Hóa học - Chapter 19: Amines

In the presence of a strong base, primary amides react with chlorine or bromine to form shortened amines, with the loss of the carbonyl carbon atom. This reaction, called the Hofmann rearrangement, is used to synthesize primary and aryl amines.

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Chapter 19Copyright © 2010 Pearson Education, Inc.Organic Chemistry, 7th Edition L. G. Wade, Jr.AminesChapter 19*Biologically Active AminesThe alkaloids are an important group of biologically active amines, mostly synthesized by plants to protect them from being eaten by insects and other animals. Many drugs of addiction are classified as alkaloids.Chapter 19*Biological Activity of AminesDopamine is a neurotransmitter.Epinephrine is a bioregulator.Niacin, Vitamin B6, is an amine.Alkaloids: nicotine, morphine, cocaineAmino acidsChapter 19*Classes of AminesPrimary (1): Has one alkyl group bonded to the nitrogen (RNH2).Secondary (2): Has two alkyl groups bonded to the nitrogen (R2NH).Tertiary (3): Has three alkyl groups bonded to the nitrogen (R3N). Quaternary (4): Has four alkyl groups bonded to the nitrogen and the nitrogen bears a positive charge(R4N+).Chapter 19*Examples of AminesPrimary(1º)Secondary(2º)Tertiary(3º)Chapter 19*Common NamesChapter 19*Amine as SubstituentOn a molecule with a higher priority functional group, the amine is named as a substituent. Chapter 19*IUPAC NamesName is based on longest carbon chain.-e of alkane is replaced with -amine.Substituents on nitrogen have N- prefix.3-bromo-1-pentanamineN,N-dimethyl-3-hexanamineNH2CH2CH2CHCH2CH3BrCH3CH2CHCH2CH2CH3N(CH3)2Chapter 19*Aromatic AminesIn aromatic amines, the amino group is bonded to a benzene ring. Parent compound is called aniline.Chapter 19*Heterocyclic AminesWhen naming a cyclic amine the nitrogen is assigned position number 1.Chapter 19*Structure of AminesNitrogen is sp3 hybridized with a lone pair of electrons. The angle is less than 109.5º.Chapter 19*Interconversion of Chiral AminesNitrogen may have three different groups and a lone pair, but enantiomers cannot be isolated due to inversion around N.Chapter 19*Chiral AminesAmines whose chirality stems from the presence of chiral carbon atoms. Inversion of the nitrogen is not relevant because it will not affect the chiral carbon.Chapter 19*Chiral Amines (Continued)Quaternary ammonium salts may have a chiral nitrogen atom if the four substituents are different. Inversion of configuration is not possible because there is no lone pair to undergo nitrogen inversion.Chapter 19*Chiral Cyclic AminesIf the nitrogen atom is contained in a small ring, for example, it is prevented from attaining the 120° bond angle that facilitates inversion. Such a compound has a higher activation energy for inversion, the inversion is slow, and the enantiomers may be resolved.Chapter 19*Boiling PointsN—H less polar than O—H.Weaker hydrogen bonds, so amines will have a lower boiling point than the corresponding alcohol.Tertiary amines cannot hydrogen-bond, so they have lower boiling points than primary and secondary amines.Chapter 19*Solubility and OdorSmall amines (< 6 Cs) are soluble in water.All amines accept hydrogen bonds from water and alcohol.Branching increases solubility.Most amines smell like rotting fish.1,5-pentanediamine or cadaverineNH2CH2CH2CH2CH2CH2NH2Chapter 19*Basicity of AminesLone pair of electrons on nitrogen can accept a proton from an acid.Aqueous solutions are basic to litmus.Ammonia pKb = 4.74Alkyl amines are usually stronger bases than ammonia. Increasing the number of alkyl groups decreases solvation of ion, so 2 and 3 amines are similar to 1 amines in basicity.Chapter 19*Reactivity of AminesChapter 19*Base-Dissociation Constant of AminesAn amine can abstract a proton from water, giving an ammonium ion and a hydroxide ion. The equilibrium constant for this reaction is called the base-dissociation constant for the amine, symbolized by Kb.Chapter 19*Base Dissociation of an AmineAlkyl groups stabilize the ammonium ion, making the amine a stronger base.Chapter 19*Alkyl Group Stabilization of AminesAlkyl groups make the nitrogen a stronger base than ammonia. Chapter 19*Resonance EffectsAny delocalization of the electron pair weakens the base.Chapter 19*Protonation of PyrroleWhen the pyrrole nitrogen is protonated, pyrrole loses its aromatic stabilization.Therefore, protonation on nitrogen is unfavorable and pyrrole is a very weak base.Chapter 19*Hybridization EffectsPyridine is less basic than aliphatic amines, but it is more basic than pyrrole because it does not lose its aromaticity on protonation.Chapter 19*Ammonium SaltsIonic solids with high melting points. Soluble in water. No fishy odor.Chapter 19*Purifying an AmineChapter 19*Phase Transfer CatalystsChapter 19*CocaineCocaine is usually smuggled and “snorted” as the hydrochloride salt. Treating cocaine hydrochloride with sodium hydroxide and extracting it into ether converts it back to the volatile “free base” for smoking.Chapter 19*IR SpectroscopyN—H stretch between 3200–3500 cm-1.Two peaks for 1 amine, one for 2.Chapter 19*NMR Spectroscopy of AminesNitrogen is not as electronegative as oxygen, so the protons on the a-carbon atoms of amines are not as strongly deshielded.Chapter 19*NMR SpectrumChapter 19*Alpha Cleavage of AminesThe most common fragmentation of amines is a-cleavage to give a resonance-stabilized cation—an iminium ion. Chapter 19*Fragmentation of Butyl Propyl AmineChapter 19*MS of Butyl Propyl AmineChapter 19*Reaction of Amines with Carbonyl Compounds Chapter 19*Electrophilic Substitution of Aniline—NH2 is strong activator, ortho- and para-directing.Multiple alkylation is a problem.Protonation of the amine converts the group into a deactivator (—NH3+).Attempt to nitrate aniline may burn or explode.Chapter 19*Protonation of Aniline in Substitution ReactionsStrongly acidic reagents protonate the amino group, giving an ammonium salt. The —NH3+ group is strongly deactivating (and meta-allowing). Therefore, strongly acidic reagents are unsuitable for substitution of anilines.Chapter 19*Electrophilic Substitution of PyridineStrongly deactivated by electronegative N.Substitutes in the 3-position.Electrons on N react with electrophile.Chapter 19*Electrophilic Aromatic Substitution of PyridineChapter 19*Electrophilic Aromatic Substitution of Pyridine (Continued)Attack at the 2-position would have an unfavorable resonance structure in which the positive charge is localized on the nitrogen. Substitution at the 2-position is not observed.Chapter 19*Nucleophilic Substitution of PyridineDeactivated toward electrophilic attack.Activated toward nucleophilic attack.Nucleophile will replace a good leaving group in the 2- or 4-position.Chapter 19*Mechanism for Nucleophilic SubstitutionAttack at the 3-position does not have the negative charge on the nitrogen, so substitution at the 3-position is not observed.Chapter 19*Alkylation of Amines by Alkyl HalidesEven if just one equivalent of the halide is added, some amine molecules will react once, some will react twice, and some will react three times (to give the tetraalkylammonium salt).Chapter 19*Examples of Useful AlkylationsExhaustive alkylation to form the tetraalkylammonium salt. Reaction with large excess of NH3 to form the primary amine.CH3CH2CHCH2CH2CH3N(CH3)3CH3CH2CHCH2CH2CH3NH23CH3INaHCO3+_ICH3CH2CH2BrNH3 (xs)CH3CH2CH2NH2 +NH4BrChapter 19*Acylation of AminesPrimary and secondary amines react with acid halides to form amides.This reaction is a nucleophilic acyl substitution.Chapter 19*Acylation of Aromatic AminesWhen the amino group of aniline is acetylated, the resulting amide is still activating and ortho, para-directing. Acetanilide may be treated with acidic (and mild oxidizing) reagents to further substitute the ring. The acyl group can be removed later by acidic or basic hydrolysis.Chapter 19*Show how you would accomplish the following synthetic conversion in good yield.An attempted Friedel–Crafts acylation on aniline would likely meet with disaster. The free amino group would attack both the acid chloride and the Lewis acid catalyst.Solved Problem 1SolutionChapter 19*We can control the nucleophilicity of aniline’s amino group by converting it to an amide, which is still activating and ortho, para directing for the Friedel–Crafts reaction. Acylation, followed by hydrolysis of the amide, gives the desired product.Solved Problem 1 (Continued)Solution (Continued)Chapter 19*Formation of SulfonamidesPrimary or secondary amines react with sulfonyl chloride.Chapter 19*Synthesis of SulfanilamideChapter 19*Biological Activity of SulfanilamideSulfanilamide is an analogue of p-aminobenzoic acid. Streptococci use p-aminobenzoic acid to synthesize folic acid, an essential compound for growth and reproduction. Sulfanilamide cannot be used to make folic acid.Bacteria cannot distinguish between sulfanilamide and p-aminobenzoic acid, so it will inhibit their growth and reproduction.Chapter 19*Hofmann EliminationA quaternary ammonium salt has a good leaving group—a neutral amine.Heating the hydroxide salt produces the least substituted alkene.Chapter 19*Exhaustive Methylation of AminesAn amino group can be converted into a good leaving group by exhaustive elimination: Conversion to a quaternary ammonium salt that can leave as a neutral amine. Methyl iodide is usually used.Chapter 19*Conversion to the Hydroxide SaltThe quaternary ammonium iodide is converted to the hydroxide salt by treatment with silver oxide and water.The hydroxide will be the base in the elimination step.Chapter 19*Mechanism of the Hofmann EliminationThe Hofmann elimination is a one-step, concerted E2 reaction using an amine as the leaving group.Chapter 19*Regioselectivity of the Hofmann EliminationThe least substituted product is the major product of the reaction—Hofmann product.Chapter 19*E2 MechanismChapter 19*Predict the major product(s) formed when the following amine is treated with excess iodomethane, followed by heating with silver oxide.Solving this type of problem requires finding every possible elimination of the methylated salt. In this case, the salt has the following structure:Solved Problem 2SolutionChapter 19*The green, blue, and red arrows show the three possible elimination routes. The corresponding products areThe first (green) alkene has a disubstituted double bond. The second (blue) alkene is monosubstituted, and the red alkene (ethylene) has an unsubstituted double bond. We predict that the red products will be favored.Solved Problem 2 (Continued)Solution (Continued)Chapter 19*Oxidation of AminesAmines are easily oxidized, even in air.Common oxidizing agents: H2O2 , MCPBA.2 Amines oxidize to hydroxylamine (—NOH)3 Amines oxidize to amine oxide (R3N+—O-) Chapter 19*Preparation of Amine OxidesTertiary amines are oxidized to amine oxides, often in good yields. Either H2O2 or peroxyacid may be used for this oxidation.Chapter 19*Cope RearrangementE2 mechanism.The amine oxide acts as its own base through a cyclic transition state, so a strong base is not needed.Chapter 19*Predict the products expected when the following compound is treated with H2O2 and heated.Oxidation converts the tertiary amine to an amine oxide. Cope elimination can give either of two alkenes. We expect the less hindered elimination to be favored, giving the Hofmann product.Solved Problem 3SolutionChapter 19*Formation of Diazonium SaltsPrimary amines react with nitrous acid (HNO2) to form dialkyldiazonium salts.The diazonium salts are unstable and decompose into carbocations and nitrogen.Chapter 19*Diazotization of an AmineStep 1: The amine attacks the nitrosonium ion and forms N-nitrosoamine.Step 2: A proton transfer (a tautomerism) from nitrogen to oxygen forms a hydroxyl group and a second N-N bond.Chapter 19*Diazotization of an Amine (Continued)Step 3: Protonation of the hydroxyl group, followed by the loss of water, gives the diazonium ion.Chapter 19*Arenediazonium SaltsBy forming and diazotizing an amine, an activated aromatic position can be converted into a wide variety of functional groups.Chapter 19*Reactions of Arenediazonium SaltsChapter 19*The Sandmeyer ReactionChapter 19*Formation of N-NitrosoaminesSecondary amines react with nitrous acid (HNO2) to form N-nitrosoamines.Secondary N-nitrosoamines are stable and have been shown to be carcinogenic in lab animals.Chapter 19*Reductive Amination: 1º AminesPrimary amines result from the condensation of hydroxylamine (zero alkyl groups) with a ketone or an aldehyde, followed by reduction of the oxime. LiAlH4 or NaBH3CN can be used to reduce the oxime.Chapter 19*Reductive Amination: 2º AminesCondensation of a ketone or an aldehyde with a primary amine forms an N-substituted imine (a Schiff base). Reduction of the N-substituted imine gives a secondary amine.Chapter 19*Reductive Amination: 3º AminesCondensation of a ketone or an aldehyde with a secondary amine gives an iminium salt. Iminium salts are frequently unstable, so they are rarely isolated. A reducing agent in the solution reduces the iminium salt to a tertiary amine.Chapter 19*Show how to synthesize the following amines from the indicated starting materials.(a) N-cyclopentylaniline from aniline (b) N-ethylpyrrolidine from pyrrolidine(a) This synthesis requires adding a cyclopentyl group to aniline (primary) to make a secondary amine. Cyclopentanone is the carbonyl compound. (b) This synthesis requires adding an ethyl group to a secondary amine to make a tertiary amine. The carbonyl compound is acetaldehyde. Formation of a tertiary amine by Na(AcO)3BH reductive amination involves an iminium intermediate, which is reduced by (sodium triacetoxyborohydride).Solved Problem 3SolutionChapter 19*Synthesis of 1º Amines by Acylation–ReductionAcylation of the starting amine by an acid chloride gives an amide with no tendency toward overacylation. Reduction of the amide by LiAlH4 gives the corresponding amine.Chapter 19*Synthesis of 2º Amines by Acylation–ReductionAcylation–reduction converts a primary amine to a secondary amine. LiAlH4, followed by hydrolysis, can easily reduce the intermediate amide to the amine.Chapter 19*Synthesis of 3º Amines by Acylation–ReductionAcylation–reduction converts a secondary amine to a tertiary amine. Reduction of the intermediate amide is accomplished with LiAlH4.Chapter 19*Show how to synthesize N-ethylpyrrolidine from pyrrolidine using acylation–reduction.This synthesis requires adding an ethyl group to pyrrolidine to make a tertiary amine. The acid chloride needed will be acetyl chloride (ethanoyl chloride). Reduction of the amide gives N-ethylpyrrolidine.Compare this synthesis with Solved Problem 19-5(b) to show how reductive amination and acylation–reduction can accomplish the same result.Solved Problem 4SolutionChapter 19*The Gabriel SynthesisThe phthalimide ion is a strong nucleophile, displacing the halide or tosylate ion from a good SN2 substrate. Heating the N-alkyl phthalimide with hydrazine displaces the primary amine, giving the very stable hydrazide of phthalimide.Chapter 19*Reduction of AzidesAzide ion, N3-, is a good nucleophile.React azide with unhindered 1 or 2 halide or tosylate (SN2).Alkyl azides are explosive! Do not isolate.Chapter 19*Reduction of NitrilesNitrile (CN) is a good SN2 nucleophile.Reduction with H2 or LiAlH4 converts the nitrile into a primary amine.Chapter 19*Reduction of Nitro CompoundsThe nitro group can be reduced to the amine by catalytic hydrogenation or by an active metal and H+.Commonly used to synthesize anilines.Chapter 19*The Hofmann Rearrangement of AmidesIn the presence of a strong base, primary amides react with chlorine or bromine to form shortened amines, with the loss of the carbonyl carbon atom. This reaction, called the Hofmann rearrangement, is used to synthesize primary and aryl amines.Chapter 19*Mechanism of the Hofmann Rearrangement: Steps 1 and 2Chapter 19*Mechanism of the Hofmann Rearrangement: Steps 3 and 4

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