Carboxylic Acid Derivatives (Chapter 20) Flashcards

Question

*Mechanism:* Acyl Halide Anhydride Synthesis

Answer 1

1. The carbonyl Oxygen (of the carboxylic acid) **attacks the carbonyl Carbon** (of the acyl halide). 2. An **intramolecular proton transfer** occurs to protonate the Oxygen of the halide-bonded Carbon. 3. π-electron rearrangement forms an oxacarbenium intermediate and **eliminates the halide**. 4. Deprotonation of the oxacarbenium Oxygen **forms an anhydride compound**.

Answer 2

Acyl Halide Anhydride Synthesis ## Footnote The Carboxylic Acid Dehydration reaction can also be used to synthesize anhydrides (from carboxylic acids), but the mechanism is **less efficient** than Acyl Halide Anhydride Synthesis.

Answer 3

R—O—OH, Δ | R—O—OH = Carboxylic Acid

Answer 4

Carboxylic acids are *less reactive* (i.e. less susceptible to nucleophilic attack) than acyl halides in addition-elimination mechanisms, so carboxylic acid dehydration occurs less readily. ## Footnote Acyl halides are very reactive in addition-elimination mechanisms due to the carbonyl Carbon's highly electrophilic character.

Answer 5

Organocuprate Ketone Synthesis

Answer 6

(Alkenyl) Organocuprate Ketone Synthesis

Answer 7

R₂CuLi | R₂CuLi = Organocuprate ## Footnote Acyl halide ketone synthesis with an *alkenyl organocuprate* will yield an **ɑ,β-unsaturated ketone**.

Answer 8

Hard Organometallic Acyl Halide Addition

Answer 9

1. R—MgBr / R—Li 2. H₂SO₄, H₂O

Answer 10

The *hard* organometallic reagents are **highly nucleophilic** (i.e. more nucleophilic than organocuprate reagents), so they will add to the acyl chloride reagent *and* the ketone intermediate product. ## Footnote Two equivalents of the *hard* organometallic reagent are consumed during Hard Organometallic Acyl Halide Addition since the organometallic is reactive enough to add to the less-electrophilic carbonyl Carbon of the ketone intermediate.

Answer 11

* R—MgBr (Grignard) * R—Li (Organolithium)

Answer 12

Acyl Halide Reduction

Answer 13

LiAl(OtBu)₃ | LiAl(OtBu)₃ = Lithium Tri-(*t*-Butoxy) Aluminum Hydride

Answer 14

## Footnote LiAl(OtBu)₃ is a *bulky* (i.e. less reactive) hydride reagent.

Answer 15

The LiAl(OtBu)₃ reductant is a *less reactive reducing agent*, so it is unable to add to the less-electrophilic aldehyde product.

Answer 16

The carbonyl Carbons of an anhydride are neighbored by a resonance-donating Oxygen atom. (The strong electron donation effect of the Oxygen reduces the carbonyl Carbon's electrophilic character.) | Less Electrophilic = Less Reactive

Answer 17

* Acyl Halides * Anhydrides

Answer 18

* **Leaving Group** The leaving group for Acyl Halide AE is a *halide anion*, whereas the leaving group for Anhydride AE is a *carboxylate anion*. * **Kinetics**: The Acyl Halide AE reaction is *fast*, whereas the Anhydride AE reaction is *slower*. * **Thermodynamics:** The Acyl Halide AE reaction is *highly exothermic*, whereas the Anhydride AE reaction is *slightly exothermic*. | AE = Addition-Elimination

Answer 19

Anhydride Hydrolysis ## Footnote The Anhydride Hydrolysis reaction creates **two carboxylic acid molecules** per every one anhydride molecule.

Answer 20

## Footnote (CH₃OC)—O—(COCH₃)

Answer 21

Non-Catalyzed Anhydride Esterification

Answer 22

* **Acyl Halide Esterification:** Reaction requires alchol (R—OH) *and* Triethylamine (N—Et₃) to occur. * **Anhydride Esterification:** Reaction requires *only* alcohol (R—OH) occur. ## Footnote In Acyl Halide Esterification, the Triethylamine must be added to neutralize the strongly acidic H—X byproduct. In Anhydride Esterification, a strongly acidic byproduct is *not formed*, so no Triethylamine is necessary.

Answer 23

The carboxylic acid byproduct is **not sufficiently electrophilic** to react with the alcohol reagent without an acid catalyst. ## Footnote An acid catalyst would activate the carboxylic acid byproduct to allow the alcohol reagent to add to the CA's carbonyl Carbon. (The result of this addition reaction is the formation of an ester.)

Answer 24

Acid-Catalyzed Anhydride Esterification ## Footnote The acid-catalyzed Anhydride Esterification reaction produces **two ester equivalents** per every one anhydride equivalent.

Answer 25

R—OH, H₂SO₄

Answer 26

The acid catalyst activates the (unobserved) carboxylic acid intermediate byproduct (by protonating the carbonyl Carbon) to allow the alcohol reagent to add to the CA's carbonyl Carbon. ## Footnote The *carboxylic acid* and *alcohol* addition reaction results in ester formation.

Answer 27

Acid-Catalyzed Anhydride Esterification ## Footnote Esterification from a *cyclic anhydride* requires the addition of **heat**.

Answer 28

Non-Catalyzed Anhydride Esterification ## Footnote Esterification from a *cyclic anhydride* requires the addition of **heat**.

Answer 29

* **Acyclic Anhydride:** Addition of heat is *not* required. * **Cyclic Anhydride:** Addition of heat is *required*.

Answer 30

Acyl Halides > Anhydrides > Esters > Amides ## Footnote **Addition-Elimination Mechanisms:** Acyl Halides react *very quickly*, Anhydrides react *slower*, and Esters react *slowly*, and Amides react *extremely slowly*.

Answer 31

Ester Hydrolysis | Reverisble ## Footnote * The Ester Hydrolysis reaction forms *carboxylic acid* and ***alcohol***. * The **addition of heat** and a **strong catalyst** are required for the reaction to occur.

Answer 32

Ester Hydrolysis | Reversible ## Footnote * The Ester Hydrolysis reaction forms *carboxylic acid* and ***alcohol***. * The **addition of heat** and a **strong catalyst** are required for the reaction to occur.

Answer 33

An excess of alcohol (i.e. a product of Ester Hydrolysis) induces the reformation of ester.

Answer 34

H₂O, H₂SO₄, Δ ## Footnote A *strong catalyst* (acid or base) is required for Ester Hydrolysis to occur.

Answer 35

H₂O, NaOH, Δ ## Footnote A *strong catalyst* (acid or base) is required for Ester Hydrolysis to occur.

Answer 36

* **Acyl Halide Hydrolysis:** *No* catalyst *nor* heating is required. * **Anhydride Hydrolysis:** *No* catalyst *nor* heating is required. * **Ester Hydrolysis:** *Strong catalyst* and *heating* are required. * **Amide Hydrolysis:** *Strong catalyst* and *heating* are required.

Answer 37

1. Protonation of Ester Carbonyl Carbon 2. Addition of H₂O to Ester Carbonyl Carbon 3. Intramolecular Proton Transfer to Protonate Ester Oxygen 4. Formation of Oxacarbenium Ion to Eliminate Alcohol 5. Deprotonation of Oxacarbenium Ion to Form Carboxylic Acid.

Answer 38

* **Leaving Group:** The acid-catalyzed mechanism creates an *alcohol leaving group*, wherease the base-catalyzed mechanism creates an *alkoxide anion leaving group*. * **Nucleophile:** The acid-catalyzed mechanism uses an *H₂O nucleophile*, wherease the base-catalyzed mechanism uses a *hydroxide nucleophile*. ## Footnote The *alkoxide leaving group* of the base-catalyzed mechanism is a **poor leaving group**. This alkoxide elimination is promoted by the *instability of the tetrahedral intermediate* and the *formation of a resonance-stabilized carboxylate anion*.

Answer 39

1. Attack of OH^– at Ester Carbonyl Carbon 2. π-Electron Rearrangement to Eliminate Alkoxide Anion 3. Protonation of Alkoxide to Form Carboxylate Anion

Answer 40

Leaving of the alkoxide anion (1) converts the *unstable tetrahedral intermediate to a carboxylic acid* and (2) results in a *final carboxylate anion*. ## Footnote * The tetrahedral intermediate is destabilized by steric crowding (electron-electron repulsion between the Oxygens) about the central Carbon. * The alkoxide anion becomes protonated (immediately following its elimination) to form the more stable carboxylate anion.

Answer 41

Transesterification | Reverisble ## Footnote Transesterification is **reversible** via large excess of reagent (i.e. alcohol and/or ester).

Answer 42

Transesterification | Reversible ## Footnote Transesterification is *reversible* via large excess of reagent (i.e. alcohol and/or ester).

Answer 43

R—OH (Excess), H₂SO₄

Answer 44

1. The acid catalyst *protonates* the ester's carbonyl Carbon to "activate" the ester. 2. The alcohol reagent attacks the ester at the carbonyl Carbon to yield a *tetrahedral intermediate*. 3. The cationic alkoxy substituent's Hydrogen is transferred to the ether Oxygen. 4. π-electron rearrangement at the hydroxyl group creates an *oxacarbenium ion* and eliminates an *alcohol*. 5. Deprotonation of the oxacarbenium ion's Hydrogen yields an *ester*.

Answer 45

1. The alkoxide reagent attacks the ester's carbonyl Carbon to yield a *tetrahedral intermediate*. 2. π-electron rearrangement at the oxide group creates an *ester* and eliminates an *alkoxide anion*.

Answer 46

Nitrogen's π electrons are *more easily donated* (i.e. are *farther* from the Nitrogen nucleus) than Oxygen's π electrons, so N-nucleophiles are more reactive/nucleophilic that O-nucleophiles.

Answer 47

Ester Amidification ## Footnote * The Ester Amidification reaction requires **heat** to occur. * An **alcohol byproduct** is formed/eliminated during the reaction.

Answer 48

Amine, Δ ## Footnote Ester Amidification can occur *only* with **0°/1°/2° Amines**. (Reactions of esters with 3° Amines do NOT occur.)

Answer 49

Hard Organometallic Ester Addition ## Footnote The Hard Organometallic Ester Addition reaction requires **two equivalents of organometallic reagent** to complete the conversion to a 3° alcohol.

Answer 50

1. 2 R—MgBr / R—Li, THF 2. H₂O, H₂SO₄ | THF = Tetrahydrofuran ## Footnote The organometallic reagent must add to the ester/ketone *twice* to complete conversion to the 3° alcohol.

Answer 51

The ketone intermediate product is *more electrophilic* than the ester reagent, so the ketone will *readily be attacked* by the organometallic reagent. ## Footnote The resonance-donating Oyxgen of the ester reagent causes the ester to be *less electrophilic* than the ketone intermediate.

Answer 52

LiAlH₄ Reduction

Answer 53

1. LiAlH₄ 2. H₂O, H₂SO₄

Answer 54

Partial Reduction

Answer 55

1. DIBAL, (Low Temperature) 2. H₂O, H₂SO₄ | DIBAL = Diisobutylaluminum Hydride

Answer 56

Diisobutylaluminum Hydride ## Footnote DIBAL is a bulky (i.e. less reactive) hydride reagent that facilitates *partial* ester reduction.

Answer 57

1. A Hydride ion attacks the ester's carbonyl Carbon to yield a tetrahedral oxide intermediate. 2. π-Electron rearrangement at the oxide ion *and* elimination of an alkoxide ion yields an aldehyde product. 3. A Hydride ion attacks the aldehyde's carbonyl Carbon to yield a tetrahedral oxide intermediate. 4. Protonation of the oxide ion yields a 1° alcohol.

Answer 58

* Organocuprates * Carboxylic Acids

Answer 59

Carboxylic Acids

Answer 60

Amide Hydrolysis ## Footnote * The Amide Hydrolysis reaction requires **high temperatures** AND **strong acid/base catalyst**. * The reaction forms a *carboxylic acid* and an ***ammonium/amine***.

Answer 61

H₂O, H₂SO₄, Δ ## Footnote The *acid-catalyzed* Amide Hydrolysis mechanism yields an **ammonium ion byproduct**. (The eliminated amine group is protonated by the strong acid catalyst to form a cationic ammonium compound.)

Answer 62

H₂O, NaOH, Δ ## Footnote The *base-catalyzed* Amide Hydrolysis mechanism yields an **amine byproduct**. (The eliminated NH₂^– group is protonated by the carboxylic acid product to form an amine compound.)

Answer 63

Leaving of the azanide anion (1) converts the *unstable tetrahedral intermediate to a carboxylic acid* and (2) results in a *final carboxylate anion*. ## Footnote * The tetrahedral intermediate is destabilized by steric crowding (electron-electron repulsion between the Oxygens/Nitrogen) about the central Carbon. * The azanide anion becomes protonated (immediately following its elimination) to form the resonance-stabilized carboxylate anion.

Answer 64

* **Acid-Catalyzed:** Protonation of the amine byproduct to form the *ammonium cation/salt*. * **Base-Catalyzed:** Deprotonation of the carboxylic acid product to form the *carboxylate ion* and *amine*.

Answer 65

LiAlH₄ Reduction ## Footnote During LiAlH₄ Reduction, the carbonyl group (of the amide) is converted to a methylene group.

Answer 66

A **hydroxide ion** (OH^–) serves as the leaving group during LiAlH₄ Reduction of amides due to it being *more stable* than the azanide ion (NH₂^–). ## Footnote LiAlH₄ Reduction of Amides *differs* from other addition-elimination reactions in that the carbonyl bond (C=O) is broken (instead of the C—Heteroatom bond).

Answer 67

Partial Reduction ## Footnote The Partial Reduction reaction requires the DIBAL reductant (i.e. a weak/bulky hydride reagent).

Answer 68

The *partial reduction* of esters AND amides results in an **aldehyde product**. ## Footnote Both partial reduction mechanisms are identical, except for the production of different tetrahedral intermediates prior to acid workup.

Answer 69

* **Hybridization:** sp²-hybridized * **Geometry:** Linear * **Reactivity:** Relatively Electrophilic

Answer 70

Nitrile Hydrolysis ## Footnote * The Nitrile Hydrolysis reaction requires **high temperatures** AND **strong acid/base catalyst**. * The reaction forms a *carboxylic acid* and an ***ammonium/amine***. * An **amide intermediate** is always formed prior to H₂O-facilitated hydrolysis.

Answer 71

H₂O, H₂SO₄, Δ ## Footnote The *acid-catalyzed* Nitrile Hydrolysis mechanism yields an **ammonium ion byproduct**. (The eliminated amine group is protonated by the strong acid catalyst to form a cationic ammonium compound.)

Answer 72

H₂O, NaOH, Δ ## Footnote The *base-catalyzed* Nitrile Hydrolysis mechanism yields an **amine byproduct**. (The eliminated NH₂^– group is protonated by the carboxylic acid product to form an amine compound.)

Answer 73

* **Products:** Identical Products (i.e. Carboxylic Acid and Ammonium/Amine). * **Reagents:** Identical Reagents (i.e. Water + Strong Acid + Heat) * **Mechanism:** Nitrile Hydrolysis involves the conversion to an Amide intermediate prior to H₂O-facilitated hydrolysis. ## Footnote The two hydrolysis mechanisms are identical, except that Nitrile Hydrolysis involves the nitrile-to-amide conversion prior to hydrolysis. (The Amide Hydrolysis mechanism does *not* involve any conversion prior to H₂O-facilitated hydrolysis.)

Answer 74

1. The nitrile's **Nitrogen is protonated** by the strong acid catalyst (to "activate" the central Carbon). 2. **Nucleophilic H₂O** adds to the nitrile's central Carbon (to form an imine compound). 3. The **Oxygen is deprotonated** (by H₂O) to form a hydroxyl-substituted imine. 4. The Oxygen's **π electrons rearrange** to form an oxocarbenium ion *while* the C=N π electrons attack the acid to **protonate the Nitrogen** (and form an amine). 5. The **Oxocarbenium ion is deprotonated** (by H₂O) to form the Amide.

Answer 75

1. The **hydroxide catalyst attacks** the nitrile's central Carbon (to form an imine anion). 2. An **intramolecular proton transfer** occurs to protonate the Nitrogen *and* deprotonate the hydroxyl group (to form an amide anion). 3. The **Nitrogen is protonated** (by H₂O) to form the Amide.

Answer 76

1. R—MgBr 2.

Answer 77

1. R—MgBr 2. H₂O, H₂SO₄ ## Footnote Since the acidic workup step occurs *after* the removal of the grignard reagent (from the reaction mixture), the resulting ketone compound does *not* experience grignard addition.

Answer 78

Grignard Addition ## Footnote A **single grignard addition step** occurs during Grignard Nitrile Addition because the imine anion product is stable (i.e. not sufficiently electrophilic to be attacked by the grignard reagent).

Answer 79

LiAlH₄ Reduction

Answer 80

Cleavage of the C—N bond would create an azanide ion (NH₂^–) leaving group, which is extremely unstable in non-basic conditions.

Answer 81

* **LiAlH₄ Reduction:** The C—N bond is *not* cleaved (i.e. H₂ equivalents are being added across the C—N bond). * **Partial Reduction:** The C—N bond *is* cleaved (during the hydrolysis step).

Answer 82

Deprotonation (of α-Hydrogen) ## Footnote The α-Hydrogen can be deprotonated under **basic conditions** only.

Answer 83

* E2 Elimination of Alkyl Halide (if 2°/3° Alkyl Halide) * Claisen Condensation (i.e. Ester Enolate Self-Alkylation) * Ester Hydrolysis (due to Basic Conditions) ## Footnote A more effective mechanism for ester α-alkylation is **β-Ketoester α-Alkylation**.

Answer 84

NaOCH₂CH₃, R—X ## Footnote The alkyl halide reagent is limited to **0°/1° alkyl halides** and **allyl halides**. (Any other alkyl halide reagent may undergo E2 elimination within the basic conditions.)

Answer 85

Ester Enolate Alkylation ## Footnote The Ester Enolate Alkation reaction is an **ineffective α-alkylation mechanism** due to multiple potential side reactions.

Answer 86

The carbonyl Carbon of acyl halides is highly electrophilic, so the basic catalyst/solvent will attack the Carbon *before deprotonating the α-Hydrogen*.

Answer 87

Deprotonation of the the amide α-Hydrogen will create a highly unstable enolate ion (due to the strong electron-donating character of the Nitrogen).

Answer 88

9–13 | Highly Acidic ## Footnote * **β-Diketone:** pK_a = 9 * **β-Ketoester:** pK_a = 11 * **β-Diester:** pK_a = 13

Answer 89

β-Diketone > β-Ketoester > β-Diester > Ketone > Ester

Answer 90

* **β-Dicarbonyl Compounds:** Deprotonation is *favored* (due to the stability of the conjugate base). * **Monocarbonyl Compounds:** Deprotonation is *unfavorable* (due to the relative instability of the ester/ketone enolate).

Answer 91

The electron-donating character of the ether Oxygen *destabilizes* the conjugate base of ester compounds.

Answer 92

20–25 | Relatively Acidic ## Footnote * **Ketone:** pK_a = 20 * **Ester:** pK_a = 25

Answer 93

Partial Reduction

Answer 94

* Cyanide SN2 Addition to Alkyl Halide * CuCN Addition to Aryldiazonium Salt * Cyanide Addition to Aldehyde/Ketone * Cyanide Addtion to α,β-Unsaturated Aldehyde/Ketone

Answer 95

The Nitrogen of the imine anion product is *minimally electronegative*, so the imine's central Carbon is not sufficiently nucleophilic to experience Grignard attack. ## Footnote The Nitrogen is *less electronegative* than an Oxygen, so it cannot "activate" the central Carbon to enable grignard addition.

Answer 96

A single-step reduction facilitated by a weak/bulky hydride reagent. ## Footnote LiAlH₄ Reduction, in constrast to Partial Reduction, is a double-step reduction facilitated by a strong reducing agent.

Carboxylic Acid Derivatives (Chapter 20) Flashcards

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