The figure below includes the spectra of (E)-1,3,5-hexatriene and benzene. Each of these molecules has three conjugated double bonds, and their spectra is dominated by $\pi \rightarrow \pi^{*}$ transitions.

Using the particle in a 1D box and the particle on a ring models to describe $\pi$ electrons in each of these molecules, indicate which spectrum corresponds to which compound?

Note that while these models can provide useful rough estimates and trends, they cannot be expected to produce accurate values.

Data:
We can consider that for both molecules the average bond distance is $1.4 \AA$. For (E)$1,3,5$-hexatriene we will neglect the bond angles.
The mass of the electron is $m_{e}=9.109 \times 10^{-31} \mathrm{~kg}$.
The particle in a 1D box and particle on a ring energy levels are given respectively by:
$$
\begin{array}{r}
E_{n}=\frac{h^{2} n^{2}}{8 m L^{2}} \
E_{m_{\ell}}=\frac{h^{2} m_{\ell}^{2}}{8 \pi^{2} m r^{2}}
\end{array}
$$

Question

The figure below includes the spectra of (E)-1,3,5-hexatriene and benzene. Each of these molecules has three conjugated double bonds, and their spectra is dominated by $\pi \rightarrow \pi^{*}$ transitions.

Using the particle in a 1D box and the particle on a ring models to describe $\pi$ electrons in each of these molecules, indicate which spectrum corresponds to which compound?

Note that while these models can provide useful rough estimates and trends, they cannot be expected to produce accurate values.

Data:
We can consider that for both molecules the average bond distance is $1.4 \AA$. For (E)$1,3,5$-hexatriene we will neglect the bond angles.
The mass of the electron is $m_{e}=9.109 \times 10^{-31} \mathrm{~kg}$.
The particle in a 1D box and particle on a ring energy levels are given respectively by:
$$
\begin{array}{r}
E_{n}=\frac{h^{2} n^{2}}{8 m L^{2}} \
E_{m_{\ell}}=\frac{h^{2} m_{\ell}^{2}}{8 \pi^{2} m r^{2}}
\end{array}
$$

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Determine the length of the box for (E)-1,3,5-hexatriene**: For (E)-1,3,5-hexatriene, which can be modeled as a particle in a 1D box, the length of the box ($L$) is the length of the conjugated system. Since we are neglecting bond angles, we can simply add the bond distances together. There are 4 bonds in the conjugated system, each with a length of $1.4 ext{ Å}$. ‖ ‖ 2 ⋮ **Calculate the length of the box**: $L = 4 imes 1.4 ext{ Å} = 5.6 ext{ Å} = 5.6 imes 10^{-10} ext{ m}$ ‖ ‖ 3 ⋮ **Determine the circumference of the ring for benzene**: For benzene, modeled as a particle on a ring, the circumference ($2\pi r$) is the length of the conjugated system. Benzene has 6 carbon atoms, with each C-C bond considered to be $1.4 ext{ Å}$. ‖ ‖ 4 ⋮ **Calculate the circumference of the ring**: $2\pi r = 6 imes 1.4 ext{ Å} = 8.4 ext{ Å} = 8.4 imes 10^{-10} ext{ m}$; thus, $r = \frac{8.4 imes 10^{-10} ext{ m}}{2\pi}$ ‖ ‖ 5 ⋮ **Relate the energy levels to the wavelength of light absorbed**: The energy difference between the ground state and the first excited state corresponds to the wavelength of light absorbed. For the particle in a 1D box, this is between $E_1$ and $E_2$, and for the particle on a ring, it is between $E_{m_{\ell}=0}$ and $E_{m_{\ell}=1}$. ‖ ‖ 6 ⋮ **Calculate the energy difference for (E)-1,3,5-hexatriene**: Using the formula for the particle in a 1D box, $E_n = \frac{h^2 n^2}{8 m L^2}$, we find the energy difference $\Delta E = E_2 - E_1 = \frac{h^2}{8 m L^2}(2^2 - 1^2)$. ‖ ‖ 7 ⋮ **Calculate the energy difference for benzene**: Using the formula for the particle on a ring, $E_{m_{\ell}} = \frac{h^2 m_{\ell}^2}{8 \pi^2 m r^2}$, we find the energy difference $\Delta E = E_{m_{\ell}=1} - E_{m_{\ell}=0} = \frac{h^2}{8 \pi^2 m r^2}(1^2 - 0^2)$. ‖ ‖ 8 ⋮ **Convert the energy difference to wavelength**: The wavelength of light absorbed ($\lambda$) is related to the energy difference by $\lambda = \frac{hc}{\Delta E}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\Delta E$ is the energy difference. ‖ ‖ 9 ⋮ **Compare the calculated wavelengths with the given spectra**: The molecule with the larger box or ring circumference will have a smaller energy difference and thus absorb light at a longer wavelength. ‖ ‖ 10 ⋮ **Identify the spectra**: Since benzene has a larger circumference than (E)-1,3,5-hexatriene, it will absorb at a longer wavelength. Therefore, the spectrum with the peak at 250 nm corresponds to benzene, and the one with the peak at 175 nm corresponds to (E)-1,3,5-hexatriene. ‖ ∻Answer∻ ⚹ The spectrum with the peak at 175 nm corresponds to (E)-1,3,5-hexatriene, and the spectrum with the peak at 250 nm corresponds to benzene. ⚹ ∻Key Concept∻ ⚹ The particle in a 1D box and the particle on a ring models are used to estimate the wavelengths of light absorbed by conjugated systems based on their geometry. ⚹ ∻Explanation∻ ⚹ The length of the conjugated system affects the energy levels of the $\pi$ electrons. A longer system, such as benzene's ring, results in closer energy levels and absorption at a longer wavelength compared to a shorter system like (E)-1,3,5-hexatriene. ⚹◊How can the concept of conjugated double bonds in (E)-1,3,5-hexatriene and benzene be related to their respective spectra dominated by $\pi ightarrow \pi^{*}$ transitions?⍭ Generate me a similar question◊

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