1. Overview
Infrared (IR) spectroscopy is a quick way to find out which functional groups a molecule contains. Each type of bond absorbs infrared radiation in its own range of wavenumbers, listed in the Data Booklet. This section shows you how to read a spectrum, match its absorptions to bonds and turn that evidence into a clear answer, for example to check whether a reaction has worked. IR is often used alongside mass spectrometry (22.2).
Key Definitions
- Infrared spectroscopy: an analytical technique in which infrared radiation is passed through a sample and the wavenumbers it absorbs are recorded; the absorptions show which bonds, and so which functional groups, the molecules contain.
- Infrared spectrum: a graph of transmittance (in %) against wavenumber (in cm⁻¹), with wavenumber decreasing from left to right; each absorption appears as a dip in the line.
- Wavenumber: the number of waves in one centimetre of the radiation, equal to 1 divided by the wavelength in cm and measured in cm⁻¹; a larger wavenumber means a higher frequency.
- Transmittance: the percentage of the infrared radiation at a given wavenumber that passes through the sample; 100% means none is absorbed.
- Characteristic absorption range: the range of wavenumbers in which a particular bond in a particular functional group absorbs, as listed in the Data Booklet, for example 1710 to 1750 cm⁻¹ for C=O in esters.
- Functional group: the atom or group of atoms in an organic molecule that gives it its typical chemical reactions, for example the hydroxy group, -OH, in alcohols.
Content
What an infrared spectrum shows
Covalent bonds are always vibrating: they stretch and bend. A bond can absorb infrared radiation of particular frequencies, which makes it vibrate more strongly. Different bonds absorb at different frequencies, so the absorptions show which bonds, and so which functional groups, are present.
An infrared spectrum has two axes:
- Wavenumber (horizontal), in cm⁻¹. The scale runs from about 4000 cm⁻¹ on the left to about 500 cm⁻¹ on the right, so the numbers decrease from left to right.
- Transmittance (vertical), in %. Where nothing is absorbed, the line runs near the top.
Each absorption is a dip pointing down from the top line, though it is still usually called a peak. Describe it by its position (the wavenumber at the bottom of the dip), its strength (deep or shallow) and its width (sharp or broad). On some spectra the scale changes at 2000 cm⁻¹, so read positions from the numbers on the axis, not by measuring along it.
The Data Booklet table
These are the ranges you must use, sorted here from high to low wavenumber so they match a spectrum read from left to right. The booklet names the group: hydroxy is the -OH of alcohols, carbonyl is the C=O of aldehydes and ketones, and carboxyl is the -COOH of carboxylic acids.
| Bond (functional group) | Range / cm⁻¹ |
|---|---|
| O-H (hydroxy) | 3200 to 3600 |
| N-H (amine, amide) | 3300 to 3500 |
| O-H (carboxyl) | 2500 to 3000 |
| C-H (alkane) | 2850 to 2950 |
| C≡N (nitrile) | 2200 to 2250 |
| C=O (ester) | 1710 to 1750 |
| C=O (carbonyl, carboxyl) | 1670 to 1740 |
| C=O (amide) | 1640 to 1690 |
| C=C (alkene, aromatic compound) | 1500 to 1680 |
| C-O (hydroxy, ester) | 1040 to 1300 |
The booklet gives positions only. The typical shapes below are general chemistry, not booklet data, but they help you recognise each absorption:
- O-H absorptions are broad. In liquids and solids, hydrogen bonding between molecules gives the O-H bonds a range of strengths, so they absorb over a wide range.
- C=O usually gives one of the deepest, sharpest dips, so it is easy to spot. C=C is often much weaker, and N-H is usually narrower and shallower than O-H.
- C-H appears in almost every organic compound, so on its own it tells you very little.
Alcohol or carboxylic acid? Two kinds of O-H
Both alcohols and carboxylic acids contain O-H, but their absorptions are in different places:
- In an alcohol (hydroxy group), O-H absorbs at 3200 to 3600 cm⁻¹ as a broad dip.
- In a carboxylic acid (carboxyl group), O-H absorbs lower, at 2500 to 3000 cm⁻¹, and the dip is even broader. The sharp C-H absorptions often show as small notches on top of it.
A carboxylic acid also contains C=O, so its spectrum always shows a strong C=O absorption at 1670 to 1740 cm⁻¹ as well. A broad dip near 3000 cm⁻¹ without a C=O absorption is not a carboxylic acid.
Telling the C=O compounds apart
The three C=O ranges overlap. A C=O absorption at 1725 cm⁻¹, for example, fits a carbonyl, a carboxyl and an ester. For a compound with only one functional group, decide using the other absorptions:
| Compound | Also look for |
|---|---|
| aldehyde or ketone | no O-H absorption at all |
| carboxylic acid | very broad O-H, 2500 to 3000 cm⁻¹ |
| ester | strong C-O, 1040 to 1300 cm⁻¹, and no O-H |
| amide | N-H, 3300 to 3500 cm⁻¹ |
The booklet table cannot tell an aldehyde from a ketone: both give only a C=O absorption, in the same range. Use a chemical test instead, such as Tollens' reagent (17.1).
Other overlaps to watch
- N-H and O-H. The N-H range, 3300 to 3500 cm⁻¹, lies inside the hydroxy O-H range, 3200 to 3600 cm⁻¹. The formula often decides: no nitrogen means no N-H.
- C=C and C=O. The C=C range reaches 1680 cm⁻¹, so it overlaps the amide C=O range and the bottom of the carbonyl range. A C=O dip is usually much deeper.
- Water. A wet sample shows a broad O-H absorption even if the compound has no O-H, so samples are dried first.
- Isomers with the same groups, such as two alcohols with the same formula, absorb in the same booklet ranges, so the table alone cannot tell them apart.
How to analyse a spectrum
- Start from what you are told. No oxygen in the formula means no O-H, C=O or C-O; no nitrogen means no N-H or C≡N.
- Work above 1500 cm⁻¹, from left to right. For each deep or distinctive dip, read the wavenumber at its lowest point.
- Match each value to a bond, then to a group. Write both: "the absorption at 1712 cm⁻¹ is due to C=O, so there is a carbonyl, carboxyl or ester group". Then use the other absorptions to decide which.
- Use absences too. No strong dip at 1640 to 1750 cm⁻¹ means no C=O, so no aldehyde, ketone, carboxylic acid, ester or amide. No broad dip between 2500 and 3600 cm⁻¹ means no O-H.
- Below 1500 cm⁻¹, use only the C-O range (1040 to 1300 cm⁻¹), and only as support: alcohols, esters and carboxylic acids all contain C-O. This crowded region is often called the fingerprint region (a common term, but not one you need to define); you are not expected to assign its other dips.
- Put the evidence together and check that your groups fit the formula.
Following a reaction. To show that a reaction has happened, compare the reactant and the product. Ignore bonds that both contain, such as C-H. Name a bond that only one of them has, give its range, and say whether its absorption appears or disappears. When a secondary alcohol is oxidised to a ketone, for example, the broad O-H absorption at 3200 to 3600 cm⁻¹ disappears and a strong C=O absorption at 1670 to 1740 cm⁻¹ appears.
Worked example 1: one compound, two functional groups
Question: Compound P contains carbon, hydrogen and oxygen only. Its infrared spectrum has these main absorptions:
- a broad absorption centred at 3410 cm⁻¹
- sharp absorptions at 2940 and 2880 cm⁻¹
- a strong, sharp absorption at 1705 cm⁻¹
- a strong absorption at 1060 cm⁻¹
Identify the bond responsible for each absorption and the functional groups in P. Explain why P is not a carboxylic acid.
Step 1: match each absorption to a bond.
| Absorption / cm⁻¹ | Bond (range / cm⁻¹) |
|---|---|
| 3410, broad | O-H, hydroxy (3200 to 3600) |
| 2940 and 2880 | C-H (2850 to 2950) |
| 1705, strong | C=O, carbonyl or carboxyl (1670 to 1740) |
| 1060 | C-O (1040 to 1300) |
P has no nitrogen, so the absorption at 3410 cm⁻¹ cannot be N-H and the C=O cannot be in an amide. 1705 cm⁻¹ is also below the ester range (1710 to 1750 cm⁻¹).
Step 2: rule out a carboxylic acid. A carboxyl O-H would give a very broad absorption at 2500 to 3000 cm⁻¹, but P has only the sharp C-H absorptions there.
Answer: P contains a hydroxy group (the broad O-H at 3410 cm⁻¹, supported by the C-O at 1060 cm⁻¹) and a carbonyl group, as in an aldehyde or ketone (the C=O at 1705 cm⁻¹). It is not a carboxylic acid: its O-H is in the hydroxy range and there is no very broad absorption at 2500 to 3000 cm⁻¹. One structure that fits is CH₃COCH₂CH₂CH₂OH; IR alone cannot fix the exact structure.
Worked example 2: which oxidation product?
Question: Hexan-1-ol, CH₃(CH₂)₄CH₂OH, is warmed with acidified potassium dichromate(VI) and the organic product is collected and dried. It could be hexanal or hexanoic acid, possibly mixed with unreacted hexan-1-ol. The main absorptions in its spectrum are at 2930 and 2860 cm⁻¹ (sharp) and 1728 cm⁻¹ (strong, sharp). There is no broad absorption anywhere between 2500 and 3600 cm⁻¹. Identify the product.
Step 1: predict the key absorptions above 1500 cm⁻¹.
| Compound | O-H | C=O |
|---|---|---|
| hexan-1-ol | broad, 3200 to 3600 | none |
| hexanal | none | 1670 to 1740 |
| hexanoic acid | very broad, 2500 to 3000 | 1670 to 1740 |
All three contain C-H bonds (2850 to 2950 cm⁻¹), so the absorptions at 2930 and 2860 cm⁻¹ cannot separate them.
Step 2: compare with the spectrum.
- The strong absorption at 1728 cm⁻¹ is C=O (1670 to 1740 cm⁻¹), so the product is hexanal or hexanoic acid. (1728 cm⁻¹ is also in the ester range, but no ester is among the possible products.)
- There is no very broad absorption at 2500 to 3000 cm⁻¹, so there is no carboxyl O-H: it is not hexanoic acid.
- There is no broad absorption at 3200 to 3600 cm⁻¹, so there is no hydroxy O-H: no hexan-1-ol is detected.
Answer: The product is hexanal, with no hexan-1-ol detected. This is the expected result when the aldehyde is distilled off as it forms; heating under reflux would oxidise it further to hexanoic acid (16.1).
Key Equations
This outcome needs no calculations, but two relationships explain the axes.
Wavenumber ( = wavelength of the radiation, in cm; wavenumber in cm⁻¹): A larger wavenumber means a shorter wavelength and a higher frequency.
Transmittance ( = intensity of the radiation entering the sample; = intensity of the radiation leaving it): A dip down to 30% transmittance means that only 30% of the radiation at that wavenumber gets through the sample.
Common Mistakes to Avoid
- Mistake: naming the functional group when the question asks for the bond, or the other way round. Fix: for an absorption at 2240 cm⁻¹ the bond is C≡N and the group is nitrile; "nitrile" alone names the group, not the bond.
- Mistake: mixing up the two O-H ranges. Fix: hydroxy O-H is at 3200 to 3600 cm⁻¹; carboxyl O-H is at 2500 to 3000 cm⁻¹ and always comes with C=O. A broad dip at 3350 cm⁻¹ with no C=O absorption points to an alcohol, not an acid.
- Mistake: using C-H as evidence for a group. Fix: almost every organic compound absorbs at 2850 to 2950 cm⁻¹. In Worked example 2, hexan-1-ol, hexanal and hexanoic acid all do.
- Mistake: deciding the group from the C=O position alone. Fix: 1725 cm⁻¹ fits carbonyl, carboxyl and ester. Check for a very broad O-H (acid), or for C-O with no O-H (ester), before you decide.
- Mistake: reading the wavenumber axis as if it increased to the right. Fix: numbers fall from left to right. A dip just to the right of the 2000 mark is below 2000 cm⁻¹, for example a C=O at 1715 cm⁻¹, not above it.
Exam Tips
- Identify a bond from a spectrum: give the bond, the wavenumber you read and the matching range, for example "C=O, absorption at 1705 cm⁻¹, in the range 1670 to 1740 cm⁻¹". If the question says to use absorptions above 1500 cm⁻¹, stay in that region.
- Explain in a short chain: absorption at this wavenumber, so this bond, so this group. Add a missing absorption to rule out the obvious alternative, as in Worked example 2.
- Predict differences between two spectra: use bonds found in only one compound and say whether each absorption appears or disappears. If asked why two compounds are hard to tell apart, name the bonds they share and their ranges.
- Use the table you are given. When a question prints its own table, use its ranges, even if a few differ slightly from the Data Booklet's.
- Multiple choice "which compound gives this spectrum": test each option for absorptions that must be there and for absorptions that must be absent. The absences usually rule out the wrong options fastest.