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Amines

Amines. Chapter 23. Structure & Classification. Amines are classified as: 1°, 2°, or , 3° amines: Amines in which there are 1, 2, or 3 alkyl or aryl groups. Structure & Classification. Amines are further divided into aliphatic, aromatic, and heterocyclic amines:

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Amines

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  1. Amines Chapter 23

  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: • Aliphaticamine: 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. Nomenclature • Aliphatic amines: replace the suffix -e of the parent alkane by -amine.

  6. Nomenclature • The IUPAC system retains the name aniline.

  7. Nomenclature • Among the various functional groups, -NH2 is one of the lowest in order of precedence. Amine vs alcohol Amine vs acid

  8. 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.

  9. Nomenclature • When four groups are bonded to nitrogen, the compound is named as a salt of the corresponding amine.

  10. Chirality of Amines • 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 three different groups bonded to N is chiral and exists as a pair of enantiomers and, in principle, can be resolved.

  11. Chirality of Amines • In practice, however, they cannot be resolved because they undergo inversion, which converts one enantiomer to the other.

  12. Chirality of Amines • Pyramidal inversion is not possible with quaternary ammonium ions, and their salts can be resolved.

  13. Physical Properties • Amines 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 (3.0 - 2.1 =0.9) is less than that between O and H (3.5 - 2.1 = 1.4). Using bp as an indication of H bonding Increasing strength

  14. Basicity • All 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.

  15. Basicity • Using values of pKa, we can compare the acidities of amine conjugate acids with other acids.

  16. Basicity-Aliphatic Amines • Aliphatic Amines • note that pKa + pKb = 14 Stronger bases

  17. Basicity-Aromatic Amines Weaker bases Intermediate

  18. Basicity-Aromatic Amines • Aromatic amines are considerably weaker bases than aliphatic amines.

  19. 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.

  20. 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 substituents • Electron-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.

  21. Example: Basicity-Aromatic Amines 3-nitroaniline is a stronger base than 4-Nitroaniline. Cannot do this kind of resonance in 3 nitroaniline

  22. Basicity-Aromatic Amines • Heterocyclic aromatic amines are weaker bases than heterocyclic aliphatic amines.

  23. 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).

  24. Basicity-Aromatic Amines • Imidazole Which N lone pair is protonated? The one which is not part of the aromatic system.

  25. Basicity-Guanidine • Guanidine is the strongest base among neutral organic compounds. • Its basicity is due to the delocalization of the positive charge over the three nitrogen atoms.

  26. Reaction with Acids • All amines, whether soluble or insoluble in water, react quantitatively with strong acids to form water-soluble salts.

  27. Reaction with acids • Separation and purification of an amine and a neutral compound.

  28. 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 imines • reduction of imines to amines • reduction of amides to amines by LiAlH4 • reduction of nitriles to a 1° amine • nitration of arenes followed by reduction of the NO2 group to a 1° amine

  29. 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

  30. Preparation via Azides • Alkylation of azide ion. Overall Alkyl Halide  Alkyl amine

  31. Example: Preparation via Azides • Alkylation of azide ion. Note retention of configuration, trans  trans

  32. Reaction with HNO2 • Nitrous 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.

  33. 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.

  34. 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.

  35. Secondary Amines with HNO2 • 2° Aliphatic and aromatic amines react with NO+ to give N-nitrosamines. carcinogens Mechanism:

  36. RNH2 with HNO2 • 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 N2 to give a carbocation. • Diazonium ion: An RN2+ or ArN2+ ion

  37. 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.

  38. 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.

  39. 1° RNH2 with HNO2 • Tiffeneau-Demjanov reaction: Treatment of a -aminoalcohol with HNO2 gives a ketone and N2.

  40. 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

  41. Pinacol Rearrangement: an example of stabilization of a carbocation by an adjacent lone pair. Overall:

  42. Mechanism Reversible protonation. Elimination of water to yield tertiary carbocation. This is a protonated ketone! 1,2 rearrangement to yield resonance stabilized cation. Deprotonation.

  43. 1° Primary Amines with HNO2 (Aromatic) • The -N2+ group of an arenediazonium salt can be replaced in a regioselective manner by these groups.

  44. 1° ArNH2 with HNO2 • A 1° aromatic amine converted to a phenol.

  45. 1° ArNH2 with HNO2 Problem:What reagents and experimental conditions will bring about this conversion?

  46. 1° ArNH2 with HNO2 • Problem: Show how to bring about each conversion.

  47. Hofmann Elimination • Hofmann elimination: Thermal decomposition of a quaternary ammonium hydroxide to give an alkene. • Step 1: Formation of a 4° ammonium hydroxide.

  48. Hofmann Elimination • Step 2: Thermal decomposition of the 4° ammonium hydroxide.

  49. Hofmann Elimination • Hofmann 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.

  50. Hofmann Elimination • The 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.

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