IB Chemistry HLTopic 3 — Classification of MatterPaper 1 & 2HL only~13 min read
Structural Analysis of Molecules
No single instrument identifies a compound. Mass spectrometry gives you a mass but not a shape. Infrared gives you bonds but not how many. NMR maps the hydrogens but says nothing about the oxygens. Put all three together, though, and there is usually exactly one structure left standing. This page is about doing that reliably, in order, under exam pressure.
📘 What you need to know
Mass spectrometry gives relative molecular mass, isotope information, and structural clues from the fragments.
Infrared identifies which bonds and functional groups are present — and, just as usefully, which are absent.
Proton NMR gives the number of hydrogen environments, how many hydrogens are in each, and how many hydrogens sit on the neighbouring atom.
These techniques are rarely used in isolation. Each one rules out some possibilities, and you need all three to be confident.
Work in a fixed order: mass first, bonds second, hydrogens third, then assemble and check.
Always test your final answer back against every piece of data. If one peak is unexplained, the structure is wrong.
Data booklet sections 20 (infrared), 21 (NMR shifts) and 22 (mass spectrometry fragments) are all provided.
What each technique can and cannot do
The “but not” rows matter as much as the others. A question that gives you all three spectra is testing whether you know which one to trust for each part of the answer.
A method that works every time
🧩 Identifying an unknown from combined spectra
Get the relative molecular mass from the highest m/z peak. If you are given a molecular formula, check it matches.
Check the hydrogen budget. Compare the number of hydrogens with CnH2n+2. Fewer than that means a double bond or a ring somewhere.
Read the infrared for the functional group. Ask in order: is there a C=O? Is there an O–H? Is it broad or narrow? Write down what the group must be, and what it cannot be.
Count the NMR signals and integrations and check the hydrogens add up to the formula.
Use the splitting to join the pieces. Each multiplicity tells you what is on the next atom along, so build fragments and then link them.
Use the mass spectrum fragments as confirmation. Subtract each from the molecular ion and check the losses fit the structure you have built.
Test your answer against every peak. One unexplained signal means you have the wrong structure.
Step 3 is where most marks are won or lost, and the reason is that absence of a peak is powerful evidence. “No broad absorption above 3200, so there is no O–H, so it is not an alcohol or an acid” eliminates half the possibilities in one sentence. Examiners award that reasoning explicitly.
Working one through
Here is the shape of a full combined-spectra question. Notice how each technique narrows the field rather than giving the answer outright.
Each stage removes candidates rather than proving the answer. Methyl propanoate has the same formula and also shows a C=O with no O–H, so it survives the infrared step — it is the NMR that finally separates the two.
Worked examples
WORKED EXAMPLE
A compound has molecular formula C4H8O2. Its infrared spectrum has a strong sharp absorption at 1740 cm−1 and nothing above 3200 cm−1. Its proton NMR shows a 3H singlet at 2.0 ppm, a 2H quartet at 4.1 ppm and a 3H triplet at 1.3 ppm. Deduce the structure.
Step 1: check the mass and the hydrogen budgetMr = (4 × 12) + (8 × 1) + (2 × 16) = 88A saturated four-carbon chain would be C4H10. We have 8 H, so two are missing — there is one double bond or ring.Step 2: read the infrared1740 cm⁻¹ is in the C=O range 1700–1750So the missing pair of hydrogens is accounted for by the C=O. Nothing above 3200 means no O–H, which rules out a carboxylic acid and any alcohol. With two oxygens and a C=O but no O–H, it is an ester.Step 3: read the NMR integrations3 + 2 + 3 = 8 H ✓ consistent with the formulaStep 4: use the splitting to build fragmentsA 2H quartet next to a 3H triplet is a CH2CH3 group. The 2H quartet at 4.1 ppm is strongly deshielded, so that CH2 is bonded to an oxygen.fragment 1: –O–CH2CH3A 3H singlet has no neighbouring hydrogens at all, so that methyl is attached to the carbonyl carbon.fragment 2: CH3CO–Step 5: join them and check against the formulaCH3CO–O–CH2CH3 = C4H8O2 ✓ethyl ethanoate, CH3COOCH2CH3The alternative ester, methyl propanoate, would give a 3H singlet near 3.7 for its OCH3 and a 2H quartet near 2.3. Both numbers are wrong here, so it is ruled out.
WORKED EXAMPLE
Two colourless liquids both have Mr = 58. One gives a single NMR signal; the other gives three, one of them near 9.7 ppm. Both show a strong infrared peak near 1720 cm−1. Identify them and say which technique was decisive.
Step 1: find a formula with Mr 58C3H6O: 36 + 6 + 16 = 58 ✓Step 2: use the infraredA strong peak near 1720 is a C=O in both, so both are carbonyl compounds. Infrared has not separated them at all.Step 3: use the NMR signal countOne signal means every hydrogen is equivalent, which needs a symmetrical molecule.one signal → propanone, CH3COCH3Step 4: use the shift near 9.7 ppmAlmost nothing except an aldehyde proton appears that far to the left, and the booklet gives R–CHO as 9.4–10.0.three signals with one at 9.7 → propanal, CH3CH2CHOpropanone and propanal; NMR was decisiveThis is the standard warning about functional group isomers. Same mass, same C=O, same infrared region — only NMR can tell an aldehyde from a ketone here.
WORKED EXAMPLE
An unknown gives molecular ion peaks of almost equal height at m/z 124 and 126, a broad infrared absorption at 3350 cm−1, and an NMR spectrum with a 2H triplet, a 2H triplet and a 1H singlet. Suggest a structure.
Step 1: interpret the twin molecular ion peaksTwo peaks two units apart and of near-equal height is the bromine pattern, from bromine-79 and bromine-81.one Br atom, so Mr = 124 using bromine-79Step 2: interpret the infrareda broad band at 3350 cm⁻¹ is in the alcohol O–H rangeSo the molecule contains a bromine and an OH. No C=O is mentioned, so there is no carbonyl.Step 3: work out what is left after the Br and the O124 − 79 = 45, and 45 − 16 = 2929 of carbon and hydrogen is C2H5, so the formula is C2H5BrO.Step 4: use the NMR to place the atomsTwo separate 2H triplets means two CH2 groups, each with the other as its only neighbour. The 1H singlet is the OH.2 + 2 + 1 = 5 H ✓ consistent with C2H5BrOThe two CH2 groups are bonded to each other, with the Br on one end and the OH on the other.2-bromoethanol, BrCH2CH2OHCheck it back: 124 and 126 explained by bromine, 3350 by the O–H, and both triplets by two adjacent CH2 groups. Every piece of data is accounted for, which is what tells you the answer is right.
Quick reference
Evidence you are given
What it means
Technique
highest m/z peak
relative molecular mass
mass spectrometry
twin peaks 2 apart, near equal
one bromine atom
mass spectrometry
twin peaks 2 apart, about 3:1
one chlorine atom
mass spectrometry
loss of 18
water lost, so an alcohol
mass spectrometry
strong sharp band 1700–1750
a C=O is present
infrared
broad band 3200–3600
an alcohol O–H
infrared
very broad band 2500–3000
a carboxylic acid O–H
infrared
no band above 3200
no O–H, so not an alcohol or acid
infrared
signal near 9.7 ppm
an aldehyde proton
proton NMR
2H quartet plus 3H triplet
an ethyl group
proton NMR
3H singlet
a methyl with no hydrogen neighbours
proton NMR
6H doublet
two equivalent methyls on a single CH
proton NMR
💡 Exam tip
Always start with the molecular mass. Without it you cannot check anything else, and it is usually the easiest mark on the question.
State the absences explicitly. “No absorption above 3200, so no O–H” is a scoring statement in its own right.
Reference the data booklet section as you go. Quoting the range you used shows the examiner where your number came from.
Check the NMR integrations sum to the formula before you build anything. If they do not, you have misread a value.
Build fragments, then join them. Do not try to guess the whole molecule at once from a jumble of data.
Finish by re-reading every peak against your answer. An unexplained signal is the exam telling you to think again.
⚠ Common mix-up
Relying on one technique. Isomers share a molecular mass and often share a functional group. You need the combination.
Using the base peak as the relative molecular mass. The molecular ion is the highest m/z, not the tallest peak.
Ignoring evidence that does not fit. If one peak is unexplained, the structure is wrong, not the peak.
Assuming C=O means aldehyde. Ketones, acids, esters and amides all have one, and infrared cannot choose between them.
Confusing NMR splitting with integration. Splitting counts the neighbours; integration counts the hydrogens in that signal.
Forgetting oxygens and halogens are invisible to proton NMR. They carry no hydrogens, so only the mass and the infrared reveal them.
Skipping the final check. Most lost marks on these questions are structures that explain most of the data but not all of it.
That completes Classifying Organic Compounds by Functional Group — twelve pages from reading a formula to identifying an unknown from its spectra. Up next: the first page of Topic 4.
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