Combined Techniques: Worked Problems II
This lesson extends the combined techniques approach to more complex molecules, including aromatic compounds and molecules with multiple functional groups. The worked examples here are more challenging, reflecting the standard expected in AQA A-Level exam questions.
Worked Example 5: Unknown Compound E (C₈H₉NO)
Given data:
Mass spectrum: M⁺˳ at m/z = 135. Base peak at m/z = 93. Other peaks at 77, 65, 43.
IR spectrum: Sharp absorption at 1685 cm⁻¹. Broad absorption at 3300–3500 cm⁻¹ (two peaks).
¹³C NMR: 6 peaks (including peaks near 120–140 ppm and one at ~168 ppm).
¹H NMR:
- δ = 2.2 ppm, singlet, 3H
- δ = 7.1–7.6 ppm, multiplet, 5H
- δ = 8.5 ppm, broad singlet, 1H (disappears with D₂O)
Step 1: Mr = 135. C₈H₉NO. Check: 8(12) + 9(1) + 14 + 16 = 96 + 9 + 14 + 16 = 135. ✔
Step 2: IHD = (16 + 2 + 1 − 9) / 2 = 5. Five degrees of unsaturation. Benzene ring (4) + one more.
Step 3: IR:
- 1685 cm⁻¹: C=O in the amide range (1630–1690 cm⁻¹).
- 3300–3500 cm⁻¹ with two peaks: N–H of a secondary amide (–CONHR) or primary amide (–CONH₂).
Wait — two N–H peaks typically indicate a primary amide (NH₂), but a secondary amide (–CONH–) shows only one N–H stretch. The ¹H NMR shows only 1H for the NH peak, so this is a secondary amide (–CONHR). The two IR peaks might be an overtone pattern or the Amide I and Amide II bands.
Step 4: ¹H NMR:
- δ = 2.2, singlet, 3H: CH₃ group. Singlet means no adjacent H. The chemical shift of 2.2 suggests it is adjacent to C=O or N.
- δ = 7.1–7.6, multiplet, 5H: monosubstituted benzene ring (C₆H₅).
- δ = 8.5, broad singlet, 1H, disappears with D₂O: N–H proton.
Step 5: Total H from NMR: 3 + 5 + 1 = 9. Matches C₈H₉NO.
Step 6: Assemble:
- C₆H₅– (phenyl) + –NHCO– (secondary amide) + CH₃ = C₆H₅NHCOCH₃
- Or: CH₃CONHC₆H₅ (N-phenylethanamide, also known as acetanilide).
Verification:
- MS: m/z = 93: loss of 135 − 93 = 42 (CH₂CO or ketene). Actually, m/z = 93 = C₆H₅NH₂⁺ (aniline cation plus H, m/z = 93) or C₆H₅NH⁺ (not quite right). Better: loss of COCH₃ (43) from 135 gives 92 (C₆H₅NH), but the peak is at 93. This could be [C₆H₅NH₂]⁺ (protonated amine fragment) = 93. ✔
- m/z = 77: C₆H₅⁺ (phenyl cation) ✔
- m/z = 43: CH₃CO⁺ (acetyl cation) ✔
- ¹³C: 6 peaks: CH₃ (~24 ppm), C=O (~168 ppm), and 4 aromatic carbons (C-1, C-2/C-6, C-3/C-5, C-4). ✔
Answer: N-phenylethanamide (acetanilide), C₆H₅NHCOCH₃.
Worked Example 6: Unknown Compound F (C₄H₈O)
Given data:
Mass spectrum: M⁺˳ at m/z = 72. Base peak at m/z = 57. Other peaks at 43, 29, 27.
IR spectrum: Strong absorption at 1715 cm⁻¹. No broad O–H.
¹H NMR:
- δ = 0.9 ppm, triplet, 3H
- δ = 1.6 ppm, sextet, 2H
- δ = 2.4 ppm, triplet, 2H
- δ = 9.8 ppm, triplet, 1H
Step 1: Mr = 72. C₄H₈O.
Step 2: IHD = (8 + 2 − 8) / 2 = 1. One C=O.
Step 3: IR at 1715 cm⁻¹: could be aldehyde (~1720–1740 cm⁻¹) or ketone (~1705–1725).
Step 4: ¹H NMR:
- δ = 9.8, triplet, 1H: aldehyde H. The triplet splitting is crucial — it means the CHO proton has 2 neighbouring H atoms. This identifies the compound as an aldehyde, not a ketone.
- δ = 2.4, triplet, 2H: CH₂ adjacent to both C=O and another CH₂. The chemical shift ~2.4 fits CH₂ next to C=O.
- δ = 1.6, sextet, 2H: CH₂ with 5 neighbours (2 from one side + 3 from other, n+1 = 6 → sextet).
- δ = 0.9, triplet, 3H: CH₃ with 2 neighbours.
Step 5: Chain: CH₃–CH₂–CH₂–CHO = butanal.
Verification:
- MS: m/z = 57: loss of 72 − 57 = 15 (CH₃). Fragment = C₃H₅O⁺ or CHO–CH₂–CH₂⁺. But m/z = 57 as the base peak is consistent with loss of CH₃. Actually, for butanal, loss of H (72−71=1) gives m/z = 71 (M−1, loss of H from CHO). The base peak at 57 represents loss of 15 from 72 (CH₃ loss) giving CH₂CH₂CHO⁺. This is plausible. Also m/z = 43 could be C₃H₇⁺ or CH₃CH₂CH₂⁺ (but that would be 43), and m/z = 44 would be CH₃CH₂CHO⁺ (44, the McLafferty rearrangement product).
- m/z = 29: CHO⁺ ✔
- m/z = 43: C₃H₇⁺ (propyl cation) or CH₃CO⁺ (43) → likely C₂H₃CO⁺ fragment.
Answer: Butanal, CH₃CH₂CH₂CHO.
Worked Example 7: Unknown Compound G (C₉H₁₀O₂)
Given data:
Mass spectrum: M⁺˳ at m/z = 150. Peaks at 122, 105, 77, 43.
IR spectrum: Strong absorption at 1745 cm⁻¹. No broad O–H.
¹³C NMR: 7 peaks (one at ~170 ppm, several between 125–135 ppm, one at ~21 ppm).
¹H NMR:
- δ = 2.3 ppm, singlet, 3H
- δ = 7.1 ppm, doublet, 2H
- δ = 7.4 ppm... wait, let me present this more carefully.
Let’s reconsider with a simpler aromatic ester. C₉H₁₀O₂: Mr = 150. IHD = (18 + 2 − 10)/2 = 5. Benzene (4) + C=O (1) = 5.
Revised ¹H NMR:
- δ = 2.3 ppm, singlet, 3H
- δ = 7.1–8.1 ppm, multiplet, 5H
- δ = 3.9 ppm, singlet...