Ethene absorbs ultraviolet light most strongly at 171 nm. (A nanometer, nm, is a billionth of a meter.) Buta-1,3-diene, with two double bonds in conjugation, absorbs at 217 nm, and hexa-1,3,5-triene, with three, absorbs at 258 nm.
UV-Vis spectroscopy is the measurement behind those numbers. Visible light runs from about 400 to 700 nm, and ultraviolet light has shorter wavelengths. The instrument shines both kinds of light through a sample and records which wavelengths the sample absorbs. For an organic chemist it answers one question well: how far a conjugated π system extends. It turns the orbital picture from Conjugated systems into a number you can read off an instrument.
What the measurement is
Light arrives in single packets of energy called photons. The shorter the wavelength, the more energy each photon carries. A molecule absorbs a photon when the photon's energy lifts an electron from a filled molecular orbital into an empty one.
In ethene, the cheapest such jump takes an electron from the filled π orbital to the empty antibonding one, written π* and read "pi star". These two are ethene's HOMO and LUMO, the frontier orbitals from Conjugated systems. The jump is written π → π*, and ethene absorbs most strongly the photons whose energy matches the gap between the two orbitals.
An absorption is not a single sharp line but a band spread over a range of wavelengths. The wavelength at the top of the band is called λmax ("lambda max"). UV-Vis reports this value, and the tables below list it.
Every polyene, a chain of alternating double and single bonds, works the same way. Its HOMO is the highest filled bonding π orbital, and its LUMO is the lowest empty antibonding one. So its main band is also a π → π* band, and the wavelength of that band tells you the size of the HOMO–LUMO gap. Use the rule in this direction:
Smaller gap → lower energy → longer wavelength absorbed.
More conjugation, smaller gap, longer wavelength
Conjugated systems built the full ladder of π orbitals. Here only the gap between its HOMO and LUMO matters. Ethene's two p orbitals give two π orbitals, one filled and one empty. Buta-1,3-diene's four p orbitals give four, and hexa-1,3,5-triene's six give six.
Each extra double bond in the conjugation brings two more p orbitals, and so two more π orbitals. The ladder does grow taller, but much more slowly than it gains rungs, so the rungs crowd closer together. The HOMO climbs, the LUMO drops, and the gap between them narrows with every double bond added. β-Carotene, the orange pigment of carrots, has eleven conjugated double bonds. That gives it eleven filled π orbitals and the narrowest gap of any compound on this page.
The measured values follow the gap. Learn them as a trend rather than as exact figures:
| Compound | Conjugated C=C | λmax (approx.) |
|---|---|---|
| Ethene | 1 | 171 nm |
| Buta-1,3-diene | 2 | 217 nm |
| Hexa-1,3,5-triene | 3 | 258 nm |
| β-Carotene | 11 | ~450 nm |
The part of a molecule that absorbs the light is called its chromophore. In all four compounds above, the chromophore is the π system. An isolated C=C absorbs below about 200 nm, further into the ultraviolet than ordinary instruments reach. On an ordinary instrument, a simple alkene therefore shows no band at all.
Conjugate two double bonds and the band moves into the range the instrument covers. That is why UV-Vis works as a detector of conjugation rather than as a general-purpose method.
Two isomeric dienes have the formula C₆H₁₀. One absorbs at 227 nm, and the other shows nothing above 200 nm. Which is conjugated?
The one at 227 nm. A conjugated diene has the smaller HOMO–LUMO gap, so it absorbs at the longer wavelength. In the other isomer the double bonds are isolated, so it behaves as two separate alkenes, each absorbing below 200 nm like any simple C=C.
What the spectrum does not tell you is where the double bonds sit. It says only whether they are conjugated. Other methods place them: IR, from IR & Mass Spectrometry, and NMR, from NMR Spectroscopy later in the course.
Why conjugation eventually produces color
β-Carotene absorbs most strongly near 450 nm, which is blue light. Yet carrots look orange, not blue. White light contains every visible color. When a compound removes blue from it, the light left over reaches your eye as orange.
So you see the light that was not absorbed. It is called the complementary color of the one absorbed. The same reasoning explains the colors of dyes, of acid–base indicators and of retinal, the light-absorbing part of the eye's visual pigment.
As conjugation extends, λmax keeps rising until it passes about 400 nm and enters the visible region. From that point on the compound absorbs visible light and looks colored. Most organic compounds are colorless because their π systems stop short of that point. When you see a strongly colored organic compound, look first for extended conjugation.
Beer–Lambert: how much is absorbed
Take a solution of a diene and dilute it to half its concentration. Its band stays at the same λmax, but it is only half as tall. The wavelength tells you what kind of π system is there; the amount absorbed tells you how much of it is there. The Beer–Lambert law links the amount absorbed to the concentration:
A = εcl
- A is the absorbance, a unitless number the instrument reports. The larger it is, the less light gets through the sample.
- ε (epsilon) is the molar absorptivity. It is a constant for one compound at one wavelength, and it measures how strongly that compound absorbs. Conjugated systems have large ε values, often 10,000 or more.
- c is the concentration in mol L−1, and l is the path length through the sample in cm, usually 1 cm.
Because absorbance is directly proportional to concentration, UV-Vis measures how much of a compound is present, not just what kind. IR and NMR, the structure-finding methods of IR & Mass Spectrometry and NMR Spectroscopy, are rarely used this way in a teaching lab. UV-Vis is the standard way to follow the concentration of a colored or conjugated compound while a reaction runs.
A solution of a conjugated diene in a 1.00 cm cell gives an absorbance of 0.63 at its λmax, where ε = 21,000 L·mol−1·cm−1. What is the concentration?
Rearrange before substituting. A = εcl, so c = A / (εl). With the algebra done first, the units are easy to check.
Substitute. c = 0.63 / (21,000 × 1.00) = 3.0 × 10−5 mol L−1.
Check the units. L·mol−1·cm−1 × cm leaves L·mol−1, and a unitless absorbance divided by that gives mol·L−1. If your answer comes out in anything else, you have the expression upside down.
Check the size. That is 30 micromolar, a few milligrams in a liter. A large ε is what makes this possible: UV-Vis can follow a reaction at concentrations far too low for a routine NMR spectrum.
The trap: above an absorbance of about 1.5, most instruments stop responding linearly, because almost no light reaches the detector. If a reading comes back at 2.8, dilute the sample and measure again rather than trusting the arithmetic.
Not every band is a π → π* band
Acetone, (CH₃)₂C=O, has only one double bond, a C=O. Like any lone double bond, it has its π → π* band below 200 nm. Yet acetone also shows a weak band near 280 nm, far to the right of where one double bond absorbs. That band comes from a different kind of jump.
Oxygen's lone pairs are not part of any bond. They sit in non-bonding orbitals, called n orbitals for short. An n orbital lies higher in energy than the π bonding orbital. Lifting an electron from n into the same empty π* therefore costs less energy, and less energy means a longer wavelength. This jump is written n → π*.
What gives the band away is how weak it is. The lone pairs lie in the plane of the carbonyl group, while the π* orbital stands above and below that plane. The two orbitals meet at right angles and barely overlap, so the jump between them is nearly forbidden: it rarely happens. An n → π* band has an ε of only 10–100, against 10,000 and up for a π → π* band.
So read a long-wavelength band by its strength as well as its position:
- Long λmax with a large ε — an extended conjugated system.
- Long λmax with a tiny ε — a lone pair, almost always a carbonyl.
Substituents shift λmax too, and by predictable amounts
Hexa-2,4-diene is buta-1,3-diene with a methyl group on each end. It absorbs at about 227 nm, 10 nm longer than buta-1,3-diene. The methyl groups add no conjugation, yet they still move the band.
Groups that donate electron density into the π system raise the HOMO a little, which narrows the gap and moves the band to longer wavelength. Alkyl groups do this weakly. Heteroatoms with lone pairs, as in OR, NR2 and SR, can do it more strongly, especially NR2 and SR. A group that shifts the band this way without absorbing on its own is called an auxochrome. Auxochromes are why a dye's color can be tuned by attaching an amino or a hydroxy group to its conjugated system rather than by lengthening the chain.
These shifts are regular enough to tabulate. The Woodward–Fieser rules start from a base value for the kind of diene and add a fixed increment for each feature around it. Four terms describe those features:
- A homoannular diene has both double bonds in the same ring (homo, same; annulus, ring).
- A heteroannular diene has its two double bonds in two different rings that share a bond.
- A ring residue is a ring carbon bonded to a carbon of the diene but not part of the diene itself. It counts like an alkyl group.
- An exocyclic double bond has one carbon inside a ring and the other outside it (exo, outside). Where two rings share a bond, a double bond can lie inside one ring and still be exocyclic to the other.
| Contribution | Increment |
|---|---|
| Base: acyclic diene | 217 nm |
| Base: diene spanning two rings (heteroannular) | 214 nm |
| Base: diene inside one ring (homoannular) | 253 nm |
| Each alkyl group or ring residue on the diene | +5 nm |
| Each double bond extending the conjugation | +30 nm |
| Each exocyclic double bond | +5 nm |
This course uses 217 nm as the base for an acyclic diene. Some tables use 214 nm instead, the same base as a heteroannular diene; the increments are the same.
Predict λmax for the heteroannular diene in the figure.
Pick the base. The two double bonds sit in different rings, so start from 214 nm.
Count the ring residues. Walk along the four diene carbons and note every ring carbon bonded to them that is not part of the diene. There are three, marked in the figure: 3 × 5 = +15 nm.
Look for exocyclic double bonds. The double bond in the right-hand ring starts at a carbon the two rings share and ends at one only the right-hand ring owns. So it lies inside the right-hand ring but is exocyclic to the left-hand one: +5 nm.
Add. 214 + 15 + 5 = 234 nm. A steroid with this same diene pattern, cholesta-3,5-diene, absorbs at about 235 nm.
Even if you never use the table, take two lessons from it. The first is about the +30 nm for each extra conjugated C=C. The measured step from buta-1,3-diene to hexa-1,3,5-triene in the first table is 217 to 258 nm, which is +41 nm, not +30.
The increment is not wrong. It is an empirical average, taken from measured compounds rather than from theory. The rules were fitted to substituted ring dienes, where they usually land within about 5 nm. On a simple parent polyene they do worse: they predict 247 nm for hexatriene, 11 nm short of the measured value. Cyclohexa-1,3-diene in the figure misses by 7 nm. Use the rules to predict the shift between two related compounds, not to reproduce the parent hydrocarbons.
The second lesson is that +5 nm for a substituent is small. That is why UV-Vis is a blunt instrument for anything except measuring how long the conjugation is.
What carries forward
UV-Vis sits in this chapter, and not with the other spectroscopic methods, because it measures conjugation directly. There are three more methods, and each answers a different question. IR & Mass Spectrometry, the chapter before this one, covers two of them, and NMR Spectroscopy, later, the third. IR identifies functional groups, mass spectrometry gives the mass of a molecule and its fragments, and NMR builds the carbon skeleton.