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Amines Chemical / Biological / Neurological Activity

Amines Chemical / Biological / Neurological Activity. Measures of Basicity. The basicity of amines may be measured and compared by using any of these values: 1) K b 2) p K b 3) K a of conjugate acid 4) p K a of conjugate acid. +. –. ••. ••. R 3 N. H. OH. OH. R 3 N. H.

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Amines Chemical / Biological / Neurological Activity

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  1. Amines Chemical / Biological / Neurological Activity

  2. Measures of Basicity • The basicity of amines may be measured and compared by using any of these values: • 1) Kb • 2) pKb • 3) Ka of conjugate acid • 4) pKa of conjugate acid

  3. + – •• •• R3N H OH OH R3N H •• •• •• •• Basicity Constant (Kb) and pKb • Kb is the equilibrium constant for the reaction: + + [R3NH+][HO–] Kb = [R3N] and pKb = - log Kb

  4. + R3N + R3N H H+ •• Ka and pKa of Conjugate Acid • Ka is the equilibrium constant for the dissociation of the conjugate acid of the amine: [R3N][H+] Ka = [R3NH+] and pKa = - log Ka

  5. Relationships between acidity and basicity constants Ka Kb = 10-14 pKa + pKb = 14

  6. Basicity of Amines in Aqueous Solution • Amine Conj. Acid pKa • NH3 NH4+ 9.3 • CH3CH2NH2 CH3CH2NH3+ 10.8 CH3CH2NH3+ is a weaker acid than NH4+;therefore, CH3CH2NH2 is a stronger base than NH3.

  7. Effect of Structure on Basicity • 1. Alkylamines are slightly stronger bases than ammonia. • 2. Alkylamines differ very little in basicity.

  8. Basicity of Amines in Aqueous Solution • Amine Conj. Acid pKa • NH3 NH4+ 9.3 • CH3CH2NH2 CH3CH2NH3+ 10.8 • (CH3CH2)2NH (CH3CH2)2NH2+ 10.9 • (CH3CH2)3N (CH3CH2)3NH+ 11.1 Notice that the difference separating a primary,secondary, and tertiary amine is only 0.3 pK units.

  9. Effect of Structure on Basicity • 1. Alkylamines are slightly stronger bases than ammonia. • 2. Alkylamines differ very little in basicity. • 3. Arylamines are much weaker bases than ammonia.

  10. Basicity of Amines in Aqueous Solution • Amine Conj. Acid pKa • NH3 NH4+ 9.3 • CH3CH2NH2 CH3CH2NH3+ 10.8 • (CH3CH2)2NH (CH3CH2)2NH2+ 10.9 • (CH3CH2)3N (CH3CH2)3NH+ 11.1 • C6H5NH2 C6H5NH3+ 4.6

  11. •• •• H OH NH2 •• – •• OH NH3 •• •• Decreased basicity of arylamines • Aniline (reactant) is stabilized by conjugation of nitrogen lone pair with ring p system. • This stabilization is lost on protonation. + + +

  12. Decreased basicity of arylamines • Increasing delocalization makes diphenylamine a weaker base than aniline, and triphenylamine a weaker base than diphenylamine. C6H5NH2 (C6H5)2NH (C6H5)3N 3.8 x 10-10 6 x 10-14 ~10-19 Kb

  13. Effect of Substituents on Basicity of Arylamines • 1. Alkyl groups on the ring increase basicity, but only slightly (less than 1 pK unit). • 2. Electron withdrawing groups, especially ortho and/or para to amine group, decrease basicity and can have a large effect.

  14. X NH2 X NH3+ Basicity of Arylamines • X pKb pKa • H 9.4 4.6 • CH3 8.7 5.3 • CF3 11.5 2.5 • O2N 13.0 1.0

  15. – •• •• O O •• •• •• + + •• N NH2 N NH2 O O •• •• •• •• – – •• •• p-Nitroaniline • Lone pair on amine nitrogen is conjugated with p-nitro group—more delocalized than in aniline itself. Delocalization lost on protonation. +

  16. Effect is Cumulative • Aniline is 3800 times more basic thanp-nitroaniline. • Aniline is ~1,000,000,000 times more basic than 2,4-dinitroaniline.

  17. •• N N •• H Heterocyclic Amines is more basic than piperidine pyridine Kb = 1.6 x 10-3 Kb = 1.4 x 10-9 (an alkylamine) (resembles anarylamine inbasicity)

  18. N H •• N •• N •• Heterocyclic Amines is more basic than imidazole pyridine Kb = 1 x 10-7 Kb = 1.4 x 10-9

  19. N H •• N •• + H H N H N •• N N •• H Imidazole • Which nitrogen is protonated in imidazole? H+ H+ +

  20. N H •• N •• + H N H •• N Imidazole • Which nitrogen is protonated in imidazole? (HINT: Resonance is the key.) H+

  21. N H •• N •• + H H N H N H N N •• Imidazole • Protonation in the direction shown gives a stabilized ion. H+ + ••

  22. Question Which of the following amines is more basic? • A) B) • C) D)

  23. Preparation of Amines by Reduction

  24. Preparation of Amines by Reduction • Almost any nitrogen-containing compound canbe reduced to an amine, including: • azidesnitrilesnitro-substituted benzene derivatives amides

  25. CH2CH2Br CH2CH2N3 1. LiAlH4 2. H2O CH2CH2NH2 (89%) Synthesis of Amines via Azides • SN2 reaction, followed by reduction, gives a primary alkylamine. NaN3 (74%) Azides may also bereduced by catalytichydrogenation.

  26. Question • What is the product of the reaction shown? • A) B) • C) D)

  27. Question • Identify compound C formed in the synthetic sequence below. • A) (R)-2-octanamine B) (S)-2-octanamine • C) (R)-2-octanol D) octane

  28. H2 (100 atm), Ni CH3CH2CH2CH2CH2NH2 (56%) Synthesis of Amines via Nitriles • SN2 reaction, followed by reduction, gives a primary alkylamine. NaCN CH3CH2CH2CH2Br CH3CH2CH2CH2CN (69%) Nitriles may also bereduced by lithiumaluminum hydride.

  29. H2 (100 atm), Ni CH3CH2CH2CH2CH2NH2 (56%) Synthesis of Amines via Nitriles • SN2 reaction, followed by reduction, gives a primary alkylamine. NaCN CH3CH2CH2CH2Br CH3CH2CH2CH2CN (69%) The reduction alsoworks with cyanohydrins.

  30. Question • What is the major organic product of the synthesis shown? • A) C6H5CH2CN • B) C6H5CH2CHO • C) C6H5CH2CH2NH2 • D) C6H5CH2NH2

  31. NO2 Cl 1. Fe, HCl 2. NaOH NH2 Cl (95%) Synthesis of Amines via Nitroarenes HNO3 Cl H2SO4 Nitro groups may alsobe reduced with tin (Sn)+ HCl or by catalytichydrogenation. (88-95%)

  32. Question • Which one of the following is produced when m-nitroacetophenone is treated with Sn and HCl followed by NaOH? • A) B) • C) D)

  33. Question • Starting with benzene, which of the sequences below will produce p-methylaniline as the major product of the reaction? • A) 1. HNO3, H2SO4; 2. CH3Cl, AlCl3; 3. Fe, HCl; 4. NaOH • B) 1. HNO3, H2SO4; 2. Fe, HCl; 3. NaOH; 4. CH3Cl, AlCl3 • C) 1. CH3Cl, AlCl3; 2. HNO3, H2SO4; 3. Fe, HCl; 4. NaOH • D) 1. CH3Cl, AlCl3; 2. HNO3, H2SO4; 3. H2

  34. O O COH CN(CH3)2 1. LiAlH4 2. H2O CH2N(CH3)2 (88%) Synthesis of Amines via Amides 1. SOCl2 2. (CH3)2NH (86-89%) Only LiAlH4 is anappropriate reducingagent for this reaction.

  35. Question • Identify the product of the synthesis shown. • A) C6H5NH2 • B) C6H5CH=NH • C) C6H5CH2NH2 • D) C6H5C(=O)NH2 LiAlH4 5. H2O

  36. Preparation and Reactions of Amines

  37. The Gabriel Synthesis of Primary Amines

  38. Question • What is the product of the Gabriel synthesis shown? • A) diethyl ether • B) ethanol • C) ethyl amine • D) CH3CH2NHNH2

  39. Reductive Amination

  40. R R O NH C C R' R' Synthesis of Amines via Reductive Amination In reductive amination, an aldehyde or ketoneis subjected to catalytic hydrogenation in thepresence of ammonia or an amine. • The aldehyde or ketone equilibrates with theimine faster than hydrogenation occurs. fast + + NH3 H2O

  41. R R O NH C C R' R' R R' NH2 C H Synthesis of Amines via Reductive Amination The imine undergoes hydrogenation fasterthan the aldehyde or ketone. An amine is the product. fast + + NH3 H2O H2, Ni

  42. O + CH3(CH2)5CH H2N CH3(CH2)5CH2NH CH3(CH2)5CH N Example: Primary amines give secondary amines NaBH3CN or H2, Ni ethanol via:

  43. O CH3CH2CH2CH N H N CH2CH2CH2CH3 Example: Secondary amines give tertiary amines + H2, Ni, ethanol (93%)

  44. Question • How would you accomplish the conversion of propanal into N-ethyl-N-methylpropanamine? • A) NH3, NaBH3CN; CH3I; CH3CH2I • B) CH3NH2, NaBH3CN; CH3COCl, pyridine; LiAlH4; H2O • C) CrO3, H2SO4; SOCl2, pyridine; 2 equiv CH3NH2; CH3I • D) CH3CH2NH2, H2, Ni; (CH3CO)2O, pyridine; NaBH4

  45. Quarternary Amines Can Undergo an E2 Elimination Reaction

  46. The Hofmann Elimination

  47. The Hofmann Elimination • a quaternary ammonium hydroxide is the reactantand an alkene is the product • is an anti elimination • the leaving group is a trialkylamine • the regioselectivity is opposite to the Zaitsev rule.

  48. – CH2N(CH3)3 I Ag2O H2O, CH3OH + – CH2N(CH3)3 HO Quaternary Ammonium Hydroxides are prepared by treating quaternary ammmoniumhalides with moist silver oxide

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