Chapter 6, Problem 6.41P

Interpretation Introduction

(a)

Interpretation:

The products for the given proton transfer reaction are to be drawn and the favored equilibrium side with numerical factor is to be determined.

Concept introduction:

The proton transfer reactions favor the side opposite the stronger acid. Larger the ${\text{pK}}_{\text{a}}$, weaker is the acid. The stronger acid is the one whose negatively charged conjugate base is more stable. Delocalizing a charge via resonance increases the stability of the charged species. The atom with a negative charge is stabilized when it is adjacent to electron withdrawing groups. Each difference of $\text{1}$ in ${\text{pK}}_{\text{a}}$ values between the two acids corresponds to a power of $10$ by which that side of the reaction is favored.

Answer to Problem 6.41P

The products for the given reaction are:

The equilibrium is favored to product side by a factor of ${\text{10}}^{\text{2}\text{.7}}=5.0\times {10}^2$.

Explanation of Solution

The given reaction is:

The ion ${\text{HO}}^{-}$ is the strongest base that can exist. It deprotonates the amide and forms a conjugate base with negative charge on nitrogen.

In the conjugate base formed, the negative charge on nitrogen is delocalized through the electron withdrawing resonance effect of carbonyl group. Thus, amide is a stronger acid than water, and hence, the equilibrium is favored to the product side.

The ${\text{pK}}_{\text{a}}$ of ${\text{H}}_{\text{2}}\text{O}$ is $\text{15}\text{.7}$. According to Appendix $\text{A}$, the ${\text{pK}}_{\text{a}}$ of amide functional group is $\text{17}$. As benzamide has an aromatic ring directly attached to ${\text{CONH}}_2$ functional group, we estimate its ${\text{pK}}_{\text{a}}$ to be $\text{13}$. The difference in those ${\text{pK}}_{\text{a}}$ values is $\text{15}\text{.7-13 = 2}\text{.7}$. Thus, benzamide is the weaker acid by a factor of ${\text{10}}^{\text{2}\text{.7}}=5.0\times {10}^2$.

Conclusion

The favored equilibrium side with numerical value is determined on the basis of stronger acid and ${\text{pK}}_{\text{a}}$.

Interpretation Introduction

(b)

Interpretation:

The products for the given proton transfer reaction are to be drawn and the favored equilibrium side with the numerical factor is to be determined.

Concept introduction:

The proton transfer reactions favor the side opposite the stronger acid. Larger the ${\text{pK}}_{\text{a}}$, weaker is the acid. Each difference of $\text{1}$ in ${\text{pK}}_{\text{a}}$ values between the two acids corresponds to a power of $10$ by which that side of the reaction is favored.

Answer to Problem 6.41P

The products for the given reaction are:

The equilibrium is favored to the reactant side by a factor of ${\text{10}}^{\text{23}}=1.0\times {10}^{23}$.

Explanation of Solution

The given reaction is:

In the given reaction, ${\text{Cl}}^{-}$ abstracts a proton from the oxygen atom of an alcohol and gives the following products:

$\text{HCl}$ is the strongest acid that can exist and tends to donate its proton. Alcohol is a weaker acid than $\text{HCl}$. Hence, the equilibrium is favored to the reactant side.

${\text{pK}}_{\text{a}}$ of $\text{HCl}$ is $\text{-7}$. According to Appendix $\text{A}$, the ${\text{pK}}_{\text{a}}$ of secondary alcohol, $\text{2-propanol}$ is $\text{16}\text{.5}$. As the cyclopentanol does not have any nearby electronegative atoms or adjacent double bonds, we estimate its ${\text{pK}}_{\text{a}}$ to be $\text{16}\text{.5}$ too. The difference in those ${\text{pK}}_{\text{a}}$ values is $\text{16-(-7) = 23}$. Thus, the cyclopentanol is the weaker acid by a factor of ${\text{10}}^{\text{23}}=1.0\times {10}^{23}$.

Conclusion

The favored equilibrium side with numerical value is determined on the basis of the stronger acid and ${\text{pK}}_{\text{a}}$.

Interpretation Introduction

(c)

Interpretation:

The products for the given proton transfer reaction are to be drawn and the favored equilibrium side with numerical factor is to be determined.

Concept introduction:

The proton transfer reactions favor the side opposite the stronger acid. Larger the ${\text{pK}}_{\text{a}}$, weaker is the acid. The stronger acid is the one whose negatively charged conjugate base is more stable. Delocalizing a charge via resonance increases the stability of the charged species. The atom with the negative charge is stabilized when it is adjacent to electron withdrawing groups. Each difference of $\text{1}$ in ${\text{pK}}_{\text{a}}$ values between the two acids corresponds to a power of $10$ by which that side of the reaction is favored.

Answer to Problem 6.41P

The products for the given reaction are:

The equilibrium is favored to product side by a factor of ${\text{10}}^{\text{23}}=1.0\times {10}^{23}$.

Explanation of Solution

The given reaction is:

In the given reaction, cyclopentadiene acts as an acid and the negatively charge nitrogen abstracts a proton from diisopropylamine to give the following products:

On the product side, the negative charge on carbon is a resonance stabilized by a conjugated double bond; such stabilization of the negative charge is not possible on the reactant side where the negative charge is on nitrogen bonded to two electron donating isopropyl groups. The acid is stronger when its conjugate base is stable, therefore, cyclopentadiene is a stronger acid than $\text{N, N-dimethylamine}$. Hence, the equilibrium is favored to the product side.

According to Appendix $\text{A}$, ${\text{pK}}_{\text{a}}$ of cyclopentadiene is $\text{16}$ and that of secondary amine, dimethylamine is $\text{38}$. As the diisopropylamine does not have any nearby electronegative atoms or adjacent double bonds, we estimate its ${\text{pK}}_{\text{a}}$ to be, $\text{38}$ too. The difference in those ${\text{pK}}_{\text{a}}$ values is $\text{38-16 = 22}$. Thus, cyclopentanol is the weaker acid by a factor of ${\text{10}}^{\text{22}}=1.0\times {10}^{22}$.

Conclusion

The favored equilibrium side with numerical value is determined on the basis of stronger acid and ${\text{pK}}_{\text{a}}$.

Interpretation Introduction

(d)

Interpretation:

The products for the given proton transfer reaction are to be drawn and the favored equilibrium side with numerical factor is to be determined.

Concept introduction:

The proton transfer reactions favor the side opposite the stronger acid. Larger the ${\text{pK}}_{\text{a}}$, weaker is the acid. The stronger acid is the one whose negatively charged conjugate base is more stable. The increase in effective electronegativity of an atom bearing negative charge increases the stability of charged species. The effective electronegativity of an atom increases in the order: ${\text{sp}}^{\text{3}}{\text{ < sp}}^{\text{2}}<\text{sp}$. Each difference of 1 in ${\text{pK}}_{\text{a}}$ values between the two acids corresponds to a power of $10$ by which that side of the reaction is favored.

Answer to Problem 6.41P

The products for the given reaction are:

The equilibrium is favored to the product side by a factor of ${\text{10}}^{10}=1.0\times {10}^{10}$.

Explanation of Solution

The given reaction is:

In the given reaction, the hydride ion abstracts the terminal proton of an alkyne to give the following products:

As the effective electronegativity of $\text{sp}$ carbon is greater, the negative charge is better stabilized. This makes the terminal alkyne a strong acid, and equilibrium is favored to right side.

According to Appendix $\text{A}$, ${\text{pK}}_{\text{a}}$ of ${\text{H}}_{\text{2}}$ is $35$ and that of alkyne, acetylene is $\text{25}$. As diisopropylamine does not have any nearby electronegative atoms or adjacent double bonds, we estimate its ${\text{pK}}_{\text{a}}$ to be $\text{25}$ too. The difference in those ${\text{pK}}_{\text{a}}$ values is $\text{35-25 = 10}$. Thus, cyclopentanol is the weaker acid by a factor of ${\text{10}}^{10}=1.0\times {10}^{10}$.

Conclusion

The favored equilibrium side with numerical value is determined on the basis of stronger acid and ${\text{pK}}_{\text{a}}$.

Interpretation Introduction

(e)

Interpretation:

The products for the given proton transfer reaction are to be drawn and the favored equilibrium side with numerical factor is to be determined.

Concept introduction:

The proton transfer reactions favor the side opposite the stronger acid. Larger the ${\text{pK}}_{\text{a}}$, weaker is the acid The proton is more acidic when attached to a positively charged atom than when attached to a neutral atom. The charged species are significantly less stable than the uncharged one. Each difference of $\text{1}$ in ${\text{pK}}_{\text{a}}$ values between the two acids corresponds to a power of $10$ by which that side of the reaction is favored.

Answer to Problem 6.41P

The products for the given reaction are:

The equilibrium is favored to the product side by a factor of ${\text{10}}^{6.45}=2.82\times {10}^6$.

Explanation of Solution

The given reaction is:

In the given reaction, the propanoate ion abstracts the proton of hydronium ion to give the following products:

${\text{H}}_{\text{3}}{\text{O}}^{\oplus }$ is more acidic than carboxylic acid as the oxygen atom bonded to acidic proton has positive charge whereas, in carboxylic acid, oxygen to which the acidic proton is attached is neutral. The proton is more acidic when the atom to which it is attached is positively charged than when that atom is uncharged. Therefore, the equilibrium is favored to the product side.

According to Appendix $\text{A}$, ${\text{pK}}_{\text{a}}$ of ${\text{H}}_{\text{3}}{\text{O}}^{\oplus }$ is $-1.7$ and that of typical carboxylic acid, ethanoic acid is $4.75$. As the propanoic acid does not have any nearby electronegative atoms or adjacent double bonds, we estimate its ${\text{pK}}_{\text{a}}$ to be $\text{4}\text{.75}$ too. The difference in those ${\text{pK}}_{\text{a}}$ values is $\text{4}\text{.75-(-1}\text{.7) = 6}\text{.45}$. Thu, cyclopentanol is the weaker acid by a factor of ${\text{10}}^{6.45}=2.82\times {10}^6$.

Conclusion

The favored equilibrium side with numerical value is determined on the basis of stronger acid and ${\text{pK}}_{\text{a}}$.

Interpretation Introduction

(f)

Interpretation:

The products for the given proton transfer reaction are to be drawn and the favored equilibrium side with numerical factor is to be determined.

Concept introduction:

The proton transfer reactions favor the side opposite the stronger acid. Larger the ${\text{pK}}_{\text{a}}$, weaker is the acid. The negative charge is more stable at an electronegative atom. Thus, the stability of a conjugate base increases with an increase in the electronegativity of an atom bearing negative charge. Each difference of $\text{1}$ in ${\text{pK}}_{\text{a}}$ values between the two acids corresponds to a power of $10$ by which that side of the reaction is favored.

Answer to Problem 6.41P

The products for the given reaction are:

The equilibrium is favored to the product side. Benzene is the weaker acid by a factor of ${\text{10}}^{\text{27}}=1.0\times {10}^{27}$.

Explanation of Solution

The given reaction is:

In the given reaction, the ${\text{sp}}^{\text{2}}$ hybridized carbon ion abstracts the proton attached to oxygen in given propanol and gives the following products:

As the oxygen atom is more electronegative than carbon, the negative charge on oxygen is more stable as compared to carbon. Thus, an anion on the right side, having negative charge on oxygen, is more stable than the anion on the left side where the negative charge is on carbon. Therefore, propanol is more acidic than benzene, and hence, the reaction is favored to the product side.

According to Appendix $\text{A}$, ${\text{pK}}_{\text{a}}$ of benzene is $\text{43}$ and that of primary alcohol ethanol is $\text{16}$. As the propanol does not have any nearby electronegative atoms or adjacent double bonds, we estimate its ${\text{pK}}_{\text{a}}$ to be $\text{16}$ too. The difference in those ${\text{pK}}_{\text{a}}$ values is $\text{43-16 = 27}$. Thus, benzene is the weaker acid by a factor of ${\text{10}}^{\text{27}}=1.0\times {10}^{27}$.

Conclusion

The favored equilibrium side with numerical value is determined on the basis of stronger acid and the ${\text{pK}}_{\text{a}}$.

Interpretation Introduction

(g)

Interpretation:

The products for the given proton transfer reaction are to be drawn and the favored equilibrium side with numerical factor is to be determined.

Concept introduction:

The proton transfer reactions favor the side opposite the stronger acid. Larger the ${\text{pK}}_{\text{a}}$, weaker is the acid. The stronger acid is the one whose negatively charged conjugate base is more stable. Each difference of $\text{1}$ in ${\text{pK}}_{\text{a}}$ values between the two acids corresponds to a power of $10$ by which that side of the reaction is favored.

Answer to Problem 6.41P

The products for the given reaction are:

The equilibrium is favored to the product side. Benzene is the weaker acid by a factor of ${\text{10}}^{\text{30}\text{.25}}=1.8\times {10}^{30}$.

Explanation of Solution

The given reaction is:

In the given reaction, the hydride ion abstracts the proton from carboxylic acid and gives the following products:

The conjugate base formed with a negative charge on the oxygen atom is better stabilized by the resonance effect. This makes the carboxylic acid the stronger acid, and the equilibrium is favored to the product side.

According to Appendix $\text{A}$, ${\text{pK}}_{\text{a}}$ of ethanoic acid is $\text{4}\text{.75}$ and that of ${\text{H}}_{\text{2}}$ is $35$. The difference in those ${\text{pK}}_{\text{a}}$ values is $\text{35-4}\text{.75 = 30}\text{.25}$. Thus, benzene is the weaker acid by a factor of ${\text{10}}^{\text{30}\text{.25}}=1.8\times {10}^{30}$.

Conclusion

The favored equilibrium side with numerical value is determined on the basis of stronger acid and the ${\text{pK}}_{\text{a}}$.