Q1(e). Intramolecular aldol condensation

Answer in brief

Cyclization followed by dehydration gives the fused bicyclo[3.3.0] enone; the surviving carbonyl is the original side-chain carbonyl.

Key points

• Hydroxide removes an α-H on the butyl side of the side-chain carbonyl; the enolate attacks the ring carbonyl intramolecularly, forming a five-membered ring.
• Protonation of the alkoxide gives the fused β-hydroxy ketone; E1cB dehydration then gives the C=C conjugated with the C=O.
• The three carbons beyond the enolate carbon remain a propyl substituent on the new enone. All eleven substrate carbons are retained; only water is eliminated.
• The new ring shares the original ring edge. It is a fused five/five system, not a six-membered Robinson-annulation product.