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

Q: 2019 HSC HSC: Compare 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.

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:** 1. $^1H$ 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). 2. $^1H$ 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.

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 OpusReviewed by Better Tuition Academy10 min answerUpdated 2026-05-18

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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	IMATH_10 (ppm)
IMATH_11 (next to other C only)	0.9
IMATH_12 (saturated chain)	1.2 to 1.4
IMATH_13 next to C=C or C=O	2.0 to 2.5
IMATH_14 next to C=O (ketone, ester)	2.0 to 2.3
IMATH_15 next to O (alcohol, ester O)	3.5 to 4.5
IMATH_16 next to halogen	3.0 to 4.0
IMATH_17 (alkene)	5.0 to 6.5
Aromatic IMATH_18	6.5 to 8.0
Aldehyde IMATH_19	9.5 to 10.0
Carboxylic acid IMATH_20	10 to 12
Alcohol $-OH$ , amine IMATH_22	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.

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	IMATH_30 (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.

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.

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.

In one sentence

Proton NMR identifies hydrogen environments by chemical shift, counts them by integration, and reveals adjacent protons by the $n+1$ splitting rule, while carbon-13 NMR counts and classifies carbon environments across a 0 to 220 ppm range; together they give a near-complete map of an organic structure when combined with the molecular mass from a mass spectrum.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

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.**

IMATH_17 ppm: $CH_3$ of ethyl group (saturated, no neighbouring electronegative atom).
IMATH_19 ppm: $CH_3$ of acetyl (next to C=O, slightly shifted).
IMATH_21 ppm: $OCH_2$ (oxygen pulls density off, large shift).
IMATH_23 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:

IMATH_2 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).
IMATH_5 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.