Amines : Page 1

Amines Organic Bases • Amines are trivalent, with a tetrahedral electron pair geometry, although the geometry of a neutral amine is a

trigonal pyramidal, since unlike nuclei, the exact positions of an electron pair cannot be defined. An exception is an ammonium salt that is surrounded by 4 atoms (nuclei).

1 Nomenclature IUPAC priority: acid > ester > amide > nitrile > aldehyde > ketone > alcohol > amine > alkene > alkyne > halide • Amines are named as for alcohols, except that you replace -ol with -amine, amine suffix. • When an anime is a substituent it is an amino substituent.

• Number the chain as usual to give the amine functional group the lowest number. • As mentioned above, amines are named as alcohols except that "amine" is used in place of "ol" and so instead of propan-ol we have propan-amine (above), instead of pentanol we have pentanamine, etc.

• Find the longest chain that contains the maximum number of functional groups (alkene and amine) = four. • Number to give the highest priority functional group amine the lowest number. • N comes before 3 when specifying the two methyl substituents.

• Remember, the common name aniline has been incorporated into IUPAC nomenclature, for example, aniline (above). • Some common names for amines

* You should know these two, piperidine seen in the Stork reaction, pyridine a common organic base.

N

1° Amine 2° Amine 3° Amine QuaternaryAmmonium Salt

R HH N

R R'H N

R R'R''

R1

NR2 R3

R4

trigonal pyramidal

tetrahedral

CH3CH2CH2NH2

1-propanamine(2S)-pentanamine

NH2(S)H

123 1 2 3 45

N-ethyl-N,3-dimethylbut-2-en-1-aminelongest chain 123

4 N

Me

Me

Et

Br

NH2

3-bromo-2-chloroaniline

Cl 12

3 #1 by definition

diethylamine diisopropylamine

putrescine

NH

HN

H2N NH2 NH2H2Ncadaverine

N HNN

piperidine pyrrolepyridine 2,6-lutidine

(amines can stink!)

* *HN

Amines : Page 2

2 Amines as Bases • Deciding the stronger of two acids, e.g. H-A1 and H-A2:

• The stronger acid has the more stable conjugate base anion, lower energy electrons in the base. • The stronger acid corresponds to the more favorable "reaction", left to right. : • Deciding the stronger of two bases, e.g. B1 and B2

• There are no anions here, but the stronger base still has the higher energy, more chemically reactive electrons. • The stronger base is stronger because the energy of the electrons go down more by getting into a bond. • The stronger base has the weaker conjugate acid since it has the stronger bond. Example:

Acidity is Measured in terms of pKa, How is Basicity Measured? • Basicity can be measured in terms of the corresponding pKb, however, in organic chemistry we often don't do that, we specify basicity in terms of the acidity of the conjugate acid. Acidity of an Amine in terms pf pKa:

Compare with (and don't confuse with) Basicity of an Amine in terms of pKa of the conjugate acid.

Trends in Basicity?

H A1 H+ + – A1

H A2 H+ + – A2

higher energy electronsstronger baseweaker acid

B1 H–B1

B2 H–B2+ H+

+ H+higher energy electrons

stronger base weaker conjugate acid

+ H+

+ H+

more stable cation

less stable cation

weaker base

stronger base

more favorable

lessfavorable

D group, raises energy of

electrons on N, more reactive

W group, lowers energy of

electrons on N, less reactive

D group stabilizes positive charge on N

W group destabilizes positive charge on N

RN

R

H

H3CO

RN

R

H

O2N

RN

R

H3CO

RN

R

O2N

Amineacid

strength

+ H3ONH3pKa ~35weak acid

NH2strongbase

+ H2O

Aminebase

strength

+ H3ONH4pKa ~9

moderate acid

NH3moderate

base

+ H2O

CH3NH2

(CH3)2NH

(CH3)3N

CH3NH3

(CH3)2NH2

(CH3)3NH

increasing cationstability

but

decreasing solvation

effects cancel!!

+ H3O NH4 pKa = 9.2NH3 + H2O

+ H3O + H2O

+ H3O + H2O

+ H3O + H2O

pKa = 10.9

pKa = 10.7

pKa = 9.8

Amines : Page 3

• There are no clear trends among the aliphatic amines because of the competing effects of alkyl group donation and reduced solvation of the larger cations. But

• Minor resonance delocalization (i.e. partial bonding) stabilizes (lowers energy) of the non-bonding electrons in aromatic N, these electrons are less reactive. • Aromatic amines are thus much less basic than aliphatic amines:

• There is a clear decrease in basicity with decreasing energy of non-bonding electrons as the hybridization of the orbitals gains more s character and less p character. • The protonated nitrile is about the strongest acid that we discuss this entire course! 4 Reactions of Amines 4.1 Amines a Nucleophiles Recall: • The reaction below does not go, H2O is not a strong enough nucleophile.

Reminder:

However:

• NH3 is a stronger enough nucleophile that this SN2 reaction goes! • MeNH2 is an even stronger nucleophile, the reaction keeps going! • One its own, this is clearly not very useful, due to "over-alkylation", except in the Hofmann elimination (see next reaction).

NH2 + H3O + H2ONH3

pKa ~ 42 other resonance contributors

NH2

NCH3CH3

H3CN CH3C N

sp3sp2

sp

"pKa" ~9.8 "pKa" ~5.3 "pKa" ~ -10!!!

decreasing basicity

XH3C IH

HO CH3 I

SN2H

HO

H HO

H HNH

H O

weakstrong intermediatenucleophilicity

H

HN H3C I

HH N CH3 IH

H

H HNH

H

HN CH3

H3C ICH3

H N CH3H

etc. etc.

CH3H3C N CH3

CH3

"exhaustive" alkylation product

H HNH

H3C

HN CH3

slightly stronger nucleophile

SN2

Amines : Page 4

4.2 Hofmann Elimination Revist a Concept we Have Seen Many Times Now:

This reaction also doesn't work because of the poor leaving group):

However, this one does because we have a much better leaving group:

Example:

Overall:

• The reaction is overall E2 elimination, forming an alkene with a neutral amine as the leaving group. • However, the least substituted alkene is the major product (Hofmann product), not the more substituted Saytzeff (Zaitsev) alkene. • "XS" is a commonly used abbreviation for "excess".

XC CH

OHCC

+ OH2

+ OHBase

no reaction : poor leaving group

reaction goes : better leaving group

C CH

OH2

CCBase

E2

E2

XC CH

NH2

CC + NH2Base

no reaction : poor leaving group

E2

+ NR3

reaction goes : better leaving group

C CH

NMe3

CCBase E2

C CH

NH2

excess Me-Iexhaustivemethylation

NH2H3C NMe3H3CNMe3H2C

HOH

CH2Excess Ag2O

H2O

Heat

converts from iodideto hydroxide anion = good base

CH3I

I

+ AgI (ppt.) E2exhaustivemethylation

1. XS CH3I2. Ag2O/H2O

3. Heat

NH2H3C

MajorHofmann

MinorSayetzeff

+

excess

Amines : Page 5

Explanation for Hofmann Product: A steric effect.

• The –NMe3+ group is large (bulky), it is roughly the same size as a t-butyl group • Therefore the E2 conformation for Hofmann elimination is lower in energy than the E2 conformation for Sayetzeff elimination for steric reasons (there are fewer higher energy gauche interactions). • Therefore elimination in the Hofmann conformation is more likely because the structure spends more time in that lower energy conformation. Example Problem 1

• There are two possible elimination products once the ammonium ion has been form in the reaction with CH3-I, the reaction that forms the least substituted (Hofmann) alkene is fastest, the Hofmann alkene is the major alkene product, together with its associated amine. • In some cases the amine leaving group will be a substantial major organic structure on its own. Example Problem 2

NH2

Major Hofmann Minor Sayetzeff

+

H

NMe3

H3CH2C

H

HH

H

H

NMe3

H3C

H

HS

CH3

HHHS

1. XS CH3I2. Ag2O/H2O

3. Heat

lowerenergy

conformation

H and leaving group are anti-

coplanar gaucheinteraction

bulky bulky

NH

N

N1. Excess CH3I

+

+ Major, Hofmann

Minor, Sayetzeff

HeatN

2. Ag2O/H2O

HHO

HeatN

HO H

HN N+2. Ag2O/H2O3. Heat

NHHO1. Excess CH3I

mono-substituted alkene formed

Amines : Page 6

Example Problem 3

3 Synthesis of Amines Amines are synthesized in Reduction Reactions, several of which we have already seen:

• In reality there are lots of different ways of doing these reductions (many different sets of reagents/conditions work), but here we use reagents/conditions that we already know for simplicity. There is no need to learn even more new reagents at this point (except one!). 3.1 Synthesis of amines using methods we know Again, we are now using reactions we already know in a synthetic context. Examples • As we have now seen many times, bromides are very convenient starting materials to start building new molecules, here we combine an SN2 reaction with reduction of a nitrile to make a primary amine.

• A 1° amine can therefore be made from a nitrile, where one carbon atom is added compared to the starting alkyl bromide. • 1°, 2° or 3° amines can be made by reducing an amide, which in turn comes from an acid chloride, which in turn comes from a carboxylic acid.

NH2N

Me

MeMe

1. Excess CH3I+

2. Ag2O/H2O3. Heat

NMe3

HHO

OR...

NO2 NH2/Pd/C

addition of H, removal of O

RCO

NHR' R C NHR'amide

1. LiAlH4

2. H3O

nitro

addition of H, removal of O

R C N R C Nnitrile

1. LiAlH4

2. H3Oaddition of H

seen

before

RCN

RIMINE

1. LiAlH4 2. H3Oaddition of H

H

H

H H

H H H

H

R'

RCN

R

R' H

H

amine

amine

amine

amineH2/Pd/C

new

BUT

related!

Br Na+ –CNC

NNH21. LiAlH4

2. H3O

NBS/hν

SN2 must add THIS carbon atomadds carbon atom

Amines : Page 7

3.2 Synthesis of Substituted Amines by Reduction of an Imine: Reductive Amination • Addition of nitrogen to a carbonyl to form an imine, followed by reduction forms an amine:

• So now we have a new reaction name: Reductive Amination, but is basically chemistry we already know • We already knew that we could react an amine with an aldehyde/ketone to form an imine (also known as a Schiff base), and it is no great surprise that the C=N bond of the imine (Schiff base) can be reduced by addition of H2 addition similarly to a C=O bond, we can even use the same reducing agents we already know! • Reduction of C=N is actually somewhat easier than reduction of C=O, N is less electronegative and the C-N pi-bond is thus a bit weaker than the C-O p-bond. Indirect Reductive Amination combines these reactions in sequence: Example 1

Example 2

1. LiAlH4

2. H3O

COH

OCCl

OSOCl2N

ONPhH

H

NHCH3

CH3

1° amine

2° amine

HPh

N

O

H3CCH3

NH

Ph

NH3C

CH3

1. LiAlH4

2. H3O

2° amide 2° amine

3° amide 3° amine

/H (cat.)

RC

R

NR'

RC

R

NR' H

H 2° amine

imine

RC

R

O

RC

R

OH

HOR...1. LiAlH4 2. H3O

H2/Pd/C

RNH2

OR...1. LiAlH4 2. H3O

H2/Pd/Creductiveamination

HCO

H (cat.) H2/Pd/CH

NCH3

CH2

NCH3H

+1° amine

CH3-NH2

HCO

1. CH3CH2NH2/H+ cat.

2. LiAlH43. H3O+

NH

Amines : Page 8

Direct Reductive Amination combines formation of the imine and the reduction in one reaction:

• "In Situ" means formed in the reaction, reaction "in situ" means that the imine reacts as it is formed. • Here is a final new reducing agent, NaBH3CN Recall NaBH4:

Compare NaBH3CN:

• The -CN withdrawing group makes the –BH3CN anion less nucleophilic, it reacts slower with carbonyls (C=O) and imines (C=N), but the reaction with C=O is slow enough that it does not happen under reductive amination conditions, Na+ –BH3CN is a selective reducing agent for imines. • Remember, "no reaction" doesn’t really mean no reaction at all, reactivity is not black and white, it means that the reaction with the C=O is now slow enough and the reaction with C=N fast enough to be useful in the reaction. • Another subtle point, EtOH may not be required as a solvent to provide the protons that go with the hydride, since the acid that is used to catalyze the reaction may also supply these protons. Examples

• Recall: Heuristics are skills students develop that allow them to solve the problems more quickly than by working them out from first principles, by skipping steps that become obvious. • Heuristics can't be taught, but here we will point you in the right direction to develop heuristics for reductive aminations.

CO

H

NaBH3CN

CN H

CHN HH

reduced in situ

+ NH3ammonia

1° amine

new!!

organic acid

HBH

HH

LB/Nucleophile

CO

CN

H

slower

faster

strongerπ-bondO more

electronegative

Na+ –BH4

HBH

HCNwithdrawing

LESS nucleophilic, less reactive

CO

CN

Hweakerπ-bondN less

electronegativeXNa+ –BH3CN

2° amine

ON

+

HN

NaBH3CN

1° amine

HN

iminereduced in situ

H

HTFA

or other organic acid

EtOH

2° amineH

ON+ NaBH3CN

1° amineH H

N

iminiumreduced in situ

HCN

HTsOH

or other organic acid

EtOH

Amines : Page 9

Example 1: Give the product of the following indirect reductive amination:

• Identify the atoms involved in the bond-forming reaction, don't forget basic Lewis acid/base nucleophile/electrophile principles! The electrons come from the nitrogen and make the new bond to the C=O carbon. • A N-C single bond forms so eliminate the oxygen. Example 2: Give the product of the following direct reductive amination:

• Note that the selective reducing agent NaBH3CN is included in the reagent/conditions in a "one pot" reaction. Example in Reverse • How can we make the provided amine using a reductive amination?

• Break bond "backwards", the benzene ring cannot carry a C=O, the carbonyl must be the fragment on the "right" • You can use either the direct or the indirect reductive amination methods in a synthesis problem. 5 Some Final Synthesis Problems Common Methods of Synthesizing Amines:

NHEt

H

O+ NEt

H

1. HCl cat.

2. H2/Pd/CLALB

nucleophilic atom, where the electrons "come from"

new N-Cbond forms

hereelectrophilic carbon accepts the electrons to make the new bond

from theamine

from theketone

NHH

H H

O+

H

H2N

NaBH3CN

TsOH

EtOH

LB

LAnew N-C bond

XNH2

O+NH

NaBH3CN O H2N+

obviously can't have thisA B

can't make bond A

CAN make bond B

H+

EtOH

reductive amination

amide reduction

nitrile reduction

1°, 2° or 3°

1°, 2° or 3°

1° only + 1 carbon atom1C 2C

OCH3NH2/H+ (cat.)

NaBH3CNNH

N

O1. LiAlH4

2. H3O+N

1. NaCN

2. LiAlH4Br

3. H3O+

NH2

Amines : Page 10

Example Problem 1 (difficulty = moderate):

• Many amines are biologically active, many act on the central nervous system, many drugs contain amine functional groups. • MDMA = N,a-dimethyl-1,3-benzodioxole-5-ethenamine = Ecstacy : 5g ~ 20 years!! • This is a secondary amine, which can in principle be synthesized either by reduction of an amide or by reductive amination, however, in this case only reductive amination would work. • Trying to make the amine via amide reduction in this case obviously doesn't work because the relevant amide can't exist.

Example Problem 2 (difficulty = moderate):

• In this case the strategy again is to identify that we need to make a tertiary amine, which requires either a reductive amination or reduction of a tertiary amide, in this case either of these two routes would work, as shown.

HN

(±)

H3C NH2

TsOH CatNaBH3CN

O

O

O

O

O

O OH

O

O O

O

O

Osamething

PCC

1. H2O/Hg(OAc)22. NaBH4

reductiveamination

MDMA

HNO

O (±)

HNO

O O

1. LiAlH4

can't exist2. H3O+

amide reductioncan't work here

CO2H

COCl CO

N

N

1. LiAlH42. H3O+

HNSOCl2

1. KMnO4/–OH/boil

2. H3O+

CO

H

HN H+ cat.NaBH3CN

Br

OH

Na+ –OH

(SN2) PCC

NBS/hν

Amines : Page 11

Example Problem 3 (fairly easy):

• In this case we need to make a primary amine band we need to add one carbon atom compared to the starting structure, the nitrile reduction method works well here Example Problem 4 (moderate difficulty): Synthesize methamphetamine (crystal meth) from benzene

• Again, the strategy again is to note that we need to make a tertiary amine, which requires either a reductive amination or reduction of a tertiary amide, in this case only the reductive amination works. • This is not the synthesis that Walter White used, this is just an organic chemistry exercise! Example Problem 5 (difficult)

C

1. LiAlH42. H3O+

Br NaCNN

NH2

NBS/hν

HN

(±)

H3C NH2

TsOH Cat

O

OH

PCC

Br2/FeBr3 NaBH3CN

meth

Br

MgBr1.

2. H3O+

O (±)

(±)

Mg.THF

NH

OOH

Br

MgBr

PCC

NBS/hν

Mg.THF

2. H3O+

1. H

O

NH

HH+ cat.NaBH3CN

NHO

no good amide route

Amines : Page 12

• This one is difficult mainly because it includes many steps, but the strategy is actually quite straightforward. • We need to make a secondary amine, and there is no obvious amide reduction route, therefore we should think about reductive amination. • We learned to do reductive aminations "backwards", and the obvious N-C bond to make is the one indicated by the dashed line. • After that, it is straightforward reactions we learned earlier in the course. 6 Summary of Reactions Involving Amines

You have seen parts of some of these reactions in earlier sections!

1. NaCN2. LiAlH4

Br3. H3O+

NH2

NO2 H2/Pd/C NH2

1. NH3/H+ (cat.)O NH2

amphetamine

OCH3NH2/H+ (cat.)NaBH3CN/EtOH

NH

H(CH3)2NH/H+ (cat.)

O N

2. H2/Pd/C

INDIRECT reductive amination

NO 1. LiAlH4

2. H3O+N

amine synthesis

HN 1. XS CH3I

2. Ag2O / H2O/Heat

N

1°

2°

3°

Hofmann eliminationNaBH3CN/EtOH

Y / N

Y / N

Y / N

Y / N

Y / N

Y / N

Y / N

DIRECT reductive amination