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A LevelChemistryCIEExam Questions37. Analytical Techniques (AL)37.4 Proton (1H) NMR Spectroscopy

Proton (1H) NMR Spectroscopy (CIE A Level Chemistry)

Exam Questions

1 hour7 questions
Easy
Medium
Hard
1a2 marks

This question is about the NMR analysis of various organic compounds.

Name and draw the structure of the chemical that is commonly used as a standard in NMR spectroscopy.

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1b3 marks

Fig. 1.1 shows the structures of compounds A, B and C.

labelled-isomers-of-pentane

Fig. 1.1

Compound A is pentane, with the chemical formula C5H12. Compound B is 2-methylbutane and compound C is 2,2-dimethylpropane, which are both isomers of pentane.

State the number of hydrogen peaks that would be expected in low resolution 1H-NMR spectrum of each isomer.

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1c3 marks

More structural details can be deduced using high resolution 1H NMR.

Explain why the methyl groups in 2-methylbutane, compound B, give a doublet splitting pattern while the methyl groups in 2,2-dimethylpropane, compound C, give a singlet splitting pattern.

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1d3 marks

Carbon-13 NMR is also commonly used to distinguish chemicals.

Predict the number of peaks in the carbon-13 NMR spectra of compounds A, B and C.

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2a
Sme Calculator3 marks

Compound X is a carbohydrate.

Compound X contains 62.1% C, 10.3% H and 27.6% O by mass.

i)
Show that the empirical formula of compound X is C3H6O. 
 
[2]
 
ii)
The empirical formula of compound X is C3H6O and the Mr is 58.0.
 
Deduce the molecular formula of compound X. You must explain your reasoning.
 
molecular formula ..............................
 
................................................................................
 
[1]

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2b2 marks

There are several possible isomers of compound X.

Draw the structures of two isomers of compound X that contain a carbonyl group.

 isomer 1

 

 

 
 

isomer 2

 

 

 

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2c3 marks

A sample of a different isomer of compound X is cyclopropanol, which was analysed by NMR spectroscopy.

i)
Predict the number of peaks in the carbon-13 NMR spectrum of cyclopropanol.
 
[1]
 
ii)
Cyclopropanol was dissolved in CDCl3 and the proton NMR spectrum of this solution was recorded as shown in Fig. 2.1.
 
cyclopropanol-proton-nmr
 
Fig. 2.1
 
Suggest one change that can be made to the solvent used for the proton NMR and how this will affect the spectrum produced.
 
[2]

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1a2 marks

Methyl cinnamate, C10H10O2, is a white crystalline solid used in the perfume industry.

The proton NMR spectrum of methyl cinnamate in the solvent CDCl3 is shown in Fig. 1.1.

methyl-cinnamate-high-res-proton-nmr

Fig. 1.1

i)
Explain why CDCl3 is used as a solvent instead of CHCl3.
 
[1]
 
ii)
Explain why TMS is added to give the small peak at chemical shift δ = 0.
 
[1]

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1b3 marks

The structure of methyl cinnamate is shown in Fig. 1.2.

 

methyl-cinnamate-skeletal

 

Fig. 1.2

 

The data in Table 1.1 should be used in answering this question.

  

Identify the proton environment that gives rise to the peak at a chemical shift of 3.8 ppm in Fig. 1.1. Explain your answer.

 
Table 1.1
 
Environment of proton  Example  chemcial shift range, δ / ppm
alkane   –CH3, –CH2–, >CH– 0.9 – 1.7
alkyl next to C=O  CH3–C=O,–CH2–C=O, >CH–C=O 2.2 – 3.0
alkyl next to aromatic ring  CH3–Ar, –CH2–Ar, >CH–Ar 2.3 – 3.0
alkyl next to electronegative atom  CH3–O,–CH2–O, –CH2–Cl 3.2 – 4.0
attached to alkene  =CHR  4.5 – 6.0
attached to aromatic ring  H–Ar  6.0 – 9.0
aldehyde  HCOR  9.3 – 10.5
alcohol  ROH  0.5 – 6.0
phenol  Ar–OH  4.5 – 7.0
carboxylic acid  RCOOH 9.0 – 13.0
alkyl amine  R–NH–  1.0 – 5.0
aryl amine  Ar–NH2  3.0 – 6.0
amide  RCONHR  5.0 – 12.0
 

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1c2 marks

Proton NMR spectroscopy can be used to distinguish between isomers of C6H12O2.

Draw the two esters with formula C6H12O2 that each have only two peaks, both singlets, in their 1H NMR spectra. The relative peak areas are 3:1 for both esters.

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1d5 marks

The proton NMR spectrum of another isomer of C6H12O2 is shown in Fig. 1.3.

 
c6h12o2-isomer-1h-nmr
 
Fig. 1.3
 
The integration values for the peaks in the proton NMR spectrum of this isomer are given in Table 1.2.
 
Table 1.2
 

Chemical shift, δ/ppm

3.8

3.5

2.6

2.2

1.2

Integration value

0.6

0.6

0.6

0.9

0.9

Splitting pattern

triplet

quartet

triplet

singlet

triplet
 
i)
Deduce the simplest ratio of the relative numbers of protons in each environment in the isomer.
 
[1]
 
ii)
The data in Table 1.1 should be used in answering this question.
 
Describe and explain the splitting patterns of the peaks at δ = 3.5 and δ = 1.2.
 
splitting pattern at δ = 3.5 ...................................................................................................
 
reason for splitting pattern at δ = 3.5 ..................................................................................
 
splitting pattern at δ = 1.2 ...................................................................................................
 
reason for splitting pattern at δ = 1.2 ..................................................................................
 
[4]

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1e1 mark

Four isomers of C6H12O2, A, B, C and D, are shown in Fig. 6.4. 

c6h12o2-abcd

Fig. 1.4

The C-13 NMR spectrum of one of the four isomers of C6H12O2 is shown in Fig. 1.5.

c6h12o2-abcd-13c-nmr

Fig. 1.5

The data in Table 1.3 should be used in answering this question.

 

Identify which of the four isomers, A, B, C or D of C6H12O2 produced the C-13 NMR spectrum shown in Fig 1.5.

 
Table 1.3
 
Hybridisation of the carbon atom Environment of carbon atom Example  Chemical shift range δ/ppm
sp3  alkyl  CH3–, CH2–, –CH<, >C<  0 – 50
sp3   next to alkene / arene –C–C=C, –C–Ar  25 – 50
sp3   next to carbonyl / carboxyl C–COR, C–O2R  30 – 65
sp3    next to halogen C–X 30 – 60
sp3   next to oxygen  C–O 50 – 70
sp2  alkene or arene  >C=C<, cie-ial-data-table-arene 110 – 160
sp2  carboxyl  R−COOH, R−COOR  160 – 185
sp2  carbonyl  R−CHO, R−CO−R  190 – 220
sp  nitrile  R−C≡N  100 – 125 
 

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2a2 marks

Ethane-1,2-diol, C2H6O2, can be distinguished from ethanedioic acid, C2H2O4, by a number of analytic techniques including MS, IR and NMR

 

Fig. 2.1 (spectrum A) and Fig. 2.2 (spectrum B) show the mass spectra of ethane-1,2-diol and ethanedioic acid.

 

q1a_21-1_ib_hl_medium_sq
 
Fig. 2.1 (spectrum A)
 
q1a2_21-1_ib_hl_medium_sq
 
Fig. 2.2 (spectrum B)
 

Complete Table 2.1 to suggest which compound is responsible for each spectrum? Explain your answer.

 
Table 2.1
 
Spectrum  Organic compound Explanation
A    
B    
 

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2b2 marks

The IR spectra of ethane-1,2-diol, C2H6O2, and ethanedioic acid dihydrate, C2H2O4.2H2O, are shown in Fig. 2.3 (Spectrum C) and Fig. 2.4 (Spectrum D).

 
q1b_21-1_ib_hl_medium_sq
 
Fig. 2.3 (spectrum C)
 
q1b2_21-1_ib_hl_medium_sq
 
Fig. 2.4 (spectrum D)
 

The data in Table 2.3 should be used in answering this question.

 

Complete Table 2.2 to suggest which compound is responsible for each spectrum? Explain your answer.

 
Table 2.2
 
Spectrum  Organic compound Explanation
C    
D    
 
Table 2.2
 
Bond  Functional groups containing
the bond		 Characteristic infrared absorption
range (in wavenumber) / cm–1
C−O  hydroxy, ester  1040 – 1300
C=C  aromatic compound, alkene  1500 – 1680
C=O  amide
carbonyl, carboxyl
ester		 1640 – 1690
1670 – 1740
1710 – 1750
C≡N  nitrile  2200 – 2250
C−H  alkane  2850 – 2950
N−H   amine, amide 3300 – 3500
O−H  carboxyl
hydroxy		 2500 – 3000
3200 – 3600
 

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2c3 marks

The proton NMR spectrum of ethane-1,2-diol is shown in Fig. 2.5.

 

Describe and explain the splitting patterns of the spectrum.

 
 
q1c_21-1_ib_hl_medium_sq
 
Fig. 2.5

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2d2 marks

Suggest the number of proton NMR peaks and splitting pattern for ethanedioic acid.

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3a1 mark

Lidocaine is used as a local anaesthetic. The structure of lidocaine is shown in Fig. 3.1.

lidocaine-structure

Fig. 3.1

A sample of lidocaine was analysed by carbon-13 NMR spectroscopy.

Predict the number of peaks in the carbon-13 NMR spectrum of lidocaine.

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3b3 marks

Lidocaine was dissolved in CDCl3 and the proton NMR spectrum of this solution was recorded as shown in Fig. 3.2.

6oXanL-P_lidocaine-proton-nmr-sketchFig. 3.2

Using Table 3.2 to complete Table 3.1 for the chemical shifts δ 1.2 ppm, 3.5 ppm and 5.5 ppm.

Table 3.1

δ / ppm environment
of proton		 number of 1H atoms
responsible for the peak		 splitting pattern
1.2 terminal methyl groups next to CH2 6 triplet
2.3      
3.0      
7.1 - 7.4 attached to the aromatic ring 3 overlapping peaks
9.0      

Table 3.2

Environment of proton  Example  chemical shift range, δ / ppm
alkane   –CH3, –CH2–, >CH– 0.9 – 1.7
alkyl next to C=O  CH3–C=O,–CH2–C=O, >CH–C=O 2.2 – 3.0
alkyl next to aromatic ring  CH3–Ar, –CH2–Ar, >CH–Ar 2.3 – 3.0
alkyl next to electronegative atom  CH3–O,–CH2–O, –CH2–Cl 3.2 – 4.0
attached to alkene  =CHR  4.5 – 6.0
attached to aromatic ring  H–Ar  6.0 – 9.0
aldehyde  HCOR  9.3 – 10.5
alcohol  ROH  0.5 – 6.0
phenol  Ar–OH  4.5 – 7.0
carboxylic acid  RCOOH 9.0 – 13.0
alkyl amine  R–NH–  1.0 – 5.0
aryl amine  Ar–NH2  3.0 – 6.0
amide  RCONHR  5.0 – 12.0

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3c1 mark

Explain the splitting pattern for the absorption at δ 1.2 ppm.

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1a
Sme Calculator3 marks

Compound P is a naturally occurring chemical found in strawberries, apples and Parmesan cheese.

The percentage by mass is carbon 58.82%, hydrogen 9.80% and oxygen 31.38%.

The mass spectrum of compound P is recorded in Fig. 1.1.

HpfzOW6M_6-9-q1a-ocr-a-as--a-level-hard-sq

Fig. 1.1

Determine the molecular formula of compound P. Show your working.

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1b4 marks

Table 1.1 shows the results of qualitative tests performed on compound P. 

Table 1.1

Test Observation
Addition of H2O Forms separate layers
Na2CO3 (aq) No visible change
2,4-DNPH No visible change
Tollens' reagent No visible change

 

Analyse the potential functional groups in compound P. Explain your answers.

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1c3 marks

The carbon-13 (13C) NMR spectrum of compound P is shown in Fig. 1.2. 

6-9-q1c-ocr-a-as--a-level-hard-sq

Fig. 1.2

Table 1.2 

Hybridisation of the carbon atom Environment of carbon atom Example  Chemical shift range
δ / ppm
sp3  alkyl  CH3–, CH2–, –CH<, >C<  0 – 50
sp3   next to alkene / arene –C–C=C, –C–Ar  25 – 50
sp3   next to carbonyl / carboxyl C–COR, C–O2R  30 – 65
sp3    next to halogen C–X 30 – 60
sp3   next to oxygen  C–O 50 – 70
sp2  alkene or arene  >C=C<, cie-ial-data-table-arene 110 – 160
sp2  carboxyl  R−COOH, R−COOR  160 – 185
sp2  carbonyl  R−CHO, R−CO−R  190 – 220
sp  nitrile  R−C≡N  100 – 125


Identify the functional group(s) present in compound P using your answer in (b) and information from Fig. 1.2 and Table 1.2. Explain your answer.

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1d3 marks

The high-resolution proton NMR spectrum of compound P was recorded as shown in Fig. 1.3.

6-9_q1d-ocr-a-as--a-level-hard-sq

Fig. 1.3

Table 1.3

Environment of proton  Example  chemical shift range, δ / ppm
alkane   –CH3, –CH2–, >CH– 0.9 – 1.7
alkyl next to C=O  CH3–C=O,–CH2–C=O, >CH–C=O 2.2 – 3.0
alkyl next to aromatic ring  CH3–Ar, –CH2–Ar, >CH–Ar 2.3 – 3.0
alkyl next to electronegative atom  CH3–O,–CH2–O, –CH2–Cl 3.2 – 4.0
attached to alkene  =CHR  4.5 – 6.0
attached to aromatic ring  H–Ar  6.0 – 9.0
aldehyde  HCOR  9.3 – 10.5
alcohol  ROH  0.5 – 6.0
phenol  Ar–OH  4.5 – 7.0
carboxylic acid  RCOOH 9.0 – 13.0
alkyl amine  R–NH–  1.0 – 5.0
aryl amine  Ar–NH2  3.0 – 6.0
amide  RCONHR  5.0 – 12.0
 

Suggest the structure of compound P using your answers to (a), (b) and (c) and information from Fig. 1.3 and Table 1.3. Explain your answer.

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2a1 mark

A chemist prepares and analyses some esters.

The chemist prepares an ester by reacting propan-2-ol with ethanoic anhydride.

Using structural formulae, write an equation for the reaction of propan-2-ol and ethanoic anhydride.

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2b3 marks

 A sample contains a mixture of two esters contaminated with an alkane and an alcohol.

The chemist attempts to separate the four organic compounds in the mixture using gas chromatography. The stationary phase in the gas chromatograph column is a liquid alkane.

i)
How does a liquid stationary phase separate the organic compounds in a mixture?
 
[1]
 
ii)
Predict the separation of these four compounds using the alkane stationary phase, including relative retention times. Explain your answer.
 
[2]

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2c8 marks

Gas chromatography is often used in conjunction with other techniques such as mass spectrometry and NMR spectroscopy.

An ester is isolated from a perfume by gas chromatography and then analysed. 

The percentage by mass is carbon 66.63%, hydrogen 11.18% and oxygen 22.19%.

The mass spectrum and high-resolution proton NMR spectrum of the ester are recorded in Fig. 2.1 and Fig. 2.2 respectively.

cie-ial-8-1h-q6-mass-spectrum

Fig. 2.1

cie-ial-8-1h-q6-proton-nmr-spectrum

Fig. 2.2

Table 2.1

Environment of proton  Example  chemical shift range, δ / ppm
alkane   –CH3, –CH2–, >CH– 0.9 – 1.7
alkyl next to C=O  CH3–C=O,–CH2–C=O, >CH–C=O 2.2 – 3.0
alkyl next to aromatic ring  CH3–Ar, –CH2–Ar, >CH–Ar 2.3 – 3.0
alkyl next to electronegative atom  CH3–O,–CH2–O, –CH2–Cl 3.2 – 4.0
attached to alkene  =CHR  4.5 – 6.0
attached to aromatic ring  H–Ar  6.0 – 9.0
aldehyde  HCOR  9.3 – 10.5
alcohol  ROH  0.5 – 6.0
phenol  Ar–OH  4.5 – 7.0
carboxylic acid  RCOOH 9.0 – 13.0
alkyl amine  R–NH–  1.0 – 5.0
aryl amine  Ar–NH2  3.0 – 6.0
amide  RCONHR  5.0 – 12.0
 

Use all of the information to draw the structure of the ester. Explain your answer.

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