Analytical techniques

📚The chemistry

22.1 Infrared spectroscopy

Covalent bonds vibrate — they stretch and bend — at particular frequencies, and a bond absorbs infrared radiation of the same frequency. An infrared spectrum plots transmittance (%) against wavenumber (cm⁻¹), so absorptions appear as dips pointing downwards. Each type of bond absorbs in a characteristic range, so the dips reveal which bonds — and therefore which functional groups — the molecule contains.

The ranges you use are printed in the data section:

How to read a spectrum:

• Start above 1500 cm⁻¹. Below that is the fingerprint region, full of overlapping absorptions that are hard to assign individually (apart from C–O at 1040–1300).
• Almost every organic molecule has C–H absorptions near 2900. Ignore them unless the question is about them.
• A strong, sharp dip at about 1700 means C=O. Then decide which kind by looking for O–H.
• O–H in an alcohol: a broad dip at 3200–3600, broad because of hydrogen bonding.
• O–H in a carboxylic acid: a very broad dip at 2500–3000, often overlapping the C–H peaks, together with C=O near 1700.

So the three classic cases are: O–H (3200–3600) only → alcohol; C=O only → aldehyde, ketone or ester; very broad O–H (2500–3000) and C=O → carboxylic acid.

22.2 Mass spectrometry

In a mass spectrometer, molecules are ionised to positive ions, which are separated by their mass-to-charge ratio, m/e. Almost all ions have a 1+ charge, so m/e is simply the mass of the ion. The spectrum plots relative abundance against m/e. You do not need to know how the instrument works.

Isotopes and relative atomic mass

The mass spectrum of an element shows one peak per isotope, with heights proportional to their abundances. The relative atomic mass is the weighted mean:

Boron has ¹⁰B (19.9%) and ¹¹B (80.1%): A_r = (10 × 19.9 + 11 × 80.1)/100 = 10.8. If abundances are given as peak heights rather than percentages, divide by the sum of the heights instead of 100.

The molecular ion

When a molecule loses one electron, it forms the molecular ion, M⁺:

The molecular ion peak is the peak with the highest m/e (ignoring the small [M+1] and [M+2] peaks beside it). Its m/e is the M_r of the compound.

Fragmentation

The molecular ion often breaks apart. Each break gives a positive ion and a neutral radical; only the ion is detected, so only it gives a peak:

The m/e values of the fragments tell you which groups are present. Common ones:

A difference between the M⁺ peak and a large fragment peak is just as useful: a loss of 15 means a CH₃ group left, a loss of 17 an OH, a loss of 29 an ethyl group or CHO.

The [M+1] peak: counting carbon atoms

About 1.1% of carbon atoms are ¹³C. A molecule with n carbon atoms therefore has about 1.1n% chance of containing one ¹³C, which gives a small peak one unit above the molecular ion: the [M+1]⁺ peak. The more carbons, the bigger it is relative to M⁺:

The [M+2] peak: chlorine and bromine

Chlorine and bromine have two common isotopes, each 2 mass units apart, so a molecule containing one of them shows two molecular ion peaks, M and M+2:

• One Cl: ³⁵Cl and ³⁷Cl are roughly 3 : 1, so the M : M+2 peaks are in the ratio 3 : 1.
• One Br: ⁷⁹Br and ⁸¹Br are roughly 1 : 1, so M : M+2 is 1 : 1 — two peaks of almost equal height.
• With two Cl atoms the pattern is M : M+2 : M+4 = 9 : 6 : 1; with two Br, 1 : 2 : 1.

✏️Worked example

Compound X contains only C, H and O.
• Its infrared spectrum shows a very broad absorption from 2500 to 3000 cm⁻¹ and a strong, sharp absorption at 1715 cm⁻¹.
• Its mass spectrum has the molecular ion peak at m/e 74 with relative abundance 20.0, and an [M+1]⁺ peak with relative abundance 0.66. There are fragment peaks at m/e 29 and 45.
(a) Identify the functional group from the IR spectrum. (b) Calculate the number of carbon atoms. (c) Identify X, and the fragments at 29 and 45. (d) Write an equation for the formation of the fragment at m/e 45.

(a) A very broad O–H absorption at 2500–3000 together with C=O at 1715 is characteristic of a carboxylic acid, –COOH.

(b)

\[ n = \frac{100 \times 0.66}{1.1 \times 20.0} = 3.0 \]

X contains three carbon atoms.

(c) A three-carbon carboxylic acid with M_r 74 is propanoic acid, CH₃CH₂COOH (3 × 12 + 6 × 1 + 2 × 16 = 74). m/e 29 is C₂H₅⁺, and m/e 45 is COOH⁺ — the two halves of the molecule, from breaking the C–C bond next to the carboxyl group.

(d)

\[ \mathrm{[CH_3CH_2COOH]^{+} \rightarrow [COOH]^{+} + {\bullet}CH_2CH_3} \]

Check it. The two fragments add up to the molecule: 29 + 45 = 74. The IR has no separate absorption at 3200–3600, so there is no alcohol O–H as well; and an ester of M_r 74 (methyl ethanoate) would show C=O but no broad O–H, so it is ruled out.

Putting the charge on both fragments, or on neither. Fragmentation gives one positive ion and one neutral radical, and only the ion is detected. An equation with two cations, or with no charge, cannot explain why you see a peak. And in the [M+1] formula, the abundances go the right way up: [M+1] on top.