University:
Rutgers University Newark
Course:
CHEM 335 | Organic Chemistry
Academic year:

2024
Views:

141

Pages:

95
Author:

Mahitab S.

CHAPTER 23 AMINES ©2018 Cengage Learning. All Rights Reserved. Chapter Outline • (23.1) • (23.2) • (23.3) • • • • • • • (23.4) (23.5) (23.6) (23.7) (23.8) (23.9) (23.10) Structure and classification Nomenclature Chirality of amines and quaternary ammonium ions Physical properties Basicity Reactions with acids Preparation Reaction with nitrous acid Hofmann elimination Cope elimination ©2018 Cengage Learning. All Rights Reserved. Structure and Classification of Amines NH3 ©2018 Cengage Learning. All Rights Reserved. HefrocyclicAmines NitrogenispartoftheRing Structure and Classification of Amines should beconnection between nandRing should beconnection between nandRing Final • Heterocyclic amines: Amines in which nitrogen is one of the atoms of a ring • An amine is classified as heterocyclic aromatic when the nitrogen is part of an aromatic ring ©2018 Cengage Learning. All Rights Reserved. b should beconnection between nandRing Example 23.1 - Types of Amines heterocyclic 20 Aliphatic Amine Inyo 2 Aromatic faiphat Heterocyclic 3 Aliphatic Heterocyclic • Classify each amino group in these alkaloids according to type (primary, secondary, tertiary, aliphatic, aromatic, heterocyclic) • Solution a. A secondary aliphatic heterocyclic amine b. A tertiary aliphatic heterocyclic amine and a heterocyclic aromatic amine c. A tertiary aliphatic heterocyclic amine ©2018 Cengage Learning. All Rights Reserved. Example 23.1 - Types of Amines • Alkaloids are basic nitrogen-containing compounds of plant origin, many of which are physiologically active when administered to humans • Ingestion of coniine, isolated from water hemlock, can cause weakness, labored respiration, paralysis, and eventually death • Coniine is the toxic substance in the “poison hemlock” used in the death of Socrates • In small doses, nicotine is an addictive stimulant • In larger doses, it causes depression, nausea, and vomiting; in still larger doses, it is a deadly poison • Solutions of nicotine in water are used as insecticides • Cocaine is a central nervous system stimulant obtained from the leaves of the coca plant ©2018 Cengage Learning. All Rights Reserved. Example • Select the best classification for the following molecule: 1. 2. 3. 4. 5. 2 Aliphaticamine 1˚ aliphatic amine 2˚ aliphatic amine 3˚ aliphatic amine Aromatic amine Heterocyclic aromatic amine ©2018 Cengage Learning. All Rights Reserved. Example • Which of the following is not represented in the molecule shown? 1. 2. 3. 4. 5. Aliphatic amine Aromatic amine Heterocyclic aromatic amine Secondary amine A pyridine ring ©2018 Cengage Learning. All Rights Reserved. Nomenclature of Amines - Systematic Names • Aliphatic amines • Suffix -e of the parent alkane is replaced with -amine I 2 propanamine I 23 1phenylethanamine 1,6Hexandiamine • IUPAC system retains the common name aniline for C6H5NH2 • Simple derivatives of aniline are named using the prefixes o-, m-, and p- or numbers i n to locate substituents z z 43 para ©2018 Cengage Learning. All Rights Reserved. 43 par 3 meta Nomenclature of Amines - Systematic Names • 2° and 3° amines are commonly named as N-substituted primary amines • Unsymmetrical amines • (1) Largest group is considered as the parent amine • (2) Smaller group(s) bonded to nitrogen are named, and (3) their location is indicated by the prefix N in order to indicate that they are bonded to nitrogen N aniline methyl N N Dimethyl cycolopentanamine ©2018 Cengage Learning. All Rights Reserved. Nomenclature of Amines - Systematic Names • Among the various functional groups discussed in the text, —NH2 is one of the lowest in order of precedence • Each of the following compounds contains a functional group of higher precedence than the amino group • Amino group is indicated by the prefix amino 43 2 Aminoethanol 2 I 2 Amino 3methylbutanol ©2018 Cengage Learning. All Rights Reserved. y 4 3 2 Aminobengor Example 23.2 - Amine Nomenclature • Write systematic names for these amines I 1 Hexanamine z 3 diamine 4Butane y • Solution a. 1-Hexanamine b. 1,4-Butanediamine c. 2-Phenylethanamine ©2018 Cengage Learning. All Rights Reserved. I 2phenylethanamine Nomenclature of Amines - Common Names • Common names for most aliphatic amines are derived by listing the alkyl groups bonded to nitrogen in alphabetical order in one word ending with the suffix -amine • Named as alkylamines • When four groups are bonded to nitrogen, the compound is named as a salt of the corresponding amine • The ending -amine (or -aniline, pyridine, and so on) is replaced with -ammonium (or anilinium, pyridinium, and so on), and the name of the anion is added ©2018 Cengage Learning. All Rights Reserved. Nomenclature of Amines - Common Names • Quaternary (4°) ammonium ion: Contains a nitrogen atom bonded to any combination of four alkyl or aryl groups • Compounds containing these ions have properties characteristic of salts ©2018 Cengage Learning. All Rights Reserved. Example 23.4 - Amine Nomenclature • Write the IUPAC name and, where possible, a common name for each compound (next slide) 2 I trans 2Amino 1 2 34 1 phenyl3 Aminebutane cycolohexand ©2018 Cengage Learning. All Rights Reserved. y Example 23.4 - Amine Nomenclature phenyl Colts a. Diphenylamine b. trans-2-Aminocyclohexanol c. (S)-1-phenyl-2-propanamine • • • Common name - Amphetamine The dextrorotatory isomer of amphetamine (shown here) is a central nervous system stimulant and is manufactured and sold under several trade names The salt with sulfuric acid is marketed as Dexedrine sulfate ©2018 Cengage Learning. All Rights Reserved. Example • Determine the IUPAC name for the following molecule: propyl 1. 2. X 3. 4. 5. Ethylpropylphenylamine N-phenyl-N-ethylpropanamine N-ethyl-N-phenylpropanamine N-ethyl-N-propylaniline N-phenyl-N-ethylaniline ©2018 Cengage Learning. All Rights Reserved. Chirality of Amines • Geometry of a nitrogen atom bonded to three other atoms or groups of atoms is trigonal pyramidal • (1) sp3 hybridized N atom is at the apex of the pyramid • (2) Three groups bonded to the N atom extend downward to form the triangular base of the pyramid b leftside • (3) If the unshared pair of electrons on N is considered as a fourth group, then the arrangement of groups around N is approximately tetrahedral Trigonalpyramidal IThreegroupsbondedtotheN ©2018 Cengage Learning. All Rights Reserved. pairof electrons Unshared lowestpriority Chirality of Amines • An amine with three different groups bonded to N is chiral, can exist as a pair of enantiomers, and is resolvable in principle • In assigning configuration, the unshared pair of electrons has the lowest priority • In special cases, enantiomers cannot be resolved as they undergo rapid interconversion by pyramidal inversion ©2018 Cengage Learning. All Rights Reserved. Chirality of Amines • Pyramidal inversion: Rapid oscillation of a nitrogen atom from one side of the plane of the three atoms bonded to it to the other side of that plane • Converts the chiral amine to a racemic mixture • Pyramidal inversion is not possible with quaternary ammonium ions, and their salts can be resolved ©2018 Cengage Learning. All Rights Reserved. Chirality of Amines • Phosphorus forms trivalent compounds called phosphines, which have trigonal pyramidal geometry (the same as amines, Ch.24) • Activation energy for pyramidal inversion of trivalent phosphorus compounds is greater than it is for trivalent compounds of nitrogen, with the result that a number of chiral phosphines can be resolved P N ©2018 Cengage Learning. All Rights Reserved. Physical Properties of Amines • Polar compounds • Both 1° and 2° amines form intermolecular hydrogen bonds • An N—H · · · N hydrogen bond is weaker than an O—H · · · O hydrogen bond 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) • All classes of amines form hydrogen bonds with water • More soluble in water than hydrocarbons of comparable molecular weight ©2018 Cengage Learning. All Rights Reserved. Table 23.1 - Physical Properties of Selected Amines ©2018 Cengage Learning. All Rights Reserved. Allamines weak bases Basicity of Amines • All amines are weak bases, and aqueous solutions of amines are basic • Curved arrows are used in reactions to emphasize that the unshared pair of electrons on N forms a new covalent bond with H and displaces hydroxide ion • It is common to discuss their basicity by referencing the acid ionization constant of the corresponding conjugate acid ©2018 Cengage Learning. All Rights Reserved. Table 23.2 - Acid Strengths of the Conjugate Acids of Selected Amines 1 ©2018 Cengage Learning. All Rights Reserved. Aniline andpyridine Around 5 Table 23.2 - Acid Strengths of the Conjugate Acids of Selected Amines Aniline and pyridine Y around ©2018 Cengage Learning. All Rights Reserved. 5 Example 23.5 - Amines: Acid/Base Equilibria • Predict the position of equilibrium for this acid-base reaction • Solution • Use the approach seen in Section 4.4 to predict the position of equilibrium in acid-base reactions • Equilibrium favors reaction of the stronger acid with the stronger base to give the weaker acid and weaker base ©2018 Cengage Learning. All Rights Reserved. Basicity - Aliphatic Amines • Share the same base strength, pKa of the conjugate acid 10–11 • Slightly stronger bases than ammonia • Increase in basicity compared with ammonia can be attributed to the greater stability of an alkylammonium ion • Example: Comparison of RCH2NH3+ with NH4+ • Greater stability arises from the: • Electron-releasing effect of alkyl groups • Resulting partial delocalization of the positive charge from nitrogen onto carbon in the alkylammonium ion ©2018 Cengage Learning. All Rights Reserved. Basicity - Aromatic Amines • Considerably weaker bases than aliphatic amines • Factors that contribute to decreased basicity • (1) Resonance stabilization of the free base form of aromatic amines ©2018 Cengage Learning. All Rights Reserved. Basicity - Aromatic Amines • (2) The greater electron-withdrawing inductive effect of sp2-hybridized carbons of the aromatic ring compared with the sp3-hybridized carbons of aliphatic amines • Electron-releasing groups, such as alkyl groups, increase the basicity of aromatic amines • Electron-withdrawing groups, such as nitro and carbonyl groups, decrease the basicity of aromatic amines • Decrease in basicity of halogen substitution is caused by the electron-withdrawing inductive effect of the electronegative halogen EDG increasethe basicity ©2018 Cengage Learning. All Rights Reserved. EWG decreasethebasicity Basicity - Aromatic Amines • Decrease in basicity on nitro substitution is caused by a combination of resonance and inductive effects as seen by comparing the pKa values of the following nitroanilines: • In the case of para and ortho substitutions, delocalization of the lone pair on the amino nitrogen involves the carbons of the aromatic ring and the oxygen of the nitro group ©2018 Cengage Learning. All Rights Reserved. Basicity - Heterocyclic Aromatic Amines • Weaker bases than aliphatic heterocyclic amines • In pyridine, the unshared pair of electrons on N is not part of the aromatic sextet • Pyridine is a weaker base than aliphatic amines because the free electron pair on N lies in a relatively electronegative sp2 hybrid orbital, whereas in aliphatic amines, the unshared pair lies in an sp3 hybrid orbital ©2018 Cengage Learning. All Rights Reserved. Basicity - Heterocyclic Aromatic Amines • Imidazole has two nitrogen atoms, each with an unshared pair of electrons • Positive charge on the imidazolium ion is delocalized on both nitrogen atoms of the ring • Stronger base than pyridine ©2018 Cengage Learning. All Rights Reserved. sp2 EP Basicity - Heterocyclic Aromatic Amines • Pyrrole is an aromatic heterocycle • Not as basic as pyridine or imidazole • Lacks base strength • Protonated pyrrole cannot maintain aromaticity because the protonated nitrogen would be sp3 hybridized • There would be only 4 π electrons remaining (in violation of two of Hückel’s aromaticity rules) • Loss of aromaticity limits the ability of pyrrole to accept a proton ©2018 Cengage Learning. All Rights Reserved. Guanidine Strongest base Guanidinumion weakerAcid Basicity - Guanidine • Guanidine is the strongest base among neutral organic compounds • Guanidinium ion is a weaker acid than almost any other protonated amine • Basicity is attributed to the delocalization of the positive charge over the entire functional group ©2018 Cengage Learning. All Rights Reserved. Example 23.6 - Relative Amine Basicity • Select the stronger base in each pair of amines than Aliphatic ismore stronger Aromatic ©2018 Cengage Learning. All Rights Reserved. Example 23.6 - Relative Amine Basicity a. Morpholine (B) is the stronger base (conjugate acid pKa 8.2) • • It has a basicity comparable to that of secondary aliphatic amines Pyridine (A), a heterocyclic aromatic amine (pKa 5.25), is considerably less basic than aliphatic amines b. Tetrahydroisoquinoline (C) has a basicity comparable to that of secondary aliphatic amines (pKa ~ 10.8) and is the stronger base • Tetrahydroquinoline (D) has a basicity comparable to that of N-substituted anilines (pKa ~ 4.4) and is the weaker base ©2018 Cengage Learning. All Rights Reserved. Example 23.6 - Relative Amine Basicity c. Benzylamine (F) is the stronger base (pKa 9.6) • • Its basicity is comparable to that of other aliphatic amines The basicity of o-toluidine (E), an aromatic amine, is comparable to that of aniline (pKa 4.6) ©2018 Cengage Learning. All Rights Reserved. Example • If the pKb of an amine is 9.54, what is the pKa of its conjugate acid? b 1. 2. 3. 4. 5. 6.46 4.46 −9.54 Not enough information to determine pKa is unrelated to pKb ©2018 Cengage Learning. All Rights Reserved. Around 5 Example do • Arrange the above from the strongest to the weakest base: 1. 2. 3. 4. 5. A, B, C B, C, A C, A, B A, C, B B, A, C ©2018 Cengage Learning. All Rights Reserved. Example • Amine A is more basic than amine B 3 1. 2. True False y meta 2 para Metais more basicthanortho para Acidic Aromaticcompoundsaremore ©2018 Cengage Learning. All Rights Reserved. Example • The equilibrium lies to the right in the above reaction 1. 2. True False left side b downward ©2018 Cengage Learning. All Rights Reserved. Example • Select the most acidic compound from the choices provided 1. 2. 4. 3. 5. Aromatic ©2018 Cengage Learning. All Rights Reserved. Example deprotonati • If a protonated amine with a pKa of 10 is placed in a solution of pH 12, the predominant form of the amine in solution will be the protonated form 1. 2. True False ©2018 Cengage Learning. All Rights Reserved. Reactions with Acids • All amines, whether soluble or insoluble in water, react quantitatively with strong acids to form watersoluble salts Amines Sty watersoluble salt ©2018 Cengage Learning. All Rights Reserved. Figure 23.2 - Separation and Purification of an Amine and a Neutral Compound • Basicity of amines and the solubility in water of amine salts can be used to separate water-insoluble amines from water-insoluble, nonbasic compounds ©2018 Cengage Learning. All Rights Reserved. Example 23.7 - Amine Acid-Base Reactions • Complete each acid-base reaction and name the salt formed a. Et2NH + HCl → EtaNH CI 3100 b. PhCH2NH2 + CH3COOH →phataNH3 CH • Solution a. Et2NH2+Cl− (diethylammonium chloride) b. PhCH2NH3+ CH3COO− (benzylammonium acetate) ©2018 Cengage Learning. All Rights Reserved. Example 23.8 - Amino Acid Protonation wind • Following are two structural formulas for (S)-serine, one of the building blocks of proteins: • Is (S)-serine better represented by structural formula A or B? ©2018 Cengage Learning. All Rights Reserved. Example 23.8 - Amino Acid Protonation • Structural formula A contains both an amino group (a base) and a carboxyl group (an acid) • Proton transfer from the stronger acid (—COOH) to the stronger base (—NH2) gives an internal salt • Therefore, B is the better representation for (S)-serine • Within the field of amino acid chemistry, the internal salt represented by B is called a zwitterion (Section 27.2) ©2018 Cengage Learning. All Rights Reserved. Preparation - Alkylation of Ammonia and Amines • Treating a haloalkane with ammonia or an amine by SN2 mechanism gives an alkylammonium salt • Such alkylations give mixtures of products through a series of proton-transfer and nucleophilic substitution reactions ©2018 Cengage Learning. All Rights Reserved. Preparation - Alkylation of Ammonia and Amines • Final product is a tetraalkylammonium halide • NOT a useful laboratory method for the preparation of more complex amines • However, 1° amines are easily prepared because ammonia is inexpensive and can be used in large excess • Other amines can be prepared in this way if the nucleophilic amine is inexpensive enough to be used in large excess ©2018 Cengage Learning. All Rights Reserved. Preparation - Alkylation of Azide Ion x• NaN3 or KN3 eliminates the problem of overalkylation • (1) Alkyl azides are prepared from sodium or potassium azide and a 1° or 2° haloalkane by an SN2 reaction • (2) Azides are reduced to 1° amines by a variety of reducing agents, including LiAlH4 ©2018 Cengage Learning. All Rights Reserved. Preparation - Alkylation of Azide Ion • Azide ion can be used for stereoselective ring opening of epoxides • Reduction of the resulting β-azidoalcohol gives a βaminoalcohol Resonance ©2018 Cengage Learning. All Rights Reserved. Resonance Example 23.10 - Amine Synthesis • Show how to convert 4-methoxybenzyl chloride to each amine MIT MEIN't i nat Ey i II f inaan ÉI ©2018 Cengage Learning. All Rights Reserved. Egon Example 23.10 - Amine Synthesis a. Two methods might be used • 1. Alkylation using a large molar excess of NH3 to reduce the extent of overalkylation • 2. Nucleophilic displacement of chloride using azide ion (from NaN3) followed by LiAlH4 reduction of the azide • Nucleophilic displacement by azide is more convenient on a laboratory scale b. Nucleophilic displacement of chloride by cyanide ion is followed by reduction of the cyano group with lithium aluminum hydride (1C homologation) ©2018 Cengage Learning. All Rights Reserved. Reaction of Amines with HNO2 • Nitrous acid, a weak acid, is prepared by treating aqueous solution of NaNO2 with H2SO4 or HCl • In its reactions with amines, nitrous acid: • Participates in proton-transfer reactions • Is a source of the nitrosyl cation, a weak electrophile produce Nitrosyl cation ©2018 Cengage Learning. All Rights Reserved. Formation of the Nitrosyl Cation • Step 1: Add a proton (protonation) • Protonation of the OH group of nitrous acid gives an oxonium ion • Step 2: Break a bond to give stable molecules or ions • Loss of H2O gives the nitrosyl cation (key step) ©2018 Cengage Learning. All Rights Reserved. Reaction of Tertiary Amines with HNO2 • (1) 3° aliphatic amines • Protonated to form water-soluble salts, whether watersoluble or water-insoluble • (2) 3° aromatic amines protonated • Form salts with nitrous acid salt HNoz • Alternative pathway: Electrophilic aromatic substitution • NO+, a very weak electrophile, reacts with aromatic rings containing strongly activating ortho-para directing groups • When treated with nitrous acid, compounds undergo nitrosation to give blue or green aromatic nitroso compounds ©2018 Cengage Learning. All Rights Reserved. Reaction of Secondary Amines with HNO2 • 2° aliphatic and 2° aromatic amines react with nitrous acid to give N-nitrosamines • Steps in reaction • Step 1: Make a new bond between a nucleophile and an electrophile: Reaction of the 2° amine (a nucleophile) with NO+ (an electrophile) gives an N-nitrosammonium ion ©2018 Cengage Learning. All Rights Reserved. Reaction of Secondary Amines with HNO2 • Step 2: Take away a proton • Proton transfer to solvent gives the N-nitrosoamine ©2018 Cengage Learning. All Rights Reserved. Reaction of Primary Aliphatic Amines with HNO2 generate ©2018 Cengage Learning. All Rights Reserved. a positivecharge Reaction of Primary Aliphatic Amines with HNO2 • Step 1: Make a new bond between a nucleophile and an electrophile (key step) • Reaction of a 1° amine with the with nitrosyl cation • Step 2: Take a proton away • Step 3: Keto-enol tautomerism ©2018 Cengage Learning. All Rights Reserved. Reaction of Primary Aliphatic Amines with HNO2 • Step 4: Add a proton: Diazotic acid is protonated • Step 5: Protonated diazotic acid loses H2O to give a diazonium ion • Step 6: Diazonium ion loses N2 to give a carbocation ©2018 Cengage Learning. All Rights Reserved. Reaction of Primary Aliphatic Amines with HNO2 • Aliphatic diazonium ions are unstable and lose N2 to give carbocations and nitrogen gas • Carbocation may: 1. Lose a proton to give an alkene 2. React with a nucleophile to give a substitution product 3. Rearrange to a more stable carbocation and then react further by Steps 1 or 2 • Since the reaction gives a mixture of products, it is generally not useful • Exception: Tiffeneau-Demjanov reaction (next slide) ©2018 Cengage Learning. All Rights Reserved. Tiffeneau-Demjanov Reaction ©2018 Cengage Learning. All Rights Reserved. Tiffeneau-Demjanov Reaction • Step 2: Break a bond to give stable molecules or ions and 1,2-shift (key step) • Step 3: Take a proton away to complete the reaction ©2018 Cengage Learning. All Rights Reserved. Reaction of Primary Aromatic Amines with HNO2 • 1° aromatic amines react with nitrous acid to form arenediazonium salts • The —NH2 group of can be replaced in a regioselective manner by the following groups: Aromatic amine b aromatic diazonium 4118 ©2018 Cengage Learning. All Rights Reserved. Reaction of Primary Aromatic Amines with HNO2 • 1° aromatic amines can be converted to phenols by forming the arenediazonium salt in aq. sulfuric acid and then heating NHz OH weuse nights • Intermediate in the decomposition of an arenediazonium ion in water is an aryl cation, which then undergoes reaction with water to form the phenol ©2018 Cengage Learning. All Rights Reserved. Example 23.12 - Multi-Step Synthesis • What reagents and experimental conditions will bring about each step in the conversion of toluene to 4hydroxybenzoic acid? 43 Ht 42504 If important NHz OH 42504 weUse Nanos heat cha cook w weuse Nz Ip KHz ©2018 Cengage Learning. All Rights Reserved. NII Example 23.12 - Multi-Step Synthesis • Step 1 • Nitration of the aromatic ring using HNO3 in H2SO4 followed by separation of the ortho and para isomers gives 4nitrotoluene • Step 2 • Oxidation at the benzylic carbon using K2Cr2O7 in H2SO4 gives 4-nitrobenzoic acid • Step 3 • Catalytic reduction of the nitro group to an amino group using H2 in the presence of Ni or another transition metal catalyst gives 4-aminobenzoic acid ©2018 Cengage Learning. All Rights Reserved. Example 23.12 - Multi-Step Synthesis • Alternatively, reduction of the nitro group to a primary amine can be brought about using Zn, Sn, or Fe in aqueous HCl followed by aqueous NaOH • Step 4 • Reaction of the aromatic amine with NaNO2 in aqueous H2SO4 followed by heating gives 4-hydroxybenzoic acid ©2018 Cengage Learning. All Rights Reserved. Reaction of Primary Aromatic Amines with HNO2 • Show how to bring about each conversion pad ©2018 Cengage Learning. All Rights Reserved. Schiemann and Sandmeyer Reactions • Schiemann reaction - Common method for the introduction of fluorine onto an aromatic ring • Thought to involve an aryl cation intermediate • Sandmeyer reaction - Treatment of a 1° aromatic amine with nitrous acid followed by heating with HCl/ CuCl, HBr/CuBr, or KCN/CuCN • Results in replacement of the diazonium group by —Cl, — Br, or —CN, respectively • Fails with CuI or CuF ©2018 Cengage Learning. All Rights Reserved. Sandmeyer Reaction - Example ©2018 Cengage Learning. All Rights Reserved. Sandmeyer-like Reactions of Primary Aromatic Amines • Treating an arenediazonium ion with iodide ion: Best and most convenient method for introducing iodine onto an aromatic ring • Treating an arenediazonium ion with hypophosphorous acid: Results in reduction of the diazonium group and its replacement by —H ©2018 Cengage Learning. All Rights Reserved. Example • The following transformation can be accomplished as shown 1. 2. True False ©2018 Cengage Learning. All Rights Reserved. Example • What is the intermediate that results from the treatment of aniline with HNO2 and H2SO4? 1. 2. 3. 4. 5. A diazonium salt An aryl radical A quaternary ammonium salt An isocyanate An aryl azide ©2018 Cengage Learning. All Rights Reserved. Hofmann Elimination • When treated with a strong base, a 4° ammonium halide undergoes β-elimination by an E2 mechanism to give the less substituted alkene as the major product • Example: (1) Preparation of a 4° ammonium hydroxide • (2) Thermal decomposition of the 4° ammonium hydroxide ©2018 Cengage Learning. All Rights Reserved. Hofmann Elimination • Mechanism - Take a proton away and break a bond to give stable molecules or ions • Following reactions occur simultaneously: • Removal of a β-hydrogen by base • Collapse of the electron pair of the C—H bond to become the π bond of the alkene • Loss of the trialkylamino group ©2018 Cengage Learning. All Rights Reserved. Hofmann Elimination ©2018 Cengage Learning. All Rights Reserved. Hofmann Elimination • Explanations for the formation of the less stable C=C bond • Hofmann elimination is governed largely by steric factors, namely the bulk of the —NR3+ group • Hydroxide ion preferentially approaches and removes the least hindered hydrogen and gives the least substituted alkene • Bulky bases, such as (CH3)3CO−K+, give Hofmann elimination from haloalkanes ©2018 Cengage Learning. All Rights Reserved. Example 23.14 - Hofmann Elimination • Draw a structural formula of the major alkene formed in each βelimination • Thermal decomposition of the quaternary ammonium hydroxide in (a) follows the Hofmann rule and gives 1-octene as the major product • E2 elimination from an alkyl iodide in (b) by sodium methoxide follows Zaitsev’s rule and gives trans-2-octene as the major product ©2018 Cengage Learning. All Rights Reserved. Example - Hofmann Elimination • Identify the structure of A based on the products that result from a Hofmann elimination + 1. 2. 4. 3. 5. ©2018 Cengage Learning. All Rights Reserved. Cope Elimination • Thermal decomposition of an amine oxide to an alkene • Example: (1) Oxidation of a 3° amine gives an amine oxide • (2) If the amine oxide with one β-hydrogen is heated, it undergoes thermal decomposition to form an alkene and an N,N-dialkylhydroxylamine ©2018 Cengage Learning. All Rights Reserved. Cope Elimination • Syn stereoselective and concerted • Mechanism: Within the same molecule, take a proton away and break a bond to give stable molecules or ions • Transition state involves: • Planar or nearly planar arrangement of the five participating atoms • Cyclic flow of three pairs of electrons ©2018 Cengage Learning. All Rights Reserved. Example 23.15 - Cope Elimination • When 2-dimethylamino-3-phenylbutane is treated with hydrogen peroxide and then made to undergo a Cope elimination, the major alkene formed is 2-phenyl-2-butene a. How many stereoisomers are possible for 2dimethylamino-3-phenylbutane? 2 b. How many stereoisomers are possible for 2-phenyl-2butene? I c. Suppose that the starting amine is the 2R,3S isomer • What is the configuration of the product? ©2018 Cengage Learning. All Rights Reserved. Example 23.15 - Cope Elimination a. There are two chiral centers in the starting amine • Four stereoisomers are possible: two pair of enantiomers b. There is one carbon-carbon double bond about which stereoisomerism is possible • Two stereoisomers are possible: one E,Z pair c. Following is a stereodrawing of the 2R,3S stereoisomer showing a syn conformation of the dimethylamino group and the β-hydrogen • Cope elimination on this stereoisomer gives S (E)-2-phenyl-2-butene so R ©2018 Cengage Learning. All Rights Reserved. Example • The major alkaloid present in tobacco leaves is nicotine, whose structure is shown below • Which will be the major ammonium ion formed when nicotine is treated with one equivalent of a strong acid? 1. 2. ©2018 Cengage Learning. All Rights Reserved. Example • Which of the following statements is true regarding the following two molecules? aromatic 1. y 2. x 3. 4. 5. x Aliphatic Both A and B are aromatic Both A and B are aliphatic amines A is more basic than B B is more basic than A Both A and B are planar molecules ©2018 Cengage Learning. All Rights Reserved. Example • Select the order of chemical shift (most downfield → most upfield) for the carbon atoms in the following molecule: 1. 2. 3. 4. 5. b, c, d, a a, d, b, c a, b, d, c c, b, d, a c, d, b, a ©2018 Cengage Learning. All Rights Reserved. Problem 23.54 (a) - Ritter reaction • The Ritter reaction is especially valuable for the synthesis of 3° aliphatic amines • In fact, there are few alternatives routes to them • The reaction is illustrated by the first step in the following sequence • In the second step, the Ritter product is hydrolyzed to the 3° amine • Propose a mechanism for the Ritter reaction ©2018 Cengage Learning. All Rights Reserved. Problem 23.54 (a) - Ritter reaction ©2018 Cengage Learning. All Rights Reserved. Problem 23.54 (a) - Ritter reaction ©2018 Cengage Learning. All Rights Reserved. Problem 23.54 (a) - Ritter reaction ©2018 Cengage Learning. All Rights Reserved. Problem 23.54 (b) - Ritter reaction • What is the product of the following sequence? • Solution • The product of this Ritter reaction is an amide, which is then reduced to an amine: ©2018 Cengage Learning. All Rights Reserved.