Proton and carbon-13 NMR spectroscopy

Nuclei in a magnet

A hospital MRI scanner places a patient inside a magnet tens of thousands of times stronger than the Earth's field and pulses radio waves through them. The hydrogen nuclei in water and fat absorb those radio waves, and the scanner maps where. An NMR spectrometer in a chemistry laboratory does the same to a few milligrams of a compound in a narrow glass tube, and asks a different question: not where the hydrogen nuclei are, but what each one is bonded to.

Nuclei with an odd number of nucleons, such as ¹H and ¹³C, have a property called spin and behave like tiny magnets. In a strong external field each one lines up either with the field or against it, and the two arrangements differ slightly in energy. Radio waves of exactly the right frequency flip a nucleus from the lower energy state to the higher one, and that absorption is what is recorded.

The gap between the two states is tiny, so the radiation that matches it is in the radio-frequency range, hundreds of megahertz, and a stronger magnet gives a larger gap and a higher frequency: a '400 MHz' spectrometer is one whose magnet makes hydrogen nuclei absorb near 400 MHz. Carbon-12, with six protons and six neutrons, has an even number of nucleons and no spin, so it gives no signal at all.

The frequency is not the same for every hydrogen in a molecule. Electrons around a nucleus set up a small field of their own that opposes the applied one, shielding the nucleus. An electronegative atom nearby, such as oxygen, draws electron density away, so a hydrogen on a carbon bonded to oxygen is less shielded and absorbs at a higher shift. Each different chemical environment gives its own peak.

Chemical environment

A set of atoms in a molecule that are bonded to identical surroundings, so they absorb at the same chemical shift.

Chemical shift

The position of an NMR absorption relative to the TMS peak, δ, measured in parts per million (ppm).

TMS

Tetramethylsilane, Si(CH₃)₄, the reference compound whose single peak is defined as δ = 0.

Nuclear spin

The property of some nuclei, including every nucleus with an odd number of nucleons such as ¹H or ¹³C, that lets them align with or against an applied magnetic field.

Shielding

The reduction in the magnetic field felt by a nucleus caused by the electrons around it; a less shielded nucleus absorbs at a higher chemical shift.

Shifts are measured from tetramethylsilane, TMS. Its twelve hydrogens are all equivalent, as are its four carbons, so it gives one sharp peak in each kind of spectrum. Silicon is less electronegative than carbon, so those nuclei are more shielded than in almost any organic compound and the TMS peak sits to the right of everything else. TMS is also inert, non-toxic and volatile, boiling at 27 °C, so it is easily removed from the sample afterwards.

The sample is dissolved in a solvent with no ¹H atoms, so that the solvent gives no peaks of its own in the proton spectrum, which would swamp the sample's: the solvent molecules outnumber the sample's by hundreds to one. Deuterium, ²H, has a nucleus with spin, but it absorbs at a frequency far from that of ¹H, so a deuterated solvent is invisible in a proton spectrum. Each choice has its reason.

Say 'no ¹H atoms' or 'no hydrogen-1' for the solvent, not 'no hydrogen': CDCl₃ contains deuterium, which is hydrogen.

Carbon-13 NMR: counting carbon environments

About 1.1% of carbon atoms are carbon-13, and that is enough for a spectrum. Each carbon environment gives one peak, and the spectra met at A level show no splitting. Peak heights are not proportional to the number of carbons, so a carbon-13 spectrum is read for two things only: how many peaks, and where.

The lack of splitting follows from that 1.1%. Two carbon-13 atoms are almost never bonded to each other in the same molecule, so one carbon peak is not split by another carbon, and the splitting that attached hydrogens would cause is switched off by the instrument as the spectrum is recorded.

Count environments by looking for symmetry. In propan-2-ol, CH₃CH(OH)CH₃, the two methyl carbons are bonded to identical surroundings and give one peak; the carbon bearing the OH gives a second. Three carbons, two peaks.

The test for equivalence is whether swapping two carbons, by rotating or reflecting the molecule, gives the same molecule. Propanone has two methyl carbons that swap in this way, so it has two peaks, one at about 30 ppm and the C=O at about 205 ppm. Propan-1-ol, CH₃CH₂CH₂OH, has three carbons with three different sets of neighbours and gives three peaks. Benzene, with six identical carbons in a ring, gives just one.

Counting peaks separates isomers that every other quick test confuses. Pentan-2-one, CH₃COCH₂CH₂CH₃, and pentan-3-one, CH₃CH₂COCH₂CH₃, are both C₅H₁₀O ketones with a C=O near 210 ppm. Pentan-2-one has no symmetry and gives five peaks. Pentan-3-one is symmetrical about its C=O: its two CH₃ carbons are equivalent, and so are its two CH₂ carbons, and it gives three peaks.

Describe the drawing

A carbon-13 NMR spectrum of butan-2-ol with δ from 220 on the left to 0 on the right: four single lines and a small TMS line at 0. The carbon bonded to oxygen is at 69, the CH₂ at 32, the CH₃ on the C–OH carbon at 23 and the CH₃ of the ethyl group at 10. Above, the displayed formula colours each carbon to match its line, and a note says peak heights are not used in carbon-13 NMR.

Proton NMR: shift and integration

A proton NMR spectrum gives more than a carbon-13 spectrum. Each hydrogen environment gives a peak, and the peaks carry three pieces of information: the shift, the area and the splitting.

Hydrogens are counted by environment, not one by one. Propan-2-ol, (CH₃)₂CHOH, has eight hydrogens but only three environments: the six hydrogens of the two equivalent methyl groups, the one on the central carbon, and the one on the oxygen. Its spectrum has three peaks. Propan-1-ol, CH₃CH₂CH₂OH, with the same formula, has four environments and four peaks, which is enough to tell the two isomers apart before any shift is read.

The shift says what the hydrogens are near. The data sheet gives ranges: R–CH₃ at 0.7–1.2 ppm, H–C–C=O at 2.1–2.6, H–C–O at 3.1–3.9, RCOO–C–H at 3.7–4.1, an aldehyde CHO at 9.0–10.0 and a carboxylic acid COOH at 10.0–12.0. An O–H in an alcohol can appear anywhere from 0.5 to 5.0 ppm, depending on concentration and solvent.

The area under each peak is proportional to the number of hydrogens in that environment. The spectrometer draws an integration trace that steps up at each peak; the heights of the steps give the ratio. In ethanol the steps are in the ratio 3:2:1, for CH₃, CH₂ and OH.

Integration ratio

The ratio of the areas under the peaks in a proton NMR spectrum, equal to the ratio of the numbers of hydrogens in each environment.

Splitting and the n + 1 rule

Under high resolution most peaks are split into a cluster of lines. The splitting comes from the hydrogens on the neighbouring carbon atoms, whose small magnetic fields add to or subtract from the field the observed hydrogens feel. A hydrogen environment with n hydrogens on the adjacent carbons gives a peak split into n + 1 lines. This is the n + 1 rule, and n counts the hydrogens on every adjacent carbon together, on both sides.

Describe the drawing

Four columns, each showing two bonded carbons with one hydrogen circled on the left carbon and zero, one, two or three amber hydrogens on the carbon next door. Beneath each, the peak for the circled hydrogen: a singlet of one line, a doublet of two equal lines, a triplet in the ratio 1:2:1, and a quartet in the ratio 1:3:3:1. A banner reads n hydrogens on adjacent carbons give n + 1 lines.

The lines in a cluster are not of equal height. Each neighbouring hydrogen is aligned either with or against the field, and the lines record how many ways each total can arise. One neighbour gives a doublet of 1:1; two give a triplet of 1:2:1, since the middle line can arise in two ways; three give a quartet of 1:3:3:1. A symmetrical cluster with the tallest lines in the middle is the sign of a genuine multiplet.

Two rules stop the n + 1 rule giving wrong answers. Equivalent hydrogens do not split each other, so the three hydrogens of a CH₃ group count only as neighbours to something else. And the hydrogen of an O–H group usually exchanges rapidly between molecules, so it neither splits nor is split: an alcohol O–H is normally a singlet.

Splitting reaches only to the next carbon. A CH₃ bonded to a C=O carbon, which carries no hydrogens, is a singlet whatever lies beyond it, and so is a CH₃ bonded to an ester oxygen. In propan-2-ol the six methyl hydrogens have one neighbour, the CH, and give a doublet of area 6. The CH has six neighbours and is split into seven lines. A doublet of area 6 beside a small multiplet of area 1 is the sign of a (CH₃)₂CH group.

Describe the drawing

A proton NMR spectrum of ethanol with δ from 10 on the left to 0 on the right. A cyan quartet of four close lines at 3.69 is the CH₂, a single slightly broadened line at 2.61 is the OH, and an amber triplet of three lines at 1.22 is the CH₃, with TMS at 0. A stepped integration trace rises by 2, then 1, then 3 across the three peaks, and expanded insets show the quartet in the ratio 1:3:3:1 and the triplet 1:2:1.

Spin-spin splitting

The splitting of an NMR peak into lines by the hydrogens on adjacent carbon atoms, n neighbouring hydrogens giving n + 1 lines.

Deuterated solvent

A solvent such as CDCl₃ in which ¹H atoms are replaced by deuterium, so that it gives no peaks in a proton NMR spectrum.

From formula and spectrum to structure

A structure question gives a molecular formula and one or two spectra. Work in a fixed order. Count the peaks, which gives the environments. Turn the integration ratio into actual numbers of hydrogens, using the formula. Read each splitting as a statement about the neighbouring carbon. Use the shifts to place each group next to O, next to C=O or in a chain. Then draw a structure and check it against every piece of data.

Other boards: D₂O exchange

OCR and Cambridge add a way of proving which peak belongs to an O–H or N–H hydrogen. The sample is shaken with a few drops of D₂O and the spectrum run again. The O–H hydrogen exchanges with deuterium to give O–D, which does not absorb in a proton spectrum, so the O–H peak disappears; an N–H peak behaves the same way, and the C–H peaks are unchanged. In propan-1-ol the singlet at 2.3 ppm goes and the other three peaks stay, which fixes it as the O–H.

CH₃CH₂CH₂OH(l) + D₂O(l) ⇌ CH₃CH₂CH₂OD(l) + HDO(l)The O–H hydrogen exchanges; the C–H hydrogens do not.