Analytical techniques

22.1

Infrared spectroscopy

Syllabus

analyse an infrared spectrum of a simple molecule to identify functional groups (see the Data section for the functional groups required)

Source: Cambridge International syllabus

Infrared spectroscopy 红外光谱 helps you find the functional group 官能团 in a molecule. Each kind of bond soaks up (absorbs) infrared radiation at its own range of frequencies. Where the bond shows strong absorption 吸收, the spectrum has a dip.

A benchtop infrared spectrometer in a laboratory An infrared spectrometer shines infrared through a sample and records which wavenumbers its bonds absorb

The position is measured in wavenumber 波数 (in $\text{cm}^{-1}$). You are given a data table, so you do not memorise the numbers. You just match the dips to bonds:

a broad dip around $3200$–$3650\ \text{cm}^{-1}$ → an O–H bond in an alcohol.
a dip around $1700\ \text{cm}^{-1}$ → a C=O bond (aldehyde, ketone, acid or ester).
a broad dip $2500$–$3000\ \text{cm}^{-1}$ together with a C=O dip → a carboxylic acid.

This is useful for checking a reaction. For example, if propene has been turned into propan-2-ol, the C=C dip should be gone and an O–H dip should appear.

An infrared spectrum with transmittance dipping at a broad O-H band and a sharp C=O band An infrared spectrum: each bond gives a dip at its own wavenumber. A broad O–H dip together with a C=O dip identifies a carboxylic acid

Explore

IR spectroscopy lab

Match an absorption to the bond or functional group it reveals.

Vocabulary Train

English	Chinese	Pinyin
infrared spectroscopy	红外光谱	hóng wài guāng pǔ
functional group	官能团	guān néng tuán
absorption	吸收	xī shōu
wavenumber	波数	bō shù

Exercise sheet

22.2

Mass spectrometry

Syllabus

analyse mass spectra in terms of $m/e$ values and isotopic abundances (knowledge of the working of the mass spectrometer is not required)
calculate the relative atomic mass of an element given the relative abundances of its isotopes, or its mass spectrum
deduce the molecular mass of an organic molecule from the molecular ion peak in a mass spectrum
suggest the identity of molecules formed by simple fragmentation in a given mass spectrum
deduce the number of carbon atoms, $n$, in a compound using the $[M + 1]^+$ peak and the formula
$$n = \frac{100 \times \text{abundance of } [M + 1]^+ \text{ ion}}{1.1 \times \text{abundance of } M^+ \text{ ion}}$$
deduce the presence of bromine and chlorine atoms in a compound using the $[M + 2]^+$ peak

Source: Cambridge International syllabus

In mass spectrometry 质谱, a molecule is turned into ions and sorted by its mass-to-charge ratio 质荷比 ($m/e$). The spectrum is a set of peaks at different $m/e$ values.

A scientist in a laboratory loading a small sample into a large modern mass spectrometer A modern mass spectrometer: the sample is loaded at the front, then the machine ionises it and sorts the ions by their mass-to-charge ratio to give the spectrum

Relative atomic mass from isotopes

An element's isotope 同位素 mixture gives several peaks. From the isotopic abundance 同位素丰度 (how common each isotope is) you can find the relative atomic mass 相对原子质量 — a weighted average:

$$A_r = \frac{\sum (\text{isotope mass} \times \text{abundance})}{\sum \text{abundance}}$$

For example, chlorine is 75% $^{35}\text{Cl}$ and 25% $^{37}\text{Cl}$, giving $A_r = \dfrac{35 \times 75 + 37 \times 25}{100} = 35.5$.

The molecular ion and fragmentation

The peak at the highest $m/e$ (the molecular ion 分子离子 peak, or molecular ion peak 分子离子峰, $M^+$) gives the relative molecular mass of the whole molecule.

The molecule also breaks into smaller pieces — this is fragmentation 碎裂. The gap between two peaks tells you the mass of the lost piece, so you can suggest each fragment 碎片. For example, a loss of 15 means a $\text{CH}_3$ group was lost, and a loss of 29 means $\text{CHO}$ or $\text{C}_2\text{H}_5$.

A mass spectrum of ethanol with peaks at several m/e values, the molecular ion at 46 marked and a gap of 15 from 46 to 31 labelled as the loss of a methyl group A mass spectrum: the highest-$m/e$ peak is the molecular ion ($M^+$, the $M_r$); the gaps between peaks give the masses of the lost fragments

The [M + 1] and [M + 2] peaks

a small [M + 1] peak comes from the $^{13}\text{C}$ isotope. The number of carbon atoms $n$ is:

$$n = \frac{100 \times \text{abundance of } [M + 1]^+}{1.1 \times \text{abundance of } M^+}$$

an [M + 2] peak shows chlorine or bromine. One chlorine gives an $[M + 2]$ peak about one third the height of $M^+$ (from $^{37}\text{Cl}$); one bromine gives an $[M + 2]$ peak about the same height as $M^+$ (from $^{81}\text{Br}$).

Two pairs of mass-spectrum peaks two units apart: a chlorine pair in a 3 to 1 ratio and a bromine pair in a 1 to 1 ratio An $[M+2]$ peak two mass units above $M^+$ shows a halogen: one chlorine gives a $3:1$ ratio, one bromine a $1:1$ ratio

Worked example. A compound shows a molecular ion at $m/e = 108$ and a peak of almost equal height at $m/e = 110$. Its infrared spectrum shows no broad absorption near $3300\ \text{cm}^{-1}$. Deduce its identity. Two peaks two units apart with roughly equal heights are the $1:1$ signature of one bromine atom (from $^{79}\text{Br}$ and $^{81}\text{Br}$); one chlorine would have given a $3:1$ ratio instead. Take the bromine away from the molecular ion: $108 - 79 = 29$, which fits a $\text{C}_2\text{H}_5$ fragment. The absence of a broad peak near $3300\ \text{cm}^{-1}$ rules out an $\text{O-H}$, so there is no alcohol group. The compound is bromoethane, $\text{C}_2\text{H}_5\text{Br}$. Read the $[M+2]$ ratio rather than merely noting the peak: $1:1$ means bromine, $3:1$ means chlorine.