Chapter 21 (Amines)

Question 1

Amine

Answer 1

A Nitrogen-based compound possessing one/two/three alkyl or aryl groups (bonded to the Nitrogen).

The reactivity of Amines differs greatly from other Nitrogen-containing compounds (i.e. Enamines, Amides, Imines, Iminiums, Ammoniums).

Question 2

Enamine

Answer 2

A compound possessing an amino group adjacent to an alkene group.

Question 3

Amide

Answer 3

A compound possessing an amino group adjacent to a carbonyl group.

The Nitrogen lone-pair electrons are in conjugation with the π electrons of the carbonyl group.

Question 4

Imine

Answer 4

A compound possessing a Carbon-Nitrogen double bond.

Question 5

Iminium Ion

Answer 5

A cationic compound possessing a Carbon-Nitrogen double bond and a quadruple-substituted Nitrogen.

Question 6

Ammonium

Answer 6

A cationic Nitrogen compound with four alkyl/aryl substituents (bonded to the Nitrogen).

Question 7

Amino Group

Answer 7

—NR₂

R = Alkyl/Aryl/Hydrogen

Question 8

Geometry: Simple Amines

Answer 8

Tetrahedral Geometry (sp³-Hybridized Nitrogen)

Question 9

Geometry: Aryl Amines

Answer 9

Tetrahedral Geometry (sp³-Hybridized Nitrogen)

The aryl amine displays a tetrahedral-like geometry despite the weak conjugation between the Nitrogen lone pair and the aromatic π electrons. (This conjugation is significantly weaker than the conjugation present in amides, so the aryl amine is tetrahedral-like overall.)

Question 10

Geometry: Amides

Answer 10

Planar Geometry (sp²-Hybridized Nitrogen)

The planar geometry of amides is caused by the strong conjugation between the Nitrogen lone pair and the carbonyl π electrons.

Question 11

Why are Alkyl Amines more nucleophilic than Aryl Amines?

Answer 11

The Nitrogen lone-pair electrons of alkyl amines are better able to undergo nucleophilic attack due to the absence of conjugation to delocalize the electrons.

The Nitrogen lone-pair electrons of aryl amines are in conjugation with the aromatic π electrons, so they are less nucleophilic/reactive.

Question 12

Reactivity: Enolate

Answer 12

β-Carbon Serves as Nucleophile

Question 13

Reactivity: Amide

Answer 13

Carbonyl Carbon Serves as Electrophile

The Nitrogen lone pair cannot serve as the nucleophile.

Question 14

Reactivity: Imine

Answer 14

Imine Carbon Serves as Electrophile

Imine Carbon = C=N

Question 15

Reactivity: Ammonium

Answer 15

Hydrogen Serves as Acid

Acid-Base Reactions

Question 16

Reactivity: Iminium

Answer 16

Iminium Carbon Serves as Electrophile

Imine Carbon = C=N⁺

Question 17

Chirality of Amines

Answer 17

An amine is chiral if it possess three unique substituents (bonded to the Nitrogen atom).

Chiral amines are not optically active since the two enantiomers cannot be separated/isolated. (The enantiomers undergo rapid transversion through a trigonal planar transition state.)

Question 18

Why are chiral amines not optically active?

Answer 18

The two enantiomers of chiral amines undergo rapid interconversion (through a trigonal planar intermediate state), so each individual enantiomer cannot be isolated.

Question 19

pK_a: Amines

Answer 19

pK_a ≈ 35

Amines are weak acids and moderate bases.

Question 20

pK_a: Ammonium

Answer 20

pK_a ≈ 10

Ammoniums are moderately acidic (i.e. more acidic than water/alcohol and less acidic than carboxylic acids).

Question 21

Acidity of Amines

Answer 21

Weakly Acidic

The conjugate bases of amines (e.g. LDA, NaNH₂) are strongly basic (due to the N—H bond being relatively strong/stable).

Question 22

Basicity of Amines

Answer 22

Moderately Basic

The conjugate acids of amines (i.e. Ammoniums) are moderately acidic (due to the instability of a quadruple-substituted cationic Nitrogen compound).

Question 23

Factors Determining Basicity of Amines

3 Factors

Answer 23

• Conjugation
• Hybridization
• Aromaticity

• The Nitrogen lone pair is less basic if it is in conjugation with a π system.
• The Nitrogen lone pair is less basic if it is possesses a lower hybridization.
• The Nitrogen lone pair is less basic if it is within an aromatic system.

Question 24

Basicity of Amines: Conjugation

Answer 24

• The Nitrogen lone pair is more basic if no conjugation is present (since the lone-pair electrons are more capable of attacking/reacting).
• The Nitrogen lone pair is less basic if it is in conjugation with a π system (since the lone-pair electrons are less reactive/nucleophilic).

Question 25

**Basicity of Amines:** Hybridization

Answer 25

* The Amine is *more basic* if it is **sp³-hybridized** (since the lone-pair electrons *not* stabilized by the less electronegative Nitrogen). * The Amine is *less basic* if it is **sp-hybridized** (since the lone-pair electrons *are* stabilized by the more electronegative Nitrogen).

Question 26

**Basicity of Amines:** Aromaticity

Answer 26

* The Amine is *more basic* if it is **outside of an aromatic system** (since the lone-pair electrons are more concentrated/reactive). * The Amine is *less basic* if it is **within an aromatic system** (since the lone-pair electrons are highly delocalized throughout the aromatic ring).

Question 27

Amine ⟶ Alkyl Amine

Answer 27

SN2 Alkyl Halide Alkylation ## Footnote The SN2 Alkyl Halide Alkylation reaction will produce a **mixture** of 1° amines, 2° amines, and 3° amines.

Question 28

**Reagents:** SN2 Alkyl Halide Alkylation | Starting Material = **Amine**

Answer 28

0°/1° Alkyl Halide ## Footnote The Nitrogen lone-pair (of the amine) undergoes SN2 attack at the alkyl halide Carbon to yield an Alkyl Amine (and a Halogen anion leaving group).

Question 29

Why is the SN2 Alkyl Halide Alkylation reaction **not** useful/effective for creating Alkyl Amines?

Answer 29

The initial 1° Amine product will undergo further alkylation reactions to yield a **mixture** of 1° Amines, 2° Amines, and 3° Amines.

Question 30

Nitrile ⟶ 1° Amine | R—CN ⟶ R—NH₂

Answer 30

LiAlH₄ Reduction

Question 31

Alkyl Azide ⟶ 1° Amine | R—N₃ ⟶ R—NH₂

Answer 31

LiAlH₄ Reduction

Question 32

**Reagents:** LiAlH₄ Reduction

Answer 32

1. LiAlH₄ 2. H₂SO₄, H₂O

Question 33

Alkyl Halide ⟶ 1° Amine | R—X ⟶ R—NH₂

Answer 33

1. SN2 Nitrile/Azide Synthesis 2. LiAlH₄ Reduction ## Footnote * **Nitrile Intermediate:** The final reduced product has *one more Carbon* than the reagent Alkyl Halide. * **Azide Intermediate:** The final reduced product has the *same number of Carbons* as the reagent Alkyl Halide.

Question 34

Alkyl Halide ⟶ Alkyl Azide | R—X ⟶ R—N₃

Answer 34

SN2 Azide Synthesis

Question 35

Alkyl Halide ⟶ Nitrile | R—X ⟶ R—CN

Answer 35

SN2 Nitrile Synthesis

Question 36

**Reagents:** SN2 Azide Synthesis | Starting Material = Alkyl Halide

Answer 36

NaN₃

Question 37

**Reagents:** SN2 Nitrile Synthesis | Starting Material = Alkyl Halide

Answer 37

NaCN

Question 38

Amide ⟶ Amine

Answer 38

LiAlH₄ Reduction ## Footnote The order/substitution of the Amine product corresponds to the order/substitution of the Amide reagent.

Question 39

Aldehyde/Ketone ⟶ Alkyl Amine

Answer 39

Reductive Amination ## Footnote The order/substitution of the Alkyl Amine product is **one greater** than that of the Amine reagent.

Question 40

**Reagents:** Reductive Amination | Starting Material = **Aldehyde/Ketone**

Answer 40

* Amine, NaBH₃CN * Amine, H₂, Ni(s) ## Footnote * The **amine** reagent can be a 0° Amine, 1° Amine, or 2° Amine. * All reagents can be added together at **one time**.

Question 41

**Mechanism:** Reductive Amination

Answer 41

1. **Nucleophilic addition** of the Amine to the carbonyl Carbon (of the aldehyde/ketone) to yield an Imine/Iminium. 2. **Reduction** of the imine/iminium Carbon-Nitrogen double bond (C=N) to yield an Amine. ## Footnote * **Step 1:** An imine will form if the reagent amine is a 0° Amine or 1° Amine. An iminium ion will form if the reagent amine is a 2° Amine. * **Step 2:** Reduction of the imine/iminium compound involves the addition of two Hydrogens across the C=N double bond.

Question 42

**Stability/Reactivity:** NaBH₃CN vs. NaBH₄

Answer 42

* The NaBH₃CN reductant is **more stable** under weakly acidic conditions. * The NaBH₃CN reductant is **less reactive** due to the electron-withdrawing Nitrile group.

Question 43

NaBH₃CN

Answer 43

Sodium Cyanoborohydride

Question 44

**Reductive Amination:** Reduction Selectivity

Answer 44

The reductive agents used in Reductive Amination (i.e. NaBH₃CN *or* H₂/Nickel) **react faster** with the imine/iminium **C=N bond** that with the aldehyde/ketone C=O bond. ## Footnote Since the aldehyde/ketone C=O bond will *not* react with the reductive agents (i.e. the reductive agents will always **preferentially react with the imine/iminium** C=N bond), all three reagents can be **added together** to the reaction mixture to yield the desired alkyl amine product.

Question 45

Ketone/Aldehyde ⟶ β-Aminocarbonyl Compound | β-Aminocarbonyl = β-Aminoketone or β-Aminoaldehyde

Answer 45

Mannich Reaction ## Footnote The Mannich Reaction requires **acidic conditions** to occur.

Question 46

1°/2° Amine ⟶ β-Aminocarbonyl Compound | β-Aminocarbonyl = β-Aminoketone or β-Aminoaldehyde

Answer 46

Mannich Reaction ## Footnote The Mannich Reaction requires **acidic conditions** to occur.

Question 47

**Reagents:** Ketone Mannich Reaction | β-Aminoketone Product

Answer 47

1. Ketone, Aldehyde, Amine, H₂SO₄, Δ 2. NaOH, H₂O ## Footnote The **amine** reagent must be a 1° Amine or 2° Amine.

Question 48

**Reagents:** Aldehyde Mannich Reaction | β-Aminoaldehyde Product

Answer 48

1. Aldehyde, Aldehyde, Amine, H₂SO₄, Δ 2. NaOH, H₂O ## Footnote The **amine** reagent must be a 1° Amine or 2° Amine.

Question 49

**Mechanism:** Ketone Mannich Reaction | β-Aminoketone Product

Answer 49

1. **Nucleophilic addition** of the Amine to the aldehyde's carbonyl Carbon to form an *Iminium cation* intermediate. 2. **Acid-catalyzed enolization** of the ketone via protonation of the ketone's carbonyl Oxygen. 3. **Nucleophilic attack** of the enol's α-Carbon (via π-electron rearragement) to the iminium's C=N Carbon to form a β-Aminooxocarbenium intermediate. 4. **Intramolecular proton transfer** to deprotonate the Oxocarbenium and protonate the Nitrogen. 5. **Base-mediated workup** to deprotonate the ammonium group to form a β-Aminoketone product.

Question 50

**Mechanism:** Aldehyde Mannich Reaction | β-Aminoaldehyde Product

Answer 50

1. **Nucleophilic addition** of the Amine to the reactive aldehyde's carbonyl Carbon to form an *Iminium cation* intermediate. 2. **Acid-catalyzed enolization** of the non-reactive aldehyde via protonation of that aldehyde's carbonyl Oxygen. 3. **Nucleophilic attack** of the enol's α-Carbon (via π-electron rearragement) to the iminium's C=N Carbon to form a β-Aminooxocarbenium intermediate. 4. **Intramolecular proton transfer** to deprotonate the Oxocarbenium and protonate the Nitrogen. 5. **Base-mediated workup** to deprotonate the ammonium group to form a β-Aminoaldehyde product.