Organic Chemistry · Conjugated Compounds and Ultraviolet Spectroscopy
Ultraviolet Spectroscopy
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In 30 seconds
Ultraviolet (UV) spectroscopy measures how much ultraviolet light — roughly 200–400 nm, just beyond the violet end of the visible spectrum — a sample absorbs. A molecule absorbs a photon only when the photon's energy matches the gap between an occupied and an empty molecular orbital; the absorbed energy promotes an electron into a higher orbital, an electronic transition Promotion of an electron from an occupied to an empty molecular orbital Full entry →. The key transitions for organic chemists are π → π\* (alkenes, aromatics, carbonyls) and n → π\* (lone-pair electrons, especially carbonyls).
The quantitative heart of the method is the Beer–Lambert law A = εb c: absorbance is proportional to concentration and path length Full entry →, A = εb c: absorbance is directly proportional to concentration. That single relationship makes UV a quantitative tool — measure the absorbance, know the extinction coefficient, and you can calculate how much compound is in solution. The instrument (UV lamp, monochromator, sample cell, detector) is simple in principle, but practical details — quartz cells, solvent choice — matter because glass and many solvents absorb UV light themselves.
Why this matters
UV spectroscopy is one of the most widely used analytical techniques in science:
- Quantitative analysis. The Beer–Lambert law gives concentrations of drugs, dyes, and contaminants from a single absorbance reading; HPLC detectors and pharmaceutical quality control rely on it.
- Detecting conjugation. Where a molecule absorbs (λ_max Wavelength of maximum absorbance Full entry →) immediately reveals how many double bonds are conjugated — the topic that follows.
- Biochemistry. DNA and proteins absorb at 260 and 280 nm; the 260/280 ratio estimates nucleic-acid purity in seconds.
- Reaction monitoring. Watching a UV band grow or shrink tracks a reaction without stopping it.
General lab-safety principle: UV radiation can damage eyes and skin, so never look directly into a UV lamp or beam, keep sample compartments closed while the lamp is on, and shield the skin. (This is a general principle; follow your institution's specific protocols.)
The college version
Core Concepts
Electronic transitions and what absorbs UV
Molecules absorb UV light when an electron jumps between orbitals. The possible transitions, in order of increasing wavelength of practical interest:
- σ → σ\* — needs very high energy (vacuum UV, below ~150 nm); only sigma-bonded molecules like alkanes show it, outside the range of ordinary instruments.
- n → σ\* — lone-pair electrons promoted into σ*; alkyl halides, ethers, and amines near 180–220 nm.
- π → π\* — the double-bond transition; strong (ε typically 103–105 L mol-1 cm-1) and the main event for alkenes, dienes, aromatics, and carbonyls.
- n → π\* — a lone pair promoted into π*; the classic weak carbonyl band near 280 nm with ε ~10–100, because the transition is symmetry-forbidden.
Because only π and lone-pair (n) electrons absorb in the practical range, UV selectively probes unsaturation, conjugation, and heteroatom lone pairs — saturated hydrocarbons are effectively transparent.
The Beer–Lambert law
The fraction of light absorbed depends on how many absorbing molecules the beam encounters. The Beer–Lambert law states:
A = εb c
where A is absorbance (unitless), ε is the molar absorptivity (extinction coefficient, L mol-1 cm-1), b is the path length through the sample (cm), and c is concentration (mol/L). Absorbance is also defined directly from the measured light intensities:
A = log10(I0I) = -log10(T)
where I0 is the incident intensity, I the transmitted intensity, and T = I/I0 the transmittance. Absorbance is linear in concentration (within the limits noted below), so a calibration curve of A versus c is a straight line through the origin.
Practical instrument details
A UV spectrophotometer has four parts: a light source (deuterium lamp for UV, tungsten-halogen for visible), a monochromator to select wavelength, a sample cell, and a detector. Most instruments are double-beam: light splits so one beam passes through the sample and the other through a reference cell of pure solvent, and the instrument reports the difference — automatically subtracting solvent absorption.
Two practical rules trip up exam takers:
- Cells must be quartz. Ordinary glass (and plastic) absorbs below about 300 nm; quartz transmits down to ~190 nm, so UV measurements use quartz cells.
- The solvent must be transparent at the wavelengths of interest. Common choices and their approximate lower cutoffs: water ~190 nm, hexane ~195 nm, 95% ethanol ~205 nm. Acetone, which absorbs strongly in the UV, is a poor choice.
What a spectrum shows
A UV spectrum plots absorbance versus wavelength, usually 200–400 nm. Two numbers summarize each band: λ_max (position) and ε_max (intensity). Strong bands (ε 104–105) belong to allowed transitions like π → π*; weak bands (ε ~10–100) betray forbidden ones like n → π*. Comparing λ_max against known values identifies a chromophore The part of a molecule responsible for its absorption (C=C, C=O, benzene ring) Full entry →; comparing ε confirms the assignment.
How It Works / Step-by-Step Process
To measure and interpret a UV spectrum:
- Choose the solvent — transparent at the wavelengths of interest (water, hexane, or 95% ethanol; never acetone).
- Prepare the blank (pure solvent) in the reference cell and the sample in a quartz cell.
- Zero the instrument — record the baseline with the blank.
- Scan 200–400 nm; record λ_max and the absorbance there.
- Convert to concentration using the Beer–Lambert law: c = A/(εb), using the published ε for the chromophore.
- Interpret: the position (λ_max) identifies the chromophore and its conjugation; the magnitude of ε tells you the transition type (strong π → π* vs weak n → π*).
Common Confusions
| Do not confuse | With | Difference |
|---|---|---|
| Absorbance | Transmittance | A = -log10(T); A is linear in concentration, T is not |
| Molar absorptivity ε | Absorbance A | ε is a fixed property of the molecule; A depends on ε, b, and c |
| Weak n → π* band | Strong π → π* band | Carbonyl ~280 nm (ε ~10–30) is n → π*; the alkene π → π* band is ~1000× stronger |
| Glass cuvette | Quartz cuvette | Glass absorbs below ~300 nm; quartz is required for UV |
| σ → σ* in ordinary UV range | Vacuum UV below ~150 nm | Alkanes appear transparent in 200–400 nm; σ → σ* needs special instruments |
| Absorbance of the sample alone | Absorbance of sample + solvent | Double-beam instruments subtract the solvent (blank) automatically |
Eli explains
The same idea, in plain words
Explain it like I’m 10
A UV spectrophotometer is like a light checkpoint at a tunnel: it shines an invisible beam through a liquid sample and counts how much light comes out the other side. Molecules that "catch" the light swallow it and jump to a higher energy state — like a kid catching a ball and climbing to the top of the playground. The more molecules, the more light swallowed, so the machine counts molecules by how much light disappears — one reading tells you how much medicine is in a pill.
Worked example
Example 1: Concentration from a single absorbance reading
A compound has ε= 1.5 × 104 L mol-1 cm-1 at its λ_max. A solution in a 1.00 cm quartz cell gives A = 0.450 at that wavelength. What is the concentration?
Rearrange the Beer–Lambert law to solve for concentration:
c = Aεb
Substituting:
c = 0.450(1.5 × 104 L mol-1 cm-1)(1.00 cm) = 3.0 × 10-5 mol/L
Unit check: (L mol-1 cm-1)(cm) = L mol-1, and the unitless absorbance divided by L mol-1 leaves mol/L — exactly the units of concentration. The reading was possible only because the cell was quartz; glass would have absorbed the 200–300 nm light itself.
Example 2: Transmittance, and the energy of a UV photon
(a) A sample transmits 25.0% of the incident light at its λ_max. What is its absorbance?
Transmittance is a fraction, T = 0.250. Using A = -log10(T):
A = -log10(0.250) = 0.602
(b) What energy does a single photon of 280 nm light carry, and what is that energy per mole of photons?
The photon energy is E = hc/λ, with Planck's constant h = 6.626 × 10-34 J · s and the speed of light c = 2.998 × 108 m/s:
E = (6.626 × 10-34 J · s)(2.998 × 108 m/s)280 × 10-9 m = 7.09 × 10-19 J
Per mole, multiply by Avogadro's number, NA = 6.022 × 1023 mol-1:
Eper mole = (7.09 × 10-19 J)(6.022 × 1023 mol-1) = 4.27 × 105 J/mol ≈ 427 kJ/mol
Comparable to typical C–C bond energies (~350 kJ/mol), a UV photon can break bonds and drive photochemistry — why prolonged UV exposure damages DNA (and why UV safety matters), and why sunscreens absorb UV before it reaches skin.
Key takeaways
- UV spectroscopy probes electronic transitions: π → π* (strong, ε 103–105) and n → π* (weak, ε ~10–100) dominate the practical 200–400 nm range.
- Beer–Lambert law: A = εb c. Absorbance is linear in concentration; ε is intrinsic to the molecule, b is the cell path length.
- Absorbance and transmittance: A = -log10(T). A is unitless.
- Quartz cells (not glass) and UV-transparent solvents (water, hexane, ethanol) are required below ~300 nm.
- Saturated hydrocarbons absorb only via σ → σ* in the vacuum UV — transparent in the ordinary range, so UV selectively "sees" unsaturation and conjugation.
- Carbonyls show a weak n → π* band near 280 nm and a strong π → π* band at shorter wavelength.
- General safety: never look into a UV source; UV can damage eyes and skin.
Check yourself
6 review questions from the chapter. Try each one, then open the answer.
Which electronic transition gives the weak absorption band of a carbonyl near 280 nm, and why is it weak?
Show answer
The n → π* transition — a carbonyl lone-pair electron promoted into the π* orbital. It is weak (ε ~10–30) because the transition is symmetry-forbidden.
A solution has absorbance 0.602 at its λ_max. What percentage of the light is transmitted?
Show answer
T = 10-A = 10-0.602 = 0.250, so 25.0% of the light is transmitted.
Why must UV measurements use quartz cells rather than glass?
Show answer
Glass absorbs UV light below about 300 nm; quartz transmits down to ~190 nm, so glass would block the very light the instrument needs to measure.
A compound with ε= 2.0 × 104 L mol-1 cm-1 in a 1.00 cm cell gives A = 0.800. What is the concentration?
Show answer
c = A/(εb) = 0.800/((2.0 × 104)(1.00)) = 4.0 × 10-5 mol/L.
Why do saturated hydrocarbons like hexane appear transparent in ordinary UV spectroscopy?
Show answer
Their only possible transition is σ → σ*, which requires vacuum UV below ~150 nm — outside the 200–400 nm range of an ordinary instrument — so they absorb nothing measurable there.
Roughly how much energy (kJ/mol) does a 300 nm photon carry? (Use E = NA hc/λ.)
Show answer
E = NA hc/λ= (6.022 × 1023)(6.626 × 10-34)(2.998 × 108)/(300 × 10-9) = 3.99 × 105 J/mol ≈ 399 kJ/mol.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- electronic transition
- Promotion of an electron from an occupied to an empty molecular orbital
- chromophore
- The part of a molecule responsible for its absorption (C=C, C=O, benzene ring)
- λ_max
- Wavelength of maximum absorbance
- molar absorptivity, ε
- Intrinsic absorption strength of a molecule at a given wavelength (L mol-1 cm-1)
- Beer–Lambert law
- A = εb c: absorbance is proportional to concentration and path length
- transmittance, T
- Fraction of incident light that passes through the sample (I/I0)
- n → π* transition
- Lone-pair electron promoted into a π* orbital
- π → π* transition
- π electron promoted into a π* orbital
- n→π transition
- Low-energy electronic transition of the carbonyl oxygen lone pair into the π orbital.
- π→π transition
- Promotion of a π electron into an antibonding π orbital
Sources & references
This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.
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