Part 2: Infrared spectra and combined evidence

Use the Edexcel wavenumber table to infer and predict functional groups, then combine IR with mass spectra and chemical tests to justify a structure.

Why particular bonds absorb infrared radiation

Covalent bonds vibrate rather than remaining at one fixed length. They absorb infrared radiation when its frequency matches an allowed vibrational energy change. An IR-active vibration changes the molecule’s dipole moment. The relevant idea is a changing dipole during the vibration, not simply whether the whole molecule is permanently polar.

Different bond types and chemical environments absorb in characteristic ranges. A stronger bond generally vibrates at a higher frequency for comparable masses, while heavier bonded atoms tend to lower the frequency. Use the supplied correlation table rather than guessing from strength alone: neighbouring groups and hydrogen bonding shift and broaden absorptions.

Wavenumber is the reciprocal of wavelength, normally in cm⁻¹. Most exam plots show wavenumber decreasing from left to right. On a transmittance spectrum an absorption is a downward trough because less radiation passes through. An absorbance plot shows upward peaks. Name the feature as an absorption and quote its wavenumber so your interpretation does not depend on drawing direction.

Use Pearson’s ranges, then inspect the shape

The ranges below are from the 8CH0 Issue 3 data table, printed page 48, and focus on the groups in Topic 7. In an examination, use the supplied table even if another revision source uses a broader generic range. A wavenumber without a bond assignment, or a bond name without an observed range, leaves the evidence-to-inference link incomplete.

Do not identify every absorption in a complicated spectrum. Begin with the strongest diagnostic regions and use the question’s candidates or molecular formula. An absorption near 1715 cm⁻¹ supports a carbonyl but does not alone prove a ketone; acids and other carbonyl environments can overlap.

Separate broad O–H, very broad acid O–H and N–H

An alcohol commonly gives a broad O–H band in the higher-wavenumber region because different hydrogen-bonding environments give a range of stretching energies. A carboxylic acid often has a very broad O–H envelope extending down towards 2500 cm⁻¹ as well as a C=O absorption near 1700–1725 cm⁻¹. It is the combination that strongly supports COOH.

An amine containing N–H can absorb around 3300–3500 cm⁻¹. Its bands are usually narrower than a hydrogen-bonded alcohol O–H feature. Primary amines often have two N–H stretching absorptions, but compare the supplied spectrum and table rather than requiring perfectly resolved bands. An absorption from C–H near 2900 cm⁻¹ does not confirm an amine.

Pearson 8CH0/02 June 2023 Q4(c)(iii) used an ethylamine IR spectrum to ask for the diagnostic amine absorption. The mark scheme distinguishes that feature from common alkane absorptions and a carbonyl region. This supports choosing a group-specific feature rather than simply the deepest trough.

A repeatable route through an unfamiliar spectrum

First inspect whether a broad or very broad O–H band is present. Next inspect the carbonyl region, then C=C and any useful aldehyde C–H or N–H features. Record absences cautiously: a predicted strong O–H or C=O band absent from a clean spectrum can help reject a candidate, but a weak alkene band may be hard to see.

Below about 1500 cm⁻¹ lies a crowded fingerprint region with many coupled vibrations. Matching the whole pattern to a reliable reference spectrum under suitable conditions can support identification. Do not infer that every tiny fingerprint trough can be assigned from a short functional-group table, or that IR alone proves a sample is pure.

Swipe horizontally to view the whole diagram.

Predict what should change during a reaction

If propan-2-ol is oxidised to propanone, the alcohol O–H absorption should disappear and a strong C=O absorption should appear in the ketone range. The C–H absorptions remain because carbon–hydrogen bonds are still present. This is a stronger prediction than “the spectrum changes”.

If ethanol is fully oxidised to ethanoic acid, do not predict the disappearance of every O–H feature: an acid also has O–H, but its broad range and the added C=O distinguish it from the starting alcohol. A mixture of reactant and product can retain features of both. IR monitoring gives evidence of functional-group change, not automatically the percentage yield.

Hydrogenation of an alkene should remove the corresponding C=C and alkene C–H signals where present, while alkane C–H remains. An alkene with no H attached to its double-bond carbons cannot have an alkene C–H stretch from those atoms, so connect each predicted absorption to an actual bond.

Worked identification with three independent clues

An illustrative pure unknown is one of butan-1-ol, butanal, butan-2-one or ethanoic acid. Its molecular-ion peak is m/z 72, IR has a strong band at 1712 cm⁻¹ with no broad O–H, and warmed Fehling’s solution gives no red precipitate. The mass rejects butan-1-ol (74) and ethanoic acid (60); butanal and butan-2-one both have C₄H₈O and mass 72.

The carbonyl absorption and absence of O–H fit the two remaining candidates. The negative warmed Fehling’s test, under working controlled test conditions, favours the ketone: butan-2-one. A fragment at m/z 43 assigned to CH₃CO⁺ supports that structure. The justified final structure is CH₃COCH₂CH₃.

Write the chain of evidence: observation → bond or ion → eliminated/retained candidate → final structure. If the functional-group tests were not available, the mass and one carbonyl band alone would not uniquely distinguish these isomers. “Consistent with” is appropriate when evidence supports but does not uniquely prove a proposal.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.

Sources