William H. Brown & Christopher S. Foote 22-1

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1 William. Brown & Christopher S. Foote 22-1

2 Chapter

3 Structure & Classification Amines are classified as 1,, 2, 2, or, 3 3 amines amines in which 1, 2, or 3 hydrogens of N 3 are replaced by alkyl or aryl groups C 3 C 3 - N 2 C 3 - N C 3 - N Methylamine (a 1 amine) Dimethylamine (a 2 amine) C 3 C 3 Trimethylamine (a 3 amine) 22-3

4 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 C 3 C 3 N 2 N- C 2 - N- C 3 Aniline (a 1 aromatic amine) N-Methylaniline (a 2 aromatic amine) Benzyldimethylamine (a 3 aliphatic amine) 22-4

5 Structure & Classification heterocyclic amine an amine in which nitrogen is one of the atoms of a ring N N Pyrrolidine Piperidine (heterocyclic aliphatic amines) N N Pyrrole Pyridine (heterocyclic aromatic amines) 22-5

6 Structure & Classification Example classify each amino group by type C 3 N O (a) (b) N (c) N C N 3 (S)-(+)-Coniine (S)-(-)-Nicotine Cocaine OC 3 O O 22-6

7 Structure & Classification Aliphatic amines replace the suffix -e of the parent alkane by -amine N 2 N 2 2 N N 2 C Propanamine (S)-1-Phenylethanamine 1,6-exanediamine 22-7

8 Nomenclature The IUPAC system retains the name aniline N 2 N 2 N 2 N 2 Aniline NO 2 4-Nitroaniline (p- Nitroaniline) C 3 OC 3 4-Methylaniline (p-toluidine) 3-Methoxyaniline (m- Anisidine) 22-8

9 Nomenclature Among the various functional groups discussed in the text, -N 2 is one of the lowest in order of precedence 2 N O O 2 N COO 2-Aminoethanol N 2 (S)-2-Amino-3-methyl- 1-butanol 4-Aminobenzoic acid 22-9

10 Nomenclature Common names for most aliphatic amines are derived by listing the alkyl groups bonded to nitrogen in one word ending with the suffix - amine C 3 N 2 N 2 N N Methylamine tert- Butylamine Dicyclopentylamine Triethylamine 22-10

11 Nomenclature When four groups are bonded to nitrogen, the compound is named as a salt of the corresponding amine + ( C 3 ) 4 N + Cl - Tetramethylammonium chloride Cl - O - NC 2 ( C 2 ) 1 2 C 3 C 2 N( C 3 ) 3 + Tetradecylpyridinium chloride (Cetylpyridinium chloride) Benzyltrimethylammonium hydroxide 22-11

12 Chirality of Amines if we 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 3 different groups bonded to N is chiral and exists as a pair of enantiomers and, in principle, can be resolved N 3 C C 2 C 3 (S)-Ethylmethylamine N C 3 C 3 C 2 (R)-Ethylmethylamine 22-12

13 Chirality of Amines in practice, however, they cannot be resolved because they undergo pyramidal inversion, which converts one enantiomer to the other C 3 N C 2 C 3 unhybridized 2p orbital sp 3 hybrid N C 3 C 2 C 3 Planar transition state C 3 C 2 C 3 N sp 3 hybrid (S)-Ethylmethylamine (R)-Ethylmethylamine 22-13

14 Chirality of Amines pyramidal inversion is not possible with quaternary ammonium ions, and their salts can be resolved 3 C Ph Cl - N Cl - Ph N C 3 R enantiomer S enantiomer 22-14

15 Physical Properties Amines are polar compounds, and both 1 and 2 amines form intermolecular hydrogen bonds N-----N hydrogen bonds are weaker than O-----O hydrogen bonds because the difference in electronegativity between N and ( =0.9) is less than that between O and ( = 1.4) C 3 C 3 C 3 N 2 C 3 O MW (g/mol) bp ( C)

16 NMR Spectroscopy 1 -NMR the chemical shift of amine hydrogens is variable, and may be found in the region 0.5 to 5.0 depending on the solvent, concentration, and temperature they generally appear as singlets 13 C-NMR NMR carbons bonded to nitrogen are generally shifted approximately 20 ppm downfield relative to their signal in an alkane of comparable structure 22-16

17 IR Spectroscopy 1 and 2 amines show N- stretching absorption in the region cm -1 1 amines show two bands in this region 2 amines show only one band in this region 22-17

18 Basicity All amines are weak bases, and aqueous solutions of amines are basic + C 3 - N + - O- C 3 - N- - O- Methylamine Methylammonium hydroxide + - [C K b = K = = 4.37 x 10-4 eq [ 2 O] 3 N 3 ][O ] [C 3 N 2 ] 22-18

19 Basicity it is more common to discuss the basicity of amines by reference to the acid ionization constant of the corresponding conjugate acid C 3 N O C 3 N O + [ C 3 N 2 ] [ 3 O + ] K = = 2.29 x a pk + a = [ C 3 N 3 ] for any acid-conjugate base pair pk a + pk b =

20 Basicity using values of pk a, we can compare the acidities of amine conjugate acids with other acids C 3 N 2 (stronger base) + C 3 COO pk a 4.76 (stronger acid) + C 3 N 3 pk a (weaker acid) + C 3 COO - pk eq = (weaker base) K eq = 7.6 x

21 Basicity-Aliphatic Amines Aliphatic Amines Amine Structure pk a pk b Ammonia Primary Amines methylamine ethylamine cyclohexylamine N 3 C 3 N 2 C 3 C 2 N 2 C N Secondary Amines dimethylamine diethylamine ( C 3 ) 2 N ( C 3 C 2 ) 2 N Tertiary Amines trimethylamine triethylamine ( C 3 ) 3 N ( C 3 C 2 ) 3 N

22 Basicity-Aromatic Amines Aromatic amines Amine Structure pk a pk b aniline N methylaniline C 3 N chloroaniline Cl N nitroaniline O 2 N N

23 Basicity-Aromatic Amines aromatic amines are considerably weaker bases than aliphatic amines N O N 3 + O - pk a = Cyclohexylamine Cyclohexylammonium hydroxide N O N 3 + O - pk a = 4.63 Aniline Anilinium hydroxide 22-23

24 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 N N N N Interaction of the electron pair on nitrogen with the pi system of the aromatic ring 22-24

25 Basicity-Aromatic Amines resonance delocalization of the electron pair on nitrogen by interaction with the pi system of the aromatic ring decreases basicity unhybridized 2p orbital of N N nitrogen is sp 2 hybridized 22-25

26 Basicity-Aromatic Amines the greater electron-withdrawing inductive effect of the sp 2 carbon of an aromatic amine compared with the sp 3 carbon of an aliphatic amine also decreases basicity Electron-releasing, 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 22-26

27 Basicity-Aromatic Amines 4-nitroaniline is a weaker base than 3-nitroaniline O 2 N N 2 O 2 N N 2 3-Nitroaniline pk b Nitroaniline pk b O - O + N N 2 + N + N 2 O O

28 Basicity-Aromatic Amines eterocyclic aromatic amines are weaker bases than heterocyclic aliphatic amines N N N Piperidine Pyridine Imidazole pk a pk a 5.25 pk a 6.95 N 22-28

29 Basicity-Aromatic Amines in pyridine, the unshared pair of electrons on N is not part of the aromatic sextet N an sp 2 hybrid orbital; the electron pair in this orbital is not a part of the aromatic sextet nitrogen is sp 2 hybridized pyridine is a weaker base than heterocyclic aliphatic amines because the free electron pair on N lies in an sp 2 hybrid orbital (33% s character) and is held more tightly to the nucleus than the free electron pair on N in an sp 3 hybrid orbital (25% s character) 22-29

30 Basicity-Aromatic Amines Imidazole This electron pair is a part of the aromatic sextet This electron pair is not a part of the aromatic sextet N N Imidazole Aromaticity is maintained when imidazole is protonated + 2 O N+ + O - N Imidazolium ion 22-30

31 Basicity-Guanidine Guanidine is the strongest base among neutral organic compounds N N N C N O 2 N C N 2 + O - pk a = 13.6 Guanidine Guanidinium ion its basicity is due to the delocalization of the positive charge over the three nitrogen atoms + N 2 N 2 N N C N2 2 N C N 2 2 N C N2 Three equivalent contributing structures 22-31

32 Reaction with Acids All amines, whether soluble or insoluble in water, react quantitatively with strong acids to form water-soluble salts O O N 2 + Cl 2 O O (R)-(-)-Norepinephrine (only slightly soluble in water) O O N + 3 Cl - O (R)-(-)-Norepinephrine hydrochloride (a water-soluble salt) 22-32

33 Preparation We have already covered these methods nucleophilic ring opening of an epoxide by ammonia and amines (11.9B) addition of ammonia and 1 and 2 amines to aldehydes and ketones to give an imine followed by reduction of the imine to an amine (16.10) reduction of an amide by LiAl 4 (18.11B) reduction of a nitrile to a 1 amine (18.11C) ofmann rearrangement of a 1 amide (18.12) nitration of an arene followed by reduction of the NO 2 group to a 1 amine (21.1B) 22-33

34 Preparation Alkylation of ammonia and amines by S N 2 unfortunately, such alkylations give mixtures of products through a series of proton transfer and nucleophilic substitution reactions C 3 Br + N 3 + S N 2 C 3 Br N 3 C 3 N 3 + Br - Methylammonium bromide C 3 N + 3 Br - + (C 3 ) 2 N + 2 Br - + (C 3 ) 3 N + Br - + (C 3 ) 4 N + Br

35 Preparation via Azides Alkylation of azide ion N N N N RN R N N N - 3 Azide ion (a good nucleophile) An alkyl azide C 2 Cl Benzyl chloride K + - N 3 (phase-transfer catalyst) 1. LiAl 4 C 2 N O C 2 N 2 Benzyl azide Benzylamine 22-35

36 Preparation via Azides alkylation of azide ion Cycloh exene ArCO 3 O 1,2-Epoxycyclohexan e 1. K + N O O N 3 1. LiAl O O N 2 t rans-2-azidocycloh exanol t rans-2-aminocycloh exanol 22-36

37 Reaction with NO 2 Nitrous acid is a weak acid, most commonly prepared by treating aqueous NaNO 2 aqueous 2 SO 4 or Cl NO 2 2 O 3 O NO 2 K a = 4.26 x 10-4 In its reactions with amines, it participates in proton-transfer reactions pk a = 3.37 is a source of the nitrosyl cation, NO +, a weak electrophile 22-37

38 Reaction with NO 2 NO + is formed in the following way O N O O N O + + O + N O N O Nitrosyl cation as a hybrid of two contributing structures we study the reactions of NO 2 with 1, 2, and 3 aliphatic and aromatic amines 22-38

39 Amines with NO 2 3 aliphatic amines, whether water-soluble or waterinsoluble, are protonated to form water-soluble salts 3 aromatic amines NO + is a weak electrophile and, as such, participates in EAS N 1. NaNO 2, Cl, 0-5 C N N= O 2. NaO, 2 O N,N-Dimethylaniline N,N-Dimethyl-4- nitrosoaniline 2 aliphatic and aromatic amines react with NO + to give N-nitrosoamines N- + NO 2 N-N=O + 2 O Piperidine N-N itrosopip erid in e 22-39

40 RN 2 with NO 2 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 an RN 2+ or ArN 2+ ion 22-40

41 1 RN 2 with NO 2 Formation of a diazonium ion Step 1 reaction of a 1 amine with the nitrosyl cation + R-N 2 + N O R-N-N=O A 1 aliphatic amine An N-nitrosamine keto-enol tautomerism R-N=N-O- A diazotic acid Step 2 protonation followed by loss of water R-N=N-O- A diazotic acid + R-N N + R N N + N N A diazonium ion + O- - 2 O R + A carbocation 22-41

42 1 RN 2 with NO 2 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 1 and/or 2 (5.2%) Cl N 2 NaNO 2, Cl 0-5 o C O (13.2%) + (25%) O + (25.9%) (10.6%) 22-42

43 1 RN 2 with NO 2 Tiffeneau-Demjanov reaction treatment of a aminoalcohol with NO 2 gives a ketone and N 2 O C 2 N 2 A -aminoalcohol O + NO O + N 2 Cycloheptanone 22-43

44 1 RN 2 with NO 2 reaction with NO + gives a diazonium ion concerted loss of N 2 and rearrangement followed by proton transfer gives the ketone O O- NO C 2 N 2 2 C 2 + N N -N 2 O + C 2 (A diazonium ion) + O proton transfer to 2 O C 2 O A resonance-stabilized cation Cycloheptanone 22-44

45 1 ArN 2 with NO 2 The -N 2+ group of an arenediazonium salt can be replaced in a regioselective manner by these groups Ar- N 2 + ( - N 2 ) 2 O BF 4 Cl, CuCl Br, CuBr KCN, CuCN KI 3 PO 2 Ar- O Ar- F Ar- Cl Ar- Br Ar- CN Ar- I Ar- Schiemann reaction Sandmeyer reaction 22-45

46 1 ArN 2 with NO 2 A 1 aromatic amine can be converted to a phenol N 2 Br 1. NaNO 2, 2 SO 4, 2 O 2. 2 O, heat O Br C 3 2-Bromo-4- methylaniline C 3 2-Bromo-4- methylphenol 22-46

47 1 ArN 2 with NO 2 Problem what reagents and experimental conditions will bring about this conversion? C 3 C 3 COO COO (1) (2) (3) (4) COO NO 2 NO 2 N 2 O 22-47

48 1 ArN 2 with NO 2 Problem Show how to bring about each conversion C 3 Cl (5) C 3 N 2 (6) C 3 C N (7) C 3 C 2 N 2 (8) C 3 C 3 N 2 (9) Cl Cl Cl Cl 22-48

49 ofmann Elimination ofmann elimination thermal decomposition of a quaternary ammonium hydroxide to give an alkene Step 1 formation of a 4 ammonium hydroxide C 3 I O C 2 -N- C 3 + Ag 2 O C 3 (Cyclohexylmethyl)trimethylammonium iodide Silver oxide C 3 O + - C 2 -N- C 3 + AgI C 3 (Cyclohexylmethyl)trimethylammonium hydroxide 22-49

50 ofmann Elimination Step 2 thermal decomposition of the 4 ammonium hydroxide C + 3 C 2 -N- C 3 C 3 (Cyclohexylmethyl)trimethylammonium hydroxide O C 2 + ( C 3 ) 3 N + 2 O Methylenecyclohexane Trimethylamine 22-50

51 ofmann Elimination ofmann elimination is regioselective - the major product is the least substituted alkene C 3 heat + N(C 3 ) 3 O - C 2 + (C 3 ) 3 N + 2 O ofmann s s rule any -elimination that occurs preferentially to give the least substituted alkene as the major product is said to follow ofmann s rule 22-51

52 ofmann Elimination O - C C 3 C 2 C + N( C 3 ) 3 E2 reaction (concerted elimination) O C 3 C 2 C C N( C 3 ) 3 the regioselectivity of ofmann 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 (C 3 ) 3 CO - K + give largely ofmann elimination with alkyl halides 22-52

53 Cope Elimination Cope elimination thermal decomposition of an amine oxide Step 1 oxidation of a 3 amine gives an amine oxide C 2 N-C O 2 - O + C 2 N-C O C 3 C 3 An amine oxide Step 2 if the amine oxide has at least one -hydrogen, it undergoes thermal decomposition to give an alkene O C C 2 N-C 3 C 2 + ( C 3 ) 2 NO C 3 Methylene cyclohexane N,N-Dimethylhydroxylamine 22-53

54 Cope Elimination Cope elimination shows syn stereoselectivity but little or no regioselectivity mechanism a cyclic flow of electrons in a sixmembered transition state C C C 3 N + - O C 3 Transition state heat C O C N C 3 C 3 an alkene N,N-dimethylhydroxylamine 22-54

55 Prob From each pair, select the stronger base. N 2 C 2 N 2 N 2 (a) or (b) or O 2 N C 3 3 C N 2 (c) N or N (d) O 2 NCN 2 or N 2 NCN 2 (e) N or N (f) ( C 3 C 2 C 2 ) 3 N or N(C 3 )

56 Prob Calculate the ratio of morpholine to its conjugate acid at p 7.0. At what p are the concentrations of morpholine and its conjugate acid equal? O + N + O N + 3 O + 2 O pk a = 8.33 Morpholinium ion Morpholine 22-56

57 Prob Which of the nitrogens of pyridoxamine is the stronger base? Explain. O C 2 N 2 C 2 O 3 C N Pyridoxamine (Vitamin B 6 ) 22-57

58 Prob Which of the nitrogens in epibatidine is the stronger base? ow many stereocenters are present in epibatidine? Cl N N Epibatidine 22-58

59 Prob Complete each acid-base reaction and predict whether equilibrium lies to the left or right. (a) C 3 COO Acetic acid + N Pyridine (b) O + ( C 3 C 2 ) 3 N Phenol Triethylamine (c) PhC C + N 3 Phenylacetylene Ammonia 22-59

60 Prob (cont d) Complete each acid-base reaction and predict whether equilibrium lies to the left or right. (d) PhC C + [ ( C 3 ) 2 C] 2 N - Li + Phenylacetylene O PhCO - Na + + Sodium benzoate Lithium diisopropylamide (LDA) (e) ( C 3 C 2 ) 3 N + Cl - Triethylammonium chloride (f) C 3 PhC 2 CN 2 1-Phenyl-2- propanamine (Amphetamine) + O O C 3 CCO 2-ydroxypropanoic acid (Lactic acid) 22-60

61 Prob (cont d) Complete each acid-base reaction and predict whether equilibrium lies to the left or right. C 3 (g) PhC 2 CN 3 + Cl - Amphetamine hydrochloride + NaCO 3 Sodium bicarbonate (h) O + ( C 3 ) 4 N + O - Phenol Tetrmethylammonium hydroxide 22-61

62 Prob Provide an explanation for fact that quinuclidine is a considerably stronger base than triethylamine. N Quinuclidine pk a 10.6 N Triethylamine pk a

63 Prob Devise a chemical procedure to separate a mixture of these three compounds and recover each in pure form. 3 C NO 2 3 C N 2 3 C O 4-Nitrotoluene 4-Methylaniline (p-toluidine) 4-Methylphenol (p-cresol) 22-63

64 Prob Show how to convert each compound to benzylamine. O O (a) (b) N (c) O N 2 O O Cl (d) (e) O (f) O 22-64

65 Prob Propose a structural formula for acetylcholine chloride. O ( C + 3 ) 3 N C 3 COC 2 C 2 Cl C ClNO 2 acetylcholine chloride 22-65

66 Prob Show how the hydroxylated compound can be transformed into an alkyldiazonium ion, an active carcinogen, in the presence of an acid catalyst. N N= O O + O O 2 N N= O cyt P-450 N + 2 N-Nitrosopiperidine 2-ydroxy-Nnitrosopiperidine An alkyl diazonium ion (a carcinogen) 22-66

67 Prob Analyze the mechanism of each rearrangement, and list their similarities and differences. Nitrous acid deamination of a -aminoalcohol O N 2 NaNO 2, Cl O N O Pinacol rearrangement O O 2 SO 4 CO + 2 O 22-67

68 Prob Propose a mechanism for this conversion. Account for the stereospecificity of the reaction. O O 2 N O O NaNO 2, Cl 0-5 C OOC O O + N 2 (S)-Glutamic acid 22-68

69 Prob Propose structural formulas for compound A and B. N C 3 ( C N) 1. C 3 I 2. Ag 2 O, 2 O 3. heat C N (A) 4. C 3 I 5. Ag 2 O, 2 O 6. heat C (B) 22-69

70 Prob Propose a structural formula for C C 3 I, 2 moles 2. 2 O 2 3. heat C N

71 Prob Propose a mechanism for pyrolysis of this ester. O O C 3 C 2 C 2 C 2 OCC 500 C 3 C3 C 2 C= C + 2 C 3 CO Butyl acetate 1-Butene Acetic acid 22-71

72 Prob Show how propranil can be synthesized from benzene and propanoic acid. Cl Cl ( 2 ) ( 1 ) ( 3 ) NO 2 Cl Cl Cl ( 4 ) Cl ( 5 ) Cl Cl NO 2 N 2 NCC 2 C 3 O Propranil 22-72

73 Prob Show to bring about each step in this synthesis. C 3 C 3 C 3 (1) (2) (3) NO 2 N 2 C 3 C 3 (4) CN C 2 N

74 Prob Show how to bring about this synthesis. C 3 C 3 O O 22-74

75 Prob Show how to bring about the conversion of phenol to 4- methoxybenzylamine. O OC 3 OC 3 (1) (2) (3) NO 2 OC 3 OC 3 (4) (5) OC 3 N 2 CN C 2 N

76 Prob Show how to prepare methylparaben from toluene. O C OC 3 O Methyl p-hydroxybenzoate (Methylparaben) C

77 Prob Show how to synthesize this 3 amine from benzene. O O O N Br 22-77

78 Prob Show how to prepare N-methylmorpholine from the named starting materials. O O O C 3 N 2 + N N O C 3 C 3 N-Methylmorpholine Methylamine Ethylene oxide 22-78

79 Prob Given this retrosynthetic analysis, propose a synthesis for tridemorph, a systemic agricultural fungicide. C 3 O N C 3 ( C 2 ) 10 COO + C 3 C= C 2 3 C ( C 2 ) 12 C 3 Tridemorph Dodecanoic acid (Lauric acid) Propene 22-79

80 Prob Propose a mechanism for this example of a Ritter reaction. What would be the product of a Ritter reaction using acetonitrile? O CN 2 SO 4 O N-C- Ritter product 2 O KO N

81 Prob Show how each of these diamines can be prepared from acetonitrile. 2 N N 2 1,3-Propanediamine 2 N N 2 1,4-Butanediamine C2 = C- C N Acrylonitrile 22-81

82 Prob Given this retrosynthetic analysis, propose a synthesis for the intravenous anesthetic propofol. O O O O 2,6-Diisopropylphenol (Propofol) NO 2 NO 2 Phenol 22-82

83 Prob Given this retrosynthetic analysis, propose a synthesis for propoxyphene. N O Ph Ph O Propoxyphene N (1) (2) O Ph Ph N (3) (4) Ph O Ph O Ph O 1-Phenyl-1-propanone (Propiophenone) 22-83

84 End Chapter

896 Chapter 21 Amines H H N R R R N R H R N R H O H R 3 N CH 2 NH 2 NHCH 3

896 Chapter 21 Amines H H N R R R N R H R N R H O H R 3 N CH 2 NH 2 NHCH 3 896 Chapter 21 Amines 21.20 Summary Section 21.1 Section 21.2 Section 21.3 Section 21.4 Alkylamines are compounds of the type shown, where R, R, and R are alkyl groups. ne or more of these groups is an