Chapter 10, Problem 10.3P

(a)

Interpretation Introduction

Interpretation:

The average dipole moment of 1,2-dichloroethane has to be determined if all the given three conformations are equal.

Concept Introduction:

Polar covalent bond involves an unequal sharing of electrons between two bonded atoms; it creates a partial separation of charge and a dipole moment.

Covalent bonds are polar, if there is a difference in electronegativity between the bonded atoms.

Calculation of the dipole moment of polyatomic molecules:

  $\text{μ}\text{=}{\left({\text{μ}}^{\text{2}}{}_{\text{1}}\text{+}{\text{μ}}^{\text{2}}{}_{\text{2}}\text{+}{\text{2μ}}_{\text{1}}{\text{μ}}_{\text{2}}\text{cosθ}\right)}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}$

Explanation of Solution

The average dipole moment of the molecule can be calculated by the equation given below:

  $\begin{array}{l}{\text{μ}}_{\text{average}}\text{=}\frac{\text{1}}{\text{N}}\left({\text{n}}_{\text{2}}{\text{μ}}_{\text{2}}\text{+}{\text{n}}_{\text{3}}{\text{μ}}_{\text{3}}\text{+}{\text{n}}_{\text{4}}{\text{μ}}_{\text{4}}\right)\\ \\ {\text{n}}_{\text{2}}{\text{,n}}_{\text{3}}\text{and}{\text{n}}_{\text{4}}-\text{weightings}\text{for}\text{each}\text{conformation}\\ \\ \text{N,}\text{total}\text{weighting}\text{=}{\text{n}}_{\text{2}}{\text{+n}}_{\text{3}}+{\text{n}}_{\text{4}}\end{array}$

Given that the dipole moment of each $\text{C}-\text{Cl}$  is $1.5\text{D}$ and also ${\text{n}}_{\text{2}}{\text{= n}}_{\text{3}}={\text{n}}_{\text{4}}$.

From the conformations, it is clear that conformation 2 is non polar and there will be no contribution to the mean dipole moment.

Thus, the mean dipole moment can be calculated by the below equation,

  $\begin{array}{c}{\text{μ}}_{\text{average}}\text{=}\frac{\text{1}}{3}\left({\text{μ}}_{\text{2}}\text{+}{\text{μ}}_{\text{3}}\text{+}{\text{μ}}_{\text{4}}\right)\\ \\ =\frac{\text{1}}{3}\left({\text{μ}}_{\text{3}}\text{+}{\text{μ}}_{\text{4}}\right)\end{array}$

${\text{μ}}_{\text{3}}\text{and}{\text{μ}}_{\text{4}}$ can be calculated using the equation,

  ${\text{μ}}_{\text{res}}\text{=}{\left({\text{μ}}^{\text{2}}{}_{\text{1}}\text{+}{\text{μ}}^{\text{2}}{}_{\text{2}}\text{+}{\text{2μ}}_{\text{1}}{\text{μ}}_{\text{2}}\text{cosθ}\right)}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}$

It can be rearranged as follows:

  ${\text{μ}}_{\text{res}}\text{=}{2}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\text{μ}{\left(1\text{+}\text{cosθ}\right)}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}$

  $\begin{array}{c}{\text{μ}}_{\text{3}}\text{=}{\text{2}}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\left(\text{1}\text{.5}\right){\left(\text{1}\text{+}\text{cos}\text{60}\right)}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\\ \\ \text{=}\text{2}\text{.59}\end{array}$

  $\begin{array}{c}{\text{μ}}_4\text{=}{\text{2}}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\left(\text{1}\text{.5}\right){\left(\text{1}\text{+}\text{cos}\text{60}\right)}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\\ \\ \text{=}\text{2}\text{.59}\end{array}$

Therefore, the mean dipole moment is,

  $\begin{array}{c}{\text{μ}}_{\text{average}}\text{=}\frac{\text{1}}{\text{3}}\left({\text{μ}}_{\text{3}}\text{+}{\text{μ}}_{\text{4}}\right)\\ \\ \text{=}\frac{\text{1}}{\text{3}}\left(\text{2}\text{.59}\text{+}\text{2}\text{.59}\right)\\ \\ \text{=}\text{1}\text{.72}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The dipole moment of 1,2-dichloroethane has to be determined if all only second conformation exist.

Concept Introduction:

Polar covalent bond involves an unequal sharing of electrons between two bonded atoms; it creates a partial separation of charge and a dipole moment.

Covalent bonds are polar, if there is a difference in electronegativity between the bonded atoms.

Calculation of the dipole moment of polyatomic molecules:

  $\text{μ}\text{=}{\left({\text{μ}}^{\text{2}}{}_{\text{1}}\text{+}{\text{μ}}^{\text{2}}{}_{\text{2}}\text{+}{\text{2μ}}_{\text{1}}{\text{μ}}_{\text{2}}\text{cosθ}\right)}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}$

Explanation of Solution

From the conformations, it is clear that conformation 2 is non polar and there will be no contribution to the mean dipole moment.

Therefore, the dipole moment will be zero for the compound, if only conformation 2 exist.

(c)

Interpretation Introduction

Interpretation:

The dipole moment of 1,2-dichloroethane has to be determined if the given conformations occur with probabilities in the ratio $\text{2:1:1}$.

Concept Introduction:

Polar covalent bond involves an unequal sharing of electrons between two bonded atoms; it creates a partial separation of charge and a dipole moment.

Covalent bonds are polar, if there is a difference in electronegativity between the bonded atoms.

Direction of dipole moment in a molecule is can be represented as follows,

Explanation of Solution

The average dipole moment of the molecule can be calculated by the equation given below:

  $\begin{array}{l}{\text{μ}}_{\text{average}}\text{=}\frac{\text{1}}{\text{N}}\left({\text{n}}_{\text{2}}{\text{μ}}_{\text{2}}\text{+}{\text{n}}_{\text{3}}{\text{μ}}_{\text{3}}\text{+}{\text{n}}_{\text{4}}{\text{μ}}_{\text{4}}\right)\\ \\ {\text{n}}_{\text{2}}{\text{,n}}_{\text{3}}\text{and}{\text{n}}_{\text{4}}-\text{weightings}\text{for}\text{each}\text{conformation}\\ \\ \text{N,}\text{total}\text{weighting}\text{=}{\text{n}}_{\text{2}}{\text{+n}}_{\text{3}}+{\text{n}}_{\text{4}}\end{array}$

Given the ratio for the conformation is $\text{2:1:1}$ which indicates that ${\text{n}}_{\text{2}}\text{= 2}{\text{;n}}_{\text{3}}=1;{\text{n}}_{\text{4}}=1$.

Thus, the mean dipole moment can be calculated by the below equation,

  $\begin{array}{c}{\text{μ}}_{\text{average}}\text{=}\frac{\text{1}}{\left(2+1+1\right)}\left({\text{μ}}_{\text{2}}\text{+}{\text{μ}}_{\text{3}}\text{+}{\text{μ}}_{\text{4}}\right)\\ \\ =\frac{\text{1}}{\left(2+1+1\right)}\left({\text{μ}}_{\text{3}}\text{+}{\text{μ}}_{\text{4}}\right)\end{array}$

${\text{μ}}_{\text{3}}\text{and}{\text{μ}}_{\text{4}}$ can be calculated using the equation,

  ${\text{μ}}_{\text{res}}\text{=}{\left({\text{μ}}^{\text{2}}{}_{\text{1}}\text{+}{\text{μ}}^{\text{2}}{}_{\text{2}}\text{+}{\text{2μ}}_{\text{1}}{\text{μ}}_{\text{2}}\text{cosθ}\right)}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}$

It can be rearranged as follows:

  ${\text{μ}}_{\text{res}}\text{=}{2}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\text{μ}{\left(1\text{+}\text{cosθ}\right)}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}$

  $\begin{array}{c}{\text{μ}}_{\text{3}}\text{=}{\text{2}}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\left(\text{1}\text{.5}\right){\left(\text{1}\text{+}\text{cos}\text{60}\right)}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\\ \\ \text{=}\text{2}\text{.59}\end{array}$

  $\begin{array}{c}{\text{μ}}_4\text{=}{\text{2}}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\left(\text{1}\text{.5}\right){\left(\text{1}\text{+}\text{cos}\text{60}\right)}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\\ \\ \text{=}\text{2}\text{.59}\end{array}$

The mean dipole moment if the conformations are present in a ratio of $\text{2:1:1}$,

  $\begin{array}{c}{\text{μ}}_{\text{average}}\text{=}\frac{\text{1}}{\left(2+1+1\right)}\left({\text{μ}}_{\text{3}}\text{+}{\text{μ}}_{\text{4}}\right)\\ \\ =\frac{\text{1}}{\left(2+1+1\right)}\left(2.59\text{+}2.59\right)\\ \\ =1.3\end{array}$

(d)

Interpretation Introduction

Interpretation:

The dipole moment of 1,2-dichloroethane has to be determined if the given conformations occur with probabilities in the ratio $\text{1:2:2}$.

Concept Introduction:

Polar covalent bond involves an unequal sharing of electrons between two bonded atoms; it creates a partial separation of charge and a dipole moment.

Covalent bonds are polar, if there is a difference in electronegativity between the bonded atoms.

Direction of dipole moment in a molecule is can be represented as follows,

Explanation of Solution

The average dipole moment of the molecule can be calculated by the equation given below:

  $\begin{array}{l}{\text{μ}}_{\text{average}}\text{=}\frac{\text{1}}{\text{N}}\left({\text{n}}_{\text{2}}{\text{μ}}_{\text{2}}\text{+}{\text{n}}_{\text{3}}{\text{μ}}_{\text{3}}\text{+}{\text{n}}_{\text{4}}{\text{μ}}_{\text{4}}\right)\\ \\ {\text{n}}_{\text{2}}{\text{,n}}_{\text{3}}\text{and}{\text{n}}_{\text{4}}-\text{weightings}\text{for}\text{each}\text{conformation}\\ \\ \text{N,}\text{total}\text{weighting}\text{=}{\text{n}}_{\text{2}}{\text{+n}}_{\text{3}}+{\text{n}}_{\text{4}}\end{array}$

Given the ratio for the conformation is $\text{1:2:2}$ which indicates that ${\text{n}}_{\text{2}}\text{= 1}{\text{;n}}_{\text{3}}=2;{\text{n}}_{\text{4}}=2$.

Thus, the mean dipole moment can be calculated by the below equation,

  $\begin{array}{c}{\text{μ}}_{\text{average}}\text{=}\frac{\text{1}}{\left(1+2+2\right)}\left({\text{μ}}_{\text{2}}\text{+}{\text{μ}}_{\text{3}}\text{+}{\text{μ}}_{\text{4}}\right)\\ \\ =\frac{\text{1}}{\left(1+2+2\right)}\left({\text{μ}}_{\text{3}}\text{+}{\text{μ}}_{\text{4}}\right)\end{array}$

${\text{μ}}_{\text{3}}\text{and}{\text{μ}}_{\text{4}}$ can be calculated using the equation,

  ${\text{μ}}_{\text{res}}\text{=}{\left({\text{μ}}^{\text{2}}{}_{\text{1}}\text{+}{\text{μ}}^{\text{2}}{}_{\text{2}}\text{+}{\text{2μ}}_{\text{1}}{\text{μ}}_{\text{2}}\text{cosθ}\right)}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}$

It can be rearranged as follows:

  ${\text{μ}}_{\text{res}}\text{=}{2}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\text{μ}{\left(1\text{+}\text{cosθ}\right)}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}$

  $\begin{array}{c}{\text{μ}}_{\text{3}}\text{=}{\text{2}}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\left(\text{1}\text{.5}\right){\left(\text{1}\text{+}\text{cos}\text{60}\right)}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\\ \\ \text{=}\text{2}\text{.59}\end{array}$

  $\begin{array}{c}{\text{μ}}_4\text{=}{\text{2}}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\left(\text{1}\text{.5}\right){\left(\text{1}\text{+}\text{cos}\text{60}\right)}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\\ \\ \text{=}\text{2}\text{.59}\end{array}$

The mean dipole moment if the conformations are present in a ratio of $\text{1:2:2}$,

  $\begin{array}{c}{\text{μ}}_{\text{average}}\text{=}\frac{\text{1}}{\left(1+2+2\right)}\left({\text{μ}}_{\text{3}}\text{+}{\text{μ}}_{\text{4}}\right)\\ \\ =\frac{\text{1}}{\left(1+2+2\right)}\left(2.59\text{+}2.59\right)\\ \\ =1.036\end{array}$