Organic Chemistry · Structure Determination: Mass Spectrometry and Infrared Spectroscopy
Spectroscopy and the Electromagnetic Spectrum
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In 30 seconds
Spectroscopy Study of how matter absorbs or emits electromagnetic radiation Full entry → is the study of how matter interacts with Electromagnetic radiation Waves of electric and magnetic fields traveling at c Full entry →. Every spectroscopic method in organic chemistry — UV-visible, infrared (IR), and nuclear magnetic resonance (NMR) — rests on one idea: a molecule (or nucleus) absorbs a Photon A discrete packet of light energy, E = hν only when its energy exactly matches the gap between two allowed energy states. The set of energies absorbed, plotted as a spectrum, is a fingerprint of molecular structure.
Electromagnetic radiation travels as waves; the complete family of such waves is the electromagnetic spectrum, ordered from highest to lowest energy: gamma rays, X-rays, ultraviolet (UV), visible, infrared, microwaves, and radio waves. Organic chemists exploit three windows, each matched to a different energy jump:
- UV-visible light moves electrons between orbitals (π → π* absorptions of conjugated molecules).
- Infrared light excites bond vibrations — stretches and bends.
- Radio waves flip the spin orientations of nuclei such as 1H and 13C (NMR).
Because these photon energies differ by orders of magnitude, each technique reports on a different structural level; together they assemble a full structure from a few milligrams.
Why this matters
Spectroscopy is how modern chemists identify molecules: no single test reveals a whole structure the way a set of spectra does. This topic is the quantitative backbone of Chapters 12 and 13 — the same equations that convert a wavenumber to photon energy are used again for every IR band and NMR frequency. The physics also runs everyday instruments: IR spectrometers identify drugs and water contaminants, and photon-matching explains why greenhouse gases absorb Earth's outgoing infrared. Master E = hc/λ once, and every later spectroscopy chapter gets easier.
The college version
Core Concepts
Photons: radiation as packets of energy
Light behaves as a wave but delivers energy in discrete packets called photons, whose energy is proportional to frequency:
E = hν
where E is in joules, ν the frequency in hertz, and h Planck's constant, 6.626 × 10-34 J·s. The proportionality is the quantization: a photon carries only the energy fixed by its frequency.
One equation links wavelength, frequency, and energy
All electromagnetic radiation travels at the speed of light, c = 2.998 × 108 m/s, so wavelength λ (m) and frequency ν (Hz) are connected by:
c = λν
Combining with E = hν gives the two most-used forms:
E = hν= hcλ
Because c is constant, wavelength and frequency are inversely related. Given any one of λ, ν, or E, you can find the other two — that is the entire arithmetic of this topic.
The regions of the spectrum and what they excite
| Region | Approximate range | Molecular event excited |
|---|---|---|
| UV | 10–400 nm | Valence electron transitions |
| Visible | 400–700 nm | Electron transitions (color) |
| Infrared | 700 nm–1 mm (4000–400 cm⁻¹) | Bond vibrations |
| Radio | > 30 cm | Nuclear spin flips (NMR) |
For structure determination, remember three boxes: UV-vis → electrons, IR → vibrations, radio/NMR → nuclear spins. A radio photon carries ~10⁹× less energy than a UV photon — which is why NMR instruments need enormous magnets.
Wavenumber: the chemist's unit for IR
Infrared spectroscopists rarely quote wavelength; they use wavenumber ν̃, the reciprocal of the wavelength in centimeters:
ν̃ = 1λ (cm)
in units of cm⁻¹. Wavenumber is directly proportional to frequency and energy:
E = hcν̃
so 3000 cm⁻¹ is higher in energy than 1200 cm⁻¹; mid-IR spectra run 4000→400 cm⁻¹.
Quantized energy means selective absorption
A molecule cannot absorb just any photon. Its Energy levels Discrete allowed energies of a molecule or nucleus Full entry → (electronic, vibrational, rotational, nuclear spin) are discrete, and Absorption Molecule gains a photon's energy and jumps to a higher state Full entry → occurs only when hν equals a level spacing. That is why spectra are sharp bands rather than a continuous smear: each band marks a specific transition.
Common Confusions
| Do not confuse | With | Difference |
|---|---|---|
| Wavenumber (ν̃, cm⁻¹) | Frequency (ν, Hz) | ν̃ = 1/λ in cm; ν= c/λ in Hz; convert via ν= cν̃ after changing cm⁻¹ to m⁻¹ (×100) |
| High wavenumber | Low energy | Opposite — bigger cm⁻¹ means higher frequency and energy |
| Wavelength order | Frequency order | They run opposite: as λ increases, ν and E decrease |
| Absorption | Emission | Absorption: molecule gains photon energy (dip in transmitted light); emission: molecule releases energy as light |
| IR radiation | Ionizing radiation | IR and radio cannot ionize atoms or break bonds; X-rays and gamma rays can |
Eli explains
The same idea, in plain words
Explain it like I’m 10
Imagine a molecule is a piano with only a few keys, and light is a box of hammers — each hammer hits one exact note. The molecule can only accept a hammer whose note matches one of its keys; the list of accepted notes is its "spectrum," and it reveals what the molecule is made of. Infrared makes bonds wiggle, UV makes electrons jump, and radio waves flip nuclear magnets.
Worked example
Example 1: From wavelength to frequency and energy (green light)
Green light has a wavelength of 500 nm. Find its frequency and photon energy.
Step 1 — convert to meters: 500 nm × 10-9 m1 nm = 5.00 × 10-7 m.
Step 2 — frequency from c = λν:
ν= cλ = 2.998 × 108 m/s5.00 × 10-7 m = 6.00 × 1014 Hz
Step 3 — photon energy:
E = hν= (6.626 × 10-34 J·s)(6.00 × 1014 s-1) = 3.98 × 10-19 J
Per mole: (3.98 × 10-19 J)(6.022 × 1023 mol-1) ≈ 240 kJ/mol — enough to promote a π → π* transition in a conjugated dye, which is why colored molecules absorb visible light.
Example 2: Converting an IR band between units
A carbonyl stretch appears at 1715 cm⁻¹. Express it as a wavelength in micrometers and as a frequency in hertz.
Wavelength:
λ= 1ν̃ = 11715 cm-1 = 5.83 × 10-4 cm
Convert: 5.83 × 10-4 cm × 104 μm1 cm = 5.83 μm.
Frequency: convert cm⁻¹ to m⁻¹ (1715 × 100 = 1.715 × 105 m-1), then:
ν= cν̃ = (2.998 × 108 m/s)(1.715 × 105 m-1) = 5.14 × 1013 Hz
Unit check: m/s × m-1 = s-1 = Hz. ✓
Example 3: Why IR light cannot break bonds
Using E = hcν̃ for the 1715 cm⁻¹ photon:
E = (6.626 × 10-34 J·s)(2.998 × 108 m/s)(1.715 × 105 m-1) = 3.41 × 10-20 J
Per mole: 3.41 × 10-20 × 6.022 × 1023 ≈ 20.5 kJ/mol — about 20× less than the ~413 kJ/mol needed to break a typical C–H bond, so IR cannot rupture bonds. A UV photon at 250 nm carries hc/λ ≈ 479 kJ/mol, comparable to bond energies — why UV drives photochemistry and why UV exposure damages DNA. Safety principle: IR is non-ionizing and non-destructive; UV sources require eye and skin protection.
Key takeaways
- Apply c = λν and E = hc/λ without prompting; write the formula first, then substitute with units.
- Energy order along the spectrum: gamma > X-ray > UV > visible > IR > microwave > radio.
- Technique mapping: UV-vis → electrons; IR → vibrations; NMR → nuclear spins.
- Wavenumber in cm⁻¹ is proportional to energy: 4000 cm⁻¹ > 400 cm⁻¹.
- Wavelength and frequency run opposite: longer λ = lower ν = lower E.
- Absorption is quantized: a photon is absorbed only if its energy matches an allowed transition, so spectra have discrete peaks.
- Conversion shortcuts: cm⁻¹ to m⁻¹ = ×100; Hz to cm⁻¹ = ÷ c = 2.998 × 1010 cm/s.
Check yourself
5 review questions from the chapter. Try each one, then open the answer.
Which region of the electromagnetic spectrum excites bond vibrations, and in what wavenumber range?
Show answer
Infrared, roughly 4000–400 cm⁻¹ (mid-IR).
Convert the O–H stretch at 3300 cm⁻¹ to a wavelength in micrometers.
Show answer
λ= 1/ν̃ = 1/(3300 cm-1) = 3.03 × 10-4 cm = 3.03 μm.
Which photon is more energetic: one at 600 nm (visible) or one at 1200 cm⁻¹ (IR)? Show the energy of each.
Show answer
600 nm: E = hc/λ= (6.626 × 10-34)(2.998 × 108)/(6.00 × 10-7) = 3.31 × 10-19 J. 1200 cm⁻¹: E = hcν̃ = 2.38 × 10-20 J. The visible photon is about 14× more energetic.
Why do spectra show discrete absorption bands rather than continuous absorption?
Show answer
Because energy levels are quantized — a photon is absorbed only when its energy matches an allowed transition.
Rank gamma rays, IR, and radio waves in order of increasing frequency.
Show answer
Radio < IR < gamma (frequency increases in that order).
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- Spectroscopy
- Study of how matter absorbs or emits electromagnetic radiation
- Electromagnetic radiation
- Waves of electric and magnetic fields traveling at c
- Photon
- A discrete packet of light energy, E = hν
- Frequency (ν)
- Wave cycles per second, in Hz
- Wavelength (λ)
- Distance between wave crests, in m
- Wavenumber (tildeν)
- 1/λ in cm⁻¹
- Planck's constant (h)
- 6.626 × 10-34 J·s
- Absorption
- Molecule gains a photon's energy and jumps to a higher state
- Energy levels
- Discrete allowed energies of a molecule or nucleus
Sources & references
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