22. Analytical techniques
- Syllabus
- 9701–2028–2029
- Section
- 22
- Level
- AS
Exam analysis
Most tested topics
22.1 Infrared spectroscopy
- Syllabus
- 9701–2028–2029
- Topic
- 22.1
- Level
- AS
Match IR absorptions in combination to identify functional groups
| Bond / functional-group context | Official absorption range / cm⁻¹ |
|---|---|
| C–O in hydroxy or ester | 1040–1300 |
| C=C in aromatic compound or alkene | 1500–1680 |
| C=O in amide | 1640–1690 |
| C=O in carbonyl or carboxyl | 1670–1740 |
| C=O in ester | 1710–1750 |
| C≡N in nitrile | 2200–2250 |
| O–H in carboxyl | 2500–3000 |
| C–H in alkane | 2850–2950 |
| O–H in hydroxy | 3200–3650 |
| N–H in amine or amide | 3300–3500 |
- Read the wavenumber axis in cm⁻¹. 2. Match clear absorptions to the supplied ranges. 3. Combine bonds required by one functional group. 4. Check that expected companion absorptions are present and that a proposed group does not conflict with major absent features.
| Spectrum evidence | Supported deduction |
|---|---|
| O–H 2500–3000 plus C=O 1670–1740 | carboxylic acid |
| C=O 1710–1750 plus C–O 1040–1300, without O–H | ester |
| C≡N 2200–2250 | nitrile |
| O–H 3200–3650 plus C–O 1040–1300 | alcohol / hydroxy group |
Several ranges overlap: a peak near 1680 cm⁻¹ cannot by itself distinguish every C=C and C=O context. Use a set of present and absent absorptions; IR normally identifies functional groups rather than a unique complete carbon skeleton.
22.2 Mass spectrometry
- Syllabus
- 9701–2028–2029
- Topic
- 22.2
- Level
- AS
Read mass-spectrum position and height as different evidence
| Spectrum feature | Meaning |
|---|---|
| horizontal m/e value | ion mass divided by its positive charge |
| vertical relative abundance | amount of that ion relative to the chosen reference peak |
| base peak | most abundant detected ion, assigned 100 |
| peaks separated by isotope mass differences | ions with the same structure but different isotopes |
For the commonly shown singly charged ions, m/e equals the ion's nominal mass. Decide whether a peak belongs to the molecular ion, an isotopic partner or a fragment by combining its position with the rest of the pattern; peak height alone cannot assign identity.
The base peak is not automatically M⁺, and one m/e value can fit more than one possible formula. Instrument operation is explicitly not required; interpretation of mass/charge and abundance is.
Relative atomic mass is an isotope-abundance weighted mean
Ar=∑relative abundance∑(isotopic mass×relative abundance)
Multiply every isotope mass by its abundance, add those contributions, then divide by the total abundance. If abundances are percentages, the denominator is 100; arbitrary relative peak heights may have a different total.
Ar=10035×75+37×25=35.5
The weighted mean must lie between the lightest and heaviest isotope masses and closer to the more abundant isotope. Do not take an unweighted mean unless the abundances are equal.
The molecular-ion peak gives the intact molecule's mass
The molecular ion M⁺ is the intact molecule after loss of one electron. For charge +1, its m/e value equals the molecule's relative molecular mass, so identify M⁺ before reading the molecular mass.
Look for a plausible high-mass parent peak and its M+1 or M+2 isotopic partners, then check that lower-mass peaks can be fragments of that parent. The tallest peak is the base peak and need not be M⁺.
M+ at m/e=46⇒Mr=46(z=+1)
An M+1 or M+2 isotope peak is not a second molecular mass, and a fragment such as m/e 31 cannot replace the identified parent. Some molecular-ion peaks are weak, so the full pattern must support the assignment.
Assign a simple fragment by mass, atoms, charge and complementary loss
- Calculate candidate formulas whose nominal mass equals the fragment m/e (for +1 ions). 2. Keep only candidates that can be cut from the given parent structure. 3. Subtract the fragment atoms/mass from M to identify a plausible neutral or radical loss. 4. Check that one fragment carries the positive charge and all parent atoms are accounted for.
| Ethanol evidence | Assignment | Accounting |
|---|---|---|
| M⁺ at 46 | C₂H₆O⁺ | intact parent ion |
| peak at 31 | CH₂OH⁺ | 46 − 31 = 15, consistent with loss of CH₃• |
| peak at 29 | C₂H₅⁺ | 46 − 29 = 17, consistent with loss of •OH |
A matching arithmetic mass is necessary but not sufficient: m/e 29 could represent different formulas in another molecule. Use the supplied parent connectivity and a chemically possible complementary loss rather than memorising one universal peak list.
Use the M+1/M abundance ratio to count carbon atoms
The [M+1]⁺ peak is mainly produced when one ¹²C in the molecular ion is replaced by naturally occurring ¹³C. Because ¹³C is about 1.1% of carbon, the [M+1]⁺ abundance grows approximately in proportion to the number of carbon atoms.
n=1.1×abundance of M+100×abundance of [M+1]+
n=1.1×50100×5.5=10
Use peak heights or areas on the same relative-abundance scale; do not use their m/e positions. The result should be close to a chemically plausible whole-number carbon count.
Use the molecular-ion M⁺ abundance, not the base peak unless it happens to be M⁺. If the answer is far from an integer, recheck the M/M+1 assignment before rounding.
M and M+2 abundance ratios distinguish chlorine from bromine
| One halogen atom | Main isotope pair | Approximate M : M+2 abundance | Inference |
|---|---|---|---|
| chlorine | ³⁵Cl : ³⁷Cl | 3 : 1 | chlorine present |
| bromine | ⁷⁹Br : ⁸¹Br | 1 : 1 | bromine present |
Find two molecular-ion cluster peaks separated by two m/e units and compare their heights. The +2 separation shows the isotope mass difference; the ratio identifies whether the element's heavier isotope is about one-third or about equal in abundance to the lighter isotope.
More than one Cl or Br atom produces a larger cluster (M, M+2, M+4, …), so a simple 3:1 or 1:1 pair is the one-atom pattern. Inspect the whole cluster before deciding.
An M+2 peak by itself does not identify the halogen. Use its abundance relative to M, and do not confuse this +2 isotope spacing with the +1 peak used for carbon counting.