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

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Syllabus coverage

  • 22.1 4 questions
  • 22.2 6 questions
  • 37.1 1 question
  • 37.2 1 question
  • 37.3 1 question
  • 37.4 2 questions

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, M+\text{M}^+, for the molecular mass and interprets fragmentation peaks to piece together structural fragments; the ratio of the [M+1]+[\text{M}+1]^+ peak gives the number of carbon atoms present, n=100×abundance of [M+1]+1.1×abundance of M+n = \dfrac{100 \times \text{abundance of } [\text{M}+1]^+}{1.1 \times \text{abundance of } \text{M}^+}, and a distinctive [M+2]+[\text{M}+2]^+ 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 RfR_f 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

Multiple choice AS 1 mark

A mass spectrometer is used to analyse a small, volatile organic compound, J. The mass spectrum shows a molecular ion peak at m/z=108m/z = 108 and a second peak, of almost identical height, at m/z=110m/z = 110. No peak of comparable height is seen at m/z=106m/z = 106 or at m/z=112m/z = 112.

Which conclusion is correctly supported by this data?

Question 2

Structured AS 9 marks

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 m/z=88m/z = 88 and a prominent fragment ion peak at m/z=57m/z = 57.

Its infrared spectrum shows a strong, sharp absorption at 1735 cm11735\ \text{cm}^{-1}, a further strong absorption at 1200 cm11200\ \text{cm}^{-1}, and no absorption anywhere in the ranges 250025003000 cm13000\ \text{cm}^{-1} or 320032003600 cm13600\ \text{cm}^{-1}.

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, MrM_r, 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 m/z=57m/z = 57 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

Structured A2 9 marks

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 8.0 cm8.0\ \text{cm} from the baseline. Spot P has travelled 6.4 cm6.4\ \text{cm}, spot Q has travelled 4.0 cm4.0\ \text{cm}, and spot R has travelled 1.6 cm1.6\ \text{cm}.

(a) Calculate the RfR_f 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 1.81.8 minutes for P, 3.53.5 minutes for Q and 6.26.2 minutes for R, and with peak areas in the ratio 3:8:93 : 8 : 9 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

Structured A2 10 marks

Compound M has molecular formula C5H10O2\text{C}_5\text{H}_{10}\text{O}_2 (Mr=102M_r = 102) and is a colourless liquid with a fruity smell.

Its carbon-13 NMR spectrum shows five distinct peaks (five different carbon environments).

Its proton (1H^1\text{H}) NMR spectrum shows four signals, with the data below:

δ\delta / 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 δ\delta / ppm
alkane, CH3-\text{CH}_3, CH2-\text{CH}_2- 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, CH3COOCH(CH3)2\text{CH}_3\text{COOCH}(\text{CH}_3)_2, 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 1H^1\text{H} 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 δ=1.65\delta = 1.65 appears as a multiplet rather than as a simple triplet or quartet. [2]

Question 5

Multiple choice A2 1 mark

In the proton (1H^1\text{H}) 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

Multiple choice AS 1 mark

The mass spectrum of an organic compound, X, containing only carbon and hydrogen, shows a molecular ion peak at m/z=128m/z = 128 with relative abundance 100, and a smaller peak at m/z=129m/z = 129 (the [M+1]+[\text{M}+1]^+ peak) with relative abundance 9.9. No peak of significant height appears at m/z=130m/z = 130.

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

Structured AS 8 marks

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 m/z=116m/z = 116 and a prominent fragment ion peak at m/z=99m/z = 99.

Its infrared spectrum shows a strong, very broad absorption across the range 250025003000 cm13000\ \text{cm}^{-1}, a strong, sharp absorption at 1715 cm11715\ \text{cm}^{-1}, and no absorption anywhere in the range 320032003600 cm13600\ \text{cm}^{-1}.

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, MrM_r, 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 m/z=99m/z = 99 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

Structured AS 7 marks

The infrared spectrum of a colourless liquid, compound Z, shows two separate medium-intensity absorptions at 3300 cm13300\ \text{cm}^{-1} and 3380 cm13380\ \text{cm}^{-1}, together with a C–H absorption in the range 285028502960 cm12960\ \text{cm}^{-1}. There is no absorption anywhere in the range 165016501750 cm11750\ \text{cm}^{-1}, and no absorption in the range 320032003600 cm13600\ \text{cm}^{-1} 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, NH2-\text{NH}_2 3300–3500 two absorptions (symmetric and asymmetric N–H stretch)
N–H secondary amine, NH-\text{NH}- 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 320032003600 cm13600\ \text{cm}^{-1} region distinguishes it from both a secondary amine and an alcohol. [3]

(b) Explain why the absence of any absorption in the range 165016501750 cm11750\ \text{cm}^{-1} rules out compound Z being a primary amide, R-CONH2\text{R-CONH}_2, even though an amide also contains N–H bonds. [2]

(c) A second compound, X, gives only one absorption in the 330033003500 cm13500\ \text{cm}^{-1} 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

Multiple choice AS 1 mark

The mass spectrum of a ketone, compound A, shows a molecular ion peak at m/z=86m/z = 86 and a prominent fragment ion peak at m/z=71m/z = 71. No peak of significant height is seen at m/z=68m/z = 68, m/z=69m/z = 69 or m/z=57m/z = 57.

Which of the following is the best interpretation of the fragment ion peak at m/z=71m/z = 71?

Question 10

Structured AS 9 marks

Compound V is a colourless liquid containing carbon, hydrogen, oxygen and chlorine only. Its mass spectrum shows three peaks close together at high m/zm/z: at m/z=126m/z = 126, m/z=128m/z = 128 and m/z=130m/z = 130, in the approximate height ratio 9:6:19 : 6 : 1. No other peak of comparable height appears above m/z=130m/z = 130.

Its infrared spectrum shows a strong, sharp absorption at 1715 cm11715\ \text{cm}^{-1}, and no absorption anywhere in the ranges 250025003000 cm13000\ \text{cm}^{-1} or 320032003600 cm13600\ \text{cm}^{-1}.

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 35Cl^{35}\text{Cl} (approximately 76%) and 37Cl^{37}\text{Cl} (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 MrM_r. [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 m/z=77m/z = 77, 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]