GCSE Chemistry
Random

Click random to get a fresh chapter.

A Level Chemistry - How to Solve: Structure Determination Using IR, Mass Spectrometry, and NMR Combined

How to Solve: Structure Determination Using IR, Mass Spectrometry, and NMR Combined

Complete Guide

---

Introduction

"Mastering IR, Mass Spec, and NMR together lets you crack any ‘unknown compound’ exam question—worth up to 15% of your A-Level Chemistry paper. Real-world chemists use these exact steps to identify drugs, pollutants, and even crime-scene evidence."

---

WHAT YOU NEED TO KNOW FIRST

1. Basic organic functional groups (alcohols, carbonyls, alkenes, etc.) and their structures.
2. How to read simple spectra (IR peaks, NMR splitting patterns, Mass Spec fragments).
3. Molecular formula from mass spec (M⁺ peak and isotopic ratios).

---

KEY TERMS & FORMULAS

1. Mass Spectrometry (MS)

• M⁺ peak: Highest m/z peak (molecular ion). MEMORISE THIS → Gives molecular mass.
• Base peak: Tallest peak (most stable fragment).
• M+1 peak: Small peak 1 unit above M⁺ (due to ¹³C isotopes). Formula: [ \text{Number of carbons} = \frac{\text{Height of M+1 peak}}{\text{Height of M⁺ peak}} \times 100 ] (Given on exam sheet, but understand how to use it.)

2. Infrared Spectroscopy (IR)

• Key wavenumbers (cm⁻¹):
• O-H (alcohol): 3200–3600 (broad)
• C=O (carbonyl): 1700–1750 (sharp)
• C-H (alkane): 2850–3000
• C=C (alkene): 1620–1680
• N-H (amine): 3300–3500 (sharp) MEMORISE THESE RANGES.

3. Nuclear Magnetic Resonance (NMR)

• Chemical shift (δ, ppm):
• TMS (reference): 0 ppm
• Alkyl (CH₃, CH₂): 0.5–2.5 ppm
• Alcohol (OH): 1–5 ppm (broad)
• Alkene (C=C-H): 4.5–6.5 ppm
• Aromatic (benzene ring): 6.5–8.5 ppm
• Aldehyde (CHO): 9–10 ppm

• Carboxylic acid (COOH): 10–12 ppm MEMORISE THESE SHIFTS.

• Splitting (n+1 rule): [ \text{Number of peaks} = n + 1 \quad \text{(where } n = \text{number of adjacent hydrogens)} ] (Given on exam sheet, but apply it correctly.)

• Integration: Area under peak = number of hydrogens.

---

STEP-BY-STEP METHOD

Follow these 7 steps for every structure determination question.

1. Find the molecular formula from Mass Spec
2. Identify the M⁺ peak (highest m/z).
3. Use the M+1 peak to estimate carbon count (if given).

4. Subtract known fragments (e.g., loss of CH₃ = -15, OH = -17).

5. Calculate degrees of unsaturation (DU) [ \text{DU} = \frac{2C + 2 - H - X + N}{2} ] (C = carbons, H = hydrogens, X = halogens, N = nitrogens)

6. DU = 0: No rings/double bonds (alkane).
7. DU = 1: 1 ring or 1 double bond.
8. DU = 2: 2 double bonds, 1 triple bond, or 1 ring + 1 double bond.

9. DU ≥ 4: Likely benzene ring (DU = 4 for benzene).

10. Identify functional groups from IR

11. Look for key peaks (e.g., C=O at 1700 cm⁻¹, O-H at 3300 cm⁻¹).

12. Cross out impossible groups (e.g., no C=O? Not a ketone/aldehyde).

13. Analyse NMR chemical shifts

14. Match peaks to expected ppm ranges (e.g., 7–8 ppm = aromatic).

15. Note integration (e.g., 3H = CH₃, 2H = CH₂).

16. Determine splitting patterns (n+1 rule)

17. Singlet (s): No adjacent hydrogens (e.g., OH, CH₃ next to O).
18. Doublet (d): 1 adjacent H (e.g., CH next to CH).
19. Triplet (t): 2 adjacent H (e.g., CH₂ next to CH₂).

20. Multiplet (m): Complex splitting (e.g., aromatic rings).

21. Piece together fragments

22. Combine MS fragments, IR groups, and NMR signals to build possible structures.

23. Check symmetry (e.g., identical CH₃ groups = same NMR signal).

24. Verify with all data

25. Does the structure match all spectra?
26. Does it fit the molecular formula and DU?
27. Are NMR integrations correct?

---

WORKED EXAMPLES

Example 1 – Basic: Ethanol (C₂H₆O)

Given: - Mass Spec: M⁺ = 46, M+1 = 2.2% of M⁺. - IR: Broad peak at 3300 cm⁻¹, sharp peak at 2900 cm⁻¹. - NMR: 3H (t, 1.2 ppm), 2H (q, 3.7 ppm), 1H (s, 2.6 ppm).

Step-by-Step Solution:
1. Molecular formula from MS: - M⁺ = 46 → C₂H₆O (mass = 46). - M+1 = 2.2% → (2.2/1.1) ≈ 2 carbons (matches C₂H₆O).

1. Degrees of unsaturation (DU): [ \text{DU} = \frac{2(2) + 2 - 6}{2} = 0 \quad \text{(No rings/double bonds)} ]

2. IR analysis:

3. 3300 cm⁻¹ (broad) → O-H (alcohol).

4. 2900 cm⁻¹ → C-H (alkane).

5. NMR analysis:

6. 3H (t, 1.2 ppm): CH₃ next to CH₂ (triplet = 2 adjacent H).
7. 2H (q, 3.7 ppm): CH₂ next to CH₃ (quartet = 3 adjacent H).

8. 1H (s, 2.6 ppm): OH (singlet = no adjacent H).

9. Structure:

10. CH₃-CH₂-OH (ethanol).

What we did and why: - Used MS for molecular formula. - IR confirmed O-H (alcohol). - NMR splitting showed CH₃-CH₂ group. - DU = 0 ruled out double bonds.

---

Example 2 – Medium: Propanone (C₃H₆O)

Given: - Mass Spec: M⁺ = 58, base peak = 43. - IR: Sharp peak at 1715 cm⁻¹, no O-H peak. - NMR: 6H (s, 2.1 ppm).

Step-by-Step Solution:
1. Molecular formula from MS: - M⁺ = 58 → C₃H₆O (mass = 58). - Base peak = 43 → Loss of CH₃ (58 - 15 = 43).

1. Degrees of unsaturation (DU): [ \text{DU} = \frac{2(3) + 2 - 6}{2} = 1 \quad \text{(1 double bond or ring)} ]

2. IR analysis:

3. 1715 cm⁻¹ → C=O (carbonyl).

4. No O-H → Not an alcohol/acid.

5. NMR analysis:

6. 6H (s, 2.1 ppm): Two identical CH₃ groups (no splitting = no adjacent H).

7. Structure:

8. CH₃-CO-CH₃ (propanone/acetone).
9. Symmetry: Both CH₃ groups are identical → same NMR signal.

What we did and why: - MS gave molecular formula and CH₃ loss (base peak = 43). - IR confirmed C=O (no O-H). - NMR showed two identical CH₃ groups (6H singlet). - DU = 1 → Double bond (C=O).

---

Example 3 – Exam-Style: Unknown Compound (C₄H₈O₂)

Given: - Mass Spec: M⁺ = 88, M+1 = 3.3% of M⁺, base peak = 43. - IR: Sharp peak at 1740 cm⁻¹, broad peak at 3000 cm⁻¹. - NMR: 3H (t, 1.2 ppm), 2H (q, 4.1 ppm), 3H (s, 2.0 ppm).

Step-by-Step Solution:
1. Molecular formula from MS: - M⁺ = 88 → C₄H₈O₂ (mass = 88). - M+1 = 3.3% → (3.3/1.1) ≈ 3 carbons (but formula has 4 → likely 1³C isotope). - Base peak = 43 → Loss of COOH (45) or C₃H₇ (43).

1. Degrees of unsaturation (DU): [ \text{DU} = \frac{2(4) + 2 - 8}{2} = 1 \quad \text{(1 double bond or ring)} ]

2. IR analysis:

3. 1740 cm⁻¹ → C=O (ester/carboxylic acid).

4. 3000 cm⁻¹ (broad) → O-H (carboxylic acid) or C-H (alkane).

5. NMR analysis:

6. 3H (t, 1.2 ppm): CH₃ next to CH₂ (triplet = 2 adjacent H).
7. 2H (q, 4.1 ppm): CH₂ next to CH₃ (quartet = 3 adjacent H).

8. 3H (s, 2.0 ppm): CH₃ next to C=O (singlet = no adjacent H).

9. Structure:

10. CH₃-CH₂-COOH (propanoic acid) → No (NMR has 3H singlet).

11. CH₃-CO-O-CH₂-CH₃ (ethyl acetate).

    • CH₃ (s, 2.0 ppm): Next to C=O (no splitting).
    • CH₂ (q, 4.1 ppm): Next to O (deshielded).
    • CH₃ (t, 1.2 ppm): Next to CH₂.

12. Verify:

13. MS: M⁺ = 88 (C₄H₈O₂), base peak = 43 (CH₃CO⁺).
14. IR: 1740 cm⁻¹ (ester C=O), 3000 cm⁻¹ (C-H).
15. NMR: Matches ethyl acetate.

What we did and why: - MS gave formula and fragmentation clues (base peak = 43 → CH₃CO⁺). - IR suggested ester (1740 cm⁻¹, no broad O-H). - NMR showed CH₃ (s) next to C=O and CH₂-CH₃ group. - DU = 1 → C=O double bond.

---

COMMON MISTAKES

MISTAKE	WHY IT HAPPENS	CORRECT APPROACH
Ignoring M+1 peak	Forgetting to check carbon count.	Always calculate carbons from M+1/M⁺ ratio.
Misidentifying IR peaks	Confusing C=O (1700) with C=C (1650).	MEMORISE key ranges (C=O is higher than C=C).
Wrong NMR splitting	Applying n+1 rule to OH or NH protons.	OH/NH protons don’t split (broad singlets).
Forgetting degrees of unsaturation	Missing rings/double bonds.	Always calculate DU before proposing structures.
Overcomplicating symmetry	Assuming all CH₃ groups are identical.	Check NMR integration (e.g., 6H = two identical CH₃).

---

EXAM TRAPS

TRAP	HOW TO SPOT IT	HOW TO AVOID IT
"Hidden" functional groups	IR shows no O-H, but NMR has a broad 1H peak.	OH protons can be broad in NMR (e.g., carboxylic acids).
Isomeric structures	Two possible structures fit the data.	Check MS fragments (e.g., base peak = 43 → CH₃CO⁺, not CH₃CH₂⁺).
NMR integration errors	3H peak assumed to be CH₃, but could be 3 identical OH.	Cross-check with IR (e.g., no O-H peak? Not an alcohol).

---

1-MINUTE RECAP

"Here’s how to ace this in under 60 seconds:
1. Mass Spec first: Find M⁺ for molecular mass, use M+1 for carbons, and note fragments.
2. Calculate DU: Rings/double bonds? DU = 0? No unsaturation.
3. IR next: C=O? O-H? N-H? Cross out impossible groups.
4. NMR last: Match shifts to groups, count hydrogens (integration), and apply n+1 splitting.
5. Piece it together: Combine all clues—MS fragments, IR groups, NMR signals.
6. Double-check: Does your structure match all spectra? If not, try again!

Remember: Examiners love testing symmetry (identical CH₃ groups) and hidden functional groups (esters vs. acids). Practice with past papers—you’ve got this!"

---