Putting the analysis together: structure from spectra

An order to work in

An unknown organic compound is identified by four techniques, each answering a different question about it: how heavy the molecule is and what it is made of, which bonds it contains, how many kinds of carbon it has, and which hydrogens sit next to which. None of them alone gives the structure. Taken in a fixed order, each one rules out structures the one before left standing.

The order is the one in the flow chart. First the molecular formula, from a precise mass or from combustion data with the molecular ion. Then the degree of unsaturation, which says how many rings and double bonds the formula allows. Then infrared, for the functional group. Then carbon-13 NMR, for the number of carbon environments. Then proton NMR, for how the fragments are joined. Last, a structure is proposed and checked against every piece of data, because a structure that fits four results and contradicts a fifth is wrong.

Describe the drawing

A vertical flow chart of six boxes joined by downward arrows: molecular formula from a precise M⁺ or combustion data; degree of unsaturation from (2C + 2 + N − H − X) ÷ 2; infrared, for C=O, O–H and C–O; carbon-13 NMR, one peak per carbon environment; proton NMR, by shift, integration and n + 1 splitting; and propose and check. Left of each box is the question it answers and right of it what it rules out. A dashed coral arrow runs from the last box back up to infrared, labelled any contradiction: think again.

Gas chromatography sometimes comes before all of this. If the sample is a mixture, every spectrum is a mixture too, and GC separates it first; in GC–MS each separated component goes straight into a mass spectrometer.

The formula and what it allows

A high-resolution mass spectrometer gives the mass of the molecular ion to four decimal places, and the precise relative atomic masses (¹H 1.0078, ¹²C 12.0000, ¹⁴N 14.0031, ¹⁶O 15.9949) make each formula's mass unique. A molecular ion at 58.0417 is C₃H₆O, not C₄H₁₀ at 58.0780 or C₂H₆N₂ at 58.0530.

Combustion analysis gets there by another method. A known mass of the compound is burned in excess oxygen and the carbon dioxide and water are weighed. All the carbon ends up in the CO₂ and all the hydrogen in the H₂O, and whatever mass is left over is oxygen. That gives the empirical formula; the molecular ion peak in an ordinary mass spectrum then gives the multiple.

The degree of unsaturation follows from the formula. Every ring and every double bond removes two hydrogens from the saturated, open-chain formula CₙH₂ₙ₊₂. For a compound containing C, H and O, the count is (2C + 2 − H) ÷ 2. Oxygen does not change it; each nitrogen adds one to the bracket, and each halogen counts as a hydrogen. C₃H₆O has (6 + 2 − 6) ÷ 2 = 1: one C=O, one C=C or one ring. C₈H₈O has (16 + 2 − 8) ÷ 2 = 5, and five is a strong hint of a benzene ring, which accounts for four, plus one more double bond.

The term itself is not required, and no mark depends on it. The count is worth the ten seconds because it rules out whole families of structure before a spectrum is read.

Empirical formula

The simplest whole-number ratio of the atoms of each element in a compound.

Molecular ion

The ion formed when a molecule loses one electron, M⁺; its m/z gives the relative molecular mass.

Degree of unsaturation

The number of rings plus double bonds in a molecule, found from (2C + 2 + N − H − X) ÷ 2, where a triple bond counts as two.

What each spectrum is read for

In the infrared spectrum, look first between 1680 and 1750 cm⁻¹. A strong absorption there is C=O. Then look from 2500 to 3550 cm⁻¹: a broad absorption from 3230 to 3550 is the O–H of an alcohol, and a very broad one from 2500 to 3000, under the C–H peaks, is the O–H of a carboxylic acid. A C=O with no O–H is an aldehyde, a ketone or an ester; a strong C–O near 1200 as well points to the ester. The fingerprint region below 1500 is for matching against a database, not for assigning bonds.

In the carbon-13 spectrum, count the peaks: each is one carbon environment. Then place them. Above 160 ppm is a C=O, with esters and acids at 160 to 185 and aldehydes and ketones at 190 to 220. From 110 to 160 are aromatic and alkene carbons. From 50 to 90 is a carbon bonded to oxygen. Below 50 is an alkyl carbon.

The proton spectrum carries three readings for every peak. The shift says what the hydrogens are near: a CH₃ on a chain near 1, a hydrogen on a carbon next to C=O at 2.1 to 2.6, a hydrogen on a carbon bonded to an ester oxygen at 3.7 to 4.1, a ring hydrogen near 7 to 8, an aldehyde CHO at 9 to 10, a carboxylic acid at 10 to 12. The integration gives the ratio of hydrogens in each environment. The splitting follows the n + 1 rule: a peak split into n + 1 lines has n hydrogens on the neighbouring carbons.

The n + 1 rule counts hydrogens on adjacent carbon atoms only. A CH₃ bonded to an oxygen, to a C=O carbon or to a benzene ring carbon with no hydrogen has no neighbouring hydrogens to couple with, and gives a singlet. Singlets are as informative as multiplets: each one marks a group cut off from other hydrogens.

Fingerprint region

The part of an infrared spectrum below 1500 cm⁻¹, whose pattern is unique to a compound and identifies it by comparison with a database.

Chemical shift

The position of an NMR peak, δ in ppm, measured relative to TMS at 0 ppm; it depends on the atom's chemical environment.

Chemical environment

A set of atoms that are bonded to identical surroundings, so they absorb at the same chemical shift and give one peak.

Integration ratio

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

n + 1 rule

A proton NMR peak is split into n + 1 lines, where n is the number of hydrogens on the adjacent carbon atoms.

Three analyses

The second analysis starts from combustion data, and the proton spectrum has to decide between two esters with the same formula and the same functional group.

Describe the drawing

Three spectra of methyl propanoate stacked on one figure, with its structure at the top, the OCH₃ in cyan, the CH₂ in amber and the CH₃ on the ethyl end in coral. A, infrared, transmittance against wavenumber: a deep sharp C=O dip at 1740, a strong C–O dip at 1200 and nothing broad above 3000, so no O–H. B, carbon-13: four lines at 175 for C=O, 52 for OCH₃, 27 for CH₂ and 9 for CH₃, with TMS at 0. C, proton NMR: a singlet at 3.67 for OCH₃, a quartet at 2.32 for CH₂ and a triplet at 1.14 for CH₃, with an integration trace rising in steps of 3, 2 and 3. A note says ethyl ethanoate would show the singlet at 2.0 and the quartet at 4.1.

Where analyses go wrong

Symmetry is the first thing to look for when a count of peaks seems wrong. Equivalent carbons give one peak, so a molecule can have far fewer peaks than carbons. Every isomer of C₄H₁₀O has four carbons, but butan-1-ol gives four carbon-13 peaks, ethoxyethane gives two and 2-methylpropan-2-ol gives two. Ethoxyethane, CH₃CH₂OCH₂CH₃, is the same on either side of its oxygen, so its two CH₂ carbons are one environment and its two CH₃ carbons are another.

Describe the drawing

Three isomers of C₄H₁₀O drawn at the same scale as displayed formulae with every bond and hydrogen shown, each carbon ringed in a colour for its environment. Butan-1-ol has four carbons in four colours: four peaks. Ethoxyethane has a dashed mirror line through its oxygen, both CH₂ carbons in one colour and both CH₃ carbons in another: two peaks. 2-Methylpropan-2-ol has three methyl carbons in one colour around a central carbon in another, with a curved arrow for the threefold rotation about the C–O bond: two peaks.

A ring makes symmetry harder to spot. A benzene ring with two different groups opposite each other, at positions 1 and 4, has only two kinds of ring hydrogen and four kinds of ring carbon; with the groups at 1 and 2 it has four kinds of ring hydrogen and six kinds of ring carbon. The number of aromatic peaks tells the substitution pattern.

An integration ratio is a ratio, not a count. Peaks in the ratio 3 : 1 for a compound with eight hydrogens mean 6H and 2H, not 3H and 1H. The integration always has to be scaled to the number of hydrogens in the molecular formula.

The hydrogen of an O–H or N–H causes two problems. Its shift is unreliable, anywhere from 0.5 to 5 ppm for an alcohol, depending on the concentration and the solvent. And it usually appears as a singlet, because it exchanges between molecules too fast to couple with its neighbours, and it does not split their peaks either. The CH₂ next to the OH in ethanol is a quartet from the CH₃ alone. Infrared is the surer test for an O–H: a broad absorption at 3230 to 3550 cm⁻¹.

Last comes the check. Draw the proposed structure, count its carbon environments and hydrogen environments, predict the shift and splitting of each, and compare every prediction with the data. Any reading the structure cannot explain means the structure is wrong.

Equivalent atoms

Atoms that are interchanged by the symmetry of the molecule, so they share an environment and give a single NMR peak.

Other boards: D₂O, calibration and pollution monitoring

OCR and Cambridge add a test that settles which peak belongs to an O–H or N–H. Shaking the sample with a few drops of D₂O replaces that hydrogen with deuterium, which does not absorb in a proton spectrum, so the peak disappears when the spectrum is run again. A peak that is present in CDCl₃ and gone after a D₂O shake is an O–H or N–H. The other side of the same fact: if a spectrum was run in D₂O, an O–H is not missing from the molecule just because it is missing from the spectrum, and the infrared is where to look for it.

CH₃CH₂OH(l) + D₂O(l) ⇌ CH₃CH₂OD(l) + HDO(l)The OH hydrogen exchanges with deuterium; the C–H hydrogens do not.

OCR also uses gas chromatography to measure how much of a component is present. Standards of known concentration are run under the same conditions and their peak areas plotted against concentration, giving an external calibration graph. The unknown's peak area is then read off the line as a concentration.

OCR examines infrared outside the laboratory as well, in monitoring air pollution. An exhaust analyser passes infrared through the gas and measures the absorbance at a wavenumber where only one pollutant absorbs: near 2140 cm⁻¹ for carbon monoxide and near 1880 cm⁻¹ for nitrogen monoxide. The more of the gas present, the more radiation it absorbs, so the absorbance gives its concentration. A breathalyser works on the same principle, measuring the absorbance of the C–H bonds of ethanol in exhaled breath.

External calibration

Finding a concentration by measuring standards of known concentration, plotting the response against concentration, and reading the unknown from the graph.

D₂O exchange

Shaking a sample with D₂O replaces the hydrogen of O–H and N–H groups with deuterium, so their proton NMR peak disappears.