C NMR spectrum of ethylbenzene, shown in Fig. 1.1.
Fig. 1.1
ii) The data in Table 1.1 should be used in answering this question. One of the carbon atoms in the structure of ethylbenzene shown in Fig. 1.1 is labelled with an asterisk (*). Suggest a C-13 chemical shift range for this carbon environment.
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< | 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 |
Compound A contains the elements carbon and hydrogen forming an aromatic ring along with the elements oxygen and nitrogen. Part of the mass spectrum of A is shown in Fig. 2.2.
Fig. 2.2
Give the identity of the molecular ion that gives rise to the peak at m/e = 76 in Fig. 2.2.
Suggest the structures of the three possible dinitrobenzene isomers of A that contain a benzene ring.
The C-13 NMR spectrum of compound A has four peaks. Identify the structure of A. Explain your reasoning by labelling the different carbon environments in all the structures drawn in part (ii).
An unknown alcohol was analysed and found to contain 64.9% carbon, 13.5% hydrogen, and the rest oxygen. The results of mass spectrometry found the mass of the alcohol to be 74.12 g/mol. Determine the molecular formula of the unknown alcohol. Show your working.
The unknown alcohol can exist as four possible isomers.
i) Using your answer to part (a), sketch the four possible isomers of the unknown alcohol.
ii) For each isomer, deduce the number of chemical peaks expected in the 13C spectrum.
The 13C NMR spectra of one of the isomers is shown in Fig. 3.1.
Fig. 3.1
Deduce which isomer produced the spectrum shown in Fig. 3.1. Explain your answer with reference to Table 3.1.
Three hydrocarbons, L, M, and N, have the molecular formula C8H10. Information about the number of peaks seen in the carbon-13 (13C) NMR spectrum of the three isomers is shown in Table 1.1.
Number of peaks | L | M | N |
---|---|---|---|
3 | 5 | 4 |
Suggest structures for compounds L, M, and N.
Complete Table 1.1 to give details of the proton NMR spectra for isomers L, M, and N.
Number of peaks | Relative Peak Area | |
---|---|---|
L | M | N |
Explain which of the three isomers, L, M, or N has the highest melting point.
A chemist analyses a naturally occurring compound, K.
The percentage composition by mass is carbon 70.58%; hydrogen 5.92%; and oxygen 23.50%. The mass spectrum of the compound is shown in Fig. 5.1.
Determine the molecular formula of the compound K. Show your working.
The results of qualitative tests performed on compound K are shown in Table 5.1.
Test | Observation |
---|---|
Acidity | pH 5.0 |
Na2CO3 (aq) | No reaction |
2,4-DNPH | Orange precipitate |
Tollens' reagent | No reaction |
Identify the functional groups present in compound K. Explain your answer.
The carbon-13 (13C) NMR of compound K is shown in Fig. 5.2.
Fig. 5.2
Suggest the structure of compound K using your answers to (a) and (b) along with information from Fig. 5.2 and Table 5.1. Explain your answer.
--------------------------------------------------------------------------------
1a
2 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.
1b
3 marks
Fig. 1.1 shows the structures of compounds A, B, and C.
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.
1c
3 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.
1d
3 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.
2a
3 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]
2b
2 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:
2c
3 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.
Suggest one change that can be made to the solvent used for the proton NMR and how this will affect the spectrum produced. [2]
1a
2 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.
(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]
1b
3 marks
The structure of methyl cinnamate is shown in 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 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
1c
2 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.
1d
5 marks
The proton NMR spectrum of another isomer of C6H12O2 is shown in 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]
1e
1 mark
Four isomers of C6H12O2, A, B, C, and D, are shown in Fig. 6.4.
The C-13 NMR spectrum of one of the four isomers of C6H12O2 is shown in Fig. 1.5.
Identify which of the four isomers, A, B, C, or D of C6H12O2 produced the C-13 NMR spectrum shown in Fig 1.5.
2a
2 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.
Complete Table 2.1 to suggest which compound is responsible for each spectrum? Explain your answer.
2b
2 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).
Complete Table 2.2 to suggest which compound is responsible for each spectrum? Explain your answer.
2c
3 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.
2d
2 marks
Suggest the number of proton NMR peaks and splitting pattern for ethanedioic acid.
3a
1 mark
Lidocaine is used as a local anaesthetic. The structure of lidocaine is shown in 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.
3b
3 marks
Lidocaine was dissolved in CDCl3 and the proton NMR spectrum of this solution was recorded as shown in Fig. 3.2.
Using Table 3.2, 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
- 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
3c
1 mark
Explain the splitting pattern for the absorption at δ 1.2 ppm.
1a
3 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.
Determine the molecular formula of compound P. Show your working.
1b
4 marks
Table 1.1 shows the results of qualitative tests performed on compound P.
- Test: Addition of H2O - Observation: Forms separate layers
- Test: Na2CO3(aq) - Observation: No visible change
- Test: 2,4-DNPH - Observation: No visible change
- Test: Tollens' reagent - Observation: No visible change
Analyse the potential functional groups in compound P. Explain your answers.
1c
3 marks
The carbon-13 (13C) NMR spectrum of compound P is shown in 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="" li="">
- 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<: 110="" 160="" li="">
- 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.
1d
3 marks
The high-resolution proton NMR spectrum of compound P was recorded as shown in Fig. 1.3.
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.
2a
1 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.
2b
3 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]
2c
8 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.
Use all of the information to draw the structure of the ester. Explain your answer.