OCR A Level Chemistry A: Analytical techniques (H432)

This guide teaches 4.2.4 Analytical techniques from the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. It is part of Module 4, Core organic chemistry, and all of 4.2.4(a) to (h) is taught here (H432 is the full A level). The assessment overview in the specification gives Component 02 (Synthesis and analytical techniques) as assessing modules 1, 2, 4 and 6, and Component 03 (Unified chemistry) as assessing all six modules. This section has no PAG link.

Related topics are gathered on the OCR A Level Chemistry hub. Mark outcomes off on the printable H432 checklist, and find gaps first with the free diagnostics.

Outcomes in 4.2.4

Outcome	What you need to do	Additional guidance in the specification
4.2.4(a)	Explain that IR radiation makes covalent bonds vibrate more and absorb energy	None
4.2.4(b)	Link IR absorption by gases with C=O, O–H and C–H bonds (e.g. CO₂, H₂O, CH₄) to suspected global warming and changes in energy use	HSW9,10,11,12: evidence has prompted renewable energy policies
4.2.4(c)	Identify (i) an alcohol, (ii) an aldehyde or ketone, (iii) a carboxylic acid from an IR spectrum	M3.1; IR data on the Data Sheet in examinations; C–H peak near 3000 cm⁻¹ in most organic compounds
4.2.4(d)	Interpret and predict IR spectra of familiar or unfamiliar substances from supplied data	M3.1; specification functional groups only (6.3.2(e)); HSW3,5
4.2.4(e)	Describe IR monitoring of polluting gases (e.g. CO, NO from cars) and IR breathalysers	HSW12: evidence for law courts, e.g. drink driving
4.2.4(f)	Find the molecular ion peak and hence the molecular mass	M3.1; single charges only; no organic halogen spectra; specification compounds only (6.3.2(e)); small M+1 peak from carbon-13; HSW3,5
4.2.4(g)	Analyse fragmentation peaks to identify parts of a structure	M3.1; suggest structures of fragment ions; HSW3,5
4.2.4(h)	Deduce structures from (i) elemental analysis, (ii) mass spectra, (iii) IR spectra	M3.1; 2.1.3(c); specification functional groups only; no organic halogen mass spectra; HSW3,5,6

4.2.4(a): What infrared radiation does to bonds

Covalent bonds are always vibrating, stretching and bending like springs. Each bond’s natural frequencies depend on the atom masses and the bond strength.

When IR radiation of exactly the right frequency meets a bond, the bond absorbs that energy and vibrates more. An IR spectrum plots percentage transmittance against wavenumber in cm⁻¹ (about 4000 on the left to 500 on the right), so each absorption is a downward peak.

4.2.4(b): Greenhouse gases and global warming

The Earth’s surface, warmed by sunlight, gives out energy as infrared radiation. Some atmospheric gases have bonds that absorb this IR:

• carbon dioxide, CO₂, through its C=O bonds
• water vapour, H₂O, through its O–H bonds
• methane, CH₄, through its C–H bonds.

The bonds absorb IR and vibrate more, then re-emit the energy in all directions, some of it back towards the surface. Less of it escapes to space, so the air near the surface is kept warmer.

Rising levels of these gases, mainly CO₂ from fossil fuels and CH₄ from farming and landfill, are linked to global warming (the specification says a suspected link). Acceptance of the evidence has moved governments towards policies that change energy use: more electricity from renewable sources such as wind and solar, less burning of coal, oil and gas, and better energy efficiency.

In answers, name the absorbing bond in each gas; “trap heat” alone is not enough.

4.2.4(c): Identifying alcohols, carbonyls and acids

In examinations, the specification says IR absorption data will be provided on the Data Sheet. For practice on this page, use these typical values:

Bond	Where it is found	Wavenumber / cm⁻¹	Appearance
O–H	alcohols	3200–3600	broad
O–H	carboxylic acids	2500–3300	very broad
C–H	almost all organic compounds	2850–3100	sharp, medium
C=O	aldehydes, ketones, carboxylic acids	1630–1820	strong, sharp
C=C	alkenes	1620–1680	medium
C–O	alcohols, carboxylic acids	1000–1300	strong

Most organic compounds give a peak near 3000 cm⁻¹ from C–H bonds. That peak identifies nothing, so never use it as evidence for a functional group.

The three patterns you must recognise:

1. Alcohol: broad O–H at 3200–3600 cm⁻¹, and no C=O peak.
2. Aldehyde or ketone: strong C=O at about 1700 cm⁻¹, and no broad O–H.
3. Carboxylic acid: strong C=O at about 1700 cm⁻¹ and a very broad O–H across 2500–3300 cm⁻¹, often overlapping the C–H peaks.

Worked example 1: three unlabelled liquids

Three bottles hold pentan-1-ol, cyclohexanone and 2-methylbutanoic acid, but the labels have fallen off. The main absorptions are:

• Liquid A: 2870–2960 cm⁻¹ and 1715 cm⁻¹ only.
• Liquid B: a very broad band from 2500 to 3300 cm⁻¹, and 1710 cm⁻¹.
• Liquid C: a broad band at 3330 cm⁻¹, 2930 cm⁻¹ and 1060 cm⁻¹.

Step 1. Set aside 2850–3100 cm⁻¹: all three have C–H bonds.

Step 2. Look for C=O near 1700 cm⁻¹. A and B have it; C does not.

Step 3. Look for O–H. B has the very broad acid O–H; C has the alcohol O–H; A has none.

Conclusion. A is cyclohexanone, B is 2-methylbutanoic acid and C is pentan-1-ol.

4.2.4(d): Interpreting and predicting unfamiliar spectra

The same data let you work forwards (structure to spectrum) or backwards (spectrum to functional group). To predict:

1. Draw the structure and list every bond type except C–C.
2. Match each bond to a range in the data supplied.
3. Say which peaks you expect and which you do not expect. A missing peak is evidence too.

Worked example 2: predicting a spectrum

Predict the main IR absorptions of prop-2-en-1-ol, CH₂=CHCH₂OH.

• O–H in an alcohol: broad, 3200–3600 cm⁻¹.
• C=C in an alkene: 1620–1680 cm⁻¹.
• C–O: 1000–1300 cm⁻¹.
• C–H: 2850–3100 cm⁻¹ (present but not diagnostic).
• No C=O, so no strong peak at 1630–1820 cm⁻¹ apart from the weaker C=C.

The C=C and C=O ranges overlap slightly; a C=O peak is usually much stronger.

4.2.4(e): Pollution monitoring and breathalysers

Each gas absorbs IR at its own wavenumbers, so an instrument can pick out one gas in a mixture, and the size of the absorption measures its concentration.

• Air pollution. Roadside and exhaust monitors use IR absorption to measure gases such as carbon monoxide, CO, and nitrogen monoxide, NO, from car engines.
• Breathalysers. Modern evidential breathalysers pass IR through a breath sample and measure absorption by ethanol, for example where its C–H bonds absorb (close to 2950 cm⁻¹) and where its C–O bond absorbs (close to 1050 cm⁻¹). Measuring at more than one wavenumber helps the instrument tell ethanol from other substances in breath. The amount absorbed gives the ethanol concentration.

Readings from these instruments can be used as evidence in court, which is the HSW12 point.

4.2.4(f): The molecular ion and molecular mass

In a mass spectrometer, an organic molecule M loses one electron to give the molecular ion, M⁺. On this course all ions are singly charged, so m/z equals the mass of the ion.

The molecular ion peak is the peak with the highest m/z, apart from a very small peak one unit above it. Its m/z value equals the relative molecular mass.

That small peak is the M+1 peak. About 1.1% of carbon atoms are carbon-13, so a few molecules are one unit heavier than the rest. Do not mistake it for the molecular ion. The specification states that you will not be expected to interpret mass spectra of organic halogen compounds.

4.2.4(g): Fragmentation

Molecular ions can break into a positive ion and a neutral radical:

M⁺ → X⁺ + Y•

Only the positive ion is detected, so only X⁺ gives a peak. Common fragment ions:

Fragment ion	m/z
CH₃⁺	15
C₂H₅⁺, CHO⁺	29
CH₂OH⁺	31
C₃H₇⁺, CH₃CO⁺	43
COOH⁺	45
C₄H₉⁺, C₂H₅CO⁺	57

Two fragments can share an m/z value, so use the rest of the evidence to choose. Write a fragment ion with its positive charge and the lost piece as a radical with a dot.

Worked example 3: interpreting peaks

A carbonyl compound gives peaks at m/z 15, 43, 71, 86 and a very small peak at 87.

• Molecular ion: m/z 86, so Mr = 86. The peak at 87 is the M+1 peak.
• m/z 71: 86 − 71 = 15, so CH₃ has been lost. The ion is C₄H₇O⁺.
• m/z 43: CH₃CO⁺ or C₃H₇⁺. Both have mass 43.
• m/z 15: CH₃⁺.

For 3-methylbutan-2-one, one fragmentation that gives m/z 43 is:

[CH₃COCH(CH₃)₂]⁺ → CH₃CO⁺ + •CH(CH₃)₂

Breaking on the other side of the C=O, the same ion can lose •CH₃ and give (CH₃)₂CHCO⁺ at m/z 71.

4.2.4(h): Combining analytical data

Work in this order:

1. Elemental analysis gives the empirical formula (method from 2.1.3(c)).
2. Mass spectrum gives Mr. Divide Mr by the empirical formula mass to get the molecular formula.
3. IR spectrum gives the functional group(s).
4. Fragment peaks show how the atoms are arranged.
5. Check that your structure fits every piece of data.

Worked example 4: from percentages to structure

Compound Z contains carbon, hydrogen and oxygen only: 69.8% C, 11.6% H and 18.6% O by mass. Its mass spectrum is the one in Worked example 3. Its IR spectrum shows a strong peak at 1715 cm⁻¹, C–H peaks near 2900 cm⁻¹, and no broad absorption above 2500 cm⁻¹. Take Ar(C) = 12.0, Ar(H) = 1.0 and Ar(O) = 16.0.

Empirical formula. Assume 100 g.

Element	Moles in 100 g	÷ 1.16
C	69.8 ÷ 12.0 = 5.82	5.02
H	11.6 ÷ 1.0 = 11.6	10.0
O	18.6 ÷ 16.0 = 1.16	1.00

Empirical formula C₅H₁₀O, mass 86.0.

Molecular formula. Mr = 86 and 86 ÷ 86.0 = 1, so Z is C₅H₁₀O.

Functional group. C=O at 1715 cm⁻¹ with no O–H, so Z is an aldehyde or a ketone.

Fragments. m/z 43 fits CH₃CO⁺ and m/z 71 fits loss of CH₃, so Z is best explained as a methyl ketone, CH₃CO–C₃H₇. An aldehyde would usually show CHO⁺ at 29, which is absent. Pentan-3-one has no CH₃ on the carbonyl carbon, so it cannot give CH₃CO⁺.

Conclusion. Z is pentan-2-one or 3-methylbutan-2-one; both fit all the data given. Telling them apart needs NMR spectroscopy, which you meet in 6.3.2 Spectroscopy. Saying that two isomers fit is the honest answer.

Common errors

• Using the C–H peak near 3000 cm⁻¹ as evidence for an alcohol or acid.
• Writing “O–H” without saying whether it is the alcohol or acid range.
• Naming a carboxylic acid from the C=O peak alone, without the very broad O–H.
• Reading the M+1 peak as the molecular ion and giving Mr one unit too high.
• Giving one structure when the data fit two isomers.

Next steps

• Condensed tables and a quick self-test: Analytical techniques revision notes
• Original questions with mark-by-mark answers: Analytical techniques practice questions
• Empirical and molecular formulae: formula calculations from 2.1.3
• Ten-minute course checks: Marlbridge diagnostics

Official syllabus

Source document: OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 4.2.4 Analytical techniques, learning outcomes (a) to (h). Published by OCR.