Questions on proton and carbon-13 NMR spectroscopy

Question 1

Explain how the NMR data in the table show that compound W, C₄H₈O₂, is methyl propanoate and not its isomer ethyl ethanoate.

Accepted Answer

Indicative content

the carbon-13 peak at 175 ppm is the C=O of an ester or acid, and with no proton peak at 10–12 ppm there is no COOH, so W is an ester
the integration 3 + 2 + 3 = 8 accounts for all eight hydrogens of C₄H₈O₂, and four carbon-13 peaks for four carbon environments
the quartet of 2H and the triplet of 3H show an ethyl group, the CH₂ split by three neighbours and the CH₃ by two
the quartet at 2.3 ppm is in the H–C–C=O range, so the CH₂ of the ethyl group is bonded to the carbonyl carbon
the singlet of 3H at 3.7 ppm is in the RCOO–C–H range, a CH₃ bonded to the ester oxygen with no neighbouring hydrogens
in ethyl ethanoate the quartet would be near 4.1 ppm, next to O, and the singlet near 2.1 ppm, next to C=O, so the data fit only methyl propanoate, CH₃CH₂COOCH₃

Levels

Level 3 (5–6 marks) the ester is identified from the carbon-13 and the absence of an acid proton, every proton signal is assigned with its splitting and its shift range, the integration is checked against the formula, and the answer shows where the signals of ethyl ethanoate would differ; the structure follows from the data rather than being asserted
Level 2 (3–4 marks) the quartet and triplet are read as an ethyl group and the singlet as OCH₃; the shift ranges are not used to place the groups either side of the ester link, or ethyl ethanoate is never compared
Level 1 (1–2 marks) a quartet and a triplet are said to show an ethyl group and the compound is then named ethyl ethanoate, or it is named correctly with no data behind it

Question 2

Deduce the number of peaks in the carbon-13 NMR spectrum of each of the four isomers of C₃H₆Cl₂: 1,1-dichloropropane, 1,2-dichloropropane, 1,3-dichloropropane and 2,2-dichloropropane.

Accepted Answer

B1 1,1-dichloropropane, CHCl₂CH₂CH₃: three different carbons, 3 peaks B1 1,2-dichloropropane, CH₂ClCHClCH₃: three different carbons, 3 peaks B1 1,3-dichloropropane, ClCH₂CH₂CH₂Cl: the two CH₂Cl carbons are equivalent, 2 peaks B1 2,2-dichloropropane, CH₃CCl₂CH₃: the two CH₃ carbons are equivalent, 2 peaks

Question 3

Explain why a sample for proton NMR is dissolved in CDCl₃ rather than CHCl₃, why CCl₄ is sometimes used instead, and why D₂O is chosen for a sample of glucose.

Accepted Answer

B1 CDCl₃ has deuterium, ²H, in place of ¹H, and deuterium does not absorb at the frequency of ¹H B1 so the solvent, present in large excess, gives no peak that could swamp or be confused with the sample's peaks B1 CCl₄ contains no hydrogen at all, so it gives no proton peak, though it dissolves only non-polar compounds well B1 glucose is polar and dissolves in D₂O, which has no ¹H, where it would barely dissolve in CDCl₃ or CCl₄

Question 4

Explain the splitting of each peak in the proton NMR spectrum of propanal, CH₃CH₂CHO, using the n + 1 rule.

Accepted Answer

B1 the n + 1 rule: hydrogens with n hydrogens on the adjacent carbons give a peak split into n + 1 lines B1 the CH₃ has two neighbouring hydrogens, on the CH₂, so it is a triplet B1 the CH₂ has four neighbouring hydrogens, three on the CH₃ and one on the CHO, so it is split into five lines B1 the CHO hydrogen, at 9.0–10.0 ppm, has two neighbouring hydrogens on the CH₂, so it is a triplet

Question 5

Describe the proton NMR spectrum of methyl butanoate, CH₃CH₂CH₂COOCH₃, giving the number of peaks and the shift region, splitting and integration of each.

Accepted Answer

B1 four peaks, in the integration ratio 3 : 2 : 2 : 3 B1 the OCH₃ is a singlet of 3H at about 3.7 ppm, in the RCOO–C–H range B1 the CH₂ bonded to the C=O carbon is a triplet of 2H at about 2.3 ppm, in the H–C–C=O range B1 the middle CH₂ is a sextet of 2H, with five neighbours, and the end CH₃ a triplet of 3H at about 0.9 ppm

Question 6

Evaluate a student's conclusion that an isomer of C₄H₁₀O is butan-1-ol because its carbon-13 NMR spectrum has four peaks.

Accepted Answer

B1 butan-1-ol has four carbon environments, so four peaks is consistent with it B1 butan-2-ol and methoxypropane also have four carbon environments, so the count alone does not decide B1 methoxypropane would show two peaks in the C–O range, 50–90 ppm, whereas each butanol shows one B1 the proton NMR spectrum is needed to separate the butanols: butan-1-ol has a CH₂ triplet of 2H near 3.6 ppm, where butan-2-ol has a single CH hydrogen of area 1

Question 7

Deduce the number of hydrogens in each environment, and the structure, of compound X, C₅H₁₀O₂, from its proton NMR spectrum summarised in the table.

Accepted Answer

M1 the steps 1.0 : 1.0 : 1.5 : 1.5 are in the ratio 2 : 2 : 3 : 3, a total of 10 parts for 10 hydrogens B1 two quartet and triplet pairs show two separate ethyl groups B1 the CH₂ quartet at 4.1 ppm is bonded to the ester oxygen and the CH₂ quartet at 2.3 ppm to the carbonyl carbon A1 X is ethyl propanoate, CH₃CH₂COOCH₂CH₃

Question 8

Deduce the structure of compound Y, C₄H₈O₂, from the proton NMR data in the table, giving the evidence for each part of the structure.

Accepted Answer

B1 the integration 2 + 3 + 3 accounts for all eight hydrogens, and with no peak at 10.0–12.0 ppm there is no COOH, so Y is an ester with both oxygens in the –COO– group B1 the quartet of 2H and the triplet of 3H show an ethyl group, each peak split by the hydrogens on the neighbouring carbon by the n + 1 rule B1 the CH₂ at 4.1 ppm is in the RCOO–C–H range, 3.7–4.1 ppm, so the ethyl group is bonded to the ester oxygen; the singlet of 3H at 2.0 ppm is a CH₃ on the carbonyl carbon, with no neighbouring hydrogens to split it B1 Y is ethyl ethanoate, CH₃COOCH₂CH₃; methyl propanoate is rejected because its quartet would lie near 2.3 ppm and its singlet near 3.7 ppm

Question 9

Give three reasons why tetramethylsilane, TMS, is used as the standard in NMR spectroscopy.

Accepted Answer

B1 all its hydrogens, and all its carbons, are equivalent, so it gives a single sharp peak B1 its peak is at a lower shift than almost all organic compounds, so it does not overlap their peaks B1 it is inert and volatile, so it does not react with the sample and is easily removed afterwards; allow non-toxic for inert

Question 10

Identify the number of peaks in the carbon-13 NMR spectrum of pentan-3-one, 2,2-dimethylpropane and butanone.

Accepted Answer

B1 pentan-3-one, CH₃CH₂COCH₂CH₃, gives 3 peaks B1 2,2-dimethylpropane, C(CH₃)₄, gives 2 peaks B1 butanone, CH₃COCH₂CH₃, gives 4 peaks

Question 11

Explain the splitting of the two peaks given by the ethyl group, CH₃CH₂–, in a proton NMR spectrum.

Accepted Answer

B1 the CH₃ hydrogens have two hydrogens on the adjacent carbon, so their peak is a triplet B1 the CH₂ hydrogens have three hydrogens on the adjacent carbon, so their peak is a quartet B1 by the n + 1 rule: 2 + 1 = 3 lines and 3 + 1 = 4 lines, with integration 3:2

Question 12

Suggest why the proton NMR spectrum of ethanoic acid, CH₃COOH, shows only two peaks, both singlets.

Accepted Answer

B1 there are two hydrogen environments, the CH₃ and the COOH, with areas in the ratio 3 : 1 B1 the CH₃ is bonded to the carbonyl carbon, which has no hydrogens, so n = 0 and it is a singlet near 2.1 ppm B1 the COOH hydrogen, at 10.0–12.0 ppm, exchanges rapidly and has no hydrogens on its neighbour, so it is also a singlet

Question 13

Identify compound P, C₄H₈O₂, from its proton NMR spectrum.

Accepted Answer

B1 the singlet at 11.9 ppm is in the 10.0–12.0 range of a carboxylic acid COOH B1 the doublet of 6H and the septet of 1H show a (CH₃)₂CH group, whose CH at 2.6 ppm is next to C=O B1 P is 2-methylpropanoic acid, (CH₃)₂CHCOOH

Question 14

Suggest the structure of an isomer of C₄H₈O whose proton NMR spectrum is summarised in the table.

Accepted Answer

B1 the quartet of 2H and triplet of 3H show an ethyl group, CH₃CH₂– B1 the singlet of 3H at 2.1 ppm is a CH₃ next to C=O with no hydrogens on its neighbour, and the CH₂ at 2.4 ppm is also next to C=O B1 the compound is butanone, CH₃COCH₂CH₃

Question 15

Explain why each carbon environment in a carbon-13 NMR spectrum gives a single line rather than a split peak.

Accepted Answer

B1 only about 1.1% of carbon atoms are carbon-13 B1 so two carbon-13 atoms are almost never bonded to each other in one molecule, and one carbon peak is not split by another B1 the splitting that attached hydrogens would cause is switched off by the instrument as the spectrum is recorded

Question 16

Compare the information given by the carbon-13 and the proton NMR spectra of an organic compound.

Accepted Answer

B1 both give one peak per chemical environment, of carbon and of hydrogen respectively, with shifts read from TMS B1 only proton NMR gives an integration ratio for the number of atoms in each environment; carbon-13 peak heights are not used B1 only proton NMR shows splitting, which gives the number of hydrogens on adjacent carbons

Question 17

Suggest the structure of an isomer of C₃H₇Cl whose proton NMR spectrum has two peaks: δ 4.1, septet, 1H, and δ 1.5, doublet, 6H.

Accepted Answer

B1 a doublet of 6H and a septet of 1H show a (CH₃)₂CH group, six equivalent hydrogens split by one neighbour and one hydrogen split by six B1 the CH at 4.1 ppm is at a high shift because it is on the carbon bonded to the electronegative chlorine atom B1 the compound is 2-chloropropane, (CH₃)₂CHCl

Question 18

Explain why one peak in the proton NMR spectrum of ethanol disappears when the sample is shaken with D₂O.

Accepted Answer

B1 the hydrogen of the O–H group exchanges with deuterium from D₂O, forming O–D B1 deuterium does not absorb in a proton NMR spectrum, so the O–H peak disappears

Question 19

Deduce the number of hydrogens in each environment of compound V, C₄H₈O₂, whose integration trace has steps of 2.4 cm, 1.6 cm and 2.4 cm.

Accepted Answer

M1 the ratio 2.4 : 1.6 : 2.4 simplifies to 3 : 2 : 3, a total of 8 parts A1 the formula has 8 hydrogens, so the environments hold 3, 2 and 3 hydrogens

Question 20

Deduce the ratio of peak areas in the proton NMR spectrum of 2-methylpropan-1-ol, (CH₃)₂CHCH₂OH.

Accepted Answer

M1 four environments: the two equivalent CH₃ groups, the CH, the CH₂ and the OH A1 the areas are in the ratio 6 : 1 : 2 : 1, in any order that matches the groups

Question 21

Explain why hydrogens on a carbon bonded to an oxygen atom absorb at a higher chemical shift than the hydrogens of an alkyl CH₃ group.

Accepted Answer

B1 oxygen is electronegative and draws electron density away from the carbon and its hydrogens B1 the hydrogens are less shielded, so they absorb at a higher chemical shift

Spectrum	δ / ppm	Splitting	Integration
Proton	3.7	Singlet	3
Proton	2.3	Quartet	2
Proton	1.1	Triplet	3
Carbon-13	175, 52, 27, 9	not split	not used

δ / ppm	Splitting	Integration step / cm
4.1	Quartet	1.0
2.3	Quartet	1.0
1.3	Triplet	1.5
1.1	Triplet	1.5

δ / ppm	Splitting	Integration
4.1	Quartet	2
2.0	Singlet	3
1.3	Triplet	3

δ / ppm	Splitting	Integration
2.4	Quartet	2
2.1	Singlet	3
1.1	Triplet	3