Analytical Techniques

This lesson covers the main analytical techniques used to identify organic compounds: mass spectrometry, infrared (IR) spectroscopy, nuclear magnetic resonance (NMR) spectroscopy (both ¹H and ¹³C), chromatography, and elemental analysis. You will learn how to use each technique individually and, crucially, how to combine data from multiple techniques to identify an unknown compound systematically. This material aligns with the AQA and OCR A specifications for A-Level Chemistry.

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Mass Spectrometry

A mass spectrometer ionises molecules, separates the resulting ions by their mass-to-charge ratio (m/z), and detects them. The output is a mass spectrum.

Key Definition: The molecular ion (M⁺) is formed when a molecule loses one electron. Its m/z value gives the relative molecular mass of the compound.

Key Features of a Mass Spectrum

• The molecular ion peak (M⁺) is the peak with the highest m/z value (ignoring the small M+1 peak from ¹³C). This gives the relative molecular mass of the compound.
• Fragmentation patterns arise when the molecular ion breaks apart into smaller pieces. Common fragments and their m/z values include:
  • m/z = 15: CH₃⁺
  • m/z = 17: OH⁺
  • m/z = 29: CHO⁺ or C₂H₅⁺
  • m/z = 31: CH₃O⁺ (or CH₂OH⁺)
  • m/z = 43: CH₃CO⁺ or C₃H₇⁺
  • m/z = 45: CHO₂⁺ (or C₂H₅O⁺)
  • m/z = 77: C₆H₅⁺ (phenyl cation)
• The base peak is the tallest peak in the spectrum (the most abundant ion, assigned 100% relative abundance).
• The difference between the M⁺ peak and a fragment peak tells you the mass of the fragment that was lost. For example, M⁺ − 15 = loss of CH₃; M⁺ − 29 = loss of CHO or C₂H₅; M⁺ − 45 = loss of OC₂H₅.

High-resolution mass spectrometry can distinguish between ions with the same nominal mass but different exact masses (e.g. CO has exact mass 27.9949 and C₂H₄ has exact mass 28.0313), allowing the molecular formula to be determined precisely.

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Infrared (IR) Spectroscopy

IR spectroscopy measures the absorption of infrared radiation by bonds in molecules. Different bonds absorb at different wavenumbers (measured in cm⁻¹), producing an absorption spectrum. A bond absorbs IR radiation when the frequency of the radiation matches the natural vibrational frequency of the bond, and the vibration causes a change in dipole moment.

Key Absorptions

Bond	Wavenumber / cm⁻¹	Appearance	Functional group
O–H (alcohol)	3230–3550	Broad	Alcohol
O–H (carboxylic acid)	2500–3300	Very broad, often overlaps C–H	Carboxylic acid
N–H	3300–3500	Medium, sometimes two peaks	Amine or amide
C–H	2850–3100	Medium/strong	Alkane, alkene, arene
C≡N	2200–2260	Medium/sharp	Nitrile
C=O	1630–1820	Strong, sharp	Aldehyde, ketone, carboxylic acid, ester, amide
C=C	1620–1680	Medium	Alkene
C–O	1000–1300	Strong	Alcohol, ester, ether

The region below 1500 cm⁻¹ is the fingerprint region — it is unique to each compound and can be used to confirm the identity of a substance by comparison with a database reference spectrum.

Exam Tip: The C=O absorption is the single most useful peak in IR spectroscopy — it is always strong and sharp, and its exact position helps identify the type of carbonyl compound: aldehydes ~1720–1740 cm⁻¹, ketones ~1705–1725 cm⁻¹, carboxylic acids ~1700–1725 cm⁻¹, esters ~1735–1750 cm⁻¹, amides ~1630–1690 cm⁻¹.

Greenhouse Gases and IR Absorption

Molecules that absorb IR radiation (such as CO₂, H₂O, and CH₄) are greenhouse gases. They absorb IR radiation emitted by the Earth's surface and re-emit it in all directions, contributing to the warming of the atmosphere. Only molecules whose vibrations cause a change in dipole moment can absorb IR — this is why N₂ and O₂ (symmetric diatomics) are not greenhouse gases.

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Elemental Analysis and Empirical Formulae

Elemental analysis (also called combustion analysis) determines the percentage composition by mass of each element in a compound. From these data, the empirical formula can be calculated.

Key Definition: The empirical formula is the simplest whole-number ratio of atoms of each element in a compound. The molecular formula shows the actual number of atoms of each element in one molecule.

Worked Example: Calculating an Empirical Formula

A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Determine its empirical formula.

Element	% by mass	÷ Aᵣ	Moles	÷ smallest	Ratio
C	40.0	÷ 12.0	3.33	÷ 3.33	1
H	6.7	÷ 1.0	6.7	÷ 3.33	2
O	53.3	÷ 16.0	3.33	÷ 3.33	1

Empirical formula: CH₂O

If the relative molecular mass (from mass spectrometry) is 60, then: molecular formula mass / empirical formula mass = 60 / 30 = 2. So the molecular formula is (CH₂O)₂ = C₂H₄O₂, which is ethanoic acid (CH₃COOH).

Exam Tip: Always check that your ratio gives whole numbers. If you get a ratio like 1 : 1.5 : 1, multiply all values by 2 to get 2 : 3 : 2. If you get 1 : 1.33 : 1, multiply by 3 to get 3 : 4 : 3.

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Nuclear Magnetic Resonance (NMR) Spectroscopy

NMR spectroscopy detects the magnetic environments of specific nuclei (usually ¹H or ¹³C) in a molecule. When placed in a strong magnetic field, certain nuclei (those with an odd mass number, like ¹H and ¹³C) absorb radio-frequency radiation and 'flip' between spin states. The exact frequency of absorption depends on the electron environment around the nucleus.

Reference Standard: TMS

The reference standard is tetramethylsilane (TMS), Si(CH₃)₄, which is assigned a chemical shift of 0 ppm. All chemical shifts are measured relative to TMS.

Why TMS is used:

• It produces a single, strong peak (all 12 H atoms and all 4 C atoms are equivalent).
• It absorbs at a lower chemical shift than most organic protons/carbons (appears at the far right of the spectrum).
• It is inert — it does not react with the sample or the solvent.
• It is volatile (boiling point 27 °C) — it can be easily removed after the measurement.

¹³C NMR

• Each peak represents a different carbon environment.
• The number of peaks tells you the number of non-equivalent carbon environments in the molecule.
• ¹³C NMR spectra are simpler to interpret than ¹H NMR because peaks are not split (in routine broadband-decoupled spectra).
• Peak heights are not directly proportional to the number of carbon atoms (unlike ¹H NMR integration), so peak heights should not be used for counting carbons.

¹³C Chemical Shift Ranges

Chemical shift (δ / ppm)	Carbon environment	Example
0–40	C–C (alkyl, saturated)	R–CH₃, R–CH₂–R
20–50	C–N, C–Br, C–Cl	CH₃NH₂, CH₃Cl
50–90	C–O (alcohols, ethers)	R–CH₂OH, R–O–R'
100–150	C=C (aromatic, alkene)	Benzene carbons, C=C
110–160	Aromatic carbons	C₆H₆ (δ ≈ 128)
160–185	C=O (carboxylic acids, esters, amides)	RCOOH, RCOOR'
190–220	C=O (aldehydes, ketones)	RCHO, RCOR'

¹H NMR

¹H NMR provides four key pieces of information: