Analytical Techniques: Chemistry 9701 (Cambridge International AS & A Level)
Syllabus 22.1, 22.2, 37.1, 37.2, 37.3, 37.4 · Strand 4 Analytical Chemistry
- Questions
- 10
- Total marks
- 56
- Tier mix
- 10 Core
0 of 10 questions completed
Syllabus coverage
- 22.1 4 questions completed
- 22.2 6 questions completed
- 37.1 1 question completed
- 37.2 1 question completed
- 37.3 1 question completed
- 37.4 2 questions completed
This topic (syllabus ref 22.1 to 22.2 at AS, extended by 37.1 to 37.4 at A Level) is about identifying a molecule’s structure from physical data rather than from its chemical reactions. Infrared spectroscopy picks out bond-specific absorptions to reveal which functional groups are present, while mass spectrometry reads the molecular ion peak, , for the molecular mass and interprets fragmentation peaks to piece together structural fragments; the ratio of the peak gives the number of carbon atoms present, , and a distinctive peak signals a bromine or chlorine atom.
Chromatography separates a mixture based on differing affinity for a stationary versus a mobile phase: thin-layer chromatography gives an value (distance travelled by spot ÷ distance travelled by solvent front), while gas/liquid chromatography gives a retention time, both dependent on polarity and interaction with the stationary phase. NMR spectroscopy is the most structurally powerful technique here: a carbon-13 spectrum counts distinct carbon environments, while a proton spectrum additionally reveals relative proton counts from peak area and neighbouring-proton counts from splitting patterns via the n + 1 rule, referenced against the TMS standard.
Original worked problems below apply every technique in full.
Question 1
A mass spectrometer is used to analyse a small, volatile organic compound, J. The mass spectrum shows a molecular ion peak at and a second peak, of almost identical height, at . No peak of comparable height is seen at or at .
Which conclusion is correctly supported by this data?
Question 2
Compound L is a colourless, fruity-smelling liquid used as a flavouring in the food industry.
In its mass spectrum, compound L shows a molecular ion peak at and a prominent fragment ion peak at .
Its infrared spectrum shows a strong, sharp absorption at , a further strong absorption at , and no absorption anywhere in the ranges – or –.
You may use the following characteristic infrared absorption ranges:
| Bond | Functional group | Wavenumber / cm⁻¹ |
|---|---|---|
| C–O | hydroxy, ester | 1040–1300 |
| C=O | ester | 1710–1750 |
| O–H | carboxylic acid | 2500–3000 |
| O–H | alcohol | 3200–3600 |
(a) State the relative molecular mass, , of compound L. [1]
(b) Identify the functional group present in compound L using the infrared data above, and explain how this data rules out both a carboxylic acid and an alcohol as possibilities. [3]
(c) The peak at is formed by the loss of a single neutral fragment from the molecular ion. Calculate the relative mass of the fragment lost, and suggest its formula. [3]
(d) Suggest a structural formula for compound L that is fully consistent with all of the data above, and give its name. [2]
Question 3
A student separates a mixture of three carboxylic acids, P, Q and R, first by thin-layer chromatography (TLC) and then by gas-liquid chromatography (GLC).
In the TLC experiment, a polar silica stationary phase and a non-polar mobile solvent are used. After development, the solvent front has travelled from the baseline. Spot P has travelled , spot Q has travelled , and spot R has travelled .
(a) Calculate the value of each of P, Q and R. [3]
(b) State which of P, Q or R interacts most strongly with the stationary phase in this TLC experiment. Explain your answer. [2]
The same mixture is then analysed by GLC, using a high-boiling-point non-polar liquid (on a solid support) as the stationary phase and an unreactive gas as the mobile phase. Three peaks are obtained, with retention times of minutes for P, minutes for Q and minutes for R, and with peak areas in the ratio for P : Q : R.
(c) (i) Calculate the percentage composition, by moles, of the original mixture. [2]
(c) (ii) Explain, in terms of interaction with the stationary phase, why R has the longest retention time in this experiment. [2]
Question 4
Compound M has molecular formula () and is a colourless liquid with a fruity smell.
Its carbon-13 NMR spectrum shows five distinct peaks (five different carbon environments).
Its proton () NMR spectrum shows four signals, with the data below:
| / ppm | Relative peak area (integration) | Splitting pattern |
|---|---|---|
| 0.95 | 3H | triplet |
| 1.65 | 2H | multiplet |
| 2.30 | 2H | triplet |
| 3.65 | 3H | singlet |
You may use the following typical chemical shift ranges:
| Environment of proton | / ppm |
|---|---|
| alkane, , | 0.9–1.7 |
| alkyl next to C=O | 2.2–3.0 |
| alkyl next to an electronegative atom (e.g. O) | 3.2–4.0 |
(a) A structural isomer of compound M, isopropyl ethanoate, , shows only four carbon-13 environments rather than five. Explain why isopropyl ethanoate gives one fewer carbon-13 environment than compound M, and state what the five separate environments observed for M tell you about whether M's carbon skeleton is branched in the same way. [3]
(b) Using the chemical shift ranges, integrations and splitting patterns given, assign each of the four signals to a specific proton environment in compound M. Hence deduce the structural formula of compound M, and give its name. [5]
(c) Explain, using the n + 1 rule, why the signal at appears as a multiplet rather than as a simple triplet or quartet. [2]
Question 5
In the proton () NMR spectrum of a compound, one signal is a quartet integrating for 2H, and a second signal is a triplet integrating for 3H. No other signals are present, and there is no evidence of an exchangeable O–H or N–H proton.
Which structural fragment is consistent with this pattern?
Question 6
The mass spectrum of an organic compound, X, containing only carbon and hydrogen, shows a molecular ion peak at with relative abundance 100, and a smaller peak at (the peak) with relative abundance 9.9. No peak of significant height appears at .
Given that carbon-13 makes up approximately 1.1% of all naturally occurring carbon atoms, which of the following is the correct interpretation of this data?
Question 7
Compound Y is a colourless liquid with a sharp, unpleasant odour, used industrially as a chemical intermediate.
In its mass spectrum, compound Y shows a molecular ion peak at and a prominent fragment ion peak at .
Its infrared spectrum shows a strong, very broad absorption across the range –, a strong, sharp absorption at , and no absorption anywhere in the range –.
You may use the following characteristic infrared absorption ranges:
| Bond | Functional group | Wavenumber / cm⁻¹ |
|---|---|---|
| O–H | carboxylic acid | 2500–3000 |
| O–H | alcohol | 3200–3600 |
| C=O | carboxylic acid | 1700–1725 |
| C=O | ester | 1710–1750 |
(a) State the relative molecular mass, , of compound Y. [1]
(b) Identify the functional group present in compound Y using the infrared data above, and explain how this data rules out an alcohol as a possibility. [3]
(c) The peak at is formed by the loss of a single neutral fragment from the molecular ion. Calculate the relative mass of the fragment lost, and suggest its formula. [2]
(d) Given that compound Y has an unbranched carbon skeleton, suggest a structural formula for compound Y that is fully consistent with all of the data above, and give its name. [2]
Question 8
The infrared spectrum of a colourless liquid, compound Z, shows two separate medium-intensity absorptions at and , together with a C–H absorption in the range –. There is no absorption anywhere in the range –, and no absorption in the range – other than the two peaks already stated.
You may use the following characteristic infrared absorption data:
| Bond | Functional group | Wavenumber / cm⁻¹ | Typical peak shape |
|---|---|---|---|
| N–H | primary amine, | 3300–3500 | two absorptions (symmetric and asymmetric N–H stretch) |
| N–H | secondary amine, | 3300–3500 | one absorption |
| O–H | alcohol (hydrogen bonded) | 3200–3600 | one broad absorption |
| C=O | carbonyl (aldehyde, ketone, acid, ester, amide) | 1650–1750 | one strong, sharp absorption |
| C–H | alkyl | 2850–2960 | one or more absorptions |
(a) Identify the functional group present in compound Z, and explain how the number of absorptions in the – region distinguishes it from both a secondary amine and an alcohol. [3]
(b) Explain why the absence of any absorption in the range – rules out compound Z being a primary amide, , even though an amide also contains N–H bonds. [2]
(c) A second compound, X, gives only one absorption in the – region. Suggest why this single absorption alone is not, by itself, enough to decide whether X is a secondary amine or an alcohol, and state what further infrared evidence would help distinguish between them. [2]
Question 9
The mass spectrum of a ketone, compound A, shows a molecular ion peak at and a prominent fragment ion peak at . No peak of significant height is seen at , or .
Which of the following is the best interpretation of the fragment ion peak at ?
Question 10
Compound V is a colourless liquid containing carbon, hydrogen, oxygen and chlorine only. Its mass spectrum shows three peaks close together at high : at , and , in the approximate height ratio . No other peak of comparable height appears above .
Its infrared spectrum shows a strong, sharp absorption at , and no absorption anywhere in the ranges – or –.
You may use the following characteristic infrared absorption ranges:
| Bond | Functional group | Wavenumber / cm⁻¹ |
|---|---|---|
| C=O | ketone | 1705–1725 |
| O–H | carboxylic acid | 2500–3000 |
| O–H | alcohol | 3200–3600 |
(a) Using the abundances of the chlorine isotopes (approximately 76%) and (approximately 24%), explain why a molecule containing two chlorine atoms gives three peaks (M⁺, [M+2]⁺ and [M+4]⁺) in the approximate ratio 9:6:1, and hence state the number of chlorine atoms in compound V and its value of . [3]
(b) Identify the functional group present in compound V using the infrared data, and explain how this data rules out both a carboxylic acid and an alcohol as possibilities. [2]
(c) A prominent fragment ion peak also appears at , formed by loss of a single neutral radical from the molecular ion. Calculate the mass of this radical and suggest its formula. [2]
(d) Given that compound V is symmetrical about its carbonyl carbon, suggest a structural formula for compound V that is fully consistent with all of the data above, and give its name. [2]