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school. “Paper 42” means Paper 4, variant 2.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Proton (1H) NMR · Core · Paper 4 style · 4 marks
¹H chemical shifts (δ/ppm): R–CH₃, R–CH₂–R 0.9–1.7; H–C–C=O or H–C attached to a benzene ring 2.0–3.0; H–C–O (alcohols, ethers, esters) 3.3–4.8; H–C–Cl 3.0–4.2; R–OH 0.5–6.0 (variable); benzene ring H 6.0–9.0; –CHO 9.3–10.5; –COOH 9.0–13.0.
(a)Ethyl ethanoate, CH₃COOCH₂CH₃, is the fruity solvent in nail-polish remover. Predict its ¹H NMR spectrum: for each peak give the approximate δ, the relative number of protons and the splitting pattern.[3]
B1 each peak with δ range, integration and splitting.
(b) It is next to a CH₃ group with 3 H; n + 1 rule → 4 lines (quartet).
B1.
Q2Proton (1H) NMR · Core · Paper 4 style · 3 marks
(a)State why tetramethylsilane, Si(CH₃)₄, is used as the reference in ¹H NMR.[2]
(b)Explain why samples are dissolved in CDCl₃ rather than CHCl₃.[1]
Show answer
(a) All 12 H are equivalent giving one strong sharp peak, defined as δ = 0, upfield of most organic protons; it is inert, non-toxic and volatile (easily removed).
B1 single peak at δ 0 (12 equivalent H); B1 any one property: inert / volatile / non-toxic.
(b) CDCl₃ contains no ¹H, so the solvent gives no peak that would interfere.
B1.
Q3Proton (1H) NMR · Core · Paper 4 style · 4 marks
¹H chemical shifts (δ/ppm): R–CH₃, R–CH₂–R 0.9–1.7; H–C–C=O or H–C attached to a benzene ring 2.0–3.0; H–C–O (alcohols, ethers, esters) 3.3–4.8; H–C–Cl 3.0–4.2; R–OH 0.5–6.0 (variable); benzene ring H 6.0–9.0; –CHO 9.3–10.5; –COOH 9.0–13.0.
(a)Predict the ¹H NMR spectrum of 2-chloropropane, CH₃CHClCH₃: number of peaks, relative areas, splitting and approximate δ.[3]
(b)Predict the number of peaks for 1-chloropropane.[1]
Show answer
(a) 2 peaks: δ ≈ 1.5, 6H, doublet (two equivalent CH₃ next to one H); δ ≈ 4.2, 1H, septet (CH next to six H, bonded to Cl)
B1 two peaks ratio 6:1; B1 doublet and septet; B1 δ values in range.
(b) 3
B1.
Q4Proton (1H) NMR · Standard · Paper 4 style · 4 marks
¹H chemical shifts (δ/ppm): R–CH₃, R–CH₂–R 0.9–1.7; H–C–C=O or H–C attached to a benzene ring 2.0–3.0; H–C–O (alcohols, ethers, esters) 3.3–4.8; H–C–Cl 3.0–4.2; R–OH 0.5–6.0 (variable); benzene ring H 6.0–9.0; –CHO 9.3–10.5; –COOH 9.0–13.0.
(a)Compound K, C₃H₆O, shows a single peak at δ 2.1 in its ¹H NMR spectrum. Identify K and explain.[2]
(b)Isomer L shows three peaks with relative areas 1 : 2 : 3; the 1H peak is at δ 9.8. Identify L.[1]
(c)Predict the splitting of the CH₃ peak of L.[1]
Show answer
(a) Propanone, CH₃COCH₃: both CH₃ groups are equivalent, next to C=O (δ 2.0–3.0), and have no H on the adjacent carbon so the peak is a singlet.
B1 propanone; B1 one environment, singlet, δ consistent with H–C–C=O.
(b) Propanal, CH₃CH₂CHO
B1.
(c) Triplet (next to CH₂)
B1.
Q5Proton (1H) NMR · Standard · Paper 4 style · 6 marks
(a)The ¹H NMR spectrum of ethanol shows peaks at δ 1.2 (3H), δ 2.6 (1H) and δ 3.7 (2H). Assign each peak and give its splitting pattern.[3]
(b)A few drops of D₂O are added and the spectrum is run again. State and explain the change.[2]
(c)Suggest why the O–H peak is not split, and does not split the CH₂ peak.[1]
(b) The δ 2.6 peak disappears: the O–H proton exchanges with deuterium (CH₃CH₂OH + D₂O ⇌ CH₃CH₂OD + HOD) and D does not absorb in this region.
B1 OH peak disappears; B1 exchange with D, which gives no ¹H signal.
(c) The O–H proton exchanges rapidly between molecules, so no coupling is seen.
B1.
Q6Proton (1H) NMR · Standard · Paper 4 style · 5 marks
¹H chemical shifts (δ/ppm): R–CH₃, R–CH₂–R 0.9–1.7; H–C–C=O or H–C attached to a benzene ring 2.0–3.0; H–C–O (alcohols, ethers, esters) 3.3–4.8; H–C–Cl 3.0–4.2; R–OH 0.5–6.0 (variable); benzene ring H 6.0–9.0; –CHO 9.3–10.5; –COOH 9.0–13.0.
(a)Ester N, C₄H₈O₂, has peaks at δ 1.1 (triplet, 3H), δ 2.3 (quartet, 2H) and δ 3.7 (singlet, 3H). Identify N, explaining each peak.[4]
(b)Explain why ethyl ethanoate, also C₄H₈O₂, would give a quartet at a different δ value.[1]
Show answer
(a) Methyl propanoate, CH₃CH₂COOCH₃. δ 3.7 singlet 3H: CH₃–O with no neighbouring H. δ 2.3 quartet 2H: CH₂ next to C=O and next to CH₃. δ 1.1 triplet 3H: CH₃ next to CH₂.
B1 identity; B1 each peak assigned (×3).
(b) In ethyl ethanoate the CH₂ is bonded to O (H–C–O, δ ≈ 4.1), not to C=O.
B1.
Q7Proton (1H) NMR · Standard · Paper 4 style · 5 marks
¹H chemical shifts (δ/ppm): R–CH₃, R–CH₂–R 0.9–1.7; H–C–C=O or H–C attached to a benzene ring 2.0–3.0; H–C–O (alcohols, ethers, esters) 3.3–4.8; H–C–Cl 3.0–4.2; R–OH 0.5–6.0 (variable); benzene ring H 6.0–9.0; –CHO 9.3–10.5; –COOH 9.0–13.0.
(a)Compare the ¹H NMR spectra of propan-1-ol and propan-2-ol: give the number of peaks and their ratio for each.[2]
(b)Give the splitting of the peak for the H on the carbon bonded to O in each alcohol (ignore coupling to O–H).[2]
(c)A forensic lab in Islamabad needs to decide which alcohol is in a sample. Name one other feature that would distinguish them.[1]
(b) Propan-1-ol CH₂–O: triplet (next to CH₂). Propan-2-ol CH–O: septet (next to two CH₃).
B1 each.
(c) The 6H doublet at δ ≈ 1.2 occurs only for propan-2-ol (or a 3H triplet only for propan-1-ol).
B1.
Q8Proton (1H) NMR · Standard · Paper 4 style · 4 marks
¹H chemical shifts (δ/ppm): R–CH₃, R–CH₂–R 0.9–1.7; H–C–C=O or H–C attached to a benzene ring 2.0–3.0; H–C–O (alcohols, ethers, esters) 3.3–4.8; H–C–Cl 3.0–4.2; R–OH 0.5–6.0 (variable); benzene ring H 6.0–9.0; –CHO 9.3–10.5; –COOH 9.0–13.0.The integration trace of a ¹H NMR spectrum gives step heights of 2.4 cm (δ 2.1), 1.6 cm (δ 2.5) and 2.4 cm (δ 1.1).
(a)Calculate the simplest whole-number ratio of protons for the three peaks.[1]
(b)The compound is C₄H₈O and the peak at δ 2.1 is a singlet. Identify the compound and assign the other two peaks.[3]
Show answer
(a) 3 : 2 : 3
B1 2.4 : 1.6 : 2.4 ÷ 0.8.
(b) Butanone, CH₃COCH₂CH₃. δ 2.1 singlet 3H: CH₃–C=O. δ 2.5 quartet 2H: CH₂ next to C=O. δ 1.1 triplet 3H: CH₃ of ethyl.
B1 butanone; B1 quartet = CH₂; B1 triplet = CH₃.
Q9Proton (1H) NMR · Challenge · Paper 4 style · 6 marks
¹H chemical shifts (δ/ppm): R–CH₃, R–CH₂–R 0.9–1.7; H–C–C=O or H–C attached to a benzene ring 2.0–3.0; H–C–O (alcohols, ethers, esters) 3.3–4.8; H–C–Cl 3.0–4.2; R–OH 0.5–6.0 (variable); benzene ring H 6.0–9.0; –CHO 9.3–10.5; –COOH 9.0–13.0.¹H NMR of M: δ 1.25 (doublet, 6H); δ 2.9 (septet, 1H); δ 7.2 (multiplet, 5H).
(a)Use the data to identify compound M, C₉H₁₂. Explain how each peak supports your answer.[4]
(b)Propylbenzene is an isomer of M. State the number of peaks (excluding aromatic H) and their splitting in propylbenzene.[2]
Show answer
(a) (1-methylethyl)benzene (cumene), C₆H₅CH(CH₃)₂. δ 7.2 (5H): monosubstituted benzene ring. δ 1.25 doublet 6H: two equivalent CH₃ next to one CH. δ 2.9 septet 1H: CH next to six H (two CH₃) and attached to the ring (δ 2–3).
B1 identity; B1 5H aromatic, C₆H₅–; B1 6H doublet → (CH₃)₂CH; B1 septet 1H at 2.9 → CH attached to ring next to 6 H.
Q10Proton (1H) NMR · Challenge · Paper 4 style · 5 marks
¹H chemical shifts (δ/ppm): R–CH₃, R–CH₂–R 0.9–1.7; H–C–C=O or H–C attached to a benzene ring 2.0–3.0; H–C–O (alcohols, ethers, esters) 3.3–4.8; H–C–Cl 3.0–4.2; R–OH 0.5–6.0 (variable); benzene ring H 6.0–9.0; –CHO 9.3–10.5; –COOH 9.0–13.0.
(a)Ketone P, C₅H₁₀O, shows only two peaks: δ 1.05 (triplet) and δ 2.45 (quartet), with area ratio 3 : 2. Identify P and explain.[3]
(b)Pentan-2-one is an isomer of P. State the number of peaks in its spectrum and identify the singlet.[2]
Show answer
(a) Pentan-3-one, CH₃CH₂COCH₂CH₃. Two equivalent ethyl groups: CH₂ next to C=O (δ 2.45, quartet, next to CH₃) and CH₃ (δ 1.05, triplet, next to CH₂); 6H : 4H = 3 : 2.
B1 identity; B1 symmetry → two environments; B1 splitting/δ explained.
(b) 4 peaks; singlet (3H) at δ ≈ 2.1 for CH₃–C=O
B1; B1.
Q11Proton (1H) NMR · Challenge · Paper 4 style · 5 marks
¹H chemical shifts (δ/ppm): R–CH₃, R–CH₂–R 0.9–1.7; H–C–C=O or H–C attached to a benzene ring 2.0–3.0; H–C–O (alcohols, ethers, esters) 3.3–4.8; H–C–Cl 3.0–4.2; R–OH 0.5–6.0 (variable); benzene ring H 6.0–9.0; –CHO 9.3–10.5; –COOH 9.0–13.0.Compound Q, C₈H₈O₂, is a white solid. ¹H NMR: δ 2.4 (singlet, 3H); δ 7.3 (2H); δ 7.9 (2H); δ 12.5 (singlet, 1H; disappears when D₂O is added).
(a)Deduce the structure of Q. Explain each peak.[4]
(b)Explain why the 2-methyl isomer would show a different aromatic region.[1]
Show answer
(a) 4-methylbenzoic acid, CH₃C₆H₄COOH. δ 12.5 (1H, disappears with D₂O): COOH. Two 2H peaks at 7.3 and 7.9: a 1,4-disubstituted ring (two pairs of equivalent ring H). δ 2.4 singlet 3H: CH₃ attached to the ring, no neighbouring H.
B1 identity; B1 COOH; B1 para-disubstituted ring from two 2H peaks; B1 CH₃ on ring.
(b) Its four ring H are all in different environments (four 1H signals / complex multiplet), not two pairs.
B1.
Q12Mass spectrometry · Core · Paper 1 style · 1 mark
(a)The mass spectrum of an organic compound has two molecular ion peaks, at m/e 122 and 124, of almost equal height. Which element does the compound contain?[1]
Show answer
(a) A bromine
⁷⁹Br and ⁸¹Br occur in about a 1 : 1 ratio, giving M and M+2 peaks of equal height. Chlorine (³⁵Cl : ³⁷Cl ≈ 3 : 1) would give M+2 one-third the height of M.
Q13Mass spectrometry · Core · Paper 2 style · 3 marks
(a)Gallium, used in LED lights, has isotopes ⁶⁹Ga (60.1 %) and ⁷¹Ga (39.9 %). Calculate the relative atomic mass of gallium to 3 significant figures.[2]
(b)The mass spectrum of gallium shows peaks at m/e 69 and 71. Give the ratio of their heights as a ratio of small whole numbers.[1]
Show answer
(a) 69.8
M1 (69 × 60.1 + 71 × 39.9)/100; A1 69.8.
(b) 3 : 2
B1 60.1 : 39.9 = 1.51 : 1 ≈ 3 : 2.
Q14Mass spectrometry · Standard · Paper 1 style · 1 mark
(a)In the mass spectrum of a hydrocarbon the molecular ion peak has a relative height of 42.0 and the M+1 peak has a relative height of 1.85. How many carbon atoms are in one molecule?[1]
Show answer
(a) A 4
n = (100 × 1.85)/(1.1 × 42.0) = 4.00 ≈ 4. Three carbons would give an M+1 height of only about 1.39, and five about 2.31.
Q15Mass spectrometry · Standard · Paper 2 style · 5 marks
(a)Butanone, CH₃COCH₂CH₃, gives peaks at m/e 72, 57, 43 and 29. Give the formula of the ion responsible for each peak.[4]
(b)Write an equation for the fragmentation of the molecular ion that gives the peak at m/e 43.[1]
Q16Mass spectrometry · Standard · Paper 1 style · 1 mark
(a)The mass spectrum of ethanol, CH₃CH₂OH, has a large peak at m/e 31. Which ion causes it?[1]
Show answer
(a) A CH₂OH⁺
Breaking the C–C bond of the molecular ion gives CH₂OH⁺ (12 + 2 + 16 + 1 = 31). CH₃CH₂⁺ has m/e 29; CH₃O⁻ is a negative ion, which is not detected. D also has m/e 31 but would need 7 H atoms; ethanol has only 6, so it cannot form.
Q17Mass spectrometry · Challenge · Paper 2 style · 4 marks
(a)Give the m/e values of the three molecular ion peaks of dichloromethane, CH₂Cl₂. (C = 12, H = 1, Cl = 35 or 37)[1]
(b)Show that the relative heights of these peaks are 9 : 6 : 1, given that ³⁵Cl : ³⁷Cl = 3 : 1.[3]
Q18Mass spectrometry · Challenge · Paper 2 style · 5 marks
A compound contains carbon, hydrogen and bromine only. Its mass spectrum shows peaks at m/e 136 and 138 of equal height (relative height 100 each), and an M+1 peak at m/e 137 of relative height 4.4.
(a)Explain what the two molecular ion peaks of equal height tell you.[1]
(b)Use the M+1 peak to calculate the number of carbon atoms.[1]
(c)Deduce the molecular formula, showing that it fits m/e 136.[2]
(d)A large peak at m/e 57 is present. Identify the ion.[1]
Show answer
(a) The compound contains one bromine atom (⁷⁹Br and ⁸¹Br, 1 : 1).
Q19Infrared spectroscopy · Core · Paper 1 style · 1 mark
(a)IR absorption ranges (cm⁻¹): O–H in alcohols 3200–3600 (broad); O–H in carboxylic acids 2500–3000 (very broad); C–H 2850–2950; C=O 1670–1740; C=C 1620–1680; C–O 1040–1300.An IR spectrum shows a strong absorption at 1715 cm⁻¹ and a very broad absorption from 2500 to 3000 cm⁻¹. Which compound could it be?[1]
Show answer
(a) A propanoic acid
1715 cm⁻¹ = C=O and the very broad 2500–3000 band = O–H of a carboxylic acid. Propanone has C=O but no O–H band.
Q20Infrared spectroscopy · Core · Paper 2 style · 4 marks
IR absorption ranges (cm⁻¹): O–H in alcohols 3200–3600 (broad); O–H in carboxylic acids 2500–3000 (very broad); C–H 2850–2950; C=O 1670–1740; C=C 1620–1680; C–O 1040–1300.
(a)State one absorption, with its range, present in propan-2-ol but not in propanone.[1]
(b)State one absorption present in propanone but not in propan-2-ol.[1]
(c)Name the reagent that converts propan-2-ol into propanone and describe how IR could show the reaction is complete.[2]
Show answer
(a) O–H, 3200–3600 cm⁻¹ (broad)
B1.
(b) C=O, 1670–1740 cm⁻¹
B1.
(c) Acidified potassium dichromate(VI) (heat); complete when the O–H band has disappeared and a strong C=O band is present.
B1; B1.
Q21Infrared spectroscopy · Standard · Paper 2 style · 4 marks
IR absorption ranges (cm⁻¹): O–H in alcohols 3200–3600 (broad); O–H in carboxylic acids 2500–3000 (very broad); C–H 2850–2950; C=O 1670–1740; C=C 1620–1680; C–O 1040–1300.
(a)Ethanol is heated under reflux with acidified potassium dichromate(VI). Describe how the IR spectrum of the organic product differs from that of ethanol.[3]
(b)If the product is instead distilled off as it forms, ethanal is obtained. State how the IR spectrum of ethanal differs from that of ethanoic acid.[1]
Show answer
(a) Product ethanoic acid: new strong C=O absorption at 1670–1740 cm⁻¹; the broad alcohol O–H at 3200–3600 is replaced by a very broad acid O–H at 2500–3000 cm⁻¹.
B1 product named; B1 C=O appears; B1 O–H band changes to 2500–3000.
(b) No broad O–H absorption at all (C=O still present).
B1.
Q22Infrared spectroscopy · Standard · Paper 2 style · 4 marks
(a)Explain what happens at the molecular level when a molecule absorbs infrared radiation.[2]
(b)Carbon dioxide and methane are greenhouse gases but nitrogen is not. Suggest why.[2]
Show answer
(a) Bonds vibrate (stretch/bend) at particular frequencies; radiation of matching frequency is absorbed, increasing the vibrational energy, provided the vibration changes the dipole moment.
B1 bond vibration energy increases; B1 specific frequencies / change in dipole.
(b) CO₂ and CH₄ have polar bonds whose vibrations change the dipole, so they absorb IR emitted by the Earth. N₂ is a non-polar diatomic; its vibration causes no dipole change.
Q23Infrared spectroscopy · Challenge · Paper 2 style · 5 marks
IR absorption ranges (cm⁻¹): O–H in alcohols 3200–3600 (broad); O–H in carboxylic acids 2500–3000 (very broad); C–H 2850–2950; C=O 1670–1740; C=C 1620–1680; C–O 1040–1300.
(a)Compound R, C₃H₆O₂, shows a strong absorption at 1740 cm⁻¹ and an absorption at 1200 cm⁻¹, but no broad absorption above 2500 cm⁻¹ except C–H near 2900 cm⁻¹. Identify the functional group and give two possible structures.[3]
(b)Propanoic acid is also C₃H₆O₂. State how its IR spectrum differs from that of R.[1]
(c)Suggest a different technique that could distinguish the two esters in (a).[1]
Show answer
(a) Ester (C=O and C–O, no O–H). Methyl ethanoate CH₃COOCH₃ or ethyl methanoate HCOOCH₂CH₃
B1 ester with evidence; B1 each structure (max 2).
(b) Very broad O–H band at 2500–3000 cm⁻¹.
B1.
(c) ¹H NMR (or mass spectrometry fragments).
B1.
Q24Chromatography (TLC, GC) · Core · Paper 4 style · 4 marks
A food-safety officer in Karachi runs a TLC plate of a red sauce colouring. The solvent front moves 8.0 cm and the red spot moves 3.2 cm from the baseline.
(a)Calculate the Rf value of the red dye.[1]
(b)The permitted dye E129 has Rf 0.40 and the banned dye Sudan I has Rf 0.70 in this solvent. Calculate the distance Sudan I would travel and state which dye is present.[2]
(c)Suggest why the result should be confirmed with a second solvent.[1]
Show answer
(a) 0.40
B1 3.2/8.0.
(b) 5.6 cm; the dye is E129 (Rf matches 0.40)
B1 0.70 × 8.0 = 5.6 cm; B1 E129.
(c) Two different dyes could have the same Rf in one solvent.
B1.
Q25Chromatography (TLC, GC) · Standard · Paper 4 style · 5 marks
(a)In TLC, name the stationary phase and the mobile phase, and explain how the components of a mixture separate.[3]
(b)Amino acids are colourless. Describe how their spots can be made visible.[1]
(c)Why must the baseline be drawn in pencil and be above the solvent level?[1]
Show answer
(a) Stationary: thin layer of silica (or alumina) on a plate; mobile: solvent. Components have different affinities: those more strongly adsorbed on the polar stationary phase move more slowly; those more soluble in the solvent move further.
B1 phases; B1 different adsorption/affinity for stationary phase; B1 different solubility in solvent → different distances.
(b) Spray with ninhydrin (and warm) or use a UV lamp/iodine.
B1.
(c) Ink would dissolve and separate; spots below the solvent would dissolve into the solvent instead of moving up.
B1 either reason.
Q26Chromatography (TLC, GC) · Standard · Paper 4 style · 5 marks
A fruit-flavouring mixture is analysed by gas chromatography. Assume peak area is proportional to the amount of each component.ethyl ethanoate: retention time 2.1 min, peak area 1.8 cm²propyl ethanoate: 3.4 min, 5.4 cm²butyl ethanoate: 5.0 min, 4.8 cm²
(a)Calculate the percentage of each ester in the mixture.[2]
(b)Suggest why butyl ethanoate has the longest retention time.[2]
(b) It has the highest boiling point (largest molecule, strongest id–id forces), so it spends more time dissolved in the liquid stationary phase.
B1 highest bp / least volatile; B1 more time in (more soluble in) stationary phase.
(c) Helium (or nitrogen)
B1.
Q27Chromatography (TLC, GC) · Challenge · Paper 4 style · 6 marks
A student runs a TLC plate of a mixture of glycine and alanine. In solvent 1 one spot appears with Rf 0.30. In solvent 2 the solvent front travels 10.0 cm; spots are at 2.6 cm (glycine) and 4.5 cm (alanine).
(a)Explain why the two amino acids in a mixture give overlapping spots in solvent 1 but are separated in solvent 2.[2]
(b)Calculate the Rf of each amino acid in solvent 2.[2]
(c)In gas chromatography a sample is too involatile to be analysed. Suggest one reason, and suggest an alternative.[2]
Show answer
(a) In solvent 1 they have the same (very similar) Rf because their relative affinities for stationary and mobile phase are similar; in solvent 2 their solubilities differ, so they travel different distances.
B1 same Rf in solvent 1; B1 different relative solubility/affinity in solvent 2.
(b) glycine 0.26; alanine 0.45
B1 each.
(c) It would decompose / not vaporise at the column temperature (e.g. has strong hydrogen bonding, high bp); use TLC (or HPLC).
B1 reason; B1 alternative.
Q28Carbon-13 NMR · Core · Paper 4 style · 4 marks
(a)State the number of peaks in the ¹³C NMR spectrum of: (i) propan-1-ol; (ii) propan-2-ol; (iii) 2-methylpropan-2-ol.[3]
(b)Explain why ¹³C NMR peaks are not split by neighbouring carbon atoms.[1]
Show answer
(a) (i) 3; (ii) 2; (iii) 2
B1 each.
(b) ¹³C is only about 1 % of carbon, so two ¹³C atoms are very rarely next to each other.
B1.
Q29Carbon-13 NMR · Standard · Paper 4 style · 4 marks
(a)Predict the number of peaks in the ¹³C spectrum of each isomer of C₅H₁₂: pentane, 2-methylbutane and 2,2-dimethylpropane.[3]
(b)A sample of one isomer from a gas cylinder shows 4 peaks. Identify it.[1]
Q30Carbon-13 NMR · Challenge · Paper 4 style · 5 marks
Ester S, C₄H₈O₂, gives four ¹³C peaks at δ 14, 21, 60 and 171. Its ¹H NMR spectrum includes a singlet (3H) at δ 2.0.¹³C shifts (δ/ppm): C–C (alkyl) 5–40; C–O 50–90; C=O in esters and acids 160–185.
(a)Assign each ¹³C peak to a carbon atom and identify S.[4]
(b)Explain how the ¹H NMR information rules out methyl propanoate.[1]
