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Dicarbonyl Compounds (Chapter 23)

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Total cards: 24•Content version: v4•Public deck•Updated: 8/1/2026

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#1
Front (Question)

Ester + Ester ⟶ β-Ketoester

Back (Answer)

Claisen Condensation Irreversible (Cannot Undergo Retro-Claisen Condensation)

#2
Front (Question)

Ester’ + Ester’’ ⟶ β-Ketoester

Back (Answer)

Ester-Ester Mixed Claisen Condensation Irreversible

#3
Front (Question)

Ketone + Ester ⟶ β-Diketone

Back (Answer)

Ketone-Ester Mixed Claisen Condensation Irreversible

#4
Front (Question)

Reagents: Claisen Condensation Starting Material = Ester (w/ two α-Hydrogens)

Back (Answer)

• NaOCH2CH3, HOCH2CH3 • H2O, H2SO4 The ester reagent must possess two α-Hydrogens. (Ester reagents with zero/one α-Hydrogen will not undergo Claisen Condensation.)

#5
Front (Question)

Why must one ester possess two α-Hydrogens and the other ester possess no α-Hydrogens? Ester-Ester Mixed Claisen Condensation

Back (Answer)

• One ester must possess no α-Hydrogens to ensure it is not deprotonated for form an ester enolate. • One ester must possess two α-Hydrogens to ensure it is deprotonated to form an ester enolate. Only one ester should become the ester enolate. The ester possessing no α-Hydrogens should be in large excess to ensure it is preferentially attacked by the ester enolate (in favor of self-Claisen Condenation).

#6
Front (Question)

Why must the ester possessing no α-Hydrogens be in large excess? Ester-Ester Mixed Claisen Condensation

Back (Answer)

The β-ketoester product will be the major product only if the non-α-Hydrogen ester is in large excess. Without large excess of the non-α-Hydrogen ester, the self-Claisen Condensation produce may become the major product.

#7
Front (Question)

Reagents: Ester-Ester Mixed Claisen Condensation Starting Materials = Two Unique Esters

Back (Answer)

• NaOCH2CH3 • H2O, H2SO4 • One ester must possess two α-Hydrogens, while the other ester must possess no α-Hydrogens. • The ester possessing no α-Hydrogens must be in large excess (to ensure that it is attacked by the ester enolate nucleophile).

#8
Front (Question)

Mechanism: Ketone-Ester Mixed Claisen Condensation

Back (Answer)

• The hydride reagent deprotonates the α-Hydrogen of the ketone (to form an enolate). • The enolate nucleophile attacks the carbonyl Carbon of the ester (to form a tetrahedral intermediate). • Rearrangement of the oxide’s π-electrons forms a ketone and eliminates an alkoxide. • The intercarbonyl** α-Hydrogen is deprotonated** (to form the β-diketone enolate). • Acid workup protonates the intercarbonyl α-Carbon to yield the β-Diketone product.

#9
Front (Question)

Why is the ketone α-Hydrogen deprotonated before the ester α-Hydrogen? Ketone-Ester Mixed Claisen Condensation

Back (Answer)

The ketone α-Hydrogen is more acidic (than the ester α-Hydrogen) due to the ester’s electron-donating ether Oxygen. The ester’s ether Oxygen destabilizes the ester enolate (i.e. the conjugate base) due its electron-donating character.

#10
Front (Question)

Mechanism: Ester-Ester Mixed Claisen Condensation

Back (Answer)

• The alkoxide reagent/catalyst deprotonates the α-Hydrogen of one ester (to form an ester enolate). • The ester enolate nucleophile attacks the carbonyl Carbon of the other ester (to form a tetrahedral intermediate). • Rearrangement of the oxide’s π-electrons forms a ketone and eliminates an alkoxide. • The intercarbonyl α-Hydrogen is deprotonated (to form the β-ketoester enolate). • Acid workup protonates the intercarbonyl α-Carbon to yield the β-Ketoester product.

#11
Front (Question)

Reagents: Ketone-Ester Mixed Claisen Condensation Starting Materials = Ketone + Ester

Back (Answer)

• NaH • H2O, H2SO4

#12
Front (Question)

Mechanism: Claisen Condensation

Back (Answer)

• Deprotonation of Ester α-Hydrogen to Yield Ester Enolate • Nucleophilic Addition of Ester Enolate to Another Ester Carbonyl • π-Electron Rearrangement to Eliminate Alkoxide Group • Deprotonation of Intercarbonyl α-Hydrogen to Form β-Ketoester Enolate • Acid Workup to protonate Intercarbonyl α-Carbon. The final β-Ketoester product is stable under weakly acidic conditions (i.e. cannot under retro-Claisen Condensation).

#13
Front (Question)

Driving Force: Claisen Condensation

Back (Answer)

Final Deprotonation to Yield β-Ketoester Enolate

#14
Front (Question)

Why is the final deprotonated step of Claisen Condensation highly favorable?

Back (Answer)

The intercarbonyl α-Hydrogens are significanly more acidic than the hydroxyl Hydrogens (of the basic catalyst’s conjugate acid), so α-Hydrogen deprotonation (to form the hydroxyl group on the catalytic base) readily occurs.

#15
Front (Question)

Why are ester reagents with zero α-Hydrogens unable to undergo Claisen Condensation?

Back (Answer)

The ester enolate (formed via or α-Hydrogen deprotonation) cannot be synthesized when no α-Hydrogen is present.

#16
Front (Question)

Why are ester reagents with one α-Hydrogen unable to undergo Claisen Condensation?

Back (Answer)

A second α-Hydrogen is required for the final/second deprotonation step that “drives” the Claisen Condensation reaction toward the β-Ketoester enolate product. WIthout a second α-Hydrogen, the formed β-Ketoester cannot be deprotonated within the basic Claisen Condensation conditions. (The non-deprotonated β-Ketoester is highly unstable under basic conditions, so the overall condensation reaction is thermodynamically unfavorable.)

#17
Front (Question)

β-Ketoester ⟶ Ester + Ester

Back (Answer)

Retro-Claisen Condensation Reversible Retro-Claisen Condensation must occur under basic conditions.

#18
Front (Question)

β-Diketone ⟶ Ester + Ketone

Back (Answer)

Retro-Claisen Condensation Reversible

#19
Front (Question)

Reagents: Retro-Claisen Condensation Starting Material = β-Dicarbonyl (w/o α-Hydrogens)

Back (Answer)

NaOCH2CH3 • The reagent β-dicarbonyl can be either a β-Ketoester OR a β-Diketone. • No α-Hydrogens can be present on the reagent β-dicarbonyl to ensure alkoxide addition to a carbonyl Carbon. (f the reagent β-dicarbonyl possesses α-Hydrogens, the alkoxide will facilitate deprotonation to form a β-Dicarbonyl enolate.)

#20
Front (Question)

Why is the self-Aldol Condensation of ketones thermodynamically unfavorable?

Back (Answer)

• The carbonyl Carbons of ketones are relatively stable (which decreases their susceptibility to nucleophilic attack by an enolate). • The carbonyl Carbons of ketones experience steric hindrance (which inhibits attack by the nucleophilic enolate). Self-Aldol Condensation of ketones does not occur.

#21
Front (Question)

Diester ⟶ Cyclic β-Ketoester

Back (Answer)

Intramolecular Claisen Condensation Dieckmann Condensation The Intramolecular Claisen Condensation reaction is effective for forming five-membered/six-membered rings.

#22
Front (Question)

Why are Ketoesters unable to undergo Intramolecular Claisen Condensation?

Back (Answer)

The ketone’s alkyl substituent cannot serve as a leaving group (following intramolecular attack of the ester enolate’s α-Carbon). Since the ketone’s alkyl substituent cannot leave, the reformation of the ketone group (C=O) cannot occur.

#23
Front (Question)

Mechanism: Intramolecular Claisen Condensation Dieckmann Condensation

Back (Answer)

• The alkoxide reagent/catalyst deprotonates the α-Hydrogen of one ester (to form an ester enolate). • The ester enolate attacks the carbonyl Carbon of the other ester (to form a five-membered/six-membered ring). • Rearrangement of the oxide’s π-electrons forms a ketone and eliminates an alkoxide. • The intercarbonyl α-Hydrogen is deprotonated (to form the cyclic β-ketoester enolate). • Acid workup protonates the intercarbonyl α-Carbon to yield the cyclic β-Ketoester product.

#24
Front (Question)

Reagents: Intramolecular Claisen Condensation Dieckmann Condensation

Back (Answer)

• NaOCH2CH3 • H2O, H2SO4