Conjugated Polyenes as Boxes: The Free-Electron Model

Introduction

The particle in a box can look like a textbook abstraction, but in 1949 Hans Kuhn showed that it gives surprisingly good predictions for real molecules. In a conjugated polyene, such as buta-1,3-diene or the long chain in β-carotene, the π electrons are not tied to individual double bonds. They spread along the whole chain of overlapping p orbitals. Treat the chain as a one-dimensional box, place the π electrons in its energy levels, and you can estimate the energy of the lowest electronic transition. This free-electron model is the simplest bridge between quantum mechanics and molecular colour.

Core explanation

The physical picture. Each carbon in a conjugated chain is sp² hybridised and contributes one p orbital perpendicular to the molecular plane. These p orbitals overlap side by side along the chain to form a delocalised π system. Along the chain the potential energy felt by a π electron is roughly constant, rising steeply beyond the terminal carbons. That is the particle in a box.

Assumptions of the model.

- The π electrons move independently along a straight line; electron–electron repulsion is ignored. - The potential is zero inside the box and infinite outside. - The box length L is the length of the conjugated path, usually taken as the number of C–C bonds plus about one extra bond length to allow the electron density to extend beyond the end atoms. With an average bond length of about 140 pm, a chain of k conjugated carbon atoms gives L ≈ k × 140 pm.

Filling the levels. The box energies are Eₙ = n²h²/8mₑL². Each level can hold two electrons with opposite spins (the Pauli principle). A chain with N π electrons, where N is even, fills the levels up to n = N/2. So:

- the HOMO has n = N/2; - the LUMO has n = N/2 + 1.

The HOMO–LUMO gap. The lowest-energy electronic excitation promotes one electron from the HOMO to the LUMO:

ΔE = (h²/8mₑL²)[(N/2 + 1)² − (N/2)²] = (N + 1)h²/8mₑL²

The corresponding absorption wavelength is λ = hc/ΔE = 8mₑcL²/h(N + 1).

Trends. Adding a CH=CH unit lengthens the chain by two bond lengths and adds two π electrons. L² grows faster than N + 1, so ΔE falls and λ increases: longer conjugated systems absorb at longer wavelengths. This agrees with observation: ethene absorbs near 170 nm, buta-1,3-diene near 217 nm, and polyenes with about eleven conjugated double bonds, such as β-carotene, absorb in the visible near 450 nm.

Limitations. Real polyenes have alternating short and long bonds rather than uniform ones. This bond alternation creates an energy gap that does not shrink to zero, so the simple model increasingly overestimates λ for long polyenes. The model works best for molecules with nearly equal bond lengths along the chain, such as symmetrical cyanine dyes.

Formulae

Eₙ = n²h²/8mₑL²; HOMO n = N/2; LUMO n = N/2 + 1; ΔE = (N + 1)h²/8mₑL²; λ = 8mₑcL²/h(N + 1).

Step-by-step reasoning

To apply the free-electron model:

1. Count the π electrons N in the conjugated system. 2. Estimate the box length L from the number of bonds and a typical bond length. 3. Identify the HOMO (n = N/2) and LUMO (n = N/2 + 1). 4. Calculate ΔE = (N + 1)h²/8mₑL². 5. Convert to wavelength with λ = hc/ΔE and compare with experiment.

Visual explanation

Draw the carbon skeleton of butadiene as a zigzag of four atoms and, above it, a box extending slightly beyond each end. Stack the levels n = 1, 2, 3 and 4 inside. Put two arrows (paired spins) in n = 1 and two in n = 2. The arrow jumping from n = 2 to the empty n = 3 is the absorption of light.

Real-world analogy

Think of a string of beads on a wire. Each bead is free to slide along the wire but cannot leave it. The π electrons are like beads on the wire of the conjugated chain: free to move along its length, confined at the ends.

Real-world example

Carrots are orange because β-carotene contains a long conjugated chain. Its π system is long enough to bring the HOMO–LUMO absorption into the blue region of the visible spectrum. The molecule absorbs blue light and reflects the remaining orange-red light.

Why?

Why does a longer chain lower the transition energy? Energies in a box scale as 1/L². Extending the chain widens the box, compressing the whole ladder of levels; although the HOMO moves higher up the ladder, the rungs become closer together faster than the extra electrons can compensate.

Common misconception

"Each double bond in a polyene absorbs light separately, like an isolated ethene." Conjugation delocalises the π electrons over the whole chain, producing a new set of levels with a much smaller HOMO–LUMO gap. That is why conjugated molecules absorb at far longer wavelengths than isolated double bonds.

Worked example

Question: Estimate λmax for buta-1,3-diene, taking L = 560 pm (four bond lengths of 140 pm).

Reasoning: N = 4, so ΔE = 5h²/8mₑL². h²/8mₑL² = (6.626 × 10⁻³⁴)² / (8 × 9.109 × 10⁻³¹ × (5.60 × 10⁻¹⁰)²) = 1.92 × 10⁻¹⁹ J, so ΔE = 9.61 × 10⁻¹⁹ J. λ = hc/ΔE = (6.626 × 10⁻³⁴ × 2.998 × 10⁸)/(9.61 × 10⁻¹⁹) = 2.07 × 10⁻⁷ m.

Answer: λ ≈ 207 nm, close to the measured value of about 217 nm.

Quick check

1. For hexa-1,3,5-triene, with six π electrons, which box levels are the HOMO and the LUMO? Answer: The HOMO is n = 3 and the LUMO is n = 4, since three levels hold the six electrons in pairs.

Exam focus

Examiners frequently ask you to derive ΔE = (N + 1)h²/8mL² and to calculate λ for a given molecule. Show clearly how you count the π electrons, choose L and apply the Pauli principle. Comment on agreement with experiment and on the model's assumptions.

Advanced insight

Hückel molecular orbital theory gives a more realistic treatment of the same π system by building orbitals from atomic p orbitals and allowing for the discrete positions of the atoms. For an infinitely long chain with equal bonds, both models predict a vanishing gap and metallic behaviour. Real polyacetylene is a semiconductor instead, because bond alternation opens a gap, but doping it produces electrical conductivity high enough to earn the 2000 Nobel Prize in Chemistry.

Summary

In the free-electron model, π electrons of a conjugated chain are particles in a one-dimensional box of length L. With N π electrons filling levels in pairs, the HOMO is n = N/2 and the LUMO n = N/2 + 1, giving ΔE = (N + 1)h²/8mₑL². Longer chains have smaller gaps and absorb at longer wavelengths. The model works well for evenly bonded systems but overestimates λ for long polyenes with bond alternation.

Practice questions

1. Derive the HOMO–LUMO gap for a box containing N π electrons. Answer: ΔE = (h²/8mL²)[(N/2 + 1)² − (N/2)²] = (h²/8mL²)(N + 1). 2. Why is the box length usually taken slightly longer than the distance between the terminal carbon atoms? Answer: The π electron density extends beyond the end nuclei, so adding roughly half a bond length at each end gives a better effective length. 3. By what factor does ΔE change for a system in which N stays the same but L doubles? Answer: ΔE falls to one quarter, since it is proportional to 1/L². 4. Give one assumption of the free-electron model that limits its accuracy for long polyenes. Answer: It assumes equal bond lengths and a flat potential, ignoring the bond alternation that keeps a finite gap in real polyenes.