Nuclear magnetic resonance spectroscopyAQA A-Level Chemistry: Topic test

20 questions, 54 marks

AQA A-Level Chemistry

Nuclear magnetic resonance spectroscopy topic test

Total 54 marks

Name

Class

Date

1
Pentan-2-one and pentan-3-one are isomers with the molecular formula C₅H₁₀O. A chemist wants to tell the two ketones apart using ¹³C NMR spectroscopy.
(a)
How many peaks are there in the ¹³C NMR spectrum of pentan-3-one, CH₃CH₂COCH₂CH₃?
[1 mark]
A2
B3
C4
D5
(b)
Using the data booklet, in which range of chemical shift, δ / ppm, does the carbon atom of the C=O group in a ketone absorb?
[1 mark]
A5 to 40
B50 to 90
C110 to 160
D190 to 220
(c)
Explain how the ¹³C NMR spectra of the two isomers would allow the chemist to tell them apart.
[2 marks]
Total for question 1: 4 marks
2
A student prepares a sample of an organic liquid for ¹H NMR spectroscopy. She dissolves it in deuterated chloroform, CDCl₃, and adds a small amount of tetramethylsilane, TMS, Si(CH₃)₄, as the standard.
(a)
Why is CDCl₃ used as the solvent rather than CHCl₃?
[1 mark]
ADeuterium atoms do not give peaks in the ¹H NMR spectrum, so the solvent does not mask peaks from the sample
BCDCl₃ dissolves organic compounds but CHCl₃ does not
CCDCl₃ has a higher boiling point, so the sample is easier to recover
DCDCl₃ reacts with TMS to give a single sharp peak
(b)
A peak in the ¹H NMR spectrum of the sample has a chemical shift of δ = 2.1 ppm. What does this mean?
[1 mark]
AThe protons absorb at a frequency 2.1 ppm lower than the protons in TMS
BThe peak has an area 2.1 times that of the TMS peak
CThe protons absorb at a frequency 2.1 ppm higher than the protons in TMS
DThere are 2.1 protons in this environment
(c)
Tetrachloromethane, CCl₄, is also a suitable solvent for ¹H NMR spectroscopy. Explain why. Give one reason why TMS is a more suitable standard than hexane.
[2 marks]
Total for question 2: 4 marks
3
Compound P has the molecular formula C₄H₈O. Its ¹³C NMR spectrum has four peaks, at δ = 8, 29, 37 and 209 ppm. Its ¹H NMR spectrum has three peaks: δ = 1.0 (triplet, relative area 3), δ = 2.1 (singlet, relative area 3) and δ = 2.4 (quartet, relative area 2). Data booklet values, δ / ppm: ¹³C: C–C 5–40, C–O 50–90, C=C 90–150, C=O (ketone or aldehyde) 190–220; ¹H: R–CH₃ 0.7–1.2, R–CO–CH₃ and R–CO–CH₂– 2.1–2.6.
(a)
Explain what the ¹³C NMR spectrum shows about the structure of P.
[3 marks]
(b)
Use the ¹H NMR data to deduce the structure of P. Explain how each peak supports your answer.
[4 marks]
Total for question 3: 7 marks
4
Three non-cyclic isomers of C₄H₁₀O, labelled X, Y and Z, are analysed by NMR spectroscopy. The ¹³C NMR spectrum of X has two peaks, at δ = 15 and 66 ppm. The ¹³C NMR spectrum of Y has two peaks, at δ = 31 and 69 ppm. The ¹³C NMR spectrum of Z has four peaks, at δ = 10, 22, 32 and 69 ppm. The ¹H NMR spectrum of X has two peaks: δ = 1.2 (triplet, relative area 3) and δ = 3.5 (quartet, relative area 2). The ¹H NMR spectrum of Y has two singlets: δ = 1.3 (relative area 9) and δ = 2.0 (relative area 1). Data booklet values, δ / ppm: ¹³C: C–C 5–40, C–O 50–90; ¹H: R–CH₃ 0.7–1.2, R–O–CH₂– 3.1–3.9, R–OH 0.5–5.0.
(a)
Deduce the structures of X and Y. Explain how the ¹³C and ¹H NMR data support each structure.
[6 marks]
(b)
Z is butan-2-ol, CH₃CH(OH)CH₂CH₃. Another isomer, 2-methylpropan-1-ol, (CH₃)₂CHCH₂OH, also has the formula C₄H₁₀O. Explain how the ¹³C NMR data are consistent with Z being butan-2-ol and not X, Y or 2-methylpropan-1-ol, and predict the splitting patterns of the two CH₃ peaks in the ¹H NMR spectrum of butan-2-ol.
[6 marks]
Total for question 4: 12 marks
5
A student records the ¹H NMR spectrum of 1,1-dichloroethane, CH₃CHCl₂. The spectrum contains two peaks.
(a)
What is the ratio of the peak areas for the CH₃ hydrogens : the CHCl₂ hydrogen?
[1 mark]
A1 : 3
B3 : 1
C1 : 1
D3 : 2
(b)
How is the peak for the hydrogen atom in CHCl₂ split?
[1 mark]
ASinglet
BDoublet
CTriplet
DQuartet
(c)
Explain why the peak for the CH₃ hydrogens is a doublet.
[2 marks]
Total for question 5: 4 marks
6
Pentane, 2-methylbutane and 2,2-dimethylpropane are isomers with the molecular formula C₅H₁₂. A chemist records the ¹³C NMR spectrum of each isomer.
(a)
Which isomer gives the fewest peaks in its ¹³C NMR spectrum?
[1 mark]
A2,2-dimethylpropane
BPentane
C2-methylbutane
DThey all give the same number of peaks
(b)
How many peaks are there in the ¹³C NMR spectrum of 2-methylbutane, CH₃CH(CH₃)CH₂CH₃?
[1 mark]
A2
B3
C5
D4
(c)
Pentane gives three peaks in its ¹³C NMR spectrum. Explain why.
[2 marks]
Total for question 6: 4 marks
7
Compound W has the molecular formula C₃H₆O₂. Its ¹H NMR spectrum has three peaks, at δ = 1.1 (triplet), δ = 2.3 (quartet) and δ = 11.5 (singlet). The integration trace has step heights of 18.0 mm, 12.0 mm and 6.0 mm for the peaks at δ = 1.1, 2.3 and 11.5 respectively. Data booklet values, δ / ppm: R–CH₃ 0.7–1.2; R–CO–CH₂– 2.1–2.6; R–COOH 9.0–13.0.
(a)
Use the integration data and the molecular formula to calculate the number of hydrogen atoms responsible for each peak.
[3 marks]
(b)
Deduce the structure of W. Use the chemical shifts and splitting patterns in your answer.
[4 marks]
Total for question 7: 7 marks
8
A food chemist must confirm the identity of a flavouring ester, M, with the molecular formula C₅H₁₀O₂. The ¹³C NMR spectrum of M has five peaks, at δ = 9, 14, 27, 60 and 174 ppm. The ¹H NMR spectrum has four peaks: δ = 1.1 (triplet, relative area 3), δ = 1.2 (triplet, relative area 3), δ = 2.3 (quartet, relative area 2) and δ = 4.1 (quartet, relative area 2). Data booklet values, δ / ppm: ¹³C: C–C 5–40, C–O 50–90, C=O (ester) 160–185; ¹H: R–CH₃ 0.7–1.2, R–CO–CH₂– 2.1–2.6, R–COO–CH₂– 3.7–4.1, R–COO–CH₃ 3.7–4.1.
(a)
Use both spectra to deduce the structure of M. Explain how the evidence supports your answer.
[6 marks]
(b)
Methyl butanoate, CH₃CH₂CH₂COOCH₃, is an isomer of M. Explain why the number of peaks in the ¹³C NMR spectrum cannot distinguish M from methyl butanoate, and explain how the ¹H NMR spectrum does.
[6 marks]
Total for question 8: 12 marks

End of questions

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).