Combining Analytical Data

In real chemistry — and in Edexcel A-Level exams — you are rarely given data from a single technique. The power of analytical chemistry comes from combining data from mass spectrometry, infrared spectroscopy, ¹H NMR, and ¹³C NMR to build up a complete structural picture. This lesson teaches you a systematic approach for tackling these combined problems.

The Systematic Approach

When given data from multiple techniques, work through them in a logical order:

flowchart TD
    A["Step 1: Mass Spectrometry"] --> B["Molecular ion → Mr\nHRMS → molecular formula\nCalculate degrees of unsaturation"]
    B --> C["Step 2: IR Spectroscopy"]
    C --> D["Identify functional groups\n(O-H, N-H, C=O, C≡N, C-O)"]
    D --> E["Step 3: ¹³C NMR"]
    E --> F["Count carbon environments\nClassify by chemical shift\nCheck for molecular symmetry"]
    F --> G["Step 4: ¹H NMR"]
    G --> H["Count H environments\nIntegration → number of H\nChemical shift → H type\nSplitting → connectivity"]
    H --> I["Step 5: Propose Structure"]
    I --> J{"Consistent with\nALL data?"}
    J -->|"Yes"| K["Structure confirmed"]
    J -->|"No"| L["Revise and\nre-examine data"]
    L --> I

Step 1: Molecular Formula and Degree of Unsaturation (from Mass Spectrometry)

Start with the mass spectrum. The molecular ion peak (M⁺) gives the relative molecular mass. If high-resolution mass spectrometry data or elemental analysis data is provided, determine the molecular formula.

From the molecular formula CₙHₘOₓNy, calculate the degree of unsaturation (also called the index of hydrogen deficiency, IHD):

Degree of unsaturation = (2n + 2 − m + y) / 2

Where:

n = number of carbons
m = number of hydrogens
y = number of nitrogens
Oxygen and halogens do not affect the calculation (but halogens count as hydrogens for this formula)

What the degree of unsaturation tells you:

DoU	Structural implication	Examples
0	No rings or double bonds (saturated, open-chain)	Alkanes, alcohols, ethers
1	One double bond or one ring	Aldehydes, ketones, alkenes, cycloalkanes
2	Two double bonds, two rings, one of each, or one triple bond	Dienes, alkynes
3	Three degrees: e.g., ring + 2 double bonds, or triple bond + ring	—
4	Almost certainly a benzene ring (ring + 3 double bonds)	Aromatic compounds
5	Benzene ring + one C=O	Benzoic acid, acetophenone, benzaldehyde

Step 2: Functional Groups (from IR Spectroscopy)

Check the IR spectrum for key absorptions:

Broad O–H at 2500–3300 cm⁻¹ → carboxylic acid
Broad O–H at 3200–3600 cm⁻¹ → alcohol
N–H at 3300–3500 cm⁻¹ → amine or amide
C=O at 1680–1750 cm⁻¹ → carbonyl compound (note exact position)
No O–H with C=O → ketone, aldehyde, or ester (not acid)

Step 3: Carbon Framework (from ¹³C NMR)

Count the number of peaks — this equals the number of distinct carbon environments. Use chemical shift values to classify each carbon:

0–55 ppm: alkyl carbons
50–90 ppm: C–O or C–N
100–160 ppm: aromatic or alkene carbons
160–185 ppm: acid/ester/amide C=O
185–220 ppm: aldehyde/ketone C=O

If DEPT data is available, classify each carbon as CH₃, CH₂, CH, or quaternary.

Step 4: Hydrogen Environments and Connectivity (from ¹H NMR)

The ¹H NMR spectrum provides:

Number of peaks = number of hydrogen environments
Integration = relative number of hydrogens in each environment
Chemical shift = type of environment
Splitting pattern = number of adjacent hydrogens (n+1 rule)

Use these together to deduce the molecular connectivity.

Worked Example 1: Molecular Formula C₃H₆O

Step 1 (MS): Mr = 58. Degree of unsaturation = (2×3 + 2 − 6) / 2 = 1. One double bond or ring.

Step 2 (IR): Strong absorption at 1720 cm⁻¹. No broad O–H. The C=O without O–H indicates a ketone or aldehyde.

Step 3 (¹³C NMR): Three peaks at δ = 206, 37, and 8 ppm.

δ = 206 → aldehyde or ketone C=O
δ = 37 → alkyl carbon (possibly CH₂ next to C=O)
δ = 8 → alkyl carbon (CH₃)

Three peaks means three carbon environments — this is consistent with all three carbons being in different environments.

Step 4 (¹H NMR): Three signals:

δ = 9.8, 1H, triplet → aldehyde H (CHO) coupled to 2 adjacent H
δ = 2.4, 2H, multiplet → CH₂ next to CHO
δ = 1.0, 3H, triplet → CH₃ coupled to 2 adjacent H

Conclusion: The triplet aldehyde hydrogen coupled to a CH₂ confirms this is propanal (CH₃CH₂CHO).

If it were propanone, there would be no peak at δ = 9.8 and only one singlet (all 6 hydrogens equivalent). The ¹H NMR data decisively distinguishes propanal from propanone.

Worked Example 2: Molecular Formula C₄H₈O₂

Step 1 (MS): Mr = 88. Degree of unsaturation = (2×4 + 2 − 8) / 2 = 1. One C=O.

Step 2 (IR): Strong C=O at 1740 cm⁻¹, strong C–O at 1240 cm⁻¹, no O–H. The high C=O position and absence of O–H indicate an ester.

Step 3 (¹³C NMR): Four peaks — four different carbon environments.

Step 4 (¹H NMR):

δ = 1.1, 3H, triplet → CH₃ next to CH₂
δ = 2.0, 3H, singlet → CH₃ next to C=O (no adjacent H)
δ = 4.1, 2H, quartet → OCH₂ next to CH₃

Conclusion: The singlet CH₃ at δ 2.0 is the acetyl methyl (CH₃CO–). The triplet/quartet pattern is an ethyl group attached to oxygen (–OCH₂CH₃). The compound is ethyl ethanoate (CH₃COOCH₂CH₃).

Worked Example 3: An Aromatic Compound, C₈H₈O

Step 1 (MS): Mr = 120. Degree of unsaturation = (2×8 + 2 − 8) / 2 = 5. Four degrees from a benzene ring + one from C=O.

Step 2 (IR): C=O at 1685 cm⁻¹. Aromatic C=C at 1600 and 1500 cm⁻¹. No O–H. The C=O at 1685 cm⁻¹ is slightly low for a ketone — consistent with conjugation with the aromatic ring.

Step 3 (¹³C NMR): Five peaks. For C₈H₈O with 8 carbons but only 5 carbon environments — symmetry is present.

Step 4 (¹H NMR):

δ = 7.4–8.0, 5H, complex multiplet → monosubstituted benzene ring (C₆H₅)
δ = 2.6, 3H, singlet → CH₃ next to C=O

Conclusion: A benzene ring (C₆H₅) + COCH₃ = 120. The compound is acetophenone (C₆H₅COCH₃).

Worked Example 4: Interpreting Combined MS/IR/NMR Data for a Complete Unknown

A compound gives the following data:

HRMS: M⁺ = 102.0681, suggesting C₅H₁₀O₂
MS fragments: m/z = 57 (M−45), m/z = 43
IR: Strong C=O at 1740 cm⁻¹, strong C–O at 1180 cm⁻¹, no O–H or N–H
¹³C NMR: 4 peaks at δ = 14, 19, 36, 174 ppm
¹H NMR: δ = 0.9 (3H, t), δ = 1.2 (3H, t), δ = 2.3 (2H, q), δ = 4.1 (2H, q)

Analysis:

C₅H₁₀O₂: DoU = (10 + 2 − 10)/2 = 1. One C=O.
IR: Ester (C=O at 1740, C–O, no O–H).
MS: m/z = 57 is M − 45 (loss of OC₂H₅•), and m/z = 43 could be C₃H₇⁺ or CH₃CO⁺.
¹H NMR: Two triplet-quartet pairs indicate two separate ethyl groups. One ethyl group at δ = 2.3/0.9 is CH₃CH₂CO– (next to carbonyl). The other at δ = 4.1/1.2 is –OCH₂CH₃ (next to oxygen).
¹³C at 174 ppm confirms ester C=O. Peaks at 14 and 19 are the two CH₃ carbons; 36 is the CH₂ next to C=O.

Conclusion: The compound is ethyl propanoate (CH₃CH₂COOCH₂CH₃).

Quick-Reference: What Each Technique Tells You

Technique	Information provided	Key question it answers
Low-res MS	Mr from molecular ion peak	How heavy is the molecule?
High-res MS	Molecular formula	What is the molecular formula?
MS fragmentation	Structural fragments	What pieces does it break into?
IR	Functional groups	What bonds/groups are present?
¹³C NMR	Carbon environments	How many distinct carbons? What types?
DEPT	H atoms on each carbon	Is each C a CH₃, CH₂, CH, or quaternary?
¹H NMR shifts	H environments	What type of H (alkyl, aromatic, aldehyde)?
¹H NMR integration	Number of H per environment	How many H in each environment?
¹H NMR splitting	Connectivity	How many H on adjacent carbons?

Tips for Combined Problems

Always calculate the degree of unsaturation first. It immediately tells you about rings and double bonds and narrows the possibilities enormously.
Use IR to identify the functional group class, then use NMR to determine the exact structure within that class.
¹³C NMR tells you symmetry. If the number of ¹³C peaks is less than the number of carbons in the molecular formula, the molecule has symmetry.
The n+1 splitting pattern reveals connectivity. A triplet/quartet pair is almost always an ethyl group. A doublet/septet pair is an isopropyl group. Two singlets mean the groups are not adjacent to any H.
Check that your proposed structure is consistent with ALL the data. After proposing a structure, verify it accounts for every peak in every spectrum.
Use MS fragmentation to confirm. After proposing a structure, check that the fragment ions make sense. If your structure is ethyl ethanoate, you should see m/z = 43 (CH₃CO⁺) in the MS.

Common Mistakes in Combined Analysis

Mistake 1: Stopping after one technique gives a plausible answer. Always check against ALL the data. Two compounds may give similar IR spectra but completely different NMR spectra.

Mistake 2: Forgetting to account for all hydrogens. After proposing a structure, add up the integration values from ¹H NMR and verify they match the total number of H in the molecular formula.

Mistake 3: Ignoring the degree of unsaturation. If DoU = 4, you almost certainly have a benzene ring. If DoU = 0, there are no double bonds or rings. This single calculation eliminates huge numbers of possible structures.

Mistake 4: Not checking MS fragments against the proposed structure. If you propose a structure, its fragmentation should be consistent with the observed MS peaks.

Summary

Combining analytical data is a skill that brings together everything you have learned about MS, IR, ¹H NMR, and ¹³C NMR. The systematic approach — molecular formula, then functional groups, then carbon framework, then hydrogen connectivity — ensures you extract the maximum information from each technique and arrive at the correct structure. This integrated approach is heavily tested at A-Level.

A-Level Deep Dive: Combining Analytical Data

Spec mapping

Edexcel 9CH0 specification, Topics 18 and 19 — Modern Analytical Techniques I and II (synoptic application) require candidates to integrate data from MS, IR, ¹H NMR, ¹³C NMR, and (where relevant) chromatographic and chemical-test results to identify unknown organic compounds. The typical workflow is: MS gives Mr and formula; IR gives functional group; ¹H NMR gives connectivity (environments + integrations + splitting); ¹³C NMR confirms carbon skeleton; chemical tests resolve aldehyde/ketone or other ambiguities. Degree of unsaturation (DoU) — calculated from the molecular formula — constrains the structural possibilities (refer to the official specification document for exact wording). Examined extensively on Paper 3 (General and Practical Principles), which is heavily synoptic by design.

Worked example with full mark scheme

Question (10 marks):

An unknown organic compound G has the following data:

MS: molecular ion at m/z = 88; major fragments at m/z = 73, 43, 29.
IR: broad 2500–3300 cm⁻¹, sharp strong 1710 cm⁻¹.
¹H NMR: singlet 11.5 ppm (1H, broad), multiplet 2.4 ppm (2H, integration), multiplet 1.6 ppm (2H, integration), triplet 0.9 ppm (3H, integration).
¹³C NMR: 4 peaks at approximately δ = 13, 18, 36, 180 ppm.

(a) Calculate the molecular formula and degree of unsaturation. (2)

(b) Identify the functional group from IR. (1)

(d) Combine all data to suggest a structure for G, justifying briefly. (4)

Solution with mark scheme:

(a) Mr = 88. Trying C₄H₈O₂ (Mr = 48 + 8 + 32 = 88). DoU = (2×4 + 2 − 8) / 2 = 1, consistent with one C=O or one ring. A1 A1.

(b) Broad O–H 2500–3300 + C=O 1710 → carboxylic acid. A1.

(c)

Broad singlet 11.5 ppm (1H) = COOH proton (range 9–13). M1.
Triplet 0.9 ppm (3H) = terminal CH₃ coupled to CH₂. M1.
Multiplets 2.4 and 1.6 ppm (2H each) = two CH₂ groups; the alpha-CH₂ at 2.4 ppm (alpha to C=O), middle CH₂ at 1.6 ppm. A1.

(d) Structure: butanoic acid CH₃CH₂CH₂COOH (Mr = 88; one COOH; CH₃ triplet at 0.9; central CH₂ multiplet at 1.6; alpha-CH₂ multiplet at 2.4; COOH broad singlet at 11.5). M1.

¹³C NMR confirms 4 environments: ~13 (CH₃), ~18 (central CH₂), ~36 (alpha-CH₂), ~180 (COOH). M1.

MS fragments: [M − 15]⁺ = 73 (loss of CH₃•); [M − 45]⁺ = 43 (loss of COOH); 29 = C₂H₅⁺. All consistent with butanoic acid. M1.

The four lines of evidence (MS Mr, IR functional group, ¹H NMR connectivity, ¹³C carbon count) all converge on butanoic acid. A1.

Total: 10 marks (M6 A4).

Specimen question modelled on the Edexcel 9CH0 Paper 3 format

Question (8 marks): An unknown compound H has Mr = 102 from MS. IR shows a strong sharp peak at 1735 cm⁻¹ but no broad O–H. ¹H NMR shows: triplet 0.9 ppm (3H), multiplet 1.6 ppm (2H), triplet 2.3 ppm (2H), singlet 3.7 ppm (3H). ¹³C NMR shows 5 peaks.