NESA 2022 HSC-style practice: The proton NMR spectrum of an unknown C_4H_8O_2 compound shows three signals: a triplet at δ = 1.25 integrating to 3H, a singlet at δ = 2.05 integrating to 3H, and a quartet at δ = 4.10 integrating to 2H. The carbon-13 NMR shows four signals at δ = 14, 21, 60, 171. Deduce the structure, naming each signal.

A 6 mark answer needs the structure, an explicit explanation of each ^1H NMR feature, and assignment of each ^13C signal. **Molecular formula.** C_4H_8O_2, degree of unsaturation (2 · 4 + 2 - 8)/2 = 1. One ring or double bond. The δ = 171 carbon-13 peak indicates a carbonyl carbon, accounting for the one degree of unsaturation. **Reading ^1H NMR:** - **Triplet at 1.25 ppm, 3H.** CH_3 next to a CH_2 (n+1 = 3, so 2 neighbours). - **Singlet at 2.05 ppm, 3H.** CH_3 with no neighbouring H. Chemical shift 2.0 is typical of CH_3 next to a C=O. - **Quartet at 4.10 ppm, 2H.** CH_2 next to a CH_3 (n+1 = 4, so 3 neighbours). Chemical shift 4.1 is typical of CH_2 next to an electronegative O. The triplet at 1.25 and quartet at 4.10 are a classic **ethyl ester** pattern (-O-CH_2-CH_3). The singlet at 2.05 is an acetyl methyl (CH_3-CO-). **Structure: ethyl ethanoate CH_3COOC_2H_5.** **^13C assignment.** - δ = 14 ppm: CH_3 of ethyl group (saturated, no neighbouring electronegative atom). - δ = 21 ppm: CH_3 of acetyl (next to C=O, slightly shifted). - δ = 60 ppm: OCH_2 (oxygen pulls density off, large shift). - δ = 171 ppm: ester C=O (carbonyl carbons of esters are 165 to 175 ppm). Markers reward (1) correct molecular structure, (2) n+1 explanation for each splitting, (3) chemical shifts justified, (4) full ^13C assignment.

§-Syllabus dot point

NSWChemistrySyllabus dot point

Inquiry Question 2: How is information about the reactivity and structure of organic compounds obtained?

Investigate the processes used to analyse the structure of simple organic compounds, including proton and carbon-13 NMR

A focused answer to the HSC Chemistry Module 8 dot point on NMR spectroscopy. How spin-half nuclei resonate in a strong magnetic field, the four features of a proton NMR spectrum (number of signals, chemical shift, integration, multiplicity via the n+1 rule), carbon-13 chemical shift ranges, the role of TMS, and worked HSC past exam questions.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-22

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
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In plain English

Every hydrogen and every carbon in a molecule sits in its own little chemical "neighbourhood", and NMR is like asking each neighbourhood to sing a note back to you. Put the sample in a strong magnet, ping it with a radio pulse, and each distinct neighbourhood resonates at its own frequency, reported as a chemical shift in ppm. A proton spectrum then tells you four things at once: how many different neighbourhoods there are (number of signals), roughly what kind of neighbourhood each one is (chemical shift), how many identical protons live in it (integration, like a headcount), and who its immediate neighbours are (multiplicity, via the n+1 rule, like counting how many people are standing right next to you at a party because that changes how loud your own voice sounds). Carbon-13 NMR asks the same question of the carbon skeleton instead, across a much wider range of possible notes. Read the two spectra together and you can rebuild the molecule's structure almost like solving a jigsaw puzzle where the shift tells you what kind of piece it is and the splitting tells you which pieces interlock.

What this dot point is asking

NESA wants you to explain the principle of nuclear magnetic resonance, identify the four pieces of information in a proton NMR spectrum (number of signals, chemical shift, integration, multiplicity by the n+1 rule), use a carbon-13 NMR spectrum to count and classify carbon environments, and deduce the structure of a simple organic compound from $^1H$ and $^{13}C$ NMR together.

The answer

The physics in one paragraph

Nuclei with a non-zero spin (spin-half nuclei like $^1H$ and $^{13}C$ ) behave as tiny magnets. In a strong external magnetic field they take one of two orientations (aligned or opposed) with a small energy gap between them. Irradiating the sample with radiofrequency energy matched to that gap causes the lower-energy nuclei to flip to the higher state; this is resonance. The frequency at which resonance happens depends slightly on the local electron density around the nucleus, which differs for each chemical environment. The differences in resonance frequency are reported as a chemical shift $\delta$ in parts per million (ppm) from a reference compound (tetramethylsilane, TMS, $\delta = 0$ ).

The reference: TMS

Tetramethylsilane $(CH_3)_4Si$ has 12 equivalent protons in a single environment, all with very low resonance frequency (silicon is more electropositive than carbon, so the methyls are electron-rich and shielded). Chemical shifts are positive to the left (downfield, deshielded) and zero at TMS.

Proton NMR: four features per spectrum

1. Number of signals. Each unique proton environment gives one signal. Symmetry can make two formally different protons equivalent. For example, all three protons in $CH_3$ are equivalent. The methyl, methylene and OH of ethanol are three signals.

2. Chemical shift ( $\delta$ in ppm). Tells you the electronic environment of the proton.

Proton environment	$\delta$ (ppm)
$-CH_3$ (next to other C only)	0.9
$-CH_2-$ (saturated chain)	1.2 to 1.4
$-CH_2-$ next to C=C or C=O	2.0 to 2.5
$-CH_3$ next to C=O (ketone, ester)	2.0 to 2.3
$-CH_2-$ next to O (alcohol, ester O)	3.5 to 4.5
$-CH_2-$ next to halogen	3.0 to 4.0
$=CH-$ (alkene)	5.0 to 6.5
Aromatic $-CH$	6.5 to 8.0
Aldehyde $-CHO$	9.5 to 10.0
Carboxylic acid $-COOH$	10 to 12
Alcohol $-OH$ , amine $-NH_2$	0.5 to 5 (variable, broad)

3. Integration. The area under each signal is proportional to the number of equivalent protons in that environment. The spectrometer reports areas as a step trace; the ratio of step heights gives the proton ratio. Integration is what distinguishes a methyl (3H) from a methylene (2H) at similar chemical shift.

4. Multiplicity (splitting) and the n+1 rule. Spin-spin coupling to neighbouring protons splits each signal into a multiplet. The rule:

\text{number of peaks} = n + 1

where $n$ is the number of protons on the immediately adjacent carbon(s). So a $CH_3$ next to a $CH_2$ appears as a triplet (n = 2, peaks = 3) and the $CH_2$ next to the $CH_3$ appears as a quartet (n = 3, peaks = 4). The relative heights of the peaks follow Pascal's triangle: 1:1 (doublet), 1:2:1 (triplet), 1:3:3:1 (quartet), 1:4:6:4:1 (pentet).

Coupling is normally only seen between protons on adjacent carbons; protons on the same carbon are typically equivalent and do not split each other.

Must-know NMR facts

TMS reference: $(CH_3)_4Si$ , defined as $\delta = 0$ ppm; downfield (higher $\delta$ ) means more deshielded.
Chemical shift formula: $\delta\ (\text{ppm}) = \dfrac{\text{shift from TMS in Hz}}{\text{spectrometer frequency in MHz}}$ .
The $n+1$ rule: number of peaks in a multiplet $= n + 1$ , where $n$ counts protons on the immediately adjacent carbon(s) only.
$^{13}C$ is run proton-decoupled, so every unique carbon environment gives a clean singlet with no splitting to interpret.
Natural abundance: $^1H$ is essentially 100% NMR-active; $^{13}C$ is only 1.1%, which is why carbon-13 spectra need much longer acquisition times.

Carbon-13 NMR

$^{13}C$ NMR is run proton-decoupled as standard, which collapses all couplings and gives a singlet for each unique carbon environment. The spectrum tells you two things:

Number of signals = number of unique carbon environments.
Chemical shift classifies each carbon.

The shift range is much wider than for $^1H$ , about 0 to 220 ppm:

Carbon environment	$\delta$ (ppm)
sp $^3$ C-C, $-CH_3$ saturated	5 to 25
sp $^3$ $-CH_2-$ saturated	25 to 50
sp $^3$ C next to halogen, N	30 to 60
sp $^3$ C next to O (alcohol, ester O)	50 to 90
sp $^2$ alkene C	100 to 145
Aromatic C	110 to 160
C=O ester or acid	165 to 180
C=O aldehyde or ketone	190 to 215

Carbon-13 NMR does not have integration in the conventional sense (relaxation times differ between carbons, so peak heights are not reliable counts), but the number of peaks is a hard constraint on the structure.

An owned illustrative proton NMR spectrum shows how the three features (chemical shift, multiplicity, integration) combine to pin down a structure:

NMR structure-determination worked examples

Worked example 1: chemical shift from Hz.

A signal on a 400 MHz spectrometer appears 1240 Hz downfield of TMS.

\delta = \frac{1240\ \text{Hz}}{400\ \text{MHz}} = 3.10\ \text{ppm}

A shift of 3.10 ppm is consistent with a $-CH_2-$ or $-CH_3$ next to an electronegative atom such as O or a halogen.

Marker's note: always show the division explicitly with units (Hz over MHz) rather than quoting only the final number; this is where partial credit is earned if the final ppm value is wrong.

Worked example 2: assigning compound Y ( $C_4H_8O_2$ ) from the spectrum above.

Degree of unsaturation: $(2(4) + 2 - 8)/2 = 1$ , one carbonyl group expected.
Singlet at 3.7 ppm (3H): an isolated $OCH_3$ with zero adjacent protons ( $n = 0$ ), consistent with a methyl ester oxygen.
Quartet at 2.3 ppm (2H): a $CH_2$ next to a $CH_3$ ( $n = 3$ , so 4 peaks); the shift (2.0 to 2.5) places it alpha to the carbonyl, not next to oxygen.
Triplet at 1.3 ppm (3H): a $CH_3$ next to a $CH_2$ ( $n = 2$ , so 3 peaks).
Combining the $CH_3CH_2-$ fragment (alpha to $C=O$ ) with the isolated $OCH_3$ and the one degree of unsaturation gives methyl propanoate, $CH_3CH_2COOCH_3$ .

Marker's note: full marks require BOTH the multiplicity reasoning (stating $n$ explicitly for every signal) and the chemical shift reasoning (why 2.3 ppm is alpha to carbonyl rather than next to oxygen); a structure given with no reasoning from the spectrum earns little credit even if correct.

Reading both spectra together: a workflow

From molecular formula (from mass spectrometry), compute the degree of unsaturation $= (2C + 2 - H + N - X)/2$ .
Count $^{13}C$ peaks. That fixes the number of unique carbon environments.
Classify each $^{13}C$ peak by its chemical shift range.
Count $^1H$ peaks and their integrations. Confirm the total proton count.
Assign each $^1H$ peak by chemical shift to a type of environment.
Use multiplicity to determine which protons are adjacent.
Assemble the fragments into a structure consistent with all evidence.

Why we use NMR for structure

Question	NMR answer
How many distinct hydrogens / carbons?	Count $^1H$ / $^{13}C$ signals
How many of each type?	Integration ( $^1H$ )
What kind of environment?	Chemical shift
Which are adjacent?	Multiplicity ( $n+1$ rule in $^1H$ )
Which are aromatic, alkene, carbonyl?	Chemical shift in either spectrum

NMR is the most informative single technique for organic structure determination. Combined with mass spectrometry (molecular mass) and IR (functional groups), the three give an essentially complete picture.

Strengths and limits

Strengths. Non-destructive (sample is recovered after analysis), enormously information-rich, distinguishes isomers that IR and mass spectrometry cannot (e.g. propan-1-ol vs propan-2-ol from chemical shift and multiplicity patterns).

Limits. Needs tens of milligrams of dissolved sample (compared to nanograms for mass spec). $^{13}C$ has poor sensitivity due to 1.1% natural abundance. Solvent peaks (and water from $-OH$ ) can obscure regions of the spectrum. Magnetically equivalent groups give one peak even when their environments are subtly different.

Exam technique for NMR questions

Work in this fixed order every time: (1) compute the degree of unsaturation from the molecular formula, (2) count and classify the $^{13}C$ signals first (they are unsplit and easier to read), (3) count the $^1H$ signals and read off integration to get proton totals, (4) use multiplicity and the $n+1$ rule to work out which protons are adjacent, (5) assemble the fragments into one structure consistent with every piece of evidence, then name it. When asked to justify a chemical shift, name BOTH the type of proton/carbon AND the nearby group causing the shift (e.g. "CH2 next to the carbonyl oxygen of an ester, hence 3.5 to 4.5 ppm"), not just the number. For a chemical-shift-from-Hz calculation, always divide Hz by MHz and keep the units in your working line.

Common traps

Counting protons on the wrong carbon: The $n+1$ rule uses the number of protons on the adjacent carbon, not on the carbon itself.
Splitting equivalent protons against each other: Three equivalent protons of a $CH_3$ do not split each other. They appear as a single multiplet with shape determined by the neighbours.
Reading integration off carbon-13 NMR: Peak heights in $^{13}C$ are not reliable counts; integration is a feature of $^1H$ NMR.
Forgetting that O-H and N-H protons exchange: $-OH$ and $-NH-$ peaks are often broad and at variable position (0.5 to 5 ppm for alcohols), and they do not always show coupling to neighbours because of fast exchange.
Saying "two peaks at the same chemical shift are the same environment": Two different environments can coincidentally overlap. Use multiplicity and integration to test.

Examples in context

Example 1. Structural identification at the Bragg Crystallography Facility, UNSW. Researchers at UNSW use a 600 MHz NMR spectrometer to confirm the structure of newly synthesised pharmaceutical candidates. A typical screen records the proton NMR (chemical shift, integration, multiplicity) and the carbon-13 NMR (count of carbon environments), then combines them with mass-spec data to confirm the molecule. For a candidate with the formula $C_5H_{10}O$ , three proton signals plus three carbon signals suggests pentan-3-one (symmetric ketone). The HSC framework of "count environments, read shifts, apply $n+1$ " is exactly the analysis a PhD student runs to confirm the synthesis worked before sending the compound for biological testing.

Example 2. Ethanol identification in NSW HSC depth study. A common Stage 6 NMR task gives students the proton NMR of ethanol and asks them to assign every signal. The spectrum shows three signals: a triplet at 1.2 ppm (3 H, $CH_3$ , coupled to 2 H of $CH_2$ , multiplicity $2+1$ ), a quartet at 3.7 ppm (2 H, $CH_2$ , coupled to 3 H of $CH_3$ , multiplicity $3+1$ ), and a broad singlet at 2.6 ppm (1 H, OH, exchanges with water). The carbon-13 spectrum shows two signals at 18 and 57 ppm. NESA marking rewards correct assignment of every feature plus the $n+1$ rule logic.

Try this

Q1. State the four pieces of structural information that a proton NMR spectrum provides. [4 marks]

Cue. Number of signals (number of H environments); chemical shift (electronic environment); integration (relative number of H); multiplicity (number of adjacent H via $n+1$ rule).

Q2. A compound with molecular formula $C_3H_6O$ shows one proton signal: a singlet (6 H) at 2.1 ppm. The carbon-13 NMR shows two signals at 30 and 207 ppm. Identify the compound and justify. [3 marks]

Cue. $C_3H_6O$ with one degree of unsaturation; only 1 H environment (6 H singlet) plus two $^{13}C$ signals including a carbonyl carbon at 207 ppm; the compound is propanone ( $CH_3COCH_3$ ).

Q3. A compound $C_2H_6O$ shows two proton signals: a triplet (3 H) at 1.2 ppm and a quartet (2 H) at 3.7 ppm, plus a broad singlet (1 H) at 2.6 ppm that exchanges with $D_2O$ . (a) Identify the compound. (b) Justify the multiplicity of the triplet and quartet. (c) Explain the chemical shift of the OH signal. [2+2+1 marks]

Cue. (a) Ethanol. (b) Triplet from $CH_3$ adjacent to 2 H of $CH_2$ ( $2+1$ ); quartet from $CH_2$ adjacent to 3 H of $CH_3$ ( $3+1$ ). (c) OH is deshielded by electronegative oxygen; broad and exchangeable in $D_2O$ .

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 HSC6 marksThe proton NMR spectrum of an unknown

C_4H_8O_2

compound shows three signals: a triplet at

\delta = 1.25

integrating to 3H, a singlet at

\delta = 2.05

integrating to 3H, and a quartet at

\delta = 4.10

integrating to 2H. The carbon-13 NMR shows four signals at

\delta = 14, 21, 60, 171

. Deduce the structure, naming each signal.

Show worked answer →

A 6 mark answer needs the structure, an explicit explanation of each $^1H$ NMR feature, and assignment of each $^{13}C$ signal.

Molecular formula. $C_4H_8O_2$ , degree of unsaturation $(2 \cdot 4 + 2 - 8)/2 = 1$ . One ring or double bond. The $\delta = 171$ carbon-13 peak indicates a carbonyl carbon, accounting for the one degree of unsaturation.

Reading $^1H$ NMR:

Triplet at 1.25 ppm, 3H. $CH_3$ next to a $CH_2$ (n+1 = 3, so 2 neighbours).
Singlet at 2.05 ppm, 3H. $CH_3$ with no neighbouring H. Chemical shift 2.0 is typical of $CH_3$ next to a C=O.
Quartet at 4.10 ppm, 2H. $CH_2$ next to a $CH_3$ (n+1 = 4, so 3 neighbours). Chemical shift 4.1 is typical of $CH_2$ next to an electronegative O.

The triplet at 1.25 and quartet at 4.10 are a classic ethyl ester pattern ( $-O-CH_2-CH_3$ ). The singlet at 2.05 is an acetyl methyl ( $CH_3-CO-$ ).

Structure: ethyl ethanoate $CH_3COOC_2H_5$ .

$^{13}C$ assignment.

$\delta = 14$ ppm: $CH_3$ of ethyl group (saturated, no neighbouring electronegative atom).
$\delta = 21$ ppm: $CH_3$ of acetyl (next to C=O, slightly shifted).
$\delta = 60$ ppm: $OCH_2$ (oxygen pulls density off, large shift).
$\delta = 171$ ppm: ester C=O (carbonyl carbons of esters are 165 to 175 ppm).

Markers reward (1) correct molecular structure, (2) $n+1$ explanation for each splitting, (3) chemical shifts justified, (4) full $^{13}C$ assignment.

2019 HSC4 marksCompare proton NMR and carbon-13 NMR. State two ways the techniques provide complementary information and one reason why carbon-13 NMR spectra are typically harder to obtain than proton NMR spectra.

Show worked answer →

Both proton ( $^1H$ ) and carbon-13 ( $^{13}C$ ) NMR rely on the same physics: spin-half nuclei in a strong magnetic field absorb radiofrequency energy at frequencies determined by their electronic environment.

Complementary information:

$^1H$ NMR shows hydrogens (chemical shift 0 to 12 ppm), so it reports on the local environment of each H and on coupling to neighbouring H atoms via the n+1 rule. $^{13}C$ shows carbons directly (chemical shift 0 to 220 ppm), giving the number of distinct carbon environments and the type of each (saturated, sp $^2$ , carbonyl, aromatic).
$^1H$ NMR integration gives the relative numbers of each type of H. $^{13}C$ NMR (proton-decoupled, as usually run) gives singlets and a count of carbon environments, which fixes the carbon skeleton independently of the proton information.

Why $^{13}C$ is harder. Natural abundance of $^{13}C$ is only 1.1% (the rest is $^{12}C$ , which is NMR-silent). The signal-to-noise ratio is therefore much lower than for $^1H$ (which is essentially 100% NMR-active). A $^{13}C$ spectrum needs much longer acquisition or a more sensitive instrument.

Markers reward (1) the chemical shift range and information type for each technique, (2) integration as a feature of $^1H$ , (3) the natural abundance reason for the $^{13}C$ sensitivity problem.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marksFor ethanol,

CH_3CH_2OH

, state the number of proton NMR signals, and for each signal give its expected multiplicity using the

n+1

rule.

Show worked solution →

Number of signals. Ethanol has three distinct proton environments: the $CH_3$ , the $CH_2$ , and the $OH$ . So three signals.

Multiplicities (n+1 rule).

$CH_3$ : adjacent to the $CH_2$ (2 protons), so $n = 2$ , giving a triplet (3 peaks).
$CH_2$ : adjacent to the $CH_3$ (3 protons), so $n = 3$ , giving a quartet (4 peaks). (The adjacent $OH$ proton is not counted because exchange is fast and it does not reliably couple.)
$OH$ : exchanges rapidly with trace water, so it typically appears as a broad singlet regardless of neighbours.

Marking criteria: 1 mark for identifying three signals, 1 mark for the correct triplet/quartet reasoning with $n$ stated for each, 1 mark for correctly treating the $OH$ as an exchange-broadened singlet.

foundation3 marksThe proton NMR spectrum of a compound with molecular formula

C_4H_{10}O

shows only ONE signal, a singlet. Identify the compound and explain why only one proton signal appears.

Show worked solution →

Identifying the compound: $C_4H_{10}O$ with all protons in one environment is 2-methylpropan-2-ol (tert-butanol), $(CH_3)_3COH$ .
Why only one signal: The three methyl groups are related by the molecule's symmetry: each $CH_3$ is bonded to the same central carbon, so all nine methyl protons are chemically equivalent and give a single peak. (The $OH$ proton coincides closely or exchanges broadly and is often not resolved as a separate sharp signal in a simplified spectrum of this kind.)
Why a singlet: There are no protons on a carbon adjacent to the methyl protons (the neighbouring carbon is the quaternary central carbon, which bears no H), so $n = 0$ and the $n+1$ rule gives one peak, a singlet.

Marking criteria: 1 mark for identifying tert-butanol (or an equivalent valid symmetric $C_4H_{10}O$ structure), 1 mark for correctly invoking molecular symmetry to explain the equivalent methyl protons, 1 mark for correctly applying the $n+1$ rule (zero adjacent protons on the quaternary carbon) to explain the singlet.

core5 marksA

500\ \text{MHz}

NMR spectrometer is used to record a proton spectrum. The chemical shift scale is measured relative to TMS. A signal appears

1850\ \text{Hz}

downfield of the TMS peak. Calculate the chemical shift of this signal in ppm, to 3 significant figures.

Show worked solution →

Step 1: recall the definition of chemical shift.

\delta \,(\text{ppm}) = \frac{\text{shift from TMS (Hz)}}{\text{spectrometer frequency (MHz)}}

This works because ppm is a frequency ratio scaled by $10^6$ , and dividing Hz by MHz (i.e. by $10^6$ Hz) gives exactly that ratio.

Step 2: substitute the values.

\delta = \frac{1850\ \text{Hz}}{500\ \text{MHz}} = \frac{1850}{500}\ \text{ppm}

Step 3: compute.

\delta = 3.70\ \text{ppm}

Step 4: check significant figures. The data (1850 Hz, 500 MHz) supports 3 significant figures.

\delta = 3.70\ \text{ppm}

This shift is consistent with a $-CH_2-$ or $-CH_3$ proton adjacent to an electronegative oxygen (e.g. the $OCH_2$ of an ester or ether).

Marking criteria: 1 mark for the correct formula (shift in Hz divided by spectrometer frequency in MHz), 1 mark for correct substitution, 1 mark for the numeric calculation, 1 mark for the answer to 3 significant figures with correct units (ppm), 1 mark for a chemically sensible comment on what environment that shift suggests.

core6 marksThe owned illustrative proton NMR spectrum below is for an unknown compound Y with molecular formula

C_4H_8O_2

. Using the labelled signals (chemical shift, integration and multiplicity), deduce the structure of Y and name it.

Show worked solution →

Reading the spectrum: Three labelled signals: a triplet at $\delta \approx 1.3$ ppm (integration 3H), a singlet at $\delta \approx 3.7$ ppm (integration 3H), and a quartet at $\delta \approx 2.3$ ppm (integration 2H).
Degree of unsaturation: $C_4H_8O_2$ : $(2(4) + 2 - 8)/2 = 1$ , consistent with one carbonyl group.
Assigning the signals

Triplet, 1.3 ppm, 3H: a $CH_3$ next to a $CH_2$ ( $n = 2$ , triplet).
Quartet, 2.3 ppm, 2H: a $CH_2$ next to a $CH_3$ ( $n = 3$ , quartet); the shift (2.0 to 2.5) fits a $CH_2$ alpha to a carbonyl.
Singlet, 3.7 ppm, 3H: an isolated $OCH_3$ with no adjacent protons (singlet), shift typical of a methyl ester oxygen.

Assembling the structure. The $CH_3-CH_2-$ (triplet/quartet pair, alpha to $C=O$ ) plus an isolated $OCH_3$ singlet and the $C=O$ implied by the degree of unsaturation gives methyl propanoate, $CH_3CH_2COOCH_3$ .

Marking criteria: 1 mark for the degree of unsaturation, 1 mark per correctly assigned signal (max 3), 1 mark for correctly combining the fragments into a structure consistent with $C_4H_8O_2$ , 1 mark for the correct IUPAC name methyl propanoate.

core4 marksCompare what proton NMR and carbon-13 NMR each reveal about a molecule with the formula

C_2H_4O_2

(ethanoic acid), and explain one limitation of using carbon-13 NMR integration to count carbons.

Show worked solution →

What $^1H$ NMR reveals: Ethanoic acid, $CH_3COOH$ , has two proton environments: the $CH_3$ (singlet, no adjacent protons, $\delta \approx 2.1$ ppm) and the $COOH$ proton (broad singlet, strongly deshielded by the adjacent carbonyl and electronegative oxygen, $\delta \approx 10$ to $12$ ppm). Integration would show a 3:1 ratio of $CH_3$ to $COOH$ protons.
What $^{13}C$ NMR reveals: Two carbon environments: the $CH_3$ carbon (sp $^3$ , $\delta$ around 5 to 25 ppm) and the carbonyl carbon of the acid (sp $^2$ , $\delta$ around 165 to 180 ppm). The wide separation between the two shifts makes the carbonyl carbon unambiguous.
Limitation of $^{13}C$ integration: Peak heights in $^{13}C$ NMR are not reliable counts of the number of equivalent carbons, because different carbons relax at different rates during the pulse sequence; a carbon with 3 equivalent atoms will not necessarily give a peak three times the height of a carbon with 1. Structural conclusions from $^{13}C$ NMR therefore rely on the NUMBER of peaks and their chemical shift, not on comparing peak heights.

Marking criteria: 1 mark for correctly describing the two $^1H$ signals with shifts, 1 mark for correctly describing the two $^{13}C$ signals with shifts, 1 mark for stating the $^1H$ integration ratio, 1 mark for correctly explaining why $^{13}C$ peak height is not a reliable proton-style integration (differing relaxation times).

exam7 marksA chemist proposes that a colourless liquid, Z, with molecular formula

C_3H_6O

, is propan-2-one (acetone). Assess whether the combined proton and carbon-13 NMR evidence below supports this proposal:

^1H

NMR shows one signal, a singlet at

\delta = 2.1

ppm (integration corresponding to 6H);

^{13}C

NMR shows two signals, at

\delta = 30

ppm and

\delta = 207

ppm.

Show worked solution →

This is a 7-mark ASSESS: markers reward a judgement that is explicitly weighed against the full evidence, not just a structure.

Band 6 PLAN.

Thesis: the NMR evidence strongly supports propan-2-one, because every independent piece of data (formula, degree of unsaturation, proton count, proton environment count, carbon environment count, and both chemical shift ranges) is consistent with that single structure and no simpler alternative fits all constraints simultaneously.
Work the degree of unsaturation: $(2(3) + 2 - 6)/2 = 1$ , one ring or pi bond.
Work the $^1H$ evidence: one signal only means all 6 protons are chemically equivalent; a singlet means no protons on an adjacent carbon; $\delta = 2.1$ ppm matches a methyl next to a carbonyl.
Work the $^{13}C$ evidence: two signals means two unique carbon environments (fits a symmetric molecule with two equivalent $CH_3$ carbons and one distinct central carbon); $\delta = 207$ ppm is squarely in the ketone/aldehyde carbonyl range (190 to 215), and specifically too far downfield for an ester/acid (165 to 180), ruling those out; $\delta = 30$ ppm fits a saturated $CH_3$ alpha to a carbonyl.
Rule out alternatives explicitly: propanal ( $CH_3CH_2CHO$ ) would show three $^1H$ signals (not one) and an aldehyde proton near 9.5 to 10 ppm, which is absent; propenal derivatives would need $C=C$ signals near 5 to 6.5 ppm, also absent.
Judgement: the evidence is fully consistent with, and uniquely points to, propan-2-one; the proposal should be accepted.

Model paragraph (excerpt). The single $^1H$ signal integrating to 6H already restricts the structure to one with two equivalent methyl environments totalling six protons, and the singlet multiplicity confirms no protons sit on an adjacent carbon, exactly the pattern expected either side of a symmetric carbonyl. The $^{13}C$ carbonyl shift of 207 ppm sits inside the aldehyde/ketone window and well outside the ester/acid window, which together with the absence of any aldehydic $^1H$ signal near 9.5 to 10 ppm rules out both esters and aldehydes as competing structures. Because propan-2-one is the only $C_3H_6O$ isomer consistent with one degree of unsaturation, one proton environment, and two carbon environments with a ketonic carbonyl shift, the evidence assessed as a whole supports the chemist's proposal.

Marker's note: top-band answers (1) compute the degree of unsaturation explicitly, (2) use BOTH the proton count/multiplicity AND the carbon count/shift as independent lines of evidence, (3) explicitly rule out at least one plausible alternative structure by naming a specific feature it would show but does not, and (4) end with a clear, justified judgement rather than merely restating the data.

Marking criteria: 1 mark for the degree of unsaturation, 2 marks for correct interpretation of the $^1H$ evidence (signal count, multiplicity, shift), 2 marks for correct interpretation of the $^{13}C$ evidence (signal count, both shifts), 1 mark for explicitly ruling out at least one alternative structure with a stated reason, 1 mark for a clear, evidence-based judgement.

exam8 marksEvaluate the claim that carbon-13 NMR alone, without any proton NMR data, is sufficient to fully determine the structure of an unknown simple organic compound. Support your evaluation with a worked example based on a compound with molecular formula

C_3H_6O_2

giving three

^{13}C

signals at

\delta = 14, 61, 161

ppm.

Show worked solution →

This is an 8-mark EVALUATE: markers reward a balanced judgement (what $^{13}C$ CAN and CANNOT do) anchored to a fully worked example, not a one-sided answer.

Band 6 PLAN.

Thesis: carbon-13 NMR alone is informative but NOT sufficient on its own for full structure determination; it reliably counts and classifies carbon environments, but cannot fix how many protons sit on each carbon or which carbons are adjacent, so proton NMR data is needed to close the gap.
Work the example as far as $^{13}C$ alone allows: $C_3H_6O_2$ has degree of unsaturation $(2(3) + 2 - 6)/2 = 1$ , one carbonyl group. Three $^{13}C$ signals means three unique, non-equivalent carbon environments. $\delta = 161$ ppm is a carbonyl carbon at the lower end of the ester/acid range, consistent with a formate ester carbon ( $H-COO-$ , whose carbonyl carbon sits slightly upfield of a typical acetate-type ester due to the attached H rather than an alkyl group); $\delta = 61$ ppm fits a $CH_2$ next to an ester oxygen; $\delta = 14$ ppm fits a saturated terminal $CH_3$ .
Propose the consistent structure: ethyl formate, $HCOOCH_2CH_3$ , has exactly three distinct carbons (the formate $HC=O$ , the $OCH_2$ , and the terminal $CH_3$ ), matching both the formula $C_3H_6O_2$ and all three shifts.
State the key limitation directly: $^{13}C$ alone cannot confirm HOW MANY protons sit on each carbon (the 61 ppm carbon looks identical in a decoupled spectrum whether it carries 1H, 2H or 3H) and cannot show which carbons are adjacent to which, because $^{13}C$ is run proton-decoupled and carries no coupling information. Only the $^1H$ spectrum's integration (proton count per environment) and multiplicity ( $n+1$ adjacency, e.g. confirming the $OCH_2$ is a quartet next to the $CH_3$ , and the formate $CHO$ is a singlet) can resolve this.
Judgement: $^{13}C$ NMR narrows the possibilities dramatically (carbon count and functional-group class) but proton NMR is required to confirm the exact substitution pattern and adjacency; the two are complementary, not substitutable, so the claim that $^{13}C$ ALONE is sufficient should be rejected.

Model paragraph (excerpt). Carbon-13 NMR data for $C_3H_6O_2$ fixes the number of carbon environments at three and places one carbon firmly in the ester carbonyl region, one next to an ester oxygen, and one as a saturated terminal carbon, which is enough to propose ethyl formate, $HCOOCH_2CH_3$ , as a structure fully consistent with the formula and the shift pattern. However, the $^{13}C$ spectrum cannot by itself confirm how many hydrogens sit on the 61 ppm carbon, nor whether it is adjacent to the methyl seen at 14 ppm, because decoupling removes exactly the coupling information that would demonstrate that adjacency; only a proton spectrum showing a triplet/quartet pair (for the ethyl group) plus a separate singlet (for the formate $CHO$ ) would confirm this skeleton over any rearranged alternative. This shows carbon-13 NMR is a powerful but partial tool, most effective when read together with proton data rather than alone.

Marker's note: top-band answers (1) work the degree of unsaturation and classify every given shift, (2) propose a structure consistent with the data, while (3) explicitly stating what information decoupled $^{13}C$ CANNOT supply (proton count per carbon, and adjacency via coupling), and (4) end with an explicit accept/reject judgement on the claim rather than a neutral list of pros and cons.

Marking criteria: 1 mark for the degree of unsaturation, 2 marks for correctly classifying all three $^{13}C$ shifts, 2 marks for a structure proposal (ethyl formate or an equally well-justified alternative) consistent with the formula and shifts, 2 marks for explicitly identifying what information decoupled $^{13}C$ NMR cannot supply (proton count and adjacency), 1 mark for a clear, justified judgement on the claim.