Edexcel International A Level Chemistry: Spectroscopy and chromatography (YCH11) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

Topic 15E: Spectroscopy and chromatography, outcomes 15.17 to 15.23, is tested here, following Pearson Edexcel’s International Advanced Level in Chemistry (YCH11) specification, Issue 1, September 2017. Being IA2 material, it appears in Unit 4 (Rates, Equilibria and Further Organic Chemistry). To learn it first, use the study guide or revision notes. Other topics are on the Edexcel IAL Chemistry hub and its outcome checklist; our free diagnostics show what to practise first.

Data for all questions. Accurate relative atomic masses: H 1.0078, C 12.0000, O 15.9949, N 14.0031.

¹³C environment	δ / ppm
Alkyl C	5–45
C bonded to O	50–90
Carbonyl C (acid, ester)	160–185
Carbonyl C (aldehyde, ketone)	190–220

¹H environment	δ / ppm
Alkyl C–H	0.7–1.7
H on C bonded to C=O	2.0–2.7
H–C bonded to O or Cl	3.3–4.5
Alcohol O–H	1.0–5.0

Questions

1. State what is meant by the mobile phase and the stationary phase in chromatography. [2]

2. A compound gives a molecular ion at m/z = 87.0681. Three possible molecular formulae are C₄H₉NO, C₅H₁₃N and C₃H₅NO₂.

(a) Calculate the accurate relative molecular mass of each formula. [3] (b) Identify the molecular formula of the compound. [1]

3. Butan-2-ol, 2-methylpropan-1-ol and ethoxyethane all have the formula C₄H₁₀O.

(a) How many peaks would each isomer give in its ¹³C NMR spectrum? [3] (b) Justify your answer for 2-methylpropan-1-ol. [1]

4. (a) Give the number of peaks and their relative areas in the low resolution ¹H NMR spectrum of 2,2-dimethylpropan-1-ol, (CH₃)₃CCH₂OH. [2] (b) For 1,3-dichloropropane, ClCH₂CH₂CH₂Cl, give the number of peaks, their area ratio and the splitting of each peak in the high resolution ¹H NMR spectrum. [3]

5. Predict the high resolution ¹H NMR spectrum of 3-chlorobutan-2-one, CH₃COCHClCH₃. For each environment give the δ range from the data, the relative area and the splitting. [4]

6. Ester W has molecular formula C₆H₁₂O₂. Its high resolution ¹H NMR spectrum is:

δ / ppm	Splitting	Relative area
1.15	doublet	6
1.25	triplet	3
2.5	septet	1
4.1	quartet	2

Deduce the structure of W. Explain how each peak supports your answer. [5]

7. An orange food colouring is run on a TLC plate beside four dyes, A to D. The solvent front reaches 7.5 cm from the baseline. The colouring separates into three spots at 1.8 cm, 3.3 cm and 5.1 cm. The Rf values of the dyes, measured on the same plate, are A 0.24, B 0.44, C 0.53, D 0.68.

(a) Work out Rf for each of the three spots. [2] (b) Identify the dyes in the colouring. [1] (c) The plate is coated with silica. Suggest why dye A has the lowest Rf value. [2]

8. A student finds that a pure dye has Rf 0.62 on paper with one solvent and 0.18 with a different solvent.

(a) Explain why the Rf value changes. [2] (b) State why a known sample should be run on the same sheet as the unknown. [1]

9. A sports anti-doping laboratory analyses a urine sample by HPLC linked to a mass spectrometer.

(a) Suggest why HPLC rather than GC is used for large molecules that decompose on heating. [1] (b) Explain how HPLC separates the components of the sample. [2] (c) Explain why the mass spectrometer is needed to identify a banned substance. [2]

10. Compound Z is a neutral liquid. Its high resolution mass spectrum shows a molecular ion at m/z = 100.0887.

(a) Show that the molecular formula of Z is C₆H₁₂O rather than C₅H₈O₂ or C₅H₁₂N₂. [2] (b) The IR spectrum of Z has a strong absorption at 1710 cm⁻¹. Z gives no silver mirror when warmed with Tollens’ reagent. Deduce the functional group in Z. [2] (c) The ¹H NMR spectrum of Z shows two singlets: δ 1.1 (relative area 9) and δ 2.1 (relative area 3). Deduce the structure of Z, giving your reasoning. [3] (d) How many peaks appear in the ¹³C NMR spectrum of Z? [1] (e) Identify the ion responsible for a fragment peak at m/z = 57 in the mass spectrum of Z. [1]

11. A solvent from a fire scene is tested by GC. Two peaks, at 2.8 min and 3.4 min, match propan-2-ol and propan-1-ol run under the same conditions.

(a) Explain why the two alcohols have different retention times. [2] (b) Calculate the accurate Mr of C₃H₈O and explain why high resolution mass spectrometry alone cannot tell these two alcohols apart. [2] (c) State how the ¹³C NMR spectra of the pure alcohols differ. [1] (d) Predict the splitting and relative area of the two C–H peaks in the high resolution ¹H NMR spectrum of propan-2-ol. [2]

Answers

1. Mobile phase: the moving phase (a solvent or gas) that carries the components along [1]. Stationary phase: the phase that does not move and holds back components to different extents, such as silica or a column packing [1]. Examiner insight: “The solvent” alone is not a definition; say what each phase does, not only what it is.

2. (a) C₄H₉NO: 48.0000 + 9.0702 + 14.0031 + 15.9949 = 87.0682 [1] C₅H₁₃N: 60.0000 + 13.1014 + 14.0031 = 87.1045 [1] C₃H₅NO₂: 36.0000 + 5.0390 + 14.0031 + 31.9898 = 87.0319 [1] (b) C₄H₉NO, as it is closest to 87.0681 (difference 0.0001) [1]. Examiner insight: Give every value to four decimal places; a mass quoted to one decimal place cannot separate these formulae.

3. (a) Butan-2-ol: 4 [1]. 2-methylpropan-1-ol: 3 [1]. Ethoxyethane: 2 [1]. (b) In (CH₃)₂CHCH₂OH the two CH₃ carbons are equivalent, while the CH carbon and the CH₂OH carbon are each unique, giving three environments [1]. Examiner insight: A justification must name the equivalent carbons; “it is symmetrical” with no detail is unlikely to be enough.

4. (a) Three peaks [1], areas 9 : 2 : 1 for (CH₃)₃C, CH₂ and OH [1]. (b) Two peaks, area ratio 4 : 2 = 2 : 1 [1]. The four CH₂Cl hydrogens have two neighbours: triplet [1]. The middle CH₂ has four neighbours: quintet (5 lines) [1]. Examiner insight: The two CH₂Cl groups are equivalent, so they give one peak; count the neighbours of the middle CH₂ on both sides.

5. CH₃CO: δ 2.0–2.7, area 3, singlet (no H on the C=O carbon) [1]. CHCl: δ 3.3–4.5, area 1, quartet (three H on the adjacent CH₃) [1]. CH₃ on the CHCl carbon: δ 0.7–1.7, area 3, doublet (one adjacent H) [1]. Relative areas 3 : 1 : 3 [1]. Examiner insight: Link each δ, area and splitting to a named group; a splitting given for the wrong group gains nothing.

6. Doublet (6H) with septet (1H): an isopropyl group, (CH₃)₂CH– [1]. The septet at δ 2.5 is in the H–C next to C=O range, so the CH is bonded to the C=O [1]. Quartet (2H) with triplet (3H): an ethyl group, CH₃CH₂– [1]. The quartet at δ 4.1 shows the CH₂ is bonded to O [1]. W is (CH₃)₂CHCOOCH₂CH₃, ethyl 2-methylpropanoate (12 H, matching C₆H₁₂O₂) [1]. Examiner insight: Use δ to place each fragment on the correct side of the ester link; the reverse arrangement fits the splitting but not the shifts.

7. (a) 1.8 ÷ 7.5 = 0.24; 3.3 ÷ 7.5 = 0.44 [1]; 5.1 ÷ 7.5 = 0.68 [1]. (b) Dyes A, B and D (not C) [1]. (c) A is the most strongly attracted to the polar silica stationary phase, for example by hydrogen bonds or dipole attractions [1], and/or the least soluble in the mobile phase, so it moves the shortest distance [1]. Examiner insight: Name the phases; “A is heavier” gives no chemical reason.

8. (a) Rf depends on the balance between solubility in the mobile phase and attraction to the stationary phase [1]. The dye is less soluble in the second solvent relative to its attraction to the stationary phase, so it moves a smaller fraction of the solvent distance [1]. (b) Both then meet identical conditions (solvent, paper, temperature), so their Rf values can be compared [1]. Examiner insight: An answer that just says “different solvent” restates the question; explain how the solvent shifts the balance between the phases.

9. (a) In HPLC the sample stays in solution, so it does not have to be vaporised [1]. (b) The liquid mobile phase is pumped at high pressure through a column packed with a solid stationary phase [1]. Components spend different times attracted to the stationary phase, so they leave at different retention times [1]. (c) Different substances can share a retention time [1]. Each component’s mass spectrum is matched against a database to confirm its identity [1]. Examiner insight: Use the term retention time; “they separate” is not an explanation.

10. (a) C₆H₁₂O = 72.0000 + 12.0936 + 15.9949 = 100.0885; C₅H₈O₂ = 100.0522; C₅H₁₂N₂ = 100.0998 [1]. C₆H₁₂O is closest to 100.0887 [1]. (b) The 1710 cm⁻¹ absorption shows C=O [1]. No reaction with Tollens’ reagent rules out an aldehyde, so Z is a ketone [1]. (c) Area 9 singlet: nine equivalent H in a (CH₃)₃C– group, whose central carbon carries no H [1]. Area 3 singlet at δ 2.1: CH₃ bonded to C=O, CH₃CO– [1]. Z is CH₃COC(CH₃)₃, 3,3-dimethylbutan-2-one [1]. (d) Four peaks [1]. (e) (CH₃)₃C⁺, the C₄H₉⁺ ion [1]. Examiner insight: In part (c), explain both singlets; a fragment ion in part (e) needs a positive charge to gain credit.

11. (a) The alcohols differ in volatility (boiling temperature) and in how strongly they interact with the stationary phase [1], so one spends longer dissolved in the stationary phase and is detected later [1]. (b) C₃H₈O = 36.0000 + 8.0624 + 15.9949 = 60.0573 [1]. Both are isomers with formula C₃H₈O, so they have the same accurate mass [1]. (c) Propan-1-ol gives three peaks; propan-2-ol gives two [1]. (d) The two CH₃ groups: doublet, area 6 [1]. The CH: septet, area 1 [1]. Examiner insight: For an isomer pair, say plainly that accurate mass identifies the formula, not the structure.

Where marks are usually lost

• Quoting accurate masses to fewer than four decimal places.
• Treating “symmetrical” as a full justification of a ¹³C peak count.
• Counting the hydrogens on the carbon itself when applying the (n+1) rule.
• Giving one peak for each CH₃ group when the groups are equivalent.
• Ignoring δ when deciding which side of an ester link a group sits on.
• Measuring Rf from the plate edge instead of the baseline.
• Saying a matching retention time proves identity.

Next steps

• Spectroscopy and chromatography revision notes for a final recap
• Spectroscopy and chromatography study guide to relearn anything you missed
• Edexcel IAL Chemistry hub for the other units
• Edexcel chemistry checklist: mark outcomes 15.17 to 15.23 as done
• All free 10-minute diagnostics
• Prefer working with a teacher? Book a free trial class with our tutors

Official syllabus

Question source: Pearson Edexcel YCH11, International Advanced Subsidiary/Advanced Level in Chemistry, Issue 1 of September 2017, from Pearson Education Limited. This set covers Topic 15E: Spectroscopy and chromatography, outcomes 15.17 to 15.23, in Unit 4.