater than one, the formation of product is favored at equilibrium.

Answer to Problem 6.37P

The equatorial conformation is present in higher concentration in the given compound.

Explanation of Solution

Given:

The value of $K_{\text{eq}}$ is $23$ when $\text{R}={\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$.

The equilibrium reaction of monosubstituted cyclohexane is shown below.

Figure 1

The value of ${\text{K}}_{\text{eq}}$ is $23$ when $\text{R}={\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$. The value of $K_{\text{eq}}$ is greater than $1$. It implies that the concentration of the starting material is comparatively lower than that of the concentration of product. The formation of products is favored at equilibrium. Therefore, the equatorial conformation is present in higher concentration in the given compound.

Conclusion

The equatorial conformation is present in higher concentration in the given compound.

Interpretation Introduction

(b)

Interpretation: The $\text{R}$ group which shows the higher percentage of equatorial conformation at equilibrium is to be identified.

Concept introduction: The change in Gibbs free energy is represented by $\Delta G°$. It is a state function. The value of $\Delta G°$ depends on $K_{\text{eq}}$. The free energy change is calculated as,

$\text{Δ}G\text{°}=-2.303\text{R}T\log K_{\text{eq}}$

If the $\Delta G°$ is greater than zero and $K_{\text{eq}}$ is smaller than one, then the formation of starting material is favored at equilibrium. However, if the $\Delta G°$ is smaller than zero and $K_{\text{eq}}$ is greater than one, the formation of product is favored at equilibrium.

Answer to Problem 6.37P

The $\text{R}$ group that shows the higher percentage of equatorial conformation at equilibrium is $\text{R}=\text{C}{\left({\text{CH}}_{\text{3}}\right)}_3$.

Explanation of Solution

The values of $K_{\text{eq}}$ are $23$ and $4000$.

The both given values are greater than $1$, which indicates that in both the cases, the formation of the product is favored at equilibrium. The equilibrium constant $\left(K_{\text{eq}}\right)$ is higher for $\text{R}=\text{C}{\left({\text{CH}}_{\text{3}}\right)}_3$ than that of $\text{R}={\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$.

If the $K_{\text{eq}}$ value is smaller than one, then the formation of starting material is favored at equilibrium. However, if the $K_{\text{eq}}$ value is greater than one, the formation of product is favored at equilibrium.

Therefore, the $\text{R}$ group that shows the higher percentage of equatorial conformation at equilibrium is to $\text{R}=\text{C}{\left({\text{CH}}_{\text{3}}\right)}_3$.

Conclusion

The $\text{R}$ group that shows the higher percentage of equatorial conformation at equilibrium is to $\text{R}=\text{C}{\left({\text{CH}}_{\text{3}}\right)}_3$.

Interpretation Introduction

(c)

Interpretation: The $\text{R}$ group which shows the higher percentage of axial conformation at equilibrium is to be identified.

Concept introduction: The change in Gibbs free energy is represented by $\Delta G°$. It is a state function. The value of $\Delta G°$ depends on $K_{\text{eq}}$. The free energy change is calculated as,

$\text{Δ}G\text{°}=-2.303\text{R}T\log K_{\text{eq}}$

If the $\Delta G°$ is greater than zero and $K_{\text{eq}}$ is smaller than one, then the formation of starting material is favored at equilibrium. However, if the $\Delta G°$ is smaller than zero and $K_{\text{eq}}$ is greater than one, the formation of product is favored at equilibrium.

Answer to Problem 6.37P

The $\text{R}$ group that shows the higher percentage of axial conformation at equilibrium is $\text{R}={\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$.

Explanation of Solution

The values of $K_{\text{eq}}$ are $23$ and $4000$.

The equilibrium constant $\left(K_{\text{eq}}\right)$ is higher for $\text{R}=\text{C}{\left({\text{CH}}_{\text{3}}\right)}_3$ than that of $\text{R}={\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$. The given both values are greater than $1$, which indicates that in both cases the formation of the product is favored at equilibrium. However, the ethyl group experiences less $1,3-$ diaxial interactions due to less value of $K_{\text{eq}}$.

Therefore, the $\text{R}$ group that shows the higher percentage of axial conformation at equilibrium is $\text{R}={\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$.

Conclusion

The $\text{R}$ group that shows the higher percentage of axial conformation at equilibrium is to $\text{R}={\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$.

Interpretation Introduction

(d)

Interpretation: The $\text{R}$ group for which $\Delta G°$ is more negative to be identified.

Concept introduction: The change in Gibbs free energy is represented by $\Delta G°$. It is a state function. The value of $\Delta G°$ depends on $K_{\text{eq}}$. The free energy change is calculated as,

$\text{Δ}G\text{°}=-2.303\text{R}T\log K_{\text{eq}}$

If the $\Delta G°$ is greater than zero and $K_{\text{eq}}$ is smaller than one, then the formation of starting material is favored at equilibrium. However, if the $\Delta G°$ is smaller than zero and $K_{\text{eq}}$ is greater than one, the formation of product is favored at equilibrium.

Answer to Problem 6.37P

The value of $\Delta G^{\text{o}}$ is more negative for $\text{R}=\text{C}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$.

Explanation of Solution

Given:

The values of $K_{\text{eq}}$ are $23$ and $4000$.

The relationship between $\Delta G°$ and $K_{\text{eq}}$ is expressed as,

$\text{Δ}G\text{°}=-2.303\text{R}T\log K_{\text{eq}}$

As the value of $K_{\text{eq}}$ increases, the value of $\Delta G^{\text{o}}$ become more and more negative. The value of $K_{\text{eq}}$ is more for $\text{R}=\text{C}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$. Hence, the value of $\Delta G^{\text{o}}$ is more negative for $\text{R}=\text{C}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$.

Conclusion

The value of $\Delta G^{\text{o}}$ is more negative for $\text{R}=\text{C}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$.

Interpretation Introduction

(e)

Interpretation: The explanation corresponding to the relation of size of $\text{R}$ to the amount of axial and equatorial conformation at equilibrium is to be stated.

Concept introduction: The change in Gibbs free energy is represented by $\Delta G°$. It is a state function. The value of $\Delta G°$ depends on $K_{\text{eq}}$. The free energy change is calculated as,

$\text{Δ}G\text{°}=-2.303\text{R}T\log K_{\text{eq}}$

If the $\Delta G°$ is greater than zero and $K_{\text{eq}}$ is smaller than one, then the formation of starting material is favored at equilibrium. However, if the $\Delta G°$ is smaller than zero and $K_{\text{eq}}$ is greater than one, the formation of product is favored at equilibrium.

Answer to Problem 6.37P

The large size of $\text{R}$ group favors the formation of product that has equatorial conformation at equilibrium whereas the small size of $\text{R}$ group favors the formation of reactant that has axial conformationat equilibrium.

Explanation of Solution

The value $K_{\text{eq}}$ for $\text{R}=\text{C}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$ and $\text{R}={\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$ is $4000$ and $23$ respectively. The formation of product that has equatorial conformation is more favored as the value of $K_{\text{eq}}$ increases. The large size of $\text{R}$ group favors formation of product that has equatorial conformation at equilibrium whereas the small size of $\text{R}$ group favors formation of reactant that has axial conformationat equilibrium.

Conclusion

The large size of $\text{R}$ group favors formation of product that has equatorial conformation at equilibrium whereas the small size of $\text{R}$ group favors formation of reactant that has axial conformationat equilibrium.