Amines : Page 1

Amines Organic Bases • amines are trivalent, but with a tetrahedral arrangement of electron pairs, although the geometry of a neutral

amine is properly described as 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 • named as for alcohol, 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 • 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 (above) etc.

• find the longest chain that contains the maximum number of functional groups (alkene and amine) = 4 • 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 • some common names for amines

* you NEED TO 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 on N, these electrons are less reactive • aromatic amines are thus MUCH less basic than aliphatic amines

• 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:

• reaction does NOT go, H2O not strong enough nucleophile

However:

• NH3 is a stronger enough nucleophile that this SN2 reaction goes! • MeNH2 is an even stronger stronger nucleophile, reaction keeps going! • 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 (poor leaving group)

However, this one does (better leaving group)

Example

Overall

• the reaction is overall E2 ELIMINATION, forming an alkene with a neutral amine as the leaving group

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.) E2

exhaustivemethylation

1. XS CH3I

2. Ag2O/H2O

3. Heat

NH2H3C

MajorHofmann

MinorSayetzeff

+

Amines : Page 5

• HOWEVER, the LEAST substituted alkene is the major product (Hofmann product), not the more substituted Saytzeff (Zaitsev) alkene Why is this? There are TWO reasons, the first is a STERIC EFFECT, the second is an ELECTRONIC EFFECT Reason #1: The steric effect of the bulky -NR3 leaving group

• 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 (fewer gauche interactions) • therefore elimination in the Hofmann conformation is more likely because the structure spends more time in that lower energy conformation Reason #2: electronic effect • NH3 is a good leaving group, but not as good as Br–, I– etc. • at this point we need a more sophisticated and subtle look at the transition state that goes beyond conventional general organic chemistry, the transition state is not really "symmetrical", in that although all of the bonds are made and broken in one continuous process, as required for all concerted reactions, the bonds are not all broken and made at EXACTLY the same time • the proton "leaves" earlier than the NH3, which means that in the transition state the partial bond to the departing hydrogen atom is LONG and WEAK, but the leaving group leaves "later", which means that in the transition state the bond to the leaving group is still comparatively SHORT and STRONG, as illustrated below • because the proton leaves "earlier", a partial negative charge starts to develop on the carbon atom that the proton is attached to, and in the Hofmann elimination this carbon is less substituted: RECALL: negative charges are LESS STABLE on more substituted carbons, thus the transition state for the Hofmann elimination has the partial negative charge on the less substituted carbon, this transition state is lower in energy, it is thus formed faster (under kinetic control), the Hofmann alkene is thus the major product

NH2

Major Hofmann Minor Sayetzeff

+

H

NMe3

H3CH2C

H

HH

H

H

NMe3

H3C

H

HS

CH3

HHHS

1. XS CH3I

2. Ag2O/H2O

3. Heat

lowerenergy

conformation

H and leaving group are anti-

coplanar gaucheinteraction

bulky bulky

excess

NMe3

HHHSOHHO

NMe3

HS

OHHHHofmann

Sayetzeff

‡

NMe3

HS

HH

‡

HO

partial negative developing on 1° carbon

+ H2O + NH3

+ H2O + NH3

δ

δ

leaving "later"

leaving "later"

leaving "earlier"

leaving "earlier"

partial negative developing on 2° carbon

higherenergy ‡

lowerenergy ‡

faster

slower

Amines : Page 6

• this is a rather subtle advanced organic chemistry concept, but it is important that you are exposed to this thinking as we get towards the end of our general organic chemistry course 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

Example Problem 3

3 Synthesis of Amines Amines are made in REDUCTION REACTIONS, several of which we have ALREADY SEEN

NH

N

N1. Excess CH3I

+

+ Major, Hofmann

Minor, Sayetzeff

HeatN

2. Ag2O/H2O

HHO

HeatN

HO H

HN N+1. Excess CH3I

2. Ag2O/H2ONHHO

heat

mono-substituted alkene formed

NH2NMe

MeMe

1. Excess CH3I+

2. Ag2O/H2O3. Heat

NMe3

HHO

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

H

H

H H

H H H

H

amine

amine

amine

Amines : Page 7

• 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., no need to learn even more new reagents at this point 3.1 Amine Synthesis by Reduction of Azides: Synthesis of PRIMARY amines AGAIN, the reactions we already know in a synthetic context, AND A NEW ONE

• 1°, 2° or 3° amines can be made by reducing an amide, which in turn comes from an acid chloride, which in turn comes form a carboxylic acid • a 1° amine can be made from a nitrile, which can be made in an SN2 reaction from an appropriate alkyl bromide, ONE CARBON ATOM IS ADDED COMPARED TO THE STARTING ALKYL BROMIDE • a 1° amine can be made from an azide, which can be made in an SN2 reaction from an appropriate alkyl bromide, ZERO CARBON ATOMS ARE ADDED COMPARED TO THE STARTING ALKYL BROMIDE A New Reagent

Examples • Synthesize these two PRIMARY amines

OR...R

CN

RIMINE

1. LiAlH4 2. H3Oaddition of H

R'

RCN

R

R' H

H amineH2/Pd/C

OR...AZIDE 1. LiAlH4 2. H3O

addition of HH2/Pd/C

R CH2 N3 R CH2 NH

Hamine

new

BUT

related!

RCO

Cl RCO

NHR'R CH2 NHR'

R X

R C N

R N3 R NH2

R'NH2

Na+ –CN adds 1 carbon

does not add carbonNaN3

amide

R CH2 NH2

1. LiAlH4

2. H3O

SN2

acid chloride makes a 1°, 2° or 3° alkyl aminedoes not add a carbon

1° aminenitrilealkyl halide

azideSN2

R X1. LiAlH4

2. H3O

1. LiAlH4

2. H3O 1° amine

RCO

OH

SOCl2

acid

alkyl halideNEW

Na NNNsodium azide = Na+ –N3 =Very good nucleophilesmall "bullet" shaped

similar to the cyanide anionNC

cyanide anionazide anion

NaN3

Br

N3 NH2

NaCNC

NNH2

1. LiAlH4

2. H3O

1. LiAlH4

2. H3O

NBS/hνSN2

must add THIS carbon atom

no additional carbon atoms

adds carbon atom

Amines : Page 8

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 to form 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 REAGENTS 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

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 • NEW REDUCING AGENT, NaBH3CN Recall NaBH4:

/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

CO

H

NaBH3CN

CN H

CHN HH

reduced in situ

+ NH3ammonia

1° amine

new!!

organic acidLBLA

HBH

HH

LB/Nucleophile

CO

CN

H

slower

faster

strongerπ-bondO more

electronegative

Na+ –BH4

Amines : Page 9

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

Example of an INDIRECT reductive amination

Example of a 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

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

LBLA

2° amineH

ON+ NaBH3CN

1° amineH H

N

iminiumreduced in situ

HCN

HTsOH

or other organic acid

EtOH

LA LB

NHEt

H

O+ NEt

H

1. HCl cat.

2. H2/Pd/CLALB

NHH

H H

O+

H

H2N

NaBH3CN

TsOH

EtOH

LB

LA

XNH2

O+NH

NaBH3CN O H2N+

obviously can't have thisA B

can't make bond A

CAN make bond B

H+

EtOH

Amines : Page 10

5 Some Final Synthesis Problems 5.1 Synthesis of Amines Common Methods of Synthesizing Amines (this is not all of the methods, see Section 3 of these notes

Example Problem 1

• 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

OCH3NH2/H+ (cat.)

NaBH3CNNH

N

O1. LiAlH4

2. H3O+N

2. LiAlH4Br

3. H3O+

NH2

1. NaCN

2. LiAlH4Br

3. H3O+

NH2

reductive amination

amide reduction

azide reduction

nitrile reduction

1°, 2° or 3°

1°, 2° or 3°

1° only

1° only + 1 carbon atom1C 2C

1. NaN3

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

Amines : Page 11

Example Problem 2

• in this case 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 either of these two routes would work, as shown Example Final Exam Synthesis Problems Involving Amines Problem 1: Difficulty = "moderate": Synthesize methamphetamine (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 normal method of synthesizing meth, this is just an organic chemistry exercise! Problem 2: Difficulty = "easy"

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

HN

(±)

H3C NH2

TsOH Cat

O

OH

PCC

Br2/FeBr3 NaBH3CN

meth

Br

MgBr1.

2. H3O+

O (±)

(±)

Mg.THF

C

1. LiAlH42. H3O+

Br NaCNN

NH2

NBS/hν

Amines : Page 12

• in this case we need to make a primary amine AND we need to add one carbon atom compared to the starting structure, the nitrile reduction method works well here Problem 3: Difficulty = "difficult"

• this one is tricky because you have to construct the amine from which to make the amide • it is also tricky because the "order" of the reactants to make the amide are reversed, usually we make the acid chloride and react it with an amine on the reagent arrow, here we make an amine and have the acid chloride on the reagent arrow

N

O

Ph

N

O

PhO

H

H+ cat.NaBH3CNX

does not work for an amide

NH

O

PhCl

ignore stereochemistry

OOH

Br

MgBr

PCC

NBS/hν

Mg.THF

2. H3O+

1. H

ON

H

HH+ cat.NaBH3CN

Amines : Page 13

5.2 Different Types of Synthesis problems for the Final Exam Synthesis is hard because the strategy is often difficult and there are simply a lot of reactions to know. Synthesis problems on the final exam will be of two kinds. The first includes all of the major reactions we have learned this semester. The second kind of synthesis problem will include only a "minimal set" of reactions, that everybody should know. The idea here is these problems test your synthesis problem solving skills rather than your ability to learn a lot of reactions. There are two parts to this "minimal set: of reactions, the first are a set of basic functional group interconversions, the second includes the most basic carbon-carbon bond forming reactions, i.e. the Grignard and acetylide reactions. These two sets of reactions are summarized below, and also on the class website.

Example Synthesis Problems that Include Only Reactions from the "Minimal Set" Problem 1

PCCNBS/hν

MgBr

O

Br Mg.THF

OH

2. H3O+

1. H

O

Amines : Page 14

Problem 2

Problem 3

NBS/hν

Br

1.

O

OH

Ph

MCPBA2. H3O+

PhMgBr

Na+ –OH

Br2 Na/NH3(l)

Br

Br

Na

1. Excess NaNH2/heat2. H2O

NaNH2

Br

Amines : Page 15

6 Summary of Reactions Involving Amines Do NOT start studying by trying to memorize the reactions here! Work as many problems as you can, with this list of reactions in front of you if necessary, so that you can get through as many problems as you can without getting stuck on the reagents/conditions, and so that you can learn and practice solving reaction problems. Use this list AFTER you have worked all of the problems, and just before an exam. By then you will have learned a lot of the reagents/conditions just by using them and you will only have to memorize what you haven't learned yet. Then do the following: • Cover the entire page of reagents/conditions with a long vertical strip of paper, see if you can write down the reagents/conditions for each reaction, check to see which you get correct, if COMPLETELY correct, circle Y, if incorrect or even slightly incorrect, circle N. In this way you keep track of what you know and what you don't know. • Keep coming back to this list and so the same thing only for those reactions you circled N, until all are circled Y. Knowing the reagents/conditions on this page is INSUFFICIENT to do well on an exam since you will ALSO need to recognize how to use and solve reaction problems in different contexts, this page ONLY helps you to learn the reagents/conditions that you have not YET learned by working problems.

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

1. NaN32. LiAlH4

Br

3. H3O+

NH2

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

Y / N

DIRECT reductive amination