Q8(a). Keto–enol tautomers of acetylacetone from ¹H-NMR
In a solution, acetylacetone exhibits keto–enol tautomerism. The chemical shift (δ in ppm) and their intensity observed in the ¹H-NMR of acetylacetone in CDCl₃ is: 5.52 (360), 3.61 (90), 2.24 (270), 2.04 (2160). [O–H signal of the enol tautomer is not shown] (i) Assign the signals to the corresponding hydrogen in the tautomers. (ii) Calculate the ratio of keto–enol tautomers and identify the major tautomer present in the solution.
Answer in brief
(i) 5.52: enol vinyl =CH; 3.61: keto CH₂; 2.24: keto CH₃; 2.04: enol CH₃. (ii) keto: enol =1:8; the enol (89 %) is major.
Key points
- (i) Assignment (4 marks): keto form has two kinds of H: CH₂ between two C=O (3.61, deshielded by two carbonyls) and two CH₃ (2.24). Enol form: =CH at 5.52 (vinyl), CH₃ at 2.04; the enolic O–H (very deshielded, δ≈15.5,…
- (ii) Ratio (4 marks): divide each integral by the number of H: enol =360 per H, keto =45 per H. ([enol])/([keto])=(360)/(45)=8:keto=8:1 (or 360:90/2); % enol=8/9×100=88.9 %
- a six-membered ring is formed by the strong intramolecular O–H…O hydrogen bond, and the C=C–C=O unit is conjugated; in a non-polar solvent (CDCl₃) the less polar chelated enol is favoured (in water the keto form increases).
The full model answer
Worked out step by step, as it should go in the answer booklet. The solution sheet follows it.
CH₃-CO-CH₂-CO-CH₃ ⇌ CH₃-CO-CH=C(OH)-CH₃
| δ | intensity | assignment | H count | per H |
| 5.52 | 360 | enol =CH– (vinyl) | 1 | 360 |
| 3.61 | 90 | keto -CH₂⁻ | 2 | 45 |
| 2.24 | 270 | keto CH₃ (two equivalent) | 6 | 45 |
| 2.04 | 2160 | enol CH₃ (two equivalent, averaged by fast H-shift) | 6 | 360 |
- (i) Assignment (4 marks): keto form has two kinds of H: CH₂ between two C=O (3.61, deshielded by two carbonyls) and two CH₃ (2.24). Enol form: =CH at 5.52 (vinyl), CH₃ at 2.04; the enolic O–H (very deshielded, δ≈15.5, intramolecular H-bond) is not shown.
- (ii) Ratio (4 marks): divide each integral by the number of H: enol =360 per H, keto =45 per H. ([enol])/([keto])=(360)/(45)=8:keto=8:1 (or 360:90/2); % enol=8/9×100=88.9 %
- Why the enol dominates (2 marks): a six-membered ring is formed by the strong intramolecular O–H…O hydrogen bond, and the C=C–C=O unit is conjugated; in a non-polar solvent (CDCl₃) the less polar chelated enol is favoured (in water the keto form increases).
Coverstautomer ratioKeto–enol tautomers of acetylacetone from ¹H-NMR
All parts of Question 8
- 8(a) Keto–enol tautomers of acetylacetone from ¹H-NMR10 marks
- 8(b) π-MOs of butadiene; reactivity of dienes in the Diels–Alder reaction10 marks
- 8(c) Claisen condensation from IR; structure of a monoterpene from ozonolysis10 marks