Analytical techniques

📚The chemistry

AS analysis (topic 22) found functional groups (infrared) and molecular mass (mass spectrometry). This topic adds ways to separate mixtures (chromatography) and to map the carbon and hydrogen skeleton of a molecule (NMR). All chromatography works on one principle: components of a mixture are distributed between a stationary phase and a mobile phase, and those that interact more with the stationary phase move more slowly.

37.1 Thin-layer chromatography (TLC)

• Stationary phase: a thin layer of a polar solid such as aluminium oxide (or silica) on a solid support (a glass or plastic plate).
• Mobile phase: a solvent, polar or non-polar, which moves up the plate by capillary action.
• Baseline: the pencil line where the samples are spotted, above the solvent level so they do not simply dissolve away.
• Solvent front: how far the solvent has travelled when the plate is removed; mark it immediately.

Both distances are measured from the baseline. R_f is between 0 and 1 and has no units. Under the same conditions (same stationary phase, solvent and temperature) a substance always has the same R_f, so comparing with standards identifies components. Colourless spots are shown with UV light or a locating agent.

Explaining R_f values: a component that is strongly adsorbed onto the stationary phase (for polar aluminium oxide, a polar molecule, especially one that can hydrogen bond) moves slowly: low R_f. A component that is more soluble in the mobile phase and interacts less with the stationary phase moves further: high R_f.

37.2 Gas–liquid chromatography (GLC)

• Stationary phase: a high-boiling-point, non-polar liquid coated on a solid support, packed in or lining a long, thin, coiled column in an oven.
• Mobile phase: an unreactive carrier gas such as helium or nitrogen.
• Retention time: the time from injection to the component reaching the detector. Each component, at a fixed temperature and flow rate, has a characteristic retention time.

Explaining retention times: a component that is more soluble in the stationary liquid (interacts more strongly with it) is held back longer: longer retention time. With a non-polar stationary phase, larger, less volatile molecules with stronger id-id interactions are retained longest; volatile, small molecules emerge first.

Percentage composition: the area under each peak is proportional to the amount of that component. Percentage of a component = its peak area ÷ total of all peak areas × 100. (Where peaks are triangles, area ≈ ½ × base × height.)

37.3 Carbon-13 NMR

Each chemically different carbon environment gives one peak. Carbons are equivalent if they are in identical surroundings — related by the symmetry of the molecule. The position of the peak, its chemical shift δ (ppm), shows the type of environment; the data section gives ranges, for example sp³ alkyl 0–50, next to oxygen 50–70, alkene or arene 110–160, carboxyl or ester C=O 160–185, aldehyde or ketone C=O 190–220.

In ¹³C spectra (as usually recorded) peak height is not a reliable guide to the number of carbons, and there is no splitting to interpret: just count the peaks and read their shifts.

37.4 Proton (¹H) NMR

A proton NMR spectrum gives four pieces of information:

1. Number of peaks — the number of different proton environments.
2. Chemical shift, δ — the type of environment, from the data section ranges: alkyl 0.9–1.7; alkyl next to C=O 2.2–3.0; next to an electronegative atom (CH₂–O, CH₂–Cl) 3.2–4.0; on an alkene 4.5–6.0; on an aromatic ring 6.0–9.0; aldehyde 9.3–10.5; carboxylic acid 9.0–13.0. Electronegative atoms nearby move a peak to higher δ.
3. Relative peak area (integration) — the relative number of protons in each environment.
4. Splitting pattern — the number of protons on the adjacent carbon atom(s). By the n + 1 rule, a peak is split into n + 1 lines by n equivalent protons on neighbouring carbons: singlet (no neighbouring H), doublet (1), triplet (2), quartet (3), multiplet (more, or several different sets).

Protons on the same carbon, or equivalent protons, do not split each other. The classic ethyl pattern is a triplet (3H) and a quartet (2H): the CH₃ has two neighbours (CH₂) so it is a triplet; the CH₂ has three neighbours (CH₃) so it is a quartet.

TMS, solvents and D₂O

• Tetramethylsilane, TMS, Si(CH₃)₄, is the standard: its peak defines δ = 0. It is used because all 12 protons are equivalent (a single sharp peak), it absorbs upfield of almost all organic protons, it is inert and volatile (easily removed from the sample afterwards).
• Deuterated solvents, such as CDCl₃, are used because deuterium (²H) does not give a signal in a proton spectrum, so the solvent does not swamp the sample’s peaks.
• D₂O exchange: O–H and N–H protons have variable, sometimes broad peaks that are hard to assign. Shake the sample with D₂O: these protons exchange with deuterium (R–OH + D₂O ⇌ R–OD + HOD), and their peak disappears from a new spectrum. That identifies it. O–H protons usually appear as singlets, because rapid exchange stops them splitting or being split.

✏️Worked example

Compound X, C₃H₆O₂, has this proton NMR spectrum:
• δ 1.1, triplet, relative area 3
• δ 2.4, quartet, relative area 2
• δ 11.7, singlet, relative area 1; this peak disappears when D₂O is added.
(a) Identify X, explaining each peak. (b) How many peaks would its ¹³C spectrum show, and in which range would the peak for the C=O carbon appear? (c) Methyl ethanoate is an isomer of X. Predict its proton NMR spectrum (number of peaks, their areas and splitting).

(a)

• δ 11.7, singlet, 1H, removed by D₂O: an O–H proton; the shift is in the carboxylic acid range (9.0–13.0), so this is –COOH.
• δ 2.4, quartet, 2H: a CH₂ next to C=O (range 2.2–3.0), split into a quartet by three neighbouring H, a CH₃.
• δ 1.1, triplet, 3H: a CH₃ in an alkyl environment (0.9–1.7), split into a triplet by two neighbouring H, the CH₂.

Together: CH₃–CH₂–COOH. X is propanoic acid, which fits C₃H₆O₂, and the areas add to 6 H.

(b) Three carbons, all in different environments: 3 peaks. The carboxyl C=O appears in the 160–185 range.

(c) Methyl ethanoate, CH₃COOCH₃, has two CH₃ groups in different environments, and neither has H on the neighbouring atom (C=O on one side, O on the other): two singlets, each area 3. The CH₃ attached to O is at the higher shift, in the 3.2–4.0 range.

Check it. The areas must add up to the number of H in the formula (3 + 2 + 1 = 6), and the splitting must be mutually consistent: a triplet (2 neighbours) and a quartet (3 neighbours) always come as a pair, the signature of an ethyl group.

Counting the protons in the peak itself for the splitting. The n + 1 rule uses the H atoms on the adjacent carbon, not the ones in the group giving the peak. A CH₃ next to a CH₂ is a triplet, not a quartet. And the O–H is a singlet regardless of its neighbours.