Chapter 11, Problem 11.138QP

(a)

Interpretation Introduction

Interpretation:

Consider the compounds,

${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$(1-Pentanol) and ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$(Hexane), given statements has to be explained.

Concept introduction:

Intermolecular forces are Van der Waals forces.  They are weak and have threetypes viz., London dispersion forces, dipole-dipole forces and hydrogen bonding. Hydrogen bonding is relatively the strongest one.

• Intermolecular forces are the forces acting between molecules whereas Intramolecular forces are the forces that operate within a molecule.
• Hydrogen bonding is a special type of Dipole-dipole forces but stronger than the former.
• London dispersion forces exist in non-polar covalent compounds whereas dipole-dipole forces exist in polar covalent compounds but both are weak.
• Larger the molecular size, stronger the London dispersion force.
• Arrangement of major types of intermolecular forces in increasing order of strength:

$\text{London}\text{dispersion}\text{forces}\text{<}\text{Dipole-dipole}\text{forces}\text{<}\text{Hydrogen}\text{bonding}$

There exist no deviations in this arrangement.

Explanation of Solution

Intermolecular force in each compound

In a 1-Pentanol, the intermolecular forces are London forces, dipole-dipole forces and hydrogen bonding, whereas only London forces are present in Hexane.

(b)

Interpretation Introduction

Interpretation:

Consider the compounds,

${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$(1-Pentanol) and ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$(Hexane), given statements has to be explained.

Concept introduction:

Intermolecular forces are Van der Waals forces.  They are weak and have threetypes viz., London dispersion forces, dipole-dipole forces and hydrogen bonding. Hydrogen bonding is relatively the strongest one.

• Intermolecular forces are the forces acting between molecules whereas Intramolecular forces are the forces that operate within a molecule.
• Hydrogen bonding is a special type of Dipole-dipole forces but stronger than the former.
• London dispersion forces exist in non-polar covalent compounds whereas dipole-dipole forces exist in polar covalent compounds but both are weak.
• Larger the molecular size, stronger the London dispersion force.
• Arrangement of major types of intermolecular forces in increasing order of strength:

$\text{London}\text{dispersion}\text{forces}\text{<}\text{Dipole-dipole}\text{forces}\text{<}\text{Hydrogen}\text{bonding}$

There exist no deviations in this arrangement.

Explanation of Solution

Boiling point increases with increasing intermolecular forces. Hence, 1-Pentanol has higher intermolecular forces and hence it has higher boiling point. Thus Hexane has low boiling point due to its low intermolecular forces. That is the boiling point would be $\text{69°C}$.

(c)

Interpretation Introduction

Interpretation:

Consider the compounds,

${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$(1-Pentanol) and ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$(Hexane), given statements has to be explained.

Concept introduction:

Intermolecular forces are Van der Waals forces.  They are weak and have threetypes viz., London dispersion forces, dipole-dipole forces and hydrogen bonding. Hydrogen bonding is relatively the strongest one.

• Intermolecular forces are the forces acting between molecules whereas Intramolecular forces are the forces that operate within a molecule.
• Hydrogen bonding is a special type of Dipole-dipole forces but stronger than the former.
• London dispersion forces exist in non-polar covalent compounds whereas dipole-dipole forces exist in polar covalent compounds but both are weak.
• Larger the molecular size, stronger the London dispersion force.
• Arrangement of major types of intermolecular forces in increasing order of strength:

$\text{London}\text{dispersion}\text{forces}\text{<}\text{Dipole-dipole}\text{forces}\text{<}\text{Hydrogen}\text{bonding}$

There exist no deviations in this arrangement.

Explanation of Solution

To assign: viscosities of each compound

Viscosity increases with increasing intermolecular forces. Hence, 1-Pentanol has higher intermolecular forces and viscosity of this compound should be $\text{2}\text{.987g/(cm•s)}$ and hexane has weaker intermolecular forces, so it should have viscosity of $\text{0}\text{.313 g/(cm•s)}$.

(d)

Interpretation Introduction

Interpretation:

Consider the compounds,

${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$(1-Pentanol) and ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$(Hexane), given statements has to be explained.

Concept introduction:

Intermolecular forces are Van der Waals forces.  They are weak and have threetypes viz., London dispersion forces, dipole-dipole forces and hydrogen bonding. Hydrogen bonding is relatively the strongest one.

• Intermolecular forces are the forces acting between molecules whereas Intramolecular forces are the forces that operate within a molecule.
• Hydrogen bonding is a special type of Dipole-dipole forces but stronger than the former.
• London dispersion forces exist in non-polar covalent compounds whereas dipole-dipole forces exist in polar covalent compounds but both are weak.
• Larger the molecular size, stronger the London dispersion force.
• Arrangement of major types of intermolecular forces in increasing order of strength:

$\text{London}\text{dispersion}\text{forces}\text{<}\text{Dipole-dipole}\text{forces}\text{<}\text{Hydrogen}\text{bonding}$

There exist no deviations in this arrangement.

Explanation of Solution

To identify: boiling point of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Cl}$ lie between or higher or lower than 1-pentanol and hexane

The compound ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Cl}$ has dipole-dipole forces and London forces. Therefore boiling point of this compound should be intermediate between 1-pentanol and hexane.