About Analytical techniques in A Level Chemistry
Analytical techniques is one of the 18 main topics in our A Level Chemistry (Cambridge (CAIE) 9701) question bank, and about 3% of its real questions come from it. Tayyari Karo has 22 real questions on this topic from 2021 to 2026, from Paper 1, Paper 2, Paper 4; 12 are multiple choice and marked instantly, and the rest are structured questions with a model answer. By difficulty: 14 hard, 7 medium, 1 easy.
Work through the questions below, then open the explanation for each one. Redo the ones you missed a few days later.
Try 6 real questions
These are real past-paper questions. Think of your answer first, then open the explanation.
May/June 2025 · Paper 1 · Analytical techniques
Complete combustion of compound T produces only CO2 and water. T produces steamy fumes with PCl5. T gives no visible product with 2,4-dinitrophenylhydrazine reagent. What can be deduced with certainty from this information?
- Compound T is a carboxylic acid.
- Compound T is a hydrocarbon.
- Compound T is an alcohol.
- Compound T is not an aldehyde.
Show answer and explanation
Answer: D. Compound T is not an aldehyde.
Steamy fumes with PCl5 show an -OH (or -COOH) group is present, and no reaction with 2,4-DNPH rules out a carbonyl (aldehyde/ketone) group — so T is definitely not an aldehyde, though it could be an alcohol or carboxylic acid.
May/June 2025 · Paper 1 · Analytical techniques
Hydrocarbon X is saturated and each molecule contains one ring of carbon atoms. In its mass spectrum, the M+ peak is 14 mm high and the [M+1]+ peak is 0.77 mm high. What is the m/e value for the M+ ion of X?
- 56
- 58
- 70
- 72
Show answer and explanation
Answer: C. 70
%[M+1]/M+ = 0.77/14 × 100 = 5.5%; since each carbon contributes about 1.1% to the [M+1] peak (13C natural abundance), number of carbons ≈ 5.5/1.1 = 5. A saturated cyclic hydrocarbon (one ring) has formula CnH2n, so C5H10 gives M+ = 70.
May/June 2023 · Paper 1 · Analytical techniques
There are two naturally occurring isotopes of bromine: one with 44 neutrons, the other with 46 neutrons. Ignoring fragments, how many peaks are there in the mass spectrum of tribromomethane, 12C1HBr3?
- 2
- 3
- 4
- 6
Show answer and explanation
Answer: C. 4
With 12C and 1H fixed and 3 Br atoms each being either 79Br or 81Br, the possible total masses are 3×79, 2×79+81, 79+2×81, 3×81 — 4 distinct molecular ion peaks.
October/November 2023 · Paper 1 · Analytical techniques
The infrared spectrum of compound L shows a broad strong absorption around 3200–3600 cm⁻1 (O–H, hydroxy), essentially no absorption in the 1670–1740 cm⁻1 carbonyl region, and no absorption around 2500–3000 cm⁻1 (no carboxyl O–H). What is the structure of L?
- HOCH2COCH2OH
- HOCH2CH(OH)CHO
- HOCH2CH(OH)CH2OH
- HOCH2CH2COOH
Show answer and explanation
Answer: C. HOCH2CH(OH)CH2OH
L has O–H absorption but no C=O absorption, so it must contain only hydroxy groups and no carbonyl/carboxyl group — this matches HOCH2CH(OH)CH2OH (propane-1,2,3-triol), which has three OH groups and no C=O.
October/November 2023 · Paper 1 · Analytical techniques
In the mass spectrum of compound J, the ratio of the height of the M+1 ion peak to the height of the M+ ion peak is 4 : 91. Compound J forms a carboxylic acid when heated with acidified K2Cr2O7. What is compound J?
- butanal
- butanone
- propan-1-ol
- propanenitrile
Show answer and explanation
Answer: A. butanal
A M+1:M+ ratio of about 4:91 (~4.4%) corresponds to a compound with 4 carbon atoms (each ¹³C contributes ~1.1% per carbon); only oxidation of an aldehyde (not a ketone, alcohol needing only mild oxidation to aldehyde/acid, or nitrile) directly gives a carboxylic acid on heating with acidified K2Cr2O7 — butanal, C4H8O.
October/November 2022 · Paper 1 · Analytical techniques
In the mass spectrum of a compound, Z, the relative abundances of the M and M+1 peaks are in the ratio 13:1. What is compound Z?
- butyl butanoate
- hexan-3-one
- 2,2,3-trimethylhexane
- 3,3-dimethylpentan-1-ol
Show answer and explanation
Answer: D. 3,3-dimethylpentan-1-ol
The M+1 peak arises from ¹³C isotope abundance (~1.1% per carbon), so ratio M:(M+1) ≈ (100/1.1n):1; a ratio of 13:1 corresponds to n ≈ 7 carbon atoms, matching 3,3-dimethylpentan-1-ol (C7H16O).
