Analytical Techniques

0. What Examiners Want

Where this topic appears. Recent AS papers have asked for: a Paper 1 question matching an infrared spectrum to one of four compounds (for example which diagram shows a compound with both C=O and O–H, or which compound could give a given spectrum); a Paper 2 question using the [M+1]⁺ and M⁺ abundances of an ester to show that it has seven carbon atoms and the infrared spectrum to identify its functional group (2024 Paper 2); a comparison of the infrared spectra of two compounds above 1500 cm⁻¹ (2023 Paper 2); and a prediction of how the spectra of a monomer and its polymer differ (2022 Paper 2). At AS, no knowledge of how the instruments work is needed. NMR and chromatography belong to A Level (Topic 37).

1. Infrared Spectroscopy

Covalent bonds are always vibrating (stretching and bending). A bond absorbs infrared radiation of the frequency that matches its own vibration, so each type of bond absorbs at its own characteristic wavenumber, measured in cm⁻¹. An infrared spectrum plots transmittance (%) against wavenumber. An absorption shows as a downward peak: the deeper the peak, the more radiation absorbed. Wavenumber is plotted with the highest value on the left.

The characteristic absorption ranges you need, drawn on the same wavenumber axis as a spectrum. Each bar is one row of the Data Booklet table.

Some papers print slightly different limits for C–H (2850–2950) and hydroxy O–H (3200–3600): always use the table printed in your paper. The table is printed in the question paper, so you do not need to memorise every number, but you must know which bonds are in it and the order along the axis. A useful picture: from the left, O–H and N–H (3200–3650), then C–H (about 3000), then C≡N (about 2200), then the C=O and C=C region (1500–1750), and C–O to the right (1040–1300).

2. Reading Infrared Spectra

Read a spectrum in four steps. (1) Look above 3000 cm⁻¹ for O–H or N–H. (2) Look at 2200–2250 for C≡N. (3) Look at 1670–1750 for C=O, and 1500–1680 for C=C. (4) Check what is missing. Almost every organic spectrum has a C–H absorption near 2850–3100, so it does not tell you which compound you have.

Ethanol. A broad, deep O–H absorption near 3350 (hydroxy), the C–H peaks just below 3000 and a strong C–O peak near 1050. There is no C=O peak.

Propanone. One very strong, sharp C=O absorption near 1715. The O–H region (3200–3650) is empty, so there is no hydroxy group or carboxylic acid.

Propanoic acid. Two clues together: a very broad O–H absorption covering about 2500–3000 (carboxyl) and a strong C=O near 1710. The broad band swallows the C–H peaks.

The same group, different compounds. A carboxylic acid and an ester both have C=O, but only the acid has the very broad O–H absorption at 2500–3000. Compare the three spectra above with the next one:

Ethyl ethanoate. Strong ester C=O near 1740 (1710–1750) and a strong C–O near 1240, with no O–H absorption. It has the same C=O region as propanoic acid but no broad band at 2500–3000.

Propanenitrile. A sharp, narrow absorption near 2250 (C≡N), in an otherwise empty region of the spectrum. Nothing else absorbs between 2000 and 2400 here.

Propylamine. N–H absorptions at 3300–3500 (often two peaks, sharper and weaker than the O–H band of an alcohol) together with C–H absorptions.

But-1-ene. A C=C absorption near 1640 (1500–1680) and C–H absorptions. There is no O–H, no C=O and no C≡N, which rules those groups out.

Spectra before and after a reaction. Infrared spectra show whether a reaction has worked. When propan-1-ol is oxidised to propanoic acid, the broad 3200–3650 absorption is replaced by the very broad 2500–3000 absorption and a new C=O absorption appears at about 1710. When an alkene is polymerised (Topic 20) the C=C absorption at 1500–1680 disappears. When a halogenoalkane is converted into a nitrile (Topic 19) a sharp C≡N absorption appears near 2250.

3. Mass Spectrometry

In a mass spectrometer the sample is ionised, and the positive ions are separated according to their mass-to-charge ratio, m/e. Almost every ion has a charge of 1+, so m/e is simply the mass of the ion on the carbon-12 scale. A mass spectrum is a bar chart of relative abundance (%) against m/e. You are not asked how the instrument works.

• The molecular ion, M⁺, is the molecule with one electron removed: CH₃COCH₃ → [CH₃COCH₃]⁺ + e⁻. Its m/e is the relative molecular mass, M_r, of the molecule made from the most abundant isotopes. It is normally the peak with the highest m/e apart from the small [M+1]⁺ and [M+2]⁺ peaks.
• The base peak is the tallest peak (100% relative abundance). It need not be the molecular ion.
• Fragment ions are formed when the molecular ion breaks apart. Only the positively charged pieces are detected; the neutral pieces are not.

Propanone. The molecular ion at m/e 58 gives M_r = 58. The base peak at 43 is CH₃CO⁺, formed by loss of a CH₃ group (58 − 15 = 43). The peak at 15 is CH₃⁺. A very small peak at 59 is the [M+1]⁺ peak.

4. Isotopes and Relative Atomic Mass

Atoms of an element with different numbers of neutrons are isotopes (Topic 1). A mass spectrum of an element shows one peak for each isotope: the m/e value is the isotopic mass and the height is the relative abundance. The relative atomic mass is the weighted average:

A_r = Σ(isotopic mass × abundance) / Σ(abundances)

Magnesium. Three isotopes: ²⁴Mg (78.6%), ²⁵Mg (10.1%) and ²⁶Mg (11.3%). A_r = (24 × 78.6 + 25 × 10.1 + 26 × 11.3) / 100 = 24.3, which matches the Periodic Table in the Data Booklet.

Chlorine. ³⁵Cl (75.8%) and ³⁷Cl (24.2%) give A_r = (35 × 75.8 + 37 × 24.2) / 100 = 35.5. The ratio of the two peaks is about 3 : 1.

If the abundances are given as numbers that do not add to 100 (peak heights, for example), divide by their sum, not by 100. Isotopes have the same chemical properties because they have the same number of electrons, but different physical properties such as mass.

5. The [M+1] and [M+2] Peaks

The [M+1]⁺ peak and carbon. About 1.1% of carbon atoms are ¹³C. A molecule with n carbon atoms therefore has a small peak one unit above M⁺, whose height relative to M⁺ is about 1.1 × n %. Rearranged:

n = 100 × abundance of [M+1]⁺ / (1.1 × abundance of M⁺)

Hexane. M⁺ at m/e 86 has abundance 15.0 and the [M+1]⁺ peak at 87 has abundance 0.99, so n = 100 × 0.99 / (1.1 × 15.0) = 6.0: six carbon atoms.

Worked example. M⁺ has abundance 9.0 and [M+1]⁺ has abundance 0.50. n = 100 × 0.50 / (1.1 × 9.0) = 5.05, so the compound has 5 carbon atoms. Always round to the nearest whole number and state it.

The [M+2]⁺ peak and halogens. Chlorine and bromine each have two common isotopes, so a compound containing one atom of either gives an M⁺ peak and an [M+2]⁺ peak, two mass units higher.

The shape of the molecular-ion region identifies the halogen. One chlorine atom (³⁵Cl and ³⁷Cl, 3 : 1) gives M : [M+2] = 3 : 1. One bromine atom (⁷⁹Br and ⁸¹Br, about 1 : 1) gives two peaks of nearly equal height.

Chloroethane. M⁺ at 64 (C₂H₅³⁵Cl) and [M+2]⁺ at 66 (C₂H₅³⁷Cl) in the ratio 3 : 1. The base peak at 29 is C₂H₅⁺.

Bromoethane. M⁺ at 108 (C₂H₅⁷⁹Br) and [M+2]⁺ at 110 (C₂H₅⁸¹Br) of almost equal height. The base peak at 29 again shows C₂H₅⁺, which has lost the halogen.

Worked example. A halogenoalkane has M⁺ peaks at m/e 122 and 124 of equal height. Equal height means one bromine atom. The molar mass of the rest of the molecule is 122 − 79 = 43, which is C₃H₇ (3 × 12.0 + 7 × 1.0 = 43.0). The compound is C₃H₇Br, 1-bromopropane or 2-bromopropane.

6. Fragmentation

The molecular ion is unstable and breaks at weak bonds or to give stable ions. Each fragment peak tells you the mass of a piece of the molecule. To assign a peak, subtract it from M⁺ to find what was lost, and match the m/e value to a formula.

The m/e values that occur most often at AS, with the ion they usually represent. Check your answer by adding up the atomic masses: for example CH₃CO⁺ = 12.0 + 3 × 1.0 + 12.0 + 16.0 = 43.0.

Pentan-3-one. M⁺ at 86. The base peak at 57 is CH₃CH₂CO⁺ (86 − 29: loss of C₂H₅). The peak at 29 is C₂H₅⁺.

Ethanol. M⁺ at 46. The base peak at 31 is CH₂OH⁺ (46 − 15, loss of CH₃), the peak at 45 is [M−H]⁺, and 29 is C₂H₅⁺.

Worked example. A compound with M_r = 88 shows peaks at 43 and 45. A peak at 43 is CH₃CO⁺ (or C₃H₇⁺) and 45 could be C₂H₅O⁺ or COOH⁺. The loss from M⁺ to 43 is 88 − 43 = 45, consistent with loss of OC₂H₅ from ethyl ethanoate, CH₃COOCH₂CH₃ (M_r = 4 × 12.0 + 8 × 1.0 + 2 × 16.0 = 88.0).

7. Combining IR and Mass Spectra

Questions often give both spectra (and sometimes test results) and ask for the compound. The method is the same each time.

1. M⁺ gives M_r. The [M+1]⁺ peak gives the number of carbons; an [M+2]⁺ peak shows Cl or Br.
2. IR gives the functional group(s), including what is absent.
3. Build a structure that fits the formula and the groups.
4. Use a fragment peak or a test reaction to choose between isomers.

Worked example. Compound Q has M⁺ at m/e 74. M⁺ has abundance 15.0 and [M+1]⁺ has abundance 0.50, so n = 100 × 0.50 / (1.1 × 15.0) = 3.0: three carbons. The IR spectrum has a very broad absorption at 2500–3000 and a strong absorption at 1710. These show a carboxylic acid, so Q is CH₃CH₂COOH, propanoic acid, M_r = 3 × 12.0 + 6 × 1.0 + 2 × 16.0 = 74.0. Its isomer methyl ethanoate (also C₃H₆O₂, M_r 74) has C=O but no broad O–H band, so the IR spectrum tells the two apart. Butan-1-ol (C₄H₁₀O, also M_r 74) has no C=O and a broad O–H at 3200–3650, and [M+1] would give four carbons.

8. Quick Sheet and Checklist

Before you leave the question, check:

• Every IR peak is matched to a bond and a functional group, with the wavenumber range from the table.
• You have used what is absent from the spectrum as well as what is present.
• M⁺ is read from the molecular ion, not the base peak, and fragment ions are written with a + charge.
• The [M+1] calculation shows the substitution and is rounded to a whole number of carbons.
• A halogen is named from the ratio of M : [M+2] (3 : 1 or 1 : 1) and the isotopes are given.
• A_r calculations divide by the total abundance and agree with the Periodic Table value.
• The final structure fits the formula, the IR spectrum and the fragments.