Amines Chemical / Biological / Neurological Activity.

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Presentation transcript:

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)pK b 3)K a of conjugate acid 4) pK a of conjugate acid

Basicity Constant (K b ) and pK b K b is the equilibrium constant for the reaction: R3NR3NR3NR3N + HOH R3NR3NR3NR3N+H – OH + K b = [R 3 NH + ][HO – ] [R 3 N] pK b = - log K b and

K a and pK a of Conjugate Acid K a = [R 3 N][H + ] [R 3 NH + ] pK a = - log K a and R3NR3NR3NR3N R3NR3NR3NR3N+H + H+H+H+H+ K a is the equilibrium constant for the dissociation of the conjugate acid of the amine:

Relationships between acidity and basicity constants pK a + pK b = 14 K a K b =

AmineConj. AcidpK a NH 3 NH CH 3 CH 2 NH 2 CH 3 CH 2 NH Basicity of Amines in Aqueous Solution CH 3 CH 2 NH 3 + is a weaker acid than NH 4 + ; therefore, CH 3 CH 2 NH 2 is a stronger base than NH 3.

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

AmineConj. AcidpK a NH 3 NH CH 3 CH 2 NH 2 CH 3 CH 2 NH (CH 3 CH 2 ) 2 NH(CH 3 CH 2 ) 2 NH (CH 3 CH 2 ) 3 N(CH 3 CH 2 ) 3 NH Basicity of Amines in Aqueous Solution Notice that the difference separating a primary, secondary, and tertiary amine is only 0.3 pK units.

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.

AmineConj. AcidpK a NH 3 NH CH 3 CH 2 NH 2 CH 3 CH 2 NH (CH 3 CH 2 ) 2 NH(CH 3 CH 2 ) 2 NH (CH 3 CH 2 ) 3 N(CH 3 CH 2 ) 3 NH C 6 H 5 NH 2 C 6 H 5 NH Basicity of Amines in Aqueous Solution

+ HOH NH2NH2NH2NH2 + NH3NH3NH3NH3 +– OH Aniline (reactant) is stabilized by conjugation of nitrogen lone pair with ring  system. This stabilization is lost on protonation. Decreased basicity of arylamines

C 6 H 5 NH 2 (C 6 H 5 ) 2 NH (C 6 H 5 ) 3 N KbKbKbKb 3.8 x x ~ Increasing delocalization makes diphenylamine a weaker base than aniline, and triphenylamine a weaker base than diphenylamine.

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.

XpK b pK a H CH CF O 2 N Basicity of Arylamines X NH 2 X NH 3 +

p-Nitroaniline NH 2 O N O – + O N O – – NH Lone pair on amine nitrogen is conjugated with p-nitro group—more delocalized than in aniline itself. Delocalization lost on protonation.

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

Heterocyclic Amines N H N piperidinepyridine is more basic than K b = 1.6 x K b = 1.4 x (an alkylamine) (resembles an arylamine in basicity)

Heterocyclic Amines N imidazolepyridine is more basic than K b = 1 x K b = 1.4 x NH N

Imidazole NH N Which nitrogen is protonated in imidazole? H+H+H+H+ H+H+H+H+ NH N H + N H N H +

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

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

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

Preparation of Amines by Reduction

Almost any nitrogen-containing compound can be reduced to an amine, including: azides nitriles nitro-substituted benzene derivatives amides Preparation of Amines by Reduction

S N 2 reaction, followed by reduction, gives a primary alkylamine. Synthesis of Amines via Azides CH 2 CH 2 Br CH 2 CH 2 N 3 NaN 3 (74%) CH 2 CH 2 NH 2 (89%) 1. LiAlH 4 2. H 2 O Azides may also be reduced by catalytic hydrogenation.

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

Identify compound C formed in the synthetic sequence below. A)(R)-2-octanamineB) (S)-2-octanamine C)(R)-2-octanolD) octane Question

S N 2 reaction, followed by reduction, gives a primary alkylamine. Synthesis of Amines via Nitriles CH 3 CH 2 CH 2 CH 2 Br NaCN (69%) CH 3 CH 2 CH 2 CH 2 CN CH 3 CH 2 CH 2 CH 2 CH 2 NH 2 (56%) H 2 (100 atm), Ni Nitriles may also be reduced by lithium aluminum hydride.

S N 2 reaction, followed by reduction, gives a primary alkylamine. Synthesis of Amines via Nitriles CH 3 CH 2 CH 2 CH 2 Br NaCN (69%) CH 3 CH 2 CH 2 CH 2 CN CH 3 CH 2 CH 2 CH 2 CH 2 NH 2 (56%) H 2 (100 atm), Ni The reduction also works with cyanohydrins.

Question What is the major organic product of the synthesis shown? A)C 6 H 5 CH 2 CN B)C 6 H 5 CH 2 CHO C)C 6 H 5 CH 2 CH 2 NH 2 D)C 6 H 5 CH 2 NH 2

Synthesis of Amines via Nitroarenes HNO3HNO3HNO3HNO3 (88-95%) Cl Cl NO2NO2NO2NO2 H 2 SO 4 (95%) 1. Fe, HCl 2. NaOH Cl NH2NH2NH2NH2 Nitro groups may also be reduced with tin (Sn) + HCl or by catalytic hydrogenation.

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

Question Starting with benzene, which of the sequences below will produce p-methylaniline as the major product of the reaction? A)1. HNO 3, H 2 SO 4 ; 2. CH 3 Cl, AlCl 3 ; 3. Fe, HCl; 4. NaOH B)1. HNO 3, H 2 SO 4 ; 2. Fe, HCl; 3. NaOH; 4. CH 3 Cl, AlCl 3 C)1. CH 3 Cl, AlCl 3 ; 2. HNO 3, H 2 SO 4 ; 3. Fe, HCl; 4. NaOH D)1. CH 3 Cl, AlCl 3 ; 2. HNO 3, H 2 SO 4 ; 3. H 2

Synthesis of Amines via Amides (86-89%) COHO 1. SOCl 2 2. (CH 3 ) 2 NH CN(CH 3 ) 2 O (88%) 1. LiAlH 4 2. H 2 O CH 2 N(CH 3 ) 2 Only LiAlH 4 is an appropriate reducing agent for this reaction.

Question Identify the product of the synthesis shown. A)C 6 H 5 NH 2 B)C 6 H 5 CH=NH C)C 6 H 5 CH 2 NH 2 D)C 6 H 5 C(=O)NH 2 LiAlH4 5. H 2 O

Preparation and Reactions of Amines

The Gabriel Synthesis of Primary Amines

Question What is the product of the Gabriel synthesis shown? A)diethyl ether B)ethanol C)ethyl amine D) CH 3 CH 2 NHNH 2

Reductive Amination

The aldehyde or ketone equilibrates with the imine faster than hydrogenation occurs. Synthesis of Amines via Reductive Amination O CRR' + NH3NH3NH3NH3 fast NHNHNHNH CRR' + H2OH2OH2OH2O In reductive amination, an aldehyde or ketone is subjected to catalytic hydrogenation in the presence of ammonia or an amine.

Synthesis of Amines via Reductive Amination H 2, Ni O CRR' + NH3NH3NH3NH3 fast NHNHNHNH CRR' + H2OH2OH2OH2O NH2NH2NH2NH2RR' C H The imine undergoes hydrogenation faster than the aldehyde or ketone. An amine is the product.

Example: Primary amines give secondary amines H 2, Ni ethanol CH 3 (CH 2 ) 5 CH 2 NH + H2NH2NH2NH2N CH 3 (CH 2 ) 5 CH O NaBH 3 CN or via: N CH 3 (CH 2 ) 5 CH

Example: Secondary amines give tertiary amines H 2, Ni, ethanol (93%) + CH 3 CH 2 CH 2 CH O N H N CH 2 CH 2 CH 2 CH 3

Question How would you accomplish the conversion of propanal into N-ethyl-N-methylpropanamine? A)NH 3, NaBH 3 CN; CH 3 I; CH 3 CH 2 I B)CH 3 NH 2, NaBH 3 CN; CH 3 COCl, pyridine; LiAlH 4 ; H 2 O C)CrO 3, H 2 SO 4 ; SOCl 2, pyridine; 2 equiv CH 3 NH 2 ; CH 3 I D)CH 3 CH 2 NH 2, H 2, Ni; (CH 3 CO) 2 O, pyridine; NaBH 4

Quarternary Amines Can Undergo an E 2 Elimination Reaction

The Hofmann Elimination

a quaternary ammonium hydroxide is the reactant and an alkene is the product is an anti elimination the leaving group is a trialkylamine the regioselectivity is opposite to the Zaitsev rule.

Quaternary Ammonium Hydroxides Ag 2 O H 2 O, CH 3 OH CH 2 N(CH 3 ) 3 + HO – are prepared by treating quaternary ammmonium halides with moist silver oxide CH 2 N(CH 3 ) 3 I–

Regioselectivityheat Elimination occurs in the direction that gives the less-substituted double bond. This is called the Hofmann rule. N(CH 3 ) 3 + HO – CH 3 CHCH 2 CH 3 H2CH2CH2CH2C CHCH 2 CH 3 CH 3 CH CHCH 3 +(95%)(5%)

Regioselectivity N(CH 3 ) 3 + HHH H CH 3 CH 2 largest group is between two H atoms C H CHH CH 3 CH 2 major product

Regioselectivity N(CH 3 ) 3 +H H H CH 3 largest group is between an H atom and a methyl group C H C CH 3 H minor product CH 3

Nitrosation of Arylamines

Nitrosation of Primary Arylamines Gives aryl diazonium ions. Aryl diazonium ions are much more stable than alkyl diazonium ions. Most aryl diazonium ions are stable under the conditions of their formation (0-10°C). ArN N + RNRNRNRNN+fast slow R + + N2N2N2N2 Ar + + N2N2N2N2

Synthetic Origin of Aryl Diazonium Salts ArH Ar NO2NO2NO2NO2 Ar NH2NH2NH2NH2 Ar N N+

Synthetic Transformations of Aryl Diazonium Salts

Transformations of Aryl Diazonium Salts Ar N N+ ArH Ar OHOHOHOH ArI ArF ArBrArCl ArCN H 2 O, heat KIKIKIKI HBF 4 / heat CuCl or CuBr heat heat CuCN,heat H 3 PO 2

Question Identify the product isolated from the reaction of p-nitroaniline with NaNO 2 in H 2 SO 4 followed by the addition of potassium iodide (KI). A)nitrobenzene B)p-iodoaniline C)p-iodonitrobenzene D)p-diiodonitrobenzene

Alkaloids

Alkaloids: Naturally Occuring Bases Nitrogen Heterocycles ibogaine

Amines & Neurotransmitters

R-ethylamine dopamine mescaline

Serotonin Melatonin

Acetylcholine Epinephrine (Adrenaline)

Cathecols: epinephrine & mdma Principal sympathomimetic adrenal hormone & a controlled substancesubstance

Drug Uptake: Rank from slowest to fastest. a) injection; b) ingestion; c) inhalation; d) snorting A)a<b<c<dB) c<a<d<b C) b<d<a<cD) d<b<c<a

Drug Uptake: Rank from slowest to fastest. a) injection; b) ingestion; c) inhalation; d) snorting A)a<b<c<dB) c<a<d<b C) b<d<a<cD) d<b<c<a

morphine LSD ibogaine