Chapter 15, Problem 15.2P

Interpretation Introduction

Interpretation:

The delocalization energy of $\left(E\right)-\text{1, 3, 5}-\text{hexatriene}$ is to be calculated.

Concept introduction:

The term conjugated dienes is used when the double bonds are present alternatively in a hydrocarbon chain. The mathematical function that describes the wave like behavior of an electron in a molecule is expressed by molecular orbital.

The delocalization energy for conjugated molecules is defined as the extra stability that a molecule attains by spreading its electron over the entire molecule. The delocalization of $\pi$ electrons gives the delocalization energy.

Answer to Problem 15.2P

The delocalization energy of $\left(E\right)-\text{1, 3, 5}-\text{hexatriene}$ is $49\text{kJ}{\text{mol}}^{-\text{1}}$.

Explanation of Solution

The delocalization energy of $\left(E\right)-\text{1, 3, 5}-\text{hexatriene}$ is calculated as the difference of $\text{π-electron}$ energy of three isolated ethylene molecules and $\text{π-electrons}$ energy of $\left(E\right)-\text{1, 3, 5}-\text{hexatriene}$. The expression for the delocalization energy of $\left(E\right)-\text{1, 3, 5}-\text{hexatriene}$ is shown below.

$\begin{array}{l}\text{Delocalisation}\text{energy}\text{of}\left(E\right)-\text{1, 3, 5}-\text{hexatriene}\\ =\left(\begin{array}{l}\text{π-electron}\text{energy}\text{of}\text{BMOs}\text{of}\\ \left(E\right)-\text{1, 3, 5}-\text{hexatriene}\text{molecule}\end{array}\right)-\left(\begin{array}{l}\text{π-electron}\text{energy}\text{of}\text{BMOs}\text{of}\\ \text{three}\text{ethylene}\text{molecule}\end{array}\right)\\ \end{array}$ …(1)

The energy of ethylene bonding molecular orbital is $+1.00\text{β}$, where, the symbol $\text{β}$ is the unit of energy for $\text{π}$ molecular orbital and has value $-50\text{kJ}{\text{mol}}^{-\text{1}}$ approximately. The $\text{β}$ has a negative value by convention.

The formula for the calculation of $\text{π-electron}$ energy is given below.

$\begin{array}{l}\text{π-electron}\text{energy}\text{of}\text{BMOs}\text{of}\text{ethylene}\text{molecule}\\ =\text{Energy}\text{of}\text{ethylene}\text{molecules}\times \text{Number}\text{of}\text{electrons}\end{array}$ …(2)

The bonding molecular orbital (BMOs) of ethylene contains two electrons.

The calculation of $\text{π-electron}$ energy of bonding molecular orbitals of one ethylene molecule is given below.

$\begin{array}{rll}\text{π-electron}\text{energy of BMOs of one ethylene molecule}& =& \left(+1.00\text{β}\right)\times 2\\ & =& +2.00\text{β}\end{array}$

The calculation of $\text{π-electron}$ energy of bonding molecular orbitals of three ethylene molecule is given below.

$\begin{array}{rll}\text{π-electron}\text{energy of BMOs of three ethylene molecule}& =& \left(+2.00\text{β}\right)\times 3\\ & =& +6.00\text{β}\end{array}$ …(3)

The diagram for orbital interaction in $\left(E\right)-\text{1, 3, 5}-\text{hexatriene}$ showing the combination of six $2p$ orbitals is given below.

Figure 1

The $\pi \text{-electrons}$ energy of $\left(E\right)-\text{1, 3, 5}-\text{hexatriene}$ molecule is given by the formula shown below.

$\left(\begin{array}{l}\pi \text{-electron}\text{energy}\text{of}\text{BMOs}\text{of}\\ \left(E\right)-\text{1, 3, 5}-\text{hexatriene}\text{molecule}\end{array}\right)=\left(\begin{array}{l}\text{Energy}\text{of}\text{electrons}\text{in}{\text{π}}_{\text{1}}\\ +\text{Energy}\text{of}\text{electrons}\text{in}{\text{π}}_{\text{2}}\\ +\text{Energy}\text{of}\text{electrons}\text{in}{\text{π}}_{\text{3}}\end{array}\right)$ …(4)

The energy of electrons in ${\text{π}}_1$ bonding molecular orbital is calculated below.

$\begin{array}{rll}\text{Energy}\text{of}\text{electrons}\text{in}{\text{π}}_{\text{1}}& =& \left(+1.80\text{β}\right)\times 2\\ & =& +3.60\text{β}\end{array}$

The energy of electrons in ${\text{π}}_2$ bonding molecular orbital is calculated below.

$\begin{array}{rll}\text{Energy}\text{of}\text{electrons}\text{in}{\text{π}}_{\text{2}}& =& \left(+1.25\text{β}\right)\times 2\\ & =& +2.50\text{β}\end{array}$

The energy of electrons in ${\text{π}}_{\text{3}}$ bonding molecular orbital is calculated below.

$\begin{array}{rll}\text{Energy}\text{of}\text{electrons}\text{in}{\text{π}}_{\text{1}}& =& \left(+0.44\text{β}\right)\times 2\\ & =& +0.88\text{β}\end{array}$

Substitute the values of the energies of ${\text{π}}_1$, ${\text{π}}_2$ and ${\text{π}}_{\text{3}}$ bonding molecular orbitals in the equation (4) and calculate the $\text{π-electrons}$ energy of bonding molecular orbitals of $\left(E\right)-\text{1, 3, 5}-\text{hexatriene}$molecule.

$\begin{array}{rll}\left(\begin{array}{l}\text{π-electron}\text{energy}\text{of}\text{BMOs}\text{of}\\ \left(E\right)-\text{1, 3, 5}-\text{hexatriene}\text{molecule}\end{array}\right)& =& \left(\begin{array}{l}\left(\text{+3}\text{.60β}\right)+\left(\text{+2}\text{.50β}\right)\\ +\left(\text{+0}\text{.88β}\right)\end{array}\right)\\ & =& +6.98\text{β}\end{array}$ …(5)

Now, substitute the value from equation (3) and (5) in equation (1) in order to calculate the value of delocalization energy of the molecule $\left(E\right)-\text{1, 3, 5}-\text{hexatriene}$.

$\begin{array}{rll}\text{Delocalisation}\text{energy}\text{of}\left(E\right)-\text{1, 3, 5}-\text{hexatriene}& =& \left(+6.98\text{β}\right)-\left(+6.00\text{β}\right)\\ & =& 0.98\text{β}\end{array}$

Now, $\text{β}$ is $-50\text{kJ}{\text{mol}}^{-1}$. Substitute the value of $\text{β}$ in the above equation as shown below.

$\begin{array}{rll}\text{Delocalisation}\text{energy}\text{of}\left(E\right)-\text{1, 3, 5}-\text{hexatriene}& =& \left(+0.98\beta \right)\times \left(-50\text{kJ}{\text{mol}}^{-\text{1}}\right)\\ & =& 49\text{kJ}{\text{mol}}^{-\text{1}}\end{array}$

Therefore, the delocalization energy obtained for $\left(E\right)-\text{1, 3, 5}-\text{hexatriene}$ is $49\text{kJ}{\text{mol}}^{-\text{1}}$.

Conclusion

The molecule $\left(E\right)-\text{1, 3, 5}-\text{hexatriene}$ has the delocalization energy $49\text{kJ}{\text{mol}}^{-\text{1}}$.