CHAPTER 22 AMINES SOLUTIONS TO TEXT PROBLEMS 22.1 (b) The amino and phenyl groups are both attached to C-1 of an ethyl group C6HSCHCH H,C=CHCH,NH, 2-propen-1-amin 22.2 N, N-Dimethylcycloheptylamine may also be named as a dimethyl derivative of cycloheptanamine. N(CH3)2 22.3 Three substituents are attached to the nitrogen atom; the amine is tertiary In alphabetical order, the substituents present on the aniline nucleus are ethyl, isopropyl, and methyl. Their posi pecified as N-ethyl, 4-isopropyl, and N-methyl (CH3)2CH- N一CH2CH3 N-Ethyl-4-isopropyl-N-methylaniline 604 Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
CHAPTER 22 AMINES SOLUTIONS TO TEXT PROBLEMS 22.1 (b) The amino and phenyl groups are both attached to C-1 of an ethyl group. (c) 22.2 N,N-Dimethylcycloheptylamine may also be named as a dimethyl derivative of cycloheptanamine. 22.3 Three substituents are attached to the nitrogen atom; the amine is tertiary. In alphabetical order, the substituents present on the aniline nucleus are ethyl, isopropyl, and methyl. Their positions are specified as N-ethyl, 4-isopropyl, and N-methyl. (CH3)2CH CH2CH3 CH3 N N-Ethyl-4-isopropyl-N-methylaniline N(CH3)2 N,N-Dimethylcycloheptanamine H2C CHCH2NH2 Allylamine, or 2-propen-1-amine C6H5CHCH3 NH2 1-Phenylethylamine, or 1-phenylethanamine 604 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
AMINES 605 22.4 The electron-donating amino group and the electron-withdrawing nitro group are directly conjugated in p-nitroaniline. The planar geometry of p-nitroaniline suggests that the delocalized resonance form shown is a major contributor to the structure of the compoun NH NH 22.5 The pk, of an amine is related to the equilibrium constant kb by Kb The pk, of quinine is therefore pKb=-log(1×10-6)=6 the values of Kb and pk, for an amine and Ka and pKa of its conjugate acid are given by KXK=1×10 The values of K and pK for the conjugate acid of quinine are therefore 10-141×10 Ka 1×10 pKb=14-6 22.6 The Henderson-Hasselbalch equation described in Section 19.4 can be applied to bases such as amines, as well as carboxylic acids. The ratio [CH, NH,"I/[CH, NH, is given by CHNH3_田 The ionization constant of methylammonium ion is given in the text as 2 X 10. At pH =7 the hydrogen ion concentration is 1 X10-.Therefore CHaNH3]_1×107 CH3NH22×10-=5×103 22.7 Nitrogen is attached directly to the aromatic ring in tetrahydroquinoline, making it an arylamine, and the nitrogen lone pair is delocalized into the T system of the aromatic ring. It is less basic than tetrahydroisoquinoline, in which the nitrogen is insulated from the ring by an sp-hybridized NH (an alkylamine): more basic, (an arylar K25×10-3(pk46) kb1.0 See Learning By Modeling for the calculated charges on nitrogen Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
AMINES 605 22.4 The electron-donating amino group and the electron-withdrawing nitro group are directly conjugated in p-nitroaniline. The planar geometry of p-nitroaniline suggests that the delocalized resonance form shown is a major contributor to the structure of the compound. 22.5 The pKb of an amine is related to the equilibrium constant Kb by pKb log Kb The pKb of quinine is therefore pKb log (1 106 ) 6 the values of Kb and pKb for an amine and Ka and pKa of its conjugate acid are given by Ka Kb 1 1014 and pKa pKb 14 The values of Ka and pKa for the conjugate acid of quinine are therefore Ka 1 108 and pKa 14 pKb 14 6 8 22.6 The Henderson–Hasselbalch equation described in Section 19.4 can be applied to bases such as amines, as well as carboxylic acids. The ratio [CH3NH3 ][CH3NH2] is given by The ionization constant of methylammonium ion is given in the text as 2 1011. At pH 7 the hydrogen ion concentration is 1 107 . Therefore 5 103 22.7 Nitrogen is attached directly to the aromatic ring in tetrahydroquinoline, making it an arylamine, and the nitrogen lone pair is delocalized into the system of the aromatic ring. It is less basic than tetrahydroisoquinoline, in which the nitrogen is insulated from the ring by an sp3 -hybridized carbon. See Learning By Modeling for the calculated charges on nitrogen. N H Tetrahydroquinoline (an arylamine): less basic, Kb 1.0 109 (pKb 9.0) NH Tetrahydroisoquinoline (an alkylamine): more basic, Kb 2.5 105 (pKb 4.6) 1 107 2 1011 [CH3NH3 ] [CH3NH2] [H ] Ka [CH3NH3 ] [CH3NH2] 1 1014 1 106 1014 Kb NH2 NH2 N O O O O N Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
606 AMINES 22.8 (b) An acetyl group attached directly to nitrogen as in acetanilide delocalizes the nitrogen lone pair into the carbonyl group. Amides are weaker bases than amines CCH CCH (c) An acetyl group in a position para to an amine function is conjugated to it and delocalizes the C CH3 22.9 The reaction that leads to allylamine is nucleophilic substitution by ammonia on allyl chloride H, C=CHCH,CI+ 2NH H,C=CHCH,NH, NHC Allyl chloride Allyl chloride is prepared by free-radical chlorination of propene( see text page 371) 40°C H,C=CHCH3 CI H, C=CHCH,CI+ HCl allyl chloride Hydrogen 22.10 (b) Isobutylamine is(CH3),CHCH,NH, It is a primary amine of the type RCH,NH, and can be prepared from a primary alkyl halide by the gabriel synthesis (CH3), CHCH, Br CH, CH(CH3)2 Isobutyl bromide -Potassiophthalimide (CH3),CHCH,NH2+ Isobutylamine Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
606 AMINES 22.8 (b) An acetyl group attached directly to nitrogen as in acetanilide delocalizes the nitrogen lone pair into the carbonyl group. Amides are weaker bases than amines. (c) An acetyl group in a position para to an amine function is conjugated to it and delocalizes the nitrogen lone pair. 22.9 The reaction that leads to allylamine is nucleophilic substitution by ammonia on allyl chloride. Allyl chloride is prepared by free-radical chlorination of propene (see text page 371). 22.10 (b) Isobutylamine is (CH3)2CHCH2NH2. It is a primary amine of the type RCH2NH2 and can be prepared from a primary alkyl halide by the Gabriel synthesis. Isobutyl bromide (CH3)2CHCH2Br N-Potassiophthalimide NK O O N-Isobutylphthalimide NCH2CH(CH3)2 O O Phthalhydrazide Isobutylamine (CH3)2CHCH2NH2 H2NNH2 O O NH NH Cl2 Chlorine H2C CHCH3 Propene H2C CHCH2Cl Allyl chloride HCl Hydrogen chloride 400C 2NH3 Ammonia H2C CHCH2Cl Allyl chloride H2C CHCH2NH2 Allylamine NH4Cl Ammonium chloride O H2N C CH3 O H2N C CH3 O N H CCH3 O N H CCH3 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
AMINES 607 (c) Although tert-butylamine( CH3)3CNH, is a primary amine, it cannot be prepared by the Gabriel method, because it would require an SN2 reaction on a tertiary alkyl halide in the first step. Elimination occurs instead. ( CH3)2 CBr (CH3),C=CH,+ KBr 2-Methylpropene Phthalimide (d) The preparation of 2-phenylethylamine by the Gabriel synthesis has been described in the chemical literature C6HsCH, CH, Br+ NCH,CH, CSH 2-Phenylethyl bromide N-Potassiophthalimide N-(2-Phenylethyl)phthalimide CHCHChNh+ 2-Phenylethylamine Phthalhydrazide (e) The Gabriel synthesis leads to primary amines; N-methylbenzylamine is a secondary amine and cannot be prepared by this method. CH,一N nitrogen; a secondary amine) (f) Aniline cannot be prepared by the Gabriel method. Aryl halides do not undergo nucleophilic substitution under these conditions no reaction Bromobenzene N-Potassiophthalimide Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
(c) Although tert-butylamine (CH3)3CNH2 is a primary amine, it cannot be prepared by the Gabriel method, because it would require an SN2 reaction on a tertiary alkyl halide in the first step. Elimination occurs instead. (d) The preparation of 2-phenylethylamine by the Gabriel synthesis has been described in the chemical literature. (e) The Gabriel synthesis leads to primary amines; N-methylbenzylamine is a secondary amine and cannot be prepared by this method. ( f ) Aniline cannot be prepared by the Gabriel method. Aryl halides do not undergo nucleophilic substitution under these conditions. Br Bromobenzene no reaction N-Potassiophthalimide NK O O CH2 CH3 H N N-Methylbenzylamine (two carbon substituents on nitrogen; a secondary amine) 2-Phenylethyl bromide C6H5CH2CH2Br N-Potassiophthalimide NK O O N-(2-Phenylethyl)phthalimide NCH2CH2C6H5 O O Phthalhydrazide 2-Phenylethylamine C6H5CH2CH2NH2 H2NNH2 O O NH NH tert-Butyl bromide (CH3)2CBr N-Potassiophthalimide NK O O KBr Potassium bromide Phthalimide NH O O 2-Methylpropene (CH3)2C CH2 AMINES 607 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
608 AMINES 22.11 For each part of this problem, keep in mind that aromatic amines are derived by reduction of the corresponding aromatic nitro compound. Each synthesis should be approached from the standpoint of how best to prepare the necessary nitroaromatic compound. Ar—NH, NO, Ar-H (Ar substituted aromatic ring) (b) The para isomer of isopropylaniline may be prepared by a procedure analogous to that used for its ortho isomer in part(a) CH(CH3) CH(CH3) CH(CH3h2 NO Benzene Isopropylbenzene O-lsopropylnitro. p-Isopropylnitro- After separating the ortho, para mixture by distillation, the nitro group of p-isopropyl benzene is reduced to yield the desired CH(CH3)2 CH(CH3)2 NO NH (c) The target compound is the reduction product of 1-isopropyl-2, 4-dinitrobenzene CH(CH3)2 CH(CHi)2 reduce I-Isopropyl-2, 4- 4-1sopropyl-1, 3- dinitrobenzene This reduction is carried out in the same way as reduction of an arene that contains only a single nitro group. In this case hydrogenation over a nickel catalyst gave the desired product in 90% yield The starting dinitro compound is prepared by nitration of isopropylbenzene CH(CHa) CH(CH3)2 HNO.H Isopropylbenze 1-lsopropyl-2,4- Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
22.11 For each part of this problem, keep in mind that aromatic amines are derived by reduction of the corresponding aromatic nitro compound. Each synthesis should be approached from the standpoint of how best to prepare the necessary nitroaromatic compound. (b) The para isomer of isopropylaniline may be prepared by a procedure analogous to that used for its ortho isomer in part (a). After separating the ortho, para mixture by distillation, the nitro group of p-isopropylnitrobenzene is reduced to yield the desired p-isopropylaniline. (c) The target compound is the reduction product of 1-isopropyl-2,4-dinitrobenzene. This reduction is carried out in the same way as reduction of an arene that contains only a single nitro group. In this case hydrogenation over a nickel catalyst gave the desired product in 90% yield. The starting dinitro compound is prepared by nitration of isopropylbenzene. 80C HNO3, H2SO4 Isopropylbenzene CH(CH3)2 1-Isopropyl-2,4- dinitrobenzene (43%) CH(CH3)2 NO2 NO2 reduce CH(CH3)2 NH2 NH2 4-Isopropyl-1,3- benzenediamine CH(CH3)2 NO2 NO2 1-Isopropyl-2,4- dinitrobenzene H2, Ni; or 1. Fe, HCl; 2. HO or 1. Sn, HCl; 2. HO CH(CH3)2 NH2 CH(CH3)2 NO2 (CH3)2CHCl AlCl3 HNO3 H2SO4 CH(CH3)2 NO2 p-Isopropylnitrobenzene CH(CH3)2 NO2 o-Isopropylnitrobenzene CH(CH3)2 Benzene Isopropybenzene Ar NH2 Ar NO2 Ar H (Ar substituted aromatic ring) 608 AMINES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
AMINES 609 (d) The conversion of p-chloronitrobenzene to p-chloroaniline was cited as an example in the text to illustrate reduction of aromatic nitro compounds to arylamines. p-Chloronitrobenzene prepared by nitration of chlorobenzene Cl HNO, H,SO The para isomer accounts for 69% of the product in this reaction (30% is ortho, 1%o meta) Separation of p-chloronitrobenzene and its reduction completes the synthesis Cl Cl 1. Fe. HCl. 2. HO or 1. Sn. HCl: 2. HO N p-Chloronitrobenzene p-Chloroanilin Chlorination of nitrobenzene would not be a suitable route to the required intermediate, because it would produce mainly m-chloronitrobenzene (e) The synthesis of m-aminoacetophenone may be carried out by the scheme shown CH3 Benzene m-Nitroacetophenone The acetyl group is attached to the ring by Friedel-Crafts acylation. It is a meta director, and its nitration gives the proper orientation of substituents. The order of the first two steps cannot reversed, because Friedel-Crafts acylation of nitrobenzene is not possible( Section 12.16) Once prepared, m-nitroacetophenone can be reduced to m-nitroaniline by any of a number of reagents. Indeed, all three reducing combinations described in the text have been employed for this transformation Reducing agent m-Nitroacetophenone m-Aminoacetophenone Sn HCl 82 22.12(b) Dibenzylamine is a secondary amine and can be prepared by reductive amination of benz. HSCH CH-CH2NH, 6H5 Benzaldehyde Benzylamine Dibenzylamine Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
(d ) The conversion of p-chloronitrobenzene to p-chloroaniline was cited as an example in the text to illustrate reduction of aromatic nitro compounds to arylamines. p-Chloronitrobenzene is prepared by nitration of chlorobenzene. The para isomer accounts for 69% of the product in this reaction (30% is ortho, 1% meta). Separation of p-chloronitrobenzene and its reduction completes the synthesis. Chlorination of nitrobenzene would not be a suitable route to the required intermediate, because it would produce mainly m-chloronitrobenzene. (e) The synthesis of m-aminoacetophenone may be carried out by the scheme shown: The acetyl group is attached to the ring by Friedel–Crafts acylation. It is a meta director, and its nitration gives the proper orientation of substituents. The order of the first two steps cannot be reversed, because Friedel–Crafts acylation of nitrobenzene is not possible (Section 12.16). Once prepared, m-nitroacetophenone can be reduced to m-nitroaniline by any of a number of reagents. Indeed, all three reducing combinations described in the text have been employed for this transformation. Yield Reducing agent (%) m-Nitroacetophenone H2, Pt 94 ↓ Fe, HCl 84 m-Aminoacetophenone Sn, HCl 82 22.12 (b) Dibenzylamine is a secondary amine and can be prepared by reductive amination of benzaldehyde with benzylamine. H2, Ni C6H5CH O C6H5CH2NH2 C6H5CH2NHCH2C6H5 Benzaldehyde Benzylamine Dibenzylamine Benzene HNO3 H2SO4 reduce AlCl3 CH3CCl O Acetophenone CCH3 O m-Nitroacetophenone NO2 CCH3 O m-Aminoacetophenone NH2 CCH3 O p-Chloroaniline Cl NH2 p-Chloronitrobenzene Cl NO2 1. Fe, HCl; 2. HO or 1. Sn, HCl; 2. HO H2, catalyst; or Benzene Chlorobenzene Cl o-Chloronitrobenzene NO2 Cl p-Chloronitrobenzene Cl NO2 Cl2 FeCl3 HNO3 H2SO4 AMINES 609 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
610 AMINES (c) N,N-Dimethylbenzylamine is a tertiary amine. Its preparation from benzaldehyde requires dimethylamine, a secondary amine. HAN 6HSCH +(CH))NH C6HS CH,N(CH3)2 Benzaldehyde Dimethylamine N, N-Dimethylbenzylamine (d) The preparation of N-butylpiperidine by reductive amination is described in the text in Section 22.11. An analogous procedure is used to prepare N-benzylpiperidine C6HSCH+ CH-CH Piperidine 22.13 (b) First identify the available B hydrogens. Elimination must involve a proton from the carbon atom adjacent to the one that bears the nitrogen. (CH3)3CCH2--C-CH (CH3)3 It is a proton from one of the methyl groups, rather than one from the more sterically hindered methylene, that is lost on elimination (CH3)3CCH2-C-CH2-H OH -(CH3)3CCH,C=CH +(CH3) (1, 1,3,3-Tetramethylbutyl)- 2, 4, 4-Trimethyl-l-pentene Trimethylamine trimethylammonium only alkene formed hydroxid 70% isolated yield) (c) The base may abstract a proton from either of two B carbons. Deprotonation of the B methyl carbon yields ethylene. CH CH2-CH2TNCHCH, CH,CH3 H,C=CH,+(CH,),NCH,CH, CH,CH CH - Ethyl-N, N-dimethy lbutylammonium hydroxide Ethylene N, N-Dimethylbutylamin Deprotonation of the B methylene carbon yields 1-butene C HH-2 CH2CHCH,CH,部,· CH,CH, N(CH,)2+EC=HcH H N-Ethyl-N, N-dimethy lbutylammonium N, N-Dimethylethylamine 1-Butene Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
(c) N,N-Dimethylbenzylamine is a tertiary amine. Its preparation from benzaldehyde requires dimethylamine, a secondary amine. (d ) The preparation of N-butylpiperidine by reductive amination is described in the text in Section 22.11. An analogous procedure is used to prepare N-benzylpiperidine. 22.13 (b) First identify the available hydrogens. Elimination must involve a proton from the carbon atom adjacent to the one that bears the nitrogen. It is a proton from one of the methyl groups, rather than one from the more sterically hindered methylene, that is lost on elimination. (c) The base may abstract a proton from either of two carbons. Deprotonation of the methyl carbon yields ethylene. Deprotonation of the methylene carbon yields 1-butene. 1-Butene H2C CHCH2CH3 N-Ethyl-N,N-dimethylbutylammonium hydroxide heat (H2O) N,N-Dimethylethylamine N CH3 CH3 H CH2 CHCH2CH3 OH CH3CH2N(CH3) CH3CH2 2 Ethylene H2C CH2 N-Ethyl-N,N-dimethylbutylammonium hydroxide NCH2CH2CH2CH3 CH3 CH3 OH H CH2 CH2 heat (H2O) N,N-Dimethylbutylamine (CH3)2NCH2CH2CH2CH3 OH (1,1,3,3-Tetramethylbutyl)- trimethylammonium hydroxide (CH3)3CCH2 C H CH2 N(CH3)3 CH3 2,4,4-Trimethyl-1-pentene (only alkene formed, 70% isolated yield) (CH3)3CCH2C CH2 CH3 Trimethylamine (CH3)3N (CH3)3CCH2 C CH3 N(CH3)3 CH3 A methylene group Two equivalent methyl groups H2, Ni C6H5CH O Benzaldehyde Piperidine H N N-Benzylpiperidine C6H5CH2 N H2, Ni C6H5CH O (CH3)2NH C6H5CH2N(CH3) 2 Benzaldehyde Dimethylamine N,N-Dimethylbenzylamine 610 AMINES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
AMINES 611 The preferred order of proton removal in Hofmann elimination reactions is B CH3 B CH2>BCH. Ethylene is the major alkene formed, the observed ratio of ethylene to l-butene 22. 14(b) The pattern of substituents in 2, 4-dinitroaniline suggests that they can be introduced by dini- tration. Since nitration of aniline itself is not practical, the amino group must be protected by conversion to its N-acetyl derivative H NHCCH NHCCH CH.CCI CH,, Acetanilide 4-Dinitroacetanilide Hydrolysis of the amide bond in 2, 4-dinitroacetanilide furnishes the desired 2, 4-dinitroaniline NHCCH H,O HO NO NO, 2.4.Nitroacetanilide 2.4.Dinitroaniline c) Retrosynthetically, p-aminoacetanilide may be derived from p-nitroacetanilide CH CNH A)NH:[ CHCNH--NO p-Aminoacetanilide This suggests the CH. COCCH H,N CH, CNH CH CNH p-Nitroacetanilide arate from ortho iso Fe HC1: 2. HO Sn HCl: 2. HO CH CNH p-Aminoacetanilide Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
The preferred order of proton removal in Hofmann elimination reactions is CH3 CH2 CH. Ethylene is the major alkene formed, the observed ratio of ethylene to 1-butene being 98 : 2. 22.14 (b) The pattern of substituents in 2,4-dinitroaniline suggests that they can be introduced by dinitration. Since nitration of aniline itself is not practical, the amino group must be protected by conversion to its N-acetyl derivative. Hydrolysis of the amide bond in 2,4-dinitroacetanilide furnishes the desired 2,4-dinitroaniline. (c) Retrosynthetically, p-aminoacetanilide may be derived from p-nitroacetanilide. This suggests the sequence HNO3 H2SO4 CH3CNH O Acetanilide H2N Aniline 1. Fe, HCl; 2. HO or 1. Sn, HCl; 2. HO or H2, Pt CH3CNH NO2 O p-Nitroacetanilide (separate from ortho isomer) CH3CNH NH2 O p-Aminoacetanilide CH3COCCH3 O O CH3CNH O NH2 p-Aminoacetanilide CH3CNH O NO2 p-Nitroacetanilide 2,4-Dinitroaniline NH2 NO2 NO2 2,4-Dinitroacetanilide NHCCH3 NO2 NO2 O H2O, HO, or 1. H2O, H 2. HO NH2 Aniline Acetanilide NHCCH3 O 2,4-Dinitroacetanilide NHCCH3 NO2 NO2 O HNO3 H2SO4 CH3CCl O or CH3COCCH3 O O AMINES 611 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
612 AMINES 22.15 The principal resonance forms of N-nitrosodimethylamine are H,CH co: HC、 O: All atoms(except hydrogen) have octets of electrons in each of these structures. Other resonance orms are less stable because they do not have a full complement of electrons around each atom. 22.16 Deamination of 1, 1-dimethylpropylamine gives products that result from 1, 1-dimethylpropyl cation. Because 2, 2-dimethylpropylamine gives the same products, it is likely that 1, I-dimethyl propyl cation is formed from 2, 2-dimethylpropylamine by way of its diazonium ion. A carbocation arrant gement is indicated CH CH3 H CH CCHN CH C-CH CH3 CCH, CH3 2. 2-Dimethylpropylamine . 2-Dimethylpropyldiazo 1, l-Dimethylpropyl Once formed, 1, 1-dimethylpropyl cation loses a proton to form an alkene or is captured by water to give an alcohol CH3 H,C=CCH, CH, +(CH),C=CHCH Methyl-l-butene 2-Methyl-2-butene CH, CCHCH 1, l-Dimethylpropyl (CH3),CCH,CH3 2-Methyl-2-butanol 22.17 Phenols may be prepared by diazotization of the corresponding aniline derivative. The probler simplifies itself, therefore, to the preparation of m-bromoaniline. Recognizing that arylamines are ultimately derived from nitroarenes, we derive the retrosynthetic sequence of intermediates OH NH, NO, Bromophenol Bromoaniline m-bromonitrobenzene Nitrobenzene The desired reaction sequence is straightforward, using reactions that were discussed previously in the text NO NO NH L NaNO. Hs Oa O, 2. H.O. heat Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
22.15 The principal resonance forms of N-nitrosodimethylamine are All atoms (except hydrogen) have octets of electrons in each of these structures. Other resonance forms are less stable because they do not have a full complement of electrons around each atom. 22.16 Deamination of 1,1-dimethylpropylamine gives products that result from 1,1-dimethylpropyl cation. Because 2,2-dimethylpropylamine gives the same products, it is likely that 1,1-dimethylpropyl cation is formed from 2,2-dimethylpropylamine by way of its diazonium ion. A carbocation rearrangement is indicated. Once formed, 1,1-dimethylpropyl cation loses a proton to form an alkene or is captured by water to give an alcohol. 22.17 Phenols may be prepared by diazotization of the corresponding aniline derivative. The problem simplifies itself, therefore, to the preparation of m-bromoaniline. Recognizing that arylamines are ultimately derived from nitroarenes, we derive the retrosynthetic sequence of intermediates: The desired reaction sequence is straightforward, using reactions that were discussed previously in the text. Br OH Br NH2 Br NO2 NO2 HNO3 H2SO4 Br2 Fe 1. Fe, HCl 2. NaOH 2. H2O, heat 1. NaNO2, H2SO4 H2O, 0–5C OH Br Br NH2 Br NO2 NO2 m-Bromophenol m-Bromoaniline m-Bromonitrobenzene Nitrobenzene H H2O CH3CCH2CH3 CH3 1,1-Dimethylpropyl cation (CH3)2CCH2CH3 OH 2-Methyl-2-butanol H2C CCH2CH3 CH3 2-Methyl-1-butene (CH3)2C CHCH3 2-Methyl-2-butene HONO N2 CH3C CH2 N CH3 CH3 N 2,2-Dimethylpropyldiazonium ion CH3CCH2CH3 CH3 1,1-Dimethylpropyl cation CH3CCH2NH2 CH3 CH3 2,2-Dimethylpropylamine N N O H3C H3C N N O H3C H3C 612 AMINES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
AMINES 613 22.18 The key to this problem is to recognize that the iodine substituent in m-bromoiodobenzene is derived from an arylamine by diazotization The preparation of m-bromoaniline from benzene has been described in Problem 22 17. All that remains is to write the equation for its conversion to m-bromoiodobenzene I NaNO, HCl. H, B m-Bromoaniline m-Bromoiodobenzene 22 19 The final step in the preparation of ethyl m-fluorophenyl ketone is shown in the text example im- mediately preceding this problem, therefore all that is necessary is to describe the preparation of 7-aminophenyl ethyl ketone CCHCH CCHCH CCHCH Ethyl m-fluorophenyl aminophenyl ethyl Ethyl m-nitrophenyl Recalling that arylamines are normally prepared by reduction of nitroarenes, we see that ethyl m-nitrophenyl ketone is a pivotal synthetic intermediate. It is prepared by nitration of ethyl phenyl ketone, which is analogous to nitration of acetophenone, shown in Section 12.16 The preparation of ethyl phenyl ketone by Friedel-Crafts acylation of benzene is shown in Sec tion 12.7 CCH.CH CCH,CH3 Ethyl m-nitrophenyl Ethyl phenyl ketone Reversing the order of introduction of the nitro and acyl groups is incorrect. It is possible to nitrate ethyl phenyl ketone but not possible to carry out a Friedel-Crafts acylation on nitrobenzene, owing to the strong deactivating influence of the nitro group 22.20 Direct nitration of the prescribed starting material cumene (isopropylbenzene) is not suitable, because isopropyl is an ortho, para-directing substituent and will give the target molecule Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
22.18 The key to this problem is to recognize that the iodine substituent in m-bromoiodobenzene is derived from an arylamine by diazotization. The preparation of m-bromoaniline from benzene has been described in Problem 22.17. All that remains is to write the equation for its conversion to m-bromoiodobenzene. 22.19 The final step in the preparation of ethyl m-fluorophenyl ketone is shown in the text example immediately preceding this problem, therefore all that is necessary is to describe the preparation of m-aminophenyl ethyl ketone. Recalling that arylamines are normally prepared by reduction of nitroarenes, we see that ethyl m-nitrophenyl ketone is a pivotal synthetic intermediate. It is prepared by nitration of ethyl phenyl ketone, which is analogous to nitration of acetophenone, shown in Section 12.16. The preparation of ethyl phenyl ketone by Friedel–Crafts acylation of benzene is shown in Section 12.7. Reversing the order of introduction of the nitro and acyl groups is incorrect. It is possible to nitrate ethyl phenyl ketone but not possible to carry out a Friedel–Crafts acylation on nitrobenzene, owing to the strong deactivating influence of the nitro group. 22.20 Direct nitration of the prescribed starting material cumene (isopropylbenzene) is not suitable, because isopropyl is an ortho, para-directing substituent and will give the target molecule NO2 CCH2CH3 O Ethyl m-nitrophenyl ketone CCH2CH3 O Ethyl phenyl ketone F CCH2CH3 NH2 NO2 Ethyl m-nitrophenyl ketone Ethyl m-fluorophenyl ketone m-Aminophenyl ethyl ketone CCH2CH3 O CCH2CH3 O O I Br m-Bromoiodobenzene Br NH2 m-Bromoaniline 1. NaNO2, HCl, H2O 2. KI I Br Br NH2 m-Bromoiodobenzene m-Bromoaniline AMINES 613 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website