Chapter 3, Problem 37PP

(a)

Interpretation Introduction

Interpretation:

Conjugate bases of given acids are needed to be drawn.

Concept introduction:

According to Bronsted-Lowry concept, acid is a proton donor. The deprotonated acids are called conjugate base. The stability of this conjugate base determines the location of deprotonation in an acid.

In a conjugate base where the negative charge is carried by a more electronegative atom will show more stability. The greater electronegativity makes lone pair closer to the nucleus hence more stabilized.

The conjugate base having resonance will have greater stability as compared to conjugate base having single anion. Because of the resonance conjugate base will get resonance stabilization.

Answer to Problem 37PP

Explanation of Solution

The most acidic proton of the given acid (a) is $\text{O-H}$ proton, because deprotonation at oxygen gives a conjugate base having negative charge on oxygen atom, it is very much stable as compared to negative charge on carbon atom which is resulted by the deprotonation of any $\text{C}-\text{H}$

(b)

Interpretation Introduction

Interpretation:

Conjugate bases of given acids are needed to be drawn.

Concept introduction:

According to Bronsted-Lowry concept, acid is a proton donor. The deprotonated acids are called conjugate base. The stability of this conjugate base determines the location of deprotonation in an acid.

In a conjugate base where the negative charge is carried by a more electronegative atom will show more stability. The greater electronegativity makes lone pair closer to the nucleus hence more stabilized.

The conjugate base having resonance will have greater stability as compared to conjugate base having single anion. Because of the resonance conjugate base will get resonance stabilization.

Answer to Problem 37PP

Explanation of Solution

The most acidic hydrogen atom presented in the given acid is $\text{C}-\text{H}$ located adjacent to $\text{C}\text{=}\text{O}$. The deprotonation of this $\text{C}-\text{H}$ will results in the formation of a conjugate base will have resonance stabilization. This is shown below,

.

No other $\text{C}-\text{H}$ deprotonation will give a conjugate base stable than above shown.

(c)

Interpretation Introduction

Interpretation:

Conjugate bases of given acids are needed to be drawn.

Concept introduction:

According to Bronsted-Lowry concept, acid is a proton donor. The deprotonated acids are called conjugate base. The stability of this conjugate base determines the location of deprotonation in an acid.

In a conjugate base where the negative charge is carried by a more electronegative atom will show more stability. The greater electronegativity makes lone pair closer to the nucleus hence more stabilized.

The conjugate base having resonance will have greater stability as compared to conjugate base having single anion. Because of the resonance conjugate base will get resonance stabilization.

Answer to Problem 37PP

Explanation of Solution

Deprotonation of ${\text{NH}}_3$ gives ${}^{-}{\text{NH}}_2$,

(d)

Interpretation Introduction

Interpretation:

Conjugate bases of given acids are needed to be drawn.

Concept introduction:

According to Bronsted-Lowry concept, acid is a proton donor. The deprotonated acids are called conjugate base. The stability of this conjugate base determines the location of deprotonation in an acid.

In a conjugate base where the negative charge is carried by a more electronegative atom will show more stability. The greater electronegativity makes lone pair closer to the nucleus hence more stabilized.

The conjugate base having resonance will have greater stability as compared to conjugate base having single anion. Because of the resonance conjugate base will get resonance stabilization.

Answer to Problem 37PP

    

Explanation of Solution

Deprotonation of ${\text{H}}_3{\text{O}}^{+}$ gives ${\text{H}}_2\text{O}$,

(e)

Interpretation Introduction

Interpretation:

Conjugate bases of given acids are needed to be drawn.

Concept introduction:

According to Bronsted-Lowry concept, acid is a proton donor. The deprotonated acids are called conjugate base. The stability of this conjugate base determines the location of deprotonation in an acid.

In a conjugate base where the negative charge is carried by a more electronegative atom will show more stability. The greater electronegativity makes lone pair closer to the nucleus hence more stabilized.

The conjugate base having resonance will have greater stability as compared to conjugate base having single anion. Because of the resonance conjugate base will get resonance stabilization.

Answer to Problem 37PP

Explanation of Solution

The most acidic proton of the given acid (e) is $\text{O-H}$ proton, because the deprotonation at oxygen gives a conjugate base having negative charge on oxygen atom. It is more stable than the negative charge on carbon atom which is resulted by the deprotonation of any $\text{C}-\text{H}$. There is resonance stabilization after deprotonation of $\text{O-H}$ in the conjugate base.

(f)

Interpretation Introduction

Interpretation:

Conjugate bases of given acids are needed to be drawn.

Concept introduction:

According to Bronsted-Lowry concept, acid is a proton donor. The deprotonated acids are called conjugate base. The stability of this conjugate base determines the location of deprotonation in an acid.

In a conjugate base where the negative charge is carried by a more electronegative atom will show more stability. The greater electronegativity makes lone pair closer to the nucleus hence more stabilized.

The conjugate base having resonance will have greater stability as compared to conjugate base having single anion. Because of the resonance conjugate base will get resonance stabilization.

Answer to Problem 37PP

Explanation of Solution

The most acidic proton of the given acid (e) is ${}^{-}{\text{NH}}_2$ proton, because deprotonation at nitrogen atom gives a conjugate base having negative charge on nitrogen atom, it is very much stable as compared to negative charge on carbon atom which is resulted by the deprotonation any $\text{C}-\text{H}$ in the above compound.

(g)

Interpretation Introduction

Interpretation:

Conjugate bases of given acids are needed to be drawn.

Concept introduction:

According to Bronsted-Lowry concept, acid is a proton donor. The deprotonated acids are called conjugate base. The stability of this conjugate base determines the location of deprotonation in an acid.

In a conjugate base where the negative charge is carried by a more electronegative atom will show more stability. The greater electronegativity makes lone pair closer to the nucleus hence more stabilized.

The conjugate base having resonance will have greater stability as compared to conjugate base having single anion. Because of the resonance conjugate base will get resonance stabilization.

Answer to Problem 37PP

Explanation of Solution

Deprotonation of ${\text{NH}}_4{}^{+}$ gives ${\text{NH}}_3$,