Proton (¹H) & Carbon-13 (¹³C) NMR Splitting Patterns, TMS, IR Spectroscopy & GC-MS

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Infrared (IR) Spectroscopy & The Fingerprint Region

Covalent bonds vibrate (stretch and bend) and absorb infrared radiation at specific wavenumbers (cm⁻¹, proportional to frequency) that depend on bond strength and atomic masses. Key diagnostic absorptions: 1. O-H alcohol: broad peak at 3230-3550 cm⁻¹ (broadened by intermolecular hydrogen bonding). 2. O-H carboxylic acid: very broad peak at 2500-3000 cm⁻¹ overlapping C-H. 3. C=O carbonyl (aldehyde, ketone, carboxylic acid, ester): sharp, strong peak at 1680-1750 cm⁻¹. The region below 1500 cm⁻¹ is the fingerprint region, containing complex, unique skeletal vibrations matched against a computer database to identify a specific compound and check purity.

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Carbon-13 (¹³C) NMR & Molecular Symmetry

Each peak in a proton-decoupled ¹³C NMR spectrum corresponds to one unique chemical environment of carbon atoms in the molecule (chemically equivalent carbons related by a plane or axis of symmetry give only one shared peak). For example, benzene (C₆H₆) gives 1 peak; 1,4-dimethylbenzene (C₈H₁₀) gives 3 peaks (4 symmetrical ring CH carbons, 2 symmetrical ring C-CH₃ carbons, and 2 symmetrical CH₃ carbons); 1,2-dimethylbenzene gives 4 peaks; and 1,3-dimethylbenzene gives 5 peaks.

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Proton (¹H) NMR: Chemical Shift, Integration & The (n + 1) Splitting Rule

1. Number of signals = number of non-equivalent ¹H environments. 2. Chemical shift (δ in ppm) identifies the type of proton environment from the Data Booklet (e.g. R-CH₃ ~0.7-1.2 ppm; H-C-C=O ~2.1-2.6 ppm; H-C-O ~3.1-3.9 ppm; R-CHO ~9.0-10.0 ppm; R-COOH ~10.0-12.0 ppm). 3. Integration ratio = ratio of the number of ¹H atoms in each environment. 4. Spin-spin coupling (n + 1 rule): if a proton environment has n equivalent protons on adjacent carbon atoms, its peak splits into (n + 1) sub-peaks: 0 adjacent H = singlet; 1 adjacent H (-CH-) = doublet (1:1); 2 adjacent H (-CH₂-) = triplet (1:2:1); 3 adjacent H (-CH₃) = quartet (1:3:3:1). Protons in an ethyl group (-CH₂CH₃) always give a quartet (weight 2) and a triplet (weight 3).

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Tetramethylsilane (TMS) Standard, Deuterated Solvents (CDCl₃) & D₂O Shake

Tetramethylsilane, Si(CH₃)₄, is added as the internal calibration standard at δ = 0.0 ppm because: 1. All 12 protons (and 4 carbons) are in the exact same environment, giving a single sharp, intense peak. 2. Si is less electronegative than C, so its protons are highly shielded and appear upfield at δ = 0 ppm, away from almost all organic proton peaks. 3. It is non-toxic, chemically inert, and volatile (b.p. 27°C, easily evaporated to recover the sample). Samples are dissolved in CCl₄ or deuterated solvents like CDCl₃ because deuterium (²H) has an even mass number (no net nuclear spin) and does not produce peaks that would swamp the ¹H spectrum. Shaking a sample with D₂O causes acidic -OH or -NH protons to exchange with deuterium (-OD / -ND), causing their singlet peaks to DISAPPEAR from the ¹H NMR spectrum (and -OH/-NH protons never split adjacent protons).

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Thin-Layer (TLC), Column & Gas Chromatography (GC) Principles

All chromatography separates mixtures by partitioning components between a stationary phase and a mobile phase: 1. Thin-Layer Chromatography (TLC) & Column Chromatography: Stationary phase = polar silica (SiO₂) or alumina (Al₂O₃) powder (adsorbs polar molecules strongly via hydrogen bonds/dipole forces); Mobile phase = liquid organic solvent. More polar components adsorb more strongly to the polar stationary phase and move slower (smaller Rf in TLC; elute later from a column). 2. Gas Chromatography (GC): Stationary phase = high-boiling liquid (or solid) coated inside a long coiled capillary column in a thermostatted oven; Mobile phase = unreactive carrier gas (Helium or Nitrogen). Retention time depends on boiling point (volatility) and solubility in the liquid stationary phase.

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GC-MS (Gas Chromatography-Mass Spectrometry) & High-Resolution MS

Coupling Gas Chromatography directly to a Mass Spectrometer (GC-MS) separates a complex mixture (e.g. athlete blood/urine samples, forensics, environmental pollutants) in the GC capillary column so each pure component enters the mass spectrometer sequentially at its retention time, where its mass spectrum and fragmentation pattern are matched against a library for definitive identification. High-resolution mass spectrometry measures m/z to 4 or 5 decimal places using exact isotopic masses (¹H = 1.0078, ¹²C = 12.0000, ¹⁴N = 14.0031, ¹⁶O = 15.9949) to distinguish compounds that share the same nominal integer Mr (e.g. C₃H₈O vs C₂H₄O₂, both nominal Mr = 60).

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1. Q1

   Three structural isomers, P, Q, and R, all have the molecular formula C₄H₈O₂ and all show a strong infrared absorption at 1740 cm⁻¹ but no absorption above 3100 cm⁻¹. In their proton (¹H) NMR spectra: Isomer P has two singlet peaks (integration ratio 3:1 is not possible for C₄H₈O₂; rather, Isomer P has a triplet (3H), a quartet (2H), and a singlet (3H), where the quartet is at δ 4.1 ppm; Isomer Q also has a triplet (3H), a quartet (2H), and a singlet (3H), where the singlet is at δ 3.7 ppm; Isomer R has a doublet (6H), a multiplet/septet (1H), and a singlet (1H) at δ 8.1 ppm). Identify the functional group present in P, Q, and R, and deduce the structure and IUPAC name of each isomer. (6 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. The strong IR peak at 1740 cm⁻¹ (C=O) with no O-H peak above 3100 cm⁻¹ proves that P, Q, and R are all esters (not carboxylic acids) [1 mark]. 2. In both P and Q, the triplet (3H) and quartet (2H) indicate an ethyl group (-CH₂CH₃), while the singlet (3H) indicates an isolated methyl group (-CH₃) attached to C=O or O [1 mark]. 3. In Isomer P, the -CH₂- quartet is at δ 4.1 ppm (typical of H-C-O, attached directly to the single-bonded ester oxygen), and the -CH₃ singlet is at δ 2.1 ppm (H-C-C=O), so P is ethyl ethanoate (CH₃COOCH₂CH₃) [1 mark]. 4. In Isomer Q, the -CH₃ singlet is at δ 3.7 ppm (H-C-O, attached directly to the ester oxygen) and the -CH₂- quartet is at δ 2.3 ppm (H-C-C=O), so Q is methyl propanoate (CH₃CH₂COOCH₃) [1 mark]. 5. In Isomer R, the doublet (6H) and septet/multiplet (1H) indicate a branched 1-methylethyl (isopropyl) group, -CH(CH₃)₂ [1 mark]. 6. Combined with the 1H singlet at δ 8.1 ppm (H-COO-), Isomer R is 1-methylethyl methanoate, HCOOCH(CH₃)₂ [1 mark].

2. Q2

   Propane-1-ol (CH₃CH₂CH₂OH), propane, and propanoic acid all have three carbon atoms, and propane-1-ol and ethanoic acid both have a nominal relative molecular mass (Mr) of 60. Explain how high-resolution mass spectrometry can distinguish between propane-1-ol (C₃H₈O) and ethanoic acid (C₂H₄O₂) without relying on fragmentation, using the accurate atomic masses: ¹H = 1.0078, ¹²C = 12.0000, ¹⁶O = 15.9949. (3 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Both C₃H₈O and C₂H₄O₂ have the same integer nominal Mr of 60, so low-resolution mass spectrometry shows an M⁺ peak at m/z = 60 for both [1 mark]. 2. Accurate molecular mass of propane-1-ol (C₃H₈O) = (3 × 12.0000) + (8 × 1.0078) + 15.9949 = 36.0000 + 8.0624 + 15.9949 = 60.0573 [1 mark]. 3. Accurate molecular mass of ethanoic acid (C₂H₄O₂) = (2 × 12.0000) + (4 × 1.0078) + (2 × 15.9949) = 24.0000 + 4.0312 + 31.9898 = 60.0210; high-resolution mass spectrometry resolves these two distinct m/z values to 4 decimal places [1 mark].

3. Q3

   Predict the number of peaks in the ¹³C NMR spectrum and describe the splitting pattern and integration ratio of the peaks in the ¹H NMR spectrum of: (a) pentan-3-one (CH₃CH₂COCH₂CH₃), and (b) 2-bromo-2-methylpropane ((CH₃)₃CBr). (4 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. (a) Pentan-3-one has a plane of symmetry through the C=O group, so its ¹³C NMR spectrum has 3 peaks (the two CH₃ carbons are equivalent, the two CH₂ carbons are equivalent, plus the C=O carbon) [1 mark]. 2. In the ¹H NMR spectrum of pentan-3-one, there are 2 peaks in a 3:2 (or 6:4) integration ratio: a triplet (6H, from the two equivalent -CH₃ groups adjacent to -CH₂-) and a quartet (4H, from the two equivalent -CH₂- groups adjacent to -CH₃) [1 mark]. 3. (b) In 2-bromo-2-methylpropane, all three -CH₃ groups are chemically equivalent by symmetry, so its ¹³C NMR spectrum has only 2 peaks (one for the 3 equivalent CH₃ carbons and one for the central quaternary C-Br carbon) [1 mark]. 4. Its ¹H NMR spectrum consists of a single singlet peak (integration 9H) because all 9 protons are equivalent and the adjacent quaternary carbon has 0 attached protons (0 + 1 = 1, singlet) [1 mark].

4. Q4

   A mixture of phenylamine (C₆H₅NH₂) and methylbenzene (C₆H₅CH₃) was separated by Thin-Layer Chromatography (TLC) using a silica-gel plate and a non-polar hexane/ethyl ethanoate solvent mixture. State and explain which compound has the smaller Rf value, and explain why gas chromatography (GC) alone cannot conclusively prove the identity of a banned steroid in an athlete's blood sample without also linking it to a mass spectrometer (GC-MS). (4 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. The stationary phase in TLC is polar silica (SiO₂, containing surface -OH groups) [1 mark]. 2. Phenylamine has a polar -NH₂ group that forms strong hydrogen bonds with the polar silica stationary phase, whereas methylbenzene is non-polar and only forms weak van der Waals forces [1 mark]. 3. Therefore, phenylamine adsorbs much more strongly to the stationary phase, travels a shorter distance up the plate, and has the smaller Rf value [1 mark]. 4. In Gas Chromatography alone, two different compounds can happen to have the exact same retention time under the same column conditions; coupling to a mass spectrometer (GC-MS) produces a unique molecular ion peak and fragmentation fingerprint for each separated peak, providing definitive identification [1 mark].

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