8.2 · Conjugation & colour

Conjugated π systems

Alternating double and single bonds let π electrons spread out.

By the end you should be able to:

Identify conjugated π systems and count their π electrons
Use the particle-in-a-box model to relate conjugation length to the HOMO–LUMO gap and λmax

Interactive

Conjugation and colour

Lengthen a conjugated chain and watch the HOMO–LUMO gap shrink and the absorbed colour shift.

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Key idea

What makes a π system conjugated

π bonds are conjugated when they alternate with single bonds, so that neighbouring π bonds are separated by exactly one single bond. Every atom along the chain is $sp^2$ and has a p orbital, the p orbitals overlap continuously, and the π electrons are delocalized over the whole system.

1,3-Butadiene, $\ce{CH2=CH-CH=CH2}$ : conjugated; 4 π electrons spread over 4 carbons.
1,4-Pentadiene, $\ce{CH2=CH-CH2-CH=CH2}$ : not conjugated. The $sp^3$ CH₂ has no p orbital and breaks the chain, so the molecule behaves like two separate ethene units ( $\lambda_{\max}$ = 178 nm, close to ethene's 171 nm).
Benzene: three alternating C=C in a ring; 6 π electrons delocalized around the ring.
A C=O can be part of the system: acrolein, $\ce{CH2=CH-CH=O}$ , has 4 conjugated π electrons.

Method

Finding the conjugated system and counting π electrons

Mark every double bond (C=C, C=O, N=N).
Join double bonds that are separated by exactly one single bond. An $sp^3$ atom (such as a CH₂) between them ends the system.
Count 2 π electrons for each double bond in the largest connected system. (In this course, lone pairs on O or N are not counted.)
For a linear polyene, the number of conjugated carbons equals the number of π electrons.

Examples: styrene (vinyl group on a benzene ring) has 8 conjugated π electrons; naphthalene 10; 1,4-cyclohexadiene only 2, because its two C=C are separated by CH₂ groups.

Key idea

The particle-in-a-box model of a polyene

Treat the delocalized π electrons as particles free to move along the conjugated chain but not beyond it: a one-dimensional box of length $L$ .

Energy levels: $E_n = \dfrac{n^2 h^2}{8 m_e L^2}$ with $n$ = 1, 2, 3, …; each level holds 2 electrons.
With $N$ π electrons, levels 1 to $N/2$ are full: the HOMO is $n = N/2$ and the LUMO is $n = N/2 + 1$ .
The lowest-energy absorption promotes one electron from the HOMO to the LUMO.
Course convention: $L$ = (number of conjugated carbon atoms) × 0.140 nm.

Longer conjugation means a longer box, more closely spaced levels, a smaller HOMO–LUMO gap and a longer $\lambda_{\max}$ :

Polyene	Conjugated C	π electrons	Measured $\lambda_{\max}$ (nm)
ethene	2	2	171
1,3-butadiene	4	4	217
1,3,5-hexatriene	6	6	258
1,3,5,7-octatetraene	8	8	290
β-carotene (11 C=C)	22	22	450

Formula

HOMO–LUMO gap and absorbed wavelength

With $n_H = N/2$ for the HOMO:

\Delta E = E_{n_H+1} - E_{n_H} = \frac{\left[(n_H+1)^2 - n_H^2\right] h^2}{8 m_e L^2} = \frac{(2n_H + 1)\,h^2}{8 m_e L^2} = \frac{(N+1)\,h^2}{8 m_e L^2}

\lambda = \frac{hc}{\Delta E} = \frac{8 m_e c L^2}{(N+1)\,h}

Use $h = 6.626 \times 10^{-34}$ J s, $c = 2.998 \times 10^{8}$ m/s, $m_e = 9.109 \times 10^{-31}$ kg, and $L$ in metres. Because $N$ grows in step with $L$ , the gap shrinks roughly as $1/L$ and $\lambda$ grows roughly in proportion to the chain length.

Worked example

Worked example: 1,3,5-hexatriene

$\ce{CH2=CH-CH=CH-CH=CH2}$ has 6 conjugated carbons and 6 π electrons.

$L = 6 \times 0.140\ \text{nm} = 0.840\ \text{nm} = 8.40 \times 10^{-10}\ \text{m}$ .
HOMO $n = 3$ , LUMO $n = 4$ , so $\Delta E = (4^2 - 3^2)\dfrac{h^2}{8 m_e L^2} = 7\,\dfrac{h^2}{8 m_e L^2}$ .
$\dfrac{h^2}{8 m_e L^2} = \dfrac{(6.626 \times 10^{-34})^2}{8\,(9.109 \times 10^{-31})\,(8.40 \times 10^{-10})^2} = 8.54 \times 10^{-20}\ \text{J}$ .
$\Delta E = 7 \times 8.54 \times 10^{-20} = 5.98 \times 10^{-19}\ \text{J}$ .
$\lambda = \dfrac{hc}{\Delta E} = \dfrac{(6.626 \times 10^{-34})(2.998 \times 10^{8})}{5.98 \times 10^{-19}} = 3.32 \times 10^{-7}\ \text{m} = 332\ \text{nm}$ .

The measured $\lambda_{\max}$ is 258 nm; both are in the UV, so hexatriene is colourless. The model gets the trend right and is close for butadiene (207 nm predicted, 217 nm measured), but it overestimates λ, more and more for longer chains (octatetraene: 460 nm predicted, 290 nm measured). Real polyenes alternate short C=C and longer C–C bonds, so the π electrons are not perfectly free and the gap does not shrink as fast as the model says.

Common mistake

Box-model traps

Wrong: $L$ = (number of carbons − 1) × 0.140 nm, counting bonds. Right: in this course $L$ = number of conjugated carbons × 0.140 nm (hexatriene: 0.840 nm).
Wrong: the HOMO is $n = N$ . Right: each level holds 2 electrons, so the HOMO is $n = N/2$ (hexatriene: $n = 3$ , not 6).
Wrong: counting every C=C in the molecule. Right: count only the connected conjugated system. 1,4-Pentadiene has two isolated 2-electron systems, not one 4-electron system.
Wrong: leaving $L$ in nanometres. Right: convert to metres before squaring: 0.840 nm = $8.40 \times 10^{-10}$ m.

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