When a bulky base, such as LDA, is used, it will almost always deprotonate the least hindered position. In this case, the reaction is performed under low temperatures (to prevent the thermodynamic product from forming) and we get the kinetic product. When we use a strong base that is not bulky, it will favor deprotonating the position that will provide the most stable product at equilibrium. As a result, the enolate that forms when using NaH or NaOEt is more substituted and more stable.
23-3 When acetophenone is halogenated at the α-position, the α-carbon becomes more acidic as a result of the electron-withdrawing halogen. This makes it more likely to go through α-halogenation again, until it no longer has any α-protons. When the base performs a nucleophilic attack on the ketone, the triiodomethyl group becomes a good leaving group. The resulting products are sodium benzoate and triiodomethane.
Enols and Enolate Ions
Formation and Alkylation of Enamines
Alpha Halogenation of Ketones
Alpha Bromination of Acids: The HVZ Reaction
The Aldol Condensation of Ketones and Aldehydes
Dehydration of Aldol Products
Crossed Aldol Condensations
Several answers possible. One plausible method of synthesis:
Syntheses Using β-Dicarbonyl Compounds
Possible route of synthesis:
Conjugate Additions: The Michael Reaction
Michael acceptors are generally α,β-unsaturated ketones; however, aldehydes or acid derivatives can also function as acceptors. Michael donors are weak bases/strong nucleophiles.
The Robinson Annulation