Understanding $\pi$ Electron Density in Butadiene

The question asks about the ratio of total $\pi$ electron density on the four carbon atoms of trans butadiene. Butadiene is a conjugated system, meaning it has alternating single and double bonds. In trans butadiene, the four carbon atoms form a planar structure, and the four p orbitals on these carbon atoms combine to form molecular orbitals (MOs).

We can determine the $\pi$ electron density on each carbon atom using the Hückel Molecular Orbital (HMO) theory. According to HMO theory, for a linear conjugated system like butadiene (N=4), the energies and coefficients of the $\pi$ molecular orbitals can be found.

Butadiene has 4 carbon atoms contributing 4 $\pi$ electrons (two from each double bond). These 4 electrons occupy the lowest energy $\pi$ molecular orbitals. For a system with 4 MOs, the lowest two MOs ($\psi_1$ and $\psi_2$) will be filled with 2 electrons each in the ground state.

Hückel Coefficients for Linear Polyenes

The normalized coefficients ($c_{kj}$) for the $k^{th}$ molecular orbital on the $j^{th}$ carbon atom in a linear polyene with N carbon atoms are given by the formula:

$$c_{kj} = \sqrt{\frac{2}{N+1}} \sin\left(\frac{kj\pi}{N+1}\right)$$

For butadiene, N=4. The occupied MOs are $k=1$ and $k=2$. The carbon atoms are indexed $j=1, 2, 3, 4$.

Coefficients for MO 1 (k=1):

$$c_{1j} = \sqrt{\frac{2}{4+1}} \sin\left(\frac{1 \cdot j\pi}{4+1}\right) = \sqrt{\frac{2}{5}} \sin\left(\frac{j\pi}{5}\right)$$

• $c_{11} = \sqrt{2/5} \sin(\pi/5)$
• $c_{12} = \sqrt{2/5} \sin(2\pi/5)$
• $c_{13} = \sqrt{2/5} \sin(3\pi/5) = \sqrt{2/5} \sin(\pi - 2\pi/5) = \sqrt{2/5} \sin(2\pi/5)$
• $c_{14} = \sqrt{2/5} \sin(4\pi/5) = \sqrt{2/5} \sin(\pi - \pi/5) = \sqrt{2/5} \sin(\pi/5)$

Coefficients for MO 2 (k=2):

$$c_{2j} = \sqrt{\frac{2}{4+1}} \sin\left(\frac{2 \cdot j\pi}{4+1}\right) = \sqrt{\frac{2}{5}} \sin\left(\frac{2j\pi}{5}\right)$$

• $c_{21} = \sqrt{2/5} \sin(2\pi/5)$
• $c_{22} = \sqrt{2/5} \sin(4\pi/5) = \sqrt{2/5} \sin(\pi/5)$
• $c_{23} = \sqrt{2/5} \sin(6\pi/5) = \sqrt{2/5} \sin(\pi + \pi/5) = -\sqrt{2/5} \sin(\pi/5)$
• $c_{24} = \sqrt{2/5} \sin(8\pi/5) = \sqrt{2/5} \sin(2\pi - 2\pi/5) = -\sqrt{2/5} \sin(2\pi/5)$

Calculating $\pi$ Electron Density

The total $\pi$ electron density ($\rho_j$) on the $j^{th}$ carbon atom is the sum of the squares of the coefficients for that atom in each occupied MO, multiplied by the number of electrons in that MO. Both MO1 and MO2 are occupied by 2 electrons ($n_1=2, n_2=2$).

$$\rho_j = \sum_{k=1}^{2} n_k |c_{kj}|^2 = 2 |c_{1j}|^2 + 2 |c_{2j}|^2$$

Electron Density on Carbon 1 ($\rho_1$):

$$\rho_1 = 2 |c_{11}|^2 + 2 |c_{21}|^2 = 2 \left(\sqrt{\frac{2}{5}} \sin\left(\frac{\pi}{5}\right)\right)^2 + 2 \left(\sqrt{\frac{2}{5}} \sin\left(\frac{2\pi}{5}\right)\right)^2$$

$$\rho_1 = 2 \cdot \frac{2}{5} \sin^2\left(\frac{\pi}{5}\right) + 2 \cdot \frac{2}{5} \sin^2\left(\frac{2\pi}{5}\right) = \frac{4}{5} \left( \sin^2\left(\frac{\pi}{5}\right) + \sin^2\left(\frac{2\pi}{5}\right) \right)$$

We know $\sin^2(\pi/5) = \frac{10 - 2\sqrt{5}}{16}$ and $\sin^2(2\pi/5) = \frac{10 + 2\sqrt{5}}{16}$.

$$\rho_1 = \frac{4}{5} \left( \frac{10 - 2\sqrt{5}}{16} + \frac{10 + 2\sqrt{5}}{16} \right) = \frac{4}{5} \left( \frac{10 - 2\sqrt{5} + 10 + 2\sqrt{5}}{16} \right) = \frac{4}{5} \cdot \frac{20}{16} = \frac{4}{5} \cdot \frac{5}{4} = 1$$

Electron Density on Carbon 2 ($\rho_2$):

$$\rho_2 = 2 |c_{12}|^2 + 2 |c_{22}|^2 = 2 \left(\sqrt{\frac{2}{5}} \sin\left(\frac{2\pi}{5}\right)\right)^2 + 2 \left(\sqrt{\frac{2}{5}} \sin\left(\frac{\pi}{5}\right)\right)^2$$

$$\rho_2 = \frac{4}{5} \left( \sin^2\left(\frac{2\pi}{5}\right) + \sin^2\left(\frac{\pi}{5}\right) \right) = \frac{4}{5} \left( \frac{10 + 2\sqrt{5}}{16} + \frac{10 - 2\sqrt{5}}{16} \right) = 1$$

Electron Density on Carbon 3 ($\rho_3$):

$$\rho_3 = 2 |c_{13}|^2 + 2 |c_{23}|^2 = 2 \left(\sqrt{\frac{2}{5}} \sin\left(\frac{2\pi}{5}\right)\right)^2 + 2 \left(-\sqrt{\frac{2}{5}} \sin\left(\frac{\pi}{5}\right)\right)^2$$

$$\rho_3 = \frac{4}{5} \left( \sin^2\left(\frac{2\pi}{5}\right) + \sin^2\left(\frac{\pi}{5}\right) \right) = \frac{4}{5} \left( \frac{10 + 2\sqrt{5}}{16} + \frac{10 - 2\sqrt{5}}{16} \right) = 1$$

Electron Density on Carbon 4 ($\rho_4$):

$$\rho_4 = 2 |c_{14}|^2 + 2 |c_{24}|^2 = 2 \left(\sqrt{\frac{2}{5}} \sin\left(\frac{\pi}{5}\right)\right)^2 + 2 \left(-\sqrt{\frac{2}{5}} \sin\left(\frac{2\pi}{5}\right)\right)^2$$

$$\rho_4 = \frac{4}{5} \left( \sin^2\left(\frac{\pi}{5}\right) + \sin^2\left(\frac{2\pi}{5}\right) \right) = \frac{4}{5} \left( \frac{10 - 2\sqrt{5}}{16} + \frac{10 + 2\sqrt{5}}{16} \right) = 1$$

Ratio of $\pi$ Electron Densities

The $\pi$ electron densities on the four carbon atoms are $\rho_1 = 1$, $\rho_2 = 1$, $\rho_3 = 1$, and $\rho_4 = 1$.

$$\rho_1 : \rho_2 : \rho_3 : \rho_4 = 1 : 1 : 1 : 1$$

This result is characteristic of neutral, unsubstituted linear polyenes within the simple Hückel method, where the total $\pi$ electron density on each carbon atom is calculated to be exactly 1.

Therefore, the total $\pi$ electron density on the four carbon atoms of trans butadiene are in the ratio 1 : 1 : 1 : 1.