Combined Techniques: Worked Problems I

In real chemical analysis and in A-Level exam questions, you rarely use just one technique in isolation. The most powerful approach is to combine data from mass spectrometry, infrared spectroscopy, and NMR spectroscopy to determine the complete structure of an unknown compound. This lesson presents a systematic approach and several fully worked examples.

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Systematic Approach to Structure Determination

Follow these steps when given spectroscopic data for an unknown compound:

Step 1: Determine the Molecular Formula

• From high-resolution mass spectrometry: the exact molecular mass gives the molecular formula directly.
• From low-resolution MS: the molecular ion peak gives Mr. Use this with percentage composition data (if given) to find the molecular formula.
• If given the molecular formula, proceed to Step 2.

Step 2: Calculate the Degree of Unsaturation (Index of Hydrogen Deficiency, IHD)

IHD = (2C + 2 + N − H − X) / 2

where C = number of carbons, H = number of hydrogens, N = number of nitrogens, X = number of halogens. Oxygen does not affect the calculation.

IHD	Significance
0	No double bonds or rings
1	One C=C or one C=O or one ring
2	Two of the above
3	Could include C≡C or C≡N
4	Likely a benzene ring (3 C=C + 1 ring)

Exam Tip: A benzene ring contributes 4 to the IHD (three double bonds + one ring). If IHD ≥ 4, always consider the possibility of an aromatic ring.

Step 3: Analyse the IR Spectrum

• Identify functional groups from characteristic absorptions.
• Key absorptions: O–H broad (alcohol 3200–3600, acid 2500–3300), C=O sharp (1680–1750), C–O (1000–1300), N–H (3300–3500).

Step 4: Analyse the ¹³C NMR Spectrum

• Count the number of carbon environments.
• Note chemical shifts to identify carbon types (alkyl, C–O, aromatic, C=O).

Step 5: Analyse the ¹H NMR Spectrum

• Count the number of hydrogen environments.
• Note chemical shifts (what type of H?).
• Read integration ratios (how many H in each environment?).
• Analyse splitting patterns (how many H on adjacent carbons?).
• Perform D₂O shake to identify OH/NH.

Step 6: Assemble the Structure

• Combine all the information to propose a structure.
• Check that your proposed structure is consistent with ALL the spectroscopic data.

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Worked Example 1: Unknown Compound A (C₃H₆O)

Given data:

Mass spectrum: M⁺˳ at m/z = 58. Base peak at m/z = 43. Other peaks at 29, 27, 15.

IR spectrum: Strong, sharp absorption at 1720 cm⁻¹. No broad O–H absorption.

¹H NMR: Two peaks:

• δ = 2.1 ppm, singlet, integration 3H
• δ = 0.97 ppm, singlet... wait, that cannot be right for C₃H₆O.

Let me reconsider. C₃H₆O has Mr = 58. IHD = (6 + 2 − 6) / 2 = 1. One degree of unsaturation.

Actually, the molecular formula should give us the correct analysis. C₃H₆O with IHD = 1: the unsaturation must be a C=O (confirmed by IR at 1720).

¹H NMR:

• Peak at δ = 2.1 ppm, singlet, 3H
• Peak at δ = 9.8 ppm, singlet, 1H
• Peak at δ = 1.0 ppm, triplet...

No — for C₃H₆O with a C=O, we have propanal or propanone:

• Propanone (CH₃COCH₃): Mr = 58, IHD = 1, C=O at ~1715 cm⁻¹. ¹H NMR: one singlet at δ = 2.1 ppm (6H). But C₃H₆O only has 6 hydrogens, and propanone accounts for all of them.
• Propanal (CH₃CH₂CHO): Mr = 58, IHD = 1, C=O at ~1730 cm⁻¹. ¹H NMR: triplet at δ ≈ 1.0 (3H, CH₃), quartet at δ ≈ 2.4 (2H, CH₂), triplet at δ ≈ 9.8 (1H, CHO).

From the mass spectrum: base peak at m/z = 43 = CH₃CO⁺. This is characteristic of a methyl ketone.

IR at 1720 cm⁻¹: consistent with a ketone.

If the ¹H NMR shows a single peak (singlet at δ = 2.1 ppm, 6H), the compound is propanone.

Verification:

• Mr = 58 ✔
• IHD = 1 (C=O) ✔
• IR: C=O at 1720 cm⁻¹ ✔
• MS: m/z = 43 (CH₃CO⁺, loss of 15 = CH₃) ✔
• MS: m/z = 15 (CH₃⁺) ✔
• ¹H NMR: singlet, δ = 2.1, 6H (two equivalent CH₃ groups) ✔

Answer: Propanone (acetone), CH₃COCH₃.

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Worked Example 2: Unknown Compound B (C₄H₈O₂)

Given data:

Mass spectrum: M⁺˳ at m/z = 88. Significant peaks at m/z = 43, 45, 29.

IR spectrum: Strong absorption at 1740 cm⁻¹. No broad O–H absorption.

¹H NMR:

• δ = 1.3 ppm, triplet, 3H
• δ = 2.0 ppm, singlet, 3H
• δ = 4.1 ppm, quartet, 2H

Step 1: Mr = 88. Molecular formula C₄H₈O₂.

Step 2: IHD = (8 + 2 − 8) / 2 = 1. One degree of unsaturation (likely C=O).

Step 3: IR at 1740 cm⁻¹ is in the ester C=O range (1735–1750). No broad O–H → not a carboxylic acid or alcohol.

Step 4: Three ¹H NMR environments, total 8H (3+3+2 = 8). Consistent with C₄H₈O₂.