Chapter 2.SE, Problem 65AP

Interpretation Introduction

a) CH₃(CH₂)₂₉CH₃, a component found in paraffin wax is a solid at room temperature while octane is a liquid.

Interpretation:

The observation that, CH₃(CH₂)₂₉CH₃, a component found in paraffin wax is a solid at room temperature while octane is a liquid is to be explained.

Concept introduction:

Even in non polar molecules, the attractive dispersion forces are possible, caused by temporary dipoles arising due to the costant change in electon distribution wihin the molecule. These temporary dipoles, though have a fleeting existence are constantly changing. But their cumulative effect may be strong enough to hold the molecule close together so that a substance can be a solid or liquid.

To explain:

The observation that, CH₃(CH₂)₂₉CH₃, a component found in paraffin wax is a solid at room temperature while octane is a liquid.

Interpretation Introduction

b) CH₃CH₂CH₂OH has a higher boiling point than CH₄

Interpretation:

The observation CH₃CH₂CH₂OH has a higher boiling point than CH₄.

Concept introduction:

An attractive interaction between a hydrogen bonded to an electronegative oxygen or nitrogen atom and an unshared electron pair on another oxygen or nitrogrn atom is termed as hydrogen bonding. It is a very strong dipole-dipole interaction that leads to molecular association. These molecules associated through hydrogen bonding will have a high boiling point.

To explain:

The observation CH₃CH₂CH₂OH has a higher boiling point than CH₄.

Interpretation Introduction

c) CH₃CO₂H, which is found in vinegar, will dissolve in water but not in oil- for simplicity you may assume oil is CH₃(CH₂)₄CH₃

Interpretation:

The observation that CH₃CO₂H, which is found in vinegar, will dissolve in water but not in oil with molecular formula CH₃(CH₂)₄CH₃.

Concept introduction:

Like dissolves like. A polar molecule will dissolve in another polar solvent. The solubility can also be explained using the formation of hydrogen bonding between the two substances.