Chapter 2, Problem 2.49P

Draw the products of each reaction. Use the $\text{p}{K}_{\text{a}}$ table in Appendix A to decide if the equilibrium favors the starting materials or products.

a. ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\rightleftarrows$

b.

c.

d. $\text{H}-\text{C}\equiv \text{C}-\text{H}+{\text{CH}}_{\text{3}}{\text{CH}}_2^{-}{\text{Li}}^{+}\rightleftarrows$

Interpretation Introduction

(a)

Interpretation: The products of the given reaction are to be drawn. If the equilibrium favors the starting materials or a product is to be predicted.

Concept introduction: According to Bronsted-Lowry theory, when an acid donates a proton the species formed is known as conjugate base and when the base accepts a proton the species formed is known as conjugate acid.

In a reaction which strongly favors the formation of products, the base used to remove proton from the acid should be stronger than the base formed when the proton is removed. In a reaction which favors the products, equilibrium will favors the formation of the weaker acid or weaker base. The $\text{p}{K}_{\text{a}}$ value of the products is higher than the $\text{p}{K}_{\text{a}}$ value of the reactants.

Answer to Problem 2.49P

The products of the given reaction are ${\text{CH}}_3{\text{NH}}_3^{+}$ and ${\text{HSO}}_4^{-}$. The equilibrium favors the product formation.

Explanation of Solution

The complete reaction between methyl amine and ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is shown below.

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\rightleftharpoons {\text{CH}}_3{\text{NH}}_3^{+}+{\text{ HSO}}_4^{-}\\ \text{ Base Acid Conjugate Conjugate}\\ \text{ acid base}\end{array}$

The $\text{p}{K}_{\text{a}}$ value of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $-9$ and the $\text{p}{K}_{\text{a}}$ value of ${\text{CH}}_3{\text{NH}}_3^{+}$ is $10.7$. The equilibrium favors the formation of a weaker acid with a higher $\text{p}{K}_{\text{a}}$ value. The $\text{p}{K}_{\text{a}}$ value of an acid is lower than the $\text{p}{K}_{\text{a}}$ value of a conjugate acid. Hence, equilibrium favors the product formation.

Conclusion

The products of the given reaction are ${\text{CH}}_3{\text{NH}}_3^{+}$ and ${\text{HSO}}_4^{-}$. The equilibrium favors the product formation.

Interpretation Introduction

(b)

Interpretation: The products of the given reaction are to be drawn. If the equilibrium favors the starting materials or a product is to be predicted.

Concept introduction: According to Bronsted-Lowry theory, when an acid donates a proton the species formed is known as conjugate base and when the base accepts a proton the species formed is known as conjugate acid.

In a reaction which strongly favors the formation of products, the base used to remove proton from the acid should be stronger than the base formed when the proton is removed. In a reaction which favors the products, equilibrium will favors the formation of the weaker acid or weaker base. The $\text{p}{K}_{\text{a}}$ value of the products is higher than the $\text{p}{K}_{\text{a}}$ value of the reactants.

Answer to Problem 2.49P

The products of the given reaction are ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COONa}$ and $\text{HCl}$. The equilibrium favors the starting materials.

Explanation of Solution

The complete reaction between propanoic acid and sodium chloride is shown below.

$\begin{array}{l}\text{NaCl}+{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH}\rightleftharpoons \text{HCl }+{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COONa}\\ \text{ Base Acid Conjugate Conjugate}\\ \text{ acid base}\end{array}$

The $\text{p}{K}_{\text{a}}$ value of propanoic acid is $4.9$ and the $\text{p}{K}_{\text{a}}$ value of $\text{HCl}$ is $-7$. The equilibrium favors the formation of a weaker acid with a higher $\text{p}{K}_{\text{a}}$ value. The $\text{p}{K}_{\text{a}}$ value of an acid is greater than the $\text{p}{K}_{\text{a}}$ value of a conjugate acid. Hence, equilibrium favors the starting materials.

Conclusion

The products of the given reaction are ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COONa}$ and $\text{HCl}$. The equilibrium favors the starting materials.

Interpretation Introduction

(c)

Interpretation: The products of the given reaction are to be drawn. If the equilibrium favors the starting materials or a product is to be predicted.

Concept introduction: According to Bronsted-Lowry theory, when an acid donates a proton the species formed is known as conjugate base and when the base accepts a proton the species formed is known as conjugate acid.

In a reaction which strongly favors the formation of products, the base used to remove proton from the acid should be stronger than the base formed when the proton is removed. In a reaction which favors the products, equilibrium will favors the formation of the weaker acid or weaker base. Or the $\text{p}{K}_{\text{a}}$ value of the products is higher than the $\text{p}{K}_{\text{a}}$ value of the reactants.

Answer to Problem 2.49P

The products of the given reaction are ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ and ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{ONa }$. The equilibrium favors the starting materials.

Explanation of Solution

The complete reaction between phenol and sodium hydrogen carbonate is shown below.

$\begin{array}{l}{\text{NaHCO}}_3+{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}\rightleftharpoons {\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}+{\text{ C}}_{\text{6}}{\text{H}}_{\text{5}}\text{ONa}\\ \text{ Base Acid Conjugate Conjugate}\\ \text{ acid base}\end{array}$

The $\text{p}{K}_{\text{a}}$ value of phenol is $10$ and the $\text{p}{K}_{\text{a}}$ value of ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ is $6.4$. The equilibrium favors the formation of a weaker acid with a higher $\text{p}{K}_{\text{a}}$ value. The $\text{p}{K}_{\text{a}}$ value of an acid is greater than the $\text{p}{K}_{\text{a}}$ value of a conjugate acid. Hence, equilibrium favors the starting materials.

Conclusion

The products of the given reaction are ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ and ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{ONa }$. The equilibrium favors the starting materials.

Interpretation Introduction

(d)

Interpretation: The products of the given reaction are to be drawn. If the equilibrium favors the starting materials or a product is to be predicted.

Concept introduction: According to Bronsted-Lowry theory, when an acid donates a proton the species formed is known as conjugate base and when the base accepts a proton the species formed is known as conjugate acid.

In a reaction which strongly favors the formation of products, the base used to remove proton from the acid should be stronger than the base formed when the proton is removed. In a reaction which favors the products, equilibrium will favors the formation of the weaker acid or weaker base. Or the $\text{p}{K}_{\text{a}}$ value of the products is higher than the $\text{p}{K}_{\text{a}}$ value of the reactants.

Answer to Problem 2.49P

The products of the given reaction are ${\text{CH}}_{\text{3}}-{\text{CH}}_{\text{3}}$ and $\text{H}-\text{C}\equiv \text{C}-\text{Li}$. The equilibrium favors the product formation.

Explanation of Solution

The complete reaction between ethyne and ethyl lithium is shown below.

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{Li}+\text{H}-\text{C}\equiv \text{C}-\text{H}\rightleftharpoons {\text{CH}}_{\text{3}}-{\text{CH}}_{\text{3}}+\text{ H}-\text{C}\equiv \text{C}-\text{Li}\\ \text{ Base Acid Conjugate Conjugate}\\ \text{ acid base}\end{array}$

The $\text{p}{K}_{\text{a}}$ value of ethyne is $25$ and the $\text{p}{K}_{\text{a}}$ value of ${\text{CH}}_{\text{3}}-{\text{CH}}_{\text{3}}$ is $50$. The equilibrium favors the formation of a weaker acid with a higher $\text{p}{K}_{\text{a}}$ value. The $\text{p}{K}_{\text{a}}$ value of an acid is lower than the $\text{p}{K}_{\text{a}}$ value of a conjugate acid. Hence, equilibrium favors the product formation.

Conclusion

The products of the given reaction are ${\text{CH}}_{\text{3}}-{\text{CH}}_{\text{3}}$ and $\text{H}-\text{C}\equiv \text{C}-\text{Li}$. The equilibrium favors the product formation.