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Questions on putting the analysis together: structure from spectra
A method for identifying an organic compound: molecular formula from high-resolution mass spectrometry or combustion data, the degree of unsaturation, infrared for the functional group, carbon-13 NMR for carbon environments, proton NMR for shift, integration and splitting, and a final check; three worked analyses of C₃H₆O, C₄H₈O₂ and C₈H₈O; the pitfalls of symmetry, O–H peaks and integration ratios; and, for other boards, D₂O exchange, external calibration in gas chromatography, and infrared monitoring of carbon monoxide and nitrogen monoxide in exhaust gases.
20 original questions · 66 marks · the putting the analysis together: structure from spectra notes · Organic synthesis · Flashcards for this unit
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Identify compound G, C₈H₈O₂, from the data in the table, explaining how each piece of evidence supports your structure.
Technique Data High-resolution MS M⁺ at m/z 136.0522 Infrared Strong absorptions at 1725 and 1280 cm⁻¹; weak absorptions just above 3000 cm⁻¹; nothing broad Carbon-13 NMR Six peaks: 167, 133, 130, 129.6, 128 and 52 ppm Proton NMR δ 8.0, 2H; δ 7.4 to 7.6, 3H; δ 3.9, singlet, 3H Mark scheme
Indicative content
- C₈H₈O₂ matches the precise mass and has a degree of unsaturation of 5, a benzene ring plus one double bond
- infrared: C=O at 1725 and C–O at 1280 cm⁻¹ with no broad O–H, so an ester rather than a carboxylic acid
- carbon-13: 167 ppm is an ester C=O and 52 ppm a carbon bonded to oxygen; four peaks from 128 to 133 are ring carbons
- four ring environments and five ring hydrogens (2H + 3H) show a benzene ring with a single substituent
- the 3H singlet at 3.9 ppm is an OCH₃ on the ester oxygen, with no neighbouring hydrogens
- G is methyl benzoate, C₆H₅COOCH₃; phenyl ethanoate is ruled out because its CH₃ would sit next to C=O near 2.3 ppm
Levels
- Level 3 (5–6 marks) every piece of evidence is used: the formula and its unsaturation, the infrared showing an ester, the carbon-13 shifts, the ring hydrogens and environments showing one substituent, and the OCH₃ singlet, ending with methyl benzoate and the reason phenyl ethanoate is excluded
- Level 2 (3–4 marks) the ester and the benzene ring are identified and methyl benzoate given; one piece of evidence goes unused or phenyl ethanoate is not excluded
- Level 1 (1–2 marks) an ester or a ring is spotted from one piece of data, and the structure is often given as phenyl ethanoate or as a ketone
Calculate the molecular formula of a compound of carbon, hydrogen and oxygen from the combustion data in the table.
Measurement Value Mass of compound burned 0.580 g Mass of CO₂ formed 1.320 g Mass of H₂O formed 0.540 g m/z of the molecular ion 58 Mark scheme
- M1 C: 1.320 ÷ 44.0 = 0.0300 mol, mass 0.360 g; H: 0.540 ÷ 18.0 = 0.0300 mol H₂O, so 0.0600 mol H, mass 0.0600 g
- M1 O: 0.580 − 0.360 − 0.0600 = 0.160 g, so 0.160 ÷ 16.0 = 0.0100 mol; error carried forward from the C and H masses
- M1 C : H : O = 0.0300 : 0.0600 : 0.0100 = 3 : 6 : 1, so the empirical formula is C₃H₆O, mass 58.0
- A1 the empirical mass equals Mr, so the molecular formula is C₃H₆O
Deduce the structure of compound J, C₅H₁₀O₂, an ester, from the proton NMR data in the table, showing how the integration is scaled.
δ / ppm Splitting Relative integration 4.1 quartet 1.0 2.3 quartet 1.0 1.25 triplet 1.5 1.1 triplet 1.5 Mark scheme
- B1 the integrations sum to 5 and the formula has 10 hydrogens, so the ratio scales by 2 to 2H : 2H : 3H : 3H
- B1 two quartets of 2H and two triplets of 3H are two separate ethyl groups, each CH₂ split by a CH₃ and each CH₃ by a CH₂
- B1 the CH₂ at 4.1 ppm is bonded to the ester oxygen and the CH₂ at 2.3 ppm is next to the C=O
- B1 J is ethyl propanoate, CH₃CH₂COOCH₂CH₃; accept a correct displayed formula
Deduce the number of peaks in the carbon-13 and in the proton NMR spectra of 1,4-dimethylbenzene and of 1,2-dimethylbenzene, explaining the symmetry in each.
Mark scheme
- B1 1,4-dimethylbenzene, carbon-13: 3 peaks, for the two equivalent CH₃ carbons, the two ring carbons bearing methyls and the four equivalent ring CH carbons
- B1 1,4-dimethylbenzene, proton: 2 peaks, the six methyl hydrogens and the four ring hydrogens, in the ratio 6 : 4
- B1 1,2-dimethylbenzene, carbon-13: 4 peaks, because the mirror plane between the methyls pairs C1 with C2, C3 with C6 and C4 with C5, plus the CH₃ carbons
- B1 1,2-dimethylbenzene, proton: 3 environments, CH₃ 6H, H3 and H6 2H, H4 and H5 2H; allow 2 peaks if the ring hydrogens are said to overlap
Evaluate a student's conclusion that compound K, C₅H₁₀O, is pentan-2-one because it absorbs strongly at 1715 cm⁻¹ and its carbon-13 spectrum has five peaks, one of them at 209 ppm.
Mark scheme
- B1 the absorption is a C=O and the 209 ppm peak an aldehyde or ketone carbon; five peaks fit pentan-2-one, whose five carbons are all different
- B1 pentan-3-one is ruled out, because its symmetry gives only three carbon peaks
- B1 pentanal also has five carbon environments and a C=O in the same ranges, so these data do not separate it from pentan-2-one; allow 2-methylbutanal in place of pentanal, since it too has five carbon environments
- B1 the conclusion is not yet justified: proton NMR would decide, with a 3H singlet near 2.1 ppm for pentan-2-one or a 1H peak near 9.8 ppm for pentanal
Deduce the structure of compound H, C₃H₆O₂, from the data in the table, explaining how the data rule out its isomers propanoic acid and ethyl methanoate.
Technique Data Infrared Strong absorptions at 1745 and 1240 cm⁻¹; nothing broad between 2500 and 3550 cm⁻¹ Carbon-13 NMR Three peaks: 171, 52 and 21 ppm Proton NMR δ 3.7, singlet, 3H; δ 2.1, singlet, 3H Mark scheme
- B1 C=O at 1745 and C–O at 1240 cm⁻¹ with no broad O–H show an ester, so propanoic acid, whose O–H would absorb very broadly from 2500 to 3000 cm⁻¹, is ruled out
- B1 three carbon-13 peaks for three carbons mean no two carbons are equivalent: 171 ppm is the ester C=O, 52 ppm a carbon bonded to oxygen and 21 ppm an alkyl carbon
- B1 two singlets of 3H are two CH₃ groups with no neighbouring hydrogens: the one at 3.7 ppm, in the RCOO–C–H range, is bonded to the ester oxygen and the one at 2.1 ppm, in the H–C–C=O range, to the carbonyl carbon
- B1 H is methyl ethanoate, CH₃COOCH₃; ethyl methanoate, HCOOCH₂CH₃, is ruled out because it would show a quartet and a triplet from its ethyl group and a 1H singlet near 8 ppm for the HCOO hydrogen
Calculate the precise masses of C₄H₈O, C₃H₄O₂ and C₅H₁₂ from the atomic masses in the table, and use them to identify a compound whose molecular ion is at m/z 72.0573.
Atom Precise relative mass ¹H 1.0078 ¹²C 12.0000 ¹⁶O 15.9949 Mark scheme
- M1 C₄H₈O = 48.0000 + 8.0624 + 15.9949 = 72.0573
- M1 C₃H₄O₂ = 36.0000 + 4.0312 + 31.9898 = 72.0210 and C₅H₁₂ = 60.0000 + 12.0936 = 72.0936
- A1 only C₄H₈O matches 72.0573 to four decimal places, so the molecular formula is C₄H₈O
Calculate the degree of unsaturation of C₈H₈O₂ and of C₄H₇N, and state what the first value suggests about the structure.
Mark scheme
- M1 degree of unsaturation = (2C + 2 + N − H) ÷ 2
- A1 C₄H₇N: (8 + 2 + 1 − 7) ÷ 2 = 2, for example the triple bond of a nitrile
- A1 C₈H₈O₂: (16 + 2 − 8) ÷ 2 = 5, which suggests a benzene ring (four) and one C=O
Explain how shaking a sample of propan-1-ol with D₂O identifies which peak in its proton NMR spectrum belongs to the O–H, and why the other peaks are unchanged.
Mark scheme
- B1 the O–H hydrogen exchanges with deuterium, giving O–D
- B1 deuterium does not absorb in the proton NMR spectrum, so the O–H peak disappears when the spectrum is run again
- B1 hydrogens bonded to carbon do not exchange with D₂O, so the CH₃ and CH₂ peaks stay where they were
Explain why ethoxyethane, CH₃CH₂OCH₂CH₃, gives two peaks in its carbon-13 NMR spectrum, while its isomer butan-1-ol gives four.
Mark scheme
- B1 ethoxyethane is symmetrical about its oxygen atom
- B1 so its two CH₂ carbons are equivalent and give one peak, and its two CH₃ carbons are equivalent and give another
- B1 in butan-1-ol each carbon has different neighbours, so there are four environments and four peaks
Explain the splitting of the peaks at δ 4.0 and δ 1.6 in the proton NMR spectrum of propyl ethanoate, CH₃COOCH₂CH₂CH₃, whose peaks are given in the table, and why the peak at δ 2.0 is a singlet.
δ / ppm Relative number of hydrogens 4.0 2 2.0 3 1.6 2 0.9 3 Mark scheme
- B1 the CH₂ at 4.0 ppm is bonded to the ester oxygen and has one neighbouring CH₂ with two hydrogens, so it is a triplet
- B1 the CH₂ at 1.6 ppm has a CH₂ on one side and a CH₃ on the other, five neighbouring hydrogens, so it is split into six lines; accept 'multiplet'
- B1 the CH₃ at 2.0 ppm is bonded to the C=O carbon, which carries no hydrogens, so it is not split
Explain why the molecular formula is found before any spectrum is interpreted when identifying an unknown compound.
Mark scheme
- B1 the formula fixes the numbers of carbons, hydrogens and other atoms that every structure must contain
- B1 it gives the degree of unsaturation, which limits how many rings and double bonds the structure can have
- B1 it allows integration ratios to be scaled to real numbers of hydrogens and the carbon-13 peak count to be compared with the number of carbons
Identify compound E, C₃H₆O, from the data in the table, giving the evidence for each part of its structure.
Technique Data Infrared Strong absorption at 1730 cm⁻¹; nothing broad above 3000 cm⁻¹ Carbon-13 NMR Three peaks: 203, 37 and 6 ppm Proton NMR δ 9.8, triplet, 1H; δ 2.4, multiplet, 2H; δ 1.1, triplet, 3H Mark scheme
- B1 the 1730 cm⁻¹ absorption and the 203 ppm carbon peak show an aldehyde or ketone C=O, with no O–H
- B1 the 1H peak at 9.8 ppm is an aldehyde CHO hydrogen, so E is an aldehyde, and three carbon peaks mean three different carbons
- B1 the CH₃ triplet and CH₂ multiplet are an ethyl group next to the CHO, so E is propanal, CH₃CH₂CHO
Suggest a structure for compound F, C₄H₁₀O, from the data in the table, and explain how the data rule out an alcohol.
Technique Data Infrared Strong absorption at 1120 cm⁻¹; nothing broad between 3230 and 3550 cm⁻¹ Carbon-13 NMR Two peaks: 66 and 15 ppm Proton NMR δ 3.5, quartet, 4H; δ 1.2, triplet, 6H Mark scheme
- B1 there is no broad O–H absorption and no peak that could be an O–H, so F is not an alcohol; the 1120 cm⁻¹ absorption is C–O
- B1 four carbons give only two carbon peaks, so the molecule is symmetrical, with one carbon bonded to O at 66 ppm
- B1 a quartet of 4H and a triplet of 6H are two equivalent ethyl groups on oxygen, so F is ethoxyethane, CH₃CH₂OCH₂CH₃
Suggest how proton NMR alone distinguishes butanone, CH₃COCH₂CH₃, from butanal, CH₃CH₂CH₂CHO.
Mark scheme
- B1 butanal shows a peak at 9 to 10 ppm for the aldehyde hydrogen, which butanone does not
- B1 butanone shows a 3H singlet near 2.1 ppm, a CH₃ next to the C=O with no neighbouring hydrogens
- B1 butanone has three hydrogen environments in the ratio 2 : 3 : 3, butanal four in the ratio 1 : 2 : 2 : 3
Compare the information given by a carbon-13 NMR spectrum with that given by a proton NMR spectrum of the same compound.
Mark scheme
- B1 both give one peak per environment, and the shift of each peak shows what the atoms are bonded to
- B1 proton NMR also gives an integration ratio of the numbers of hydrogens, while carbon-13 peak heights do not give numbers of carbons
- B1 proton NMR shows splitting by neighbouring hydrogens, which shows how groups are joined; carbon-13 spectra at this level show single lines, but they detect carbons with no hydrogens, such as C=O
Describe what an infrared spectrum can show about an unknown compound, and one thing it cannot show.
Mark scheme
- B1 absorptions at characteristic wavenumbers show which bonds are present, such as C=O at 1680 to 1750 cm⁻¹
- B1 the width and position of an O–H absorption distinguish an alcohol (3230 to 3550 cm⁻¹) from a carboxylic acid (2500 to 3000 cm⁻¹)
- B1 it cannot show where in the molecule a group is, or how many carbons there are; the fingerprint region is matched to a database rather than assigned
Calculate the concentration of ethanol in a sample from the gas chromatography data in the table, where peak area is proportional to concentration. The sample gives an ethanol peak area of 3650.
Standard concentration / g dm⁻³ Peak area 0.100 2710 0.200 5400 0.300 8120 Mark scheme
- M1 the standards give a straight line through the origin with a gradient of 5400 ÷ 0.200 = 27 000 area units per g dm⁻³
- M1 concentration of the sample = 3650 ÷ 27 000; allow a value read from a plotted calibration line
- A1 concentration = 0.135 g dm⁻³, to three significant figures
Explain why an infrared breathalyser measures the ethanol in a driver's breath by its absorption near 2950 cm⁻¹ rather than at the O–H absorption near 3300 cm⁻¹.
Mark scheme
- B1 exhaled breath is saturated with water vapour, whose O–H bonds absorb strongly across 3200–3600 cm⁻¹, so a reading there would measure the water as well as the ethanol
- B1 near 2950 cm⁻¹ the absorption is by the C–H bonds of ethanol, which the other gases in breath do not have, so the absorbance rises with the ethanol concentration alone
Suggest why an external calibration graph is used to find a concentration from gas chromatography, rather than the peak area alone.
Mark scheme
- B1 the detector's response per gram differs from one compound to another and from one instrument setting to another, so a peak area is not itself a concentration
- B1 standards of the same compound run under the same conditions give the area produced by a known concentration, so the unknown can be read from the line
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