Spectroscopy Challenge Solution

Yesterday’s challenge asked you to identify an unknown compound using mass spec, IR, ¹H NMR and ¹³C NMR — exactly the sort of problem OCR loves to set.

Let’s now work through the evidence logically and methodically, just as you’d do in an exam.

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🔍 Step 1: Mass Spectrometry – The Molecular Mass

• Molecular ion peak at m/z = 89 Now this wasn't clear so I helped a bit.

• This strongly suggests a relative molecular mass of 89

A common OCR exam move here is to ask:

What biologically important molecules have Mr ≈ 89?

Keep that in mind — we’ll come back to it.

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🔍 Step 2: IR Spectrum – Functional Groups

Key absorptions:

• Broad peak at 2500–3300 cm⁻¹
  → characteristic of an O–H stretch in a carboxylic acid

• Strong absorption at ~1700 cm⁻¹
  → C=O stretch, consistent with a carboxyl group

• Absorption near 1550 cm⁻¹
  → consistent with N–H bending, suggesting an amine group

📌 Conclusion so far:
The compound contains both a carboxylic acid and an amine → this immediately points towards an amino acid.

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🔍 Step 3: ¹H NMR – Proton Environments

Observed signals:

• ~1.5 ppm (3H)
  → a CH₃ group, likely adjacent to a carbon atom

• ~3.8 ppm (1H)
  → a CH group attached to electronegative atoms (N or O)

• Broad signal at ~8–10 ppm
  → exchangeable protons, consistent with –NH₂ / –NH₃⁺ and –COOH

This is textbook amino acid behaviour in proton NMR.

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🔍 Step 4: ¹³C NMR – Carbon Environments

Only three carbon signals, meaning three different carbon environments:

• ~175 ppm
  → carboxylic acid carbonyl carbon

• ~50 ppm
  → carbon attached to –NH₂

• ~20 ppm
  → methyl carbon

This perfectly matches a simple amino acid with a methyl side chain.

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🧠 Final Identification

Putting everything together:

• Mr = 89

• Contains –COOH and –NH₂

• Three carbon environments

• Methyl side chain

✅ The compound is alanine.

This is how we teach the students to identify chemicals from spectra at Hemel Private Tuition.