IChO 2014 · Theory problem 1

Particles in a box: polyenes

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In quantum mechanics, the movement of \(\pi\) electrons along a neutral chain of conjugated carbon atoms may be modeled using the ‘particle in a box’ method. The energy of the \(\pi\) electrons is given by the following equation:

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

where \(n\) is the quantum number (\(n=1,2,3,\ldots\)), \(h\) is Planck’s constant, \(m\) is the mass of electron, and \(L\) is the length of the box which may be approximated by \(L=(k+2)\times1.40\ \text{Å}\) (\(k\) being the number of conjugated double bonds along the carbon chain in the molecule). A photon with the appropriate wavelength \(\lambda\) may promote a \(\pi\) electron from the highest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO). An approximate semi-empirical formula based on this model which relates the wavelength \(\lambda\), to the number of double bonds \(k\) and constant \(B\) is as follows:

\[ \lambda(\mathrm{nm})=B\times\frac{(k+2)^2}{(2k+1)} \qquad\text{Equation 1} \]

1.1 Using this semi-empirical formula with \(B=65.01\ \mathrm{nm}\) calculate the value of the wavelength \(\lambda\) (nm) for octatetraene (\(\ce{CH2=CH-CH=CH-CH=CH-CH=CH2}\)).

1.2 Derive equation 1 (an expression for the wavelength \(\lambda\) (nm) corresponding to the transfer of an electron from the HOMO to the LUMO) in terms of \(k\) and the fundamental constants, and hence calculate theoretical value of the constant \(B_{\mathrm{calc.}}\).

1.3 We wish to synthesize a linear polyene for which the excitation of a \(\pi\) electron from the HOMO to the LUMO requires an absorption wavelength of close to 600 nm. Using your expression from part 2, determine the number of conjugated double bonds (\(k\)) in this polyene and give its structure. [If you did not solve Part 2, use the semi-empirical Equation 1 with \(B=65.01\ \mathrm{nm}\) to complete Part 3.]

1.4 For the polyene molecule found in Part 3, calculate the difference in energy between the HOMO and the LUMO, \(\Delta E\), (\(\mathrm{kJ\,mol^{-1}}\)).

In case that Part 3 was not solved, take \(k=5\) to solve this problem.

The model for a particle in a one-dimensional box can be extended to a three dimensional rectangular box of dimensions \(L_x\), \(L_y\) and \(L_z\), yielding the following expression for the allowed energy levels:

\[ E_{n_x,n_y,n_z} =\frac{h^2}{8m} \left( \frac{n_x^2}{L_x^2}+ \frac{n_y^2}{L_y^2}+ \frac{n_z^2}{L_z^2} \right) \]

The three quantum numbers \(n_x\), \(n_y\), and \(n_z\) must be integer values and are independent of each other.

1.5 Give the expressions for the three different lowest energies, assuming that the box is cubic with a length of \(L\).

1.6 Levels with the same energy are said to be degenerate. Draw a sketch showing all the energy levels, including any degenerate levels, that correspond to quantum numbers having values of 1 or 2 for a cubic box.

Official English text of the problem.

Answers

  • 1.1: \(\boxed{\lambda=260.0\ \mathrm{nm}}\).
  • 1.2: With \(a=1.40\times10^{-10}\ \mathrm m\), \[ \boxed{\lambda=\frac{8mca^2}{h}\frac{(k+2)^2}{2k+1}}, \qquad \boxed{B_{\mathrm{calc}}=\frac{8mca^2}{h}=64.6\ \mathrm{nm}}. \]
  • 1.3: \(\boxed{k=15}\), giving \(\lambda\approx602\ \mathrm{nm}\). The molecule is triaconta-1,3,5,7,9,11,13,15,17,19,21,23,25,27,29-pentadecaene, \(\ce{C30H32}\), with structure \(\ce{CH2=CH-(CH=CH)13-CH=CH2}\). SMILES: C=C-C=C-C=C-C=C-C=C-C=C-C=C-C=C-C=C-C=C-C=C-C=C-C=C-C=C-C=C.
  • 1.4: \(\boxed{\Delta E\approx1.99\times10^2\ \mathrm{kJ\,mol^{-1}}}\).
  • 1.5: The three lowest distinct energies are \[ \boxed{\frac{3h^2}{8mL^2},\quad\frac{6h^2}{8mL^2},\quad\frac{9h^2}{8mL^2}}. \]
  • 1.6: Writing \(A=h^2/(8mL^2)\), the requested levels are \(\boxed{3A,\ 6A,\ 9A,\ 12A}\), with spatial degeneracies \(\boxed{1,\ 3,\ 3,\ 1}\), respectively.

Solution

1.1

Octa-1,3,5,7-tetraene (octatetraene), SMILES C=CC=CC=CC=C, contains four conjugated double bonds, so \(k=4\). Therefore,

\[ \lambda =(65.01\ \mathrm{nm})\frac{(4+2)^2}{2(4)+1} =(65.01\ \mathrm{nm})\frac{36}{9} =260.04\ \mathrm{nm}. \]

To four significant figures,

\[ \boxed{\lambda=260.0\ \mathrm{nm}}. \]

1.2

Each conjugated double bond contributes two \(\pi\) electrons, giving \(2k\) electrons altogether. Each spatial orbital accommodates two electrons of opposite spin. In the ground state, the levels \(n=1,\ldots,k\) are therefore filled:

\[ n_{\mathrm{HOMO}}=k, \qquad n_{\mathrm{LUMO}}=k+1. \]

The excitation energy is

\[ \begin{aligned} \Delta E &=E_{k+1}-E_k\\ &=\frac{h^2}{8mL^2}\left[(k+1)^2-k^2\right]\\ &=\frac{h^2(2k+1)}{8mL^2}. \end{aligned} \]

Equating this to the photon energy,

\[ \frac{hc}{\lambda}=\frac{h^2(2k+1)}{8mL^2}, \]

gives

\[ \lambda=\frac{8mcL^2}{h(2k+1)}. \]

Let \(a=1.40\ \text{Å}=1.40\times10^{-10}\ \mathrm m\). Substituting \(L=(k+2)a\),

\[ \boxed{ \lambda=\underbrace{\frac{8mca^2}{h}}_{B_{\mathrm{calc}}} \frac{(k+2)^2}{2k+1} }. \]

The problem does not tabulate fundamental constants. Using the standard values

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

we obtain

\[ B_{\mathrm{calc}} =\frac{8mca^2}{h} =6.46249\times10^{-8}\ \mathrm m =64.6249\ \mathrm{nm}. \]

Thus, to three significant figures,

\[ \boxed{B_{\mathrm{calc}}=64.6\ \mathrm{nm}}. \]

1.3

Using the theoretical constant, with additional digits retained during calculation,

\[ 600=64.6249\frac{(k+2)^2}{2k+1}. \]

Define \(r=600/64.6249\). Rearrangement gives

\[ k^2+(4-2r)k+(4-r)=0. \]

The positive root is

\[ k=r-2+\sqrt{r(r-3)}\approx14.923. \]

The number of double bonds must be an integer. Checking nearby values,

\[ \lambda(14)=570.5\ \mathrm{nm},\qquad \lambda(15)=602.5\ \mathrm{nm},\qquad \lambda(16)=634.5\ \mathrm{nm}. \]

Hence,

\[ \boxed{k=15}. \]

The structure is

\[ \boxed{\ce{CH2=CH-(CH=CH)13-CH=CH2}}, \]

an unbranched chain of 30 carbon atoms with 15 alternating double bonds and molecular formula \(\ce{C30H32}\).

Its name is triaconta-1,3,5,7,9,11,13,15,17,19,21,23,25,27,29-pentadecaene.

SMILES: C=C-C=C-C=C-C=C-C=C-C=C-C=C-C=C-C=C-C=C-C=C-C=C-C=C-C=C-C=C

The model approximates the conjugated chain as one-dimensional. It does not specify double-bond stereochemistry, so the SMILES leaves \(E/Z\) configurations unspecified.

1.4

For \(k=15\), the HOMO and LUMO have \(n=15\) and \(n=16\), respectively, and

\[ L=(15+2)(1.40\ \text{Å}) =23.8\ \text{Å} =2.38\times10^{-9}\ \mathrm m. \]

The energy gap per molecule is

\[ \begin{aligned} \Delta E_{\mathrm{molecule}} &=\frac{h^2(16^2-15^2)}{8mL^2}\\ &=\frac{31h^2}{8m(2.38\times10^{-9}\ \mathrm m)^2}\\ &=3.29717\times10^{-19}\ \mathrm J. \end{aligned} \]

Using \(N_{\mathrm A}=6.02214076\times10^{23}\ \mathrm{mol^{-1}}\),

\[ \begin{aligned} \Delta E_{\mathrm{molar}} &=N_{\mathrm A}\Delta E_{\mathrm{molecule}}\\ &=1.98560\times10^5\ \mathrm{J\,mol^{-1}}\\ &=198.560\ \mathrm{kJ\,mol^{-1}}. \end{aligned} \]

Therefore,

\[ \boxed{\Delta E\approx1.99\times10^2\ \mathrm{kJ\,mol^{-1}}}. \]

Using the target wavelength of exactly \(600\ \mathrm{nm}\) gives the close approximation \(199.4\ \mathrm{kJ\,mol^{-1}}\); the calculation above uses the actual model prediction for \(k=15\).

1.5

For a cubic box, \(L_x=L_y=L_z=L\), so

\[ E_{n_x,n_y,n_z} =\frac{h^2}{8mL^2}(n_x^2+n_y^2+n_z^2). \]

The quantum numbers are positive integers: zero would give an identically zero wavefunction rather than an allowed state. The three smallest distinct sums of squares are

\[ 1^2+1^2+1^2=3,\qquad 2^2+1^2+1^2=6,\qquad 2^2+2^2+1^2=9. \]

Consequently,

\[ \boxed{ \begin{aligned} E_{1,1,1}&=\frac{3h^2}{8mL^2},\\ E_{2,1,1}=E_{1,2,1}=E_{1,1,2} &=\frac{6h^2}{8mL^2},\\ E_{2,2,1}=E_{2,1,2}=E_{1,2,2} &=\frac{9h^2}{8mL^2}. \end{aligned} } \]

1.6

Let \(A=h^2/(8mL^2)\). Restricting each quantum number to 1 or 2 gives \(2^3=8\) spatial states:

Energy Quantum-number triples \((n_x,n_y,n_z)\) Spatial degeneracy
\(12A\) \((2,2,2)\) 1
\(9A\) \((2,2,1),\ (2,1,2),\ (1,2,2)\) 3
\(6A\) \((2,1,1),\ (1,2,1),\ (1,1,2)\) 3
\(3A\) \((1,1,1)\) 1

The sketch has an upward energy axis and four equally spaced heights at \(3A\), \(6A\), \(9A\), and \(12A\). From bottom to top, there are one, three, three, and one horizontal level lines. Degenerate lines lie side by side at the same height, each labelled with its quantum-number triple from the table.

These degeneracies count spatial states; electron spin is not included.

The solution was checked against the official solution.