Analytical Techniques (A Level)

0. What Examiners Want

This guide covers Topic 37 of Cambridge International AS & A Level Chemistry 9701 (2025–27) and follows the learning outcomes below, word for word. The right-hand column shows where each one is taught. Section 6 combines the techniques with mass spectrometry and infrared spectroscopy (Topic 22) in the way Paper 4 does.

1. Thin-Layer Chromatography

Chromatography separates a mixture into its components. Every method has a stationary phase that stays still and a mobile phase that moves through or over it, carrying the sample. Components that spend more time in the mobile phase travel further. In thin-layer chromatography (TLC) the stationary phase is a thin layer of a solid, usually aluminium oxide (alumina, Al₂O₃) spread on a glass or plastic plate. The mobile phase is a liquid solvent, which can be polar or non-polar, and it rises up the plate by capillary action.

A spot of the mixture is placed on a pencil line, the baseline, and the plate stands in a little solvent in a covered tank. When the solvent has risen almost to the top, the plate is removed and the height reached is marked at once: this is the solvent front. Colourless spots are located with ultraviolet light or by spraying with a locating agent (for example ninhydrin for amino acids).

The R_f value is the distance the spot moves divided by the distance the solvent front moves, both measured from the baseline to the centre of the spot.

Why compounds travel different distances

Alumina is a polar solid. A polar molecule, such as propanone, is attracted strongly to its surface and is adsorbed; it spends little time in the solvent and moves a short distance, so it has a low R_f. A non-polar molecule, such as hexane, is attracted only weakly to the surface and dissolves well in a non-polar solvent such as methylbenzene, so it moves far and has a high R_f. TLC is normally an adsorption method, although a layer of water on the alumina can also let the solutes partition between the water and the solvent. Switching to a more polar solvent carries every spot higher, because the solvent competes with the solutes for the surface.

R_f depends on two things together: how strongly the solute is attracted to the stationary phase, and how soluble it is in the mobile phase.

Worked example. A spot of a dye moves 4.6 cm on a TLC plate while the solvent front moves 8.0 cm. Calculate the R_f value. A second compound has R_f = 0.80 under the same conditions. Which is the more polar? [3]

R_f = 4.6 ÷ 8.0 = 0.58 [1]. The compound with the lower R_f (0.58) is the more polar [1], because it is adsorbed more strongly on the polar aluminium oxide and is less soluble in the (non-polar) solvent [1].

2. Gas–Liquid Chromatography

In gas–liquid chromatography (GLC) a vaporised sample is carried through a long, coiled column by an unreactive carrier gas, the mobile phase, such as helium or nitrogen. The column holds the stationary phase: a non-volatile, high-boiling-point, non-polar liquid coated on a solid support such as tiny silica particles. The oven keeps the column hot enough for every component to stay as a gas. Components that interact more strongly with the liquid take longer to come out of the column.

The sample is injected as a vapour. The detector at the end of the column sends a signal to the recorder, which draws the chromatogram.

The detector records a peak as each component leaves the column. The time from injection to the maximum of a peak is its retention time. Each compound has its own retention time under fixed conditions (same column, carrier gas, flow rate and temperature), so the peaks are identified by comparing retention times with those of known compounds.

Explaining retention times. In the usual non-polar column, the order depends on how strongly each component is attracted to the stationary liquid. The stronger the intermolecular forces with the non-polar liquid (mainly induced dipole–dipole forces, which increase with the number of electrons and the length of the chain), the longer the component stays in the column and the longer its retention time. A component with a lower boiling point or weaker interaction spends more time in the gas and emerges earlier. A more polar stationary phase would hold polar components for longer and could change the order of the peaks.

The area under each peak is proportional to the amount of that component, so the percentage composition comes from the peak areas.

Percentage composition. The area under a peak is proportional to the amount of that component, provided that every component gives a peak, all are separated and the detector responds equally to each. The peaks are roughly triangular, so area = ½ × base × height; many instruments print the areas. If the peaks are narrow and have similar widths, the heights can be used instead.

GLC is limited to samples that can be vaporised without decomposing, and similar compounds can have almost the same retention time. When it is linked to a mass spectrometer (GC–MS), each separated component goes straight into the spectrometer and is identified from its own mass spectrum.

Worked example. A chromatogram of a mixture of three esters shows peaks of area 12, 30 and 18 (arbitrary units). Calculate the percentage of the ester with the area 30. [2]

Total area = 12 + 30 + 18 = 60 [1]. Percentage = 30 ÷ 60 × 100 = 50% [1].

Worked example. Hexane, heptane and octane are separated on a non-polar column. Predict the order in which they leave the column and explain it. [3]

Hexane, then heptane, then octane [1]. Octane has the most electrons and the strongest induced dipole–dipole forces with the non-polar liquid stationary phase [1], so it is held longest and has the longest retention time; hexane has the weakest and emerges first [1].

3. How NMR Works

Nuclear magnetic resonance (NMR) spectroscopy uses nuclei that behave like tiny magnets. Nuclei with an odd mass number, such as ¹H and ¹³C, spin; ¹²C does not. In a strong magnetic field a spinning nucleus lines up with the field (lower energy) or against it (higher energy). The gap, ΔE, corresponds to radio-wave energy. A nucleus absorbs this energy and flips to the higher level, and this absorption is detected as a peak.

Without a field there is one level. In a field there are two, and nuclei absorb radio waves of energy ΔE to flip between them.

The electrons around a nucleus shield it a little from the field, so the size of ΔE depends on the chemical environment of the nucleus. Atoms in different environments absorb at different places. The position of a peak is its chemical shift, δ, measured in parts per million (ppm) from a reference compound. Equivalent atoms (in identical environments) give a single signal.

TMS. The reference is tetramethylsilane, Si(CH₃)₄, set at δ = 0. It is chosen because it is inert, volatile (so it is easy to remove) and mixes well with organic compounds; its 12 H atoms are all equivalent, so it gives one sharp peak; and that peak is at a lower δ than almost every other proton. The same compound is the reference for ¹³C spectra.

The scale runs from right (TMS, δ = 0) to left. Larger δ is called downfield.

Solvents. A sample is dissolved in a solvent for NMR. An ordinary solvent with H atoms, such as CHCl₃, would give its own large ¹H peak. A deuterated solvent, such as CDCl₃, contains ²H (deuterium) in place of ¹H. Deuterium does not absorb in the ¹H NMR range, so the solvent gives no signal that could hide or be mistaken for a signal of the sample. (Tetrachloromethane, CCl₄, has no H at all, but CDCl₃ dissolves more compounds.)

4. Carbon-13 NMR

About 1% of carbon atoms are ¹³C, which is enough to give a spectrum. In a standard ¹³C spectrum each different carbon environment gives one vertical line, and there is no splitting. The line heights are not proportional to the number of carbon atoms, so they cannot be used for counting. The small signal near δ = 77 comes from the solvent CDCl₃ and is ignored.

Counting lines. Two carbon atoms are equivalent if they are in identical surroundings, usually because of symmetry. Draw the molecule and ask for each carbon: what is it bonded to, and does another carbon have the same set of neighbours? Each set of equivalent atoms gives one line.

Same colour means the same environment. Propan-1-ol has three different carbons; propan-2-ol and propanone each have two; benzene has one.

¹³C shifts: the carbonyl carbon is far downfield (190–220), well away from the alkyl carbons (0–50). Use the data table in the exam.

Each line is matched to a type of carbon using the shift. A carbon bonded to an electronegative atom (O, halogen) is deshielded and appears at a larger δ; a C=O carbon is the largest of all. An aldehyde or ketone carbonyl (190–220) is distinguished from an ester or carboxylic acid carbon (160–185); an alkene or arene carbon (110–160) is distinguished from an alkyl carbon (0–50).

The tallest line is not the carbon that is most plentiful. In ethylbenzene the ring gives four of the six environments: the carbon bonded to the ethyl group and three groups of equivalent CH carbons.

Worked example. State the number of lines in the ¹³C NMR spectrum of (i) propan-1-ol, (ii) propan-2-ol, (iii) benzene, (iv) 1,4-dimethylbenzene. [4]

(i) 3, three different carbons [1]. (ii) 2, the two CH₃ carbons are equivalent [1]. (iii) 1, all six carbons are equivalent [1]. (iv) 3: CH₃, the ring carbons bonded to CH₃, and the ring CH carbons [1].

Worked example. The ¹³C NMR spectrum of an organic compound with the formula C₃H₆O has two lines, at δ = 30 and δ = 206. Deduce its structure. [3]

Two lines mean two carbon environments [1]. δ = 206 is in the range 190–220, so it is a C=O group of an aldehyde or ketone [1]. δ = 30 is alkyl carbon; two equivalent CH₃ groups on the C=O fit the formula, so the compound is propanone, (CH₃)₂C=O [1]. (Propanal, CH₃CH₂CHO, would give three lines.)

5. Proton NMR

A ¹H (proton) NMR spectrum gives three pieces of information about every set of equivalent H atoms: where the signal is (chemical shift), how big it is (area) and what it looks like (splitting). The number of signals equals the number of different proton environments.

Chemical shift. The shift shows the environment of the proton. H atoms next to electronegative atoms or groups (O, halogen, C=O, a benzene ring) are deshielded and have a larger δ. The chart gives the typical ranges; O–H and N–H shifts are variable because they depend on solvent and concentration.

¹H shifts. Blue bars are protons on carbon next to other groups, green are alkene and arene protons, red are O–H and N–H protons.

Peak area. The area under a signal is proportional to the number of H atoms in that environment. The ratio of the areas (given as numbers, integration steps or labels) is the ratio of the H atoms, and the areas add up to the total number of H in the molecule (or to a multiple of it).

Splitting and the n + 1 rule. The magnetic fields of H atoms on adjacent carbon atoms affect each other (spin–spin coupling). A signal from H atoms whose adjacent carbon has n H atoms is split into n + 1 lines. The syllabus names singlet (n = 0), doublet (1), triplet (2) and quartet (3); a signal of more than four lines is a multiplet. The relative heights follow Pascal’s triangle.

The splitting pattern shows how many H atoms are on the carbon next door.

Equivalent H atoms do not split each other. The CH₂ in CH₃CH₂– is split only by the H atoms of the CH₃, and the CH₃ only by those of the CH₂. If the neighbouring carbon has no H (for example C=O or a quaternary carbon) the signal is a singlet.

In ethanol: CH₃ next to CH₂ gives a triplet; CH₂ next to CH₃ gives a quartet; the O–H gives a singlet.

O–H and N–H protons. These protons exchange rapidly with H atoms of other molecules and with traces of water or acid, so they are normally not split and give a single peak at a variable shift. To identify one, a few drops of D₂O are added and the sample is shaken. The H of the O–H or N–H is swapped for deuterium: R–OH + D₂O ⇌ R–OD + HOD. Deuterium does not absorb in the ¹H range, so that signal disappears. Signals from C–H protons stay where they were.

Compare the two spectra: only the signal that vanishes belongs to an O–H (or N–H) proton.

Worked example. Predict the ¹H NMR spectrum of propan-2-ol, (CH₃)₂CHOH: the number of signals, the relative areas, the type of proton and the splitting of each. [5]

Three signals [1]. (1) δ about 1.2, relative area 6, two equivalent CH₃ next to CH, doublet (n = 1) [1]. (2) δ about 4.0, area 1, the CH next to O, split by 6 H into a multiplet (septet) [1]. (3) δ about 2 (variable), area 1, the O–H, a singlet [1]. It disappears when D₂O is added [1].

Worked example. Predict the number of signals, their areas and their splitting in the ¹H NMR spectrum of propan-1-ol, CH₃CH₂CH₂OH. [4]

Four signals, area ratio 3 : 2 : 2 : 1 [1]. CH₃ (δ about 0.9): triplet, next to CH₂ [1]. Middle CH₂ (δ about 1.6): multiplet, next to 5 H [1]. OCH₂ (δ about 3.6) is a triplet; the O–H is a singlet [1].

6. Putting It Together

Paper 4 gives the molecular formula (from combustion data or the mass spectrum), a spectrum and often some chemistry. Work through the data in the same order each time.

Use all the evidence, and check the structure against every signal.

Worked example. An ester, used as a solvent in a glue, has the ¹H NMR spectrum shown. Identify it. [5]

Each signal gives its environment, its H count and its neighbours. The pieces CH₃CH₂– and CH₃CO– are joined through the ester oxygen.

δ 1.3, area 3, triplet: CH₃ next to CH₂ [1]. δ 2.2, area 3, singlet: CH₃ next to C=O, with no H on the next atom [1]. δ 3.9, area 2, quartet: O–CH₂ next to CH₃ [1]. The pieces CH₃CO– and –O–CH₂CH₃ make an ester [1]: CH₃COOCH₂CH₃, ethyl ethanoate [1].

Worked example. Compound P (C₄H₈O₂) has a ¹H NMR spectrum with signals at δ 1.1 (triplet, 3), 2.3 (quartet, 2) and 3.7 (singlet, 3). Its ¹³C spectrum has four lines. Identify P and state why it is not its isomer ethyl ethanoate. [4]

δ 1.1 triplet 3H: CH₃CH₂– [1]. δ 2.3 quartet 2H: CH₂ next to the CH₃ and to C=O [1]. δ 3.7 singlet 3H: CH₃–O with no neighbouring H [1]. So P is methyl propanoate, CH₃CH₂COOCH₃; ethyl ethanoate would have a CH₃ singlet near 2.2, and an OCH₂ quartet near 3.9 [1].

7. Quick Sheet and Checklist

• [ ] I can define stationary phase, mobile phase, baseline, solvent front and R_f, calculate R_f and explain differences in R_f using adsorption and solubility.
• [ ] I can describe GLC, define retention time, explain retention times and calculate a percentage composition from peak areas.
• [ ] I can count the lines in a ¹³C spectrum, match each to a carbon environment using the shift table and use them to choose between isomers.
• [ ] I can use shifts, areas and the n + 1 rule to deduce a structure from a ¹H spectrum, and predict the spectrum of a given molecule.
• [ ] I can state why TMS, CDCl₃ and D₂O are used and describe the effect of D₂O on a spectrum.