Syllabus topics (RTMNU)
- Basic principles
- Concept of nuclear spin
- Equivalent and non-equivalent protons
- Reference compound TMS
- Factors affecting chemical shift
- Nuclear Magnetic Resonance (NMR) spectroscopy
- Proton Magnetic Resonance spectroscopy
- Nuclear shielding and deshielding
- Chemical shift
- Spin-spin splitting and Coupling constant
- Areas of signals
- Interpretation of NMR spectra of organic molecules and structure elucidation of simple organic molecules on the basis of NMR data such as ethyl bromide, ethanol, acetaldehyde, 1,2-dibromoethane, ethyl acetate, toluene, acetophenone, acetyl acetone
Basic Principles of NMR and Concept of Nuclear Spin
Nuclear Magnetic Resonance (NMR) spectroscopy is the study of the absorption of radio-frequency radiation by certain nuclei placed in a strong magnetic field. It tells us the number of different kinds of protons (or carbons), their chemical environment and their neighbours — the best tool for finding the structure of organic compounds.
Nuclear spin: nuclei with an odd number of protons or neutrons spin and have a spin quantum number I ≠ 0, e.g. 1H, 13C, 19F, 31P (all I = ½). 12C and 16O have I = 0 and are NMR-inactive. A spinning charged nucleus behaves like a tiny magnet.
In an external magnetic field (B0) a nucleus with spin I takes (2I + 1) orientations. A proton (I = ½) has two: α-spin (aligned with the field, lower energy) and β-spin (opposed, higher energy). The energy gap ΔE = hγB0/2π increases with field strength. When radio waves of exactly this energy (ν = γB0/2π, the resonance condition) are applied, the proton "flips" from α to β — this absorption is nuclear magnetic resonance. For protons, a field of 1.41 tesla needs 60 MHz and 2.35 tesla needs 100 MHz.
Equivalent and Non-Equivalent Protons
Equivalent protons are protons in the same chemical environment; they absorb at the same position and give one signal. Non-equivalent protons give separate signals. So the number of signals = number of sets of equivalent protons. Examples: CH4, C6H6, CH3OCH3 — 1 signal each; CH3CH2Br — 2 signals; CH3CH2OH — 3 signals; (CH3)2CHCl — 2 signals.
Shielding, Deshielding, Chemical Shift and Reference Compound TMS
Electrons around a proton circulate in the applied field and create a small opposing field. This shielding means a higher applied field is needed for resonance — the signal moves upfield (right, low δ). When electron density around the proton is reduced (e.g. by an electronegative atom), the proton is deshielded and absorbs downfield (left, high δ).
Chemical shift is the position of a signal relative to a reference, expressed in parts per million (ppm):
Example: a signal 220 Hz downfield from TMS on a 100 MHz instrument has δ = 220 ÷ 100 = 2.2 ppm. The same proton is 132 Hz away on a 60 MHz instrument, still δ 2.2 — δ does not depend on the instrument.
Tetramethylsilane (TMS), (CH3)4Si, is the reference (δ = 0) because: (1) its 12 protons are equivalent and give one sharp signal; (2) silicon is less electronegative than carbon, so its protons are highly shielded and appear upfield of almost all organic protons; (3) it is chemically inert; (4) volatile (b.p. 27 °C), easy to remove; (5) soluble in most organic solvents.
Factors Affecting Chemical Shift
- Electronegativity (inductive effect): electronegative atoms pull electrons away and deshield nearby protons: CH3F (δ 4.3) > CH3Cl (3.0) > CH3Br (2.7) > CH3I (2.2). The effect falls off with distance.
- Magnetic anisotropy (π-electron circulation): the ring current of benzene strongly deshields aromatic protons (δ 6.5–8); aldehyde proton δ 9–10; alkene protons δ 4.5–6. Acetylenic protons are shielded (δ 2–3) because they lie along the axis of the cylindrical π-cloud.
- Hydrogen bonding: H-bonded protons (OH, NH) are deshielded; their δ varies with concentration, solvent and temperature (alcohol OH δ 1–5; carboxylic acid OH δ 10–13).
- Hybridisation of the carbon bearing the proton, and van der Waals deshielding in crowded molecules.
| Type of proton | Approximate δ (ppm) |
|---|---|
| R–CH3 | 0.9 |
| R–CH2–R | 1.3 |
| CH3–C=O | 2.0–2.6 |
| Ar–CH3 | 2.3 |
| CH2–X (Br, Cl), CH2–O | 3.3–4.2 |
| Alkene C=C–H | 4.5–6.0 |
| Aromatic Ar–H | 6.5–8.0 |
| Aldehyde –CHO | 9–10 |
| Carboxylic acid –COOH | 10–13 |
Spin–Spin Splitting, Coupling Constant and Area of Signals
Spin–spin splitting: a proton's signal is split by protons on adjacent carbons (usually up to three bonds away), because their spins add to or subtract from the field it feels. (n + 1) rule: a signal is split into (n + 1) peaks by n equivalent neighbouring protons: 0 → singlet, 1 → doublet (1:1), 2 → triplet (1:2:1), 3 → quartet (1:3:3:1), 4 → quintet. Relative intensities follow Pascal's triangle. Equivalent protons do not split each other; protons on O–H often do not split because of fast exchange.
Coupling constant (J): the distance between adjacent peaks of a multiplet, in hertz. It does not depend on the applied field. Protons that split each other have the same J (e.g. about 7 Hz for CH3–CH2), which helps to match signals.
Area of signals (integration): the area under each signal is proportional to the number of protons producing it. The integration curve (step heights) gives the proton ratio, e.g. 3 : 2 in ethyl bromide.
Interpretation of NMR Spectra of Simple Molecules
| Compound | Signals: δ (ppm), multiplicity, protons |
|---|---|
| Ethyl bromide CH3CH2Br | 2 signals: 1.7 triplet (3H, CH3); 3.4 quartet (2H, CH2Br); ratio 3:2 |
| Ethanol CH3CH2OH | 3 signals: 1.2 triplet (3H); 3.7 quartet (2H, CH2O); about 2.6 singlet (1H, OH, position variable). In very pure ethanol OH appears as a triplet and CH2 as a multiplet. |
| Acetaldehyde CH3CHO | 2 signals: 2.2 doublet (3H, CH3); 9.8 quartet (1H, CHO); ratio 3:1 |
| 1,2-Dibromoethane BrCH2CH2Br | 1 signal: 3.6 singlet (4H) — all protons equivalent, so no splitting |
| Ethyl acetate CH3COOCH2CH3 | 3 signals: 2.0 singlet (3H, CH3CO); 4.1 quartet (2H, OCH2); 1.25 triplet (3H, CH3) |
| Toluene C6H5CH3 | 2 signals: 2.3 singlet (3H, Ar–CH3); about 7.2 (5H, aromatic, near-singlet/multiplet) |
| Acetophenone C6H5COCH3 | 2.6 singlet (3H, COCH3); 7.4–7.6 multiplet (3H, meta + para) and about 7.95 multiplet (2H, ortho, deshielded by C=O) |
| Acetylacetone CH3COCH2COCH3 ⇌ enol | Exists mainly as the enol (H-bonded). Enol: about 15.5 broad singlet (1H, enolic OH), 5.5 singlet (1H, =CH–), 2.0 singlet (6H, 2 CH3). Keto form: 3.6 singlet (2H, CH2), 2.2 singlet (6H). Both sets appear because tautomerism is slow on the NMR time scale. |
Structure elucidation example: A compound C2H4O shows a doublet at δ 2.2 (3H) and a quartet at δ 9.8 (1H). The δ 9.8 signal shows –CHO; the doublet/quartet pair shows CH3 next to CH. Structure: CH3CHO (acetaldehyde).
Likely exam questions
- (Long, 8 marks) Explain the basic principle of NMR spectroscopy. What is chemical shift? Discuss the factors affecting chemical shift.
- (Long, 8 marks) What is spin–spin splitting? Explain the (n + 1) rule and coupling constant. Interpret the PMR spectra of ethyl bromide and ethyl acetate.
- (Short, 4 marks) Why is TMS used as a reference compound in NMR?
- (Short, 4 marks) Distinguish between shielding and deshielding of protons.
- (Short, 4 marks) Predict the number of signals, their multiplicity and relative areas in the PMR spectrum of acetaldehyde.