Analytical Techniques

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Cambridge International AS & A Level Chemistry (9701)

Analytical Techniques

The big idea: Infrared spectroscopy tells you which bonds, and so which functional groups, a molecule has. Mass spectrometry tells you its mass, how many carbons it has, whether it contains Cl or Br, and which pieces it breaks into.

Summary — What This Chapter Covers

Identify functional groups from IR absorptions using the Data booklet ranges.
Read mass spectra as m/e values and abundances; calculate A_r from isotope data.
Molecular ion M⁺ gives the molecular mass; fragment peaks suggest pieces of the molecule.
[M+1]⁺ gives the number of carbons; [M+2]⁺ shows Cl or Br.

1. Infrared Spectroscopy

22.1

Bonds absorb at characteristic wavenumbers

Covalent bonds vibrate (stretch and bend). A bond absorbs infrared radiation at the frequency of its vibration, which depends on the bond strength and the masses of the atoms. An IR spectrum plots transmittance against wavenumber (cm⁻¹); each absorption shows as a dip. Match the dips above about 1500 cm⁻¹ to the Data booklet values.

Bond	Found in	Wavenumber / cm⁻¹
C–O	hydroxy, ester	1040–1300
C=C	aromatic compound, alkene	1500–1680
C=O	amide; carbonyl, carboxyl; ester	1640–1690; 1670–1740; 1710–1750
C≡N	nitrile	2200–2250
C–H	alkane	2850–2950
N–H	amine, amide	3300–3500
O–H	carboxyl; hydroxy	2500–3000 (very broad); 3200–3600 (broad)

Alcohol: broad O–H at 3200–3600 + C–O at 1040–1300; no C=O.
Aldehyde or ketone: strong C=O near 1670–1740; no broad O–H.
Carboxylic acid: very broad O–H at 2500–3000 overlapping C–H + C=O.
Ester: C=O at 1710–1750 + C–O at 1040–1300; no O–H.

Worked Example

Compound A, C₃H₆O₂, has strong absorptions at 1740 cm⁻¹ and 1200 cm⁻¹ and none between 2500 and 3600 cm⁻¹ except near 2900 cm⁻¹. Identify the functional group and suggest structures.

Practice Question

Ethanol is heated under reflux with acidified dichromate. How would the IR spectrum of the product differ from that of ethanol?

2. Mass Spectrometry

22.2

Reading a mass spectrum

A mass spectrum plots relative abundance against m/e (mass-to-charge ratio). The ions detected are positive ions with charge +1, so m/e equals the mass of the ion. For an element, each peak is an isotope and its height is that isotope’s relative abundance.

Relative atomic mass A_r = Σ(isotope mass × abundance) / Σ(abundance)

Worked Example

Magnesium’s mass spectrum has peaks at m/e 24, 25 and 26 with relative abundances 78.99, 10.00 and 11.01. Calculate A_r(Mg).

Molecules: M⁺ and fragments

The molecular ion, M⁺, is the molecule with one electron removed. Its peak is the one at the highest m/e (apart from the small [M+1]⁺ and [M+2]⁺ peaks), and its m/e is the relative molecular mass.

Molecular ions can break into a positive fragment ion and a neutral radical: M⁺ → X⁺ + Y•. Only the charged piece is detected. The gap between M⁺ and a fragment peak tells you what was lost.

m/e	Likely fragment
15	CH₃⁺
29	C₂H₅⁺ or CHO⁺
31	CH₂OH⁺
43	C₃H₇⁺ or CH₃CO⁺
45	COOH⁺

The [M+1]⁺ peak: counting carbons

¹³ ⁺ ⁺

Number of carbon atoms ⁺ ⁺

⁺

Halogen atoms	Isotopes	Peak pattern

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Analytical Techniques