NMR & chromatography: A Level Chemistry questions with answers
Free NMR & chromatography questions for A Level Chemistry (9701), part of Inorganic & organic. Each question has the answer and the marking points an examiner would look for.
Compound Y, C3H6O2, has the following proton NMR spectrum: δ 1.2 (3H, triplet), δ 2.4 (2H, quartet), δ 11.7 (1H, singlet). The peak at δ 11.7 disappears when D2O is added.
(a)
Identify Y, explaining each feature of the spectrum.
[5]
(b)
An isomer of Y, Z, shows only two singlets in its proton NMR spectrum. Identify Z and give the ratio of the peak areas.
[2]
(c)
How many peaks would be seen in the 13C NMR spectrum of each of Y and Z?
[2]
(d)
In thin-layer chromatography of a mixture, a spot travels 3.6 cm and the solvent front 8.0 cm. Calculate the Rf value and explain why spots are often viewed under UV light.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
propanoic acid CH3CH2COOH; triplet: CH3 next to CH2; quartet: CH2 next to CH3; δ 11.7 COOH proton; D2O exchange confirms O–H
propanoic acid B1 triplet explained by n + 1 rule B1 quartet explained B1 δ 11.7 is carboxylic acid O–H B1 disappears with D2O (exchangeable proton) B1
5
(b)
methyl ethanoate CH3COOCH3; 1 : 1
methyl ethanoate B1 1 : 1 B1
2
(c)
Y 3, Z 3
3 B1 3 B1
2
(d)
0.45; many organic compounds are colourless; they fluoresce/absorb UV
0.45 B1 colourless compounds made visible B1
2
Question 2Structured · 7 marksA* challenge
An ester, C₃H₆O₂, has the proton NMR spectrum: δ 8.0 (1H, singlet), δ 4.2 (2H, quartet) and δ 1.3 (3H, triplet).
(a)
Identify the ester, explaining how the splitting patterns support your answer.
[3]
(b)
State why tetramethylsilane is used in NMR.
[1]
(c)
Explain how D₂O could be used to identify an O–H proton in an isomeric acid.
[2]
(d)
In thin-layer chromatography the ester spot moved 2.4 cm and the solvent front moved 7.5 cm. Calculate Rf.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
ethyl methanoate, HCOOCH₂CH₃
ethyl methanoate A1 HCOOCH₂CH₃ A1 CH₂ next to CH₃ gives quartet and triplet (n + 1 rule) M1
3
(b)
reference, δ = 0
reference standard, single peak at δ = 0 B1
1
(c)
O–H peak disappears
D exchanges with the O–H proton M1 so the O–H peak disappears M1
2
(d)
0.32
0.32 B1
1
Question 3Structured · 12 marks
This question is about the proton NMR spectrum of ethyl ethanoate, CH₃COOCH₂CH₃.
(a)
State the number of different proton environments in ethyl ethanoate and their relative peak areas.
[2]
(b)
Predict the splitting pattern of each peak and explain using the n + 1 rule.
[3]
(c)
Explain why the CH₂ protons have a larger chemical shift than the CH₃ protons of the ethyl group.
[2]
(d)
Explain why tetramethylsilane (TMS) is used as a standard and why CDCl₃ is used as the solvent.
[3]
(e)
Describe how thin-layer chromatography and the Rf value could be used to check the purity of the ester.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
3; 3 : 2 : 3
three environments B1 ratio 3 : 2 : 3 B1
2
(b)
CH₃CO singlet (no adjacent H); OCH₂ quartet (3 adjacent H); CH₂CH₃ triplet (2 adjacent H)
CH₃ next to C=O: singlet (no neighbouring H) B1 CH₂: quartet (three H on adjacent carbon) B1 CH₃ of ethyl: triplet (two H on adjacent carbon) B1
3
(c)
adjacent to electronegative O, deshielded
the CH₂ is bonded to an electronegative oxygen atom B1 which withdraws electron density (deshields), so the signal is downfield B1
2
(d)
TMS: single peak, 12 equivalent H, inert, defined as 0; CDCl₃ contains no ¹H, so no solvent peak
TMS gives a single strong peak (12 equivalent H) and is inert / volatile B1 it is defined as δ = 0 B1 CDCl₃ contains no ¹H atoms so does not give a peak B1
3
(e)
single spot if pure; compare Rf with authentic sample in same solvent
pure sample gives a single spot B1 compare Rf with an authentic sample run under the same conditions B1
2
Question 4Structured · 12 marks
This question is about the proton NMR spectrum of propan-1-ol, CH₃CH₂CH₂OH.
(a)
Predict the number of peaks, their relative areas and splitting patterns.
[4]
(b)
Explain how adding D₂O helps identify the O–H peak.
[2]
(c)
Explain why the O–H peak is usually a singlet.
[1]
(d)
Explain how the two isomers can be distinguished by ¹³C NMR.
[2]
(e)
Gas chromatography of a mixture of the two alcohols gives two peaks. Explain how the mixture is separated and how the composition is found.
number of environments correct B1 relative areas correct B1 splitting patterns from n + 1 rule B2
4
(b)
OH proton exchanges with D, peak disappears
the O–H proton exchanges with deuterium B1 D does not show in ¹H NMR so the peak disappears B1
2
(c)
rapid exchange between molecules
the OH proton exchanges rapidly, so coupling to neighbouring protons is not observed B1
1
(d)
propan-1-ol 3 peaks; propan-2-ol 2 peaks
propan-1-ol has three carbon environments B1 propan-2-ol has two (symmetry) B1
2
(e)
partition between mobile gas and stationary liquid; different retention times; peak area proportional to amount
components partition differently between the mobile (carrier gas) and stationary (liquid) phases B1 so they have different retention times B1 the relative peak areas give the proportions B1
3
Question 5Structured · 5 marks
Proton NMR spectroscopy is used to identify ethanol, CH₃CH₂OH.
(a)
State the number of peaks and their splitting patterns in the high-resolution spectrum (ignoring O–H coupling).
[3]
(b)
Explain why D₂O is added to identify the O–H peak.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
3 peaks: triplet, quartet, singlet
three peaks B1 CH₃ triplet, CH₂ quartet B1 OH singlet B1
3
(b)
H exchanged for D; peak disappears
the O–H proton exchanges with deuterium B1 the peak disappears B1
2
Question 6Structured · 12 marks
Compound J contains 54.5% carbon, 9.1% hydrogen and 36.4% oxygen by mass, and has Mr = 88. Apart from C–H absorptions, the only absorption in its infrared spectrum above 1500 cm⁻¹ is a strong one at 1740 cm⁻¹; there is also a strong absorption at 1180 cm⁻¹. Fig. 6.1 shows its proton NMR spectrum. The number above each peak is the relative number of hydrogen atoms.
Fig. 6.1
(a)
Calculate the empirical formula and the molecular formula of J.
[3]
(b)
Use the infrared data to identify the functional group in J. Explain your answer.
[2]
(c)
Use Fig. 6.1 to identify J. Give its structure and explain the splitting of each peak.
[4]
(d)
State the number of peaks in the ¹³C NMR spectrum of J.
[1]
(e)
Explain why the spectrum is recorded with a solvent such as CDCl₃ and with TMS added.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
C₂H₄O; C₄H₈O₂
moles C 4.54, H 9.1, O 2.28 C1 ratio 2 : 4 : 1, C₂H₄O A1 Mr of C₂H₄O = 44, so C₄H₈O₂ A1
3
(b)
ester
C=O at 1740 cm−1B1 no O–H, so not a carboxylic acid: ester B1
2
(c)
propyl methanoate, HCOOCH₂CH₂CH₃
Any 4 of: HCOOCH₂CH₂CH₃; singlet at δ 8.1: H–C(=O)O (H on a carbon bonded to two O atoms; allow by elimination), no H on adjacent carbon; triplet at 4.1: OCH₂ next to CH₂; sextet at 1.7: CH₂ next to five H (CH₂ and CH₃); triplet at 1.0: CH₃ next to CH₂ (n + 1 = 3) B4
4
(d)
4
four (all carbon environments different) B1
1
(e)
no ¹H signal from solvent; TMS reference δ = 0
CDCl₃ has no ¹H atoms so gives no peaks B1 TMS is the reference / δ = 0 (single sharp peak) B1
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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in