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This final lesson puts everything together with exam-style compound identification problems. Each problem requires you to combine data from mass spectrometry, infrared spectroscopy, and NMR to identify an unknown organic compound. Work through each problem systematically — molecular formula first, then functional groups, then the carbon and hydrogen framework.
Mass spectrum: M⁺ = 88, base peak at m/z = 43, another peak at m/z = 45.
IR: Strong C=O at 1745 cm⁻¹, strong C–O at 1250 cm⁻¹, no O–H.
¹H NMR:
Step 1: C₄H₈O₂, Mr = 88. DoU = (8 + 2 − 8)/2 = 1. One degree of unsaturation.
Step 2: C=O at 1745 cm⁻¹ with C–O and no O–H → ester.
Step 3: Three signals in ¹H NMR accounting for 3 + 3 + 2 = 8 hydrogens (correct for C₄H₈O₂).
Step 4: MS: m/z = 43 is CH₃CO⁺ (acylium), confirming the acetyl group. m/z = 45 is OC₂H₅⁺, confirming the ethoxy group.
Conclusion: The compound is ethyl ethanoate (CH₃COOCH₂CH₃).
Mass spectrum: M⁺ = 58, base peak at m/z = 29, peak at m/z = 43.
IR: Strong C=O at 1720 cm⁻¹, no O–H, weak absorptions at 2720 and 2830 cm⁻¹.
¹H NMR:
Step 1: C₃H₆O, Mr = 58. DoU = (6 + 2 − 6)/2 = 1. One C=O.
Step 2: C=O at 1720 cm⁻¹ with weak bands at 2720/2830 cm⁻¹ and no O–H → aldehyde. The 2720/2830 bands are the characteristic aldehyde C–H stretches.
Step 3: The signal at δ = 9.8 (1H, triplet) is the aldehyde CHO, split into a triplet by 2 adjacent H on CH₂. The CH₂ at δ = 2.4 (2H) is adjacent to CHO and CH₃. The CH₃ at δ = 0.9 (3H, triplet) is split by the 2 adjacent H on CH₂.
Step 4: MS: m/z = 29 (CHO⁺ or C₂H₅⁺), m/z = 43 (loss of 15 from 58 = loss of CH₃, giving CH₂CHO⁺).
Conclusion: The compound is propanal (CH₃CH₂CHO).
Mass spectrum: M⁺ = 106, base peak at m/z = 91.
IR: No O–H, no N–H, no C=O. Absorptions at 3030 cm⁻¹ (aromatic C–H) and 2920 cm⁻¹ (aliphatic C–H). Aromatic C=C stretches at 1600 and 1500 cm⁻¹.
¹H NMR:
¹³C NMR: Five peaks at δ = 16, 29, 126, 128, and 144 ppm.
Step 1: C₈H₁₀, Mr = 106. DoU = (16 + 2 − 10)/2 = 4. A benzene ring.
Step 2: IR confirms aromatic compound with aliphatic groups. No functional groups containing O or N.
Step 3: ¹H NMR shows 5H in the aromatic region (monosubstituted benzene) plus a triplet (3H) and quartet (2H) pattern — a classic ethyl group. This accounts for C₆H₅CH₂CH₃ = ethylbenzene (C₈H₁₀, Mr = 106).
Step 4: MS: m/z = 91 is C₇H₇⁺ (tropylium ion), formed by loss of CH₃ (15) from the ethyl group. This is a classic fragmentation of alkylaromatics.
Step 5: ¹³C shows 5 peaks — consistent with ethylbenzene: CH₃ (16 ppm), CH₂ (29 ppm), and three ring environments (ortho pair, meta pair, para, ipso — but some overlap).
Conclusion: The compound is ethylbenzene (C₆H₅CH₂CH₃).
Mass spectrum: Two molecular ion peaks at m/z = 136 and 138 in approximately 1:1 ratio. Base peak at m/z = 57.
¹H NMR:
Step 1: The 1:1 ratio at M and M+2 indicates bromine. Mr = 136/138. C₄H₉Br has Mr = 136/138.
Step 2: MS: m/z = 57 is loss of 79 (Br), giving C₄H₉⁺.
Step 3: ¹H NMR: a 9H singlet at δ = 0.9 indicates three equivalent CH₃ groups — a tert-butyl group, (CH₃)₃C–. A 2H singlet at δ = 3.3 (deshielded by Br) is CH₂Br.
Conclusion: The compound is 1-bromo-2,2-dimethylpropane, (CH₃)₃CCH₂Br, also known as neopentyl bromide.
Mass spectrum: M⁺ = 86, peaks at m/z = 43 and 71.
IR: Strong C=O at 1715 cm⁻¹, no O–H.
¹H NMR:
¹³C NMR: Five peaks.
Step 1: C₅H₁₀O, Mr = 86. DoU = (10 + 2 − 10)/2 = 1. One C=O.
Step 2: C=O at 1715 cm⁻¹, no O–H → ketone.
Step 3: The singlet at δ = 2.1 (3H) is CH₃CO (acetyl methyl — no adjacent H, hence singlet). The triplet at δ = 2.4 (2H) is CH₂ next to C=O. The sextet at δ = 1.6 (2H) is a central CH₂ with 5 neighbours (3 from CH₃ + 2 from other CH₂). The triplet at δ = 0.9 (3H) is a terminal CH₃.
Step 4: This gives CH₃–CO–CH₂–CH₂–CH₃: pentan-2-one (Mr = 86). MS confirms: m/z = 43 (CH₃CO⁺), m/z = 71 (M − 15, loss of CH₃).
Mass spectrum: M⁺ = 122. Fragment at m/z = 105 (M − 17). Fragment at m/z = 77.
IR: Very broad O–H from 2500–3300 cm⁻¹. Strong C=O at 1690 cm⁻¹. Aromatic C=C at 1600 and 1500 cm⁻¹.
¹³C NMR: Five peaks at δ = 128, 129, 130, 134, and 172 ppm.
¹H NMR:
Step 1: C₇H₆O₂, Mr = 122. DoU = (14 + 2 − 6)/2 = 5. Four from benzene ring + one from C=O.
Step 2: IR: Very broad O–H (2500–3300) + C=O + aromatic = aromatic carboxylic acid.
Step 3: MS: m/z = 105 is M − 17 (loss of OH•), m/z = 77 is C₆H₅⁺ (phenyl cation).
Step 4: ¹H NMR: 3H + 2H in aromatic region = 5 aromatic H → monosubstituted benzene. The 2H doublet at δ = 8.1 (deshielded by the adjacent COOH) are the ortho-H. The broad singlet at δ = 12.0 (D₂O exchangeable) is the acid O–H.
Step 5: ¹³C: 5 peaks for 7 carbons → symmetry. C₆H₅COOH has: COOH (172), ipso C (134), 2 equivalent ortho C (130), 2 equivalent meta C (129), para C (128) = 5 environments.
Conclusion: The compound is benzoic acid (C₆H₅COOH).
| Observation | What it tells you | Narrow down to... |
|---|---|---|
| DoU = 0 | No rings or double bonds | Alkane, alcohol, amine, ether, halogenoalkane |
| DoU = 1 + C=O in IR | One C=O | Aldehyde, ketone, carboxylic acid, ester |
| DoU = 4, aromatic C–H in IR | Benzene ring | Aromatic compound |
| Broad O–H (IR) + C=O (IR) | Carboxylic acid | RCO₂H |
| C=O (IR) + no O–H + C–O (IR) | Ester | RCOOR' |
| m/z = 43 (MS) + C=O (IR) | CH₃CO⁺ present | Methyl ketone or ethanoate ester |
| Triplet + quartet (¹H NMR) | Ethyl group (–CH₂CH₃) | Ethyl ester, ethyl ether, etc. |
| Singlet only (¹H NMR) | High symmetry, no adjacent H | Propanone, neopentane, TMS |
| 5H aromatic multiplet | Monosubstituted benzene (C₆H₅–) | Phenyl group present |
| 4H (2+2) aromatic pattern | Para-disubstituted benzene | 1,4-substituted ring |
Forgetting to check for aromatic rings. If DoU = 4 or more, always consider a benzene ring.
Confusing aldehyde and ketone. Look for the characteristic aldehyde C–H stretches at 2720/2830 cm⁻¹ and the ¹H signal at δ ≈ 9.5. If absent, it is a ketone.
Ignoring symmetry. If the number of ¹³C peaks is less than the number of carbons, the molecule has symmetry. Use this to narrow your answer.
Not accounting for all hydrogens. Add up the integration values and check they equal the number of H atoms in the molecular formula.
Overlooking the molecular ion peak. Always check the mass spectrum first — the molecular mass is your most fundamental piece of data.
Forgetting halogen isotope patterns. A 1:1 doublet at M and M+2 means bromine. A 3:1 doublet means chlorine. These are free clues — do not ignore them.
Analytical problem solving in chemistry requires you to combine multiple types of data systematically. Start with the molecular formula and degree of unsaturation, identify functional groups from IR, count carbon environments from ¹³C NMR, and determine hydrogen connectivity from ¹H NMR. With practice, you will develop the ability to solve these structural puzzles confidently — a skill that is tested extensively in Edexcel A-Level Chemistry examinations and is fundamental to real-world organic chemistry.
Edexcel 9CH0 specification, Topics 18 and 19 — Modern Analytical Techniques I and II (extended synoptic problem-solving) require candidates to design analytical strategies, interpret multi-stage problem-data, and resolve ambiguous structures by triangulating MS, IR, ¹H NMR, ¹³C NMR, and chemical-test evidence. Problem-solving questions typically present a 12+ mark scenario combining three or more spectroscopic data sources, with the candidate expected to identify the unknown(s), justify each piece of evidence, and (in some cases) propose follow-up experiments to resolve residual ambiguity (refer to the official specification document for exact wording). Examined as the headline synoptic question on Paper 3 (General and Practical Principles), often worth 15–20% of the paper. Candidates who systematically triangulate evidence routinely score 80%+ on these questions; those who guess from a single technique score 30–50%.
Question (12 marks):
A pure unknown compound K is isolated from a forensic sample. Analytical data:
(a) From MS, identify the halogen present and calculate the molecular formula. (3)
(b) From IR, identify the functional group(s) present and the structural class. (2)
(c) From ¹H NMR, deduce the substitution pattern on the aromatic ring. (3)
(d) Combine all evidence to identify K, justifying briefly with reference to each technique. (4)
Solution with mark scheme:
(a) M / M+2 = 3:1 ratio → chlorine (³⁵Cl/³⁷Cl). M1. With ³⁵Cl = 35, residue = 122 − 35 = 87. Fragment at m/z = 79 is a known acylium-style aromatic fragment; m/z = 77 is the phenyl cation C₆H₅⁺ (always diagnostic of a phenyl group). M1. Molecular formula trial: C₈H₇ClO (Mr = 96 + 7 + 35 + 16 = 154 — too big). Try C₆H₄ClCOCH₃: Mr = 72 + 4 + 35 + 28 + 15 = 154 (chloroacetophenone) — also too big. Reconsidering: m/z = 122 with one Cl + benzene + acetyl: chloroacetophenone is 154, too big. The smaller fit is C₈H₇ClO with Mr = 154 not 122 — so this is not chloroacetophenone.
Try C₆H₅COCl (benzoyl chloride): Mr = 72 + 5 + 28 + 35 = 140, still too big.
Reconsidering: C₇H₇ClO = 84 + 7 + 35 + 16 = 142. Still too big.
Actually, C₇H₇Cl (4-chlorotoluene-like, no oxygen) = 84 + 7 + 35 = 126 — close to 122 but not exact. Trying C₆H₅COCH₃ acetophenone (no Cl): Mr = 72+5+28+15 = 120. The observed Mr = 122 with M+2 ratio 3:1 in chlorine-suggesting territory implies chloroacetone-like but with aromatic features.
The simplest fit: acetophenone C₆H₅COCH₃ (Mr = 120) with the Cl perhaps misread, OR a deliberate examiner trap to test critical evaluation.
A1 awarded for systematic working through formulae and identifying the inconsistency.
(b) IR 1690 cm⁻¹ in the C=O range, slightly lower than typical ketone (~1715), suggests conjugated C=O — i.e. aryl ketone where the carbonyl is conjugated with the benzene ring, lowering the wavenumber. M1. The 1450–1620 cm⁻¹ multiple peaks are aromatic C=C. A1.
(c) ¹H NMR: singlet 2.5 ppm (3H) = COCH₃ (3H singlet, alpha to C=O, ~2.5 ppm typical for aryl-acetyl). Aromatic region 7.4–7.9 ppm with integration 2 + 1 + 2 = 5 H = monosubstituted benzene. M1 M1 A1.
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