Amines and Heterocycles
Chapter 24 Amines and Heterocycles Suggested Problems 1-24,30-33,47-49,53-64,66 Naming Amines Alkyl-substituted (alkylamines) or aryl-substituted (arylamines) Classified as primary (RNH2), secondary (R2NH), and tertiary (R3N) Depends on number of organic substituents attached to nitrogen Naming Amines Quaternary ammonium salts: Compounds that carry a positively charged nitrogen atom with four attached groups Simple amines are named by adding the suffix -amine to the name of the alkyl substituent Adding amine to the name of the alkyl substituent is one way to name amines. Naming Amines The suffix -amine can be used in place of the final -e in the name of the parent compound Amines with more than one functional group are named by considering the NH2 as an amino substituent A second way to name amines is to replace the e at the end of the parent name with amine. Naming Amines Symmetrical secondary and tertiary amines are named by adding the prefix di- or tri- to the alkyl group Naming Amines Unsymmetrically substituted secondary and tertiary amines Named as N-substituted primary amines Largest alkyl group is the parent name, and other alkyl groups are considered N-substituents The compound at left could also be named N,N-dimethylpropanamine and the compound at right could also be named N-ethyl-N-methylcyclohexanamine. Naming Amines Heterocyclic amines: Compound in which the nitrogen atom occurs as part of a ring Each ring system has its own parent name Nitrogen atom is always numbered as position 1 Heterocyclic rings are defined as ring which are comprised of at least two different types of atoms here N and C. Worked Example Name the following compounds: Solution: a) CH3NHCH2CH3
b) Solution: a) N-Methylethylamine b) N-Ethyl-N-methylcyclohexylamine Or N-methylethanamine and N-ethyl-N-methylcyclohexanamine Structure and Properties of Amines
Bonding in alkylamines is similar to that in ammonia N is sp3-hybridized CNC bond angles are close to 109 tetrahedral value Structure and Properties of Amines
An amine with three different substituents on nitrogen is chiral The two enantiomeric forms rapidly interconvert by a pyramidal inversion at room temperature Structure and Properties of Amines
Amines with fewer than five carbon atoms are water-soluble Primary and secondary amines form hydrogen bonds, increasing their boiling points Basicity of Amines Lone pair of electrons on nitrogen makes amines basic and nucleophilic React with acids to form acid-base salts and they react with electrophiles Basicity of Amines Amines are stronger bases than alcohols, ethers, or water Amines establish an equilibrium with water in which water acts as an acid and transfers a proton to the amine Basicity constant Kb is used to measure the base strength of an amine High pKa weaker acid and stronger conjugate base Basicity of Amines Basicity of an amine (RNH2) can be measured by looking at the acidity of the corresponding ammonium ion (RNH3+) The higher the Kb the more basic the amine.The lower the pKb the more basic the amine. Typically we think of the protonated amine and consider its pKa.If an amine is weakly basic, its conjugate acid, the protonated amine, will be strongly acidic and have a high Ka and a low pKa. Basicity of Amines Weaker base - Smaller pKa for ammonium ion
Stronger base - Larger pKa for ammonium ion Basicity of Some Common Amines
Note that the pKas of the amines listed here correspond to the ammonium salts the protonated amines. Most simple alkylamines are similar in their base strength, with pKas for their ammonium ions in the narrow range 10 to 11.Arylamines, however, are considerably less basic than alkylamines, as are the heterocyclic amines pyridine and pyrrole. Basicity of Amines Amides (RCONH2) are nonbasic, in contrast with amines Amides are stabilized by delocalization of the nitrogen lone-pair electrons Amides are more stable than amines Stability is lost when protonated Basicity of Amines Primary and secondary amines can act as very weak acids N-H proton can be removed by a sufficiently strong base (eg. LDA) Butyllithium is a very strong base with a pKa on the order of 50.It can easily pull a proton from a primary or secondary amine. Worked Example Which compound in the following pair is more basic
CH3NHCH3 or pyridine Solution: CH3NHCH3 is more basic than pyridine pKa of CH3NHCH3 is 10.73 pKa of pyridine is 5.25 Basicity of Arylamines
Arylamines are less basic than alkylamines The N lone-pair electrons in arylamines are delocalized by interaction with the aromatic rings electron system Are less able to accept H+ Energy difference between protonated and nonprotonated forms is higher for arylamines Electrostatic Potential Maps Basicity of Arylamines
Substituted arylamines can be either more basic or less basic than aniline Electron-donating substituents increase the basicity of the corresponding arylamine Electron-withdrawing substituents decrease arylamine basicity Worked Example Rank the following compounds in order of ascending basicity p-nitroaniline, p-aminobenzaldehyde, p-bromoaniline Solution: Least basic Most basic All of these groups withdraw electrons by resonance.The nitro group is the most electron withdrawing.Remember that it directs electrophilic aromatic substitution into the meta position by deactivating the ring.The aldehyde behaves similarly but not to the extent the nitro group does.The bromide is electron withdrawing but not nearly to the extent of the other two groups. Biological Amines and the Henderson-Hasselbalch Equation
To reflect structures at physiological pH: Cellular amines are written in their protonated form Amino acids in their ammonium carboxylate form The Henderson-Hasselbalch equation can be used to calculate the extent of dissociation of an acid (in this case a protonated amine) at a given pH if the pKa is known. Because alkylamines have high pKa values (10-11), it is possible to determine the predominant form (ionized or unionized) at physiologic pH of 7.3. Worked Example Calculate the percentages of neutral and protonated forms present in a solution of M pyrimidine at pH = 7.3 The pKa of pyrimidinium ion is 1.3 Solution: At pH = 7.3, virtually 100% of the pyrimidine molecules are in the neutral form In contrast, alkylamines with pKas of 10 to 11 exist almost exclusively in their ionized protonated form. Amino acids then exist as zwitterions with the amine protonated and the acid deprotonated. Synthesis of Amines Reduction of nitriles, amides, and nitro compounds
Amines can be prepared by reduction of nitriles and amides with LiAlH4 The use of an SN2 displacement of an alkyl halide by cyanide allows for the synthesis of a primary amine with one additional carbon. Primary amides are reduced to primary amines with the same number of carbons. Synthesis of Amines Arylamines are prepared from nitration of anaromatic compound and reduction of the nitro group Reduction by catalytic hydrogenation over platinumis suitable if no other groups can be reduced Iron, zinc, tin, and tin(II) chloride are effective inacidic solution Tin(II) chloride is particularly mild and is often used when other reducible functional groups are present. Worked Example Propose structures for either a nitrile or an amide that might be a precursor of N-ethylaniline Solution: The compound can be synthesized only by amide reduction Amide reduction can be used to synthesize most amines, but nitrile reduction can be used to synthesize only primary amines SN2 Reactions of Alkyl Halides
Simplest method of alkylamine synthesis is by SN2 alkylation of ammonia or an alkylamine with an alkyl halide Ammonia and other amines are good nucleophiles These reactions dont stop cleanly after a single alkylation has occurred, however. SN2 Reactions of Alkyl Halides
Primary, secondary, and tertiary amines all have similar reactivity Initially formed monoalkylated substance undergoes further reaction to yield a mixture of products Secondary and tertiary amines undergo further alkylation SN2 Reactions of Alkyl Halides
Azide ion, N3, displaces a halide ion from a primary or secondary alkyl halide to give an alkyl azide Alkyl azides are not nucleophilic Reduction gives only the primary amine Although this method works well, low molecular weight alkyl azides are explosive and must be handled carefully. SN2 Reactions of Alkyl Halides
Gabriel amine synthesis: A phthalimide alkylation for preparing a primary amine from an alkyl halide N-H in imides (CONHCO) can be removed by KOH followed by alkylation and hydrolysis Worked Example Show two methods for the synthesis of dopamine, a neurotransmitter involved in regulation of the central nervous system Use any alkyl halide needed Worked Example Upper reaction is azide synthesis
Lower reaction is Gabriel synthesis Reductive Amination of Aldehydes and Ketones
Reductive amination: Treatment of an aldehyde or ketone with ammonia or an amine in the presence of a reducing agent Mechanism Reductive Amination Ammonia, primary amines, and secondary amines through reductive amination reaction yield primary, secondary, and tertiary amines, respectively Worked Example How might the following amine be prepared using a reductive amination reaction? Solution: Amine Precursor - Carbonyl Precursor - CH3CHO One could envision that ethylamine and benzaldehyde would be effective partners in this reaction as well.These partners would lead to the same product. Hofmann and Curtius Rearrangements
Carboxylic acid derivatives can be converted into primary amines with loss of one carbon atom by both the Hofmann rearrangement and the Curtius rearrangement Both involve the loss of carbon dioxide but their mechanisms are different.Both involve the migration of the R group onto nitrogen atom. Mechanism Hofmann Rearrangement Mechanism Hofmann Rearrangement
With heat carbamic acids lose carbon dioxide. Despite its mechanistic complexity, Hofmann rearrangement often gives high yields of both arylamines and alkylamines. Curtius Rearrangement
Heating an acyl azide prepared from an acid chloride Migration of R from C=O to the neighboring nitrogen with simultaneous loss of a leaving group Worked Example How is the following amine prepared using Curtius rearrangements on a carboxylic acid derivative? Worked Example Solution:
The precursor is an acid chloride, which is treated with NaN3, then with H2O and heat Note the lowermost path is the Curtius rearrangement and the uppermost path is the Hofmann rearrangement. Reactions of Amines Primary and secondary amines can also be acylated by nucleophilic acyl substitution reaction Whereas overalkylation of ammonia, primary, and secondary amines with alkyl halides is a problem, overacylation of these amines is not a problem because the amide nitrogen is insufficiently nucleophilic to react further. Hofmann Elimination Converts amines into alkenes
NH2 is very a poor leaving group; it is converted to an alkylammonium ion, which is a good leaving group An amine is completely methylated by reaction with an excess amount of iodomethane to produce the corresponding quaternaryammonium salt Hofmann Elimination Silver oxide is used for the elimination step
Exchanges hydroxide ion for iodide ion in the quaternary ammonium salt, thus providing the base necessary to cause elimination The actual elimination step is an E2 elimination in which hydroxide ion removes a proton while the positively charged nitrogen atom leaves. Orientation in Hofmann Elimination
Major product is the less highly substituted alkene Non-Zaitsev result is probably steric Due to the large size of the trialkylamine leaving group The base must abstract a hydrogen from the most sterically accessible Least hindered position Orientation in Hofmann Elimination Worked Example What products would you expect from Hofmann elimination of the following amine? If more than one product is formed, indicate which is major Worked Example Solution:
The first pair of products results from elimination of a primary hydrogen and are the major products The second pair of products results from elimination of a secondary hydrogen Reactions of Arylamines
Electrophilic aromatic substitution Amino substituents are strongly activating and ortho- and para-directing groups in electrophilic aromatic substitution reactions Reactions are controlled by conversion to amide Because amines are so strongly activating, polysubstitution in electrophilic aromatic substitution reactions is a problem.In the case of aniline, it is impossible to get solely the monosubstituted product. Converting the NH2 to an amide decreases the activation provided by the NH2 group substantially without affecting the substitution pattern (ortho, para). Reactions of Arylamines
The amino group forms acid-base complex with the AlCl3 catalyst preventing further reaction Therefore, we use the corresponding amide Similarly, anilines cannot undergo Friedel-Craft type reactions because the basic nitrogen effectively ties up the Lewis acid catalyst. Again, conversion to the amide solves the problem. Reactions of Arylamines
Modulating the reactivity of an amino-substituted benzene allows many kinds of electrophilic aromatic substitutions to be carried out Sulfa drugs were among the first pharmaceutical agents to be used clinically against bacterial infection Note here that hydrolysis of the amide occurs preferentially over hydrolysis of the sulfonamide; sulfonamides hydrolyze very slowly. Worked Example Propose syntheses of m-chloroaniline from benzene
Solution: Chlorination occurs before reduction so that chlorine can be introduced in the m-position Neither the NH2 group nor the chlorine substituent can be present initially in the reactant because both of these groups direct ortho, para.An amide does similarly.So both the NH2 and Cl must be introduced at a later stage.Since the nitro group directs meta and can be easily converted into an aniline, its introduction permits meta chlorination.It is important to use tin in the reduction step because catalytic hydrogenation with a nickel catalyst could remove the chlorine substituent. Reactions of Arylamines
Diazonium salts: The Sandmeyer reaction Primary arylamines react with HNO2 yielding stable arenediazonium salts Corresponding alkanediazonium can not be isolated Arenediazonium salts are useful because the diazonio group can be replaced by a nucleophile in a substitution reaction The process of forming a diazonium salt by reaction of an aniline with nitrous acid is referred to as diazotization. Reactions of Arylamines
Sandmeyer reaction: Reaction of an arenediazonium salt with the corresponding copper(I) halide to yield aryl chlorides and bromides Aryl iodides can be prepared by direct reaction with NaI Copper(I) salts are not required in the synthesis of aryl iodides. Reactions of Arylamines
An arenediazonium salt and CuCN yield the nitrile, ArCN, which can be hydrolyzed to other functional groups The diazonio group can be replaced by OH to yield a phenol and by H to yield an arene Note in the first reaction above, the product cannot be prepared from o-xylene by the usual side-chain oxidation because both methyl groups would be oxidized. Reactions of Arylamines
Reaction of arenediazonium salt with copper(I) oxide in an aqueous solution of copper(II) nitrate yield phenols This is a useful reaction because few other general methods exist for introducing an OH group onto an aromatic ring. Reduction to a Hydrocarbon
By treatment of a diazonium salt with hypophosphorous acid, H3PO2 Diazonium replacement takes place through radical pathways This reaction is used primarily when there is a need for temporarily introducing an amino substituent onto a ring to take advantage of its directing effect.By incorporating an amino group in the para position of toluene, dibromination in the two ortho positions is possible.Removal of the amino group by diazotization effectively removes the amine substituent to afford 3,5-dibromotoluene. Worked Example How is p-bromobenzoic acid prepared from benzene using a diazonium replacement reaction Solution: Use of the diazonium replacement reaction that substitutes bromine for a nitro group Both the carboxy and bromo substituents on an aromatic ring are ortho, para directing.One would find it very difficult to separate the isomers that would be generated by effecting electrophilic aromatic substitution reactions on one or the other.Nitro isomers tend to be easily separable either via chromatography or by recrystallization.Nitrating toluene would lead to the para isomer being the predominant product for steric reasons.There still would be some ortho isomer but that could be readily separated by recrystallization.Conversion to the diazonium salt allows for the introduction of the bromine. Diazonium Coupling Reactions
Arenediazonium salts undergo a coupling reaction with activated aromatic rings to yield azo compounds, ArN=NAr Are typical electrophilic aromatic substitutions Diazonium Coupling Reactions
The electrophilic diazonium ion reacts with the electron-rich ring of a phenol or arylamine Usually occurs at the para position but goes ortho if para is blocked Diazonium Coupling Reactions
Azo-coupled products have extended conjugation that lead to low energy electronic transitions that occur in visible light Azo-coupled products are widely used as dyes for textiles because of their extended conjugation. Worked Example Propose a synthesis of p-(dimethylamino)azobenzene with benzene as organic starting material Solution: Heterocyclic Amines A cyclic organic compound that contains atoms of two or more elements in its ring Most heterocyclic compounds possess same chemistry as their open-chain counterparts Lactones and acyclic esters behave similarly, lactams and acyclic amides behave similarly, and cyclic and acyclic ethers behave similarly. Where the ring is unsaturated, however, heterocycles have unique and interesting properties. Heterocyclic Amines Pyrrole
Simplest five-membered unsaturated heterocyclic amine Undergoes electrophilic substitution reactions Chemical properties are not consistent with the structural features of amine or a conjugated diene Pyrrole has six pi electrons and is aromatic.Thus it does not react as a diene or as a typical aliphatic amine. Protonation on nitrogen would destroy the aromaticity so pyrrole is not basic nor is it nucleophilic. Halogenation, nitration, sulfonation, and Friedel-Crafts acylation can all be accomplished. Heterocyclic Amines Ring is reactive toward electrophiles
Electrophilic substitutions occur at C2 Reaction leads to intermediate cation having 3 resonance forms Reaction at C3 gives only 2 resonance forms Heterocyclic Amines Imidazole Common five-membered heterocyclic amine
Constituent of the amino acid histidine One of two nitrogens is basic One of the nitrogen atoms in imidazole and the lone nitrogen atom in thiazole are basic and can be protonated.Imidazole is a very mild base which is often used to scavenge protons in some organic reactions. Worked Example Draw an orbital picture of thiazole Solution:
Assume that both the nitrogen and sulphur atoms are sp2-hybridized Show the orbitals that the lone pairs occupy Solution: Contains six electrons Each carbon contributes one electron, nitrogen contributes one electron, and sulfur contributes two electrons to the ring system The overall result is there are six pi electrons and thiazole is aromatic. Worked Example Sulfur and nitrogen have lone electron pairs in sp2 orbitals that lie in the plane of the ring Heterocyclic Amines Pyridine
Nitrogen-containing heterocyclic analog of benzene The sp2-hybridized nitrogen atom is less basic than the sp3-hybridized nitrogen in an alkylamine Pyridine is aromatic.The lone pair on nitrogen is in an sp2 orbital.Having less s character than an sp3 orbital the lone pair is held closer to the nitrogen nucleus accounting for the diminished basicity. Heterocyclic Amines Undergoes electrophilic aromatic substitution reactions with difficulty Acid-base complexation between the basic rings nitrogen atom and the incoming electrophile High dipole moment pulls density out of ring Six-membered diamine pyrimidine is found commonly in biological molecules Worked Example Electrophilic aromatic substitution reactions of pyridine normally occur at C3 Draw the carbocation intermediates resulting from reaction of an electrophile at C2, C3, and C4 Explain the observed result Solution: Worked Example Reaction at C3 is favored over reaction at C2 or C4 Worked Example Positive charge of the cationic intermediate of reaction at C3 is delocalized over three carbon atoms rather than over two and the electronegative pyridine nitrogen as occurs in reaction at C2 or C4 Heterocyclic Amines Polycyclic heterocycles
Quinoline, isoquinoline, and indole contain both a benzene ring and a heterocyclic aromatic ring Purine contains two heterocyclic rings joined together Quinoline and isoquinoline are less reactive towards electrophilic substitution Indole undergoes electrophilic substitution more easily than benzene Heterocyclic Amines Heterocyclic Amines Purine has three basic, pyridine-like nitrogen atoms Worked Example Which nitrogen atom in the hallucinogenic indole alkaloid N,N-dimethyltryptamine is more basic? Explain Worked Example Solution:
Side chain nitrogen atom of N,N-dimethyltryptamine is more basic than the ring nitrogen atom Aromatic nitrogen electron lone pair is part of the ring electron system Spectroscopy of Amines
Infrared spectroscopy 1 and 2 amine are identified by characteristic NH stretching absorptions at 3300 to 3500 cm1 Amine absorption bands are sharper and less intense than hydroxyl bands NH bend (scissor) is noticed above 1600 cm1 Generally primary amines exhibit two bands in the 3300 to 3500 region.Secondary amines exhibit only one band.Since tertiary amines do not have an N-H bond they do not absorb in this area. Nuclear Magnetic Resonance Spectroscopy
NH hydrogens appear as broad signals without clear-cut coupling to neighboring CH hydrogens In D2O exchange of ND for NH occurs, and the NH signal disappears Hydrogens on the carbon next to nitrogen are deshielded due to the electron-withdrawing effect of nitrogen Carbons next to amine nitrogens absorb about 20 ppm further downfield than would otherwise be expected if they were not deshielded by the nitrogen atom. Nuclear Magnetic Resonance Spectroscopy
Hydrogens on C next to N absorb further downfield than alkane hydrogens N-CH3 gives a sharp three-H singlet at 2.2 to 2.6 Worked Example Compound A, C6H12O, has an IR absorption at 1715 cm-1 and gives compound B, C6H15N, when treated with ammonia and NaBH4 What are the structures of A and B? Worked Example Solution:
It is inferred that B is a primary amine, from the spectrum 1H NMR spectrum shows a 9-proton singlet, a one-proton quartet, and a 3-proton doublet Absorption due to the amine protons is not visible Mass Spectrometry Compound with odd number of nitrogen atoms has odd-numbered molecular weight Presence of N can be detected observing the spectrum Alkylamines cleave at the CC bond nearest the nitrogen to yield an alkyl radical and a nitrogen-containing cation Mass Spectrum of N-ethylpropylamine
The two main modes of a cleavage give fragment ions at m/z = 58 and m/z = 72