Year 12 Chemistry Module 8 ⏱ ~40 min 5 MC · 3 Short Answer Lesson 14 of 16 Core

Proton & Carbon-13 NMR Spectroscopy

Put a molecule in a powerful magnet, tickle its nuclei with radio waves, and each chemically distinct hydrogen or carbon sings back at its own frequency, telling you how many environments there are and how the atoms are connected.

Today's hook: A ¹H NMR spectrum shows just two peaks, a triplet and a quartet, in a 3 : 2 height ratio. From three numbers (two peaks, a 3:2 ratio, and the splitting), a chemist deduces an entire ethyl group, CH₃CH₂–. How can peak count, integration and splitting together reveal not just which atoms are present, but how they are joined?
0/5TASKS
1

Connect: one molecule can contain several environments

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Before you read

Ethanol, CH₃CH₂OH, is analysed by ¹H NMR.

  • How many peaks would you expect, and why?
  • What would the relative heights (integration) of those peaks tell you?
Learning Intentions

Know

  • That ¹H and ¹³C NMR detect nuclei in different chemical environments, measured as chemical shift (ppm, relative to TMS at 0)
  • That ¹H peaks carry integration and splitting information

Understand

  • How number of peaks = number of chemical environments
  • How integration gives the ratio of H, and how the n+1 rule explains splitting

Can Do

  • Count environments, use integration ratios, and apply the n+1 rule
  • Compare ¹H and ¹³C NMR and say what each is best for
Key Terms
Chemical shift (δ, ppm)Position of a peak relative to the TMS reference (0 ppm), standardised in ppm. Higher δ = more deshielded (downfield).
Chemical (equivalent) environmentA set of nuclei in identical surroundings; each distinct environment gives one peak.
TMS (tetramethylsilane)The 0-ppm reference; its protons are highly shielded and equivalent.
IntegrationArea under a ¹H peak; proportional to the number of H in that environment.
Spin–spin coupling (splitting)Neighbouring non-equivalent H split a peak into a multiplet.
n+1 ruleA proton with n equivalent neighbouring protons gives a peak split into n+1 lines.
Where this fits: mass spectrometry (Lesson 12) gives the molecular mass and IR (Lesson 13) gives the functional groups. NMR adds the carbon and hydrogen environments and connectivity, the final pieces. Lesson 15 combines all three on worked unknowns.

Core Content

2

Explain: local electrons change resonance

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What ¹H and ¹³C NMR Share

+5 XP

Chemical shift, ppm and the TMS reference

An NMR-active nucleus (¹H or ¹³C) behaves like a tiny magnet. In a strong applied magnetic field its spin states have slightly different energies, and radio-frequency radiation flips spins ("resonance").

The frequency needed depends on the nucleus's electronic environment, measured as chemical shift (δ) in ppm relative to TMS (0 ppm). More electronegative or nearby groups deshield a nucleus, moving it to higher δ (downfield).

The key rule shared by both techniques: the number of peaks = the number of distinct chemical environments (TMS is not counted).

NMR measures chemical shift δ in ppm relative to TMS (0 ppm). Deshielding (electronegative/nearby groups) raises δ (downfield). Number of peaks = number of distinct chemical environments.

Pause, copy the highlighted shared basics into your book.

Key idea: "peaks = environments" is the single most useful starting move on any NMR spectrum.
In NMR, the number of peaks tells you the number of:
3

Represent: extract four clues from proton NMR

2

¹H (Proton) NMR: Environments, Integration, Splitting

+5 XP

Three layers of information from one spectrum

We just saw that peaks count environments. That raises a question: what more can ¹H NMR tell you beyond the number of peaks? This card answers it → integration (how many H) and splitting (who the neighbours are).

  • Environments → peaks: ethanol CH₃CH₂OH has 3 environments (CH₃, CH₂, OH) → 3 peaks.
  • Integration → ratio of H: the area under each peak gives the H ratio. For ethanol it is 3 : 2 : 1 (CH₃ : CH₂ : OH).
  • Splitting (n+1 rule): a proton with n equivalent neighbours is split into n+1 lines. In ethanol, CH₃ (next to CH₂'s 2 H) → a triplet; CH₂ (next to CH₃'s 3 H) → a quartet. (The OH proton usually appears as a singlet because its coupling exchanges away.)

Multiplicity names: 2 = doublet, 3 = triplet, 4 = quartet, more than 4 = multiplet.

A guide to typical ¹H chemical shifts (illustrative literature values, these are NOT on the NESA data sheet; an exam question prints any ¹H shifts you need):

Proton type δ (ppm), typical
R–CH₃ (alkyl)~0.9
R–CH₂–R~1.3
H–C–C=O (α to carbonyl)~2.0–2.7
H–C–O (alcohol/ether)~3.3–4.0
C=C–H (vinylic)~4.6–5.9
Ar–H (aromatic)~6–8.5
R–CHO (aldehyde)~9–10
R–COOH (carboxylic acid)~10–13

¹H NMR gives three things: peaks = H environments; integration = ratio of H; n+1 splitting = number of neighbouring H. Ethanol = 3 peaks, 3:2:1, with a triplet (CH₃) and quartet (CH₂). (¹H shifts are printed in the exam, not on the data sheet.)

Pause, copy the highlighted ¹H NMR summary into your book.

Ethanol shows a CH₃ triplet (~1.2 ppm, 3H), an OH singlet (~2.5 ppm, 1H) and a CH₂ quartet (~3.7 ppm, 2H).

Ethanol ¹H NMR: a quartet (CH₂, 3 neighbours → 4 lines), a singlet (OH), and a triplet (CH₃, 2 neighbours → 3 lines), with integration 2 : 1 : 3.

A ¹H NMR peak is split into a quartet. Using the n+1 rule, the number of equivalent neighbouring protons is:
4

Respond: use carbon NMR as a symmetry check

3

¹³C NMR: Counting Carbon Environments

+5 XP

Simpler than ¹H, one peak per carbon environment

We just saw the three layers of ¹H NMR. That raises a question: what does carbon-13 NMR add? This card answers it → a clean count of carbon environments over a much wider shift range, with shifts you read off the data sheet.

  • ¹³C NMR shows one peak per distinct carbon environment, over a much wider range (≈0–220 ppm).
  • Under standard HSC conditions there is no spin–spin splitting in ¹³C NMR (¹³C is only ~1.1% abundant, so ¹³C–¹³C coupling is negligible, and proton coupling is removed by decoupling).
  • Peak area is generally not used to get a carbon ratio (unlike ¹H integration).

The HSC ¹³C chemical-shift ranges are printed on the NESA data sheet you read shifts off it:

Type of carbon δ (ppm)
C–C (alkyl)5–40
R–C–Cl or R–C–Br10–70
R–C–C(=O) (α to carbonyl)20–50
R–C–N25–60
C–O (alcohols, ethers, esters)50–90
C=C90–150
R–C≡N (nitrile)110–125
aromatic (ring carbons)110–160
R–C(=O)– (esters, acids)160–185
R–C(=O)– (aldehydes, ketones)190–220

¹³C NMR: one peak per carbon environment, ≈0–220 ppm, no splitting and no useful integration. Read shifts off the NESA ¹³C table (e.g. alkyl C–C 5–40; C–O 50–90; C=C 90–150; carbonyl acids/esters 160–185; aldehydes/ketones 190–220).

Pause, copy the highlighted ¹³C NMR summary into your book.

Which statement about ¹³C NMR (HSC conditions) is correct?
5

Apply: make proton and carbon evidence agree

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¹H vs ¹³C NMR, Side by Side

+5 XP

Same idea, complementary detail

We just saw both techniques separately. That raises a question: when do you reach for each? This card answers it → use ¹³C to count carbon environments, ¹H to count hydrogen environments and read connectivity.

Feature ¹H NMR ¹³C NMR
Peaksnumber of H environmentsnumber of C environments
Range~0–12 ppm~0–220 ppm
Integrationgives H rationot used for ratios
Splittingn+1 rule gives neighboursnone (decoupled)

Both: peaks = environments, shift reflects deshielding, referenced to TMS. ¹H only: integration (H ratio) + n+1 splitting (neighbours), 0–12 ppm. ¹³C only: 0–220 ppm, no integration ratio, no splitting. Use ¹³C for carbon environments, ¹H for hydrogen environments + connectivity.

Misconception: the tallest proton peak represents the most hydrogens. Proton number is inferred from integrated area, not peak height; splitting can make one environment appear as several lines.

Pause, copy the highlighted comparison into your book.

Common error: trying to read integration or splitting off a ¹³C spectrum. Under HSC conditions ¹³C gives neither, only the count of carbon environments and their shifts.
In ¹H NMR, the area under a peak (integration) is proportional to the:
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Feedback: test one structure against every signal

5

Worked Example: Identifying a C₂ Compound

+5 XP

Reading an ethyl group out of two peaks

We just saw how ¹H and ¹³C complement each other. That raises a question: how do they work together on a real spectrum? This card answers it → a short worked example that turns peak patterns into a structure.

1
Given: ¹³C shows 2 peaks (2 carbon environments); ¹H shows 2 peaks, a triplet and a quartet, with integration 3 : 2.
2
Reason (¹H): a triplet (3H, 2 neighbours) and a quartet (2H, 3 neighbours) is the classic ethyl group CH₃CH₂– (CH₃ sees 2 H → n+1 = 3; CH₂ sees 3 H → n+1 = 4).
3
Reason (¹³C): 2 carbon environments are consistent with CH₃CH₂–X.
4
Answer: the data fit a CH₃CH₂–X compound such as bromoethane, CH₃CH₂Br. Mass spectrometry or IR would identify X.

A triplet + quartet (3:2) with 2 ¹³C peaks = an ethyl group CH₃CH₂–X (e.g. bromoethane). The CH₂, nearer the electronegative atom, sits further downfield (higher δ).

Pause, copy the highlighted ethyl-group pattern into your book.

Looking ahead: NMR gives environments and connectivity, but you still need the molecular mass (MS) and functional groups (IR) to finish the job, exactly the combination used in Lesson 15.
Ethanol (CH₃CH₂OH) shows how many peaks in its ¹H NMR spectrum (TMS not counted)?
🔬Predict, Then Reveal+8 XP
A compound with molecular formula C₂H₅Br is analysed by ¹H NMR. Predict the number of peaks, their integration ratio, and the splitting pattern, with reasoning.
Your predictionExpert answerCompare
P

Practice: make every NMR clue agree

Complete the Learn phase to unlock Practice.

Activities

ACTIVITY 1, Environments & Splitting

Apply "peaks = environments" and the n+1 rule.

1. How many ¹H environments and how many ¹³C environments does methanol (CH₃OH) have?

2. A ¹H peak is a triplet. How many equivalent neighbouring protons does that proton have?

3. Why does ¹³C NMR not show splitting under HSC conditions?

A2

Activity 2, Use the Data-Sheet Shifts

Read shifts off the ¹³C table.

1. A ¹³C peak appears at 175 ppm. Which type(s) of carbon could this be?

2. A ¹³C peak appears at 70 ppm. Suggest two types of carbon consistent with this.

3. Why is integration useful in ¹H NMR but not used for ratios in ¹³C NMR?

Check Your Understanding

DO

Adaptive practice

Interpret environments, integration and splitting before proposing a structure.

SA

Short Answer

1. A compound with molecular formula C₂H₅Br is analysed by ¹H NMR. Predict the number of peaks, their relative integration, and the splitting pattern, and explain your reasoning. (5 marks)

2. Give two ways in which a ¹³C NMR spectrum differs from a ¹H NMR spectrum, and state what each technique is most useful for. (3 marks)

3. A compound's ¹³C NMR shows two peaks, one at ~60 ppm and one at ~18 ppm, and its ¹H NMR shows a triplet and a quartet in a 3 : 2 ratio. Deduce a likely structure and justify each piece of evidence. (5 marks)

Show All Answers

Activity 1

1. Methanol CH₃OH has 2 ¹H environments (CH₃ and OH) and 1 ¹³C environment (one carbon).

2. A triplet means n+1 = 3, so n = 2 equivalent neighbouring protons.

3. ¹³C is only ~1.1% abundant so ¹³C–¹³C coupling is negligible, and proton coupling is removed by decoupling, so no splitting is seen.

Activity 2

1. 175 ppm falls in the carbonyl range for esters and acids (R–C(=O)– 160–185 ppm).

2. 70 ppm is consistent with C–O (alcohols/ethers/esters, 50–90) or R–C–Cl/Br (10–70).

3. ¹H integration is proportional to the number of hydrogens in each environment, giving an H ratio; in ¹³C NMR peak areas are not reliably proportional to carbon number under standard conditions, so they are not used for ratios.

Multiple Choice

Peak count: peaks count distinct hydrogen environments.

Quartet: n+1 = 4 means three neighbouring hydrogens.

Carbon NMR: HSC spectra use no splitting or integration ratios.

Integration: area is proportional to hydrogens in that environment.

Ethanol: CH₃, CH₂ and OH give three hydrogen environments.

Short Answer Model Answers

Q1 (5 marks): C₂H₅Br is bromoethane, CH₃CH₂Br, with two hydrogen environments → two peaks. The integration ratio is 3 : 2 (CH₃ : CH₂). Applying the n+1 rule: the CH₃ protons have 2 equivalent neighbours (the CH₂), so the CH₃ peak is a triplet (2+1); the CH₂ protons have 3 equivalent neighbours (the CH₃), so the CH₂ peak is a quartet (3+1). The CH₂ peak appears further downfield (higher δ) because it is closer to the electronegative bromine and is more deshielded.

Q2 (3 marks): (1) ¹³C NMR covers a much wider shift range (~0–220 ppm) than ¹H (~0–12 ppm). (2) Under HSC conditions ¹³C shows no spin–spin splitting and its peak areas are not used for ratios, whereas ¹H shows n+1 splitting and uses integration for the H ratio. ¹³C NMR is most useful for counting carbon environments; ¹H NMR is most useful for counting hydrogen environments and deducing connectivity.

Q3 (5 marks): Two ¹³C peaks mean two carbon environments. The ~18 ppm peak is an alkyl carbon (C–C, 5–40) and the ~60 ppm peak is a C–O carbon (C–O 50–90), suggesting an –OH or ether oxygen on one carbon. The ¹H triplet + quartet in 3 : 2 is an ethyl group CH₃CH₂– (CH₃ a triplet from 2 neighbours, CH₂ a quartet from 3 neighbours). Together, an ethyl group with one carbon bonded to oxygen, the structure is ethanol, CH₃CH₂OH (the OH proton would appear as a separate singlet). Each technique corroborates: 2 carbons, an ethyl pattern, and a C–O shift.

Return to Think First

Return to the ¹H NMR of ethanol, CH₃CH₂OH.

  • How many peaks, and what is the integration ratio?
  • Which peak is a triplet and which is a quartet, and why?
✓

Review: check what actually stuck

QUIZ
Quick Quiz

Review

What does the number of peaks tell you in NMR?

What do integration and the n+1 rule tell you in ¹H NMR?

How does ¹³C NMR differ from ¹H NMR under HSC conditions?

What pattern identifies an ethyl group in ¹H NMR?

Where do you get ¹³C chemical-shift ranges in the exam?