What you'll learn
Spectroscopic techniques allow chemists to identify substances and determine their structures quickly, using only tiny samples — far faster than classical chemical tests. They are used in hospitals, forensic laboratories, environmental agencies and quality-control labs throughout the Caribbean. This guide covers the electromagnetic spectrum and how different regions interact with matter, infrared (IR) spectroscopy for identifying functional groups, ultraviolet–visible (UV–Vis) spectroscopy and colorimetry, including the Beer–Lambert law for measuring concentration, and mass spectrometry for finding relative molecular mass and interpreting fragmentation patterns, including the M+1 and M+2 peaks. It shows how combining techniques reveals molecular structures.
CAPE questions typically give you spectra and ask you to identify functional groups, deduce structures, calculate concentrations from absorbance data, or explain what a technique measures and why it works.
Key terms and definitions
Electromagnetic spectrum — The range of all electromagnetic radiation, from radio waves to gamma rays.
Wavenumber — The reciprocal of wavelength, in cm⁻¹, used in IR spectroscopy.
Absorbance — A measure of how much light a sample absorbs.
Transmittance — The percentage of light that passes through a sample.
Chromophore — A group of atoms in a molecule responsible for absorbing UV or visible light.
Beer–Lambert law — A = εcl, where absorbance is proportional to concentration and path length.
Molar absorptivity (ε) — A constant that measures how strongly a substance absorbs light at a particular wavelength.
Molecular ion (M⁺) — The ion formed when a molecule loses one electron in a mass spectrometer.
Fragmentation — The breaking of the molecular ion into smaller ions.
Base peak — The tallest peak in a mass spectrum, given a relative abundance of 100.
M+1 peak — A small peak one mass unit above M⁺, due mainly to the presence of carbon-13.
Core concepts
The electromagnetic spectrum and matter
Different regions of the spectrum cause different changes in molecules, because their energy (E = hf) matches different energy gaps:
| Radiation | Effect on molecules | Technique |
|---|---|---|
| Ultraviolet / visible | electronic transitions (electrons promoted to higher energy levels) | UV–Vis spectroscopy, colorimetry |
| Infrared | bond vibrations (stretching and bending) | IR spectroscopy |
| Radio waves | changes in nuclear spin in a magnetic field | NMR spectroscopy |
Infrared spectroscopy
Covalent bonds vibrate — stretching and bending — at characteristic frequencies. When IR radiation of the same frequency passes through a sample, the bond absorbs it. An IR spectrometer records transmittance against wavenumber (usually 4000–500 cm⁻¹); absorptions appear as downward "troughs".
For a bond to absorb IR, its vibration must cause a change in dipole moment, so polar bonds absorb strongly. This is also why carbon dioxide, methane and water vapour absorb IR from the Earth — the basis of the greenhouse effect.
Key absorptions:
| Bond | Wavenumber / cm⁻¹ | Appearance |
|---|---|---|
| O–H (alcohols) | 3200–3600 | broad |
| O–H (carboxylic acids) | 2500–3300 | very broad |
| N–H (amines, amides) | 3300–3500 | medium |
| C–H | 2850–3100 | sharp |
| C=O (aldehydes, ketones, acids, esters) | 1680–1750 | strong, sharp |
| C=C | 1620–1680 | medium |
| C–O (alcohols, esters, acids) | 1000–1300 | strong |
The region below about 1500 cm⁻¹ is the fingerprint region, with a complex pattern unique to each compound. Comparing it with a database spectrum confirms identity.
Uses: identifying functional groups; monitoring reactions (for example, the disappearance of O–H and appearance of C=O when an alcohol is oxidised to a ketone); breathalysers that measure ethanol in breath using its C–H absorption; and monitoring air pollutants.
UV–visible spectroscopy
Molecules or ions with chromophores — such as C=C, C=O, conjugated systems (alternating double and single bonds) or transition-metal complexes — absorb UV or visible light as electrons are promoted to higher energy levels. A UV–Vis spectrum shows absorbance against wavelength; the wavelength of maximum absorbance (λmax) is characteristic.
Conjugation: the more extended the delocalised system, the smaller the energy gap, so the compound absorbs at longer wavelengths. Extended conjugation can shift absorption into the visible region, making a compound coloured — as in azo dyes and the pigment β-carotene in carrots and pumpkins.
Colorimetry and the Beer–Lambert law
A colorimeter measures how much light of a selected wavelength (using a filter of the complementary colour) is absorbed by a coloured solution. According to the Beer–Lambert law:
A = εcl
where A is absorbance, ε is molar absorptivity, c is concentration (mol dm⁻³) and l is the path length (cm).
Method for finding an unknown concentration:
- Prepare a series of standard solutions of known concentration.
- Measure the absorbance of each at λmax.
- Plot a calibration curve of absorbance against concentration (a straight line through the origin if the law is obeyed).
- Measure the absorbance of the unknown and read its concentration from the graph.
Uses: measuring iron in water, copper in solutions, glucose in blood (after a colour-forming reaction), and phosphates and nitrates in environmental samples.
Mass spectrometry of organic molecules
In a mass spectrometer, molecules are ionised by electron bombardment, forming the molecular ion: M → M⁺ + e⁻. The molecular ion may break into fragments:
M⁺ → X⁺ + Y•
Only the positively charged fragment is detected.
Interpreting a spectrum:
- The molecular ion peak (the peak at the highest m/z, ignoring the small M+1 peak) gives the Mᵣ.
- Fragment peaks reveal parts of the structure. Common fragments: m/z 15 (CH₃⁺), 29 (C₂H₅⁺ or CHO⁺), 43 (C₃H₇⁺ or CH₃CO⁺), 45 (COOH⁺), 77 (C₆H₅⁺).
- The base peak is the most abundant (often the most stable) fragment.
M+1 peak: about 1.1% of carbon atoms are carbon-13. The ratio of the M+1 to M peak heights gives an estimate of the number of carbon atoms:
number of carbon atoms ≈ (height of M+1 ÷ height of M⁺) × 100 ÷ 1.1
M+2 peaks: compounds containing chlorine show M⁺ and M+2 peaks in a ratio of about 3 : 1 (³⁵Cl : ³⁷Cl); those containing bromine show peaks of about equal height (⁷⁹Br : ⁸¹Br ≈ 1 : 1).
Combining techniques
In practice, chemists combine evidence:
- Mass spectrometry gives the Mᵣ and some fragments.
- IR identifies functional groups.
- NMR (beyond the main focus here) reveals the hydrogen environments.
- Elemental analysis gives the empirical formula.
Together, these allow an unknown structure to be deduced with confidence.
Worked examples
Example 1: Interpreting an IR spectrum
A compound C₃H₆O shows a strong absorption at 1715 cm⁻¹ and no broad absorption between 3200 and 3600 cm⁻¹. Suggest its identity.
The 1715 cm⁻¹ absorption indicates a C=O group; the absence of a broad O–H band rules out an alcohol or acid. C₃H₆O with C=O could be propanal or propanone. Further tests (such as Tollens' reagent, or the absence of the aldehyde C–H absorption near 2720 cm⁻¹) would distinguish them; a negative Tollens' test would indicate propanone.
Example 2: Beer–Lambert calculation
A solution in a 1.00 cm cuvette has absorbance 0.450. If ε = 900 dm³ mol⁻¹ cm⁻¹, calculate its concentration.
c = A ÷ (εl) = 0.450 ÷ (900 × 1.00) = 5.00 × 10⁻⁴ mol dm⁻³.
Example 3: Number of carbon atoms
The M⁺ peak of a compound has a height of 50 units and the M+1 peak 2.2 units. Estimate the number of carbon atoms.
n = (2.2 ÷ 50) × 100 ÷ 1.1 = 4.4 ÷ 1.1 = 4.
Example 4: Halogen identification
A halogenoalkane shows peaks at m/z 108 and 110 of almost equal height. What does this suggest?
Two peaks two units apart in a 1 : 1 ratio indicate the presence of one bromine atom (⁷⁹Br and ⁸¹Br). The compound could be bromoethane, C₂H₅Br (Mᵣ 108 with ⁷⁹Br).
Example 5: Fragmentation
The mass spectrum of propanone shows a base peak at m/z 43. Identify the fragment.
m/z 43 corresponds to CH₃CO⁺, formed when the molecular ion (m/z 58) loses a CH₃• radical.
Example 6: A combined structure problem
Compound Z has the molecular formula C₄H₈O₂. Its IR spectrum shows a strong absorption at 1740 cm⁻¹ and a strong band at 1240 cm⁻¹, but no broad absorption between 2500 and 3600 cm⁻¹. Its mass spectrum has a molecular ion at m/z 88 and a prominent peak at m/z 43. Suggest the identity of Z.
- M⁺ at m/z 88 confirms the Mᵣ of C₄H₈O₂ (4 × 12 + 8 × 1 + 2 × 16 = 88).
- The C=O absorption at 1740 cm⁻¹ and C–O at 1240 cm⁻¹, with no O–H absorption, rule out a carboxylic acid and suggest an ester.
- A peak at m/z 43 suggests the CH₃CO⁺ fragment.
Z is consistent with ethyl ethanoate, CH₃COOCH₂CH₃, the ester used as nail-varnish remover and a common fruity flavouring.
Example 7: Reading a calibration curve
Standard solutions of a dye give absorbances of 0.10, 0.20, 0.30 and 0.40 at concentrations of 1.0, 2.0, 3.0 and 4.0 × 10⁻⁵ mol dm⁻³. An unknown drink sample has absorbance 0.26. Find its concentration.
The calibration line is straight through the origin with absorbance = 0.10 per 1.0 × 10⁻⁵ mol dm⁻³. Concentration = 0.26 ÷ 0.10 × 1.0 × 10⁻⁵ = 2.6 × 10⁻⁵ mol dm⁻³. The value lies within the range of the standards, so the reading is reliable.
Example 8: Monitoring a reaction by IR
How could IR spectroscopy show that propan-2-ol has been completely oxidised to propanone?
The broad O–H absorption at 3200–3600 cm⁻¹ present in propan-2-ol should disappear, and a strong C=O absorption at about 1715 cm⁻¹ should appear. If any O–H absorption remains, some alcohol is still present.
Mass spectrum fragments to recognise
| m/z | Likely fragment |
|---|---|
| 15 | CH₃⁺ |
| 29 | C₂H₅⁺ or CHO⁺ |
| 31 | CH₂OH⁺ (primary alcohols) |
| 43 | C₃H₇⁺ or CH₃CO⁺ |
| 45 | COOH⁺ |
| 57 | C₄H₉⁺ |
| 77 | C₆H₅⁺ (benzene ring) |
The difference between the molecular ion and a fragment peak also gives information — a loss of 15 suggests a CH₃ group has broken off; a loss of 18 suggests loss of water from an alcohol.
Common mistakes and how to avoid them
- Reading IR troughs as peaks pointing up — absorptions appear as dips in transmittance.
- Confusing the alcohol and acid O–H bands — the acid O–H is much broader and lower.
- Using the base peak as the molecular ion — the M⁺ peak is at the highest m/z.
- Forgetting that only positive ions are detected.
- Choosing a colorimeter filter of the same colour as the solution — use the complementary colour.
- Extrapolating calibration curves beyond the range of standards.
How this links to your Internal Assessment
Colorimetry is the most accessible spectroscopic technique for SBAs. You could determine the concentration of iron in a local water supply or an iron supplement (after forming a coloured complex), investigate the rate of a reaction by following colour change (such as iodine with propanone), or measure the concentration of food dyes in drinks. A good report prepares accurate standards, plots a calibration curve, selects λmax or the appropriate filter, and evaluates errors such as cuvette contamination and stray light. Where IR or mass spectra are available, using them to confirm a product's identity strengthens an organic preparation SBA.
Exam technique for Spectroscopy: IR, UV and mass spectrometry
- In IR questions, state the bond, the wavenumber range and the functional group it suggests; refer to absences as well as presences.
- In mass spectra, identify M⁺, use M+1 for carbon count and M+2 for halogens, and write fragment formulae with their positive charge.
- In Beer–Lambert questions, rearrange the formula carefully and include units.
- When deducing a structure, combine all the evidence and check it fits the molecular formula.
Quick revision summary
- UV–Vis: electronic transitions; IR: bond vibrations; radio waves: nuclear spin (NMR).
- IR absorption requires a change in dipole; key bands: O–H 3200–3600 (broad), acid O–H 2500–3300, C=O 1680–1750, C–H 2850–3100; fingerprint region < 1500 cm⁻¹.
- Conjugation shifts absorption to longer wavelengths and can cause colour.
- Beer–Lambert: A = εcl; calibration curves give unknown concentrations.
- Mass spectra: M⁺ gives Mᵣ; fragments reveal structure; M+1 estimates carbons; M+2 ratios identify Cl (3:1) and Br (1:1).
- Combining MS, IR and other data reveals structures.