OCR A Level Chemistry A: Analytical techniques (H432) – Revision Notes

These notes compress 4.2.4 Analytical techniques from the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, into tables and short method steps. All of 4.2.4(a) to (h) in Module 4 is included; H432 is the full A level. The Analytical techniques study guide explains each idea in full with longer worked examples.

Exam-style questions with marked answers are in the Analytical techniques practice questions. Related topics sit on the hub for OCR A Level Chemistry; log progress with the printable H432 checklist. A short course-wide test is in the free diagnostics.

4.2.4(a): Bonds and infrared

• Covalent bonds vibrate (stretch and bend) all the time.
• A bond absorbs IR radiation only at the frequencies that match its own vibrations.
• Absorbing IR makes the bond vibrate more; the energy is taken in by the bond.
• The frequency a bond absorbs depends on the bond strength and the masses of the atoms joined.
• Spectrum layout: x-axis wavenumber in cm⁻¹ (high values on the left), y-axis percentage transmittance. Absorptions point down.

4.2.4(b): Greenhouse gases

Gas	Bond that absorbs IR	Main sources
CO₂	C=O	burning fossil fuels
H₂O (vapour)	O–H	evaporation; levels rise as the air warms
CH₄	C–H	farming (livestock, rice), landfill, gas leaks

Chain of reasoning to learn:

1. The Earth’s surface gives out IR.
2. C=O, O–H and C–H bonds in these gases absorb the IR and vibrate more.
3. The energy is re-emitted in all directions, some back towards the surface.
4. So less energy leaves the atmosphere, and the air near the ground warms.
5. More of these gases means more absorption: the suspected link to global warming.

Energy-use changes: accepting this evidence has pushed governments towards renewable energy (wind, solar, hydroelectric), away from fossil fuels, and towards better efficiency.

4.2.4(c) and (d): Reading IR spectra

In examinations the specification says IR absorption data are given on the Data Sheet. Practise with these typical values:

Look for	Range / cm⁻¹	What it tells you
Broad O–H	3200–3600	alcohol (if no C=O)
Very broad O–H	2500–3300	carboxylic acid (with C=O)
C–H	2850–3100	almost every organic compound; not diagnostic
Strong C=O	1630–1820	aldehyde, ketone or carboxylic acid
C=C	1620–1680	alkene; usually weaker than C=O
C–O	1000–1300	alcohol or acid; supporting evidence only

Method in steps: identify a functional group

1. Set aside the C–H peaks near 3000 cm⁻¹.
2. Is there a strong peak at about 1700 cm⁻¹? Yes: C=O present.
3. Is there a broad peak above about 2500 cm⁻¹? Check its range and shape.
4. Decide:
   • O–H only → alcohol
   • C=O only → aldehyde or ketone
   • C=O + very broad O–H (2500–3300) → carboxylic acid
5. Quote the wavenumber and the bond in your answer.

Method in steps: predict a spectrum

1. Draw the structure.
2. List every bond type except C–C.
3. Give the range for each bond from the data.
4. State any peak you would not expect. Absence is evidence too.

Worked reminder. Butanal oxidised under reflux to butanoic acid: the strong C=O at about 1700 cm⁻¹ stays, and a very broad O–H appears across 2500–3300 cm⁻¹. Butan-1-ol oxidised to butanal: the broad O–H at 3200–3600 cm⁻¹ goes, and C=O appears.

4.2.4(e): IR in the real world

• Each gas absorbs at its own wavenumbers, so one gas can be measured in a mixture. More gas means more absorption.
• Pollution monitoring: IR measures gases from car emissions, such as carbon monoxide (CO) and nitrogen monoxide (NO).
• Breathalysers: IR passes through breath and the absorption by ethanol’s bonds (for example C–H, and C–O) gives the ethanol concentration.
• Law courts (HSW12): IR readings are used as evidence, for example in drink-driving cases.

4.2.4(f): Molecular ion

Term	Meaning
Molecular ion, M⁺	the molecule with one electron removed
m/z	mass-to-charge ratio; equals mass here, because all ions are singly charged
Molecular ion peak	the peak at highest m/z (ignoring the tiny M+1 peak); gives Mr
M+1 peak	small peak one unit above M⁺, from molecules containing a carbon-13 atom

Not needed: mass spectra of organic halogen compounds, which the specification says you will not be expected to interpret.

4.2.4(g): Fragmentation

Rule: M⁺ → X⁺ + Y•. Only the ion X⁺ is detected. The radical Y• has no charge and gives no peak.

m/z	Fragment ion(s) to suggest	Lost from M⁺ to give this gap
15	CH₃⁺	M − 15: loss of •CH₃
29	C₂H₅⁺ or CHO⁺	M − 29: loss of •C₂H₅ or •CHO
31	CH₂OH⁺	M − 31: loss of •CH₂OH
43	C₃H₇⁺ or CH₃CO⁺	M − 43: loss of •C₃H₇ or •CH₃CO
45	COOH⁺	M − 45: loss of •COOH
57	C₄H₉⁺ or C₂H₅CO⁺	M − 57: loss of •C₄H₉ or •C₂H₅CO

Method in steps: use a fragment peak

1. Subtract the fragment m/z from the M⁺ value to find what was lost.
2. Match both the fragment and the lost piece to groups that could be in the molecule.
3. Write the fragment with its + charge, and the lost piece with a dot.
4. Look for peaks that are missing too. No peak at 29 hints at no ethyl group.

Worked reminder. A compound has M⁺ at m/z 128 and peaks at 113 and 85. 128 − 113 = 15, so •CH₃ is lost. 128 − 85 = 43, so a piece of mass 43 (•C₃H₇ or •CH₃CO) is lost. Suggest groups that fit, then check them against the IR evidence.

4.2.4(h): Combined techniques

Technique	What it gives
Elemental analysis (2.1.3(c))	empirical formula
Mass spectrum: M⁺	Mr, so molecular formula = (Mr ÷ empirical formula mass) × empirical formula
IR spectrum	functional groups
Fragment peaks	pieces of the carbon skeleton

Method in steps

1. % by mass ÷ Ar for each element, then divide by the smallest.
2. Scale to whole numbers if needed (e.g. 1 : 2.33 → × 3 → 3 : 7).
3. Mr from M⁺, then the molecular formula.
4. Functional group from IR.
5. Fragments to place groups.
6. Check the final structure against every piece of data. If two isomers still fit, say so.

Must-know distinctions

• Alcohol O–H vs acid O–H: 3200–3600 cm⁻¹ broad, against 2500–3300 cm⁻¹ very broad and joined to a C=O peak.
• C=O vs C=C: overlapping ranges; C=O is strong, C=C usually medium or weak.
• M⁺ vs M+1: M⁺ is the tall peak at the highest m/z; M+1 is the small one just above it.
• Fragment ion vs radical: the ion has a + and is detected; the radical has a dot and is not.
• Empirical vs molecular formula: elemental analysis alone gives only the ratio; you need Mr to get the molecular formula.

Quick self-test

1. State what happens to a covalent bond when it absorbs infrared radiation.
2. Which bond in methane absorbs infrared radiation?
3. A spectrum shows a broad peak at 3340 cm⁻¹ and nothing strong near 1700 cm⁻¹. Which group is present?
4. Strong absorption at 1715 cm⁻¹ plus a very broad band spanning 2500–3300 cm⁻¹: which type of compound is this?
5. Why does a peak at 2950 cm⁻¹ not identify a functional group?
6. The two peaks at highest m/z are at 114 (tall) and 115 (very small). What is Mr, and why is there a peak at 115?
7. Give the m/z of the CH₂OH⁺ ion.
8. A molecular ion at m/z 102 loses •C₂H₅. At what m/z is the fragment peak?
9. A compound contains 64.9% C, 13.5% H and 21.6% O by mass, and M⁺ is at m/z 74. Find its molecular formula.
10. Name two gases from car emissions that can be monitored by IR.
11. Suggest two ions that could give a peak at m/z 43.
12. A hydrocarbon has empirical formula CH₂ and M⁺ at m/z 84. Give its molecular formula.

Answers

1. It vibrates more; the bond absorbs the energy.
2. C–H.
3. O–H in an alcohol (no C=O).
4. A carboxylic acid.
5. Almost every organic compound has C–H bonds, which absorb near 3000 cm⁻¹.
6. Mr = 114. The 115 peak is the M+1 peak from molecules containing one carbon-13 atom.
7. m/z 31.
8. 102 − 29 = 73.
9. C 64.9 ÷ 12.0 = 5.41, H 13.5 ÷ 1.0 = 13.5, O 21.6 ÷ 16.0 = 1.35. Divide by 1.35: 4.01 : 10.0 : 1. Empirical formula C₄H₁₀O, mass 74.0, so the molecular formula is C₄H₁₀O.
10. CO and NO.
11. C₃H₇⁺ and CH₃CO⁺.
12. CH₂ has mass 14.0; 84 ÷ 14.0 = 6, so C₆H₁₂.

Where marks are usually lost

• Quoting “O–H” for a peak without saying whether it is the alcohol range (3200–3600 cm⁻¹) or the very broad acid range (2500–3300 cm⁻¹).
• Using the C–H peak near 3000 cm⁻¹ as evidence for anything.
• Identifying a carboxylic acid from C=O alone, without the very broad O–H.
• For greenhouse gases, saying the gases “trap” or “reflect” heat without naming the bond that absorbs IR and vibrates more.
• Taking the M+1 peak as the molecular ion and giving Mr one too high.
• Leaving off the + on a fragment ion, or showing the lost radical with a charge.
• Stopping at the empirical formula and never using Mr to get the molecular formula.
• Rounding a ratio such as 2.33 to 2 instead of multiplying by 3.
• Giving a structure that fits the IR but not the fragment peaks, or the other way round.

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.