Analytical techniques

How a molecule absorbs infrared radiation

Infrared (IR) spectroscopy finds the functional groups in a molecule by measuring the infrared radiation the molecule absorbs. A covalent bond is not rigid: the two bonded atoms behave like two masses joined by a spring, so the bond stretches and bends all the time. When the molecule absorbs infrared radiation of exactly the right energy, the vibration becomes more vigorous, and the spectrometer records that absorption as a dip in the spectrum.

The energy a bond absorbs depends on two things. Heavier atoms vibrate more slowly and absorb lower-energy radiation, while a stronger bond needs more energy to stretch it, so it absorbs higher-energy radiation. That single idea explains the whole pattern: a C-C single bond absorbs at about 1000-1300 cm-1, a C=C double bond at about 1620-1680 cm-1 and a C-C triple bond at about 2100-2260 cm-1, in that order, because the bonds get stronger. The same reasoning puts a C-H bond, a light atom joined by a strong bond, high in the spectrum at about 2840-3100 cm-1.

Absorptions are measured as wavenumbers, in cm-1. The wavenumber is 1 / wavelength (with the wavelength measured in cm), so it is proportional to energy: the bigger the wavenumber, the more energetic the radiation. A spectrum is plotted as percentage transmittance against wavenumber, with the scale running from about 4000 cm-1 on the left down to about 400 cm-1 on the right, so every absorption is a trough pointing downwards. Read the wavenumber from the position of the trough and read how strong the absorption is from how deep the trough goes.

Only a vibration that changes the dipole moment of the molecule absorbs infrared radiation. This is why nitrogen and oxygen, which together make up most of the air and whose vibrations are perfectly symmetrical, do not absorb infrared radiation at all, while water, carbon dioxide and almost every organic molecule do. It also means that the same functional group absorbs in almost the same place whatever the rest of the molecule is, which is what makes IR spectroscopy useful for identifying groups rather than whole structures.

• IR radiation is absorbed when it matches the energy needed to make a bond stretch or bend.
• Stronger bonds absorb at higher wavenumbers: C-C (about 1000-1300 cm-1), C=C (about 1620-1680 cm-1), C-C triple bond (about 2100-2260 cm-1).
• Heavier atoms on a bond push the absorption to a lower wavenumber.
• Wavenumber = 1 / wavelength, measured in cm-1, and is proportional to energy.
• Percent transmittance is plotted downwards, so an absorption is a trough, and a strong absorption is a deep one.

The characteristic absorptions you must know

The absorptions in the Data section are enough to identify the functional groups in an AS Level question. Learn each one with its region, but also learn whether it is broad or sharp, because shape is often what decides the answer. A broad absorption spread over hundreds of wavenumbers is hydrogen bonding in an O-H or an N-H group; a sharp, narrow absorption belongs to a carbonyl group.

Two absorptions do most of the work. A broad O-H absorption between 2500 and 3300 cm-1 shows that an alcohol or a carboxylic acid is present, but it does not by itself tell you which, because both show it. A sharp, strong C=O absorption between 1640 and 1750 cm-1 shows a carbonyl group, and its exact position moves a little: a carboxylic acid absorbs near the bottom of the range (about 1700-1725 cm-1), an aldehyde or a ketone near the middle (about 1700-1750 cm-1) and an ester near the top (about 1735-1750 cm-1). These positions shift with the rest of the molecule, so never argue from a few wavenumbers when the presence or absence of a whole band is the better evidence.

The remaining groups are identified by their own regions. An N-H bond of an amine absorbs at about 3300-3500 cm-1, is broad but weaker than an O-H, and a primary amine often shows two peaks there because it has two N-H bonds. A C=C of an alkene absorbs at about 1620-1680 cm-1 but is much weaker than a C=O, so it is easily missed. A carbon-nitrogen triple bond of a nitrile absorbs sharply at about 2200-2250 cm-1, and a C-O single bond of an alcohol or an ester gives a strong absorption at about 1000-1300 cm-1.

The wavenumber alone is never the whole answer. You are being asked to combine the bands you can see with the bands that are missing, so always run through the table below in both directions: which groups are present, and which are ruled out?

Analysing an infrared spectrum

Work through a spectrum in a fixed order so you never miss a band. First look at 3300-2500 cm-1: is there a broad O-H or N-H absorption sitting above the C-H peaks? Then look at 2000-1500 cm-1: is there a sharp, strong C=O, a weak C=C, or a sharp nitrile band at about 2200 cm-1? Finally check the fingerprint region below about 1500 cm-1, where bending vibrations of the whole skeleton overlap into a complicated pattern that is unique to the compound. That pattern is compared with a database of known spectra, because individual bands there cannot be assigned to individual bonds.

Four worked examples cover nearly every question you will meet. Ethanol shows a broad O-H at about 3200-3300 cm-1, sharp C-H peaks at about 2900 cm-1 and a strong C-O at about 1050 cm-1, with no C=O at all: an alcohol, not a carbonyl compound. Ethanoic acid shows an extremely broad O-H running from about 2500 to 3300 cm-1 that overlaps the C-H absorptions, plus a sharp strong C=O at about 1710 cm-1: an acid, because the broad O-H and the carbonyl appear together.

Propanal shows a sharp strong C=O at about 1725 cm-1, the small C-H absorptions near 2720 and 2840 cm-1 that belong to the aldehyde hydrogen, and no O-H at all. Propanone shows a C=O at about 1715 cm-1 and nothing else above the C-H region, so an aldehyde and a ketone are told apart by the shapes and the extra small bands high in the C-H region rather than by the carbonyl position alone. Ethyl ethanoate shows a C=O at about 1740 cm-1 and strong C-O absorptions at about 1200 and 1050 cm-1, with no broad O-H, which is what an ester looks like.

The logic to write down in an examination is the combination: a broad O-H with no C=O is an alcohol; a broad O-H with a sharp strong C=O is a carboxylic acid; a sharp strong C=O with no O-H is an aldehyde or a ketone; a sharp strong C=O with a strong C-O and no O-H is an ester; and an N-H with no C=O is an amine.

• Ethanol: broad O-H at about 3200-3300, C-H at about 2900, C-O at about 1050, no C=O.
• Ethanoic acid: very broad O-H from about 2500 to 3300 plus a sharp strong C=O at about 1710.
• Propanal and propanone: both show a C=O at about 1720, so check the small aldehyde C-H bands.
• Ethyl ethanoate: C=O at about 1740 with strong C-O bands and no O-H.
• Below about 1500 cm-1, identify the compound by matching the whole fingerprint region with a database.

Mass spectrometry: m/e values and isotopic abundances

A mass spectrometer ionises the sample, separates the ions according to their mass and records how many of each arrive at the detector. You are not required to know how the instrument works, but you must be able to read its output. The x-axis of a mass spectrum is the m/e value of the ion: its mass divided by its charge. Nearly every ion formed carries a single positive charge, so the m/e value is simply the mass of that ion in atomic mass units, and a peak at m/e 46 is a cation of mass 46.

The y-axis is the relative abundance of each ion, usually as a percentage of the tallest peak. That tallest peak is the base peak, and it is given the value 100; every other peak is measured relative to it. The relative abundance of a peak is proportional to how much of that ion was formed, so the heights of the peaks carry the quantitative information in the spectrum.

A mass spectrum is used in two quite different ways. The spectrum of an element has one line for each isotope, and the heights of those lines are the isotopic abundances, so the relative atomic mass can be worked out straight from them. The spectrum of a compound has a molecular ion peak that gives the relative molecular mass and a set of fragment peaks that help to identify the structure, which is what the next blocks deal with.

The relative atomic mass is a weighted mean, so you must multiply each isotope mass by its fractional abundance and add the results. Chlorine is the classic case: its spectrum has two lines, at m/e 35 and m/e 37, because chlorine is 75.8% chlorine-35 and 24.2% chlorine-37. Ar = (35 x 75.8 + 37 x 24.2) / 100 = (2653 + 895.4) / 100 = 35.5, which is why the Periodic Table quotes 35.5 for chlorine rather than a whole number. A simple mean of 35 and 37 would give 36, which is wrong because it ignores how much more abundant the lighter isotope is.

The molecular ion and relative molecular mass

When a molecule is ionised by losing a single electron, the ion that remains has the same mass as the molecule. That ion is the molecular ion, written M+, and it appears as the peak at the highest m/e value in the spectrum, with the important exception of the small isotope peaks that sit one or two units above it. Because only an electron has been removed, the m/e value of the molecular ion is numerically equal to the relative molecular mass of the compound, so reading the molecular ion off the spectrum gives you Mr at once.

Ethanol is a good example. Its molecular ion appears at m/e 46, and that agrees with the formula C2H5OH: (2 x 12) + (6 x 1) + 16 = 46. A spectrum with a molecular ion at m/e 58 fits C4H10, and 58 also fits C3H6O, so the molecular ion fixes the mass but not the formula; the infrared spectrum and the fragment peaks decide between the possibilities.

Two warnings save marks. The molecular ion is not necessarily the tallest peak. Ionising a molecule gives it surplus energy, and if it falls apart easily the molecular ion peak is small, so identify M+ by its position at the top of the spectrum and not by its height. In the spectrum of an alcohol the molecular ion is often very weak indeed. Second, do not read the [M + 1] peak, one unit higher, as the molecular ion: carbon-13 makes that smaller peak unavoidable in every compound that contains carbon.

The molecular ion also tells you whether a whole class of formula is possible. If the highest m/e value is 46, the compound cannot be C3H8O, because that is 60; if it is 74, look for an acid with two oxygens such as propanoic acid, C3H6O2 ((3 x 12) + (6 x 1) + (2 x 16) = 74). Work out what the mass allows first, then use the other evidence to choose.

Fragmentation: what the other peaks tell you

The molecular ion has surplus energy, so it breaks apart. This is fragmentation, and it produces a charged fragment and a neutral fragment. Only the charged fragment is accelerated and detected, so the spectrum shows a peak at the m/e value of the charged piece and nothing at all for the neutral piece. Every fragment peak therefore satisfies a simple sum: the m/e value of the fragment plus the mass of the neutral fragment lost equals the relative molecular mass of the compound.

Ethanol shows this clearly. Its molecular ion is at m/e 46. A peak at m/e 45 is the molecular ion that has lost a hydrogen atom. A peak at m/e 31 is the CH2OH+ ion, formed when the molecular ion loses a methyl radical, CH3, and 31 + 15 = 46. A peak at m/e 29 is the C2H5+ ion, formed when it loses an OH radical, and 29 + 17 = 46. A small peak at m/e 15 is the CH3+ ion. In the spectrum of ethanol, m/e 31 is the base peak, because the cation formed by losing a methyl radical is more stable than most of the alternatives.

Fragmentation patterns are characteristic of a class of compound, which makes them useful as well as awkward. Primary alcohols such as ethanol and propan-1-ol both give a strong peak at m/e 31, CH2OH+, and both give a peak from loss of a methyl radical, 15 below the molecular ion. Alkanes show a series of peaks spaced 14 mass units apart, because the CH2 units are lost one at a time, and a fragment peak at m/e 15, 29, 43 or 57 is a strong hint of an alkyl group.

To answer a fragmentation question, first locate the molecular ion and write down Mr. Then take the m/e value of the fragment you are asked about, work out the mass that has been lost (Mr minus the m/e value) and identify it as a group: 15 is CH3, 17 is OH, 29 is C2H5 and 31 is CH2OH. Check that the fragments you name add up to the molecular mass, because a suggestion that does not balance cannot be right.

• Fragmentation gives a charged fragment, which is detected, and a neutral fragment, which is not.
• Fragment m/e value + mass of the neutral fragment lost = relative molecular mass.
• Ethanol, Mr 46: m/e 45 (loss of H), m/e 31 (CH2OH+, loss of CH3), m/e 29 (C2H5+, loss of OH), m/e 15 (CH3+).
• A strong peak at m/e 31 is typical of a primary alcohol; a series of peaks 14 units apart is typical of an alkane.

The [M + 1] peak: counting the carbon atoms

Wherever there is carbon there is a small peak one mass unit above the molecular ion. Carbon is 98.9% carbon-12 and 1.1% carbon-13, so about one carbon atom in ninety is a carbon-13 atom, which weighs one unit more. A molecule that contains n carbon atoms therefore has a chance of about 1.1n% of containing exactly one carbon-13 atom instead of carbon-12, and that gives a peak at one unit above M+. The peak at that position is the [M + 1] peak.

Because the size of the [M + 1] peak is set by the number of carbon atoms, it can be used to count them. The relationship is given by n = 100 x (abundance of the [M + 1]+ ion) / (1.1 x abundance of the M+ ion). Put plainly: divide the height of the [M + 1] peak by 1.1% of the height of the M+ peak, having first checked that the M+ peak has been given the value 100, and you have the number of carbon atoms.

Take a compound whose molecular ion is at m/e 60 with a relative abundance of 100 and whose [M + 1] peak at m/e 61 has a relative abundance of 3.3. Then n = (100 x 3.3) / (1.1 x 100) = 3, so the molecule contains three carbon atoms. Three carbons account for 36 of the 60 mass units, so the 24 left over must come from hydrogen and oxygen, and C3H8O fits: propan-1-ol or propan-2-ol. The number is not always perfectly whole in practice, so round a value such as 3.1 to 3 rather than searching for a fractional number of atoms.

Treat the method as an estimate and say what it assumes. The small contributions to the [M + 1] peak from deuterium and from oxygen-17 are negligible, but nitrogen-15 is more abundant than carbon-13 is, and an element such as silicon or sulphur adds noticeably to the peak, so a compound that contains one of those elements needs the position and the rest of the spectrum to corroborate the count.

The [M + 2] peak: chlorine and bromine

Carbon never produces a peak two units above the molecular ion, because carbon-14 is present only in trace amounts, so a significant [M + 2] peak is evidence of something else. Chlorine and bromine are the two elements that give it, because each has two isotopes separated by two mass units and each isotope is present in a large natural abundance. Chlorine is about 75% chlorine-35 and 25% chlorine-37, and bromine is almost exactly 50% bromine-79 and 50% bromine-81.

The ratio of the two peaks is the fingerprint of the halogen. A compound with one chlorine atom has an [M + 2] peak about one third the height of the M+ peak, because chlorine-37 is one third as abundant as chlorine-35. A compound with one bromine atom has an [M + 2] peak about the same height as its M+ peak, because the two isotopes are equally abundant. Chloromethane, CH3Cl, has its molecular ion at m/e 50 and an [M + 2] peak at m/e 52 that is about one third as tall; bromomethane, CH3Br, has its molecular ion at m/e 94 and a peak at m/e 96 of very nearly equal height, which is the clearest possible signal of bromine.

Two halogen atoms give an [M + 4] peak as well, because a molecule can contain two heavy isotopes at once. The proportions follow the abundances: a compound with two chlorine atoms gives M+, [M + 2] and [M + 4] peaks in the ratio 9 : 6 : 1, while a compound with two bromine atoms gives 1 : 2 : 1. One chlorine and one bromine together give 3 : 4 : 1. You only need the single halogen case to answer most questions, but noticing an [M + 4] peak tells you at once that two halogen atoms are present.

Do not confuse the two isotope peaks. The [M + 1] peak at one unit above M+ is always there in a carbon compound and comes from carbon-13, whereas the [M + 2] peak two units above M+ is only prominent when chlorine or bromine is present. In a spectrum whose molecular ion is at an even m/e value, a small peak at one unit higher is the [M + 1] peak; a peak of comparable height two units higher is the halogen. A molecular ion at an odd m/e value, meanwhile, is typical of a compound containing an odd number of nitrogen atoms, such as an amine.