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Spectroscopy · Magnetic Resonance

NMR Spectroscopy

Nuclear Magnetic Resonance (NMR) is the most powerful single technique for determining organic structures. By placing molecules in a strong magnetic field and measuring how nuclear spins respond to RF pulses, we map the carbon-hydrogen framework directly. This lab focuses on ¹H NMR (chemical shift, integration, multiplicity, coupling) with a brief introduction to ¹³C NMR.

Theory — NMR Spectroscopy

1. The basic NMR experiment

Nuclei with non-zero spin (¹H, ¹³C, ¹⁹F, ³¹P) align with or against a strong magnetic field. An RF pulse flips them; as they relax, they emit a signal at a frequency that depends on the local electronic environment. Different chemical environments give different frequencies, so each unique proton (or carbon) appears at a different position in the spectrum. Modern NMR magnets are 7-21 Tesla (300-900 MHz for ¹H).

2. Chemical shift (δ)

Reported in parts per million (ppm) on the δ scale. TMS (Si(CH₃)₄) is the reference at δ = 0. Higher ppm = "downfield" (left); lower ppm = "upfield" (right). Electronegative atoms or pi systems deshield protons (move them downfield); alkyl environments shield them (keep them upfield).

¹H environmentδ (ppm)Notes
Saturated alkyl C-H (CH₃, CH₂, CH)0.5-2.0CH₃ ~0.9, CH₂ ~1.3, CH ~1.5
Alpha to C=O, allyl, benzyl2.0-2.6-CH₂-CO-, ArCH₂-, etc.
C-H next to N, S, halogen2.5-4.0Cl/Br ~3.5; N ~2.5
C-H next to O (-CH₂-O-)3.5-4.5Ester, ether, alcohol
Vinyl C-H (alkene)4.5-6.5
Aromatic C-H6.5-8.5Benzene ~7.3
Aldehyde C-H9.0-10.5Very distinctive
Carboxylic acid O-H10-13Very broad; often missing in CDCl₃
Alcohol/amine O-H, N-H0.5-5 (variable)Broad; exchange in D₂O

3. Integration: how many protons?

The area under each peak is proportional to the number of protons in that environment. Integration tells you relative numbers, e.g., 3:2:1 for ethanol (CH₃CH₂OH). Modern NMR shows step curves or numerical ratios.

4. Multiplicity: the n+1 rule

If a proton has n equivalent neighbours, its signal is split into n+1 peaks. Intensities follow Pascal\'s triangle.

NeighboursMultiplicityPatternCommon in
0singlet (s)1Isolated CH₃; OH; tert-butyl
1doublet (d)1:1One adjacent H
2triplet (t)1:2:1CH₃ in ethyl group
3quartet (q)1:3:3:1CH₂ in ethyl group
4quintet1:4:6:4:1Middle CH₂ in -CH₂-CH₂-CH₂-
5sextet1:5:10:10:5:1CH₂ with 5 neighbours (e.g., propyl bromide CH₂)
6septet1:6:15:20:15:6:1CH in isopropyl group

Ethyl group (-CH₂-CH₃): CH₃ has 2 neighbours \u2192 triplet (3 peaks). CH₂ has 3 neighbours \u2192 quartet (4 peaks). Classic pattern in ethyl esters, ethyl ethers, ethylbenzene.

5. Coupling constants (J)

The spacing between peaks of a multiplet is J (in Hz). Reveals geometry between coupled nuclei.

Coupling typeTypical J (Hz)
Vicinal ³J (free rotation, sp³)6-8
Vicinal cis (alkene)6-12
Vicinal trans (alkene)12-18
Aromatic ortho7-10
Aromatic meta2-3
Aromatic para0-1

6. Common diagnostic patterns

7. Brief introduction to ¹³C NMR

¹³C NMR shows one peak per unique carbon. Wider chemical shift range (0-220 ppm) gives better resolution. Spectra are typically broadband proton-decoupled (no C-H splitting), so each carbon is a singlet.

¹³C environmentδ (ppm)
Saturated alkyl C10-50
C-N30-60
C-O (alcohol, ether)50-90
Alkene C100-150
Aromatic C120-150
Carbonyl C (ester)165-175
Carbonyl C (acid)170-185
Carbonyl C (aldehyde, ketone)190-220
Memorise these landmarks for ¹H NMR Aliphatic CH₃: ~0.9 ppm
Aliphatic CH₂: ~1.2-1.5
Allyl/benzyl/α-carbonyl CH: 2.0-2.6
-OCH₃ (methyl ester): ~3.65 (singlet)
-OCH₂- (in ester R-O-CH₂-): ~4.1 (often quartet)
Aromatic H: 6.5-8.5
Aldehyde H: 9-10
Acid OH: 10-13
Combined with integration + multiplicity, these landmarks identify ~80% of common organic compounds.

8. Systematic NMR analysis approach

  1. Count signals. How many distinct chemical environments are there?
  2. Read chemical shifts. What functional groups do they suggest?
  3. Read integrations. How many protons in each environment? Express as ratio.
  4. Read multiplicities. How many neighbours does each have? Use n+1 rule.
  5. Assemble fragments. Match patterns to ethyl, isopropyl, t-butyl, methyl ester, aromatic, etc.
  6. Combine with MS/IR. NMR gives the framework; MS gives the molecular weight; IR gives functional group confirmation.

Instructions

The Simulation has four parts. Complete in order.

1
Section I — Chemical Shift Library. 8 cases. For each: identify the chemical shift region; identify the multiplicity expected; identify the integration / proton type.
2
Section II — NMR Instrument Bench. 6 unknown samples. Select the bottle, click "Acquire Spectrum", view the ¹H NMR, identify the compound. Peak labels appear after you answer.
3
Section III — Spectrum Interpretation. 8 puzzles: distinguishing isomers (ortho/meta/para; primary/secondary/tertiary alcohols), identifying compounds from full NMR data, assembling fragments.
4
Section IV — SDS & Microscale. SDS for 4 NMR materials (CDCl₃, DMSO-d₆, D₂O, TMS) = 16 questions. Then 6 microscale sample-prep scenarios.
5
Prepare your lab notebook. Use the Example Report as your template.

Prerequisite: Familiarity with organic functional groups and the IR + Mass Spec labs is helpful. NMR is the third member of the spectroscopy quartet.

Simulation

Four interactive parts. Use the ↺ Reset Simulation button to clear all answers and start over.

NMR Workbench Section I — Chemical Shift Library

Eight chemical-shift identification cases. For each: (a) chemical shift region; (b) expected multiplicity; (c) which proton type.

Score: 0 / 24

Six unknown samples. Click Acquire Spectrum, identify the compound from the ¹H NMR. Peak labels appear after you answer.

Score: 0 / 6

Eight harder puzzles: distinguishing isomers, full structural deduction.

Score: 0 / 8

Round 1 — SDS interpretation

Four common NMR solvents. Each has 4 questions.

SDS score: 0 / 16

Round 2 — Microscale sample prep

Six sample-prep scenarios.

Microscale score: 0 / 6

Team Questions

Discuss with your team before answering.

Question 1 — The n+1 rule. A proton has 3 equivalent neighbours. What multiplicity will its signal show, and what intensity ratios?
Question 2 — Ethyl pattern. Describe the ¹H NMR pattern of a generic ethyl group (-CH₂CH₃).
Question 3 — OH proton. Why is the OH peak in alcohols often broad and at variable chemical shift?
Question 4 — Singlet identification. A ¹H NMR shows a singlet at 9.7 ppm integrating to 1H. What functional group?
Question 5 — D₂O shake. What happens when you add D₂O to a sample and re-acquire the spectrum? What information does this give?
Question 6 — Why deuterated solvents? Why do we use CDCl₃ instead of CHCl₃ for NMR samples?

Example Lab Notebook Entry

Use the format below as a template.

NMR Spectroscopy — Lab Notebook Entry

Submitted by: [Student Name]

Course: Organic Chemistry I · Section: 201-A · Date: May 9, 2026

Objective

To learn the systematic interpretation of ¹H NMR spectra: chemical shift, integration, multiplicity, and coupling constants. To recognise common diagnostic patterns (ethyl, isopropyl, tert-butyl, methyl ester, aromatic, OH/NH). To deduce structure by combining NMR data with mass spectrometry and IR. To understand sample preparation requirements: deuterated solvents, TMS reference, lock signal.

Instrument bench results (Section II)

SampleCompoundKey ¹H NMR signals
1EthanolTriplet 1.20 (3H), quartet 3.70 (2H), broad singlet 2.6 (1H, OH)
2AcetoneSinglet 2.17 (6H, two equivalent methyls)
3Ethyl acetateSinglet 2.04 (3H, COCH₃), quartet 4.12 (2H, OCH₂), triplet 1.26 (3H, CH₃)
4TolueneSinglet 2.36 (3H, ArCH₃), multiplet 7.2 (5H, aromatic)
51-BromopropaneTriplet 1.04 (3H, CH₃), sextet 1.86 (2H, central CH₂), triplet 3.40 (2H, CH₂Br)
6p-XyleneSinglet 2.30 (6H, two equivalent ArCH₃), singlet 7.04 (4H, aromatic)

Discussion

NMR spectroscopy answers four questions about each unique proton: where (chemical shift), how many (integration), how many neighbours (multiplicity), and what coupling (J). The chemical shift indicates the chemical environment: aliphatic CH₃ ~0.9, allyl/benzyl/α-carbonyl CH₂ ~2-2.6, -OCH₂- ~3.5-4.5, aromatic 6.5-8.5, aldehyde 9-10, acid 10-13. The shift moves downfield with proximity to electronegative atoms or pi systems.

Integration tells you the relative number of protons in each environment. For ethanol (CH₃CH₂OH), the integration ratio is 3:2:1 corresponding to the 3 H of CH₃, 2 H of CH₂, and 1 H of OH. The ratios are reliable; absolute counts require comparison to a known reference.

Multiplicity follows the n+1 rule: if a proton has n equivalent neighbours, its signal splits into n+1 peaks with intensities matching Pascal\'s triangle. The classic ethyl pattern (-CH₂CH₃) shows the CH₃ (with 2 neighbours) as a triplet, and the CH₂ (with 3 neighbours) as a quartet. The isopropyl pattern shows a doublet (the 6 equivalent methyls) and a septet (the lone CH); the t-butyl group is a singlet integrating for 9H.

The instrument bench (Section II) demonstrated these patterns on six samples. Ethanol shows the alcohol triplet/quartet/broad-OH; acetone shows a single peak (all 6 H equivalent); ethyl acetate shows the diagnostic ester pattern (singlet OCH₃ or COCH₃, quartet OCH₂, triplet CH₃); toluene shows ArCH₃ singlet + aromatic multiplet; 1-bromopropane shows a triplet/sextet/triplet pattern characteristic of -CH₂-CH₂-CH₂X; para-xylene shows two singlets (high symmetry: two CH₃ equivalent + 4 aromatic H equivalent).

For sample preparation, the key requirement is a DEUTERATED solvent. Regular CHCl₃ would dominate the spectrum with its ¹H signal; CDCl₃ replaces ¹H with ²H (deuterium, which doesn\'t appear in ¹H NMR) and provides a "lock signal" that the spectrometer uses to maintain field stability. TMS (tetramethylsilane) is added at 0 ppm as the chemical shift reference. The choice of solvent depends on the sample: CDCl₃ for most organics; DMSO-d₆ for polar compounds and amides; D₂O for very polar/charged compounds; methanol-d₄ for OH/NH-containing compounds where exchange info is wanted.

D₂O shake is a useful diagnostic: adding a drop of D₂O to the sample exchanges OH and NH protons with deuterium, removing them from the ¹H spectrum. The remaining peaks (CH only) confirm which were the exchangeable protons. Alcohols and acids lose their OH peaks; amines lose NH peaks.

Conclusion

NMR spectroscopy is the most powerful single technique for organic structure determination, providing direct mapping of the carbon-hydrogen framework with information about connectivity through coupling. Combined with mass spectrometry (molecular weight) and IR (functional groups), NMR completes the spectroscopic identification toolkit. The next lab in this Spectroscopy quartet (UV/Vis) covers electronic transitions and is the fourth and final member.

References

1. Pavia, D. L.; Lampman, G. M.; Kriz, G. S.; Vyvyan, J. R. Introduction to Spectroscopy, 5th ed., Cengage, 2015, Ch 3-7.
2. Silverstein, R. M.; Webster, F. X.; Kiemle, D. J. Spectrometric Identification of Organic Compounds, 8th ed., Wiley, 2014, Ch 3-5.
3. SDBS — Spectral Database for Organic Compounds, AIST Japan: https://sdbs.db.aist.go.jp.
4. Friebolin, H. Basic One- and Two-Dimensional NMR Spectroscopy, 5th ed., Wiley-VCH, 2010.

Practice Questions

Work through each before peeking at the hint.

Practice 1 — Reading shifts
A ¹H NMR shows: triplet 0.9 ppm (3H), multiplet 1.3 ppm (4H), triplet 2.3 ppm (2H), singlet 11 ppm (1H, broad). What functional group is present, and what is the molecule likely to be?
Hint: Singlet at 11 ppm = COOH. Triplet 2.3 = CH₂ alpha to carbonyl. Multiplet 1.3 = central CH₂\'s. Triplet 0.9 = terminal CH₃. Pattern: CH₃-CH₂-CH₂-CH₂-COOH = pentanoic (valeric) acid (or propanoic if shorter chain). Combined with MW from mass spec, the chain length is fixed.
Practice 2 — Isopropyl recognition
A ¹H NMR has a doublet at 1.20 ppm (6H) and a septet at 4.0 ppm (1H). What functional group does this indicate?
Hint: Doublet 6H + septet 1H = isopropyl group (CH₃)₂CH-X. The 6H are the two equivalent methyls (each split by the lone CH); the 1H is the central CH (split by 6 equivalent neighbouring methyl H). Position of the septet (4.0) = isopropyl attached to O (e.g., (CH₃)₂CH-O-R \u2014 isopropyl ester or ether).
Practice 3 — Ethyl ester
An NMR shows: singlet 2.0 ppm (3H), quartet 4.1 ppm (2H), triplet 1.25 ppm (3H). What is this compound?
Hint: Singlet 3H at 2.0 = -COCH₃ (acetyl group; alpha to C=O). Quartet 2H at 4.1 = -OCH₂- (in an ester, R-O-CH₂-). Triplet 3H at 1.25 = the other CH₃ of the ethyl. Combined: CH₃-CO-O-CH₂-CH₃ = ethyl acetate (CH₃COOC₂H₅).
Practice 4 — Aromatic substitution
A monosubstituted benzene shows aromatic protons as a complex multiplet at ~7.2-7.4 (5H total). A para-disubstituted benzene with two different substituents shows two doublets (2H each) at slightly different chemical shifts. Why does the substitution pattern change the splitting?
Hint: Mono-substituted benzene has 5 aromatic H, with the ortho H (2), meta H (2), and para H (1) all slightly different but often overlapping in CDCl₃. Para-disubstituted with two different groups has 4 aromatic H in TWO equivalence classes: 2 H ortho to one substituent (and meta to the other), and 2 H ortho to the other substituent. These appear as two "doublets" (technically AA\'BB\'), each integrating for 2H. The pattern is highly diagnostic of para-substitution.
Practice 5 — Methyl ester
A compound shows a singlet at 3.65 ppm (3H), plus aromatic protons. What is the singlet, and what does it tell you?
Hint: Singlet 3H at 3.65 = -OCH₃ (the methyl of a methyl ester). The singlet (no neighbours) is because the OCH₃ group is attached to oxygen (no adjacent C-H). The position 3.65 (close to OCH₂O range) confirms an ester R-CO-O-CH₃. Combined with aromatic peaks: an aromatic methyl ester like methyl benzoate (PhCOOCH₃) or methyl salicylate.
Practice 6 — tert-Butyl
A ¹H NMR shows a singlet at 1.45 ppm integrating for 9H. What does this indicate?
Hint: Singlet 9H at 1.45 = tert-butyl group ((CH₃)₃C-). All three methyls are equivalent (related by C₃ rotation), giving a single peak. The lack of splitting confirms no adjacent C-H. Position 1.45 is typical for tert-butyl on aromatic, ester, ether. tert-Butyl is one of the easiest groups to identify by NMR.
Practice 7 — D₂O shake
A compound shows a peak at 3.5 ppm (broad, 1H). After adding a drop of D₂O and re-acquiring, this peak disappears. What does this tell you?
Hint: The peak at 3.5 ppm (broad, 1H) was an EXCHANGEABLE proton: -OH or -NH. D₂O exchanges these for deuterium, which doesn\'t appear in ¹H NMR. The position 3.5 (broader range 0.5-5) and breadth (from H-bonding heterogeneity) plus disappearance with D₂O confirms it was OH (or NH). The compound contains an alcohol or amine group.
Practice 8 — Counting protons
A compound has MW = 122 (from MS) and shows the following ¹H NMR: doublet 7.95 (2H), triplet 7.5 (1H), triplet 7.4 (2H), broad singlet 12.0 (1H). What is the compound?
Hint: 2H + 1H + 2H + 1H = 6H total, all aromatic + acid OH region. The pattern 2H/1H/2H is mono-substituted benzene aromatic (ortho-meta-para, with the para H being a triplet, the meta H also a triplet, and ortho H a doublet). Singlet 12.0 = COOH. So this is benzoic acid (PhCOOH, MW 122). The aromatic pattern with three multiplets (2:1:2 ratio) is diagnostic of mono-substituted benzene with para-H as the most-shielded triplet.
Practice 9 — Multiplicity prediction
For 1-bromopropane (CH₃CH₂CH₂Br), predict the chemical shift, multiplicity, and integration for each proton type.
Hint: 3 environments: CH₃ (3H), central CH₂ (2H), CH₂Br (2H). CH₃ (~1.0 ppm): 2 neighbours \u2192 triplet, 3H. Central CH₂ (~1.85 ppm): 5 neighbours (3 from CH₃ + 2 from CH₂Br) \u2192 sextet, 2H. CH₂Br (~3.40 ppm): 2 neighbours \u2192 triplet, 2H. The sextet on the central CH₂ is the diagnostic pattern of -CH₂-CH₂-CH₂-X.
Practice 10 — Combining techniques
An unknown shows: MS M⁺ 86; IR: strong C=O 1715, sp³ C-H 2950, no broad O-H; ¹H NMR: triplet 1.05 (3H), quartet 2.45 (2H), singlet 2.13 (3H). What is the compound?
Hint: M\u207A = 86 (try C₅H₁₀O = 86, e.g., 2-pentanone or 3-pentanone). IR: C=O 1715 + no broad O-H = ketone. NMR: singlet 3H (2.13) = CH₃CO-; triplet 3H (1.05) + quartet 2H (2.45) = ethyl group adjacent to a deshielding group (the C=O, hence the 2.45 quartet position). So: CH₃-CO-CH₂-CH₃ = 2-butanone? But MW C₄H⁸O = 72, not 86. Hmm. Reconsider: CH₃CO-CH₂CH₂CH₃ = 2-pentanone, MW 86. The NMR for 2-pentanone: CH₃CO-CH₂-CH₂-CH₃: singlet ~2.13 (3H, CH₃CO); triplet ~2.43 (2H, CH₂ alpha to CO); sextet ~1.55 (2H, central CH₂); triplet ~0.93 (3H, terminal CH₃). The given pattern fits if the central CH₂ sextet was missed/not described; alternatively, this is 3-pentanone (CH₃CH₂COCH₂CH₃) which has: triplet 1.05 (6H, two CH₃ equivalent), quartet 2.45 (4H, two CH₂ equivalent). 3-Pentanone\'s symmetry makes only TWO signals visible (3:2 integration scaled to 6:4 actual). With singlet 2.13 (3H) suggesting an extra methyl, the compound is most likely 3-methyl-2-butanone (methyl isopropyl ketone): CH₃-CO-CH(CH₃)₂, MW 86. The given NMR (triplet/quartet/singlet) doesn\'t perfectly fit either; this is a teaching question about the ambiguity \u2014 NMR + MS + IR together help narrow possibilities, but a single experiment is rarely conclusive.