Year 12 Chemistry Module 8 ⏱ ~40 min 5 MC · 3 Short Answer Lesson 15 of 16

Structure Determination

No single instrument tells you the whole molecule, but stack mass spectrometry, infrared and NMR together and an unknown's structure clicks into place like a three-way crossword. This capstone skill is one of the highest-stakes competencies in HSC Chemistry (e.g. 2023 HSC Q36, 2024 HSC Q38).

Today's hook: Two compounds, ethanoic acid and methyl methanoate, share the exact same molecular formula (C₂H₄O₂) and the same molecular ion at m/z = 60. Mass spectrometry alone cannot separate them. Yet one IR band settles it instantly. How do you combine mass, infrared and NMR evidence so that, even for identical-mass isomers, only one structure survives every test?
0/5TASKS
Before you read

An unknown has a molecular ion at m/z = 46, a broad IR O–H with no C=O, and a ¹H NMR with three peaks in a 3 : 2 : 1 ratio (a triplet, a quartet and a singlet).

  • What does each technique tell you about this molecule?
  • Putting them together, what is the most likely structure?
Learning Intentions

Know

  • What each technique contributes, MS → mass + fragments/isotopes; IR → functional groups; ¹H NMR → H environments, ratios, connectivity; ¹³C NMR → C environments

Understand

  • How to cross-check evidence between techniques to converge on one structure and rule out isomers

Can Do

  • Work systematically through a combined-spectra problem to deduce a structure, justifying each step
Key Terms
Molecular formulaThe exact atom count, fixed once the molecular mass (MS) and element clues are known.
Molecular ion (M⁺)The peak at the highest significant m/z; its value is the molecular mass, the anchor for the whole analysis.
Structural isomersCompounds with the same molecular formula but different connectivity; distinguished by NMR/IR/MS fragmentation.
ConnectivityHow atoms are joined; deduced mainly from ¹H integration and n+1 splitting.
CorroborationUsing one technique's result to confirm or eliminate another's interpretation.
Absence as evidenceA missing band or peak rules options out (e.g. no C=O rules out acids, esters, aldehydes and ketones).
Where this fits: this lesson combines the three instrumental techniques, mass spectrometry (Lesson 12), IR (Lesson 13) and NMR (Lesson 14), with the functional-group tests (Lesson 11). It is the capstone skill of Module 8 IQ2.

Core Content

1

What Each Technique Contributes

+5 XP

Four complementary views of one molecule

  • MS → molecular mass (M⁺), element clues from isotopes (Cl 3:1 M+2, Br 1:1 M+2, ¹³C M+1) and fragment masses → pieces of the structure.
  • IR → which functional groups are present or absent (C=O? broad acid O–H? alcohol O–H? C≡N?).
  • ¹H NMR → number of H environments (peaks), ratio of H (integration), neighbour counts (n+1 splitting) → connectivity.
  • ¹³C NMR → number of distinct carbon environments (a symmetry/skeleton check).
MS molecular mass + fragments IR functional groups ¹H NMR H environments + connectivity ¹³C NMR carbon environments one consistent structure

Each technique answers a different question; the correct structure is the one consistent with all of them at once.

MS = molecular mass + fragments/isotopes. IR = functional groups. ¹H NMR = H environments, ratios, connectivity. ¹³C NMR = carbon environments. The correct structure fits ALL four.

Pause, copy the highlighted toolkit summary into your book.

Which technique most directly gives the molecular mass?
2

A Systematic Workflow

+5 XP

A fixed order that always works

We just saw what each technique offers. That raises a question: in what order should you use them on an unknown? This card answers it → anchor on the mass, then layer on groups, skeleton and connectivity.

  1. Molecular mass & formula from MS (M⁺; resolve M+1/M+2 isotope traps; use isotope peaks to flag Cl/Br and the nitrogen rule for an odd mass).
  2. Functional groups from IR (C=O ~1700; acid O–H 2500–3000 + C=O; alcohol O–H 3230–3550; C≡N ~2250; remember absence is also evidence).
  3. Carbon skeleton size from ¹³C (number of carbon environments).
  4. Hydrogen layout from ¹H (peaks = H environments; integration = H ratio; n+1 = neighbours).
  5. Assemble & cross-check: propose a structure; confirm it predicts the observed mass, fragments, bands, environments, ratios and splitting. Eliminate any isomer that fails a test.

Workflow: (1) mass + formula from MS → (2) functional groups from IR → (3) carbon environments from ¹³C → (4) H environments/ratios/splitting from ¹H → (5) assemble and cross-check every structure against all the data. One contradiction = wrong structure.

Pause, copy the highlighted workflow into your book.

¹³C NMR is used in structure determination primarily to find the number of:
3

Worked Unknown A, an Alcohol

+5 XP

Four techniques, one answer

We just saw the workflow. That raises a question: what does it look like on a real unknown? This card answers it → a fully worked alcohol.

1
MS: M⁺ at m/z = 46 → molecular mass 46.
2
IR: broad O–H ~3300, C–H ~2950, C–O ~1050, and no C=O → an alcohol (not an acid/aldehyde/ketone).
3
¹³C: 2 carbon environments.
4
¹H: 3 peaks, integration 3 : 2 : 1 (a triplet, a quartet and a singlet).
5
Assemble: mass 46 with C, H, O fits C₂H₆O; the alcohol groups + 2 carbons + the ethyl (triplet/quartet) pattern give ethanol, CH₃CH₂OH. The MS fragment at m/z = 31 (loss of CH₃, leaving CH₂OH⁺) corroborates.

Unknown A: M⁺ 46 (C₂H₆O), IR broad O–H + C–O + no C=O (alcohol), 2 ¹³C environments, ¹H 3:2:1 triplet/quartet/singlet = ethanol CH₃CH₂OH; fragment 31 = CH₂OH⁺ confirms.

Pause, copy the highlighted worked unknown A into your book.

An unknown (M⁺ = 46) shows IR with broad O–H and no C=O, and ¹H NMR with a 3:2:1 triplet/quartet/singlet pattern. The compound is:
4

Worked Unknown B, Acid vs Ester

+5 XP

Same formula, one band decides

We just saw a clean alcohol. That raises a question: what about two isomers with the same mass? This card answers it → one IR band separates a carboxylic acid from its ester isomer.

1
MS: M⁺ at m/z = 60 → molecular mass 60 (fits C₂H₄O₂).
2
IR: a very broad O–H 2500–3000 plus a strong C=O ~1710 and a C–O.
3
¹H: 2 peaks, integration 3 : 1, with the 1H peak far downfield (~11–12 ppm).
4
Assemble: the very-broad O–H + C=O identifies a carboxylic acid; with mass 60 and a 3:1 H ratio (CH₃ + acidic OH) the answer is ethanoic acid, CH₃COOH. The isomer methyl methanoate (same formula) is ruled out because an ester shows no broad 2500–3000 O–H.

Unknown B: M⁺ 60 (C₂H₄O₂), very broad O–H 2500–3000 + C=O = carboxylic acid (ethanoic acid CH₃COOH); the ester isomer methyl methanoate is ruled out because it has no broad acid O–H. Same mass, one band decides.

Pause, copy the highlighted worked unknown B into your book.

Key move: when two isomers share a molecular ion, look for the technique that separates them, here, the acid's very-broad O–H that an ester lacks.
Which evidence best distinguishes ethanoic acid from methyl methanoate (both C₂H₄O₂, M⁺ = 60)?
5

Worked Unknown C, & Pitfalls

+5 XP

Using fragments and environments to pick the isomer

We just saw an IR band settle two isomers. That raises a question: what if the isomers are hydrocarbons with the same mass and the same groups? This card answers it → fragment intensities and carbon-environment counts decide.

1
MS: M⁺ at m/z = 72 with a strong peak at m/z = 57 (loss of CH₃, 15); no M+2 isotope peak.
2
IR: only C–H bands (no O–H, no C=O) → a hydrocarbon.
3
¹³C: the environment count separates the isomers, pentane has 3 carbon environments; 2-methylbutane has 4.
4
Assemble: both have M⁺ 72, but 2-methylbutane loses CH₃ to give a more stable secondary carbocation, so its m/z = 57 peak is relatively larger; combined with the ¹³C environment count, you assign the correct isomer.

Pitfalls to avoid:

  • Always state what the absence of a band/peak rules out (no C=O rules out acids, esters, aldehydes, ketones).
  • Check the proposed structure reproduces all the data, one contradiction means it is wrong.
  • Mind the isotope traps (M+1 from ¹³C; M+2 from Cl/Br) and the nitrogen rule.

Unknown C: M⁺ 72, loss of 15 → 57, only C–H (hydrocarbon). Pentane has 3 ¹³C environments; 2-methylbutane has 4, and its branching gives a relatively larger 57 peak. Always use "absence as evidence" and check every datum.

Pause, copy the highlighted worked unknown C into your book.

An unknown shows a very broad IR band 2500–3000 cm⁻¹ and a strong band ~1710 cm⁻¹. The most likely functional group is:
🔬Predict, Then Reveal+8 XP
An unknown gives: MS M⁺ = 46 and a fragment at m/z = 31; IR broad ~3300, C–H ~2950, ~1050, no ~1700; ¹³C 2 peaks; ¹H 3 peaks 3:2:1 (triplet, quartet, singlet). Deduce the structure, justifying each piece of evidence.
Your predictionExpert answerCompare

Complete the Learn phase to unlock Practice.

Activities

ACTIVITY 1, Work the Unknown

Apply the workflow step by step.

1. An unknown has M⁺ = 58 and an IR with a strong C=O ~1715 but no O–H. ¹³C shows 3 environments. What functional-group class is it, and what does the missing O–H rule out?

2. Why does identifying the molecular mass first make the rest of the analysis easier?

3. State one example of how the absence of a signal is useful evidence.

A2

Activity 2, Cross-Check the Isomers

Use one technique to separate isomers another cannot.

1. Two compounds share M⁺ = 60. How would IR tell ethanoic acid from methyl methanoate?

2. Pentane and 2-methylbutane both have M⁺ = 72. How many ¹³C environments does each have, and how does that help?

3. Why is a structure only accepted once it fits the data from all the techniques?

Check Your Understanding

MC

Multiple Choice

1. Which technique most directly gives the molecular mass of an unknown organic compound?

2. An unknown shows a very broad IR band 2500–3000 cm⁻¹ and a strong band ~1710 cm⁻¹. The most likely functional group is:

3. ¹³C NMR is used in structure determination primarily to find the number of:

4. An unknown (M⁺ = 46) shows IR with broad O–H and no C=O, and ¹H NMR with a 3:2:1 triplet/quartet/singlet pattern. The compound is:

5. Which combination of evidence best distinguishes ethanoic acid from methyl methanoate (both C₂H₄O₂, M⁺ = 60)?

SA

Short Answer

1. An unknown gives: MS molecular ion at m/z = 46 and a fragment at m/z = 31; IR with a broad band ~3300 cm⁻¹, C–H ~2950 cm⁻¹, a band ~1050 cm⁻¹ and no band near 1700 cm⁻¹; ¹³C NMR with two peaks; ¹H NMR with three peaks in a 3:2:1 ratio (triplet, quartet, singlet). Deduce the structure, justifying each piece of evidence. (6 marks)

2. Explain why mass spectrometry, infrared spectroscopy and NMR are used together rather than relying on any one technique to determine an organic structure. (4 marks)

3. Two compounds both have a molecular ion at m/z = 60 (C₂H₄O₂). Explain how you would use combined spectra to decide whether an unknown is ethanoic acid or methyl methanoate, and why the molecular ion alone is insufficient. (5 marks)

Show All Answers

Activity 1

1. The strong C=O with no O–H is a carbonyl compound without an acid/alcohol O–H, a ketone or aldehyde. The missing O–H rules out a carboxylic acid and an alcohol.

2. The molecular mass from MS fixes the molecular formula and is the value every other technique's interpretation must be consistent with, so it anchors and constrains the whole deduction.

3. No C=O band near 1700 rules out acids, esters, aldehydes and ketones, an absent signal eliminates whole classes of compound.

Activity 2

1. Ethanoic acid shows a very broad O–H at 2500–3000 plus a C=O; methyl methanoate (an ester) shows the C=O but no broad acid O–H, so the broad O–H band identifies the acid.

2. Pentane has 3 carbon environments and 2-methylbutane has 4; the different ¹³C peak counts distinguish the two isomers, and 2-methylbutane also gives a relatively larger m/z = 57 peak.

3. Different structures (especially isomers) can match some data but not all; only a structure consistent with every technique's result is uniquely supported, so one contradiction rejects a structure.

Multiple Choice

1. B mass spectrometry gives the molecular mass via M⁺.

2. C a very broad O–H (2500–3000) plus C=O is a carboxylic acid.

3. B ¹³C NMR counts carbon environments.

4. C mass 46, alcohol IR, ethyl ¹H pattern → ethanol.

5. C the acid's very broad 2500–3000 O–H (absent in the ester) decides it.

Short Answer Model Answers

Q1 (6 marks): MS gives molecular mass 46 (M⁺), consistent with C₂H₆O. IR shows a broad O–H (~3300) and C–O (~1050) with no C=O (1700 absent), so the compound is an alcohol, not an acid/aldehyde/ketone. ¹³C NMR (2 peaks) shows two carbon environments. ¹H NMR shows three environments in 3:2:1: a triplet (CH₃, 2 neighbours), a quartet (CH₂, 3 neighbours) and a singlet (OH). This ethyl group plus one OH, two carbons and mass 46 give ethanol, CH₃CH₂OH. The MS fragment at m/z = 31 corresponds to CH₂OH⁺ (loss of CH₃), consistent with the structure. All techniques corroborate ethanol.

Q2 (4 marks): Each technique supplies different, complementary information and none is sufficient alone. Mass spectrometry gives the molecular mass and element/fragment clues but not the functional groups; IR identifies functional groups (e.g. C=O, O–H) but not the full skeleton or how atoms connect; ¹H and ¹³C NMR reveal hydrogen and carbon environments, hydrogen ratios and connectivity but not directly the molecular mass. Because structural isomers can share a molecular mass or even functional groups, combining the techniques lets each result corroborate or eliminate possibilities, converging on a single consistent structure.

Q3 (5 marks): The molecular ion alone is insufficient because ethanoic acid and methyl methanoate are isomers with the same molecular formula C₂H₄O₂ and the same M⁺ = 60. To decide, use IR: ethanoic acid shows a very broad O–H band at 2500–3000 cm⁻¹ together with a strong C=O ~1710, whereas the ester shows the C=O but no broad acid O–H. ¹H NMR corroborates: the acid shows a strongly downfield acidic proton (~11–12 ppm) in a 3:1 ratio with its CH₃. The presence (acid) or absence (ester) of the very-broad O–H is the deciding evidence, so combining MS with IR/NMR identifies the compound that the molecular ion alone cannot.

Return to Think First

Return to the unknown with M⁺ = 46, a broad O–H with no C=O, and a 3:2:1 triplet/quartet/singlet ¹H NMR.

  • What does each technique contribute?
  • What is the structure, and which fragment confirms it?

Review

What does each technique contribute to a structure determination?

What is the recommended order of analysis?

How do you separate two isomers with the same molecular ion?

Why is the absence of a signal useful?

When is a proposed structure accepted?