IB Chemistry HLTopic 3 — Classification of MatterPaper 1 & 2HL only~12 min read
Infrared Spectra Interpretation
Every covalent bond behaves like a tiny spring, and every spring has a natural frequency it likes to wobble at. Shine infrared light through a molecule and the bonds absorb exactly the frequencies that match their own. So an infrared spectrum is a list of which bonds are present — and you only need to recognise about three peaks to answer most questions.
📘 What you need to know
Bonds stretch and bend. Each mode of vibration has its own natural frequency, in the infrared part of the spectrum.
When the infrared frequency matches a bond’s natural frequency, energy is absorbed and the vibration gets bigger. This is resonance.
A bond only absorbs infrared if its vibration causes a change in dipole moment. Symmetrical molecules such as O2 and N2 are infrared inactive.
The x-axis is wavenumber in cm−1, which is 1 divided by the wavelength. It runs from high to low, left to right.
Three peaks do most of the work: O–H in alcohols (3200–3600, broad), C=O in carbonyls (1700–1750, strong and sharp), O–H in acids (2500–3000, very broad).
Below about 1500 cm−1 is the fingerprint region — too complicated to assign, but unique to each compound, so it is used for database matching.
All the wavenumber ranges are in section 20 of the data booklet. You do not have to memorise the numbers.
How a bond absorbs infrared
A bond is not a rigid bar. The two atoms sit at an average distance and vibrate about it, and there are several different ways they can move. Each way has its own frequency, which is why one molecule gives you many peaks rather than one.
The amber circle is a central atom such as carbon or oxygen; the pink ones are hydrogens. Bending modes are lower in energy than stretching modes, so they show up further to the right of a spectrum.
Why greenhouse gases work. Carbon dioxide, water and methane all have polar bonds whose vibrations change the molecular dipole, so they absorb infrared radiated from the Earth’s surface and re-emit it. Nitrogen and oxygen make up most of the atmosphere but are symmetrical diatomics, so they let infrared straight through. This is the same rule, applied at planetary scale.
Where each bond absorbs
The x-axis of an infrared spectrum is wavenumber, and it is plotted backwards — 4000 cm−1 on the left, 500 on the right. That takes some getting used to. Here is roughly where each bond you need sits.
Notice that the O–H band of an acid sits much lower than that of an alcohol, and overlaps the C–H region. That is because hydrogen bonding in an acid is far stronger, which weakens the O–H bond and drops its frequency.
Bond
Found in
Wavenumber / cm−1
What it looks like
C–Cl
chloroalkanes
600–800
strong
C–O
alcohols, esters, ethers
1050–1410
strong
C=C
alkenes
1620–1680
medium to weak
C=O
aldehydes, ketones, acids, esters
1700–1750
strong and sharp
C≡C
alkynes
2100–2260
variable
O–H
carboxylic acids
2500–3000
strong, very broad
C–H
alkanes, alkenes, arenes
2850–3090
strong
N–H
primary amines
3300–3500
medium, two bands
O–H
alcohols and phenols
3200–3600
strong, broad
Shape matters as much as position
Two peaks in nearly the same place can still be told apart by how they look. Broad, rounded troughs mean hydrogen bonding is smearing the frequency out over a range. Narrow, sharp troughs mean a single well-defined vibration. Compare a ketone with an alcohol:
Schematic traces, drawn to show the difference in shape rather than to reproduce a real instrument reading. Absorptions point downwards here because the y-axis is transmittance — less light gets through where the molecule absorbs.
The pair of questions to ask, in this order: is there a strong sharp trough just below 1750? That is a C=O, so you have a carbonyl compound. Is there a broad trough above 3200? That is an O–H, so you have an alcohol. Both together points at a carboxylic acid, and neither points at a hydrocarbon or a halogenoalkane.
The fingerprint region
Below about 1500 cm−1 the spectrum becomes a forest of peaks. These come from complicated combined vibrations involving many bonds at once, and there is no realistic way to assign them one by one. That is fine, because their value lies elsewhere: the pattern is unique to each compound.
Two members of the same homologous series will show the same broad features — the same O–H, the same C–H — but no two compounds share a fingerprint region. So a computer can match an unknown spectrum against a database and identify it exactly, which is how forensic and pharmaceutical labs confirm what a sample is.
Where infrared is used
Vehicle emissions. Roadside sensors measure carbon monoxide, carbon dioxide and unburnt hydrocarbons by how much infrared each absorbs.
Breathalysers. Infrared is passed through exhaled breath, and the absorbance at the wavenumbers characteristic of ethanol gives the alcohol concentration.
Climate science. Measuring how strongly each gas absorbs infrared is how the warming effect of carbon dioxide and methane is quantified.
Worked examples
WORKED EXAMPLE
Two spectra are recorded. Spectrum A has a strong sharp absorption at 1715 cm−1 and nothing above 3100. Spectrum B has a broad absorption from 3200 to 3550 and nothing near 1700. One is propanone and one is propan-1-ol. Assign them.
Step 1: work out which bonds each compound containsPropanone, CH3COCH3, contains a C=O but no O–H. Propan-1-ol, CH3CH2CH2OH, contains an O–H but no C=O.Step 2: look up the ranges in section 20C=O: 1700–1750 cm⁻¹, strong and sharpO–H in an alcohol: 3200–3600 cm⁻¹, strong and broadStep 3: match each spectrumA’s peak at 1715 sits inside the C=O range, and A has nothing in the O–H region. B is the other way round.A is propanone; B is propan-1-olQuote the actual wavenumber and the bond it belongs to. “A has a peak so it is the ketone” earns nothing.
WORKED EXAMPLE
A spectrum shows a very broad absorption from about 2600 to 3100 cm−1 and a strong sharp one at 1710 cm−1. Deduce the class of compound.
Step 1: identify the sharp peak1710 cm⁻¹ is in the 1700–1750 range, so there is a C=OSo the compound is an aldehyde, ketone, acid or ester.Step 2: identify the broad absorptionAn alcohol O–H is 3200–3600. This one is far lower and much broader, which matches the O–H of a carboxylic acid at 2500–3000.a carboxylic acid O–H, overlapping the C–H regionStep 3: combine the two pieces of evidenceA C=O and a very broad low-frequency O–H in the same molecule is the carboxyl group, –COOH.a carboxylic acidThe clue is the position and width of the O–H, not just its presence. An alcohol with a separate ketone group would show a narrower band, and much higher up.
WORKED EXAMPLE
Explain why nitrogen makes up 78 per cent of the atmosphere but contributes nothing to the greenhouse effect.
Step 1: state the condition for absorbing infraredA molecule only absorbs infrared if the vibration produces a change in its dipole moment.Step 2: look at the bonding in nitrogenN≡N joins two identical atomsBoth atoms have the same electronegativity, so the bond is non-polar and the molecule has no dipole at all.Step 3: consider what happens when it stretchesStretching a symmetrical diatomic keeps it symmetrical, so there is still no dipole. Nothing changes, so no energy is absorbed.N₂ is infrared inactive, because it is symmetrical and has no dipole to changeCarbon dioxide is also symmetrical overall, but its asymmetric stretch and its bends do create a temporary dipole, which is why it is a greenhouse gas and nitrogen is not.
💡 Exam tip
Quote a wavenumber and name the bond. Every infrared mark scheme wants both, and neither alone is enough.
Use section 20 of the data booklet. The ranges are provided, so never guess a number from memory.
Describe the shape as well as the position. Broad, sharp, strong and weak are all doing real work in these questions.
Absence of a peak is evidence too. “No absorption near 1700, so there is no C=O” is a valid deduction and often the one being tested.
Do not try to assign the fingerprint region. Say it is used for matching against a database and move on.
Combine infrared with mass spectrometry. Infrared tells you which bonds are present; it cannot tell you the relative molecular mass.
⚠ Common mix-up
Reading the axis forwards. Wavenumber decreases from left to right, which is the opposite of every other graph you draw.
Confusing the alcohol O–H with the acid O–H. The alcohol band is 3200–3600 and reasonably broad; the acid band is 2500–3000 and very broad indeed.
Thinking every bond absorbs infrared. Only vibrations that change the dipole moment do, so symmetrical molecules are inactive.
Assuming C=O identifies the compound. Aldehydes, ketones, acids and esters all have one. You need more evidence to choose between them.
Confusing wavenumber with wavelength. Wavenumber is 1 divided by wavelength, in cm−1.
Reading troughs as peaks. Most spectra plot transmittance, so absorptions point downwards. The language is still “peak”.
Saying carbon dioxide is infrared inactive because it is symmetrical. Its asymmetric vibrations do create a dipole, which is exactly why it is a greenhouse gas.
Up next: Proton NMR Spectroscopy — mass spectrometry weighed the molecule, infrared found the bonds. NMR is the one that maps out where every hydrogen sits.
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