Chapter 9, Problem 9E.2P

(a)

Interpretation Introduction

Interpretation:

The energies of $\pi$ molecular orbitals of benzene and cyclooctatetraene have to be calculated.  Whether the degenerate energy levels are present or absent have to be commented.

Concept introduction:

Delocalization of energy refers to the extra stabilization of the molecule which is due to allowing the electrons to spread over the entire molecule.  To calculate the delocalization energy, the energy of the molecule in question is compared with a molecule in which electrons cannot spread over the molecule.  Energy levels which have the same value of energy but are separate are known as degenerate energy levels.

Answer to Problem 9E.2P

The energies of the $\pi$ molecular orbitals of benzene are $\alpha +2\beta$, $\alpha +\beta$, $\alpha +\beta$, $\alpha -\beta$, $\alpha -\beta$, $\alpha -2\beta$.  The energies of the $\pi$ molecular orbitals of cyclooctatetraene are $\alpha +2\beta$, $\alpha +1.414\beta$, $\alpha +1.414\beta$, $\alpha$, $\alpha$, $\alpha -1.414\beta$, $\alpha -1.414\beta$, $\alpha -2\beta$.

Benzene has two degenerate levels of energy, $\alpha +\beta$ and two degenerate levels of energy $\alpha -\beta$.  Cyclooctatetraene has a total of $6$ degenerate levels.  $2$ levels have the energy $\alpha +1.414\beta$, $2$ levels have the energy $\alpha$, and $2$ levels have the energy $\alpha -1.414\beta$.

Explanation of Solution

The expression for the energies $E_k$ of the $\pi$ molecular orbitals is shown below.

    $E_k=\alpha +2\beta \cos \frac{2k\pi }{N}$        (1)

Where,

• $N$ is the number of carbon atoms.
• $\alpha$ is coulomb integral.
• $\beta$ is resonance integral.
• $k=\frac{N}{2}$ for even $N$.
• $k=\frac{N-1}{2}$ for odd value of $N$.

In benzene, the number of carbon atoms, $N$ is $6$.

Therefore, the value of $k$ is calculated as shown below.

    $\begin{array}{c}k=\frac{6}{2}\\ =3\\ =0,\pm 1,\pm 2,3\end{array}$

Substitute the value of $k=0$ and $N=6$ in equation (1).

    $\begin{array}{c}{E}_0=\alpha +2\beta \cos \frac{2\times 0\times \pi }{6}\\ =\alpha +2\beta \text{as}\cos 0=1\end{array}$

Substitute the value of $k=1$ and $N=6$ in equation (1).

    $\begin{array}{c}{E}_1=\alpha +2\beta \cos \frac{2\times 1\times \pi }{6}\\ =\alpha +2\beta \cos \frac{\pi }{3}\\ =\alpha +\left(2\beta \times \frac{1}{2}\right)\text{as}\cos \frac{\pi }{3}=\frac{1}{2}\\ =\alpha +\beta \end{array}$

Substitute the value of $k=-1$ and $N=6$ in equation (1).

    $\begin{array}{c}{E}_{-1}=\alpha +2\beta \cos \frac{2\times \left(-1\right)\times \pi }{6}\\ =\alpha +2\beta \cos \left(-\frac{\pi }{3}\right)\\ =\alpha +\left(2\beta \times \frac{1}{2}\right)\text{as}\cos \left(-\frac{\pi }{3}\right)=\frac{1}{2}\\ =\alpha +\beta \end{array}$

Substitute the value of $k=2$ and $N=6$ in equation (1).

    $\begin{array}{c}{E}_2=\alpha +2\beta \cos \frac{2\times 2\times \pi }{6}\\ =\alpha +2\beta \cos \left(\frac{2\pi }{3}\right)\\ =\alpha +\left(2\beta \times \left(-\frac{1}{2}\right)\right)\text{as}\cos \left(\frac{2\pi }{3}\right)=-\frac{1}{2}\\ =\alpha -\beta \end{array}$

Substitute the value of $k=-2$ and $N=6$ in equation (1).

    $\begin{array}{c}{E}_{-2}=\alpha +2\beta \cos \frac{2\times \left(-2\right)\times \pi }{6}\\ =\alpha +2\beta \cos \left(-\frac{2\pi }{3}\right)\\ =\alpha +\left(2\beta \times \left(-\frac{1}{2}\right)\right)\text{as}\cos \left(-\frac{2\pi }{3}\right)=-\frac{1}{2}\\ =\alpha -\beta \end{array}$

Substitute the value of $k=3$ and $N=6$ in equation (1).

    $\begin{array}{c}{E}_3=\alpha +2\beta \cos \frac{2\times 3\times \pi }{6}\\ =\alpha +2\beta \cos \left(\pi \right)\\ =\alpha +\left(2\beta \times \left(-1\right)\right)\text{as}\cos \pi =-1\\ =\alpha -2\beta \end{array}$

Therefore, the energies of the $\pi$ molecular orbitals of benzene are $\alpha +2\beta$, $\alpha +\beta$, $\alpha +\beta$, $\alpha -\beta$, $\alpha -\beta$, $\alpha -2\beta$.

Benzene has two degenerate levels of energy, $\alpha +\beta$ and two degenerate levels of energy $\alpha -\beta$.

In cyclooctatetraene, the number of carbon atoms, $N$ is $8$.

Therefore, the value of $k$ is calculated as shown below.

    $\begin{array}{c}k=\frac{8}{2}\\ =4\\ =0,\pm 1,\pm 2,\pm 3,4\end{array}$

Substitute the value of $k=0$ and $N=8$ in equation (1).

    $\begin{array}{c}{E}_0=\alpha +2\beta \cos \frac{2\times 0\times \pi }{8}\\ =\alpha +2\beta \text{as}\cos 0=1\end{array}$

Substitute the value of $k=1$ and $N=8$ in equation (1).

    $\begin{array}{c}{E}_1=\alpha +2\beta \cos \frac{2\times 1\times \pi }{8}\\ =\alpha +2\beta \cos \frac{\pi }{4}\\ =\alpha +\left(2\beta \times 0.707\right)\text{as}\cos \frac{\pi }{4}=0.707\\ =\alpha +1.414\beta \end{array}$

Substitute the value of $k=-1$ and $N=8$ in equation (1).

    $\begin{array}{c}{E}_{-1}=\alpha +2\beta \cos \frac{2\times \left(-1\right)\times \pi }{8}\\ =\alpha +2\beta \cos \left(-\frac{\pi }{4}\right)\\ =\alpha +\left(2\beta \times 0.707\right)\text{as}\cos \left(-\frac{\pi }{4}\right)=0.707\\ =\alpha +1.414\beta \end{array}$

Substitute the value of $k=2$ and $N=8$ in equation (1).

    $\begin{array}{c}{E}_2=\alpha +2\beta \cos \frac{2\times 2\times \pi }{8}\\ =\alpha +2\beta \cos \left(\frac{\pi }{2}\right)\\ =\alpha +\left(2\beta \times 0\right)\text{as}\cos \left(\frac{\pi }{2}\right)=0\\ =\alpha \end{array}$

Substitute the value of $k=-2$ and $N=8$ in equation (1).

    $\begin{array}{c}{E}_{-2}=\alpha +2\beta \cos \frac{2\times \left(-2\right)\times \pi }{8}\\ =\alpha +2\beta \cos \left(-\frac{\pi }{2}\right)\\ =\alpha +\left(2\beta \times 0\right)\text{as}\cos \left(-\frac{\pi }{2}\right)=0\\ =\alpha \end{array}$

Substitute the value of $k=3$ and $N=8$ in equation (1).

    $\begin{array}{c}{E}_3=\alpha +2\beta \cos \frac{2\times 3\times \pi }{8}\\ =\alpha +2\beta \cos \left(\frac{3\pi }{4}\right)\\ =\alpha +2\beta \times \left(-0.707\right)\text{as}\cos \frac{3\pi }{4}=-0.707\\ =\alpha -1.414\beta \end{array}$

Substitute the value of $k=-3$ and $N=8$ in equation (1).

    $\begin{array}{c}{E}_{-3}=\alpha +2\beta \cos \frac{2\times \left(-3\right)\times \pi }{8}\\ =\alpha +2\beta \cos \left(-\frac{3\pi }{4}\right)\\ =\alpha +2\beta \times \left(-0.707\right)\text{as}\cos \left(-\frac{3\pi }{4}\right)=-0.707\\ =\alpha -1.414\beta \end{array}$

Substitute the value of $k=4$ and $N=8$ in equation (1).

    $\begin{array}{c}{E}_4=\alpha +2\beta \cos \frac{2\times 4\times \pi }{8}\\ =\alpha +2\beta \cos \pi \\ =\alpha +2\beta \times \left(-1\right)\text{as}\cos \pi =-1\\ =\alpha -2\beta \end{array}$

Therefore, the energies of the $\pi$ molecular orbitals of cyclooctatetraene are $\alpha +2\beta$, $\alpha +1.414\beta$, $\alpha +1.414\beta$, $\alpha$, $\alpha$, $\alpha -1.414\beta$, $\alpha -1.414\beta$, $\alpha -2\beta$.

Cyclooctateraene has a total of $6$ degenerate levels.  $2$ levels have the energy $\alpha +1.414\beta$, $2$ levels have the energy $\alpha$, and $2$ levels have the energy $\alpha -1.414\beta$.

(b)

Interpretation Introduction

Interpretation:

The delocalization energies of benzene have to be calculated and compared with hexatriene.  The conculsions have to be stated.

Concept introduction:

Delocalization of energy refers to the extra stabilization of the molecule which is due to allowing the electrons to spread over the entire molecule.  To calculate the delocalization energy, the energy of the molecule in question is compared with a molecule in which electrons cannot spread over the molecule.

Answer to Problem 9E.2P

The delocalization energy of benzene is $2\beta$.  The delocalization energy of hexatriene is $0.988\beta$.  This is less than the delocalization energy of benzene.  Benzene is molar stable than hexatriene.

Explanation of Solution

The delocalization energy of benzene is calculated by the formula below.

    $\text{Delocalization}\text{energy}=E_{\text{conjugated}}-E_{\text{isolated}\text{double}\text{bond}}$        (2)

The energies of the $\pi$ molecular orbitals of benzene are $\alpha +2\beta$, $\alpha +\beta$, $\alpha +\beta$, $\alpha -\beta$, $\alpha -\beta$, $\alpha -2\beta$.

Out of these, the energy levels, $\alpha +2\beta$, $\alpha +\beta$, $\alpha +\beta$ are occupied by two electrons each.

Therefore, the energy of the conjugated molecule is calculated below.

    $\begin{array}{c}{E}_{\text{conjugated}}=2\left(\alpha +2\beta \right)+2\left(\alpha +\beta \right)+2\left(\alpha +\beta \right)\\ =6\alpha +8\beta \end{array}$

The energy of the isolated double bond is given below.

    $\begin{array}{c}{E}_{\text{isolated}\text{double}\text{bond}}=6\left(\alpha +\beta \right)\\ =6\alpha +6\beta \end{array}$

Substitute the value of $E_{\text{conjugated}}$ and $E_{\text{isolated}\text{double}\text{bond}}$ in equation (2).

    $\begin{array}{c}\text{Delocalization}\text{energy}=6\alpha +8\beta -\left(6\alpha +6\beta \right)\\ =8\beta -6\beta \\ =2\beta \end{array}$

Therefore, the delocalization energy of benzene is $2\beta$.

The expression for the energies $E_k$ of the $\pi$ molecular orbitals of linear polyenes is shown below.

    $E_k=\alpha +2\beta \cos \frac{k\pi }{N+1}$        (3)

Where,

• $N$ is the number of carbon atoms.
• $\alpha$ is coloumb integral.
• $\beta$ is resonance integral.
• $k=\frac{N}{2}$ for even $N$.
• $k=\frac{N-1}{2}$ for odd value of $N$.

In hexatriene, the number of carbon atoms, $N$ is $6$.

Therefore, the value of $k$ is is shown below.

    $k=1,2,3,4,5,6$

Substitute the value of $k=1$ and $N=6$ in equation (3).

    $\begin{array}{c}{E}_1=\alpha +2\beta \cos \frac{1\times \pi }{6+1}\\ =\alpha +2\beta \cos \frac{\pi }{7}as\cos \frac{\pi }{7}=0.900968\\ =\alpha +\left(2\beta \times 0.900968\right)\\ =\alpha +1.802\beta \end{array}$

Substitute the value of $k=2$ and $N=6$ in equation (3).

    $\begin{array}{c}{E}_2=\alpha +2\beta \cos \frac{2\times \pi }{6+1}\\ =\alpha +2\beta \cos \frac{2\pi }{7}as\cos \frac{2\pi }{7}=0.6235\\ =\alpha +\left(2\beta \times 0.6235\right)\\ =\alpha +1.247\beta \end{array}$

Substitute the value of $k=3$ and $N=6$ in equation (3).

    $\begin{array}{c}{E}_3=\alpha +2\beta \cos \frac{3\times \pi }{6+1}\\ =\alpha +2\beta \cos \frac{3\pi }{7}as\cos \frac{3\pi }{7}=0.2225\\ =\alpha +\left(2\beta \times 0.2225\right)\\ =\alpha +0.445\beta \end{array}$

Substitute the value of $k=4$ and $N=6$ in equation (3).

    $\begin{array}{c}{E}_4=\alpha +2\beta \cos \frac{4\times \pi }{6+1}\\ =\alpha +2\beta \cos \frac{4\pi }{7}as\cos \frac{4\pi }{7}=-0.2225\\ =\alpha +\left(2\beta \times \left(-0.2225\right)\right)\\ =\alpha -0.445\beta \end{array}$

Substitute the value of $k=5$ and $N=6$ in equation (3).

    $\begin{array}{c}{E}_5=\alpha +2\beta \cos \frac{5\times \pi }{6+1}\\ =\alpha +2\beta \cos \frac{5\pi }{7}as\cos \frac{2\pi }{7}=-0.6235\\ =\alpha +\left(2\beta \times \left(-0.6235\right)\right)\\ =\alpha -1.247\beta \end{array}$

Substitute the value of $k=6$ and $N=6$ in equation (3).

    $\begin{array}{c}{E}_6=\alpha +2\beta \cos \frac{6\times \pi }{6+1}\\ =\alpha +2\beta \cos \frac{6\pi }{7}as\cos \frac{6\pi }{7}=-0.900968\\ =\alpha +\left(2\beta \times \left(-0.900968\right)\right)\\ =\alpha -1.802\beta \end{array}$

Therefore, the energies of the $\pi$ molecular orbitals of hexatriene are $\alpha +1.802\beta$, $\alpha +1.247\beta$, $\alpha +0.445\beta$, $\alpha -0.445\beta$, $\alpha -1.247\beta$, $\alpha -1.802\beta$.

Out of these, the energy levels, $\alpha +1.802\beta$, $\alpha +1.247\beta$, $\alpha +0.445\beta$ are occupied by two electrons each.

Therefore, the energy of the conjugated molecule is calculated below.

    $\begin{array}{c}{E}_{\text{conjugated}}=2\left(\alpha +1.802\beta \right)+2\left(\alpha +1.247\beta \right)+2\left(\alpha +0.445\beta \right)\\ =6\alpha +6.988\beta \end{array}$

The energy of the isolated double bond is given below.

    $\begin{array}{c}{E}_{\text{isolated}\text{double}\text{bond}}=6\left(\alpha +\beta \right)\\ =6\alpha +6\beta \end{array}$

Substitute the value of $E_{\text{conjugated}}$ and $E_{\text{isolated}\text{double}\text{bond}}$ in equation (2).

    $\begin{array}{c}\text{Delocalization}\text{energy}=6\alpha +6.988\beta -\left(6\alpha +6\beta \right)\\ =6.988\beta -6\beta \\ =0.988\beta \end{array}$

Therefore, the delocalization energy of hexatriene is $0.988\beta$.  This is less than the delocalization energy of benzene.  Therefore, benzene is more stable than hexatriene.

(c)

Interpretation Introduction

Interpretation:

The delocalization energies of cyclooctatetraene and octatetraene have to be calculated and compared.  Whether the conclusions for this pair same as that of part (b) has to be stated.

Concept introduction:

Delocalization of energy refers to the extra stabilization of the molecule which is due to allowing the electrons to spread over the entire molecule.  To calculate the delocalization energy, the energy of the molecule in question is compared with a molecule in which electrons cannot spread over the molecule.

Answer to Problem 9E.2P

The delocalization energy of cyclooctatetraene is $1.656\beta$.  The delocalization energy of octatetraene is $1.516\beta$.  This is less than the delocalization energy of cyclooctatetraene.  Cyclooctatetraene is more stable than octatetraene.  This is similar to the conclusion obtained in part (b).

Explanation of Solution

The delocalization energy of benzene is calculated by the formula below.

    $\text{Delocalization}\text{energy}=E_{\text{conjugated}}-E_{\text{isolated}\text{double}\text{bond}}$        (2)

The energies of the $\pi$ molecular orbitals of cyclooctatetraene are $\alpha +2\beta$, $\alpha +1.414\beta$, $\alpha +1.414\beta$, $\alpha$, $\alpha$, $\alpha -1.414\beta$, $\alpha -1.414\beta$, $\alpha -2\beta$.

Out of these, the energy levels, $\alpha +2\beta$, $\alpha +1.414\beta$, $\alpha +1.414\beta$, $\alpha$ are occupied by two electrons each.

Therefore, the energy of the conjugated molecule is calculated below.

    $\begin{array}{c}{E}_{\text{conjugated}}=2\left(\alpha +2\beta \right)+2\left(\alpha +1.414\beta \right)+2\left(\alpha +1.414\beta \right)+2\left(\alpha \right)\\ =8\alpha +9.656\beta \end{array}$

The energy of the isolated double bond is given below.

    $\begin{array}{c}{E}_{\text{isolated}\text{double}\text{bond}}=8\left(\alpha +\beta \right)\\ =8\alpha +8\beta \end{array}$

Substitute the value of $E_{\text{conjugated}}$ and $E_{\text{isolated}\text{double}\text{bond}}$ in equation (2).

    $\begin{array}{c}\text{Delocalization}\text{energy}=8\alpha +9.656\beta -\left(8\alpha +8\beta \right)\\ =9.656\beta -8\beta \\ =1.656\beta \end{array}$

Therefore, the delocalization energy of cyclooctatetraene is $1.656\beta$.

The expression for the energies $E_k$ of the $\pi$ molecular orbitals of linear polyenes is shown below.

    $E_k=\alpha +2\beta \cos \frac{k\pi }{N+1}$        (3)

Where,

• $N$ is the number of carbon atoms.
• $\alpha$ is coulomb integral.
• $\beta$ is resonance integral.
• $k=\frac{N}{2}$ for even $N$.
• $k=\frac{N-1}{2}$ for odd value of $N$.

In octatetraene, the number of carbon atoms, $N$ is $8$.

Therefore, the value of $k$ is is shown below.

    $k=1,2,3,4,5,6,78$

Substitute the value of $k=1$ and $N=8$ in equation (3).

    $\begin{array}{c}{E}_1=\alpha +2\beta \cos \frac{1\times \pi }{8+1}\\ =\alpha +2\beta \cos \frac{\pi }{9}as\cos \frac{\pi }{9}=0.9397\\ =\alpha +\left(2\beta \times 0.9397\right)\\ =\alpha +1.879\beta \end{array}$

Substitute the value of $k=2$ and $N=8$ in equation (3).

    $\begin{array}{c}{E}_2=\alpha +2\beta \cos \frac{2\times \pi }{8+1}\\ =\alpha +2\beta \cos \frac{2\pi }{9}as\cos \frac{2\pi }{9}=0.7660\\ =\alpha +\left(2\beta \times 0.7660\right)\\ =\alpha +1.532\beta \end{array}$

Substitute the value of $k=3$ and $N=8$ in equation (3).

    $\begin{array}{c}{E}_3=\alpha +2\beta \cos \frac{3\times \pi }{8+1}\\ =\alpha +2\beta \cos \frac{3\pi }{9}as\cos \frac{3\pi }{9}=0.5\\ =\alpha +\left(2\beta \times 0.5\right)\\ =\alpha +\beta \end{array}$

Substitute the value of $k=4$ and $N=8$ in equation (3).

    $\begin{array}{c}{E}_4=\alpha +2\beta \cos \frac{4\times \pi }{8+1}\\ =\alpha +2\beta \cos \frac{4\pi }{9}as\cos \frac{4\pi }{9}=0.1736\\ =\alpha +\left(2\beta \times 0.1736\right)\\ =\alpha +0.347\beta \end{array}$

Substitute the value of $k=5$ and $N=8$ in equation (3).

    $\begin{array}{c}{E}_4=\alpha +2\beta \cos \frac{5\times \pi }{8+1}\\ =\alpha +2\beta \cos \frac{5\pi }{9}as\cos \frac{5\pi }{9}=-0.1736\\ =\alpha +\left(2\beta \times \left(-0.1736\right)\right)\\ =\alpha -0.347\beta \end{array}$

Substitute the value of $k=6$ and $N=8$ in equation (3).

    $\begin{array}{c}{E}_3=\alpha +2\beta \cos \frac{6\times \pi }{8+1}\\ =\alpha +2\beta \cos \frac{6\pi }{9}as\cos \frac{3\pi }{9}=-0.5\\ =\alpha +\left(2\beta \times \left(-0.5\right)\right)\\ =\alpha -\beta \end{array}$

Substitute the value of $k=7$ and $N=8$ in equation (3).

    $\begin{array}{c}{E}_2=\alpha +2\beta \cos \frac{7\times \pi }{8+1}\\ =\alpha +2\beta \cos \frac{7\pi }{9}as\cos \frac{2\pi }{9}=-0.7660\\ =\alpha +\left(2\beta \times \left(-0.7660\right)\right)\\ =\alpha -1.532\beta \end{array}$

Substitute the value of $k=8$ and $N=8$ in equation (3).

    $\begin{array}{c}{E}_1=\alpha +2\beta \cos \frac{8\times \pi }{8+1}\\ =\alpha +2\beta \cos \frac{8\pi }{9}as\cos \frac{8\pi }{9}=-0.9397\\ =\alpha +\left(2\beta \times \left(-0.9397\right)\right)\\ =\alpha -1.879\beta \end{array}$

Therefore, the energies of the $\pi$ molecular orbitals of octatetraene are $\alpha +1.879\beta$, $\alpha +1.532\beta$, $\alpha +\beta$, $\alpha +0.347\beta$, $\alpha -0.347\beta$, $\alpha -\beta$, $\alpha -1.532\beta$, $\alpha -1.879\beta$.

Out of these, the energy levels, $\alpha +1.879\beta$, $\alpha +1.532\beta$, $\alpha +\beta$, $\alpha +0.347\beta$ are occupied by two electrons each.

Therefore, the energy of the conjugated molecule is calculated below.

    $\begin{array}{c}{E}_{\text{conjugated}}=2\left(\alpha +1.879\beta \right)+2\left(\alpha +1.532\beta \right)+2\left(\alpha +0.347\beta \right)+2\left(\alpha +\beta \right)\\ =8\alpha +9.516\beta \end{array}$

The energy of the isolated double bond is given below.

    $\begin{array}{c}{E}_{\text{isolated}\text{double}\text{bond}}=8\left(\alpha +\beta \right)\\ =8\alpha +8\beta \end{array}$

Substitute the value of $E_{\text{conjugated}}$ and $E_{\text{isolated}\text{double}\text{bond}}$ in equation (2).

    $\begin{array}{c}\text{Delocalization}\text{energy}=8\alpha +9.516\beta -\left(8\alpha +8\beta \right)\\ =9.516\beta -8\beta \\ =1.516\beta \end{array}$

Therefore, the delocalization energy of octatetraene is $1.516\beta$.  This is less than the delocalization energy of cyclooctatetraene.  Therefore, cyclooctatetraene is more stable than hexatriene.