An introduction to infrared (IR) spectroscopy
How chemists use the vibrations of chemical bonds to identify the functional groups in an unknown organic compound.
Infrared spectroscopy is one of the fastest, most accessible tools in the organic chemistry laboratory. In a matter of seconds it answers the important question: which functional groups are present in this molecule? This page introduces what IR light does to a molecule, how to read the spectrum it produces, and how to translate a set of signals into a chemical structure.
Why do we use IR spectroscopy?
When you are handed an unknown compound, the single most useful thing you can learn early is what functional groups it contains: is there a carbonyl? an O–H? a nitrile? IR spectroscopy answers exactly this. Different bonds absorb infrared light at different, characteristic energies, so an IR spectrum acts like a chemical fingerprint of the bonds in a molecule. It does not usually tell you the entire structure on its own, but combined with other data (such as NMR and mass spectrometry) it is a powerful diagnostic tool.
The physics: bonds behave like springs
A covalent bond is not rigid. The two atoms it joins are in constant motion, stretching toward and away from each other and bending, much like two objects connected by a spring. Each of these motions is called a vibrational mode, and each mode has a specific natural frequency.
Infrared radiation contains just the right amount of energy to excite these vibrations. When the frequency of the IR light matches the natural frequency of a particular bond vibration, the molecule absorbs that light and the vibration jumps to a higher-energy state. The frequencies that are absorbed depend on two things, exactly as a spring would predict:
Bond strength
Stronger (stiffer) bonds vibrate faster and absorb at higher wavenumbers. A triple bond absorbs at a higher frequency than a double bond, which is higher than a single bond.
Atomic mass
Lighter atoms vibrate faster and absorb at higher wavenumbers. This is why bonds to hydrogen (C–H, O–H, N–H) all appear at the high-frequency left side of the spectrum.
Not every vibration shows up: the dipole rule
A vibration is IR-active (produces a peak) only if it changes the molecule’s dipole moment. Polar bonds such as C=O and O–H absorb strongly. Perfectly symmetrical vibrations, for example the C≡C stretch of a symmetrically substituted alkyne or the bond in N₂, cause little or no change in dipole and give weak or absent signals. This is why some expected bands are surprisingly faint.
Anatomy of an IR spectrum
An IR spectrum is a plot of how much light passes through the sample against the energy of that light. Three conventions are used that could be confusing at first:
- The x-axis is wavenumber (cm⁻¹), a unit of energy proportional to frequency. It runs from 4000 on the left to 400 on the right, so numbers decrease from left to right.
- The y-axis is % transmittance. 100% means all the light passed through (no absorption). Because absorption removes light, peaks point downward, toward the bottom of the plot.
- Strong / medium / weak describes how deep a signal is; a strong absorption reaches far down toward 0% transmittance.
The four regions of the spectrum
You can interpret most spectra by remembering that the plot divides into four broad zones. Learning these first makes the detailed table far easier to use.
The fingerprint region
The dense pattern of peaks below ~1500 cm⁻¹ comes from single-bond stretches and whole-molecule bending motions. It is usually too complicated to assign peak by peak, but it is unique to each compound; two spectra with an exact match in the fingerprint region are the same substance. We use the region above 1500 cm⁻¹ to identify functional groups, and the fingerprint region to confirm identity.
Characteristic absorption frequencies
This is the reference table to keep beside you. Focus on the bold rows first; they are the highest-yield diagnostics.
| Bond / group | Wavenumber (cm⁻¹) | Intensity | Notes |
|---|---|---|---|
| O–H (alcohol) | 3200–3550 | strong, broad | Broad because of hydrogen bonding |
| O–H (carboxylic acid) | 2500–3300 | strong, very broad | Very wide; overlaps the C–H region |
| N–H (amine, amide) | 3300–3500 | medium | One band for 2°, two bands for 1° |
| C–H (sp³, alkane) | 2850–2960 | medium | Just to the right of 3000 |
| =C–H (sp², alkene/arene) | 3010–3100 | medium | Just to the left of 3000 |
| ≡C–H (sp, alkyne) | ~3300 | strong, sharp | Narrow spike |
| C–H (aldehyde) | 2720 & 2820 | weak | Two small bands; a giveaway for –CHO |
| C≡C (alkyne) | 2100–2260 | weak | Absent if internal/symmetric (no dipole change) |
| C≡N (nitrile) | 2210–2260 | medium | Sharp, in an otherwise empty region |
| C=O (carbonyl) | 1670–1780 | strong | The most useful single peak; see box below |
| C=C (alkene) | 1620–1680 | medium–weak | |
| C=C (aromatic ring) | 1450–1600 | medium | Often several bands |
| N–O (nitro, –NO₂) | 1500 & 1350 | strong | A pair of strong bands |
| C–O (alcohol, ether, ester) | 1000–1300 | strong | In the fingerprint region |
Zooming in on the carbonyl (C=O)
Because it is so strong and sits in an otherwise clear region near 1700 cm⁻¹, the carbonyl is often the first thing to look for. Its exact position hints at which kind of carbonyl you have:
How to read a spectrum, step by step
- 1
Look near 1700 cm⁻¹ for a C=O. A strong signal here means an aldehyde, ketone, ester, acid, or amide. Its absence rules all of those out.
- 2
Check 3200–3550 cm⁻¹ for O–H or N–H. A broad band indicates O–H (alcohol). A very broad band spreading over to ~2500, together with a C=O, is diagnostic for a carboxylic acid. Sharper medium bands here belong to N–H.
- 3
Read the C–H region around 3000 cm⁻¹. Absorptions just left of 3000 signal sp² C–H (alkene or aromatic); absorptions only to the right of 3000 mean a saturated (sp³) compound.
- 4
Scan 2000–2500 cm⁻¹ for triple bonds. A sharp peak here points to a nitrile or alkyne; this region is usually empty, so any peak is meaningful.
- 5
Use the fingerprint region to confirm. Don’t over-interpret individual peaks below 1500; you will usually have NMR and mass spectrometry data to help solve more complex problems.
Worked interpretation examples
Example 1. Reading Figure 1
Work through the ethanol spectrum above using the five steps. Near 1700 cm⁻¹ there is no strong signal, so we can rule out all carbonyl compounds. There is a broad band around 3350 cm⁻¹, characteristic of a hydrogen-bonded O–H, so we have an alcohol. The C–H absorptions fall just to the right of 3000, telling us the carbon skeleton is saturated (sp³). Finally, the strong band near 1050 cm⁻¹ is the C–O stretch, consistent with an alcohol. Together: a small saturated alcohol, here ethanol.
Example 2. An unknown with formula C₄H₈O₂
Suppose the spectrum shows a very broad absorption stretching from about 2500 to 3300 cm⁻¹ and a strong peak at 1710 cm⁻¹. The strong 1710 band is a carbonyl. The extremely broad band overlapping the C–H region is the hallmark of a carboxylic acid O–H. A carbonyl plus that distinctive broad O–H means a carboxylic acid; with the formula C₄H₈O₂, this is butanoic acid.
Example 3. Telling an alcohol, an aldehyde, and a ketone apart
Butan-1-ol (C₄H₁₀O), butanal, and butan-2-one (both C₄H₈O) are easy to distinguish by two regions alone:
| Compound | ~1700 (C=O)? | ~3300 (O–H)? | Verdict |
|---|---|---|---|
| Butan-1-ol | absent | broad band present | alcohol |
| Butanal | strong (~1725) + two weak C–H at 2720/2820 | absent | aldehyde |
| Butan-2-one | strong (~1715) | absent | ketone |
Practice questions
Try each before opening the answer.
1Why do peaks in an IR spectrum point downward instead of up?Show answer
The y-axis is % transmittance. When the sample absorbs light at a given wavenumber, less light passes through, so transmittance drops and the signal dips toward the bottom of the plot.
2A compound gives a strong band at ~1715 cm⁻¹ and shows no broad absorption around 3300 cm⁻¹. What functional group is present, and what is ruled out?Show answer
A carbonyl (C=O), most likely a ketone at ~1715 cm⁻¹. With no broad O–H band, an alcohol and a carboxylic acid are ruled out.
3How can IR distinguish the C–H bonds of an alkane from those of an alkene or aromatic ring?Show answer
By position relative to 3000 cm⁻¹. Alkane sp³ C–H stretches appear just below 3000 (2850–2960); alkene and aromatic sp² C–H stretches appear just above 3000 (3010–3100).
4The internal alkyne 2-butyne (CH₃–C≡C–CH₃) shows almost no C≡C absorption. Why?Show answer
The molecule is symmetrical about the triple bond, so stretching the C≡C causes no change in dipole moment. By the dipole rule, that vibration is IR-inactive.
5You see a very broad band from ~2500–3300 cm⁻¹ overlapping the C–H peaks, plus a strong band at ~1710 cm⁻¹. Identify the functional group.Show answer
A carboxylic acid. The very broad, hydrogen-bonded O–H combined with a C=O near 1710 cm⁻¹ is its signature.
Key takeaways
- IR light excites bond vibrations; the absorbed frequencies reveal which bonds, and therefore which functional groups, are present.
- The x-axis is wavenumber (cm⁻¹, 4000 → 400, decreasing left to right); peaks are downward signals in % transmittance.
- Interpret in this order: C=O near 1700 → O–H / N–H near 3300 → C–H around 3000 → triple bonds 2000–2500 → fingerprint region to confirm.
- A vibration is only visible if it changes the dipole moment, so symmetric bonds can be weak or silent.