chapter 19 ver 1

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Chapter 19 Amines

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Chapter 19 2

Introduction• Organic derivatives of ammonia.

• Many are biologically active.

!

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Chapter 19 "

#iological Activity• $eurotransmitters% dopamine

• #ioregulators% epinephrine

• &itamins% niacin' #(

• Al)aloids% nicotine' morphine' cocaine• Amino acids

!

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Chapter 19 *

Classes of Amines

• +rimary ,1°-% one C$ bond' 2 $/ bonds.

• 0econdary ,2°-% to C$ bonds' 1 $/

bond.• ertiary ,"°-% three C$ bonds' no $/

bond.

• 3uaternary ,*°-% four C$ bonds' nitrogenhas a 4 formal charge.

!

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Chapter 19 5

Classify%

N

H

CH3 C

CH3

CH3

NH2

N

CH3

CH3

CH3CH2 N

CH3

CH2CH3+

Br _

!

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Chapter 19 (

Common $ames

$ame the al)yl or aryl groups bonded to

nitrogen' then add suffi6 amine.

diethylmethylamine

cyclohe6yldimethylamine

diphenylamine !

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Chapter 19 7

Amine as 0ubstituent• On a molecule ith a higher priority

functional group' the amine is named as

a substituent.

NH2CH2CH2CH2COOH

NHCH3

OH

γ aminobutyric acid or

*aminobutanoic acid

2methylaminophenol

!

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Chapter 19 8

Aromatic Amines

Amino group is bonded to a benene ring.

+arent compound is called aniline.

!

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Chapter 19 9

/eterocyclic Amines

he nitrogen is assigned the number 1.

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Chapter 19 1:

I;+AC $ames

• $ame is based on longest carbon chain.

• e of al)ane is replaced ith amine.

• 0ubstituents on nitrogen have N prefi6.

NH2CH2CH2CHCH2CH3

Br

CH3CH2CHCH2CH2CH3

N(CH3)2

"bromo1pentanamine N 'N dimethyl"he6anamine

!

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Chapter 19 11

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Chapter 19 12

0tructure of Amines

$itrogen is sp"

hybridied ith a lone pairof electrons in an sp" orbital.

!

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Chapter 19 1"

Chirality of Amines

$itrogen may have " different groups anda lone pair' but enantiomers cannot be

isolated due to inversion around $.

!

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Chapter 19 1*

Chiral Amines• Amines ith chiral

carbon% 2butanamine.• 3uaternary ammonium

salts ith four differentgroups bonded tonitrogen.

• Amines in hich thenitrogen is restricted inrotation so it cannotinterconvert.

!

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Chapter 19 15

#oiling +oints

• $/ less polar than O/.•

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Chapter 19 1(

0olubility and Odor

• 0mall amines ,=( C- soluble in ater.

• All amines accept hydrogen bonds from

ater and alcohol.• #ranching increases solubility.

• Most amines smell li)e rotting fish.

NH2CH2CH2CH2CH2CH2 NH21'5pentanediamine or cadaverine

!

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Chapter 19 17

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Chapter 19 18

#asicity of Amines

• >one pair of electrons on nitrogen can

accept a proton from an acid

• A?ueous solutions are basic to litmus.

• Ammonia pK b *.7*

• Al)yl amines are usually stronger bases

than ammonia. Increasing the number of

al)yl groups decreases solvation of ion' so

2° and "° amines are similar to 1° amines

in basicity. !

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Chapter 19 19

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Chapter 19 2:

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Chapter 19 21

@nergy iagram

Al)yl groups are electrondonating and

stabilie the cation.

!

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Chapter 19 22

Besonance @ffects

Any delocaliation of the electron pairea)ens the base.

!

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Chapter 19 2"

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Chapter 19 2*

/ybridiation @ffects

@lectrons are held more tightly in orbitals

ith more s character' so those

compounds are ea)er bases.

!

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Chapter 19 25

Amine 0alts

• Ionic solids ith high melting points

• 0oluble in ater

• $o fishy odor

!

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Chapter 19 2(

+urifying an Amine

!

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Chapter 19 27

+hase ransfer Catalysts

!

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Chapter 19 28

IB 0pectroscopy

• $/ stretch beteen "2::"5:: cm1.

• o pea)s for 1° amine' one for 2°.

!

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Chapter 19 29

$MB 0pectrum

!

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Chapter 19 ":

Mass 0pectrum

!

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Chapter 19 "1

Beactions ith CO

Ammonia and primary amines react ith

carbonyls to give an imine ,0chiff base-.

!

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Chapter 19 "2

@lectrophilic 0ubstitution

of Aniline• $/2 is strong activator' o' pdirecting.

• May trisubstitute ith e6cess reagent.

• /4 changes $/2 to $/"4' a meta

alloing deactivator.

• Attempt to nitrate aniline may e6plode.

!

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Chapter 19 ""

Aniline 0ubstitution

!

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Chapter 19 "*

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Chapter 19 "5


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Chapter 19 "(


of +yridine

•0trongly deactivated by electronegative $.

• 0ubstitutes in the "position.

• @lectrons on $ can react ith electrophile.

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Chapter 19 "7

Mechanism for

@lectrophilic 0ubstitution Attack at the 3-position (observed)

!

Attack at the 2-position (not observed)

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Chapter 19 "8

$ucleophilic 0ubstitution

of +yridine• eactivated toard electrophilic attac).

• Activated toard nucleophilic attac).

• $ucleophile ill replace a good leaving

group in the 2 or *position.

!

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Chapter 19 "9

Mechanism for

$ucleophilic 0ubstitution Attack at the 2-position (observed)

!

Nucleophilic Attack at the 3-position (not observed)

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Chapter 19 *:

!

Al)ylation of Amines• Amines react ith 1° al)yl halides via

the 0$2 mechanism.

• Mi6tures of the mono' di' and tri

al)ylated products are obtained.

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Chapter 19 *1

;seful Al)ylations

• @6haustive al)ylation to form thetetraal)ylammonium salt.

CH3CH2CHCH2CH2CH3

N(CH3)3

CH3CH2CHCH2CH2CH3

NH23 CH3I

NaHCO3

+ _ I

• Beaction ith large e6cess of $/" to

form the primary amine.

CH3CH2CH2Br NH3 (xs)

CH3CH2CH2 NH2 + NH4Br

!

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Chapter 19 *2

Acylation of Amines

by Acid Chlorides• Amine attac)s CO' chloride ion leaves.

• +roduct is amide' neutral' not basic.

• ;seful for decreasing activity of anilinetoard electrophilic aromatic substitution.

NH2CH3 C

OCl

NHC

O

CH3

N

to remove /Cl

!

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Chapter 19 *"

SOLVED PROBLEM 19-1

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Chapter 19 **

Show how you would accomplish the following synthetic conversion in good yield.

Solution

An attempted Friedel–Crafts acylation on aniline would likely meet with disaster. The free amino

group would attack both the acid chloride and the Lewis acid catalyst.

SOLVED PROBLEM 19-1 (continued)

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Chapter 19 *5

Solution (continued)

We can control the nucleophilicity of aniline’s amino group by converting it to an amide, which is still

activating and ortho, para directing for the Friedel–Crafts reaction. Acylation, followed by hydrolysis

of the amide, gives the desired product.

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Chapter 19 *(

ormation of 0ulfonamides

• +rimary or secondary amines react ithsulfonyl chloride.

R NH2 S

O

O

R' Cl S

O

O

R' NH R

H

+Cl

_ base S

O

O

R' NH R

• 0ulfa drugs are sulfonamides

that are antibacterial agents.

!

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Chapter 19 *7

f

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Chapter 19 *8

/ofmann @limination• A ?uaternary ammonium salt has a

good leaving group a neutral amine.• /eating the hydro6ide salt produces the

least substituted al)ene.

CH3CHCH2CH2CH3

N(CH3)3+ _

IAg2O

H2O

+

CH3CHCH2CH2CH3

N(CH3)3

_

OHheat

CH2 CHCH2CH2CH3

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Chapter 19 *9

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Chapter 19 5:

@2 Mechanism

!

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Chapter 19 51

SOLVED PROBLEM 19-2

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Chapter 19 52

Predict the major product(s) formed when the following amine is treated with excess iodomethane,

followed by heating with silver oxide.

Solution

Solving this type of problem requires finding every possible elimination of the methylated salt. In this

case, the salt has the following structure:

SOLVED PROBLEM 19-2 (continued)

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Chapter 19 5"

Solution (continued)

The green, blue, and red arrows show the three possible elimination routes. The corresponding

products are

The first (green) alkene has a disubstituted double bond. The second (blue) alkene is monosubstituted,

and the red alkene (ethylene) has an unsubstituted double bond. We predict that the red products will

be favored.

O id ti f A i

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Chapter 19 5*

O6idation of Amines• Amines are easily o6idied' even in air.• Common o6idiing agents% /2O2 ' MC+#A.

• 1° Amines o6idie to nitro compounds via hydro6ylamineand nitroso groups

• 2° Amines o6idie to hydro6ylamine ,$O/-.• "° Amines o6idie to amine o6ide ,$4O-.

!

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Chapter 19 55

Cope @limination

Amine o6ides undergo elimination to formthe least substituted al)ene under milder

conditions than the /ofmann reaction.

CH2

N(CH3)2

CHCH2CH2CH3

H

O O

H

CHCH2CH2CH3

N(CH3)2

CH2 CH2 CHCH2CH2CH3

N(CH3)2

HO+

_

!

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Chapter 19 5(

SOLVED PROBLEM 19-3

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Chapter 19 57

Predict the products expected when the following compound is treated with H2O2 and heated.

Solution

Oxidation converts the tertiary amine to an amine oxide. Cope elimination can give either of two

alkenes. We expect the less hindered elimination to be favored, giving the Hofmann product.

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Chapter 19 58

$itrous Acid Beagent

• $itrous acid is produced in situ by

mi6ing sodium nitrite ith /Cl.

• he nitrous acid is protonated' loses

ater to form the nitrosonium ion.

!

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Chapter 19 59

Beaction ith $itrous Acid

• 1° Amines form diaonium salts' B$4≡$.

• Al)yldiaonium salts are unstable' but

arenediaonium salts are idely used forsynthesis.

• 2° Amines form N nitrosoamines' B2$$O'

found to cause cancer in laboratory animals.

!

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Chapter 19 (:

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Chapter 19 (1

Arenediaonium 0alts

• 0table in solution at :DE1:DC.• he 4$≡$ group is easily replaced bymany different groups.

• $itrogen gas' $2' is a byproduct.

!

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Chapter 19 (2

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Chapter 19 ("

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Chapter 19 (*

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Chapter 19 (5

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Chapter 19 ((

0ynthesis by

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Chapter 19 (7

0ynthesis by

Beductive Amination• o produce a 1° amine' react an

aldehyde or )etone ith hydro6ylamine'

then reduce the o6ime.

• o produce a 2° amine' react analdehyde or )etone ith a 1° amine'

then reduce the imine.

• o produce a "° amine' react analdehyde or )etone ith a 2° amine'

then reduce the imine salt. !

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Chapter 19 (8

@6amples

primary amine

secondary amine

tertiary amine !

C

N

H

CH3H3C

H Na(CH3COO)3BH

C

N

H

CH3H3C+

H+

HN(CH3)2C

O

HCH3COOH

SOLVED PROBLEM 19-5

Show how to synthesize the following amines from the indicated starting materials

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Chapter 19 (9

Show how to synthesize the following amines from the indicated starting materials.

(a) N -cyclopentylaniline from aniline (b) N -ethylpyrrolidine from pyrrolidine

Solution

(a) This synthesis requires adding a cyclopentyl group to aniline (primary) to make a secondary amine.Cyclopentanone is the carbonyl compound.

(b) This synthesis requires adding an ethyl group to a secondary amine to make a tertiary amine. The

carbonyl compound is acetaldehyde. Formation of a tertiary amine by Na(AcO)3BH reductive

amination involves an iminium intermediate, which is reduced by (sodium triacetoxyborohydride).

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Chapter 19 7:

AcylationBeduction

• An acid chloride reacts ith ammonia or a1° amine or a 2° amine to form an amide.

• he CO of the amide is reduced to C/2

ith lithium aluminum hydride.• Ammonia yields a 1° amine.

• A 1° amine yields a 2° amine.

• A 2° amine yields a "° amine. !

SOLVED PROBLEM 19-6

Show how to synthesize N-ethylpyrrolidine from pyrrolidine using acylation–reduction

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Chapter 19 71

Show how to synthesize N ethylpyrrolidine from pyrrolidine using acylation reduction.

Solution

This synthesis requires adding an ethyl group to pyrrolidine to make a tertiary amine. The acid chloride

needed will be acetyl chloride (ethanoyl chloride). Reduction of the amide gives N -ethylpyrrolidine.

Compare this synthesis with Solved Problem 19-5(b) to show how reductive amination and acylation–

reduction can accomplish the same result.

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Chapter 19 72

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Chapter 19 7"

@6amples

primary amine

tertiary amine

!

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Chapter 19 7*

irect Al)ylation ,1°-

• ;se a large e6cess ,1:%1- of ammonia

ith a primary al)yl halide or tosylate.

• Beaction mechanism is 0$2.

CH3CH2CH2 Br NH3

+CH3CH2CH2 NH2 NH4Br

!

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Chapter 19 75

Fabriel 0ynthesis ,1°-

• ;se the phthalimide anion as a form ofammonia that can only al)ylate once.

• Beact the anion ith a good 0$2

substrate' then heat ith hydraine.

!

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Chapter 19 7(

Aide Beduction ,1°-

• Aide ion' $"' is a good nucleophile.

• Beact aide ith unhindered 1° or 2°

halide or tosylate ,0$2-.

• Al)yl aides are e6plosiveG o not isolate.

!

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Chapter 19 77

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Chapter 19 78

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Chapter 19 79

$itrile Beduction ,1°-

• $itrile' C≡$' is a good 0$2 nucleophile.

• Beduction ith /2 or >iAl/* adds C/2$/2.

Br

NaCN

CNLiAlH4

H2O

1)

2)

CH2NH2

!

B d ti f $it

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Chapter 19 8:

Beduction of $itro

Compounds ,1°-• $O2 is reduced to $/2 by catalytic

hydrogenation' or active metal ith acid.

• Commonly used to synthesie anilines.

!

SOLVED PROBLEM 19-4

Show how you would convert toluene to 3,5-dibromotoluene in good yield.

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Chapter 19 81

Solution

Direct bromination of toluene cannot give 3,5-dibromotoluene because the methyl group activates the

ortho and para positions.

Starting with p-toluidine ( p-methylaniline), however, the strongly activating amino group directs

bromination to its ortho positions. Removal of the amino group (deamination) gives the desired

product.

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Chapter 19 82

19.1 $ame

• $Methyl1butanamine

• @thyl dimethyl amine• $@thyl1propanamine

• iethyl methyl amine

$C/"C/2

C/"

C/2C/"

>evel "

Challenging

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Chapter 19 8"

19.1 anser

• $Methyl1butanamine

• @thyl dimethyl amine

• $@thyl1propanamine


$ame each group attached to the

nitrogen.

>evel "

Challenging

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Chapter 19 8*

19.2 $ame

• +yrrolidine• +yrrole

• +iperidine

• Imidaole

• +yridine• +urine

• +yrimidine

• Indole

$

/

>evel 2

@asy

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Chapter 19 85

19.2 anser

• +yrrolidine

• +yrrole

• +iperidine• Imidaole

• +yridine

• +urine

• +yrimidine• Indole

>evel 2

@asy

+yrrolidine contains a fivemembered ringith no double bonds.

19 " Fi h h b idi i f

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Chapter 19 8(

19." Five the hybridiation for

a nitrogen in an amine.

• sp

• sp2

• sp"

• sp*

>evel 2

@asy

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Chapter 19 87

19." anser

• sp

• sp2

• sp"

• sp*

>evel 2

@asy

he nitrogen is sp3 hybridied in an amine.

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Chapter 19 88

19.*

• @thyl methyl

ammoniumbromide

• iethyl methyl

ammoniumbromide

• imethyl amine

• iethyl amine• @thyl methyl

amine

C/"$/2

1. C/"C/2#r

2. /#r

>evel "

Challenging

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Chapter 19 89

19.* anser

• @thyl methyl

ammonium bromide

•iethyl methylammonium bromide

• imethyl amine

• iethyl amine

• @thyl methyl amine

>evel "

Challenging

An ethyl group adds to the nitrogen.

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Chapter 19 9:

19.5

• $@thylpropanamide• $Methylpropanamide

• $@thylethanamide

• $Methylethanamide

O

CClC/"C/2

C/"$/2

>evel "

Challenging

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Chapter 19 91

19.5 anser

• $@thylpropanamide

• $Methylpropanamide

• $@thylethanamide

• $Methylethanamide

he methylamino group replaces the

chlorine group.

>evel "

Challenging

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Chapter 19 92

19.(

• $'$imethyl$nitrosoamine

• $'$imethyl diaonium salt

• $Methyl$nitrosoamine

• $o reaction

,C/"-2$/

/$O2

>evel "

Challenging

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Chapter 19 9"

19.( anser

• $'$imethyl$nitrosoamine

• $'$imethyl diaonium salt

• $Methyl$nitrosoamine

• $o reaction

he nitroso group replaces the hydrogen.

>evel "

Challenging

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Chapter 19 9*

19.7

• Aniline

• $itrosobenene

• #enene

• +henol

C(/

5$/

2

1. $a$O2

2. /"+O2

>evel "

Challenging

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Chapter 19 95

19.7 anser

• Aniline

• $itrosobenene

• #enene• +henol

he reaction of aniline ith sodium nitriteproduces the diaonium salt. he diaonium

salt is deaminated ith hypophosphorus acid.

>evel "

Challenging

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Chapter 19 9(

19.8

• "+entanone$ethylimine• 2+entanone$methylimine

• imethyl ethyl amine


O

C

C/2C/"C/"C/2

1. C/"$/2

2. /4

>evel "

Challenging

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Chapter 19 97

19.8 anser

• "+entanone$ethylimine

• 2+entanone$methylimine

•imethyl ethyl amine• iethyl methyl amine

he methyl amino group replaces the o6ygen.

,for this to happen I thin) the second stepshould be reduction and not protonationGGG-

>evel "

Challenging

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Chapter 19 98

19.9

• 1+ropanamine

• 2+ropanamine

• 2#utanamine

• 1#utanamine

C/"C/2C/2Cl

1. $a$"

2. /2' $i

>evel "

Challenging

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Chapter 19 99

19.9 anser

• 1+ropanamine

• 2+ropanamine

• 2#utanamine

• 1#utanamine

he aide replaces the chlorine and is

then reduced to an amine group.

>evel "

Challenging

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Chapter 19 1::

19.1:

• 1+ropanamine

• 2+ropanamine

• 2#utanamine

• 1#utanamine

C/"C/2C/2Cl

1. $aC$

2. /2' $i

>evel "

Challenging

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19.1: anser

• 1+ropanamine

• 2+ropanamine

• 2#utanamine

• 1#utanamine

he nitrile replaces the chlorine and is

then reduced to an amine group.

chapter 19 ver 1

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