OCR A Level Chemistry A: Chromatography and qualitative analysis (H432)

Section 6.3.1, on chromatography and qualitative analysis, is taught here from scratch, following the H432 Chemistry A specification from OCR (Version 3.1, May 2026, for first assessment in 2017). The section belongs to Module 6, Organic chemistry and analysis. All three lettered outcomes, with their additional guidance and the PAG6 and PAG7 links, are covered here. H432 is the full A level. Its assessment overview lists Module 6 under component 02 (Synthesis and analytical techniques); component 03 (Unified chemistry) draws on it too.

For the whole course, start at the OCR chemistry hub. Record your progress statement by statement on the printable checklist; the free diagnostics show where your gaps are.

What 6.3.1 asks of you

Statement	What you must be able to do	Additional guidance (specification)
6.3.1(a)	Interpret one-way TLC chromatograms in terms of Rf values	M3.1; PAG6; HSW3
6.3.1(b)(i)	Interpret gas chromatograms in terms of retention times	M3.1, M3.2; HSW3
6.3.1(b)(ii)	Use gas chromatograms to find the amounts and proportions of the components in a mixture	Includes making and using external calibration curves to confirm concentrations; peak integration values will be supplied
6.3.1(c)(i)–(vii)	Identify alkenes, haloalkanes, phenols, carbonyl compounds, aldehydes, primary and secondary alcohols, and carboxylic acids by test-tube tests	PAG7; HSW4; cross-references to 4.1.3(f), 4.2.2(a), 6.1.1(h), 6.1.2(a), (d), (e), 4.2.1(c) and 6.1.3(b)

M3.1 is translating between graphical, numerical and algebraic forms; M3.2 is plotting two variables from data, as in a calibration graph. PAG6 (synthesis of an organic solid) includes thin-layer or paper chromatography. PAG7 is qualitative analysis of organic functional groups.

How chromatography separates a mixture

Every chromatography method has two phases:

• a stationary phase, which stays put;
• a mobile phase, which moves through or over the stationary phase and carries the mixture with it.

Components held more strongly by the stationary phase move more slowly, so the mixture spreads out.

6.3.1(a): One-way thin-layer chromatography

In TLC the stationary phase is a thin layer of a solid, usually silica gel or alumina, on a glass, metal or plastic sheet. The mobile phase is a solvent. Components are held on the surface of the solid by adsorption.

“One-way” means the plate is developed once, in one direction, with one solvent.

Running a plate (PAG6)

1. Mark a baseline in pencil roughly 1 cm above the lower edge (ink would run).
2. Spot the sample and pure references on the line with a fine capillary.
3. Stand the plate in a covered tank, solvent below the baseline.
4. Remove the plate before the solvent reaches the top and mark the solvent front at once.
5. Locate colourless spots under an ultraviolet lamp or with iodine vapour, and ring them in pencil.

Rf value

Rf = distance moved by the component ÷ distance moved by the solvent front

Measure both from the baseline; for the spot, use its centre. Rf is a ratio, so it has no units, and it can never exceed 1.

On a polar silica plate, a more polar component is adsorbed more strongly and has a lower Rf. Rf depends on the plate, solvent and temperature, so run references on the same plate.

Worked example 1: following a reduction by TLC

A student reduces a ketone with NaBH₄ (6.1.2(b)). She spots pure ketone, pure alcohol product and the reaction mixture after 30 minutes. The solvent front is 6.40 cm above the baseline.

• Ketone reference: spot centre at 4.16 cm
• Alcohol reference: spot centre at 1.28 cm
• Reaction mixture: spots at 1.28 cm and 4.16 cm

Step 1. Calculate the Rf values. Ketone: 4.16 ÷ 6.40 = 0.65. Alcohol: 1.28 ÷ 6.40 = 0.20.

Step 2. Match the spots. The mixture lane has spots with Rf 0.20 and 0.65, the same as the two references run on the same plate.

Step 3. Conclude. Product and unreacted ketone are both present, so the reaction is not finished.

Step 4. Explain. The alcohol’s O–H group hydrogen bonds to the silica, so it is adsorbed more strongly and has the lower Rf.

Limitation. Two compounds can share an Rf value, so a match supports an identification but does not prove it.

6.3.1(b): Gas chromatography

In GC the mobile phase is an unreactive carrier gas, such as helium or nitrogen. The stationary phase is a high-boiling liquid on a solid support inside a long, thin column in an oven. The sample is injected, vaporised and carried through the column to a detector. The chromatogram shows detector response against time.

(b)(i) Retention time

The retention time is how long a component takes, from the moment of injection, to reach the detector. Read it at the top of the peak. A component that is more soluble in the liquid stationary phase, or less volatile, spends longer in the column and has a longer retention time.

Retention time also depends on the column, oven temperature and carrier gas flow rate. To identify a peak, compare it with a pure standard run under the same conditions. Two compounds can share a retention time.

(b)(ii) Amounts and proportions

Each peak’s area shows how much of that component the sample contains. The instrument integrates the peaks and gives you the areas.

Percentage of a component = (its peak area ÷ total of all peak areas) × 100

This simple method assumes the detector responds equally to each component. For an exact concentration you need a calibration curve.

Worked example 2: proportions in a fruit flavouring

A flavouring gives three peaks. Standards run under the same conditions identify them as ethyl ethanoate, ethyl butanoate and pentyl ethanoate. The integration values are 1840, 3220 and 1140.

Step 1. Total the areas. 1840 + 3220 + 1140 = 6200

Step 2. Find each percentage.

• Ethyl ethanoate: 1840 ÷ 6200 × 100 = 29.7%
• Ethyl butanoate: 3220 ÷ 6200 × 100 = 51.9%
• Pentyl ethanoate: 1140 ÷ 6200 × 100 = 18.4%

External calibration curves

An external calibration uses standards prepared and run separately from the sample.

1. Make up solutions of the pure compound at several known concentrations that span the expected value.
2. Run each standard under identical conditions and record its peak area.
3. Plot peak area (y-axis) against concentration (x-axis) and draw the line of best fit.
4. Inject the sample in the same way and find its concentration from the line.
5. If the sample was diluted before injection, multiply by the dilution factor.

Worked example 3: limonene in a citrus extract

Limonene standard / g dm⁻³	0.50	1.00	1.50	2.00
Peak area	1210	2400	3630	4810

A citrus extract was diluted by a factor of 5 and injected. Its limonene peak area was 2950.

Step 1. Plot and fit. The points fall close to a straight line through the origin; the best-fit equation is area = 2406 × concentration + 5.

Step 2. Read off. Concentration = (2950 − 5) ÷ 2406 = 1.22 g dm⁻³ in the diluted sample. On a graph, read across from 2950 to the line and down to the x-axis.

Step 3. Allow for dilution. Using the unrounded value, 1.224 × 5 = 6.12 g dm⁻³ in the original extract.

The unknown lies inside the range of the standards; reading beyond them would mean extrapolating, which is less reliable.

6.3.1(c): Tests for organic functional groups

These tests use a test-tube scale: a few drops of liquid or a little solid, with a few cm³ of reagent. Key techniques for PAG7:

• Take a new sample of the unknown for every test.
• Many organic liquids are flammable. Warm them in a hot water bath or with an electric heater, never over a naked flame.
• Record observations precisely (“orange solution turns green”), then state the inference.

Spec	Group	Reagent and conditions	Positive result
(c)(i)	Alkene C=C	Bromine water; shake, no heating needed	Orange solution decolourises
(c)(ii)	Haloalkane	Aqueous silver nitrate in ethanol, warm in a water bath	Precipitate: white (Cl), cream (Br), yellow (I)
(c)(iii)	Phenol	NaOH(aq), then separately Na₂CO₃(aq)	Reacts with NaOH; no effervescence with carbonate
(c)(iv)	Carbonyl (aldehyde or ketone)	2,4-DNP	Yellow-orange precipitate
(c)(v)	Aldehyde	Tollens’ reagent; warm (water bath)	Silver mirror
(c)(vi)	Primary or secondary alcohol, aldehyde	Acidified dichromate, warm in a water bath	Orange solution turns green
(c)(vii)	Carboxylic acid	Na₂CO₃(aq)	Effervescence; gas turns limewater milky

(i) Alkenes. Bromine adds across the C=C bond, for example CH₂=CHCH₃ + Br₂ → CH₂BrCHBrCH₃. Phenol also decolourises bromine water, but it gives a white precipitate of 2,4,6-tribromophenol as well (6.1.1(i)).

(ii) Haloalkanes. Water in the reagent hydrolyses the C–X bond, releasing halide ions: R–X + H₂O → R–OH + H⁺ + X⁻. The silver ions then precipitate the halide: Ag⁺(aq) + X⁻(aq) → AgX(s). Ethanol is the common solvent, because haloalkanes do not mix with water. Iodoalkanes give a precipitate fastest, because the C–I bond is the weakest.

(iii) Phenols. Phenol is a weak acid, neutralised by NaOH: C₆H₅OH + NaOH → C₆H₅ONa + H₂O. It is too weak to react with carbonate, so there is no fizzing. This separates a phenol from a carboxylic acid.

(iv) Carbonyl compounds. 2,4-dinitrophenylhydrazine gives a yellow-orange precipitate with aldehydes and ketones. The equation and the derivative’s structure are not required. To identify the compound, filter off the precipitate, recrystallise it, dry it and measure its melting point. A sharp melting point matching data for one derivative identifies the compound.

(v) Aldehydes. Tollens’ reagent is ammoniacal silver nitrate, made fresh. The aldehyde becomes a carboxylic acid (RCHO + [O] → RCOOH). Silver ions are reduced to silver, Ag⁺ + e⁻ → Ag, which coats the tube as a mirror. Ketones are not oxidised, so there is no change.

(vi) Primary and secondary alcohols and aldehydes. Warm with K₂Cr₂O₇/H₂SO₄. Orange Cr₂O₇²⁻ is reduced to green Cr³⁺. Tertiary alcohols and ketones resist oxidation, so the solution stays orange. Primary and secondary alcohols both turn it green; to tell them apart, distil off the product and test it with Tollens’ (an aldehyde means primary).

(vii) Carboxylic acids. For example, 2C₄H₉COOH + Na₂CO₃ → 2C₄H₉COONa + CO₂ + H₂O. Confirm the gas is carbon dioxide with limewater.

Worked example 4: sorting four colourless liquids

Bottles A to D hold butanal, butanone, butan-2-ol and 2-methylpropan-2-ol, in some order.

1. 2,4-DNP. A and C give yellow-orange precipitates, so they are the carbonyl compounds. B and D give no precipitate.
2. Tollens’ on A and C. C gives a silver mirror; A does not. C is butanal and A is butanone.
3. Acidified dichromate on B and D. B turns green; D stays orange. B is butan-2-ol (secondary) and D is 2-methylpropan-2-ol (tertiary).

Common errors

• Measuring to the top of a spot, or from the bottom of the plate, instead of from baseline to spot centre.
• Using peak height instead of peak area for amounts.
• Forgetting the dilution factor after reading a calibration graph.
• Saying 2,4-DNP shows an aldehyde. It shows a C=O in an aldehyde or ketone; Tollens’ then decides.
• Expecting dichromate to separate primary from secondary alcohols.
• Writing “turns clear” for bromine water instead of “orange to colourless”.
• Warming flammable organic liquids over a Bunsen flame.

Next steps

• Condensed tables, method boxes and a self-test: Chromatography and qualitative analysis revision notes
• Original questions with mark-by-mark answers: Chromatography and qualitative analysis practice questions
• How PAG6 and PAG7 fit the Practical Endorsement: Practical skills for the Practical Endorsement

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 6.3.1 Chromatography and qualitative analysis, learning outcomes (a), (b)(i)–(ii) and (c)(i)–(vii), with their additional guidance, PAG6 and PAG7. Published by OCR.