Analytical techniques

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.

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.

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.

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:

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 $\text{C}_2\text{H}_5$.

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:

• 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 $^{81}\text{Br}$).

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.

• On an IR spectrum quote the wavenumber and the bond (O-H broad $\sim 3200-3600$, C=O $\sim 1700$); use the data booklet.
• The peak at the highest $m/z$ is the molecular ion $\text{M}^+$ and gives the $M_r$.
• Explain fragmentation: the gap between peaks is the lost fragment (loss of 15 = $\text{CH}_3$, 29 = $\text{C}_2\text{H}_5$ or CHO).
• Learn common isotope patterns (Cl gives M and M+2 in about 3:1; M+1 from $^{13}\text{C}$).