What NMR actually measures

Certain nuclei — ¹H and ¹³C — behave like tiny magnets because they have a nuclear spin. Put the sample in a strong magnetic field and those spins line up either with or against the field. A pulse of radio waves flips them into the higher-energy alignment; as they relax back, they emit radiation that the machine detects.

The exact frequency absorbed depends on the electron density around the nucleus. Nearby electronegative atoms (O, Cl, N) pull electrons away, so the nucleus is deshielded and absorbs further to the left (a higher chemical shift, δ).

• Chemical shift, δ, is measured in parts per million (ppm) relative to a standard. It is independent of the machine's field strength, so results are comparable everywhere.
• ¹²C has no nuclear spin, so it is invisible. It is the ~1.1% of carbon that is ¹³C that gives the carbon spectrum.

Why it matters: mass spectrometry gives you the M_r and the fragments; infrared gives you the functional groups; NMR gives you the carbon skeleton — which atom is next to which. Put all three together and the structure is nailed.

TMS, and why the solvent is deuterated

Tetramethylsilane, TMS — Si(CH₃)₄ — is the reference standard, and its peak is defined as δ = 0. It is chosen because:

• all 12 of its hydrogen atoms are equivalent (and all 4 carbons are equivalent), so it gives a single sharp peak;
• silicon is less electronegative than carbon, so the protons are very shielded and the peak sits upfield of virtually every organic signal — it never overlaps with the sample;
• it is inert, non-toxic and volatile (b.p. 27 °C), so it is easily distilled off and the sample is recovered.

The solvent must contain no ordinary ¹H. A normal solvent would swamp the spectrum with a huge solvent peak. So chemists use a deuterated solvent such as CDCl₃ (deuterated trichloromethane) or CCl₄ (which has no hydrogen at all). Deuterium, ²H, does not absorb in the ¹H region of the spectrum, so it is effectively invisible.

Why TMS?

¹³C NMR — counting carbon environments

¹³C NMR is the easy one: the number of peaks = the number of different carbon environments. Two carbons are in the same environment only if they are chemically equivalent — swap them and the molecule is unchanged (usually because of symmetry).

Rough shift ranges worth knowing (they'll be in the data booklet):

• C=O in an aldehyde or ketone: δ ≈ 190–220 — always the far-left peak
• C=O in an acid, ester or amide: δ ≈ 160–185
• Aromatic ring carbons: δ ≈ 110–160
• C–O (alcohol, ester): δ ≈ 50–90; plain C–C alkyl: δ ≈ 5–55

Your turn — carbon environments

Using ¹³C to choose between isomers

¹H NMR — three pieces of information

A proton (¹H) spectrum tells you three separate things. Read all three, every time.

• 1. Chemical shift (δ) — what each set of protons is attached to. Deshielding by O, N or a ring shifts a signal to the left.
• 2. Integration — the relative area under each peak (shown as a step height, or just a number) is proportional to the number of hydrogen atoms in that environment. It gives you a ratio, so you must scale it up to fit the molecular formula.
• 3. Splitting (spin–spin coupling) — which environments are next-door neighbours.

Typical shift ranges (you get these in the data booklet — learn the shape of the pattern, not the exact numbers):

• δ 0.7–1.6 — plain alkyl H (R–CH₃, R–CH₂–R)
• δ 2.0–2.9 — H on a carbon next to a C=O (CH₃CO–)
• δ 3.1–4.3 — H on a carbon bonded to O or a halogen (–O–CH₂–, –CH₂Cl)
• δ 6.5–8.0 — H on a benzene ring
• δ 9.3–10.5 — the H of an aldehyde, –CHO
• δ 10–12 — the H of a carboxylic acid, –COOH (broad)

The n + 1 rule and spin–spin coupling

A proton "feels" the tiny magnetic fields of the protons on the adjacent carbon atoms. Those neighbours can be aligned with or against the field, which splits the signal into a group of lines.

• n = 0 → singlet · n = 1 → doublet · n = 2 → triplet · n = 3 → quartet
• Equivalent protons do not split each other — the three H of a CH₃ group are identical, so they produce one signal, not a multiplet.
• Splitting is mutual: if A splits B, then B splits A. In CH₃CH₂Br the CH₃ (2 neighbours) is a triplet and the CH₂ (3 neighbours) is a quartet.

Your turn — how many lines?

Your turn — reading the integration trace

Deducing a structure

The D₂O shake — finding OH and NH protons

Protons on O–H and N–H groups are labile: they swap rapidly between molecules. That is why they give a broad singlet at a variable chemical shift — and it is also how you identify them.

Method: run the ¹H spectrum, then shake the sample with a few drops of D₂O (deuterium oxide, "heavy water") and run it again. The labile H is exchanged for a D:

Any peak that vanishes after the D₂O shake was an OH or NH proton. (A small new HOD peak may appear at about δ 4.8.) Every other peak is unchanged, because C–H protons are not labile and do not exchange.

Exam habit: the question usually says "the peak at δ 5.4 disappeared when the sample was shaken with D₂O". Your answer: that proton is on an O–H (or N–H) group, because the labile H has exchanged with D and deuterium does not show up in a ¹H spectrum.

What does the D₂O shake prove?

Thin-layer chromatography and R_f

All chromatography separates a mixture by the competition between a stationary phase and a mobile phase. The more strongly a component is adsorbed onto the stationary phase, the more slowly it moves; the more soluble it is in the mobile phase, the further it travels.

• TLC: the stationary phase is a thin layer of silica (SiO₂) or alumina on a glass or plastic plate; the mobile phase is the solvent that rises up the plate by capillary action.
• Draw the baseline in pencil (ink would dissolve and run) and make sure the spots start above the solvent level, or they will simply dissolve off the plate.
• Colourless spots are located with UV light (the plate contains a fluorescent dye) or by using a locating agent such as iodine vapour or ninhydrin (for amino acids).

Careful: R_f is only reproducible if the solvent, temperature and stationary phase are the same. That is why you should always run a known reference alongside the unknown on the same plate rather than trusting a book value.

Your turn — calculate R_f

An impossible result

Gas chromatography and GC-MS

In gas chromatography the mobile phase is an inert carrier gas (helium or nitrogen) and the stationary phase is a thin film of a high-boiling liquid coated onto a solid support, packed into a long coiled column inside an oven.

• The sample is injected and vaporised. Each component partitions between the gas and the liquid film.
• A component that is more soluble in the stationary phase (or less volatile) spends longer dissolved in it, so it moves more slowly and has a longer retention time.
• Retention time is the time between injection and the component reaching the detector. Under fixed conditions it is characteristic of a substance, so you identify components by comparing with the retention times of known standards run on the same column.
• The area under each peak is proportional to the amount of that component — that is how GC quantifies (e.g. alcohol in blood).

Why GC-MS is so powerful: retention time alone can be ambiguous, because two different compounds can happen to have the same retention time. Feeding the separated components straight into a mass spectrometer gives you the M_r (from the molecular ion) and a unique fragmentation fingerprint, which is matched against a computer database. Uses: forensics, drug testing in sport, airport explosives screening, and analysing samples on Mars.

Your turn — the molecular ion

Why couple GC to MS?

Which technique is it?

Tap a statement, then tap the technique it belongs to.

Spotting the aldehyde

Match the observation to its meaning

Tap an item on the left, then its partner on the right.

The big ideas to know

You've now covered Topic 19: Modern Analytical Techniques II of the Edexcel A-level Chemistry (9CH0) specification. Press Finish to see your score.