Chapter 15, Problem 15.44P

Interpretation Introduction

(a)

Interpretation:

Problem 15.43 calls attention to a resonance structure of an amide that accounts for the lowering of the $\text{C=O}$ stretch frequency. The resonance structure for ${\text{methyl acetate (CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{CH}}_{\text{3}}\text{)}$ is to be drawn.

Concept introduction:

An ester has O atom having a lone pair on it. The lone pair on O atom participates in resonance with $\text{C=O}$.

Interpretation Introduction

(b)

Interpretation:

Problem 15.43 calls attention to a resonance structure of an amide that accounts for the lowering of the $\text{C=O}$ stretch frequency. An analogous resonance structure can be drawn for an ester. Based on the fact that an ester’s $\text{C=O}$ stretch frequency is higher than an amide’s, the resonance structure that has a larger contribution toward the resonance hybrid of an ester or an amide is to be determined.

Concept introduction:

The $\text{C-O}$ single bond is weaker than $\text{C=O}$ bond. The $\text{C-O}$ single bond character decreases the stiffness of $\text{C=O}$ bond. The stretching frequency of $\text{C=O}$ band lowers by the Weakening of $\text{C=O}$ bond. Oxygen is higher electronegative than nitrogen. The positive charge is less stable on O atom than N atom. In the resonance of amide, the C-O single bond character is more significant than that in ester.