Analytical techniques (A Level)

• Separate, then identify: Rf and retention time are evidence under identical conditions, never proof; the NMR spectrum is read in the order signals, areas, shifts, splitting — and n is always counted on the carbon next door.

• 37.1.1 I can describe — describe and understand the terms (a) stationary phase, for example aluminium oxide (on a solid support) (b) mobile phase; a polar or non-polar solvent (c) Rf value (d) solvent front and baseline
• 37.1.2 I can interpret — interpret Rf values
• 37.1.3 I can explain — explain the differences in Rf values in terms of interaction with the stationary phase and of relative solubility in the mobile phase
• 37.2.1 I can describe — describe and understand the terms (a) stationary phase; a high boiling point non-polar liquid (on a solid support) (b) mobile phase; an unreactive gas (c) retention time
• 37.2.2 I can interpret — interpret gas/liquid chromatograms in terms of the percentage composition of a mixture
• 37.2.3 I can explain — explain retention times in terms of interaction with the stationary phase
• 37.3.1 I can analyse — analyse and interpret a carbon-13 NMR spectrum of a simple molecule to deduce: (a) the different environments of the carbon atoms present (b) the possible structures for the molecule
• 37.3.2 I can predict — predict or explain the number of peaks in a carbon-13 NMR spectrum for a given molecule
• 37.4.1 I can analyse — analyse and interpret a proton (¹H) NMR spectrum of a simple molecule to deduce: (a) the different environments of proton present using chemical shift values (b) the relative numbers of each type of proton present from relative peak areas (c) the number of equivalent protons on the carbon atom adjacent to the one to which the given proton is attached from the splitting pattern, using the n + 1 rule (limited to singlet, doublet, triplet, quartet and multiplet) (d) the possible structures for the molecule
• 37.4.2 I can predict — predict the chemical shifts and splitting patterns of the protons in a given molecule
• 37.4.3 I can describe — describe the use of tetramethylsilane, TMS, as the standard for chemical shift measurements
• 37.4.4 I can state — state the need for deuterated solvents, e.g. CDCl₃, when obtaining a proton NMR spectrum
• 37.4.5 I can describe — describe the identification of O–H and N–H protons by proton exchange using D₂O

Units and significant figures are marked. The syllabus states that failure to quote units, the inclusion of units in quantities defined as ratios, and answers given to an inappropriate number of significant figures are all liable to be penalised. Give a calculated answer to the same number of significant figures as the least precise data, or one more; keep full precision in the working and round only at the end. A fifth of the qualification is experimental: Papers 3 and 5 test AO3 only, and their questions may be set in contexts outside the syllabus content, so the practical work in this chapter is set out as procedure, recording and evaluation rather than as theory.

• “The CH₃ of ethanol is a quartet because it has three protons.” Correct n is the number of protons on the adjacent carbon, and an O–H proton neither splits nor is split. The CH₃ sits next to a CH₂, so it is a triplet; the CH₂ sits next to the CH₃ (the O–H does not count), so it is a quartet; the O–H is a singlet.
• “Rf = distance moved by the spot ÷ the length of the plate.” Correct Divide by the distance moved by the solvent front. Both distances are measured from the baseline, the spot’s to its centre. Rf has no unit and can never exceed 1.
• “The spot with the high Rf is the polar one.” Correct On polar aluminium oxide, the polar component is held most strongly and moves least: low Rf. An explanation names both the interaction with the stationary phase and the solubility in the mobile phase.
• “Percentage composition = peak height ÷ total height × 100.” Correct Use peak areas. Later peaks are broader, so a short late peak can hold more material than a tall early one.
• “The same Rf (or retention time) proves it is the same compound.” Correct It is evidence, and only under identical conditions. Two different compounds can share a value; a second solvent or a second column settles it.
• “Methylbenzene has seven carbons, so seven carbon-13 peaks.” Correct Count environments, not atoms. The mirror plane through the CH₃ group makes two pairs of ring carbons equivalent: five peaks.
• “Methyl propanoate and ethyl ethanoate can be told apart by their splitting.” Correct Both give a 3H singlet, a 2H quartet and a 3H triplet. Only the shifts separate them: find the singlet and ask whether it lies in 3.2–4.0 (O–CH₃) or 2.2–3.0 (CH₃–C=O).
• “TMS is the solvent.” Correct TMS is the reference that defines δ = 0. The solvent is deuterated, such as CDCl₃, so that it gives no signal.
• “Rf = distance moved by the spot ÷ length of the plate.” Repair Divide by the distance moved by the solvent front, both measured from the baseline, the spot’s to its centre.
• “The Rf is 1.3.” Repair Impossible: a spot cannot move further than the solvent that carries it. Re-measure; check the front was marked at once and both distances start at the baseline.
• “I drew the baseline in pen and stood the plate in solvent up to 2 cm.” Repair Pencil, because ink runs with the solvent; and the solvent level must be below the baseline, or the sample dissolves into the pool.
• “A high Rf means the compound is polar.” Repair On a polar stationary phase the polar compound is adsorbed most strongly and has the low Rf.
• “The Rf of X is 0.4, so any spot at 0.4 is X.” Repair Only under identical conditions, with the standard on the same plate, and even then two compounds can share an Rf in one solvent.
• “In GLC the stationary phase is the gas.” Repair The gas is the mobile phase; the stationary phase is the high-boiling-point non-polar liquid on its solid support.
• “Percentage = peak height ÷ total height × 100.” Repair Peak areas. A short broad peak can hold more material than a tall narrow one.
• “The compound with the longest retention time is the most volatile.” Repair The least. It dissolves most in the stationary phase (strongest id–id forces) and spends least time in the gas.
• “Propanone has three carbons, so three carbon-13 lines.” Repair The two CH₃ carbons are equivalent by symmetry: two lines.
• “The tallest carbon-13 line is the environment with the most carbons.” Repair Carbon-13 line heights are not used. Read the number of lines and their shifts only.
• “The CH₃ of ethanol is a quartet: it has three protons.” Repair n is the number of protons on the adjacent carbon. The neighbouring CH₂ has two, so the CH₃ is a triplet.
• “Ethanol’s CH₂ is a multiplet because it is next to the CH₃ and the OH.” Repair The O–H proton exchanges too fast to split its neighbours: the CH₂ is a quartet and the O–H a singlet.
• “Integration 1 : 1 means one proton in each environment.” Repair It is a ratio; it could be 3 : 3. Scale it with the molecular formula.
• “TMS is the solvent.” Repair TMS is the reference, added in a trace to define δ = 0; the solvent is deuterated, such as CDCl₃.
• “Shaking with D₂O removes every signal.” Repair Only the O–H and N–H signals disappear; protons bonded to carbon do not exchange.

• Read the command word before you decide how much to write. This syllabus has twenty-two of them: analyse, calculate, compare, consider, contrast, deduce, define, demonstrate, describe, determine, discuss, evaluate, examine, explain, give, identify, justify, predict, show (that), sketch, state and suggest. Comment, estimate, name and outline are not among them: where a question wants something named it says identify, which the syllabus glosses as “name/select/recognise”. State and give want a fact and nothing more. Describe wants the points or the features. Explain wants the reasons and the relationships — a describe-level answer to an explain question is incomplete however well written it is. Deduce and determine want a conclusion reached from the information given, with the reasoning visible.
• A model sentence. “Q is more polar than P: it forms hydrogen bonds with the –OH groups of the aluminium oxide, so it interacts more strongly with the stationary phase, and it is less soluble in the non-polar solvent; it therefore spends a smaller fraction of its time in the mobile phase and has the lower Rf.”
• A complete explanation has four links. Name the force (id–id between the component and the non-polar stationary phase); say which component forms it more strongly and why (more electrons; a longer chain; more surface contact); conclude that this component dissolves more in the stationary phase and spends longer out of the gas; therefore it has the longer retention time.
• Interleave with the chapters that use this one. Topic 37 is the last topic, but topic 36 leans on it: when you revise a synthetic route, ask what carbon-13 and proton spectra would prove the product is the target. When you revise topics 13–21 and 29–35, predict the number of carbon-13 lines for each named compound. When you revise topic 22, re-answer worked example 6. Recalling a topic inside a new context is worth more than another pass over this chapter on its own; at A Level, Paper 4 assumes the whole of the AS content, so nothing here is ever finished with.

• Cambridge International AS & A Level Chemistry 9701 has five components. Topic 37 is A Level content, so it is examined in Papers 4 and 5. A Level content: examined in Paper 4 (A Level structured, which also requires the AS content) and, as practical context, Paper 5. AS Level candidates take Papers 1, 2 and 3; A Level candidates take all five, either staged over two years (Papers 1–3 in year one, Papers 4 and 5 in year two) or together in one series. Examinations are available in the June and November series, and in March in India.
• Across both the AS Level and the A Level the assessment objectives are weighted AO1 40% (knowledge and understanding), AO2 40% (handling, applying and evaluating information) and AO3 20% (experimental skills and investigations). AS candidates are graded a–e; A Level candidates A*–E.
• There is no multiple-choice paper on A Level content, so every Topic 37 mark is written. A Paper 4 question on this topic can hand you a plate, a chromatogram or a spectrum and asks you to calculate an Rf or a percentage, explain an Rf or a retention time in terms of the two phases, predict the number of carbon-13 peaks or a proton spectrum, or deduce a structure from several kinds of evidence, including the topic 22 infrared and mass spectra. The quick-check items in this chapter are retrieval practice, not Paper 1 items.
• Rf (no unit, two significant figures from distances read to 0.1 cm), percentage composition from peak areas, and the integration ratio of a proton spectrum. Data-section tables 6 (proton shifts) and 7 (carbon-13 shifts) are supplied in the examination together with the infrared table 8: you use them, you do not memorise them. Every shift quoted here is one of their ranges.
• Chromatography is not among the procedures the syllabus lists for Paper 3, so this topic reaches the practical fifth through Paper 5: planning a TLC identification (pencil baseline, solvent below it, closed jar, standards on the same plate, ruler read to 0.1 cm with a half-division uncertainty of 0.05 cm) and evaluating a GLC analysis (identical conditions, provisional identification, areas not heights, equal detector response assumed).
• Read the command word before you decide how much to write. This syllabus has twenty-two of them: analyse, calculate, compare, consider, contrast, deduce, define, demonstrate, describe, determine, discuss, evaluate, examine, explain, give, identify, justify, predict, show (that), sketch, state and suggest. Comment, estimate, name and outline are not among them: where a question wants something named it says identify, which the syllabus glosses as “name/select/recognise”. State and give want a fact and nothing more. Describe wants the points or the features. Explain wants the reasons and the relationships — a describe-level answer to an explain question is incomplete however well written it is. Deduce and determine want a conclusion reached from the information given, with the reasoning visible.