# Pi molecular orbitals of polyenes

As an organic chemist, I'm comfortable deriving the pi molecular orbitals of linearly conjugate systems to give the following result: In a qualitative sense, these molecular orbitals are easily arrived at. For a polyene of n-atoms, the lowest energy combination will have 0 nodes (all in-phase) and the highest energy combination will have n-1 nodes (in, out, in, out...).

Is there a way of arriving at the same result using a more rigorous, formal molecular orbital approach?

• Is Hückel MO Theory what you are looking for? Jun 26 '17 at 4:54
• Hückel MO theory will also give you different absolute values for the linear combination coefficients, whereas your drawing suggests the same coefficients throughout. Note that all you need to perform HMO calculations are a way to set up the matrix (which is easily done by hand for these cases) and a diagonalizer, such as Wolfram Alpha. Interpretation is again easy by hand. Jun 26 '17 at 5:13

These pi-type MOs are most commonly derived using Hückel MO theory. I've never been a fan of Wikipedia's technical articles, so for further reading I'd suggest using the QMUL resources online, which are very comprehensive. $\require{begingroup}\begingroup \newcommand{ket}{\left|#1\right>} \newcommand{bra}{\left<#1\right|} \newcommand{braket}{\left< #1 \right>}$

## Theory

Simple Hückel theory makes some key assumptions about the forms of the MOs as well as some of the integrals involved in a typical quantum chemical calculation:

• pi-type MOs are linear combinations of p-orbitals, and no other orbitals contribute (so-called "sigma-pi separation")
• the value of $\braket{a|b}$ (where $\ket{a},\ket{b}$ are p-type AOs on atoms $a$ and $b$) is equal to $\delta_{ab}$ ($1$ if $a = b$, $0$ otherwise)
• the value of $\braket{a|H|b}$ is

$$H_{ab} = \braket{a|H|b} = \begin{cases}\alpha \text{ if } a = b \\ \beta \text{ if atom }a\text{ is bonded to atom } b \\ 0 \text{ otherwise}\end{cases}$$

With these simplifications the secular equations

$$\mathbf{H}\mathbf{c} = E\mathbf{S}\mathbf{c}$$

can be readily solved (the derivation of the secular equations will not be presented here). Here $\mathbf{H}$ refers to the Hamiltonian matrix and $\mathbf{S}$ refers to the overlap matrix. These are defined as the matrices with elements $\mathbf{H}_{ij} = \braket{i|H|j}$ and $\mathbf{S}_{ij} = \braket{i|j}$ respectively. In simple Hückel theory the matrix $\mathbf{S}$ can be neglected as it is simply equal to the identity matrix.

$\mathbf{c}$ is a column vector and one column vector represents one MO. The elements of $\mathbf{c}$ represent the contribution of each AO to the MO. This will be clearer with an example (to follow).

In general, for $n$ atoms there will be $n$ vectors $\mathbf{c}$ that will satisfy the secular equations, and $n$ corresponding values of the energy $E$. Thus, for the allyl cation (for example), there will be three column vectors $\mathbf{c}_1, \mathbf{c}_2, \mathbf{c}_3$ and four associated energies $E_1, E_2, E_3$.

## The allyl cation Using the atom numbering scheme as shown above, and the assumptions mentioned in the previous section, we find that the Hamiltonian matrix $\mathbf{H}$ is:

$$\mathbf{H} = \begin{pmatrix} \alpha & \beta & 0 \\ \beta & \alpha & \beta \\ 0 & \beta & \alpha \end{pmatrix}$$

For example the entry $\mathbf{H}_{12} = \beta$ because atoms 1 and 2 are bonded to each other, and the entry $\mathbf{H}_{13} = 0$ because atoms 1 and 3 are not bonded to each other.

The secular equations then reduce to the eigenvalue problem

$$\begin{pmatrix} \alpha & \beta & 0 \\ \beta & \alpha & \beta \\ 0 & \beta & \alpha \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \\ c_3 \end{pmatrix} = E\begin{pmatrix} c_1 \\ c_2 \\ c_3 \end{pmatrix}$$

Obtaining the eigenvalues and eigenvectors of a matrix is a process outlined in many maths books. The crux is that the "secular determinant" must equal zero:

$$\begin{vmatrix} \alpha - E & \beta & 0 \\ \beta & \alpha - E & \beta \\ 0 & \beta & \alpha - E \end{vmatrix} = 0$$

from which one obtains the eigenvalues (here I chose an example where it's easy to do by hand, but in general you can use software like MATLAB). In increasing energy (note that $\beta < 0$), they are

\begin{align}E_1 &= \alpha + \sqrt{2}\beta & E_2 &= \alpha & E_3 &= \alpha - \sqrt{2}\beta \end{align}

and the corresponding eigenvectors

\begin{align} \mathbf{c}_1 &= \begin{pmatrix} 1/2 \\ 1/\sqrt{2} \\ 1/2 \end{pmatrix} & \mathbf{c}_2 &= \begin{pmatrix} 1/\sqrt{2} \\ 0 \\ -1/\sqrt{2} \end{pmatrix} & \mathbf{c}_3 &= \begin{pmatrix} 1/2 \\ -1/\sqrt{2} \\ 1/2 \end{pmatrix} \end{align}

This means that the lowest-energy pi-type MO, $\mathbf{c}_1$, is

$$\psi_1 = \frac{1}{2}\mathrm{p}_1 + \frac{1}{\sqrt{2}}\mathrm{p}_2 + \frac{1}{2}\mathrm{p}_3$$

where $\mathrm{p}_i$ is the 2p atomic orbital on carbon $i$. Note that the coefficients all have the same sign, indicating that the p orbitals are all in phase with each other. However, the coefficient for carbon 2 (the middle carbon) is larger than the others. Hence, a sketch of this MO should technically show that the middle p-orbital makes a larger contribution than the outer two p-orbitals. This is what TAR86 meant by their comment on your question:

Hückel MO theory will also give you different absolute values for the linear combination coefficients, whereas your drawing suggests the same coefficients throughout.

Anyway, likewise for the LUMO we have

$$\psi_2 = \frac{1}{\sqrt{2}}\mathrm{p}_1 - \frac{1}{\sqrt{2}}\mathrm{p}_3$$

Note that 1) the p orbitals have different phases (different sign of coefficient), and 2) the p orbital on carbon 2 does not contribute to the MO - this leads to the node observed in the second MO. This is why the allyl cation is electrophilic on C-1 and C-3, but not on C-2: the LUMO has zero coefficient on C-2.

All in all, considering that the allyl cation only has two electrons in the pi system, this lets us construct the MO diagram I should like to extend @orthocresol's clear answer and so obtain results for any size of conjugated system. Starting with the determinant in $\alpha -E$ and $\beta$.

$$\begin{vmatrix} \alpha - E & \beta & 0 \\ \beta & \alpha - E & \beta \\ 0 & \beta & \alpha - E \end{vmatrix} = 0$$

and substituting $(\alpha -E)/\beta = -x$ rewriting this in a more general form produces

$$D_n =\begin{vmatrix} -x & 1 & 0 & 0 & 0&\cdots\\ 1 & -x & 1 & 0 & 0&\cdots \\ 0 & 1 & -x & 1 & 0 & \cdots\\ \vdots& & & \ddots& \\ & & & & &-x \end{vmatrix} = 0$$

Evaluating the determinant gives the results

$$D_1 = -x , \quad D_2= x^2-1, \quad D_3= -x^3 +2x,\quad D_4=x^4-3x^2+1,\cdots \\ \cdots D_n=-xD_{n-1}-D_{n-2}$$

Now it is known that the solutions to these equations can be recast in a very convenient form because they are they are similar to the Chebyscheff polynomials.

The n'th term here is

$$T_n= \frac{\sin([n+1]\theta)}{\sin(\theta)}=2\cos(\theta)T_{n-1}-T_{n-2}$$

which has the same form as $D_n$ if $x=2\cos(\theta)$.

Thus

$$D_n= \frac{\sin([n+1]\theta)}{\sin(\theta)}$$

is satisfied if $\displaystyle (n+1)\theta = j\pi$ with $j=1,2,3,\cdots$.

Substituting $(\alpha -E)/\beta = -x$ and $x=2\cos(\theta)$ produces for the energy

$$E_j= \alpha + 2\beta\cos \left( \frac{j\pi}{n+1} \right)$$

where n is the number of carbon atoms and each j from 1 upwards indicates one of the solutions of the secular equation. Thus for butadiene, $n=4$ and $j=1,\,2,\,3,\,4$ only. It should also be remembered when ordering these energies that $\alpha$ and $\beta$ are negative numbers.

[In the case of circular polyenes the solutions are $\displaystyle E_j=\alpha + 2\beta \cos(2\pi j/n)$ ]