2 Structure & Classification Amines are classified as:1°, 2°, or , 3° amines: Amines in which there are 1, 2, or 3 alkyl or aryl groups.
3 Structure & Classification Amines are further divided into aliphatic, aromatic, and heterocyclic amines:Aliphatic amine: An amine in which nitrogen is bonded only to alkyl groups.Aromatic amine: An amine in which nitrogen is bonded to one or more aryl groups.
4 Structure & Classification Heterocyclic amine: An amine in which nitrogen is one of the atoms of a ring.
5 Structure & Classification Example: Classify each amino group by type.
6 NomenclatureAliphatic amines: replace the suffix -e of the parent alkane by -amine.
7 NomenclatureThe IUPAC system retains the name aniline.
8 NomenclatureAmong the various functional groups, -NH2 is one of the lowest in order of precedence.Amine vs alcoholAmine vs acid
9 NomenclatureCommon names for most aliphatic amines are derived by listing the alkyl groups bonded to nitrogen in one word ending with the suffix amine.
10 NomenclatureWhen four groups are bonded to nitrogen, the compound is named as a salt of the corresponding amine.
11 Chirality of AminesConsider 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.
12 Chirality of AminesIn practice, however, they cannot be resolved because they undergo inversion, which converts one enantiomer to the other.
13 Chirality of AminesPyramidal inversion is not possible with quaternary ammonium ions, and their salts can be resolved.
14 Physical PropertiesAmines 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 ( =0.9) is less than that between O and H ( = 1.4).Using bp as an indication of H bondingIncreasing strength
15 BasicityAll 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.
16 BasicityUsing values of pKa, we can compare the acidities of amine conjugate acids with other acids.
19 Basicity-Aromatic Amines Aromatic amines are considerably weaker bases than aliphatic amines.
20 Basicity-Aromatic Amines Aromatic amines are weaker bases than aliphatic amines because of two factors:Resonance stabilization of the free base, which is lost on protonation.
21 Basicity-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 substituentsElectron-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.
22 Example: Basicity-Aromatic Amines 3-nitroaniline is a stronger base than 4-Nitroaniline.Cannot do this kind of resonance in 3 nitroaniline
23 Basicity-Aromatic Amines Heterocyclic aromatic amines are weaker bases than heterocyclic aliphatic amines.
24 Basicity-Aromatic Amines 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).
25 Basicity-Aromatic Amines Imidazole Which N lone pair is protonated? The one which is not part of the aromatic system.
26 Basicity-GuanidineGuanidine is the strongest base among neutral organic compounds.Its basicity is due to the delocalization of the positive charge over the three nitrogen atoms.
27 Reaction with AcidsAll amines, whether soluble or insoluble in water, react quantitatively with strong acids to form water-soluble salts.
28 Reaction with acidsSeparation and purification of an amine and a neutral compound.
29 Preparation 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 iminesreduction of imines to aminesreduction of amides to amines by LiAlH4reduction of nitriles to a 1° aminenitration of arenes followed by reduction of the NO2 group to a 1° amine
30 Preparation Alkylation of ammonia and amines by SN2 substitution. Unfortunately, such alkylations give mixtures of products through a series of proton transfer and nucleophilic substitution reactions.polyalkylations
31 Preparation via Azides Alkylation of azide ion.OverallAlkyl Halide Alkyl amine
32 Example: Preparation via Azides Alkylation of azide ion.Note retention of configuration, trans trans
33 Reaction with HNO2Nitrous acid, a weak acid, is most commonly prepared by treating NaNO2 with aqueous H2SO4 or HCl.In its reactions with amines, nitrous acid:Participates in proton-transfer reactions.A source of the nitrosyl cation, NO+, a weak electrophile.
34 Reaction with HNO2 NO+ is formed in the following way. Step 1: Protonation of HONO.Step 2: Loss of H2O.We study the reactions of HNO2 with 1°, 2°, and 3° aliphatic and aromatic amines.
35 Tertiary Amines with HNO2 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.
36 Secondary Amines with HNO2 2° Aliphatic and aromatic amines react with NO+ to give N-nitrosamines.carcinogensMechanism:
37 RNH2 with HNO21° 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 N2 to give a carbocation.Diazonium ion: An RN2+ or ArN2+ ion
38 1° RNH2 with HNO2 Formation of a diazonium ion. Step 1: Reaction of a 1° amine with the nitrosyl cation.Step 2: Protonation followed by loss of water.
39 1° RNH2 with HNO2 (Aliphatic) Aliphatic diazonium ions are unstable and lose N2 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.
40 1° RNH2 with HNO2Tiffeneau-Demjanov reaction: Treatment of a -aminoalcohol with HNO2 gives a ketone and N2.
41 Mechanism of Tiffeneau-Demjanov Reaction with NO+ gives a diazonium ion.Concerted loss of N2 and rearrangement followed by proton transfer gives the ketone.Similar to pinacol rearrangement
42 Pinacol Rearrangement: an example of stabilization of a carbocation by an adjacent lone pair. Overall:
43 This is a protonated ketone! MechanismReversible protonation.Elimination of water to yield tertiary carbocation.This is a protonated ketone!1,2 rearrangement to yield resonance stabilized cation.Deprotonation.
44 1° Primary Amines with HNO2 (Aromatic) The -N2+ group of an arenediazonium salt can be replaced in a regioselective manner by these groups.
45 1° ArNH2 with HNO2A 1° aromatic amine converted to a phenol.
46 1° ArNH2 with HNO2Problem: What reagents and experimental conditions will bring about this conversion?
47 1° ArNH2 with HNO2Problem: Show how to bring about each conversion.
48 Hofmann EliminationHofmann elimination: Thermal decomposition of a quaternary ammonium hydroxide to give an alkene.Step 1: Formation of a 4° ammonium hydroxide.
49 Hofmann EliminationStep 2: Thermal decomposition of the 4° ammonium hydroxide.
50 Hofmann EliminationHofmann elimination is regioselective - the major product is the least substituted alkene.Hofmann’s rule: Any -elimination that occurs preferentially to give the least substituted alkene as the major product is said to follow Hofmann’s rule.
51 Hofmann EliminationThe regioselectivity of Hofmann elimination is determined largely by steric factors, namely the bulk of the -NR3+ group.Hydroxide ion preferentially approaches and removes the least hindered hydrogen and, thus, gives the least substituted alkene.Bulky bases such as (CH3)3CO-K+ give largely Hofmann elimination with haloalkanes.
52 Cope EliminationCope 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 -hydrogen, it undergoes thermal decomposition to give an alkene.
53 Cope EliminationCope elimination shows syn stereoselectivity but little or no regioselectivity.Mechanism: a cyclic flow of electrons in a six-membered transition state.