Chapter 3, Problem 43PP

(a)

Interpretation Introduction

Interpretation:

The more stable anion can be identified in each of the given anions.

Concept introduction:

The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.

In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized.

The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is ${\text{sp}}^{\text{3}}\text{<}{\text{sp}}^{\text{2}}\text{< }\text{sp }$.

To Identify the more stable anion present in the case

(b)

Interpretation Introduction

Interpretation:

The more stable anion can be identified in each of the given anions.

Concept introduction:

The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.

In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized.

The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is ${\text{sp}}^{\text{3}}\text{<}{\text{sp}}^{\text{2}}\text{< }\text{sp }$.

To Identify the more stable anion present in the case

(c)

Interpretation Introduction

Interpretation:

The more stable anion can be identified in each of the given anions.

Concept introduction:

The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.

In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized.

The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is ${\text{sp}}^{\text{3}}\text{<}{\text{sp}}^{\text{2}}\text{< }\text{sp }$.

To Identify the more stable anion present in the case