Loav · VCE Chemistry · Unit 4 / Area of Study 2
U4A2-03 · Expanded original 2026 teaching guide · Study guide
Two questions, two kinds of evidence
Mass spectrometry (MS) provides evidence about the mass-to-charge ratios of ions derived from a sample. Infrared spectroscopy (IR) provides evidence about bonds and functional groups through their characteristic absorption. Neither technique should be treated as a picture of an entire molecule. A useful answer distinguishes the data recorded from the structural inference made from it.
Begin a spectral problem by reading axes and identifying the sample context. Is this a pure compound or a mixture? Is a molecular formula supplied? Are all peaks displayed, or just selected diagnostic ones? Are peak positions exact measurements or schematic values? The spectra and numbers in this guide are original simplified teaching examples. They are not measured reference spectra and should not be presented as experimental evidence.
Reading a mass spectrum
The horizontal axis is m/z, mass-to-charge ratio, and the vertical axis is commonly relative abundance. In the nominal-mass questions used here, a singly charged ion has a numerical m/z equal to its nominal mass. In simple organic problems, the molecular ion M⁺• represents a molecule that has lost an electron without breaking apart. The dot indicates an unpaired electron. Many examination representations use M⁺ to label the peak; retain a positive charge when drawing a proposed ion.
The molecular-ion peak can give molecular mass when the ion has been identified correctly and its charge is known. It is not necessarily the tallest peak, and it may be weak or absent. The base peak is the most intense peak and is assigned relative abundance 100. A fragment ion may be the base peak. A peak farther right than a molecular-ion peak can arise from isotopic composition, so “the largest m/z shown is always M” is unsafe.
For nominal masses in elementary assignments, use isotope-specific integer masses such as 12 for ¹²C, 1 for ¹H and 16 for ¹⁶O. Do not confuse the resulting nominal molecular-ion mass with the average molar mass calculated using natural-isotope-weighted atomic masses. A fragment assignment must fit its mass, charge and the atoms available in the parent. A missing atom or impossible valence is not repaired by getting the arithmetic right.
Worked example 1: a molecular ion and a plausible fragment
A supplied original schematic mass spectrum for a pure oxygen-containing molecule identifies M⁺• at m/z = 88, with a base peak at m/z = 43. The proposed formula C₄H₈O₂ gives 4(12) + 8(1) + 2(16) = 88, consistent with the molecular ion. The base peak might be CH₃CO⁺: 2(12) + 3(1) + 16 = 43. This ion contains only atoms available in the proposed parent. C₃H₇⁺ also has nominal mass 43, so the peak alone does not uniquely establish the oxygen-containing fragment.
If other evidence supports an acetate ester, CH₃CO⁺ becomes a chemically plausible assignment. Do not write neutral CH₃CO as the detected positive ion. Do not claim that m/z 43 means 43% of the sample has that formula: m/z is the horizontal coordinate, not abundance or mixture composition.
Quantitative information: know what was normalised
If the base peak has intensity 800 counts and another peak has 200 counts, the latter's relative abundance is (200/800) × 100 = 25. It is 25 relative to the base peak, not necessarily 25% of all ions or 25% by mass of the original sample. Isotopic abundance patterns and ion-intensity comparisons can be useful, but a quantitative concentration determination needs an appropriate response model or calibration supplied in the question. Different compounds do not necessarily give identical detector responses.
IR: from vibration to functional group
Bonds in a molecule can vibrate. Absorption occurs at characteristic infrared frequencies for suitable vibrations, so observed bands help identify bond types. The horizontal coordinate is usually wavenumber in cm⁻¹; many spectra place high wavenumbers on the left. If the vertical axis is transmittance, a downward trough means stronger absorption. If it is absorbance, an upward peak means absorption. Read the axis rather than assuming every downward feature means “less absorption”.
Use the current data book rather than an unrelated memorised table. Its 2026 ranges include alcohol O–H at 3200–3600 cm⁻¹ and acid O–H at 2500–3500 cm⁻¹. Carbonyl ranges overlap: ester C=O is 1720–1750 cm⁻¹, acid C=O 1680–1740 cm⁻¹, aldehyde C=O 1660–1745 cm⁻¹, ketone C=O 1680–1700 cm⁻¹, and amide C=O 1630–1680 cm⁻¹. N–H in amines/amides is 3300–3500 cm⁻¹. These are characteristic ranges, not unique molecular barcodes.
Band shape helps: alcohol O–H is often broad; acid O–H is characteristically very broad. But absence of a band is useful only if the relevant region was measured with adequate sensitivity and sample preparation. A wet sample may show water-related absorption. Overlap, mixtures and weak signals limit certainty. The fingerprint region can support comparison with an authenticated reference, but a lone unfamiliar trough should not be converted into an invented functional-group assignment.
Worked example 2: support a family, not an entire structure
Original schematic data show a strong absorption at 1740 cm⁻¹ and no broad O–H band in a complete, adequate-quality spectrum. Consider ethanol, ethanoic acid and ethyl ethanoate. Ethanol lacks C=O, so it cannot explain the strong carbonyl band. Ethanoic acid should show an acid O–H feature. Ethyl ethanoate is therefore best supported among these candidates: 1740 is in the ester carbonyl range and it has no OH group. Without the candidate list, this evidence would support an ester possibility but would not uniquely identify ethyl ethanoate.
Practice with worked answers
1. A molecular ion is identified at m/z 60; the tallest peak is at 43. Which is the base peak, and what gives molecular mass?
The base peak is m/z 43 because it is tallest. The identified singly charged molecular ion at 60 gives nominal molecular mass 60. These roles do not have to coincide.
2. Check the proposed ion C₂H₅⁺ at m/z 29. Could it come from a C₂H₆O parent?
Its nominal mass is 2 × 12 + 5 = 29 and its atom inventory fits the parent. It is a plausible mass assignment, not proof of a particular fragmentation pathway or parent identity.
3. A peak has 150 counts while the base peak has 600. Calculate relative abundance.
Relative abundance = 150/600 × 100 = 25. It is 25 on a scale with base peak 100; it is not established as 25% of the original sample.
4. A spectrum has a broad 2500–3500 cm⁻¹ absorption and a carbonyl band at 1710 cm⁻¹. What is supported?
Together these support a carboxylic acid: both the acid O–H and C=O characteristic ranges match. The evidence does not establish carbon-chain length or a unique acid. A mixture could also complicate interpretation.
Diagnostic repair
“The tallest MS peak gives molecular mass, and one IR carbonyl peak identifies an ester.” Repair both clauses: identify the molecular ion rather than assuming it is tallest; use multiple bands and complementary data because carbonyl ranges overlap. In a full answer, quote the actual m/z or wavenumber and explain the relevant structural implication. School spectral analysis can use supplied data without handling unknown chemicals; any sampling practical requires teacher-approved controls.
Go deeper: test a proposal against independent constraints
Formula evidence and functional-group evidence constrain different things. A molecular-ion mass gives a total; it does not locate the atoms. IR identifies characteristic vibrations; it does not count every carbon. Using the two together helps reject impossible proposals before interpreting finer evidence. For example, an acid-like OH feature and a carbonyl would require at least two oxygen atoms for a carboxyl group. A proposed molecular formula with only one oxygen cannot describe a carboxylic acid, even if someone labels a broad feature “acid OH”. Reconsider the band assignment, sample purity or formula instead of forcing agreement.
Start a fragment check by subtracting its atom inventory from the parent, not merely its mass. A fragment containing nitrogen cannot come from a parent containing only C, H and O. If both inventories and masses are plausible, examine connectivity: could a reasonable bond cleavage leave the proposed charged group? The charged fragment is detected; an accompanying neutral piece need not give a peak. In simple cases a loss such as 18 mass units may suggest water, but that difference alone does not establish a unique mechanism. Treat plausible neutral losses as supporting, not decisive, evidence.
Worked example 3: same molecular mass, different functional group
Original schematic data identify a singly charged molecular ion at m/z = 58 for a pure compound containing C, H and O. A supplied elemental analysis establishes C₃H₆O. IR contains a strong band at 1690 cm⁻¹ and no broad OH. Two candidates are propanal and propanone. Both have nominal molecular mass 3(12) + 6 + 16 = 58, so MS molecular mass cannot distinguish them. The IR band lies within both the 2026 aldehyde and ketone ranges, so it also does not decide between them.
A student proposes CH₃CO⁺ for an observed fragment at m/z = 43. Its mass is correct and it is plausible for propanone, but fragment evidence alone is not a unique identification. Request complementary NMR: a pure propanone sample has one proton environment containing six H and two carbon environments; propanal has an aldehyde proton and three carbon environments. Data showing a six-H singlet at 2.2 ppm and two carbon-13 signals, including one at 210 ppm, support propanone. These selected shifts are within the relevant 2026 ranges and are schematic. The decision used a feature on which the candidates actually differ.
Read negative evidence carefully
“No OH band” only excludes a detectable OH contribution under the actual measurement conditions. A missing band in a cropped graph is no evidence about the region omitted. Very dilute contamination can escape detection, and a small water contamination can add an OH feature unrelated to the main compound. Compare blanks and sample history where relevant. Distinguish “the main component is consistent with an ester” from “every molecule in the sample is that ester”.
Transfer practice
5. A C₃H₆O molecular ion is at 58. A student assigns a fragment at 44 to C₂H₄O⁺. Check mass and atom inventory, then explain what remains unproved.
The nominal mass is 2(12) + 4 + 16 = 44, and C₂H₄O fits within C₃H₆O. Those checks establish arithmetic and inventory compatibility only. They do not prove that the ion has this connectivity or that this fragmentation actually occurred. Additional spectral and chemical evidence is needed.
6. Two candidate compounds have nominal mass 74: C₄H₁₀O and C₃H₆O₂. What measurement might distinguish them, and what conclusion cannot come from mass alone?
Both totals are 74: 48 + 10 + 16 and 36 + 6 + 32. A suitable IR spectrum showing carbonyl plus acid OH would support a carboxylic acid compatible with C₃H₆O₂; elemental/formula evidence could distinguish formulas directly. Nominal molecular mass alone cannot establish either formula or molecular identity.
7. A displayed IR graph covers only 1800–4000 cm⁻¹. A student says no carbonyl peak proves an alcohol. Evaluate.
Most characteristic carbonyl bands listed here occur below 1800 cm⁻¹, outside the displayed region, so their absence from this cropped graph is uninformative. An OH feature could support an O–H-containing species, but the missing spectral region must be obtained before excluding a carbonyl. Even a complete alcohol-like spectrum would not uniquely identify which alcohol.
Answer-writing routine: state the measured value, its allowed interpretation, a competing explanation, and the next discriminating observation. This produces a defensible argument rather than a chain of unsupported identifications. Qualitative interpretation is strongest when the molecular formula, required functional groups and independent spectra all agree.
