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Infrared (IR) spectroscopy is one of the most widely used techniques for identifying functional groups in organic molecules. While mass spectrometry tells you the molecular mass and gives clues about fragmentation, IR spectroscopy tells you which types of bonds are present. Together, they begin to build a detailed picture of molecular structure.
All covalent bonds in molecules are not static — they vibrate continuously. These vibrations are quantised, meaning bonds can only vibrate at specific frequencies that correspond to specific energy levels. When infrared radiation passes through a sample, bonds absorb radiation at frequencies that match their natural vibrational frequencies. The radiation that is not absorbed passes through to the detector.
By measuring which frequencies of IR radiation are absorbed by the sample, we can identify which types of bonds are present.
There are two main categories of vibration:
Stretching vibrations — the bond length changes (gets longer and shorter) while the atoms remain on the same bond axis. These can be symmetric (both atoms move in and out together) or asymmetric (one atom moves in while the other moves out).
Bending vibrations — the bond angle changes. These include scissoring (in-plane angle change), rocking (in-plane movement), wagging (out-of-plane movement), and twisting. Bending vibrations generally absorb at lower frequencies than stretching vibrations of the same bond because less energy is needed to change a bond angle than to change a bond length.
Not all bonds absorb IR radiation. For a vibration to be IR-active, it must cause a change in the dipole moment of the molecule. Symmetric molecules like O₂, N₂, and H₂ have no dipole moment, and their stretching vibrations do not change the dipole moment, so they do not absorb IR radiation. This is why O₂ and N₂ in the atmosphere are transparent to infrared.
Molecules like CO₂ do absorb IR radiation — even though the molecule is linear and has no permanent dipole, the asymmetric stretch and bending modes do create a changing dipole moment and are therefore IR-active.
An IR spectrum plots transmittance (%) on the vertical axis against wavenumber (cm⁻¹) on the horizontal axis. Wavenumber is the reciprocal of wavelength and is proportional to the frequency (and therefore the energy) of the radiation:
wavenumber (cm⁻¹) = 1 / wavelength (cm)
Key conventions:
The frequency at which a bond absorbs depends on two main factors:
Bond strength — stronger bonds vibrate at higher frequencies. A C≡C triple bond absorbs at a higher wavenumber than a C=C double bond, which absorbs at a higher wavenumber than a C–C single bond.
Atomic masses — bonds between lighter atoms vibrate at higher frequencies. The O–H and N–H bonds absorb at high wavenumber (3000–3600 cm⁻¹) because hydrogen is very light. Heavier atoms (like C–Cl) absorb at lower wavenumbers.
This can be modelled as a simple harmonic oscillator (like a spring connecting two masses). The vibrational frequency increases with a stiffer spring (stronger bond) and decreases with heavier masses.
The following table includes all the key absorptions you need for A-Level, plus additional detail for confident interpretation:
| Bond | Wavenumber (cm⁻¹) | Appearance | Found in |
|---|---|---|---|
| O–H (free, no H-bond) | 3580–3650 | Sharp, narrow | Dilute alcohol solutions, gas phase |
| O–H (H-bonded, alcohol) | 3200–3550 | Broad | Alcohols, phenols |
| O–H (H-bonded, acid) | 2500–3300 | Very broad, overlaps C–H | Carboxylic acids |
| N–H (primary amine) | 3350–3500 | Two peaks (sym + asym) | Primary amines, primary amides |
| N–H (secondary amine) | 3310–3350 | One peak | Secondary amines, secondary amides |
| C–H (sp³, alkane) | 2850–2960 | Medium to strong | Alkanes, most organics |
| C–H (sp², alkene) | 3010–3100 | Medium | Alkenes |
| C–H (sp², aromatic) | 3030–3080 | Medium | Aromatic compounds |
| C–H (sp, alkyne) | 3300 | Strong, sharp | Terminal alkynes |
| C–H (aldehyde) | 2720 and 2820 | Two weak, distinctive peaks | Aldehydes (Fermi resonance doublet) |
| C≡N | 2200–2260 | Medium, sharp | Nitriles |
| C≡C | 2100–2260 | Weak to medium | Alkynes |
| C=O (ester) | 1730–1750 | Strong, sharp | Esters |
| C=O (aldehyde) | 1720–1740 | Strong, sharp | Aldehydes |
| C=O (ketone) | 1705–1725 | Strong, sharp | Ketones |
| C=O (carboxylic acid) | 1700–1725 | Strong, sharp | Carboxylic acids |
| C=O (amide) | 1630–1680 | Strong | Amides |
| C=C (alkene) | 1620–1680 | Medium | Alkenes |
| C=C (aromatic) | ~1500 and ~1600 | Medium, two bands | Aromatic compounds |
| C–O (alcohols, ethers) | 1000–1300 | Strong | Alcohols, esters, ethers |
| C–F | 1000–1400 | Strong | Fluoroalkanes |
| C–Cl | 600–800 | Strong | Chloroalkanes |
| C–Br | 500–680 | Strong | Bromoalkanes |
The region below approximately 1500 cm⁻¹ is called the fingerprint region. It contains many overlapping absorptions from complex combinations of bending and stretching vibrations. The pattern is unique to each compound — like a fingerprint — but is very difficult to interpret by eye.
The fingerprint region is most useful for confirming identity by comparing an unknown spectrum to a reference database. If the fingerprint region of your unknown matches a known compound exactly, you can be confident they are the same substance.
For A-Level purposes, you are primarily expected to identify functional groups from absorptions above 1500 cm⁻¹, though the carbonyl absorption (C=O at around 1700 cm⁻¹) falls within the boundary region and is critically important.
The traditional instrument is a dispersive IR spectrometer, which works by splitting an IR beam into two paths — one through the sample and one through a reference — and comparing the two to determine which frequencies were absorbed.
Modern instruments almost exclusively use Fourier Transform IR (FTIR) spectrometry, which is faster and more sensitive. An FTIR spectrometer uses an interferometer to produce an interferogram (a complex waveform) that is converted into a spectrum using a mathematical process called the Fourier transform.
Samples can be analysed as gases, liquids, or solids. Liquids are often placed between salt plates (NaCl or KBr, which are transparent to IR). Solids can be ground with KBr and pressed into a disc, or analysed using attenuated total reflectance (ATR).
The Beer-Lambert law relates the absorbance of a sample to its concentration and path length:
A = εcl
Where:
This law is more commonly applied in UV-visible spectroscopy, but the principle applies to IR as well. It tells us that absorbance is directly proportional to concentration — a more concentrated sample absorbs more strongly. This allows IR spectroscopy to be used quantitatively, not just qualitatively.
The connection between IR spectroscopy and climate science is direct. Greenhouse gases — including CO₂, H₂O, CH₄, and N₂O — absorb infrared radiation emitted by the Earth's surface. The bonds in these molecules have vibrational frequencies that match the wavelengths of IR radiation emitted by the warm Earth.
O₂ and N₂, which make up 99% of the atmosphere, are not greenhouse gases because their stretching vibrations do not change the dipole moment and therefore do not absorb IR radiation.
| Gas | Key absorption(s) | Vibration type |
|---|---|---|
| CO₂ | 2349 cm⁻¹, 667 cm⁻¹ | Asymmetric stretch, bending |
| H₂O | 3657, 3756, 1595 cm⁻¹ | Symmetric stretch, asymmetric stretch, bending |
| CH₄ | 3019 cm⁻¹, 1306 cm⁻¹ | C–H stretch, C–H bend |
| N₂O | 2224 cm⁻¹, 1285 cm⁻¹ | Asymmetric stretch, symmetric stretch |
| CFCs | 1000–1200 cm⁻¹ | C–F stretches (atmospheric window) |
This is why even small increases in these gases can significantly affect the Earth's energy balance.
Mistake 1: Confusing the O–H of an alcohol with the O–H of a carboxylic acid. The acid O–H is much broader (2500–3300 cm⁻¹) and extends to lower wavenumber than the alcohol O–H (3200–3550 cm⁻¹).
Mistake 2: Forgetting to check for C=O when you see O–H. The combination of very broad O–H AND strong C=O is diagnostic for a carboxylic acid. O–H without C=O indicates an alcohol.
Mistake 3: Assuming all peaks in the fingerprint region are identifiable. The fingerprint region is useful for matching against reference spectra, not for identifying specific bonds.
IR spectroscopy identifies functional groups by detecting which IR frequencies are absorbed by bond vibrations. The position of an absorption depends on bond strength and atomic mass. The fingerprint region below 1500 cm⁻¹ is unique to each compound. The technique is non-destructive, fast, and can be used for both qualitative identification and quantitative analysis.
Edexcel 9CH0 specification, Topic 18 — Modern Analytical Techniques I, sub-strand 18.2 covers the principles of infrared (IR) spectroscopy, including bond vibrations (stretching and bending), the dependence of vibrational frequency on bond strength and reduced mass, the diagnostic functional-group region (~1500–4000 cm⁻¹) and the fingerprint region (~500–1500 cm⁻¹), and the use of characteristic absorption ranges to identify functional groups (refer to the official specification document for exact wording). Examined on Paper 2 (Core Organic and Physical Chemistry) and synoptically on Paper 3 (General and Practical Principles), with characteristic absorption ranges supplied in the Edexcel data booklet (e.g. broad O–H 2500–3300 cm⁻¹ for carboxylic acids; sharp C=O 1680–1750 cm⁻¹). The data booklet is the candidate's authority — values must be cross-checked against it, not memorised exactly.
Question (7 marks):
The IR spectrum of an unknown organic compound shows the following absorptions:
(a) Identify the functional groups present from the data-booklet ranges, justifying each. (4)
(b) Suggest a class of compound consistent with the data, and explain why an alcohol alone does not fit. (2)
(c) State why the absorption near 2900 cm⁻¹ is observed in almost all organic compounds. (1)
Solution with mark scheme:
(a) Step 1. Broad 3200–3550 cm⁻¹ is consistent with O–H of an alcohol (data-booklet range typically 3200–3550 cm⁻¹ for alcohols). M1.
Step 2. Strong, sharp peak at 1715 cm⁻¹ is consistent with C=O stretching (range 1680–1750 cm⁻¹). M1.
Step 3. Medium peak at 2900 cm⁻¹ is C–H stretching of alkyl groups (range 2850–3100 cm⁻¹). M1.
Step 4. Multiple absorptions 1000–1300 cm⁻¹ include C–O stretching (range 1000–1300 cm⁻¹) consistent with the ester or alcohol. M1.
(b) The combination of broad O–H and sharp C=O is most consistent with a carboxylic acid (where O–H is bonded to the carboxyl group and broadens further to 2500–3300 cm⁻¹) — but the broad band starts at 3200 cm⁻¹, not 2500 cm⁻¹, suggesting an alcohol with a separate C=O group, i.e. an ester or hydroxy-ketone. M1. An alcohol alone would not show the C=O at 1715 cm⁻¹. A1.
(c) Almost all organic compounds contain C–H bonds (alkyl, alkenyl, aromatic), and these vibrate in the 2850–3100 cm⁻¹ region. A1.
Total: 7 marks (M5 A2).
Question (6 marks): Explain, in terms of bond vibrations and the dipole-moment requirement, why a covalent bond absorbs IR radiation at a particular frequency. Use the example of a C=O bond in a carbonyl compound to support your answer.
Mark scheme decomposition by AO:
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