Chapter 2, Problem 2.51P

Rank the following ions in order of increasing basicity.

a. ${\text{CH}}_{\text{3}}\stackrel{-}{\text{C}}{\text{H}}_{\text{2}}{\text{, CH}}_{\text{3}}{\text{O}}^{-}{\text{, CH}}_{\text{3}}\stackrel{-}{\text{N}}\text{H}$ c.

b. ${\text{CH}}_3^{-}{\text{, HO}}^{-}{\text{, Br}}^{-}$ d.

Interpretation Introduction

(a)

Interpretation: The order of increasing basicity for the given ions is to be ranked.

Concept introduction: A strong base possesses a weak conjugate acid and a strong acid has a weak conjugate base. In the periodic table, the basicity of ions decreases across a row as the electronegativity increases on moving from left to right and down a column as the size of atom increases on moving from top to bottom.

Answer to Problem 2.51P

The increasing order of basicity for the given ions is ${\text{ CH}}_{\text{3}}{\text{O}}^{-}{\text{ < CH}}_{\text{3}}\stackrel{-}{\text{N}}{\text{H < CH}}_{\text{3}}\stackrel{-}{\text{C}}{\text{H}}_{\text{2}}$.

Explanation of Solution

The given ions are ${\text{CH}}_{\text{3}}\stackrel{-}{\text{C}}{\text{H}}_{\text{2}}{\text{, CH}}_{\text{3}}{\text{O}}^{-}{\text{, CH}}_{\text{3}}\stackrel{-}{\text{N}}\text{H}$. Oxygen is more electronegative than nitrogen. The formation of more stable anion occurs by the more electronegative atom that bears a negative charge. The more stable anion are weaker bases. Therefore, in the given anions ${\text{CH}}_{\text{3}}{\text{O}}^{-}$ is a weaker base as compared to ${\text{CH}}_{\text{3}}\stackrel{-}{\text{N}}\text{H}$. Also, the basicity of ions decreases across a row as the electronegativity increases. Hence, the increasing order of basicity for the given ions is ${\text{ CH}}_{\text{3}}{\text{O}}^{-}{\text{ < CH}}_{\text{3}}\stackrel{-}{\text{N}}{\text{H < CH}}_{\text{3}}\stackrel{-}{\text{C}}{\text{H}}_{\text{2}}$.

Conclusion

The increasing order of basicity for the given ions is ${\text{ CH}}_{\text{3}}{\text{O}}^{-}{\text{ < CH}}_{\text{3}}\stackrel{-}{\text{N}}{\text{H < CH}}_{\text{3}}\stackrel{-}{\text{C}}{\text{H}}_{\text{2}}$.

Interpretation Introduction

(b)

Interpretation: The order of increasing basicity for the given ions is to be ranked.

Concept introduction: A strong base possesses a weak conjugate acid and a strong acid has a weak conjugate base. In the periodic table, the basicity of ions decreases across a row as the electronegativity increases on moving from left to right and down a column as the size of atom increases on moving from top to bottom.

Answer to Problem 2.51P

The increasing order of basicity for the given ions is ${\text{Br}}^{-}{\text{ < HO}}^{-}{\text{ < CH}}_3^{-}$.

Explanation of Solution

The given ions are ${\text{CH}}_3^{-}{\text{, HO}}^{-}{\text{, Br}}^{-}$. The basicity of ions decreases across a row as the electronegativity increases. A strong acid has a weak conjugate base. The conjugate acid of ${\text{CH}}_3^{-}$ is ${\text{CH}}_{\text{4}}$ which is a very weak acid. The conjugate acid of ${\text{OH}}^{-}$ is ${\text{H}}_{\text{2}}\text{O}$ which is a weak acid and the conjugate acid of ${\text{Br}}^{-}$ is $\text{HBr}$ which is a strong acid. Therefore, ${\text{Br}}^{-}$ is a weak base than ${\text{OH}}^{-}$ and ${\text{OH}}^{-}$ is a weak base than ${\text{CH}}_3^{-}$. Hence, the increasing order of basicity for the given ions is ${\text{Br}}^{-}{\text{ < HO}}^{-}{\text{ < CH}}_3^{-}$.

Conclusion

The increasing order of basicity for the given ions is ${\text{Br}}^{-}{\text{ < HO}}^{-}{\text{ < CH}}_3^{-}$.

Interpretation Introduction

(c)

Interpretation: The order of increasing basicity for the given ions is to be ranked.

Concept introduction: A strong base possesses a weak conjugate acid and a strong acid has a weak conjugate base. In the periodic table, the basicity of ions decreases across a row as the electronegativity increases on moving from left to right and down a column as the size of atom increases on moving from top to bottom.

Answer to Problem 2.51P

The order of increasing basicity for the given ions is ${\text{ClCH}}_{\text{2}}{\text{COO}}^{-}{\text{ < CH}}_{\text{3}}{\text{COO}}^{-}{\text{ < CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$.

Explanation of Solution

The given ions are ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$ and ${\text{ClCH}}_{\text{2}}{\text{COO}}^{-}$. The ion ${\text{ClCH}}_{\text{2}}{\text{COO}}^{-}$, contains chlorine atom which is an electron withdrawing group and makes the ion more acidic. Therefore, it is less basic than the other two ions.

The conjugate acid of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$ is ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ which is an alcohol. Alcohols are weaker acids as compared to carboxylic acids. Therefore, their conjugate bases are stronger than the conjugate bases of carboxylic acids. Hence, the order of increasing basicity for the given ions is ${\text{ClCH}}_{\text{2}}{\text{COO}}^{-}{\text{ < CH}}_{\text{3}}{\text{COO}}^{-}{\text{ < CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$.

Conclusion

The order of increasing basicity for the given ions is ${\text{ClCH}}_{\text{2}}{\text{COO}}^{-}{\text{ < CH}}_{\text{3}}{\text{COO}}^{-}{\text{ < CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$.

Interpretation Introduction

(d)

Interpretation: The order of increasing basicity for the given ions is to be ranked.

Concept introduction: In hydrocarbons, alkanes, alkenes and alkynes, the order of basicity is predicted by their s-character. More is the s-character, less is the basicity or less is the s-character, more is the basicity.

Answer to Problem 2.51P

The order of increasing basicity for the given ions is shown in Figure 2.

Explanation of Solution

The given ions are,

Figure 1

More is the s-character of a compound, less is the basicity. Alkanes contain $25\%$ s-character, alkenes have $33\%$ s-character and alkynes have $50\%$ s-character. Hence, the correct order of increasing basicity is shown below.

Figure 2

Conclusion

The order of increasing basicity for the given ions is shown in Figure 2.