Why does one organic molecule absorb only ultraviolet radiation while another appears intensely coloured? The answer begins with the chromophore. Rather than thinking of colour as a property of the whole molecule without explanation, it is more useful to ask which electrons in the molecule can be promoted to higher-energy molecular orbitals when light is absorbed.
The underlying idea connects directly with molecular orbital theory. Electrons may occupy σ bonding orbitals, π bonding orbitals or non-bonding orbitals, n. If a photon has exactly the required energy, an electron can be promoted into an antibonding orbital such as π* or σ*. Once we understand which orbitals are present and how far apart they are in energy, the behaviour of a chromophore becomes much easier to predict.
Try these five questions before reading the chapter. Focus on the connection between molecular structure, electronic transitions, energy and wavelength.
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What is a chromophore?
A chromophore is the part of a molecule responsible for a characteristic absorption of ultraviolet or visible radiation. Common organic chromophores contain π bonds, non-bonding electron pairs, or both.
Examples include C=C, C≡C, C=O and aromatic π systems. Groups containing atoms with non-bonding electrons may also contribute because these electrons can participate in electronic transitions.
The key question is not simply “Does the molecule contain a particular functional group?” but “Which occupied and unoccupied molecular orbitals are available, and what is their energy separation?”
Some common organic chromophores
π-bonded chromophores: contain occupied π orbitals and unoccupied π* orbitals, so π → π* excitation is possible.
Chromophores containing lone pairs: may also contain non-bonding n orbitals, allowing transitions such as n → π* or n → σ*.
A chromophore does not necessarily mean visible colour
This distinction is important. A molecule can contain a chromophore yet still be colourless because its absorption occurs entirely in the ultraviolet region. Ethene, for example, contains a C=C chromophore and undergoes a π → π* transition, but the required energy is too large for absorption to occur in the visible region.
Think of a chromophore as an electronic “light absorber”. Whether we can actually see a colour depends on where its absorption lies. UV absorption is invisible to the human eye; visible absorption can produce colour.
Electronic transitions in chromophores
When a molecule absorbs a photon, an electron may be promoted from an occupied molecular orbital to a higher-energy unoccupied molecular orbital. The photon is absorbed only when its energy, hν, matches the energy difference between the two states.
Therefore, a large energy gap requires high-energy radiation of short wavelength, whereas a smaller energy gap can be crossed using lower-energy radiation of longer wavelength.
The arrows point upward because absorption promotes an electron from a lower-energy occupied orbital to a higher-energy antibonding orbital.
σ → σ* transitions
A σ electron is promoted from a bonding σ orbital into an antibonding σ* orbital. Since σ bonds are relatively strong, the σ–σ* energy separation is large. These transitions therefore require very high-energy, short-wavelength radiation.
Many simple saturated molecules absorb mainly at wavelengths below the range normally used for routine UV/visible measurements. This is why an alkane such as ethane does not usually provide useful absorption in the conventional UV/visible region.
n → σ* transitions
Molecules containing atoms with non-bonding electron pairs can undergo promotion of an electron from an n orbital to a σ* orbital. Alcohols, amines and halogenoalkanes provide familiar examples of molecules containing such non-bonding electrons.
The required energy is generally lower than for σ → σ*, although it is still relatively large.
π → π* transitions
Compounds containing π bonds can undergo π → π* excitation. Since the π–π* separation is smaller than the σ–σ* separation, these absorptions generally occur at longer wavelengths.
π → π* transitions are especially important because they are commonly allowed transitions and therefore tend to produce relatively intense absorption bands.
n → π* transitions
If a molecule contains both non-bonding electrons and an accessible π* orbital, an electron can be promoted from n to π*. Carbonyl compounds are important examples.
The n → π* energy separation is often smaller than the corresponding π → π* separation, so n → π* absorption commonly occurs at a longer wavelength. However, n → π* transitions are often weak because they are formally forbidden or only weakly allowed by the relevant electronic selection rules.
| Transition | Typical energy requirement | Typical UV/visible behaviour |
|---|---|---|
| σ → σ* | Very high | Usually very short wavelength; often outside routine UV/Vis measurements |
| n → σ* | High | Usually ultraviolet |
| π → π* | Moderate | Usually relatively intense; may move to longer wavelength with conjugation |
| n → π* | Relatively low | Usually weaker and often found at longer wavelength than the corresponding π → π* transition |
This ordering is a useful qualitative guide rather than an absolute numerical rule. Actual orbital energies depend on molecular structure, substituents, conjugation and the surrounding chemical environment.
Allowed and forbidden transitions
Not every energetically possible transition occurs with the same probability. Spectroscopists therefore distinguish between transitions that are strongly allowed and those that are formally forbidden by electronic selection rules.
Allowed transition: the probability of absorption is relatively high, so the absorption band tends to be intense. π → π* transitions are commonly strong.
Forbidden transition: the transition probability is much lower. “Forbidden” does not mean absolutely impossible; it means that the transition is weak according to the relevant selection rule. n → π* bands are commonly much weaker than π → π* bands.
Why can a forbidden transition still appear?
Real molecules are not perfectly isolated, motionless electronic systems. Molecular vibrations and coupling between different motions can relax idealised selection rules, so a formally forbidden transition may still give a measurable absorption band. Its intensity is simply much lower than that of a strongly allowed transition.
In a carbonyl compound, the π → π* band is generally much stronger than the n → π* band. Do not confuse longer wavelength with greater intensity: wavelength depends primarily on the energy separation, whereas intensity depends strongly on the probability of the transition.
Predicting electronic transitions from molecular structure
A useful first approach is to inspect the structure and identify the kinds of electrons present. You do not need to draw a complete molecular orbital diagram every time.
Step 1: Look for π bonds
Double bonds, triple bonds and aromatic systems contain π electrons. Their presence immediately suggests that π → π* excitation may be possible.
Step 2: Look for atoms with non-bonding electron pairs
Oxygen, nitrogen, sulfur and halogen atoms commonly possess non-bonding electrons. Depending on the molecular orbitals available, these electrons may undergo n → σ* or n → π* excitation.
Step 3: Ask whether the π system is conjugated
If several adjacent p orbitals overlap, the π electrons become delocalised over a larger part of the molecule. This changes the energies of the molecular orbitals and can have a major effect on the wavelength absorbed.
Structure → available orbitals → possible electronic transitions → ΔE → absorption wavelength.
Example: propanone
The C=O group contains a π bond, so there are π electrons and an antibonding π* orbital. Oxygen also contains non-bonding electron pairs. Propanone can therefore show both π → π* and n → π* transitions.
Conjugation changes the energy gap
Conjugation occurs when neighbouring p orbitals overlap continuously, allowing π electrons to become delocalised over several adjacent atoms. A familiar structural pattern is alternating single and double bonds.
As conjugation becomes more extensive, more p orbitals combine to form a larger set of delocalised π molecular orbitals. The spacing between neighbouring molecular orbital energy levels becomes smaller.
Most importantly for UV/visible absorption, the energy difference between the highest occupied molecular orbital, HOMO, and the lowest unoccupied molecular orbital, LUMO, generally decreases.
More conjugation → smaller HOMO–LUMO gap → lower-energy photon required → longer absorption wavelength.
Why does delocalisation lower the HOMO–LUMO gap?
An isolated C=C bond has a relatively simple pair of π and π* molecular orbitals. When several adjacent p orbitals interact, they generate several delocalised π molecular orbitals with different energies. As the number of interacting p orbitals increases, these molecular orbital energy levels become more closely spaced.
Consequently, the separation between the highest occupied member of the set and the lowest unoccupied member becomes progressively smaller.
Bathochromic shifts and visible colour
A movement of an absorption band towards a longer wavelength is called a bathochromic shift, often described as a red shift.
Increasing conjugation is an important cause of bathochromic shifts in organic molecules. An isolated double bond typically absorbs in the ultraviolet region. As the conjugated π system becomes progressively larger, the absorption moves to longer wavelengths.
Small or isolated π system: relatively large HOMO–LUMO energy gap, so relatively high-energy and short-wavelength radiation is required.
Extended conjugated π system: smaller HOMO–LUMO energy gap, so lower-energy and longer-wavelength radiation can be absorbed.
When does a conjugated molecule become coloured?
A molecule appears coloured only when it absorbs one or more wavelengths within the visible region strongly enough to affect the light reaching our eyes. Moderate conjugation may still produce absorption entirely within the ultraviolet region.
When conjugation becomes sufficiently extensive, the π → π* energy gap can decrease enough for absorption to move into the visible region. Long conjugated systems such as those found in β-carotene provide familiar examples.
Do not write “conjugation causes colour” as an automatic rule. The safer reasoning is: conjugation decreases ΔE and therefore shifts absorption to longer wavelength; colour appears only if significant absorption reaches the visible region.
How to compare chromophores in examination questions
Questions often present two related molecules and ask which absorbs at the longer wavelength or has the larger value of λmax. Instead of trying to memorise individual wavelengths, compare their electronic structures.
Compare the extent of conjugation first
If the principal difference between two molecules is the length of their conjugated π systems, the molecule with the greater extent of conjugation will generally have the smaller HOMO–LUMO energy gap and therefore absorb at the longer wavelength.
1,3,5-Hexatriene has a more extensively conjugated π system than ethene. Its π electrons are delocalised over a larger region, giving a smaller HOMO–LUMO energy gap. Less energy is therefore required for the relevant electronic transition, so 1,3,5-hexatriene absorbs at a longer wavelength.
Do not confuse wavelength with intensity
A spectrum tells us more than where absorption occurs. The wavelength of an absorption is determined by the energy difference between the states, while the intensity of the absorption also depends strongly on how probable the electronic transition is.
This explains why an n → π* transition can occur at a longer wavelength than a π → π* transition yet still produce a much weaker absorption band.
Common Mistakes
| Common mistake | Why it is wrong | Better approach |
|---|---|---|
| Assuming every chromophore makes a compound visibly coloured | A chromophore may absorb entirely in the ultraviolet region. | Distinguish between having an electronic absorption and absorbing within the visible region. |
| Saying that greater conjugation increases the HOMO–LUMO energy gap | Greater delocalisation generally causes the relevant molecular orbital levels to become more closely spaced. | Use the chain: more conjugation → smaller ΔE → longer λ. |
| Assuming “forbidden” means the transition cannot occur at all | Forbidden transitions can still occur weakly because ideal selection rules may be relaxed. | Describe a forbidden transition as having a much lower probability and therefore usually lower intensity. |
| Confusing absorption wavelength with absorption intensity | They describe different properties of the transition. | Relate wavelength to ΔE and intensity to transition probability. |
| Claiming a carbonyl group has only a π → π* transition | Oxygen also possesses non-bonding electrons. | Recognise that carbonyl compounds commonly show both π → π* and n → π* transitions. |
| Saying longer wavelength means higher photon energy | E = hc/λ, so energy and wavelength are inversely related. | Longer λ means lower photon energy. |
What you should be able to do
Recognise chromophores
Identify structural features such as C=C, C=O and conjugated aromatic systems that can produce characteristic UV/visible absorption.
Identify possible transitions
Use the presence of σ bonds, π bonds and non-bonding electron pairs to predict σ → σ*, n → σ*, π → π* and n → π* transitions.
Relate energy to wavelength
Apply ΔE = hc/λ to explain why a smaller molecular orbital energy gap gives absorption at a longer wavelength.
Explain conjugation
Explain how increasing π-electron delocalisation generally decreases the HOMO–LUMO gap.
Explain bathochromic shifts
Recognise a shift to longer wavelength as a bathochromic or red shift and connect it to changes in molecular electronic structure.
Compare spectra from structure
Compare related molecules and predict which should absorb at the longer wavelength by considering conjugation and the relevant electronic transition.
These ten questions test whether you can connect molecular structure with electronic transitions, absorption wavelength, intensity and conjugation.
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Frequently asked questions
Chromophores: questions students often ask
Chromophores become much easier to understand once the ideas of molecular orbitals, transition energy, conjugation and wavelength are connected rather than memorised separately.
What exactly is a chromophore?
A chromophore is the part of a molecule responsible for characteristic UV or visible absorption because its electrons can undergo an electronic transition between molecular orbitals. Common examples include C=C, C=O and conjugated aromatic π systems.
What happens when a chromophore absorbs light?
A photon is absorbed when its energy matches the separation between an occupied molecular orbital and a suitable higher-energy unoccupied orbital. An electron is then promoted to the higher-energy orbital.
Which electronic transition generally requires the least energy?
Among the commonly discussed σ → σ*, n → σ*, π → π* and n → π* transitions, n → π* often has the smallest energy requirement. The precise orbital energies nevertheless depend on the molecule, so this should be treated as a qualitative trend rather than an unbreakable numerical ordering.
What is the difference between an allowed and a forbidden transition?
An allowed transition has a relatively high transition probability and therefore normally gives a stronger absorption. A forbidden transition has a much lower probability according to the relevant selection rules, so its absorption is weak. “Forbidden” does not mean that the transition can never be observed.
What is the relationship between ΔE and the wavelength absorbed?
They are inversely related through ΔE = hc/λ. A larger energy gap requires a higher-energy, shorter-wavelength photon, while a smaller energy gap can be crossed using a lower-energy, longer-wavelength photon.
Why does propanone show both π → π* and n → π* transitions?
Its C=O group contains a π bond and therefore occupied π electrons together with an antibonding π* orbital. Oxygen also contains non-bonding electrons, n. Electrons can therefore be promoted either from π to π* or from n to π*.
Why does increasing conjugation decrease the HOMO–LUMO gap?
Increasing conjugation causes more adjacent p orbitals to interact and produces a larger set of delocalised π molecular orbitals. These energy levels become more closely spaced, so the separation between the highest occupied and lowest unoccupied orbitals generally decreases.
What is a bathochromic shift?
A bathochromic shift is a movement of an absorption band towards a longer wavelength. It is also called a red shift. Increasing the extent of conjugation is a common cause because it decreases the energy required for electronic excitation.
Does more conjugation always mean that a compound is coloured?
No. Greater conjugation generally moves absorption towards longer wavelengths, but the absorption may still remain entirely within the ultraviolet region. The compound becomes visibly coloured only when significant absorption extends into the visible region.
Why can an n → π* transition occur at a longer wavelength but still be weaker than a π → π* transition?
These observations describe different properties. The longer wavelength reflects the smaller energy gap for the n → π* transition. Its lower intensity arises because n → π* is commonly a low-probability, formally forbidden transition, whereas π → π* transitions are often strongly allowed.
Chromophore chemistry ultimately follows one continuous idea: molecular structure determines the available molecular orbitals, those orbital energies determine ΔE, and ΔE determines the wavelength that can be absorbed. Conjugation is especially important because it allows π electrons to spread over a larger system, narrows the HOMO–LUMO gap and shifts absorption towards longer wavelengths. With that relationship clear, UV/visible spectra become something you can reason from rather than simply memorise.