amazon.co.ukAnalysis: NMR, IR and Mass Spectrometry
A* revision for structure determination: mass spectrometry fragmentation, IR absorptions, proton and carbon NMR including TMS, integration, splitting and the D₂O shake, for AQA, Edexcel, OCR, WJEC and CCEA.
🎯 Learning Objectives
- Deduce relative molecular mass from the molecular ion peak and use isotope peaks (M+1, Cl, Br) to identify elements
- Interpret fragmentation patterns using common fragment losses (15, 29, 31, 43, 45)
- Use high-resolution mass spectrometry to distinguish compounds with the same integer M_r
- Identify functional groups from IR absorptions, including broadness of O-H bands and the fingerprint region
- Interpret ¹H NMR spectra using chemical shifts, integration, splitting (n+1 rule) and the D₂O shake
- Count carbon environments from ¹³C NMR and combine all techniques to determine structures with justified assignments
📋 Key Definitions and Core Concepts
Molecular ion (M⁺•): The ion formed by loss of one electron from a molecule in electron-impact MS; its m/z equals the relative molecular mass.
Fragmentation: Break-up of the molecular ion into smaller charged fragments, whose pattern identifies the molecule.
Fingerprint region: The IR region below ~1500 cm⁻¹ unique to each compound; matching it confirms identity.
Chemical shift (δ): The resonant frequency position of a nucleus relative to TMS, measured in ppm.
TMS (tetramethylsilane): The NMR standard: single sharp peak at δ = 0, inert, non-toxic, volatile.
Spin-spin splitting: Splitting of a NMR signal into n+1 peaks by n non-equivalent protons on adjacent atoms.
Integration trace: The NMR peak areas, proportional to the relative number of protons in each environment.
D₂O shake: Addition of deuterium oxide that exchanges with labile O–H/N–H protons, removing their signals — identifies OH/NH peaks.
High-resolution MS: MS measuring m/z to 4 decimal places, distinguishing compounds of the same integer mass.
Mass Spectrometry of Molecules: Molecular Ions and Fragmentation
In electron-impact mass spectrometry the molecule is bombarded with high-energy electrons, knocking out an electron to form the molecular ion M⁺•. The m/z of this peak (using z = +1) equals the relative molecular mass — the first reading to take from any spectrum. Isotopic peaks help identify elements: an M+1 peak (from ¹³C, ~1.1% per carbon) indicates the number of carbons — 7% M+1 means ~7 carbons; chlorine gives 3:1 doublets and bromine 1:1 doublets in fragments (see Atomic Structure).
Fragmentation: the molecular ion breaks into smaller cations and radicals; only charged fragments are detected. The pattern of fragment peaks identifies the molecule. Common fragments to memorise: 15 = CH₃⁺, 29 = C₂H₅⁺ or CHO⁺, 31 = CH₂OH⁺, 43 = CH₃CO⁺ or C₃H₇⁺, 45 = COOH⁺, 77 = C₆H₅⁺ (phenyl). Loss of 18 = water, loss of 28 = CO or ethene, loss of 31/32 = methanol/methoxy. A fragment at M−15 shows loss of a methyl group; at M−45 loss of COOH suggests a carboxylic acid.
High-resolution mass spectrometry measures m/z to four decimal places, distinguishing compounds with the same integer mass: C₃H₈O (58.0731), C₂H₄O₂ (58.0367) and C₂H₆N₂ (58.0531) all give M⁺ = 58 at low resolution but separate with high-resolution data. Exam questions give exact isotopic masses (¹²C = 12.0000, ¹H = 1.0078, ¹⁶O = 15.9949, ¹⁴N = 14.0031) and ask you to compute the precise M+ and match to a listed formula — add the mass contributions carefully, keeping four decimal places.
Interpreting full mass spectra — method for A* answers: (1) take M⁺ as the Mr (careful: the tallest peak is the base peak, usually a stable fragment, not the molecular ion — a small peak at high m/z may be M⁺); (2) check isotope patterns for Cl (3:1) and Br (1:1) — their presence tells you how many halogen atoms are in the molecule; (3) work out each fragment's formula from the m/z and identify the loss; (4) assemble a structure consistent with the fragments. Example: a spectrum with M⁺ = 92 and a 3:1 doublet at 89/94 would suggest one Cl; a fragment at m/z 31 (CH₂OH⁺) points to an alcohol. M+1 peaks: ~1.1% per carbon means a compound with M⁺ = 100 and M+1 at ~7.7% has about 7 carbons — quoted ratios in exams are always consistent with the formula given, so use them as a cross-check.
Infrared Spectroscopy: Absorptions and the Fingerprint Region
Bonds absorb IR radiation at characteristic wavenumbers, vibrating more strongly when the frequency matches their natural frequency. Absorptions to know: O–H (alcohol) broad 3200–3600 cm⁻¹; O–H (carboxylic acid) very broad 2500–3300 cm⁻¹; C=O sharp/strong 1680–1750 cm⁻¹; C–H 2850–3000 cm⁻¹; N–H 3300–3500 cm⁻¹; C–N 1000–1300 cm⁻¹; C=C 1620–1680 cm⁻¹; C–O 1000–1300 cm⁻¹. The fingerprint region (below 1500 cm⁻¹) is unique to each compound — a match between an unknown's spectrum and a database spectrum confirms identity.
Broadness matters: hydrogen-bonded O–H groups absorb over a wide range, so a very broad band spanning 2500–3300 identifies a carboxylic acid; a sharper band above 3200 identifies an alcohol. C=O is the most reliable diagnostic — a strong sharp peak near 1700 cm⁻¹ signals aldehyde, ketone, acid, ester, acyl chloride or amide; combine with other peaks to decide which (ester: C=O plus C–O; acid: C=O plus very broad O–H; aldehyde: C=O plus C–H aldehydic doublet ~2720/2820).
Applications: breathalysers measure IR absorption by C–H/O–H bonds of ethanol in breath; IR spectroscopy monitors pollutant gases (CO, NO, hydrocarbons) by their characteristic absorptions. Exam technique: quote the wavenumber, the bond, the shape of the absorption, and the functional group implied. When identifying an unknown from IR alone, state which absorptions are present and absent — the absence of C=O rules out all carbonyl compounds at once.
Common IR misidentifications to avoid: an O–H band alone does not distinguish alcohol from acid — check the range (broad 3200-3600 alcohol; very broad 2500-3300 acid, overlapping the C–H region). A C=O peak with no other diagnostic could be aldehyde, ketone, acid, ester, acyl chloride or amide — use supporting evidence: acid gives the very broad O–H; ester gives C–O at 1000-1300; aldehyde gives the aldehydic C–H doublet near 2720-2820 cm⁻¹. Water in a sample shows a broad O–H band — drying agents matter before running spectra of alcohols. When the question asks which absorption is 'most useful for identification', answer C=O (sharp, strong, reliable range) or the fingerprint region (unique match) — justify the choice. Remember the units: wavenumber in cm⁻¹, always quoted with the bond and shape (sharp/broad/strong).
Proton NMR: Chemical Shift, Integration, Splitting and D₂O
¹H NMR exploits nuclear spin: protons in different chemical environments resonate at different frequencies in a strong magnetic field. Chemical shift (δ) is measured in ppm relative to TMS (tetramethylsilane) at δ = 0 ppm. TMS is used because it gives a single sharp peak (12 equivalent H), is inert, non-toxic and volatile (easily removed). Higher δ means more deshielded protons — near electronegative atoms or π systems.
Chemical shift ranges: δ 0.7–1.2 alkyl C–H; δ 2.2–3.0 C–H next to C=O or C=C; δ 3.1–4.2 C–H next to O (alcohol/ester); δ 4.7–5.7 =C–H alkene; δ 6.0–9.0 aromatic H; δ 9.0–10.0 aldehyde; δ 10–12 carboxylic acid O–H; δ 1.0–5.0 alcohol O–H (variable, hydrogen-bonded).
Integration: the area under each peak (shown as an integration trace) is proportional to the relative number of protons in that environment — ratios 3:2:1 indicate groups like CH₃/CH₂/OH. Spin-spin splitting (n+1 rule): a signal is split into n+1 peaks by n non-equivalent protons on adjacent carbons — singlet (0), doublet (1), triplet (2), quartet (3). Classic patterns: triplet + quartet with 3:2 integration = ethyl group CH₃CH₂–; two equivalent methyls next to no H = singlet integrating 6H. Splitting only occurs between non-equivalent neighbours — equivalent protons do not split each other; O–H and N–H protons usually appear as singlets (rapid exchange).
D₂O shake: adding D₂O exchanges labile O–H and N–H protons (R–OH + D₂O → R–OD + HOD), so their signals disappear from the spectrum — identifying OH/NH peaks unambiguously. Exam workflow for structure determination: (1) molecular formula from MS; (2) degree of unsaturation (rings/double bonds); (3) IR identifies functional groups; (4) number of ¹H signals = number of proton environments; (5) integration gives H ratios; (6) splitting identifies neighbours; (7) assemble and check against all data.
Carbon-13 NMR and Combining Techniques
¹³C NMR is simpler: the number of distinct signals equals the number of unique carbon environments. Symmetry reduces signals — propanone CH₃COCH₃ gives 2 signals (two equivalent CH₃, one C=O), but propanal CH₃CH₂CHO gives 3. Chemical shift ranges: δ 5–50 alkyl C; δ 50–90 C–O; δ 90–150 alkene/aromatic C; δ 160–210 C=O; δ 190–220 aldehyde C=O. A peak near δ 200 with no other C=O clue suggests aldehyde/ketone; δ 170 with an ester/acid fragment suggests acid/ester/amide.
Model answer (combining all techniques): identify C₃H₆O₂ showing IR 1715 cm⁻¹, broad 2500–3300 cm⁻¹, ¹H NMR triplet δ 1.1 (3H), quartet δ 2.4 (2H), singlet δ 11.5 (1H). (1) IR: 1715 (C=O) + very broad 2500–3300 (acid O–H) → carboxylic acid −COOH. (2) ¹H NMR: δ 11.5 singlet confirms the −COOH proton. (3) Triplet δ 1.1 (3H) + quartet δ 2.4 (2H) = ethyl group CH₃CH₂−. Structure: propanoic acid CH₃CH₂COOH — consistent with Mr 74 and all data. Method marks are given for each deduction step, not just the final structure.
Sequencing logic for multi-technique questions: MS gives Mr and fragments; IR gives functional groups; ¹³C NMR counts carbon environments; ¹H NMR gives proton environments, ratios and neighbours; D₂O removes labile protons. Cross-check: the assembled structure must account for every signal — if a δ 2.4 quartet is unexplained, the structure is wrong. A* students state the evidence for each assignment explicitly ('δ 9.8 confirms CHO') because marks are awarded for justification.
Splitting-pattern library beyond the basics: triplet + quartet (3:2) = CH₃CH₂–; doublet + septet (6:1) = CH–(CH₃)₂ (isopropyl); singlet integrating 6H = two equivalent CH₃ next to no H (e.g. propanone); singlet integrating 9H = C(CH₃)₃; doublet + quartet (3:1 with one H each) = CH₃CH(OH)–. The n+1 rule applies only to protons on adjacent carbons — CH₃ next to C=O is not split by it (no H on the carbonyl carbon). Equivalent protons never split each other. Integration ratios are relative, not absolute: 3:2:1 with a 6H total means 2H/1.3H/0.7H in reality scaled — exams give whole-number ratios you divide by the smallest. When two environments overlap in shift, the exam data avoids ambiguity — trust the given splitting and integration over memory of shift tables, and state which data each assignment uses.
Interpretation Workflow, Required Practicals and Synthesis Links
The combined-techniques question (AQA Paper 3, OCR, Edexcel) rewards a fixed workflow applied under time pressure: (1) molecular formula from M⁺ (check M+1 for carbon count, Cl/Br doublets); (2) functional groups from IR (C=O, O–H ranges, fingerprint match); (3) number of ¹H signals = proton environments; number of ¹³C signals = carbon environments; (4) integration ratios give relative numbers of each H; (5) splitting (n+1) identifies adjacent groups; (6) D₂O shake locates labile O–H/N–H; (7) assemble the structure and verify every datum is explained. Write the evidence for each assignment ('δ 9.8 confirms CHO', 'triplet/quartet 3:2 confirms ethyl') — justification marks are the difference between grades. If two structures fit, find the discriminating datum: a chemical shift range, a splitting pattern, or an IR band that only one explains.
Planning synthesis with analysis: exam questions ask how you would confirm a product's identity and purity after preparation — melting point (sharp, matches data book, mixed melting point confirms identity), TLC (single spot, Rf matches standard), IR (expected absorptions present, starting-material absorptions absent — e.g. no broad acid O–H if you made the ester), and boiling point for liquids. Percentage yield and atom economy link the synthesis to analysis: yield = actual/theoretical × 100; atom economy = Mr of desired product / total Mr of all products × 100. Observing an impure product (low melting point, multiple TLC spots) means recrystallisation is needed before analysis — the techniques feed each other. Required practicals referenced: making a pure organic solid (recrystallisation, melting point), preparing an ester and purifying by distillation, and using TLC to separate plant pigments or amino acids — describe the apparatus (Buchner funnel, still head, TLC chamber) and the safety points (fume cupboard for volatile solvents) because practical-write-up marks are embedded in these questions.
💡 Worked Example
Exam-style: Identify C₃H₆O₂ showing IR absorption at 1715 cm⁻¹ and a broad band 2500-3300 cm⁻¹, and ¹H NMR with a triplet at δ 1.1 (3H), a quartet at δ 2.4 (2H) and a singlet at δ 11.5 (1H).
Mark-scheme logic: 1. IR: 1715 cm⁻¹ (C=O) plus very broad 2500-3300 cm⁻¹ (carboxylic acid O-H) → carboxylic acid (-COOH). 2. ¹H NMR: δ 11.5 singlet is the -COOH proton (acid range, singlet because exchange prevents splitting). 3. Triplet δ 1.1 (3H) and quartet δ 2.4 (2H) indicate an ethyl group CH₃CH₂- (triplet: CH₃ split by 2 H on adjacent C; quartet: CH₂ split by 3 H). Structure: propanoic acid CH₃CH₂COOH. Mark points: identify functional groups from IR; assign -COOH proton; deduce ethyl group from splitting and integration; correct final structure consistent with formula.
❓ Practice Questions & Mark Schemes
Why is TMS used as the NMR standard?
Answer: Single sharp peak at δ = 0 ppm (12 equivalent protons), inert, non-toxic, volatile (easily evaporated from the sample).
Describe the D₂O shake in ¹H NMR and what it identifies.
Answer: D₂O exchanges with labile -OH and -NH protons: R-OH + D₂O → R-OD + HOD. The affected signals disappear from the spectrum, identifying O-H/N-H peaks unambiguously.
A compound shows M⁺ = 58 at low resolution. Three candidates: C₃H₆O, C₂H₄O₂ and C₂H₆N₂. Using exact masses (C 12.0000, H 1.0078, O 15.9949, N 14.0031), calculate the precise M⁺ value of each and explain how high-resolution MS distinguishes them.
Answer: C₃H₆O: 36.0000 + 6.0468 + 15.9949 = 58.0417. C₂H₄O₂: 24.0000 + 4.0312 + 31.9898 = 58.0210. C₂H₆N₂: 24.0000 + 6.0468 + 28.0062 = 58.0530. All give M⁺ = 58 at low resolution, but high-resolution MS measures m/z to 4 d.p. and separates the three ions, identifying the exact formula.
An unknown shows fragment peaks at m/z 15, 29, 43, 58 and a strong IR peak at 1715 cm⁻¹. Suggest the compound and justify.
Answer: m/z 43 = CH₃CO⁺, 15 = CH₃⁺, 29 = C₂H₅⁺/CHO⁺, M⁺ = 58; strong C=O at 1715 → ketone. Compound: propanone (CH₃COCH₃, M_r 58) - fragments CH₃CO⁺ (43) and CH₃⁺ (15) both characteristic.
How many ¹H NMR signals and ¹³C NMR signals does propanone show, and why does propanal differ?
Answer: Propanone: 2 ¹H environments (two equivalent CH₃) and 2 ¹³C environments (equivalent CH₃ + C=O) by symmetry. Propanal: 3 ¹H and 3 ¹³C environments - no symmetry, every carbon unique.
A ¹H NMR shows a singlet integrating 9H at δ 1.0. What group is indicated and why is it a singlet?
Answer: Three equivalent CH₃ groups - a tert-butyl group C(CH₃)₃. Singlet because the adjacent central carbon bears no hydrogens, so no spin-spin splitting occurs (n+1 = 1).