Chromophores and Types of Transition

Introduction

Only part of a molecule is usually responsible for its UV-visible absorption. In propanone, for example, the methyl groups contribute almost nothing, while the C=O group produces the observed bands. A group of atoms that absorbs in the UV-visible region is called a chromophore (from the Greek for "colour bearer"). Recognising chromophores and the kind of electronic transition each one undergoes lets chemists predict where a compound will absorb and how strongly.

Core explanation

Four types of transition. Using the orbital ladder σ < π < n < π < σ , four transitions matter for organic molecules:

Transition Found in Typical λ Typical intensity (ε / dm³ mol⁻¹ cm⁻¹) --- --- --- --- σ → σ all single bonds (C–C, C–H) below 150 nm strong, but out of range n → σ atoms with lone pairs (O, N, Cl, S) 150–250 nm weak to moderate π → π C=C, C=O, C≡C, aromatic rings about 170–250 nm for isolated groups; longer when conjugated strong, 10³–10⁵ n → π C=O, N=N, NO₂, C=S about 270–350 nm weak, 10–100

Why σ transitions are rarely observed. The σ → σ gap is very large, so it needs vacuum-ultraviolet radiation that normal instruments cannot use. n → σ bands for alcohols and amines fall mostly below 200 nm. In practice, routine UV-visible spectroscopy is about compounds with π bonds.

π → π transitions. An electron in a bonding π orbital is promoted to the antibonding π orbital. The π and π orbitals occupy the same region of space, so the overlap between them is good and the transition is highly probable: bands are intense . An isolated C=C, as in ethene, absorbs at about 171 nm; benzene has bands near 180, 204 and a weaker band at 254 nm.

n → π transitions. A lone-pair electron on an oxygen or nitrogen atom is promoted into a π orbital. Because the non-bonding orbital lies higher than π, the gap to π is smaller and the band appears at a longer wavelength . However, the lone-pair orbital lies roughly at right angles to the π system, so overlap is poor and the transition is symmetry-forbidden: bands are weak . Propanone shows its n → π band near 280 nm with ε only about 15 dm³ mol⁻¹ cm⁻¹.

Auxochromes. Groups such as –OH, –OR, –NH₂ and halogens have lone pairs that can interact with a neighbouring π system. They do not absorb usefully themselves, but attached to a chromophore they shift λmax to longer wavelength and often increase the intensity. Phenol absorbs at a longer wavelength than benzene for this reason.

Solvent effects. Polar protic solvents hydrogen-bond to lone pairs, lowering the energy of the n orbital. This widens the n → π gap, shifting the band to shorter wavelength (a hypsochromic or "blue" shift). This behaviour is a useful test for identifying n → π bands.

Step-by-step reasoning

To predict the transitions in a molecule:

1. List its bond types and lone pairs. 2. Ignore σ → σ (out of range). 3. Each C=C, C=O or aromatic ring gives a π → π band: expect strong absorption. 4. Each C=O or N=O with lone pairs also gives an n → π band at longer wavelength: expect weak absorption.

Visual explanation

On an energy diagram, draw a short arrow from n up to π and a longer arrow from π up to π . The short arrow corresponds to a longer wavelength. Draw the n → π arrow thin and the π → π arrow thick to represent their very different intensities.

Real-world analogy

A chromophore is like the antenna on a radio. The rest of the radio's case does not pick up signals; only the antenna is tuned to receive them. Similarly, only the chromophore in a molecule is tuned to absorb UV-visible photons.

Real-world example

Proteins are measured routinely at 280 nm because the aromatic rings in the amino acids tryptophan and tyrosine act as chromophores there. DNA is measured at 260 nm, where the aromatic bases absorb strongly. Comparing the two readings helps biochemists check sample purity.

Why?

Why is the n → π band longer in wavelength but weaker than the π → π band? Lone-pair electrons start higher in energy, so less energy is needed to reach π . But the n orbital and π orbital point in different directions and overlap poorly, so the transition has a low probability.

Common misconception

"The whole molecule absorbs the light equally." Absorption is localised mainly in the chromophore; saturated alkyl parts barely affect λmax, although attached auxochromes and conjugation can shift it.

Worked example

Question: Butanone shows a strong band near 190 nm and a weak band at 277 nm (ε ≈ 20 dm³ mol⁻¹ cm⁻¹). Assign each band.

Reasoning: The C=O group is the chromophore. The strong, short-wavelength band must be π → π ; the weak, longer-wavelength band is characteristic of a lone pair on oxygen being promoted to π .

Answer: 190 nm is π → π ; 277 nm is n → π .

Quick check

1. Which type of transition gives very weak bands at relatively long wavelength in aldehydes and ketones? Answer: The n → π transition of a lone-pair electron on the carbonyl oxygen.

Exam focus

Know the order of orbital energies, and be able to justify which transition gives the longest wavelength (n → π ) and which gives the most intense band (π → π ). Always name the chromophore responsible, for example "the C=O group".

Advanced insight

A shift of λmax to longer wavelength is called bathochromic (red shift) and to shorter wavelength hypsochromic (blue shift). An increase in intensity is hyperchromic and a decrease hypochromic . Changing solvent polarity typically moves π → π and n → π bands in opposite directions, which is why recording spectra in two solvents helps assign them.

Summary

A chromophore is the part of a molecule responsible for UV-visible absorption, usually a π bond or a group with lone pairs next to a π bond. π → π transitions are intense and found in all unsaturated groups; n → π transitions occur at longer wavelength but are weak because of poor orbital overlap. σ → σ and most n → σ transitions lie below the usual instrument range. Auxochromes and solvents shift absorption bands.

Practice questions

1. Define the term chromophore and give two examples. Answer: A group of atoms that absorbs UV or visible radiation, such as C=C, C=O or a benzene ring. 2. Explain why ethanol shows no absorption above 220 nm. Answer: It has only σ bonds and lone pairs on oxygen, so its only transitions are σ → σ and n → σ , which occur below about 200 nm. 3. A band at 290 nm has ε = 18 dm³ mol⁻¹ cm⁻¹. Which transition is most likely responsible? Answer: An n → π transition, because it is weak and at relatively long wavelength, typical of a carbonyl group. 4. What is an auxochrome, and how does it affect an absorption band? Answer: A group with lone pairs, such as –OH or –NH₂, attached to a chromophore; it shifts λmax to longer wavelength and often increases intensity.