Q4(d). Structure of C₅H₁₀O from UV, IR, NMR and mass spectra
An organic compound (MF C₅H₁₀O) exhibits the following spectral characteristics:
UV (λ_(max): 280 nm)
IR ν_(max): 1715 cm⁻¹
¹H NMR (δ, ppm): 0.90 (3H, t), 1.60 (2H, m), 2.20 (3H, s), 2.40 (2H, t)
Mass (m/z): 86, 71, 58, 43 (100%)
Deduce the structure of the compound and explain the principal ions observed in the mass spectrum.
Answer in brief
The compound is pentan-2-one, CH₃COCH₂CH₂CH₃. MS: 86 (M⁺.), 71 (M - CH₃, acylium), 58 (McLafferty, loss of ethene), 43 (CH₃CO⁺, base peak).
Key points
- C₅H₁₀O; degrees of unsaturation = (2×5+2-10)/2 = 1: one C=O or one C=C or a ring.
- a weak band at 280 nm is the n→π* band of a saturated aldehyde or ketone (no conjugation).
- 1715 cm⁻¹ is the C=O stretch of a saturated, open-chain ketone (an aldehyde would show 1725–1730 cm⁻¹ and C–H bands near 2720 and 2820 cm⁻¹; there is no O–H).
- 2.20 (3H, s) is CH₃-CO (no neighbouring H); 2.40 (2H, t) is -CO-CH₂-CH₂⁻; 1.60 (2H, m) is -CH₂-CH₂-CH₃ (coupled to 2 + 3 H); 0.90 (3H, t) is the terminal -CH₂-CH₃. Integration 3: 2: 2: 3 = 10 H.
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
Introduction. Each spectrum gives one piece: the formula gives the unsaturation, UV and IR the functional group, NMR the carbon chain, and MS the molar mass and how the molecule breaks.
Deduction
- Formula: C₅H₁₀O; degrees of unsaturation = (2×5+2-10)/2 = 1: one C=O or one C=C or a ring.
- UV: a weak band at 280 nm is the n→π* band of a saturated aldehyde or ketone (no conjugation).
- IR: 1715 cm⁻¹ is the C=O stretch of a saturated, open-chain ketone (an aldehyde would show 1725–1730 cm⁻¹ and C–H bands near 2720 and 2820 cm⁻¹; there is no O–H).
- NMR: 2.20 (3H, s) is CH₃-CO (no neighbouring H); 2.40 (2H, t) is -CO-CH₂-CH₂⁻; 1.60 (2H, m) is -CH₂-CH₂-CH₃ (coupled to 2 + 3 H); 0.90 (3H, t) is the terminal -CH₂-CH₃. Integration 3: 2: 2: 3 = 10 H.
- Assembly: CH₃-CO-CH₂-CH₂-CH₃, pentan-2-one (methyl propyl ketone). Pentan-3-one would show only a quartet and a triplet; pentanal would show a CHO signal near δ 9.8.
Principal ions in the mass spectrum
Scheme: McLafferty rearrangement: a γ-H moves to O through a six-membered transition state and ethene is lost.
- m/z 86: molecular ion [CH₃COCH₂CH₂CH₃]⁺. (C₅H₁₀O, 86).
- m/z 71: α-cleavage on the methyl side, loss of CH₃• (15): butanoyl acylium CH₃CH₂CH₂C#O⁺ (C₄H₇O⁺).
- m/z 58: McLafferty rearrangement: a γ-H (on the terminal CH₃, C-5) transfers to the carbonyl O through a six-membered transition state, the C3–C4 (α–β) bond breaks and ethene (28) is lost, leaving the enol radical cation [CH₂=C(OH)CH₃]⁺. (C₃H₆O). An even mass from an even-mass ion is the sign of a rearrangement.
- m/z 43 (base peak): α-cleavage on the propyl side, loss of C₃H₇• (43): the acetyl cation CH₃C#O⁺, stabilised as an acylium ion. Cleavage that expels the larger radical is preferred, so 43 is stronger than 71.
Conclusion. The compound is pentan-2-one, CH₃COCH₂CH₂CH₃. MS: 86 (M⁺.), 71 (M - CH₃, acylium), 58 (McLafferty, loss of ethene), 43 (CH₃CO⁺, base peak).
Application. The McLafferty ion at m/z 58 is used to recognise methyl ketones with a γ-hydrogen; pentan-2-one itself is a flavour component and a solvent.
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All parts of Question 4
- 4(a) Which dienes cannot take part in a Diels–Alder reaction10 marks
- 4(b) Two C₇H₁₄ cycloalkanes from their base peaks10 marks
- 4(c) Maleic anhydride: fast [4+2] with butadiene, no thermal [2+2] with ethene15 marks
- 4(d) Structure of C₅H₁₀O from UV, IR, NMR and mass spectra15 marks