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Organic Chemistry 2 > Dicarbonyl Compounds (Chapter 23) > Flashcards

Dicarbonyl Compounds (Chapter 23) Flashcards

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1
Q

Ester + Ester ⟶ β-Ketoester

A

Claisen Condensation

Irreversible (Cannot Undergo Retro-Claisen Condensation)

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2
Q

Ester’ + Ester’’ ⟶ β-Ketoester

A

Ester-Ester Mixed Claisen Condensation

Irreversible

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3
Q

Ketone + Ester ⟶ β-Diketone

A

Ketone-Ester Mixed Claisen Condensation

Irreversible

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4
Q

Reagents: Claisen Condensation

Starting Material = Ester (w/ two α-Hydrogens)

A
  1. NaOCH2CH3, HOCH2CH3
  2. H2O, H2SO4

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

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5
Q

Why must one ester possess two α-Hydrogens and the other ester possess no α-Hydrogens?

Ester-Ester Mixed Claisen Condensation

A
  • 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).

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6
Q

Why must the ester possessing no α-Hydrogens be in large excess?

Ester-Ester Mixed Claisen Condensation

A

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.

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7
Q

Reagents: Ester-Ester Mixed Claisen Condensation

Starting Materials = Two Unique Esters

A
  1. NaOCH2CH3
  2. 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).
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8
Q

Mechanism: Ketone-Ester Mixed Claisen Condensation

A
  1. The hydride reagent deprotonates the α-Hydrogen of the ketone (to form an enolate).
  2. The enolate nucleophile attacks the carbonyl Carbon of the ester (to form a tetrahedral intermediate).
  3. Rearrangement of the oxide’s π-electrons forms a ketone and eliminates an alkoxide.
  4. The intercarbonyl** α-Hydrogen is deprotonated** (to form the β-diketone enolate).
  5. Acid workup protonates the intercarbonyl α-Carbon to yield the β-Diketone product.
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9
Q

Why is the ketone α-Hydrogen deprotonated before the ester α-Hydrogen?

Ketone-Ester Mixed Claisen Condensation

A

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.

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10
Q

Mechanism: Ester-Ester Mixed Claisen Condensation

A
  1. The alkoxide reagent/catalyst deprotonates the α-Hydrogen of one ester (to form an ester enolate).
  2. The ester enolate nucleophile attacks the carbonyl Carbon of the other ester (to form a tetrahedral intermediate).
  3. Rearrangement of the oxide’s π-electrons forms a ketone and eliminates an alkoxide.
  4. The intercarbonyl α-Hydrogen is deprotonated (to form the β-ketoester enolate).
  5. Acid workup protonates the intercarbonyl α-Carbon to yield the β-Ketoester product.
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11
Q

Reagents: Ketone-Ester Mixed Claisen Condensation

Starting Materials = Ketone + Ester

A
  1. NaH
  2. H2O, H2SO4
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12
Q

Mechanism: Claisen Condensation

A
  1. Deprotonation of Ester α-Hydrogen to Yield Ester Enolate
  2. Nucleophilic Addition of Ester Enolate to Another Ester Carbonyl
  3. π-Electron Rearrangement to Eliminate Alkoxide Group
  4. Deprotonation of Intercarbonyl α-Hydrogen to Form β-Ketoester Enolate
  5. 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).

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13
Q

Driving Force: Claisen Condensation

A

Final Deprotonation to Yield β-Ketoester Enolate

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14
Q

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

A

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.

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15
Q

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

A

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

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16
Q

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

A

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.)

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17
Q

β-Ketoester ⟶ Ester + Ester

A

Retro-Claisen Condensation

Reversible

Retro-Claisen Condensation must occur under basic conditions.

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18
Q

β-Diketone ⟶ Ester + Ketone

A

Retro-Claisen Condensation

Reversible

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19
Q

Reagents: Retro-Claisen Condensation

Starting Material = β-Dicarbonyl (w/o α-Hydrogens)

A

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.)
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20
Q

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

A
  • 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.

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21
Q

Diester ⟶ Cyclic β-Ketoester

A

Intramolecular Claisen Condensation

Dieckmann Condensation

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

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22
Q

Why are Ketoesters unable to undergo Intramolecular Claisen Condensation?

A

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.

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23
Q

Mechanism: Intramolecular Claisen Condensation

Dieckmann Condensation

A
  1. The alkoxide reagent/catalyst deprotonates the α-Hydrogen of one ester (to form an ester enolate).
  2. The ester enolate attacks the carbonyl Carbon of the other ester (to form a five-membered/six-membered ring).
  3. Rearrangement of the oxide’s π-electrons forms a ketone and eliminates an alkoxide.
  4. The intercarbonyl α-Hydrogen is deprotonated (to form the cyclic β-ketoester enolate).
  5. Acid workup protonates the intercarbonyl α-Carbon to yield the cyclic β-Ketoester product.
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24
Q

Reagents: Intramolecular Claisen Condensation

Dieckmann Condensation

A
  1. NaOCH2CH3
  2. H2O, H2SO4
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25
Q

β-Dicarbonyl ⟶ Alkyl-Substitued β-Dicarbonyl

β-Dicarbonyl = β-Ketoester OR β-Diketone OR β-Diester

A

β-Dicarbonyl Monoalkylation

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26
Q

β-Dicarbonyl ⟶ Dialkyl-Substitued β-Ketoester

β-Dicarbonyl = β-Ketoester OR β-Diketone OR β-Diester

A

β-Dicarbonyl Dialkylation

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27
Q

Why are 2° Alkyl Halides able to be used in β-dicarbonyl alkylation reactions?

A

The alkoxide base (used to deprotonate the α-Hydrogen) is not present during the alkylation step, so there are no side reactions competing with alkoxide SN2 addition to the 2° alkyl halide.

3° alkyl halides cannot to be used in β-dicarbonyl alkylation reactions due the steric hindrances against SN2 attack.

28
Q

Why does E2-Elimination not occur during β-dicarbonyl alkylation reactions.

A

The β-dicarbonyl anion is a weak base, so it will preferentially perform SN2 addition to 0°/1°/2° alkyl halides.

E2-Elimination will occur when 3° alkyl halides are used since SN2 addition is not possible (due to steric hindrance).

29
Q

Reagents: β-Dicarbonyl Dialkylation

A
  1. NaOCH2CH3
  2. R—X
  3. KOC(CH3)3
  4. R—I

  • The alkyl halide (R—X) can be a 0°/1°/2° Alkyl Halide or an Allyl Halide. (2° Alkyl Halides can be used in β-dicarbonyl alkylation reactions since the alkoxide base is not present during the alkylation step.)
  • The more reactive alkyl iodide (R—I) is used during the second α-alkylation step (since this second step is less favorable).
30
Q

Reagents: β-Dicarbonyl Monoalkylation

A
  1. NaOCH2CH3
  2. R—X

The alkyl halide (R—X) can be a 0°/1°/2° Alkyl Halide or an Allyl Halide. (2° Alkyl Halides can be used in β-Dicarbonyl alkylation reactions since the alkoxide base is not present during the alkylation step.)

31
Q

Why does over-alkylation not occur during β-Dicarbonyl Alkylation?

A

The basic reagent/catalyst and the alkyl halide are reacted separately with the β-Dicarbonyl reagent.

Since α-Hydrogen deprotonation and alkylation occur in unique steps, only one alkyl group adds to the β-Dicarbonyl compound at a time.

32
Q

Why is hydroxide (OH–) not used as the basic reagent/catalyst?

β-Dicarbonyl Alkylation

A

Hydroxide could add to the ester’s carbonyl Carbon to facilitate ester hydrolysis (instead of β-Dicarbonyl Alkylation).

33
Q

NaOCH2CH3 vs. KOC(CH3)3

KOC(CH3)3 = Potassium t-Butoxide

A

KOC(CH3)3 is used as a stronger alkoxide base (than NaOCH2CH3) to promote the less reactive second α-Hydrogen deprotonation.

The Potassium t-Butoxide is paired with an alkyl iodie to drive the less-favorable second α-alkylation step.

34
Q

β-Dicarbonyl ⟶ β-1,5-Tricarbonyl

β-Dicarbonyl = β-Ketoester OR β-Diketone OR β-Diester

A

Michael Addition

Michael Addition with β-dicarbonyl compounds favors 1,4-addition.

35
Q

Reagents: Michael Addition

Starting Material = β-Dicarbonyl

A

α,β-Unsaturated Carbonyl, NaOCH2CH3 (Catalyst), CH3CH2OH

The Michael Addition reaction occurs under basic conditions only.

36
Q

Mechanism: Michael Addition

A
  1. The alkoxide catalyst deprotonates the intracarbonyl α-Hydrogen (to yield a β-dicarbonyl anion).
  2. The β-dicarbonyl anion adds to the α,β-unsaturated carbonyl’s β-Carbon.
  3. The oxide’s π-electrons rearrange to form a ketone while the new ketone’s α-Carbon becomes protonated (by the ethanol reagent).
37
Q

Which carboxylic acid derivates do organocuprates react with?

A

Acyl Halides

38
Q

β-1,5-Tricarbonyl ⟶ α,β-Unsaturated Cyclohexanone

A

Intramolecular Adol Condenation

Step 2 of Robinson Annulation

The Intramolecular Aldol Condensation reaction occurs under basic conditions and high temperatures only.

39
Q

β-Dicarbonyl ⟶ α,β-Unsaturated Cyclohexanone

A

Robinson Annulation

Michael Addition + Intramolecular Adol Condensation

The Robinson Annulation reaction occurs under basic conditions and high temperatures only.

40
Q

Robinson Annulation

A

Michael Addition + Intramolecular Adol Condensation

The Robinson Annulation reaction occurs under basic conditions and high temperatures only.

41
Q

Reagents: Robinson Annulation

Starting Material = β-Dicarbonyl

A
  1. α,β-Unsaturated Carbonyl, NaOCH2CH3, CH3CH2OH
  2. NaOCH2CH3, Δ
42
Q

Reagents: Intramolecular Aldol Condensation

Step 2 of Robinson Annulation

A

NaOCH2CH3, Δ

43
Q

Mechanism: Intramolecular Aldol Condensation

Step 2 of Robinson Annulation

A
  1. The terminal α-Carbon of the non-cyclic ketone is deprotonated (by the ethoxide base) to create a terminal enolate.
  2. The terminal enolate adds to the cyclic ketone’s carbonyl Carbon to form a six-membered ring.
  3. The oxide ion is protonated (by ethanol) to yield a hydroxyl-substituted tetrahedral intermediate.
  4. The α-Carbon of the (former enolate) ketone is deprotonated to yield a cyclic enolate.
  5. Rearrangement of the enolate’s π-electrons reforms the ketone and eliminates the hydroxyl group.

The terminal α-Carbon (of the non-cyclic ketone) must possess at least two α-Hydrogens.

44
Q

Mechanism: Robinson Annulation

A
  1. Deprotonation of β-Dicarbonyl’s α-Hydrogen (to Yield β-Dicarbonyl Anion)
  2. β-Dicarbonyl Anion Attacks the α,β-Unsaturated Carbonyl’s β-Carbon (to Yield α-Substituted β-Dicarbonyl)
  3. Rearrangement of Oxide’s π-Electrons and Protonation of Ketone’s α-Carbon (to Yield β-1,5-Tricarbonyl)
  4. Deprotonation of Noncylic Ketone’s Terminal α-Hydrogen (to Yield Terminal Enolate)
  5. Terminal Enolate Attacks Cyclic Ketone (to Yield Six-Membered Ring)
  6. Protonation of Oxide (to Yield Hydroxyl-Substituted Tetrahedral Intermediate)
  7. Deprotonation of Ketone’s Alcohol-Adjacent α-Carbon (to Yield Cyclic Enolate)
  8. Rearrangement of Enolate’s π-Electrons to Reform Ketone (and Eliminate Hydroxyl)

The terminal α-Carbon (Step 4) must possess at least two α-Hydrogens.

45
Q

Decarboxylation

A

The conversion of a carboxylic acid group to a Hydrogen that involves the elimination of CO2.

46
Q

β-Ketoester ⟶ Aldehyde

A

Hydrolysis-Decarboxylation

  • The Hydrolysis-Decarboxylation reaction can be acid-catalyzed OR base-catalyzed. (The Decarboxylation step must be catalyzed by strong acid and requires heat.)
47
Q

β-Diester ⟶ Carboxylic Acid

A

Hydrolysis-Decarboxlation

  • The Hydrolysis-Decarboxylation reaction can be acid-catalyzed OR base-catalyzed. (The Decarboxylation step must be catalyzed by strong acid and requires heat.)
  • The Hydrolysis step hydrolyzes both esters to carboxylic acids.
48
Q

β-Dicarboxylic Acid ⟶ Carboxylic Acid

A

Decarboxylation

  • Decarboxylation requires an acid catalyst (i.e. H2SO4) and heat to occur.
  • Only one of the carboxylic acids is decarboxylated to an aldehyde.
49
Q

Carboxylic Acid ⟶ Aldehyde

A

Decarboxylation

Decarboxylation requires an acid catalyst (i.e. H2SO4) and heat to occur.

50
Q

Mechanism: Decarboxylation

A
  1. Intramolecular proton transfer (from the enolate to the β-carbonyl) and cleavage of Cα—CCarbonyl bond eliminates CO2 and forms an enol.
  2. Tautomerization occurs to form the aldehyde product.
51
Q

Reagents: Decarboxylation

A

H2O, H2SO4, Δ

52
Q

Reagents: Acid-Catalyzed Hydrolysis-Decarboxylation

A

H2O, H2SO4, Δ

53
Q

Reagents: Base-Catalyzed Hydrolysis-Decarboxylation

A
  1. NaOH, H2O
  2. H2O, H2SO4
  3. H2O, H2SO4, Δ
54
Q

Why does the Hydroxide attack the ester carbonyl (instead of the ketone carbonyl) during base-catalyzed hydrolysis?

Hydrolysis of β-Ketoesters

A

The ester’s alkoxide group is able to serve as the leaving group upon Hydroxide addition to the ester’s carbonyl Carbon. (The ketone does not possess a substituent that could serve as a stable leaving group following Hydroxide addition to its carbonyl Carbon.)

55
Q

β-Ketoacid

A

Carboxylic acid compound with a ketone group at the β-Carbon position

56
Q

β-Ketoester

A

Ester compound with a ketone group at the β-Carbon position

57
Q

β-Ketoester ⟶ β-Ketoacid

A

Hydrolysis

Hydrolysis can be base-catalyzed OR acid-catalyzed.

58
Q

β-Diester ⟶ β-Dicarboxylic Acid

A

Hydrolysis

Hydrolysis can be base-catalyzed OR acid-catalyzed.

59
Q

Why must a carbonyl group be present at the β-Carbon of the carboxylic acid?

Decarboxylation

A

The Decarboxylation mechanism involves π-electron rearrangement/transfer across both carbonyl groups.

A simple carboxylic acid (i.e. without the β-position carbonyl group) is not compatible for decarboxylation reaction

60
Q

Malonic Ester

A
61
Q

Acetoacetic Ester

A
62
Q

Advantages of β-Dicarbonyl Anions

A
  • Strong Nucleophile (e.g. SN2, Michael Addition)
  • Weak Base (i.e. 1°/2° Alkyl Halide Compatibility)
63
Q

Uses of DIBAL Reduction

A
  • Partial Ester Reduction (Ester ⟶ Aldehyde)
  • Partial Amide Reduction (Amide ⟶ Aldehyde)
  • Partial Nitrile Reduction (Nitrile ⟶ Aldehyde)
64
Q

Uses of NaBH4 Reduction

A
  • Ketone Reduction (Ketone ⟶ 2° Alcohol)
  • Aldehyde Reduction (Aldehyde ⟶ 1° Alcohol)
  • Complete Acyl Halide Reduction (Acyl Halide ⟶ 1° Alcohol)
  • Complete Anhydride Reduction (Anhydride ⟶ 1° Alcohol)
65
Q

Uses of LiAlH4 Reduction

A
  • Complete Ester Reduction (Ester ⟶ 1° Alcohol)
  • Complete Amide Reduction (Amide ⟶ Amine)
  • Complete Nitrile Reduction (Nitrile ⟶ Amine)
  • Complete Carboxylic Acid Reduction (Carboxylic Acid ⟶ 1° Alcohol)
  • Complete Acyl Halide Reduction (Acyl Halide ⟶ 1° Alcohol)
66
Q

Uses of LiAl(OtBu)3H

A
  • Partial Acyl Halide Reduction (Acyl Halide ⟶ Aldehyde)
  • Partial Anhydride Reduction (Anhydride ⟶ Aldehyde)

The LiAl(OtBu)3H reduction reaction follows an addition-elimination mechanism.

Organic Chemistry 2 (12 decks)

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