[8+2] cycloaddition of 2H-cyclohepta[b]furan-2-one with a silyl enol ether: azulene synthesis

Question

2H-Cyclohepta[b]furan-2-one is heated with 2-(trimethylsilyloxy)but-1-ene. Draw the complete mechanism for the formation of the azulene product, showing (a) the thermal [8+2] cycloaddition and the bridged lactone it forms, (b) loss of CO₂ to give a dihydroazulene, and (c) loss of trimethylsilanol (TMSOH) to give the final aromatic product. Identify the product, explain the regiochemistry of the cycloaddition, and explain what drives the sequence.

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[8+2] cycloaddition of 2H-cyclohepta[b]furan-2-one with a silyl enol ether: azulene synthesis — question structures

Expert answer

This is a thermal [8+2] cycloaddition, followed by two eliminations. The overall sequence is driven by loss of CO₂ and by the formation of an aromatic azulene. Step 1, [8+2] cycloaddition: The cycloheptafuranone acts as the 8π component. The eight carbons C3 to C8a carry four conjugated C=C bonds, and the silyl enol ether supplies the 2π component. The cycloaddition is concerted and suprafacial, with 10 electrons (4n+2), so the thermal reaction is allowed and the cyclic transition state is aromatic in character. Two new σ-bonds form at C3 and C8a. This closes a new five-membered ring and leaves a bridged lactone, with the –O–C(=O)– unit spanning C3 and C8a. Regiochemistry: The terminal CH₂ of the enol ether is the nucleophilic carbon, because the oxygen lone pair donates into the C=C. It bonds to the electrophilic C8a, which carries the ring oxygen. The carbon bearing OTMS and ethyl bonds to C3. This places the ethyl group at C2 of the final azulene. Step 2, loss of CO₂: The bridged lactone is strained and undergoes a retro-cycloaddition type fragmentation. The C3–C2(=O) and C8a–O bonds break, and CO₂ is expelled as a gas. This step is effectively irreversible. CO₂ is very stable, it leaves the system, and the loss restores a conjugated seven-membered ring, giving a dihydroazulene. Step 3, loss of TMSOH and aromatisation: The dihydroazulene has OTMS and ethyl on one sp³ carbon next to a CH₂. Elimination of TMSOH (OTMS from one carbon and H from the adjacent CH₂) forms the final C=C bond. This completes the fully conjugated 10π-electron bicyclic system, azulene. Aromatic stabilisation is the thermodynamic driving force for this step, even though TMSO⁻ is only a modest leaving group. Product: 2-Ethylazulene. Driving forces, in short: (i) loss of CO₂ makes the second step irreversible, and (ii) aromatisation of the azulene makes the final elimination favourable.