Question

12. In a synthesis of isopropyl pyridine, the 3-methyl-2- presence of sodium ethoxide to give a dicarbonyl compo NaOCH.CH opropyl pyridine, the 3 with ethyl formate in the 0 ethyl formate 3-methyl-2-butanone a) In the first step of the mechanism, sodiu alpha position to form an enolate. Complete the as necessary (4 points) step of the mechanism, sodium ethoxide deprotonates 3-methyl-2-buthone at the O form an enolate, Complete this reaction with curly arrows drawing in H atoms b) Why is it necessary to use sodium ethoxide and not sodium hydroxide for this reaction? points) c) Show, using curly arrows, the reaction of the enolate with ethyl formate to form the product (8 points) d) Why is the final product the sodium salt of this ester? (4 points)

Answer 1

12 a 120 Mechanism Mechanism O3 Nat Step I NO Hq Enolaté nguy Formation + CH Step-2 Nucleophilic addition + CH₃-CH2-0 y + Cay-cu-o step-3 Elimination voje = le + Cub Chao -> Step-4 Deprotonation + EłoH Step-1, step-2 and step 3 are reversible and stepy is irreversible step. Step- 4 drives the equilibriem towards product formation. 6) Sodium Hydroxide is not used as a base in this reaction because it will hydrolyze the Ethyl formate ester Csaponification, there by removing it from reaction medium coh Naot H & Nao Bit Nao LH Mechanism: Follow step-2 to step-4 removes from medium The hydrogen at the centre of dicarbonyl compound is very acidic (pkan 11 less than ethanol (apka 15.9). Thus formation of enolate is very fast and energetically favorable- Except step-4 all are reversible steps. Hence, step- 4 is the drivere of the shift of equilibrium towarcels product. The products then coming out from the medium. Hence Sodium salt,

The Mechanism is drawn in the attached image. The relevant explanations of question (b) and (d) are also written.