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 3 of 4: Solve for

k

\lambda = 600

← Previous Next →

600 = 64.6\times\dfrac{(k+2)^2}{2k+1} \Rightarrow k^2 - 14.58k + \dots = 0 \Rightarrow k_1 = 14.92,\; k_2 = -0.36 \Rightarrow k = 15

Analysis

Setting $\lambda = 600$ nm in the derived equation gives a quadratic in $k$ with roots $k_1 = 14.92$ and $k_2 = -0.36$ ; only the positive integer solution is physical, so $k = 15$ and the polyene is $\ce{CH2=CH-(CH=CH)13-CH=CH2}$ .

Common pitfall. Keep the negative root

k_2 = -0.36

out of the answer — a polyene cannot have a negative or non-integer number of double bonds.