IChO 2020 · Theory problem 5

Konya, Carrot, Beta-Carotene, Vitamin-A, Immune System, Vision

Figure 1

Mevlana (Rumi) was a great mystic and Sufi poet who lived out his days in Konya in the \(13^{\mathrm{th}}\) century. The indirect relevance of Konya to chemistry is that the city provides 65% of the country’s carrot production, from which one of the essential vitamins (vitamin A) is obtained.

Carrot is an important source of β-carotene, which gives the vegetable its orange color. This molecule is a red-orange pigment naturally found in plants and fruits and is a provitamin A carotenoid. It is converted to vitamin A, which is essential for normal growth and development, the immune system, and vision function.

Figure 2

Figure 1. Ball and stick representation of the structure of β-carotene. The gray and white spheres represent the carbon and hydrogen atoms, respectively. The numbered carbon atoms belong to the linear conjugated π-segment of the molecule.

β-Carotene has an extended polyene chain of 22 carbon atoms. It is a conjugated \(\pi\)-system, having alternating single and double bonds. Its experimental maximum absorption wavelength (\(\lambda_{\max}\)) is 455 nm. We assume that all the bonds between \(C_1\) and \(C_{22}\) are conjugated bonds. There are 22 \(\pi\)-electrons in the molecule (Figure 1).

To a crude approximation, the electrons in the \(\ce{C}\text{-}2p_z\) orbitals, which are perpendicular to the molecular plane, are assumed to move along the entire molecule, without interacting with each other. They are like independent particles confined in a molecule moving along the x-axis in one dimension. These characteristics of π-electrons make them eligible for being treated by the simplest model called the particle in one-dimensional box model.

The wave function and the energies of the quantized levels for an electron moving in a one- dimensional box with infinite potential walls are given as follows:

\[ \Psi_n(x)=\sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L} \]

here \(n\) is the quantum number, \(n=1,2,3,4,\ldots\infty\), and \(L\) is the box length.

\[ E_n=\frac{n^2h^2}{8m_eL^2} \]

In two dimension system, within the framework of independent particle approximation, the wavefunction is expressed as a product of one-dimensional wavefunctions, and the energy is expressed as a sum of one-dimensional energies. The energy levels of the two dimensional rectangular box is given as follows:

\[ E_{n_xn_y} = \left[\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}\right] \left\{\frac{h^2}{8m_e}\right\} \]

where \(n_x,n_y\) are the quantum numbers and they are positive integers. \(L_x,L_y\) are the dimensions of the box in the 2D model. They are positive numbers.

5.1 The β-carotene molecule is orange in color because:

i) It absorbs in the visible region of the electromagnetic spectrum.

ii) HOMO → LUMO transition occurs by absorption of IR photon.

iii) The spacing between the \(22^{\mathrm{nd}}\) and the \(23^{\mathrm{rd}}\) energy levels is equal to the energy of the IR photon at the orange wavelength.

iv) It absorbs green/blue light and it transmits red/yellow color.

v) It absorbs in the UV-VIS region since the molecule has no net dipole moment.

Choose which pair of the above sentences is correct?

The pairs of sentences are as follows:

☐ a) i and ii ☐ b) i and iii ☐ c) i and iv ☐ d) i and v
☐ e) ii and iii ☐ f) ii and iv ☐ g) ii and v ☐ h) iii and iv
☐ j) iii and v ☐ k) iv and v

Although it is highly unrealistic, let us assume that the conjugated segment of the molecule is linear and treated with the particle in a one-dimensional box model as shown in figure 2. In this case, the length of the box can be approximated as \(L=1.40\times n_{\mathrm C}\) (in Å), where \(n_{\mathrm C}\) is the number of carbon atoms in the conjugated segment.

Use this information to answer the questions 5.2–5.6.

Figure 3

Figure 2. Schematic representation of the conjugated line segment made up carbon atoms of β-carotene in a one-dimensional box of length L.

5.2 Calculate the energies (in J) of the lowest two energy levels.

5.3 Draw the wavefunctions of the lowest two energy levels with proper labelling the x-axis.

5.4 Sketch the energy level diagram up to \(n=4\) showing the relative spacing.

5.5 What is the total \(\pi\)-energy (in J) of the molecule?

5.6 Calculate the wavelength (in nm) at which the transition between the highest occupied and lowest unoccupied energy levels occurs.

Use the particle in a two-dimensional box model to answer questions 5.7 – 5.8.

Figure 4

Figure 3. Schematic representation of the conjugated carbon atoms of β-carotene in a two-dimensional box.

Assume that the conjugated segment is made up of conjugated bonds that are all-trans to each other. The motion of the π-electrons is studied in the two-dimensional rectangular box with the dimensions \(L_x=26.0\) Å, \(L_y=3.0\) Å (Figure 3).

5.7 Calculate the energies (in J) of the highest occupied and the lowest unoccupied energy levels and the wavelength (in nm) at which the transition between these energy levels occurs.

5.8 What should be the \(L_x\) value (in Å) in order for the molecule to absorb light at the experimental \(\lambda_{\max}=455\) nm if \(L_y\) is kept constant at 3.0 Å. (Assume that the quantum numbers for homo and lumo are the same as in the question 5.7.)

Official English text of the problem.

Answers

  • 5.1: c) i and iv.
  • 5.2: \(E_1=\boxed{6.35\times10^{-21}\ \mathrm J}\); \(E_2=\boxed{2.54\times10^{-20}\ \mathrm J}\).
  • 5.3: For \(0\leq x\leq L\), with \(L=30.8\) Å, \[ \boxed{\Psi_1(x)=\sqrt{\frac2L}\sin\frac{\pi x}{L}}, \qquad \boxed{\Psi_2(x)=\sqrt{\frac2L}\sin\frac{2\pi x}{L}}. \] The first is one positive half-wave; the second has a positive lobe followed by a negative lobe, with an internal node at \(x=L/2=15.4\) Å.
  • 5.4: Draw horizontal levels at \(E/E_1=\boxed{1,4,9,16}\), labelled \(n=1,2,3,4\). Successive gaps are in the ratio \(\boxed{3:5:7}\).
  • 5.5: \(E_{\pi,\mathrm{total}}=\boxed{6.43\times10^{-18}\ \mathrm J}\) per molecule.
  • 5.6: HOMO \(n=11\), LUMO \(n=12\); \(\boxed{\lambda=1.36\times10^3\ \mathrm{nm}}\).
  • 5.7: HOMO \((n_x,n_y)=(11,1)\), with \(\boxed{E_{\mathrm H}=1.75\times10^{-18}\ \mathrm J}\); LUMO \((12,1)\), with \(\boxed{E_{\mathrm L}=1.95\times10^{-18}\ \mathrm J}\); \(\boxed{\lambda=969\ \mathrm{nm}}\).
  • 5.8: Under the stipulated unchanged quantum numbers, \(\boxed{L_x=17.8\ \text{Å}}\).

Solution

The molecule is all-trans-β-carotene, \(\ce{C40H56}\), with SMILES CC1=C(/C=C/C(C)=C/C=C/C(C)=C/C=C/C=C(C)/C=C/C=C(C)/C=C/C2=C(C)CCCC2(C)C)C(C)(C)CCC1. This specifies the acyclic double-bond stereochemistry; it does not specify the three-dimensional conformation or the imposed box geometry.

The problem does not print the physical constants. We use

\[ h=6.62607015\times10^{-34}\ \mathrm{J\,s},\qquad m_e=9.1093837\times10^{-31}\ \mathrm{kg},\qquad c=2.99792458\times10^8\ \mathrm{m\,s^{-1}}. \]

Each spatial orbital accommodates two electrons with opposite spins. Consequently, the 22 \(\pi\)-electrons occupy the eleven lowest spatial orbitals in the ground state.

5.1 — Origin of the orange color

The measured absorption maximum, 455 nm, lies in the blue part of the visible spectrum. Absorption of blue/green light leaves the transmitted or reflected light enriched in red/yellow wavelengths, giving the orange appearance. Therefore i and iv are correct, corresponding to c.

The other statements are incorrect:

  • ii: The observed transition involves a visible photon, rather than an infrared photon.
  • iii: With two electrons per level, the one-dimensional HOMO and LUMO are levels 11 and 12. Moreover, orange light belongs to the visible spectrum, not the infrared.
  • v: Having no permanent dipole moment does not establish that a molecule absorbs in the UV-visible region. Electronic absorption depends on the transition energy and transition dipole moment.

5.2 — Lowest two one-dimensional energy levels

The prescribed box length is

\[ L=(1.40\ \text{Å})(22)=30.8\ \text{Å} =3.08\times10^{-9}\ \mathrm m. \]

Thus,

\[ E_1=\frac{h^2}{8m_eL^2} = \frac{(6.62607015\times10^{-34}\ \mathrm{J\,s})^2} {8(9.1093837\times10^{-31}\ \mathrm{kg}) (3.08\times10^{-9}\ \mathrm m)^2} = \boxed{6.35\times10^{-21}\ \mathrm J}. \]

Since \(E_n=n^2E_1\),

\[ E_2=4E_1=\boxed{2.54\times10^{-20}\ \mathrm J}. \]

5.3 — Wavefunction sketches

Let \(A=\sqrt{2/L}\). Inside the box,

\[ \Psi_1(x)=A\sin\frac{\pi x}{L}, \qquad \Psi_2(x)=A\sin\frac{2\pi x}{L}. \]

Both wavefunctions vanish at the walls and outside the box. Label the horizontal axis \(x\), using these positions:

\(x\) \(0\) \(L/4\) \(L/2\) \(3L/4\) \(L\)
\(x\), in Å 0 7.70 15.4 23.1 30.8
\(\Psi_1/A\) 0 \(1/\sqrt2\) 1 \(1/\sqrt2\) 0
\(\Psi_2/A\) 0 1 0 \(-1\) 0

For \(n=1\), draw a smooth positive half-sine wave from \(x=0\) to \(x=L\), with its maximum \(+A\) at \(L/2\). It has no internal node.

For \(n=2\), draw one full sine wave: a positive maximum \(+A\) at \(L/4\), a node at \(L/2\), and a negative minimum \(-A\) at \(3L/4\).

The overall sign of either wavefunction may be reversed without changing its physical meaning.

5.4 — Energy-level diagram

The levels obey \(E_n=n^2E_1\):

\(n\) \(E_n/E_1\) \(E_n\), in J
1 1 \(6.35\times10^{-21}\)
2 4 \(2.54\times10^{-20}\)
3 9 \(5.72\times10^{-20}\)
4 16 \(1.02\times10^{-19}\)

Draw an upward-pointing energy axis with horizontal lines at these heights. The separations increase upwards:

\[ E_2-E_1=3E_1,\qquad E_3-E_2=5E_1,\qquad E_4-E_3=7E_1. \]

Hence the successive gaps have the ratio \(\boxed{3:5:7}\).

5.5 — Total \(\pi\)-electron energy

The 22 electrons fill levels \(n=1,\ldots,11\), two electrons per level. Under the independent-electron approximation,

\[ E_{\pi,\mathrm{total}} =2\sum_{n=1}^{11}E_n =2E_1\sum_{n=1}^{11}n^2. \]

Using

\[ \sum_{n=1}^{11}n^2 =\frac{11(12)(23)}6=506, \]

we obtain

\[ E_{\pi,\mathrm{total}} =1012E_1 =1012(6.3508469\times10^{-21}\ \mathrm J) =\boxed{6.43\times10^{-18}\ \mathrm J}. \]

This is the total energy per molecule relative to the zero potential inside the box.

5.6 — One-dimensional HOMO–LUMO wavelength

The highest occupied level is \(n=11\), and the lowest unoccupied level is \(n=12\). Therefore,

\[ \Delta E=E_{12}-E_{11} =(12^2-11^2)E_1 =23E_1 =1.4606948\times10^{-19}\ \mathrm J. \]

Equating this gap to the photon energy gives

\[ \lambda=\frac{hc}{\Delta E} = \frac{(6.62607015\times10^{-34}\ \mathrm{J\,s}) (2.99792458\times10^8\ \mathrm{m\,s^{-1}})} {1.4606948\times10^{-19}\ \mathrm J}. \]

Thus,

\[ \lambda=1.35993\times10^{-6}\ \mathrm m =\boxed{1.36\times10^3\ \mathrm{nm}}. \]

This infrared prediction differs substantially from the experimental 455 nm, reflecting the limitations of the prescribed one-dimensional model.

5.7 — Two-dimensional HOMO, LUMO, and wavelength

Write the energy as

\[ E_{n_x,n_y}=A_xn_x^2+A_yn_y^2, \]

where

\[ A_x=\frac{h^2}{8m_eL_x^2} =8.91223\times10^{-21}\ \mathrm J, \qquad A_y=\frac{h^2}{8m_eL_y^2} =6.69407\times10^{-19}\ \mathrm J. \]

Here \(L_x=2.60\times10^{-9}\ \mathrm m\) and \(L_y=3.0\times10^{-10}\ \mathrm m\).

We must check the ordering of the two-dimensional states. The lowest state with \(n_y\geq2\) is \((1,2)\):

\[ E_{1,2}=A_x+4A_y =2.68654\times10^{-18}\ \mathrm J. \]

In comparison,

\[ E_{12,1}=144A_x+A_y =1.95277\times10^{-18}\ \mathrm J <E_{1,2}. \]

Therefore the first twelve spatial orbitals all have \(n_y=1\), with \(n_x=1,\ldots,12\). Filling eleven orbitals gives

\[ \boxed{\text{HOMO}=(11,1)},\qquad \boxed{\text{LUMO}=(12,1)}. \]

Their energies are

\[ E_{\mathrm H}=121A_x+A_y =1.7477873\times10^{-18}\ \mathrm J \approx\boxed{1.75\times10^{-18}\ \mathrm J}, \]

\[ E_{\mathrm L}=144A_x+A_y =1.9527686\times10^{-18}\ \mathrm J \approx\boxed{1.95\times10^{-18}\ \mathrm J}. \]

The identical \(y\)-direction contributions cancel in the gap:

\[ \Delta E=E_{\mathrm L}-E_{\mathrm H} =23A_x =2.0498129\times10^{-19}\ \mathrm J. \]

Using this unrounded gap,

\[ \lambda=\frac{hc}{\Delta E} =9.69086\times10^{-7}\ \mathrm m =\boxed{969\ \mathrm{nm}}. \]

5.8 — Length required for absorption at 455 nm

As stipulated, keep the HOMO and LUMO quantum numbers equal to \((11,1)\) and \((12,1)\). Their common \(y\)-direction energy again cancels:

\[ \Delta E=\frac{23h^2}{8m_eL_x^2}. \]

For absorption at \(\lambda=455\times10^{-9}\ \mathrm m\),

\[ \frac{hc}{\lambda}=\frac{23h^2}{8m_eL_x^2}. \]

Solving for the positive length,

\[ L_x=\sqrt{\frac{23h\lambda}{8m_ec}} = \sqrt{ \frac{23(6.62607015\times10^{-34}\ \mathrm{J\,s}) (455\times10^{-9}\ \mathrm m)} {8(9.1093837\times10^{-31}\ \mathrm{kg}) (2.99792458\times10^8\ \mathrm{m\,s^{-1}})} }. \]

Hence,

\[ L_x=1.78155\times10^{-9}\ \mathrm m =\boxed{17.8\ \text{Å}}. \]

This result uses the question’s explicit fixed-quantum-number assumption. If all two-dimensional levels were reordered and refilled at this shorter length, states with \(n_y=2\) would enter the occupied set.

The solution was checked against the official solution.