Cambridge International AS & A Level Chemistry 9701 · A Level topic 37

Analytical techniques (A Level)

What this chapter covers37.1–37.4

Forensic scientists identify drugs in a blood sample; food chemists check a flavouring for contaminants; pharmaceutical chemists confirm that a medicine is pure and has the structure intended. Two families of technique make this possible. Chromatography separates the components of a mixture and helps identify them. Nuclear magnetic resonance (NMR) spectroscopy reveals how the carbon and hydrogen atoms in a single compound are arranged. Used together with the mass and infrared spectroscopy of the AS course, they allow the structure of an unknown organic compound to be worked out.

Thin-layer chromatography37.1.1–37.1.3

Every kind of chromatography separates a mixture by distributing its components between two phases. The stationary phase stays in place; the mobile phase moves through or over it and carries the components with it. A component that is attracted strongly to the stationary phase, or that is not very soluble in the mobile phase, spends more time held back and moves slowly; a component that dissolves well in the mobile phase and interacts weakly with the stationary phase moves quickly.

In thin-layer chromatography (TLC) the stationary phase is a thin layer of a solid such as aluminium oxide or silica (silicon dioxide), spread on a solid support such as a glass or plastic sheet. The mobile phase is a solvent, which may be polar (for example water or ethanol) or non-polar (for example hexane).

1. A pencil line — the baseline — is drawn near the bottom of the plate. Pencil is used because ink would itself dissolve and separate.
2. A small spot of the mixture, and spots of known reference compounds, are placed on the baseline.
3. The plate stands in a covered chamber containing a shallow layer of solvent, below the baseline so that the spots do not dissolve into the solvent reservoir. The lid keeps the chamber saturated with solvent vapour, so the solvent does not evaporate from the plate.
4. The solvent rises up the plate, carrying the components different distances. The plate is removed before the solvent reaches the top, and the position of the solvent front is marked at once.
5. Colourless spots are made visible, for example under ultraviolet light when the plate contains a fluorescent substance, or with a chemical locating agent.

R_f values

R_f has no units and lies between 0 and 1. For a given stationary phase, solvent and temperature it is characteristic of a compound, so an unknown can be identified by comparing its R_f with those of known substances run under the same conditions — ideally on the same plate. Because different compounds can have similar R_f values in one solvent, running the plate in a second solvent, and matching in both, makes an identification much more certain.

Explaining R_f values

Differences in R_f are explained by the balance between two interactions:

• adsorption on the stationary phase — silica and alumina have polar surfaces, so polar compounds (with –OH, –COOH, –NH₂ groups, which form hydrogen bonds or strong dipole attractions) are held more strongly and have small R_f values;
• solubility in the mobile phase — a compound that dissolves well in the solvent is carried further and has a large R_f.

Changing the solvent changes the balance. With a non-polar solvent on a polar plate, non-polar compounds move furthest; a more polar solvent competes for the polar surface and moves polar compounds further too.

Gas–liquid chromatography37.2.1–37.2.3

In gas–liquid chromatography (GLC) the mobile phase is an unreactive carrier gas such as nitrogen, helium or argon. The stationary phase is a high-boiling, non-polar liquid (for example a long-chain alkane) coated on a solid support inside a long, thin, coiled column. The column is kept in an oven, so the sample is in the gas phase. A small sample is injected; the carrier gas sweeps its components through the column; each component reaches the detector at the end at a different time, giving a peak on the chromatogram.

Retention time

The retention time is the time between injecting the sample and detecting a component. Components that interact more strongly with the stationary liquid — because they are more soluble in it or have stronger intermolecular forces with it — and components that are less volatile spend longer dissolved in the stationary phase and have longer retention times. Under fixed conditions (column, temperature, gas flow rate) the retention time is characteristic of a compound, so an unknown is identified by comparing its retention time with that of a known sample run under the same conditions.

Percentage composition from a chromatogram

The area under each peak is proportional to the amount of that component. For roughly triangular peaks the area is found as ½ × base × height; when all peaks have the same width, the heights alone can be compared.

Past-paper practice · Set 37A · Thin-layer and gas–liquid chromatography

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 37A.1[3]

Answer and marking guidance

(i) J (CH₃COCO₂H) — spot 2; K (HO₂CCO₂H) — spot 3; L (CH₃CH₂COCH₂CH₃) — spot 1 ✔. (ii) the more polar the compound, the stronger its attraction to the polar stationary phase, so the shorter the distance it moves (or the reverse for less polar compounds) ✔. (iii) R_f = distance moved by the compound from the baseline ÷ distance moved by the solvent front ✔. Examiner insight: (i) discriminated well. (ii) was not well answered: the interactions between the stationary phase and the compounds were often not described, and some explained the separation by molecular mass. In (iii) many answers were imprecise — "distance of solute" rather than "distance travelled by the solute from the starting line", or no clear statement of which distance is divided by which.

Question 37A.2[2]

Answer and marking guidance

(i) SiO₂ or Al₂O₃ (silica or alumina) ✔. (ii) Cd²⁺: the R_f of M is 2.0 ÷ 5.0 = 0.40, the same as the value for Cd²⁺ in Table 6.1 ✔. Examiner insight: (i) was answered well; ethanol was a common error. (ii) was often correct.

Question 37A.3[2]

Answer and marking guidance

(i) R_f = distance travelled by the amino acid ÷ distance travelled by the solvent (front) ✔. (ii) tyrosine is more soluble in the solvent (mobile phase) used, or lysine is more attracted to the stationary phase ✔. Examiner insight: in (i) careful description was essential — the answer had to make clear which distance is divided by which; a labelled sketch helped. (ii) was found difficult: imprecise answers such as "different solubility" were common. Because lysine has the much smaller R_f, the answer needed to say that tyrosine is more soluble in the mobile phase or that lysine is more attracted to the stationary phase.

Question 37A.4[3]

Answer and marking guidance

(f)(i) TLC stationary phase: aluminium oxide or silica on a solid support; GLC mobile phase: an inert gas, e.g. N₂ ✔. (ii) S — its R_f is the same as that of the unknown amino acid in both solvents ✔. (g) mass of L = (58 ÷ 44) × 5.52 × 10⁻² = 7.28 × 10⁻² g dm⁻³; concentration = 7.28 × 10⁻² ÷ 116 = 6.27 × 10⁻⁴ mol dm⁻³ ✔. Examiner insight: (f)(i) discriminated well; some described paper chromatography rather than TLC. In (f)(ii) the best answers referred correctly to R_f values. (g) was found very difficult: it needed the concentration of L in g dm⁻³ first, and then its conversion to mol dm⁻³.

Question 37A.5[4]

Answer and marking guidance

(i) stationary: a non-volatile (non-polar) liquid; mobile: nitrogen, argon or any inert gas ✔. (ii) the time between injection and detection — the time the substance stays in the column ✔. (iii) using peak heights (or ½ × base × height), areas 30, 100, 25: % of B = 100 × 100 ÷ 155 = 64.5% ✔✔ (73.2% if areas 30, 150, 25 are used). Examiner insight: in (i) most identified a suitable mobile phase, but the stationary phase was less well known. The definition of retention time was well known. Candidates performed well in (iii), most often using areas.

Question 37A.6[3]

Answer and marking guidance

(i) mobile phase: ethyl ethanoate; stationary phase: SiO₂ (silica) or Al₂O₃ (alumina) ✔. (ii) R_f = distance travelled by the solute ÷ distance travelled by the solvent (front) ✔. (iii) X is more attracted to (adsorbed more strongly by) the stationary phase, or lidocaine dissolves better in the solvent ✔. Examiner insight: in (i) many identified ethyl ethanoate as the mobile phase; the stationary phase was less well known. (ii) was mostly correct. (iii) was challenging: "X is less soluble in the mobile phase" was a typical inadequate explanation.

How NMR works37.3, 37.4.3

Some atomic nuclei — those with an odd number of protons or of neutrons, such as ¹H and ¹³C — have a property called nuclear spin and behave like tiny magnets. Placed in a strong external magnetic field, such a nucleus can line up with the field (lower energy) or against it (higher energy). Radio-frequency radiation of exactly the right energy, ΔE, flips nuclei from the lower to the higher state; when the energy supplied matches ΔE the nuclei absorb it and flip back and forth — they are in resonance. The common isotope ¹²C has no spin and gives no signal.

The electrons around a nucleus shield it from the external field, so the field it actually experiences — and therefore the energy it absorbs — depends on its chemical environment: the atoms and bonds around it. Electronegative atoms nearby draw electron density away, reducing the shielding. Each different environment therefore gives a signal at a slightly different position.

Chemical shift and TMS

Signal positions are measured relative to a standard, tetramethylsilane (TMS), Si(CH₃)₄, as the chemical shift, δ, in parts per million (ppm). TMS is used because:

• all twelve of its hydrogen atoms, and all four carbon atoms, are in one environment, so it gives a single, sharp peak, which is defined as δ = 0;
• its signal lies away from (below) almost all the signals of organic compounds;
• it is chemically inert, non-toxic and volatile, so it does not react with the sample and is easily removed.

Carbon-13 NMR37.3.1, 37.3.2

Only about 1% of carbon atoms are ¹³C, so a molecule rarely contains more than one of them and the ¹³C signals are not split by neighbouring carbon atoms. The spectrum is a set of single lines, and the key information is:

• the number of peaks = the number of different carbon environments;
• the chemical shift of each peak, which shows the type of carbon (Table 37.1).

Carbon atoms are in the same environment — and give one peak — only if they are related by the symmetry of the molecule, i.e. they are bonded to identical groups in identical arrangements. In propanone the two CH₃ carbons are equivalent, so there are two peaks; in butanone the two CH₃ carbons are not (one is bonded to C=O, the other to CH₂), so there are four. Peak heights in ¹³C spectra are not used to count carbon atoms.

Past-paper practice · Set 37B · Carbon-13 NMR

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 37B.1[2]

Parts (a), (c) and (d) of this question are 37A.5 and 37C.1.

Answer and marking guidance

CH₃CH₂CH₂CO₂CH₃: 5; CH₃CH₂CO₂CH₂CH₃: 5; CH₃CO₂CH₂CH₂CH₃: 5; (CH₃)₂CHCO₂CH₃: 4; CH₃CO₂CH(CH₃)₂: 4 ✔✔. Examiner insight: many gained credit; the most common error was giving 3 or 4 peaks for CH₃CH₂CO₂CH₂CH₃.

Question 37B.2[2]

Answer and marking guidance

δ 27: the CH₃ carbon circled, sp³; δ 163: the COOH carbon circled, sp²; δ 192: the ketone C=O carbon circled, sp² — one mark for each correct column ✔✔.

Question 37B.3[3]

Answer and marking guidance

(ii) HCO₂H: proton 2, carbon-13 1; HO₂CCO₂H: proton 1, carbon-13 1; HO₂CCH₂CH₂CO₂H: proton 2, carbon-13 2 — three to five correct 1 mark, six correct 2 marks ✔✔. (iii) the O–H peak disappears AND the proton exchanges with deuterium ✔. Examiner insight: (ii) and (iii) discriminated well. The clearest way to show the effect of D₂O is the equation HCO₂H + D₂O → HCO₂D + HDO; it was also necessary to say that the O–H peak disappears.

Chemical shift in proton NMR37.4.1, 37.4.2

Proton NMR works on the same principle as carbon-13 NMR, but it detects ¹H nuclei, which make up almost all hydrogen atoms. A proton NMR spectrum carries four pieces of information, and a complete answer uses all four:

1. the number of peaks — the number of different hydrogen environments;
2. the chemical shift of each peak — the type of hydrogen (Table 37.2);
3. the relative peak area — the relative number of hydrogen atoms in each environment;
4. the splitting pattern of each peak — the number of hydrogen atoms on the adjacent carbon atom(s).

Hydrogen atoms bonded to the same carbon atom, or to carbon atoms related by symmetry, are equivalent. The two CH₃ groups of propanone give one peak; ethanol, CH₃CH₂OH, has three environments.

A hydrogen atom close to an electronegative atom (O, N, Cl) or to a C=O group or a benzene ring is deshielded: electron density is drawn away from it, it experiences more of the external field and its signal appears at a larger δ. Hydrogen atoms in an alkane chain, surrounded only by C–H and C–C bonds, are the most shielded and appear at small δ.

A table like this is printed in the data booklet or with the question, so you do not need to learn the numbers — but you must be able to read it. Check which hydrogen the table refers to: "alkyl next to C=O" means the CH₃ of CH₃CO–, not the C=O itself; "attached to aromatic ring" means H–Ar, while CH₃–Ar is "alkyl next to aromatic ring".

Relative peak areas37.4.1

The area under each peak is proportional to the number of hydrogen atoms responsible for it. The spectrometer usually prints an integration trace — a stepped line whose step heights are proportional to the peak areas — or states the relative areas as numbers. The areas give only a ratio: ratios of 1 : 1 could mean 1 H and 1 H, or 2 H and 2 H. Use the molecular formula to scale the ratio to the actual numbers of hydrogen atoms.

Splitting: the n + 1 rule37.4.1, 37.4.2

The small magnetic field of each proton on a neighbouring carbon atom can be aligned with or against the external field, slightly increasing or decreasing the field felt by the proton being observed. A signal is therefore split into a cluster of lines. For n equivalent hydrogen atoms on the adjacent carbon atom(s), the signal is split into n + 1 lines.

Three points avoid most errors:

• Count the H on the neighbouring carbon atoms, not on the carbon itself. In CH₃CH₂–, the CH₃ signal is a triplet (it has two neighbours) and the CH₂ signal is a quartet (three neighbours). A triplet–quartet pair therefore signals an ethyl group.
• Equivalent protons do not split one another, and protons separated by an O atom or a C=O group (no H on the carbon in between) are not split by each other: the CH₃ of CH₃CO– and of CH₃O– gives a singlet.
• O–H and N–H protons usually give singlets and do not split the signals of neighbouring C–H protons, because they exchange rapidly between molecules.

Use the names singlet, doublet, triplet, quartet and multiplet. Invented names — "duplet", "quadret", "hextet" — are not credited.

Solvents, TMS and exchange with D₂O37.4.3–37.4.5

Samples are dissolved in a solvent that contains no ¹H atoms, so that the solvent itself gives no peak that would swamp the spectrum. Deuterated solvents such as CDCl₃ are used: deuterium, ²H or D, does not absorb in the region used for proton NMR. The solvent is not the reference — a little TMS is added for that.

The peaks of O–H and N–H protons are hard to assign from chemical shift alone, because their ranges are wide and overlap with others. They are identified by proton exchange with D₂O. Two spectra are run: one in CDCl₃, and one after shaking the sample with a few drops of D₂O. The labile O–H or N–H hydrogen atoms exchange with deuterium, for example

and because D gives no signal, the O–H (or N–H) peak disappears from the second spectrum. C–H protons do not exchange, so every other peak is unchanged. A full answer states which peak disappears and why — the H is replaced by D, which does not absorb. The exchange happens with D₂O, not with CDCl₃: in CDCl₃ the O–H and N–H peaks are present.

Past-paper practice · Set 37C · Proton NMR spectroscopy

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 37C.1[2]

Part (b) of the same question is 37B.1.

Answer and marking guidance

4 peaks ✔; splitting patterns: singlet, (two) triplets and a multiplet ✔. Examiner insight: many gave the correct number of peaks, but fewer stated the splitting patterns; the commonest error was a "hextet" instead of a multiplet.

Question 37C.2[5]

Answer and marking guidance

(i) CDCl₃ does not give a signal (peak), whereas CHCl₃ would ✔. (ii) CH₃–C=O δ 1.9–2.1 and N–H δ 6.5–7 ✔; N–CH₂ δ 3–3.5 and CH₃ of the ethyl group δ 1–1.5 ✔. (iii) the peak at δ 6.6–6.8, due to N–H, disappears ✔; the H exchanges with D ✔. Examiner insight: (i) discriminated well; "CDCl₃ gives the reference peak at zero" and "CDCl₃ is a better solvent" were common wrong answers. In (ii) some gave δ values not associated with any peak in the spectrum. In (iii) some described a peak disappearing without saying which peak; H/D exchange of a labile hydrogen was seen in many answers.

Question 37C.3[3]

Answer and marking guidance

δ 1.3: CH₃ (alkane), triplet, 3 H; δ 2.2: CH₃CO (alkyl next to C=O), singlet, 3 H; δ 4.0: CH₂O (alkyl next to an electronegative atom), quartet, 2 H — one mark for every three correct entries ✔✔✔.

Question 37C.4[3]

Answer and marking guidance

Alanine, because glutamic acid would give more than two (three) peaks / proton environments ✔; alanine's CH₃ gives a doublet because there is one proton on the neighbouring carbon ✔; glutamic acid would give triplet(s) or a multiplet, or alanine's CH gives a quartet because of three neighbouring protons ✔. Examiner insight: the argument that glutamic acid would show three peaks, or a triplet, was rarely seen, but the explanations of the doublet and quartet were often clear. Imprecise wording did not score: "the doublet is caused by protons attached to CH" is weaker than "the doublet is caused by a –CH₃ group with one proton on the neighbouring carbon atom".

Question 37C.5[4]

The chemical-shift table printed with this question is Table 37.2 in this chapter.

Answer and marking guidance

(i) δ 2.6: quartet AND δ 1.1: triplet ✔. (ii) all three for 2 marks ✔✔: δ 7.1 — H attached to the aromatic ring (H–Ar); δ 3.0 — the CH₂ next to C=O; δ 2.3 — the CH₃ groups attached to the ring (CH₃–Ar). (iii) 9 ✔. Examiner insight: (i) was answered well. In (ii) most identified δ 7.1 as H–Ar ("amide" was a common error), but many reversed the assignments of δ 3.0 and δ 2.3. (iii) discriminated well; 6 and 8 were common wrong answers.

Question 37C.6[4]

The chemical-shift table printed with this question is Table 37.2 in this chapter.

Answer and marking guidance

(i) δ 1.4: doublet, 3 H, 1 adjacent H; δ 3.5: singlet, 3 H, 0; δ 4.0: quartet, 1 H, 3 — three correct 1 mark, six 2, nine 3 ✔✔✔. (ii) in CDCl₃ there is one extra peak, for the NH₂ group, because in D₂O the N–H protons exchange for D ✔. Examiner insight: in (i) variant spellings such as "duplet" and "quadret" were not credited. In (ii) many described what happens when D₂O is added, which was not asked; others wrongly said that H–D exchange happens with CDCl₃.

Question 37C.7[3]

Answer and marking guidance

(i) propanoic acid ✔. (ii) propan-1-ol would have a peak at δ 0.5–6.0 for its O–H ✔; propanal would have a peak at δ 9.3–10.5 for its CHO proton ✔. Examiner insight: (i) was mostly credited. In (ii) the clearest answers used the distinctive O–H and CHO chemical shifts; answers based on splitting patterns were possible but harder to word correctly.

Misconceptions and how the topic is assessed37.1–37.4

Self-test37.1–37.4

Ten questions on the whole chapter. Each gives its reason once you answer.

Summary

• Chromatography separates components between a stationary and a mobile phase; stronger attraction to the stationary phase or lower solubility in the mobile phase slows a component.
• TLC: silica or alumina on a solid support, a polar or non-polar solvent; R_f = spot distance ÷ solvent-front distance, both from the baseline.
• GLC: high-boiling non-polar liquid on a support, an inert carrier gas; retention time = injection to detection; peak area ∝ amount.
• NMR detects nuclei with spin, such as ¹H and ¹³C; chemical shift δ is measured from TMS (δ = 0) and increases near electronegative atoms, C=O and rings.
• Carbon-13: number of peaks = number of carbon environments; shift = type of carbon.
• Proton: number of peaks, shift, relative area and splitting (n + 1) together give the structure.
• Deuterated solvents give no proton peak; D₂O exchange removes O–H and N–H peaks.

Examination checklist

• Can I name the stationary and mobile phases in TLC and GLC?
• Can I define R_f and retention time precisely?
• Can I explain differences in R_f or retention time by interactions with the two phases?
• Can I calculate a percentage composition, and a concentration, from a chromatogram?
• Can I count carbon and proton environments in an unsymmetrical molecule?
• Can I assign each proton peak its type, number of H, splitting pattern and number of adjacent H?
• Can I explain why CDCl₃ and TMS are used, and what D₂O does?

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