Problem 1

Particles in a box: polyenes. In quantum mechanics, the movement of π electrons along a neutral chain of conjugated carbon atoms may be modeled using the 'particle in a box' method. The energy of the π electrons is

E_n = \dfrac{n^2 h^2}{8 m L^2}

, where

n

is the quantum number (

n = 1, 2, 3, \dots

h

is Planck's constant,

m

is the mass of the electron, and

L

is the length of the box, approximated by

L = (k + 2) \times 1.40\ \text{Å}

(

k

being the number of conjugated double bonds along the carbon chain). A photon of wavelength

\lambda

can promote a π electron from the HOMO to the LUMO. A semi-empirical formula relates

\lambda

k

\lambda = B \times \dfrac{(k+2)^2}{2k+1}

(Equation 1). (a) Using Equation 1 with

B = 65.01

nm, calculate

\lambda

for octatetraene,

\ce{CH2=CH-CH=CH-CH=CH-CH=CH2}

. (b) Derive Equation 1 from the particle-in-a-box expression and calculate the theoretical value

B_{\text{calc}}

. (c) Find the number of conjugated double bonds

k

and give the structure of the polyene whose HOMO–LUMO excitation requires

\lambda = 600

nm. (d) For that polyene, calculate the HOMO–LUMO energy difference

\Delta E

in kJ mol

^{-1}

Step 1 of 4: Apply Equation 1 to octatetraene

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\lambda = 65.01 \times \dfrac{(4+2)^2}{2(4)+1} = 65.01 \times \dfrac{36}{9} = 260.0\ \text{nm}

Analysis

Octatetraene has $k = 4$ conjugated double bonds. Substituting into $\lambda = B(k+2)^2/(2k+1)$ with $B = 65.01$ nm gives $\lambda = 260.0$ nm.