Aminas

23-23-11

AminesAminesChapter 23Chapter 23

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Structure & ClassificationStructure & Classification

� Amines are classified as: • 1° , 2° , or , 3° amines:1° , 2° , or , 3° amines: Amines in which there

are 1, 2, or 3 alkyl or aryl groups.

Methylamine(a 1° amine)

Dimethylamine(a 2° amine)

Trimethylamine(a 3° amine)

CH3 - NH2 CH3 - NH CH3 - N

CH3 CH3

CH3: :

:

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� Amines are further divided into aliphatic, aromatic, and heterocyclic amines:• AliphaticAliphatic amine:amine: An amine in which nitrogen is

bonded only to alkyl groups.• Aromatic amine:Aromatic amine: An amine in which nitrogen is

bonded to one or more aryl groups.

Aniline(a 1° aromatic amine)

N-Methylaniline(a 2° aromatic amine)

Benzyldimethylamine(a 3° aliphatic amine)

NH2 N-H CH2 - N- CH3

CH3 CH3

::

:

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• Heterocyclic amine:Heterocyclic amine: An amine in which nitrogen is one of the atoms of a ring.

PyrrolePiperidinePyrrolidine Pyridine(heterocyclic aliphatic amines) (heterocyclic aromatic amines)

NN N

H H

N

H

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Nomenclature Nomenclature

� Aliphatic amines: replace the suffix -ee of the parent alkane by -amineamine.

1,6-Hexanediamine(S)-1-Phenyl-ethanamine

2-Propanamine

NH2

NH2

NH2H2 N

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NomenclatureNomenclature

� The IUPAC system retains the name aniline.

3-Methoxyaniline(m-Anisidine)

4-Methylaniline(p-Toluidine)

Aniline 4-Nitroaniline(p-Nitroaniline)

NH2 NH2

CH3

OCH3

NH2

NO2

NH2

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� Among the various functional groups, -NH 2 is one of the lowest in order of precedence.

COOHH2 NOH

NH2

H2 NOH

4-Aminobenzoic acid2-Aminoethanol (S)-2-Amino-3-methyl-1-butanol

Amine vs alcohol

Amine vs acid

23-23-99


� Common names for most aliphatic amines are derived by listing the alkyl groups bonded to nitrogen in one word ending with the suffix - amineamine.

CH3 NH2 N

H

Et3 NNH2

TriethylamineDicyclopentylamineMethylamine tert-Butylamine

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� When four groups are bonded to nitrogen, the compound is named as a salt of the corresponding amine.

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Chirality of AminesChirality of Amines

• Consider the unshared pair of electrons on nitrogen as a fourth group, then the arrangement of groups around N is approximately tetrahedral.

• An amine with three different groups bonded to N is chiral and exists as a pair of enantiomers and, in principle, can be resolved.

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• In practice, however, they cannot be resolved because they undergo inversion, which converts one enantiomer to the other.

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• Pyramidal inversion is not possible with quaternary ammonium ions, and their salts can be resolved.

N

MeEt

N

MeEt

Cl- Cl-

S Enantiomer R Enantiomer

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Physical PropertiesPhysical Properties

� Amines are polar compounds, and both 1° and 2° amines form intermolecular hydrogen bonds.• N-H- - -N hydrogen bonds are weaker than O-H- -

-O hydrogen bonds because the difference in electronegativity between N and H (3.0 - 2.1 =0.9) is less than that between O and H (3.5 - 2.1 = 1.4).

bp (°C) -6.3 65.0-88.6

32.031.130.1MW (g/mol)

CH3 CH3 CH3 NH2 CH3 OH

Using bp as an indication of H bonding

Increasing strength

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BasicityBasicity

� All amines are weak bases, and aqueous solutions of amines are basic.

• It is common to discuss their basicity by reference to the acid ionization constant of the conjugate acid.CH3NH3

+ H2 O CH3 NH2 H3 O+++

[ CH3NH2 ] [H3 O+

]

[CH3 NH3+

]2.29 x 10-11==Ka pKa = 10.64

H

H

CH3 - N H- O-H

H

H

CH3 - N- H O-H

Methylammonium hydroxideMethylamine

++

-

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BasicityBasicity

• Using values of pKa, we can compare the acidities of amine conjugate acids with other acids.

CH3NH2 CH3 COOH CH3NH3+

CH3COO-

Keq = 7.6 x 105

+ +

pKa 10.64pKa 4.76(stronger

acid)(weaker

acid)

pKeq = -5.88

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Basicity-Aliphatic AminesBasicity-Aliphatic Amines

� Aliphatic Amines• note that pKa + pKb = 14

Tertiary Aminesdiethylaminedimethylamine

cyclohexylamineethylaminemethylamine

Secondary Amines

Primary AminesAmmonia

pKaStructureAmine

trimethylaminetriethylamine

9.26

10.6410.8110.66

10.7310.98

9.8110.75

pKb

4.74

3.36

3.343.19

3.273.02

4.193.25

CH3 NH2

NH3

CH3 CH2 NH2C6 H1 1 NH2

( CH3 ) 2 NH( CH3 CH2 ) 2 NH

( CH3 ) 3 N( CH3 CH2 ) 3 N

Stronger bases

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Basicity-Aromatic AminesBasicity-Aromatic Amines

NH2CH3

NH2Cl

NH2O2N

NH2

N

N

N

H

Heterocyclic Aromatic Amines

Aromatic Amines

StructureAmine

Aniline

4-Chloroaniline

4-Nitroaniline

4-Methylaniline

Pyridine

Imidazole

4.63

5.08

4.15

1.0

5.25

6.95

pKa of Conjugate Acid

Weaker bases

Intermediate

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• Aromatic amines are considerably weaker bases than aliphatic amines.

NH2 H2 O

H2 ONH2

NH3+ OH-

NH3+ OH-

Cyclohexylamine

pKa = 4.63

Aniline

pKa = 10.66+

+

Cyclohexylammoniumhydroxide

Anilinium hydroxide

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� Aromatic amines are weaker bases than aliphatic amines because of two factors:• Resonance stabilization of the free base, which is

lost on protonation.

N

HH

H

HH

H

H

..

. .. .

. .unhybridized 2p orbital of N

nitrogen is sp2 hybridized

N N NH H H H

H NH HH

+++

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Basicity-Aromatic AminesBasicity-Aromatic Amines• The greater electron-withdrawing inductive effect

of the sp2-hybridized carbon of an aromatic amine compared with that of the sp3-hybridized carbon of an aliphatic amine.

And note the effect of substituents� Electron-releasing groups, such as alkyl

groups, increase the basicity of aromatic amines.

� Electron-withdrawing groups, such as halogens, the nitro group, and a carbonyl group decrease the basicity of aromatic amines by a combination of resonance and inductive effects.

23-23-2222

Example: Basicity-Aromatic AminesExample: Basicity-Aromatic Amines

3-nitroaniline is a stronger base than 4-Nitroaniline.

NH2

O2 N

NH2O2 N

pKa 1.0pKa 2.474-Nitroaniline3-Nitroaniline

delocalization of the nitrogen lone pair onto the oxygen atoms of the nitro group

++

-

-

-

N NH2 NH2+N

O

O

O

O

Cannot do this kind of resonance in 3 nitroaniline

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� Heterocyclic aromatic amines are weaker bases than heterocyclic aliphatic amines.

Piperidine ImidazolePyridinepKa 10.75 pKa 5.25 pKa 6.95

N NH

N

NH

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• In pyridine, the unshared pair of electrons on N is not part of the aromatic sextet.

• Pyridine is a weaker base than heterocyclic aliphatic amines because the free electron pair on N lies in an sp2 hybrid orbital (33% s character) and is held more tightly to the nucleus than the free electron pair on N in an sp3 hybrid orbital (25% s character).

:N

HH

H

HH

nitrogen is sp2 hybridized

an sp2 hybrid orbital; the electronpair in this orbital is not apart of the aromatic sextet

...

.. .

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� Imidazole Which N lone pair is protonated? The one which is not part of the aromatic system.

This electron pair is not a part of the aromatic sextet

This electron pair is a partof the aromatic sextet

+ +

Aromaticity is maintained when imidazole is protonated

+

Imidazole Imidazolium ion

N

N

H

HN

N

H

H2 O OH-

:::

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Basicity-GuanidineBasicity-Guanidine

� Guanidine is the strongest base among neutral organic compounds.

• Its basicity is due to the delocalization of the positive charge over the three nitrogen atoms.

:

C

NH2

NH2H2 N H2 N C NH2

NH2

C

NH2

NH2H2 N

Three equivalent contributing structures

+ +

+

:

:

: :

:

++

Guanidine Guanidinium ion

C

NH

NH2H2 N H2 N C NH2

NH2+

H2 O OH- pKa = 13.6

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Reaction with AcidsReaction with Acids

� All amines, whether soluble or insoluble in water, react quantitatively with strong acids to form water-soluble salts.

HO

HO

NH2

OH

HClH2 O

HO

HO

NH3+ Cl-

OH

(R)-Norepinephrine hydrochloride(a water-soluble salt)

+

(R)-Norepinephrine(only slightly soluble in water)

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Reaction with acidsReaction with acids

� Separation and purification of an amine and a neutral compound.

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PreparationPreparation

� We have already covered these methods• nucleophilic ring opening of epoxides by ammonia

and amines.• addition of nitrogen nucleophiles to aldehydes

and ketones to form imines• reduction of imines to amines• reduction of amides to amines by LiAlH 4

• reduction of nitriles to a 1° amine• nitration of arenes followed by reduction of the

NO 2 group to a 1° amine

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PreparationPreparation

� Alkylation of ammonia and amines by S N2 substitution.

• Unfortunately, such alkylations give mixtures of products through a series of proton transfer and nucleophilic substitution reactions.

CH3Br NH3

CH3NH3+Br

- (CH3 ) 2NH2+Br

-(CH3 ) 3NH+Br

-(CH3 ) 4N+Br

-

+

+ + +

+ SN2

Methylammonium bromide

CH3 Br NH3 CH3 NH3+ Br -

polyalkylations

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Preparation via AzidesPreparation via Azides

� Alkylation of azide ion.

-- + + -

Azide ion (a good nucleophile)

An alkyl azide

N NN NN NRN3-

RN3:: :

: : : :

Ph CH2 ClK

+ N3

-

Ph CH2 N31. LiAlH4

2. H2OPh CH2 NH2

Benzyl chloride Benzyl azide Benzylamine

Overall Alkyl Halide Alkyl amine

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Example: Preparation via AzidesExample: Preparation via Azides

• Alkylation of azide ion.

Cyclohexene

trans-2-Amino-cyclohexanol

(racemic)

1,2-Epoxy-cyclohexane

trans-2-Azido-cyclohexanol

(racemic)

ArCO3 H 1 . K+

N3-

2 . H2 O

1 . LiAlH4

2 . H2 ON3

OH

NH2

OH

O

Note retention of configuration, trans trans

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Reaction with HNOReaction with HNO 22

� Nitrous acid, a weak acid, is most commonly prepared by treating NaNO 2 with aqueous H 2SO 4 or HCl.

� In its reactions with amines, nitrous acid: • Participates in proton-transfer reactions.• A source of the nitrosyl cation, NO +, a weak

electrophile.

HNO2 H2O H3O+NO2

-+ pKa = 3.37+

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Reaction with HNOReaction with HNO 22

� NO + is formed in the following way.• Step 1: Protonation of HONO.• Step 2: Loss of H 2O.

• We study the reactions of HNO 2 with 1° , 2° , and 3° aliphatic and aromatic amines.

H

H

N O+

OH

H+ OH N ONO OH

N O

+

+

+

+

The nitrosyl cation

(1) (2)

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Tertiary Amines with HNOTertiary Amines with HNO 22

• 3° Aliphatic amines, whether water-soluble or water-insoluble, are protonated to form water-soluble salts.

• 3° Aromatic amines: NO + is a weak electrophile and participates in Electrophilic Aromatic Substitution.

Me2 N1. NaNO2 , HCl, 0-5°C

2. NaOH, H2ON=OMe2 N

N,N-Dimethyl-4-nitrosoanilineN,N-Dimethylaniline

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Secondary Amines with HNOSecondary Amines with HNO 22

• 2° Aliphatic and aromatic amines react with NO + to give N-nitrosamines.

N-H HNO2 N-N=O H2 O

Piperidine N-Nitrosopiperidine

+ +

carcinogens

N

H

N ON

H N=OH O

H

N

N=O

H O

H

H++

+ ++• •

• •

• •• •

• •

• •

• •

• •• •

• •• •

• •

• •

(1) (2)

Mechanism:

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RNHRNH 22 with HNO with HNO 22

� 1° aliphatic amines give a mixture of unrearranged and rearranged substitution and elimination products, all of which are produced by way of a diazonium ion and its loss of N 2 to give a carbocation.

� Diazonium ion:Diazonium ion: An RN 2+ or ArN 2

+ ion

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1° RNH1° RNH 22 with HNO with HNO 22

� Formation of a diazonium ion.Step 1: Reaction of a 1° amine with the nitrosyl

cation.

Step 2: Protonation followed by loss of water.

:

:+

keto-enoltautomerism

A 1° aliphatic amine

An N-nitrosamine

R-NH2 N R-N-N=OO+ : :

:

H: :

:

A diazotic acid

R-N=N-O-H

: : ::

A diazotic acid

R-N=N-O-H

: : ::

+

A diazonium ion

++

A carbocation

O-HNR-N

H

N NR N N

H+

- H2 O

R+

•• ••

••

•• •• ••

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1° RNH1° RNH 22 with HNO with HNO 22 (Aliphatic) (Aliphatic)

� Aliphatic diazonium ions are unstable and lose N 2 to give a carbocation which may:1. Lose a proton to give an alkene.2. React with a nucleophile to give a substitution

product.3. Rearrange and then react by Steps 1 and/or 2.

(25%)

(5.2%)

(13.2%)

(25.9%) (10.6%)

0-5oC

NaNO2 , HClNH2OH

Cl

OH

+

+

23-23-4040

1° RNH1° RNH 22 with HNO with HNO 22

� Tiffeneau-Demjanov reaction:Tiffeneau-Demjanov reaction: Treatment of a β-aminoalcohol with HNO 2 gives a ketone and N 2..

CH2 NH2

OH

HNO2

O

H2 O N2

β+ + +

A β-aminoalcohol Cycloheptanone

α

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Mechanism of Tiffeneau-DemjanovMechanism of Tiffeneau-Demjanov

• Reaction with NO + gives a diazonium ion.• Concerted loss of N 2 and rearrangement followed

by proton transfer gives the ketone. :OH

CH2NH2HNO2

O-H

CH2 N N+

(A diazonium ion)

-N2

O

+ CH2

OH

CH2

O H+ proton transfer to H2O

A resonance-stabilized cation Cycloheptanone

:

:

: : :

:

:

:

Similar to pinacol rearrangement

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Pinacol Rearrangement: an example of Pinacol Rearrangement: an example of stabilization of a carbocation by an stabilization of a carbocation by an adjacent lone pair.adjacent lone pair.

Overall:

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MechanismMechanism

Reversible protonation.

Elimination of water to yield tertiary carbocation.1,2 rearrangement to yield resonance stabilized cation.

Deprotonation.

This is a protonated

ketone!

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1° 1° Primary AminesPrimary Amines with HNO with HNO 2 2 (Aromatic)(Aromatic)

� The -N 2+ group of an arenediazonium salt can

be replaced in a regioselective manner by these groups.

Ar-NH2

HNO2Ar-N2

+ (-N2 )

HCl, CuCl

H2O

HBF4

HBr, CuBr

KCN, CuCN

KI

H3PO2

Ar-I

Ar-F

Ar-H

Ar-Cl

Ar-Br

Ar-CN

Ar-OH

Schiemannreaction

Sandmeyerreaction0-5°C

23-23-4545

1° ArNH1° ArNH 22 with HNO with HNO 22

� A 1° aromatic amine converted to a phenol.

2-Bromo-4-methylaniline

2-Bromo-4-methylphenol

1. HNO2

2. H2O, heat

NH2

Br

CH3

OHBr

CH3

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Problem:Problem: What reagents and experimental conditions will bring about this conversion?

(1) (2) (3) (4)

CH3 CH3

NO2

COOH

NO2

COOH

NH2

COOH

OH

23-23-4747


� Problem:Problem: Show how to bring about each conversion.

NH2

CH3

ClCH3

CCH3

N

NH2

CH3

ClCl

CH2 NH2

CH3

CH3

ClCl

(5)

(6) (7)

(8)

(9)

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Hofmann EliminationHofmann Elimination

� Hofmann elimination:Hofmann elimination: Thermal decomposition of a quaternary ammonium hydroxide to give an alkene.• Step 1: Formation of a 4° ammonium hydroxide.

(Cyclohexylmethyl)trimethyl-ammonium hydroxide

Silveroxide

(Cyclohexylmethyl)trimethyl- ammonium iodide

+

+ H2 OAg2 O

AgI

CH2 -N- CH3

CH3

CH3

I-

+

+CH2 -N- CH3

CH3

CH3

OH-

23-23-4949


• Step 2: Thermal decomposition of the 4° ammonium hydroxide.

(Cyclohexylmethyl)trimethyl- ammonium hydroxide

TrimethylamineMethylene-cyclohexane

++CH2 ( CH3 ) 3 N H2 O

160°+CH2 -N- CH3

CH3

CH3

OH-

23-23-5050


� Hofmann elimination is regioselective - the major product is the least substituted alkene.

� Hofmann’s rule:Hofmann’s rule: Any β-elimination that occurs preferentially to give the least substituted alkene as the major product is said to follow Hofmann’s rule.

CH3

N(CH3 )3 OH- CH2 (CH3 )3N H2O++

heat+

23-23-5151


• The regioselectivity of Hofmann elimination is determined largely by steric factors, namely the bulk of the -NR 3

+ group.• Hydroxide ion preferentially approaches and

removes the least hindered hydrogen and, thus, gives the least substituted alkene.

• Bulky bases such as (CH 3) 3CO -K+ give largely Hofmann elimination with haloalkanes.

+

E2 reaction (concertedelimination)

C C

H

N( CH3 ) 3H

H H

CH

HC

H

CH3 CH2

HO-

N( CH3 ) 3

HOH

CH3 CH2

23-23-5252

Cope EliminationCope Elimination

� Cope elimination:Cope elimination: Thermal decomposition of an amine oxide.Step 1: Oxidation of a 3° amine gives an amine

oxide.

Step 2: If the amine oxide has at least one β-hydrogen, it undergoes thermal decomposition to give an alkene.

CH2 N-CH3

CH3

H2 O2

O

CH3

CH2 N-CH3 H2 O++

+

-

An amine oxide

O

CH3

CH2 N-CH3

H 100-150°CCH2 (CH3 ) 2 NOH+

N,N-Dimethyl-hydroxylamine

Methylene-cyclohexane

+

-

23-23-5353

Cope EliminationCope Elimination

• Cope elimination shows syn stereoselectivity but little or no regioselectivity.

• Mechanism: a cyclic flow of electrons in a six-membered transition state.

:O

heat+

-

Transition state

an alkene

N,N-dimethyl-hydroxylamine

C C

H NCH3

CH3N

CH3

CH3

OH

C C

:

Science

Aminas