AQA A-Level Chemistry: Organic analysis (7405) – 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.

These questions cover section 3.3.6 Organic analysis of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: 3.3.6.1 test-tube reactions with Required practical 6, 3.3.6.2 mass spectrometry and 3.3.6.3 infrared spectroscopy. It is AS content, examined on Paper 2 at AS and A-level, and A-level Paper 3 can test any content. The carbonate test for carboxylic acids draws on 3.3.9.1, which the specification marks A-level only.

Learn the ideas first in the organic analysis study guide and the revision notes. The course hub is AQA A-Level Chemistry, the printable checklist lists every specification point, and the free 10-minute diagnostics help you decide what to revise first.

Data for this set. Precise relative masses: ¹H = 1.0078, ¹²C = 12.0000, ¹⁴N = 14.0031, ¹⁶O = 15.9949. Infrared ranges (cm⁻¹): O–H (alcohols) 3230–3550; O–H (acids) 2500–3000; N–H (amines) 3300–3500; C–H 2850–3300; C≡N 2220–2260; C=O 1680–1750; C=C 1620–1680; C–O 1000–1300; C–C 750–1100.

Questions

1. State a reagent you could use to show that cyclohexene is unsaturated, and give the observation. [2]

2. Describe a test using Fehling’s solution to distinguish propanal from propanone. Give the result for each compound. [3]

3. Four unlabelled bottles contain cyclohexene, ethanoic acid, ethanal and propanone. Describe a sequence of three test-tube tests that identifies every bottle. For each test, give the reagent and the observation that identifies one compound. [6]

4. Separate samples of butan-1-ol, butan-2-ol and 2-methylpropan-2-ol are warmed with acidified potassium dichromate(VI). State what you would observe in each tube and explain the results. [4]

5. A high-resolution mass spectrum of a compound shows the molecular ion at m/z = 72.0573. The compound is C₅H₁₂, C₄H₈O or C₃H₄O₂. Show by calculation which formula is correct. [3]

6. Compound P gives a molecular ion at m/z = 88.0522. Its IR spectrum has a very broad absorption from 2500 to 3000 cm⁻¹ and a strong absorption at 1710 cm⁻¹.

(a) P is C₄H₈O₂, C₅H₁₂O or C₃H₈N₂O. Show by calculation which is its molecular formula. [2] (b) Identify the functional group in P, using the IR evidence, and suggest a structure for P. [3]

7. This question is about propan-1-ol and propanoic acid.

(a) Describe how the infrared spectra of the two compounds would differ. Refer to bonds and wavenumber ranges. [3] (b) Explain how the fingerprint region could confirm that a sample is pure propanoic acid. [2]

8. A student makes cyclohexene by the acid-catalysed elimination of water from cyclohexanol. After purification, the IR spectrum of the product shows an absorption at 1655 cm⁻¹ and a broad absorption at 3380 cm⁻¹. Explain what the spectrum shows about the product. [4]

9. Explain how the absorption of infrared radiation by carbon dioxide, methane and water vapour is linked to global warming. [4]

10. A volatile organic liquid Y gives a molecular ion at m/z = 44.0261.

(a) Y is C₃H₈, C₂H₄O or CH₄N₂. Show by calculation which is its molecular formula. [2] (b) The IR spectrum of Y has a strong absorption at 1730 cm⁻¹ and no absorption between 3230 and 3550 cm⁻¹. Identify the bond responsible and name Y. [2] (c) Describe a test-tube reaction that would confirm the functional group in Y, and write an equation for the oxidation of Y, using [O] for the oxidising agent. [3] (d) Y is heated under reflux with excess acidified potassium dichromate(VI). State one absorption that appears in the IR spectrum of the organic product but not in that of Y. [1]

11. Compound Z gives a molecular ion at m/z = 74.0366. These results are recorded:

• IR spectrum: a broad absorption at 3400 cm⁻¹, a strong absorption at 1720 cm⁻¹, and no very broad absorption across 2500–3000 cm⁻¹.
• Warming with Tollens’ reagent: a silver mirror forms.
• Warming with acidified potassium dichromate(VI): orange to green.
• The O–H group and the C=O group are on the carbon atoms at opposite ends of the chain.

(a) Z is C₄H₁₀O, C₃H₆O₂ or C₂H₆N₂O. Show by calculation which is its molecular formula. [2] (b) Use all the evidence to deduce the structure of Z. Explain each step. [7]

Answers

1. Bromine water [1]. It changes from orange to colourless [1]. [2] Examiner insight: “Goes clear” or “decolourised” with no starting colour is weaker than “orange to colourless”; give both colours.

2. Warm each compound with Fehling’s solution in a water bath [1]. Propanal: the blue solution gives a brick-red precipitate [1]. Propanone: no change, the solution stays blue [1]. [3] Examiner insight: For a “distinguish” question, give the result for both compounds; “no reaction” is a statement about chemistry, while “stays blue” is the observation asked for.

3. One correct sequence: Test 1: add sodium carbonate (or hydrogencarbonate) solution [1]; effervescence, with a gas that turns limewater cloudy, identifies ethanoic acid [1]. Test 2: warm the other three with Tollens’ reagent [1]; a silver mirror identifies ethanal [1]. Test 3: add bromine water to the last two [1]; orange to colourless identifies cyclohexene, and the bottle that stays orange is propanone [1]. [6] Examiner insight: Other valid orders can work, but each test must pick out one compound from those left; a test that two remaining compounds both pass identifies neither.

4. Butan-1-ol and butan-2-ol: orange to green [1]. 2-methylpropan-2-ol: stays orange [1]. Butan-1-ol is a primary and butan-2-ol a secondary alcohol, so both are oxidised, and orange Cr₂O₇²⁻ is reduced to green Cr³⁺ [1]. 2-methylpropan-2-ol is a tertiary alcohol, which is not easily oxidised [1]. [4] Examiner insight: Classify each alcohol; “it is tertiary” with no link to “not easily oxidised” does not explain the observation.

5. C₄H₈O: 4(12.0000) + 8(1.0078) + 15.9949 = 72.0573 [1]. C₅H₁₂: 5(12.0000) + 12(1.0078) = 72.0936; C₃H₄O₂: 3(12.0000) + 4(1.0078) + 2(15.9949) = 72.0210 [1]. Only C₄H₈O matches, so the formula is C₄H₈O [1]. [3] Examiner insight: A “show by calculation” question needs the sum for every candidate; stating the right formula alone gives no evidence.

6. (a) C₄H₈O₂: 4(12.0000) + 8(1.0078) + 2(15.9949) = 88.0522 [1]. C₅H₁₂O = 88.0885 and C₃H₈N₂O = 88.0635, so P is C₄H₈O₂ [1]. (b) The very broad absorption at 2500–3000 cm⁻¹ is O–H in a carboxylic acid [1]. The absorption at 1710 cm⁻¹ is C=O (1680–1750) [1]. P is a carboxylic acid, for example butanoic acid, CH₃CH₂CH₂COOH (2-methylpropanoic acid is also correct) [1]. [5] Examiner insight: Match each absorption to a named bond and its data range; “1710 shows an acid” skips the bond and is not a full interpretation.

7. (a) Propanoic acid shows a C=O absorption at 1680–1750 cm⁻¹, which propan-1-ol does not [1]. Propanoic acid shows a very broad O–H (acid) absorption at 2500–3000 cm⁻¹ [1]. Propan-1-ol shows a broad O–H (alcohol) absorption at 3230–3550 cm⁻¹ instead [1]. (b) The fingerprint region (below about 1500 cm⁻¹) has a pattern unique to each compound [1]. If the sample’s spectrum matches that of a known pure sample of propanoic acid exactly, the identity is confirmed and no impurity peaks are present [1]. [5] Examiner insight: “Both have O–H” is true but scores nothing here; the difference is the range and shape, so quote both ranges.

8. The absorption at 1655 cm⁻¹ is C=C (1620–1680), so cyclohexene has formed [1]. The broad absorption at 3380 cm⁻¹ is O–H (alcohol) (3230–3550) [1]. Cyclohexene has no O–H bond, so the product is not pure [1]. The impurity is likely to be unreacted cyclohexanol (or water left after drying) [1]. [4] Examiner insight: Say why the O–H absorption is unexpected (cyclohexene has no O–H); naming an impurity without that reasoning leaves the explanation incomplete.

9. The Earth’s surface emits infrared radiation [1]. Bonds in these gases (C=O in CO₂, C–H in methane, O–H in water) absorb infrared radiation [1]. The molecules re-emit the energy in all directions, some back towards the Earth, so energy is kept in the lower atmosphere [1]. Higher concentrations of the gases increase the infrared absorbed, which contributes to global warming [1]. [4] Examiner insight: Link the effect to bonds absorbing infrared; “the gases trap heat” without absorption of infrared by bonds is too vague for this specification point.

10. (a) C₂H₄O: 2(12.0000) + 4(1.0078) + 15.9949 = 44.0261 [1]. C₃H₈ = 44.0624 and CH₄N₂ = 44.0374, so Y is C₂H₄O [1]. (b) 1730 cm⁻¹ is C=O [1]. Y is ethanal, CH₃CHO [1]. (c) Warm Y with Tollens’ reagent [1]; a silver mirror forms [1] (or Fehling’s solution: blue to brick-red precipitate). CH₃CHO + [O] → CH₃COOH [1]. (d) A very broad O–H (acid) absorption at 2500–3000 cm⁻¹ [1]. [8] Examiner insight: In (d), “an O–H absorption” without the acid range does not distinguish the product from an alcohol; give the range.

11. (a) C₃H₆O₂: 3(12.0000) + 6(1.0078) + 2(15.9949) = 74.0366 [1]. C₄H₁₀O = 74.0729 and C₂H₆N₂O = 74.0479, so Z is C₃H₆O₂ [1]. (b) The broad absorption at 3400 cm⁻¹ is O–H in an alcohol (3230–3550) [1]. The strong absorption at 1720 cm⁻¹ is C=O (1680–1750) [1]. There is no very broad absorption at 2500–3000 cm⁻¹, so Z is not a carboxylic acid [1]. The silver mirror with Tollens’ reagent shows the C=O is in an aldehyde, –CHO, which is on an end carbon [1]. The dichromate result does not show the alcohol type, because the aldehyde is itself oxidised [1]. The O–H is on the carbon at the other end of the chain, so it is a primary alcohol [1]. Z is 3-hydroxypropanal, HOCH₂CH₂CHO [1]. [9] Examiner insight: In a multi-evidence structure question, tie every piece of evidence to a conclusion; a correct structure with no reasoning answers only the last step.

Where marks are usually lost

• Writing “clear” for colourless, or giving only the final colour of a test.
• Swapping the Tollens’ and Fehling’s observations.
• Using acidified dichromate(VI) while an aldehyde could still be present, so two compounds both turn green.
• Picking a molecular formula from precise masses without showing the sum for every candidate.
• Assigning a 2500–3000 cm⁻¹ absorption to an alcohol, or 3230–3550 cm⁻¹ to an acid.
• Describing the fingerprint region without saying the spectrum is compared with a known sample.

Next steps

• Organic analysis revision notes
• Organic analysis study guide
• NMR spectroscopy (A-level only)
• AQA A-Level Chemistry exam preparation
• AQA A-Level Chemistry hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA, AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: section 3.3.6 Organic analysis (3.3.6.1 to 3.3.6.3), with Required practical 6.