Q8(c). Claisen condensation from IR; structure of a monoterpene from ozonolysis
(i) The major product formed on ethyl propionate with KOH in ethanol gives a strong C=O stretching frequency at 1745 cm⁻¹ and 1725 cm⁻¹ in IR spectrum. Identify the major product and suggest the suitable mechanism. (ii) The unknown monoterpene on complete reductive ozonolysis with O₃ and Me₂S gave equimolar mixture of formaldehyde, acetone, pyruvaldehyde and malondialdehyde. Draw the structure of all the products and deduce the structure of the monoterpene.
Answer in brief
(i) Two C=O bands (ester 1745, ketone 1725) fit the β-keto ester ethyl 2-methyl-3-oxopentanoate, from Claisen condensation. (ii) The monoterpene is β-ocimene, (CH₃)₂C=CH-CH₂-CH=C(CH₃)-CH=CH₂ (3,7-dimethylocta-1,3,6-triene).
Key points
- The base (EtO⁻, present in KOH/EtOH) removes an α-H of ethyl propionate: the ester enolate (pKₐ≈25, formed in low equilibrium concentration).
- The enolate adds to the C=O of a second ester molecule: tetrahedral intermediate; loss of EtO⁻ gives ethyl 2-methyl-3-oxopentanoate.
- The product has an acidic doubly activated H (pKₐ≈11): it is deprotonated by the base. This irreversible step drives the equilibrium; acid work-up gives the β-keto ester.
- ester C=O 1745 and ketone C=O 1725 cm⁻¹: two bands, as observed. (Simple saponification would give EtCO₂⁻ with bands near 1560 and 1410 cm⁻¹.)
The full model answer
Worked out step by step, as it should go in the answer booklet. The solution sheet follows it.
(i) Claisen condensation (5 marks)
- The base (EtO⁻, present in KOH/EtOH) removes an α-H of ethyl propionate: the ester enolate (pKₐ≈25, formed in low equilibrium concentration).
- The enolate adds to the C=O of a second ester molecule: tetrahedral intermediate; loss of EtO⁻ gives ethyl 2-methyl-3-oxopentanoate.
- The product has an acidic doubly activated H (pKₐ≈11): it is deprotonated by the base. This irreversible step drives the equilibrium; acid work-up gives the β-keto ester.
- IR: ester C=O 1745 and ketone C=O 1725 cm⁻¹: two bands, as observed. (Simple saponification would give EtCO₂⁻ with bands near 1560 and 1410 cm⁻¹.)
(ii) Monoterpene (5 marks)
- C₁₀: 1 + 3 + 3 + 3 = 10 carbons, and 6 carbonyl carbons in the 4 products (formaldehyde 1, acetone 1, pyruvaldehyde 2, malondialdehyde 2) ⇒ three C=C (triene), no ring.
- Join the fragments at the carbonyl carbons (each pair of carbonyl carbons was a C=C): H₂C= (from HCHO); =C(CH₃)₂ (from acetone); =CH–CH₂–CH= (malondialdehyde); CH₃–C(=)–CH= (pyruvaldehyde).
- Assembly: (CH₃)₂C=CH-CH₂-CH=C(CH₃)-CH=CH₂ = 3,7-dimethylocta-1,3,6-triene (β-ocimene). Check: C₁₀H₁₆.
- Ozonolysis of the three C=C: Me₂C=CH– gives acetone + OHC-CH₂-CHO; –CH=C(CH₃)– gives CH₃COCHO; –CH=CH₂ gives HCHO.
CoversClaisen condensation; terpenesClaisen condensation from IRstructure of a monoterpene from ozonolysis
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