Baixe Carey - Organic Chemistry - sgchapt22 e outras Manuais, Projetos, Pesquisas em PDF para Química, somente na Docsity! 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. CoHsÇHCH, NH, 1-Phenylethylamine, or I-phenylethanamine (c) H,C=CHCH,NH, Allylamine, or 2-propen-l-amine 22.2 N,N-Dimethylcycloheptylamine may also be named as a dimethyl derivative of cycloheptanamine. (mem N.N-Dimethyleycloheptanamine 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. Io npc Sox-cnen, N-Ethyl-4-isopropyl-N-methylaniline 604 Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website Back| Forward 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. CNH: *NH, 22.5 The pK, of an amine is related to the equilibrium constant K, by Pk, = log k, The pk, of quinine is therefore 6 pk, = —log (1 X 10) the values of K, and pK, for an amine and K, and pK, of its conjugate acid are given by KXK=1x1074 and pk, + pk, = 14 The values of K, and pK, for the conjugate acid of quinine are therefore Mo 1xI104 p= 1008 1x 1004 12 Ro 1XI0 =1x 108 and pk =14-pkK,=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 [CH;NH, *]/[CH;NH,] is given by [CHNH;] [H] ICHNH,]) K, The ionization constant of methylammonium ion is given in the textas 2 X 107! At pH = 7 the hydrogen ion concentration is | X 1077. Therefore CH;NH,* 1x 107 [CHNHy') 1X107 Os [CHNH,)] 2X 10 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 7 system of the aromatic ring. It is less basic than tetrahydroisoquinoline, in which the nitrogen is insulated from the ring by an sp?-hybridized carbon. N' H Tetrahydroisoquinoline Tetrahydroquinoline (an alkylamine); more basic, (an arylamine)): less basic, K, 2,5 X 10º (pk, 4.6) K, 10X 10º (PK, 9.0) See Learning By Modeling for the calculated charges on nitrogen. MainMenu) TOC] StudyGuideTOC| 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, > Ar—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(CH;) CH(CH;), CH(CHy, NO, Ó (CHyCHCI HNO, C Y 2 C ] E, —— + AIC, HSO, NO, Benzene Isopropybenzene o-Isopropylnitro- p-Isopropylnitro- benzene benzene After separating the ortho, para mixture by distillation, the nitro group of p-isopropyl- nitrobenzene is reduced to yield the desired p-isopropylaniline. CH(CH;), CH(CH;), NO, NH, (c) The target compound is the reduction product of 1-isopropyl-2,4-dinitrobenzene. CH(CH;) CH(CH;), NO, NH, reduce —teduce NO, NH, 1Isopropy1-2,4- 4Isopropyl-1,3- dinitrobenzene benzenediamine 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(CH;), CH(CH;), O, HNO,, H;SO, ? soc NO, Isopropylbenzene L-Isopropy1-2,4- dinitrobenzene (43%) Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | 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 is prepared by nitration of chlorobenzene. cl cl cl NO, a, HNO, - —+— — — + Fecl, H;SO, NO, Benzene Chlorobenzene o-Chloronitrobenzene p-Chloronitrobenzene 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. ci ci 1. Fe, HCI; 2. HO” or Hh, cataly 1. Sn, HCI;2. HO NO, NH, p-Chloronitrobenzene p-Chloroaniline 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: o o 9 | I I 7 CcH, CcH, CcH, cHCci HNO, reduce ——— —o— E, AICI, HSO, NO, NH, Benzene Acetophenone m-Nitroacetophenone m-Aminoacetophenone The acetyl group is attached to the ring by Friedel-Crafts acylation. Itis 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 (J%) m-Nitroacetophenone H, Pt 94 4 Fe, HCI 84 m-Aminoacetophenone Sn, HCI 82 22.12 (b) Dibenzylamine is a secondary amine and can be prepared by reductive amination of benz- aldehyde with benzylamine. f Ho Ni CHCH + CHCHNH —E2D> CH.CHNHCH,CH, Benzaldehyde Benzylamine Dibenzylamine Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website 610 AMINES (c) | N,N-Dimethylbenzylamine is a tertiary amine. Its preparation from benzaldehyde requires dimethylamine, a secondary amine. | HM CHCH + (CHANH —DS CHCHNCH), 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. o I Ho, Ni CHCH + > CHCH—N N' H Benzaldehyde Piperidine N-Benzylpiperidine 22.13 (b) First identify the available 8 hydrogens. Elimination must involve a proton from the carbon atom adjacent to the one that bears the nitrogen. ; /T Two equivalent methyl groups CH, Bs. LB (CH CH —E—CHi; + A methylene group N(CHy); Itis a proton from one of the methyl groups, rather than one from the more sterically hindered methylene, that is lost on elimination. TE To “O (CH), CCH,—CLCHH (CH)CCH,C=CH, + (CHAN: NCH)s (1,1,3,3-TetramethyIbutyl)- 2,4,4-Trimethyl-I-pentene Trimethylamine trimethylammonium (only alkene formed, hydroxide 70% isolated yield) (c) The base may abstract a proton from either of two 8 carbons. Deprotonation of the 8 methyl carbon yields ethylene. cH, PULGA ENCHLCACH,CH, e H,C=CH, + (CH), NCH,CH,CH,CH, du, N-Ethyl-N,N-dimethylbutylammonium hydroxide Ethylene N.N-Dimethylbutylamine Deprotonation of the 8 methylene carbon yields 1-butene. cH, A hear N — CHCHo= CHoqqHCHCH, o CH,CH;N(CH;), + HC=CHCH;CH, CH; Ha. Ro N-Ethyl-N,N-dimethylbutylammonium N,N-Dimethylethylamine 1-Butene hydroxide Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | 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. I NH, Br Br m-Bromoiodobenzene m-Bromoaniline 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. NH, I 1. NaNO,, HCI, HO 2.KI Br Br 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 m-aminophenyl ethyl ketone. F NH, NO, qCHCH; CHE, GeHyCH, o Ethyl m-fluorophenyl m-Aminophenyl ethyl Ethyl m-nitropheny1 ketone ketone 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 Sec- tion 12.7. NO, GCHCH, o GeHca, o o o Ethyl m-nitrophenyl Ethyl phenyl ketone 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 (isopropyIbenzene) is not suitable, because isopropyl is an ortho, para-directing substituent and will give the target molecule Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website 614 AMINES m-nitrocumene as only a minor component of the nitration product. However, the conversion of 4-isopropyl-2-nitroaniline to m-isopropylInitrobenzene, which was used to illustrate reductive deamination of arylamines in the text, establishes the last step in the synthesis. CH(CH;), CH(CH;), CH(CH;), NO, NO, o NH, m-Nitrocumene 4Isopropyl-2- Cumene nitroaniline Our task simplifies itself to the preparation of 4-isopropy1-2-nitroaniline from cumene. The follow- ing procedure is a straightforward extension of the reactions and principles developed in this chapter. CH(CH;) CH(CH;) CH(CH;) o CH(CH;), HINO; 1. Fe, HCI catar HO, 2.H07 NO, NH, NHCCH, o Cumene p-Nitrocumene pisopropylaniline p-Isopropylacetanilide CH(CH;) CH(CH;) CH(CH;), HNO, Ho, NO, NO, NHCCH, NHCCH, NH, o O p-Isopropylacetanilide 4-Isopropyl-2-nitroacetanilide 4Isopropyl-2-nitroaniline Reductive deamination of 4-isopropyl-2-nitroaniline by diazotization in the presence of ethanol or hypophosphorous acid yields m-nitrocumene and completes the synthesis. 22.21 Amines may be primary, secondary, or tertiary. The C,H, N primary amines, compounds of the type C,H,NH,, and their systematic names are CH,CH,CH,CH,NH, (CH;),CHCH,NH, Butylamine Isobutylamine (I-butanamine) (2-methyl-I-propanamine) CHCHCHCH, (CH).CNH, NH, sec-Butylamine tert-Butylamine (Q-butanamine) (2-methy1-2-propanamine) Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website AMINES 615 Secondary amines have the general formula R4NH. Those of molecular formula C,H,,N are (CH.CH,), NH CHNCHCHCH, CHNCH(CHy): H H Diethylamine N-Methylpropylamine N-Methylisopropylamine (N-ethylethanamine) (N-methyl-I-propanamine) (N-methyl-2-propanamine) There is only one tertiary amine (R;N) of molecular formula C,H,,N: (CH,)NCH,CH, N.N-Dimethylethylamine (N.N-dimethylethanamine) 22.22 (a) The name 2-ethyl-1-butanamine designates a four-carbon chain terminating in an amino group and bearing an ethyl group at C-2. CHLCHCHCHNH, CHCH, 2-Ethyl-I-butanamine (b) The prefix N- in N-ethyl-1-butanamine identifies the ethyl group as a substituent on nitrogen in a secondary amine. CH,CH,CH,CH;NCH,CH, 4 N-Ethyl-I-butanamine (c) Dibenzylamine is a secondary amine. It bears two benzyl groups on nitrogen. C;H,CH,NCH,C,H, 4 Dibenzylamine (d) Tribenzylamine is a tertiary amine. (CSHsCH,);N Tribenzylamine (e) -Tetraethylammonium hydroxide contains a quaternary ammonium ion. (CH;CH,),N HO” Tetraethylammonium hydroxide (f) This compound is a secondary amine; it bears an allyl substituent on the nitrogen of cyclo- hexylamine. H O N N CH,CH=CH, N-Aliyleyclohexylamine Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website 618 Back| AMINES 22.25 (a) Basicity decreases in proceeding across a row in the periodic table. The increased nuclear charge as one progresses from carbon to nitrogen to oxygen to fluorine causes the electrons to be bound more strongly to the atom and thus less readily shared. HC: > Strongest Weakest base base K, 104 105% 10-16 35 x 104 of conjugate acid (b) The strongest base in this group is amide ion, H,N', and the weakest base is water, H,O. Ammonia is a weaker base than hydroxide ion; the equilibrium lies to the left. + NH, + HO => NH, + 0H Weaker — Weaker Stronger Stronger base acid acid base The correct order is HAN: > HOS > :NH, > H,Ô: Strongest Weakest base base (c) These anions can be ranked according to their basicity by considering the respective acidities of their conjugate acids. Base Conjugate acid K, of conjugate acid H,N” HN 1076 H,O 10716 HC=N: 72x 107! o A 4 ON HON, 2.5 x 10! (om (om The order of basicities is the opposite of the order of acidities of their conjugate acids. H,Nº > HO > :C=N: > NO; Strongest Weakest base base (d) | Acarbonyl group attached to nitrogen stabilizes its negative charge. The strongest base is the anion that has no carbonyl groups on nitrogen; the weakest base is phthalimide anion, which has two carbonyl groups. o o o Os Strongest base Weakest base Forward] MainMenu| TOC] StudyGuideTOC| StudentOLC| | MHHE Website AMINES 619 22.26 (a) An alkyl substituent on nitrogen is electron-releasing and base-strengthening: thus methyl- amine is a stronger base than ammonia. An aryl substituent is electron-withdrawing and base- weakening, and so aniline is a weaker base than ammonia. CHNH, > NH, > CGHNH, Methylamine, Ammonia: Aniline, strongest base; weakest base: K 44X KASXIOS K38X 100 PK,34 PK,47 pk,94 (b) An acetyl group is an electron-withdrawing and base-weakening substituent, especially when bonded directly to nitrogen. Amides are weaker bases than amines, and thus acetanilide is a weaker base than aniline. Alkyl groups are electron-releasing; N-methylaniline is a slightly stronger base than aniline. i C;H,NHCH, > C/HNH, > C,HNHCCH, N-methylaniline, Aniline: Acetanilide, strongest base: weakest base: K,8 x 10-10 K, 3.8 X 10710 K, 1X 10715 PK,91 Pk,94 PK, 15.0 (c) Chlorine substituents are slightly electron-withdrawing, and methyl groups are slightly electron-releasing. 2,4-Dimethylaniline is therefore a stronger base than 2,4-dichloroaniline. Nitro groups are strongly electron-withdrawing, their base-weakening effect being especially pronounced when a nitro group is ortho or para to an amino group because the two groups are then directly conjugated. NH, NH, NH, CH, ci NO, > > cH, a NO, 2.4-Dimethylaniline, 2,4-Dichloroaniline: 2,4-Dinitroaniline, strongest base: weakest base: K,8 x 10710 Ki x 102 K3 x 10º Pk,9 pk, 120 Pk, 185 (d) Nitro groups are more electron-withdrawing than chlorine, and the base-weakening effect of a nitro substituent is greater when it is ortho or para to an amino group than when it is meta toit. NH, > NH, NH, NO, > > ci NO, a a cl 34-Dichloroaniline, 4-Chloro-3-nitroaniline: 4-Chloro-2-nitroaniline, weakest base: K,8x 108 K,1x 10-18 Pk 121 PK, 15.0 MainMenu) TOC] StudyGuideTOC| Student OLC| | MHHE Website 620 AMINES (e) | According to the principle applied in part (a) (alkyl groups increase basicity, aryl groups decrease it), the order of decreasing basicity is as shown: (CH)NH > CHNHCH, > (CH),NH Dimethylamine, N-Methylaniline: Diphenylamine, strongest base; weakest base: K,S. 1x 10 K,8x 10710 K,6X 104 PK,33 pK,9.1 pk, 13.2 22.27 Nitrogen Qis the most basic and the most nucleophilic of the three nitrogen atoms of physostigmine and is the one that reacts with methyl iodide. CH CH. cH, TS o cH, p's ONA NA HC. ne N o + CHI r OCNHCH, OCNHCH, I I Physostigmine Methyl “Physostigmine methiodide” iodide The nitrogen that reacts is the one that is a tertiary alkylamine. Of the other two nitrogens, (O) is attached to an aromatic ring and is much less basic and less nucleophilic. The third nitrogen, O, is an amide nitrogen; amides are less nucleophilic than amines. 22.28 (a) Looking at the problem retrosynthetically, it can be seen that a variety of procedures are avail- able for preparing ethylamine from ethanol. The methods by which a primary amine may be prepared include o > NCH,CH, o Gabriel synthesis > CHCHN, Reduction of an azide CH,CHNH, o | > caCH Reductive amination E» i CH,CNH, Reduction of an amide Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website Back| Forward AMINES 623 (b) The carbon chain of tert-butyl chloride cannot be extended by a nucleophilic substitution reaction; the S,2 reaction that would be required on the tertiary halide would not work. The sequence employed in part (a) is therefore not effective in this case. The best route is car- boxylation of the Grignard reagent and subsequent conversion of the corresponding amide to the desired primary amine product. o | (CH,),CCH,NH, > (CH.CCNH, > (CH)CCO,H > (CHI The reaction sequence to be used is 1. Mg, diethyl ether (CH)CCl co, (CH;);CCO,H tert-Butyl HO 2.2-Dimethylpropanoie chloride acid Once the carboxylic acid has been obtained, it is converted to the desired amine by reduction of the corresponding amide. 1. SOCI, | LLAIH, (CHCCOH a (CH)CCNH, São > (CH)CCH;NH, 2,2-Dimethylpropanoie 2,2-Dimethylpropanamide 2,2-Dimethyl-1- acid propanamine (c) - Oxidation of cyclohexanol to cyclohexanone gives a compound suitable for reductive amination. CH;NH,, KCro; Ho Ni ls —E OH F5o HO = NHCH, CH;NH,, NaBH,CN Cyclohexanol Cyclohexanone N-Methyleyclohexylamine (d) The desired product is the reduction product of the cyanohydrin of acetone. oH oH I iAH, | CHCCH, São CH,CCH, CN CH,NH, Acetone 1-Amino-2-methyl- cyanohydrin 2-propanol The cyanohydrin is made from acetone in the usual way. Acetone is available by oxidation of isopropyl alcohol. oH KCr,0,, H;SO, | KCN | CH.ÇHCH, > CHCCH, —— CHÇCH, 0H CN Isopropyl Acetone Acetone alcohol cyanohydrin MainMenu) TOC] StudyGuideTOC| Student OLC| | MHHE Website 624 Back| AMINES (e) The target amino alcohol is the product of nucleophilic ring opening of 1,2-epoxypropane by ammonia. Ammonia attacks the less hindered carbon of the epoxide function. NH, CHCH-CH, . CH.CHCH,NH, o OH 1,2-Epoxypropane 1-Amino-2-propanol The necessary epoxide is formed by epoxidation of propene. o HSO, cufoou CHHCH, ar CHCH=CH, > CHCH-CH, OH Isopropyl Propene 1,2-Epoxypropane alcohol (f) The reaction sequence is the same as in part (e) except that dimethylamine is used as the nucleo- phile instead of ammonia. CHCHLH, + (CH)NH CH,CHCH,N(CH,), o OH 1,2-Epoxypropane Dimethylamine 1«(N.N-Dimethylamino)- [prepared as in part (e)] 2-propanol (g) The key to performing this synthesis is recognition of the starting material as an acetal of acetophenone. Acetals may be hydrolyzed to carbonyl compounds. a Hot I d O - C,H,CCH, + HOCH,CH,0H CH, CH, 2-Methyl-2-phenyl- Acetophenone 1,2-Ethanediol 1,3-dioxolane Once acetophenone has been obtained, it may be converted to the required product by reduc- tive amination. o I NaBH,CN C;H,CCH, + TUR N' N' H | C,H,CHCH, Acetophenone Piperidine N-(1-Phenylethy))- piperidine 22.30 (a) The reaction of alkyl halides with N-potassiophthalimide (the first step in the Gabriel synthe- sis of amines) is a nucleophilic substitution reaction. AlkyI bromides are more reactive than alkyl fluorides; that is, bromide is a better leaving group than fluoride. o o NK + FCHCHB —— NCH,CH,F O N-Potassiophthalimide 1-Bromo-2- 2-Phthalimidoethy1 fluoride fluoroethane Forward] MainMenu| TOC] StudyGuideTOC| StudentOLC| | MHHE Website Back| 625 AMINES (b) In this example one bromine is attached to a primary and the other to a secondary carbon. Phthalimide anion is a good nucleophile and reacts with alkyl halides by the S,2 mechanism. It attacks the less hindered primary carbon. o —s NOR, CH, CH: CHCH, Br 0 A ºSo o MC NT crowded E Fes egg Hs Less crowded BE 2 | Br N-4-Bromopentylphthalimide 1 4-Dibromopentane (only product, 67% yield) (c) Both bromines are bonded to primary carbons, but branching at the adjacent carbon hinders nucleophilic attack at one of them. 9 cu, — NCH CH qCHBr cH b 5 / HC TCE mor te OB, K 'Br crowded ss : crowded Br cH, CH, N-4-Bromo-3,3-dimethy Iphthalimide 1,4-Dibromo-2.2-dimethylbutane (only product, 53% yield) 22.31 (a) Amines are basic and are protonated by hydrogen halides. C;HSCHNH, + HBr C;H.CHNH, Br” Benzylamine Benzylammonium bromide (b) -Equimolar amounts of benzylamine and sulfuric acid yield benzylammonium hydrogen sulfate as the product. C,H.CH,NH, “OS0,0H C,H,CH,NH, + HOSO,0H Benzylamine Sulfuric acid Benzylammonium hydrogen sulfate (c) | Acetic acid transfers a proton to benzy lamine. | C,H,CH,NH, “OCCH, I C,H,CH,NH, + CH.COH Benzylamine Acetic acid Benzylammonium acetate Forward] MainMenu| TOC] StudyGuideTOC| StudentOLC| | MHHE Website 628 Back| AMINES The replacement reactions that can be achieved by using diazonium salts are illustrated in parts (A) through (n). In all cases molecular nitrogen is lost from the ring carbon to which it was attached and is replaced by another substituent. H'.H,0 1) E CHOH Phenol à Heels cHa Chlorobenzene q HS cAHBr Bromobenzene CuCN dO [> cen Benzenediazonium Benzonitrile hydrogen sulfate HO, DI CH Benzene m E cu Todobenzene 1.HBF, (n) 2. heat CHF Fluorobenzene (o) The nitrogens of an aryl diazonium salt are retained on reaction with the electron-rich ring of a phenol. Azo coupling occurs. can Son Benzenediazonium Phenol p-(Azophenylphenol hydrogen sulfate CHN=N: + CH,0H HsO, (p) | Azo coupling occurs when aryl diazonium salts react with N,N-dialkylarylamines. N: + CHNCH), —— can Soneto Benzenediazonium N.N-Dimethylaniline p«Azophenyl)-N.N-dimethylaniline hydrogen sulfate 22.33 (a) Amides are reduced to amines by lithium aluminum hydride. L LiAIH,, diethyl ether | CHNHCCH, ão C,H;NHCH,CH, Acetanilide N-Ethylaniline Forward] MainMenu| TOC] StudyGuideTOC| StudentOLC| | MHHE Website Back| AMINES 629 (b) (c) (d) (e) 1 Acetanilide is a reactive substrate toward electrophilic aromatic substitution. An acetamido group is ortho, para-directing. | | NHCCH, NHCCH, Í NO, HNO, Ê CHNHCCH, —so— + NO, Acetanilide o-Nitroacetanilide p-Nitroacetanilide Sulfonation of the ring occurs. NHCCH, o | so, . C;H;NHCCH, Hs + ortho isomer so,H Acetanilide p-Acetamidobenzenesulfonic acid Bromination of the ring takes place. NHCCH, I Br; . CHNHCCH, a + ortho isomer S acetic acid Br Acetanilide p-Bromoacetanilide Acetanilide undergoes Friedel-Crafts alkylation readily. o I NHCCH, | ACI, . C;ANHCCH, + (CH);CCI ———— + ortho isomer C(CH;); Acetanilide ter-Butyl piert-Butyl- chloride acetanilide Friedel-Crafts acylation also is easily carried out. f NHCCH, o | | AIC, . C;H,NHCCH,; + CH;CCI ———— + ortho isomer Ms Acetanilide Acetylchloride p-Acetamidoacetophenone Forward] MainMenu| TOC] StudyGuideTOC| StudentOLC| | MHHE Website 630 AMINES (g) Acetanilide is an amide and can be hydrolyzed when heated with aqueous acid. Under acidic conditions the aniline that is formed exists in its protonated form as the anilinium cation. | | CHNHCCH, + HO + HCL — CHNH, CI + HOCCH, Acetanilide Water Hydrogen Anilinium Acetic acid chloride chloride (h) -Amides are hydrolyzed in base. I Ho | CH,NHCCH, + NaOH - C,H,NH, + Na! “OCCH, Acetanilide Sodium Aniline Sodium acetate hydroxide 22.34 (a) The reaction illustrates the preparation of a secondary amine by reductive amination. (o + un AE (Soa Cyclohexanone Cyelohexylamine Dicyclohexylamine (70%) (b) - Amides are reduced to amines by lithium aluminum hydride. LAH, = 2.H,0,H07 T A NCH,CH, NCH,CH, CEthy-6- 6-Ethyl-6- azabicyclo[3.2.octan-7-one azabieyclo[3.2. Ioctane (c) | Treatment of alcohols with p-toluenesulfonyl chloride converts them to p-toluenesulfonate esters. í idi C,H,CH,CH,CH,0H + ne S-soci 2, cscncncnor( Sem, 0 3-Phenyl-I-propanol p-ToluenesulfonyI chloride 3-Phenylpropyl p-toluenesulfonate p-Toluenesulfonate is an excellent leaving group in nucleophilic substitution reactions. Dimethylamine is the nucleophile. csscncucnoso—/ Sen, + (CH) NH C;HSCH,CH,CH,N(CH;), 3-Phenylpropyl p-toluenesulfonate Dimethyl- N,N-Dimethyl-3-phenyl- amine L-propanamine (86%) (d) —Amines are sufficiently nucleophilic to react with epoxides. Attack occurs at the less substi- tuted carbon of the epoxide. cH,O cH,O oH £LQ ciQu. + myctcmy, ——s CHCH,NHCH(CH.,), OCH, OCH, 2-(2,5-Dimethoxyphenyl)oxirane Isopropylamine 1-(2,5-Dimethoxyphenyl)-2- (isopropylamino)ethanol (67%) Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website Back| AMINES 633 (h) Acetanilide is a reactive substrate toward electrophilic aromatic substitution. On reaction with chloroacetyl chloride, it undergoes Friedel-Crafts acylation, primarily at its para position. o o o o | | aii, I I CH.CNH + CICHCCI ——— CH,CNH CCH,CI Acetanilide Chloroacetyl p-Acetamidophenacyl chloride chloride (19-83%) Acylation, rather than alkylation, occurs. Acyl chlorides are more reactive than alkyl chlorides toward electrophilic aromatic substitution reactions as a result of the more stable intermediate (acylium ion) formed. (i) | Reduction with iron in hydrochloric acid is one of the most common methods for converting nitroarenes to arylamines. Fe Mel O Fe, HCI Br NO, qo NH, 4-Bromo-4-nitrobiphenyl 4-Amino-4'-bromobiphenyl (94%) j) | Primary arylamines are converted to aryl diazonium salts on treatment with sodium nitrite in J y ary) X aqueous acid. When the aqueous acidic solution containing the diazonium salt is heated, a phenol is formed. NaNO,. H,SO, HO, hear | O, heat Br NH ND» oH 4-Amino-4'-bromobiphenyl 4-Bromo-4-hydroxybiphenyl (85%) (k) This problem illustrates the conversion of an arylamine to an aryl chloride by the Sandmeyer reaction. + NH, N ci ON. NO, ON. ON. o, NaNO;, H,SO, 2.6-Dinitroaniline 2-Chloro-1,3- dinitrobenzene (71-74%) (1) | Diazotization of primary arylamines followed by treatment with copper(I) bromide converts them to aryl bromides. NH, Na: Br ANO te, HBr — Cumr O mm O a Br B r m-Bromoaniline m-Dibromobenzene (80-879%) (m) Nitriles are formed when aryl diazonium salts react with copper(T) cyanide. NO, NO, NO, NaNO,, HCI CuCN CX Ho CX + No) CX NH, N=N: CN o-Nitroaniline o-Nitrobenzonitrile (87%) Forward] MainMenu| TOC] StudyGuideTOC| StudentOLC| | MHHE Website 634 Back| AMINES (n) An aryl diazonium salt is converted to an aryl iodide on reaction with potassium iodide. 2.6-Diiodo- 12,3-Triiodo- 4-nitroaniline 5-nitrobenzene (94-95%) (o) — Aryl diazonium fluoroborates are converted to aryl fluorides when heated. Both diazonium salt functions in the starting material undergo this reaction. 4,4'-Bis(diazonio)biphenyl fluoroborate 4,4'-Difluorobiphenyl (82%) (p) Hypophosphorous acid (H;PO,) reduces aryl diazonium salts to arenes. NH, ON. NO, ON. NO, NO, NO, 2,4,6-Trinitroaniline 1,3,5-Trinitrobenzene (60-65%) (9) Ethanol, like hypophosphorous acid, is an effective reagent for the reduction of aryl diazo- nium salts. coH co,H co,H a] 2H a] NH, N=N: * NaNO,, HCI CH,CH,OH HO I I I 2-Amino-5- m-lodobenzoic acid iodobenzoic acid (86-93%) (1) | Diazotization of aniline followed by addition of a phenol yields a bright-red diazo-substituted phenol. The diazonium ion acts as an electrophile toward the activated aromatic ring of the phenol. HC. CH, 0H HC CH, + - He C;HSNH, C;HN=N: HSO, HC N=NC,H, Aniline Benzenediazonium 2,3,6-Trimethyl-4- hydrogen sulfate (phenylazo)phenol (98%) MainMenu) TOC] StudyGuideTOC| Student OLC| | MHHE Website AMINES 635 (s) - Nitrosation of N,N-dialkylarylamines takes place on the ring at the position para to the dialkylamino group. o NaNO,, HCI, H,0 y (CHy)> 0 (CH) N CH, CH; N, N-Dimethyl-m-toluidine 3-Methyl-4-nitroso-N, N- dimethylaniline (83%) 22.36 (a) 4-Methylpiperidine can participate in intermolecular hydrogen bonding in the liquid phase. These hydrogen bonds must be broken in order for individual 4-methylpiperidine molecules to escape into the gas phase. N-Methylpiperidine lacks a proton bonded to nitrogen and so cannot engage in intermolecular hydrogen bonding. Less energy is required to transfer a mol- ecule of N-methylpiperidine to the gaseous state, and therefore it has a lower boiling point than 4-methyIpiperidine. cH, N-Methypiperidine; no hydrogen bonding possible to other N-methyIpiperidine molecules (b) The two products are diastereomeric quaternary ammonium chlorides that differ in the con- figuration at the nitrogen atom. C(CHy; Cos CH:C1 C(CH. C(CH. CiscHiici 1/57 (CH) + cuca” (CH); I | da, CHCH cr cH ar 4tert-Butyl-N- methylpiperidine (c) -Tetramethylammonium hydroxide cannot undergo Hofmann elimination. The only reaction that can take place is nucleophilic substitution. Hc EH: Hc EH: NOcH, —— + CHÓH / / “ HC HC Tetramethylammonium Trimethylamine Methanol hydroxide Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website 638 AMINES configuration at the alcohol carbon. Conversion of the alcohol to its p-toluenesulfonate ester en- sures that the leaving group is introduced with exactly the same stereochemistry as the alcohol. Ó + nc S-soa te, H cHO O CH, oso—/ Sen, cis-2-Phenoxyeyclo- p-Toluenesulfonyl cis-2-Phenoxyeyclopentyl pentanol chloride p-toluenesulfonate Once the leaving group has been introduced with the proper stereochemistry, it can be dis- placed by a nitrogen nucleophile suitable for subsequent conversion to an amine. NaN; LAH, 2H0 D cHO os Sen, CH, N, CHO NH, cis-2-Phenoxyeyclopentyl trans-2-Phenoxyeyclo- trans-2-Phenoxyeyclo- p+oluenesulfonate pentyI azide (90%) pentylamine (As actually reported, the azide was reduced by hydrogenation over a palladium catalyst, and the amine was isolated as its hydrochloride salt in 66% yield.) (d) Recognition that the primary amine is derivable from the corresponding nitrile by reduction, CAM CHANCH:CHCH.CH;NH, > Cos CHANCH:CHCH C=N cH, cH, and that the necessary tertiary amine function can be introduced by a nucleophilic substitution reaction between the two given starting materials suggests the following synthesis. 1.LiAIH, CoHsCHaNH + BrCH,CH,CHCN —— COM CHANCH:CH:CHCN Tom Cos CHANCH;CHCH,CH;NH, cH, cH, cH, N-Methylbenzylamine 4-Bromobutanenitrile N-Benzyl-N-methyl-1,4-butanediamine Alkylation of N-methylbenzylamine with 4-bromobutanenitrile has been achieved in 92% yield in the presence of potassium carbonate as a weak base to neutralize the hydrogen bro- mide produced. The nitrile may be reduced with lithium aluminum hydride, as shown in the equation, or by catalytic hydrogenation. Catalytic hydrogenation over platinum gave the de- sired diamine, isolated as its hydrochloride salt, in 90% yield. (e) The overall transformation may be viewed retrosynthetically as follows: ArCH,N(CH;), > ArCH,Br > ArCH, E O The sequence that presents itself begins with benzylic bromination with N-bromosuccinimide. N-bromosuccinimide, ne Scan, quim no] Sense heat p-Cyanotoluene p-Cyanobenzyl bromide Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website AMINES 639 The reaction shown in the equation has been reported in the chemical literature and gave the benzylic bromide in 60% yield. Treatment of this bromide with dimethylamine gives the desired product. (The isolated yield was 83% by this method.) ne! Scumr + (CH)NH ne Sensei, p-Cyanobenzyl bromide Dimethylamine p-Cyano-N, N-dimethylbenzylamine 22.39 (a) This problem illustrates the application of the Sandmeyer reaction to the preparation of aryl cyanides. Diazotization of p-nitroaniline followed by treatment with copper(T) cyanide con- verts it to p-nitrobenzonitrile. NH, CN 1. NaNO,, HCI, HO 2.CuCN NO, NO, p-Nitroaniline p-Nitrobenzonitrile (b) An acceptable pathway becomes apparent when it is realized that the amino group in the prod- uct is derived from the nitro group of the starting material. Two chlorines are introduced by electrophilic aromatic substitution, the third by a Sandmeyer reaction. ci ci NH, NH, ci ci ci ci “o NH, NO, NO, NO, Two of the required chlorine atoms can be introduced by chlorination of the starting material, p-nitroaniline. NH, NH, cl cl cl, — ss NO, NO, p-Nitroaniline 2.6-Dichloro-4- nitroaniline The third chlorine can be introduced via the Sandmeyer reaction. Reduction of the nitro group completes the synthesis of 3,4,5-trichloroaniline. a a “a “o “o “1 NaNO, HCLHO | reduce —teduce 20d” CuCl NO, NH, 2,6-Dichloro-4- 12,3-Trichloro-5- 3,4,5-Trichloroaniline nitroaniline nitrobenzene The reduction step has been carried out by hydrogenation with a nickel catalyst in 70% yield. MainMenu) TOC] StudyGuideTOC| Student OLC| | MHHE Website 640 (c) (d) (e) 22.40 (a) (b) Back| Forward Main Menul AMINES The amino group that is present in the starting material facilitates the introduction of the bromine substituents, and is then removed by reductive deamination. NH, Br. Br Br. Br Br, acetic acid 2. ethanol NO, NO, NO, 2.6-Dibromo-4- nitroaniline (95%) 1,3-Dibromo-5- nitrobenzene (70%) p-Nitroaniline Hypophosphorous acid has also been used successfully in the reductive deamination step. Reduction of the nitro group of the 1,3-dibromo-5-nitrobenzene prepared in the preceding part of this problem gives the desired product. The customary reducing agents used for the reduc- tion of nitroarenes would all be suitable. Br. Br Br. Br Ho Ni NO, NH, 1,3-Dibromo-5-nitrobenzene [prepared from p-nitroaniline as in part (c)] 3,5-Dibromoaniline (80%) The synthetic objective is o I no Soxucen, p-Acetamidophenol This compound, known as acetaminophen and used as an analgesic to reduce fever and relieve minor pain, may be prepared from p-nitroaniline by way of p-nitrophenol. NH, oH oH 1. NaNO,, H;O, H:SO, 2. heat 1. reduce 2. acetylate NO, NO, HINCCH, p-Nitroaniline p-Nitrophenol p-Acetamidophenol Any of the customary reducing agents suitable for converting aryl nitro groups to arylamines (Fe, HCI, Sn, HCI; H,, Ni) may be used. Acetylation of p-aminophenol may be carried out with acetyl chloride or acetic anhydride. The amino group of p-aminophenol is more nucle- ophilic than the hydroxyl group and is acetylated preferentially. Replacement of an amino substituent by a bromine is readily achieved by the Sandmeyer reaction. ocH, ocH, NH, Br É 1. NaNO, HBr, 2. CuBr o-Anisidine o-Bromoanisole (88-93%) This conversion demonstrates the replacement of an amino substituent by fluorine via the Schiemann reaction. TOC| Study Guide TOC] Student OLC| | MHHE Website Back| Forward 643 AMINES Fluorobenzene is prepared from aniline by the Schiemann reaction, shown in Section 22.18. Aniline is, of course, prepared from benzene via nitrobenzene. Friedel-Crafts acylation of fluorobenzene has been carried out with the results shown and gives the required ethyl p-fluorophenyl ketone as the major product. F F o || AICI, + CH,CH,CCI q CHCH, Fluorobenzene Propanoyl Ethyl p-luorophenyl chloride ketone (86%) (c) Our synthetic plan is based on the essential step of forming the fluorine derivative from an amine by way of a diazonium salt. Br Br F NH, Oo CH, HC CH, HC CH, HC 1-Bromo-2-fluoro- 2,4-Dimethylaniline 3.5-dimethylbenzene The required substituted aniline is derived from m-xylene by a standard synthetic sequence. NO, NH, HNO, - 1. Fe, HCI o SD E SOIS a JK HC CH HC CH, HC CH, m-Xylene 1,3-Dimethyl-4- 2.4-Dimethylaniline nitrobenzene (98%) Br, Br Br F 1. NaNO,, HCI, NH, Ho, OC * 2.HBF, HC CH, he HC CH, 1-Bromo-2-fluoro- 2-Bromo-4,6-dimethylaniline 3,5-dimethylbenzene (60%) (d) In this problem two nitrogen-containing groups of the starting material are each to be replaced by a halogen substituent. The task is sufficiently straightforward that it may be confronted directly. Replace amino group by bromine: NH, Br CH, CH, 1. NaNO,, HBr, HO SE, O > Cube eo NO, NO, 2-Methyl-4-nitro-1- 1-Bromo-2-methyl-4- naphthylamine nitronaphthalene (825%) MainMenu) TOC] StudyGuideTOC| Student OLC| | MHHE Website 644 AMINES Reduce nitro group to amine: Br Br CH, CH, o 2.H0 NO, NH, 1-Bromo-2-methyl- 4-Bromo-3-methyl- 4-nitronaphthalene I-naphthylamine Replace amino group by fluorine: Br Br CH, 1. NaNO,, HCI, CH; H,0, 0-5€ +» 2.HBF, 3. hear NH, F 4-Bromo-3-methyl- 1-Bromo-4-fluoro-2- I-naphthylamine methyInaphthalene (64%) (e) -Bromination of the starting material will introduce the bromine substituent at the correct position, that is, ortho to the tert-butyl group. C(CHy, (CH), Br Bro Fe NO, NO, p-tert-Butyl- 2-Bromo-|-tert- nitrobenzene butyl-4-nitrobenzene The desired product will be obtained if the nitro group can be removed. This is achieved by its conversion to the corresponding amine, followed by reductive deamination. C(CH), CCH), CCH), Br Br Br Ho Ni 1. NaNO,, H* or other appropriate 2 HPO, reducing agent) NO, NH, 2-Bromo-I-tert- 3-Bromo-4-tert- o-Bromo-tert- butyl-4-nitrobenzene butylaniline butylbenzene (/) | The proper orientation of the chlorine substituent can be achieved only if it is introduced after the nitro group is reduced. C(CH;)s C(CH;)s (CH); (CH); DG DO The correct sequence of reactions to carry out this synthesis is shown. Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website AMINES C(CH;), Ho Ni NO, p-tert-Butyl- nitrobenzene C(CH;), NH, p-tert-Butyl- aniline C(CH), acetic anhydride ue pert-Butyl- acetanilide (CH) O m-tert-Butyl- chlorobenzene Atert-Butyl-2- chloroacetanilide hydrolysis to remove acetyl group C(CH;); qe A-tert-Butyl-2- chloroaniline Nano," “Aro, (g) The orientation of substituents in the target molecule can be achieved by using an amino group to control the regiochemistry of bromination, then removing it by reductive deamination. NH, NH, CH,CH, Br CH,CH, Br CH,CH, CH,CH, CH.CH, CH.CH, The amino group is introduced in the standard fashion by nitration of an arene followed by reduction. This analysis leads to the synthesis shown. NO, NH, CH,CH, CH,CH, CH.CH, HNO, eihanol CH,CH, CH,CH, CH,CH, m-Diethylbenzene 2,4-Diethyl-1 -nitrobenzene 2,4-Diethylaniline (75-80%) (80-90%) Br, NH, CH,CH, r CH.CH, Br 1. NaNO,, H*, H,0 ZHPO, CH,CH, CH,CH, 1-Bromo-3,5- 2-Bromo-4,6- diethylbenzene (705%) diethylaniline (40%) Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website 648 AMINES Conversion of this intermediate to the desired N-methyl-4-piperidone requires a Dieckmann cyclization followed by decarboxylation of the resulting B-keto ester. o OCH;CH, CHCH,OG 0=c ç To o qe e 1.NaOCH,CH, ii O 1.HO O) 2, E, 2.H* 2.H* CH CH, N 3 heat N N AH, dm, cH, CH, N-Methyl-4- piperidone Treatment of N-methyl-4-piperidone with the Grignard reagent derived from bromobenzene gives a tertiary alcohol that can be dehydrated to an alkene. Hydrogenation of the alkene completes the synthesis. Q CH, OH CoHs CoHs 1. diethyl ether H* “sy Ho. Pt + CHMgBr ZH ar l À l l cH, cH, cH, cH, N-Methyl-4- — Phenylmagnesium N-Methyl-4- piperidone bromide phenylpiperidine (compound A) 22.44 Sodium cyanide reacts with alkyl bromides by the Sy2 mechanism. Reduction of the cyano group with lithium aluminum hydride yields a primary amine. This reveals the structure of mescaline to be 2-(3,4,5-trimethoxyphenylethy lamine. CH.O, cH; CH.O, cHO cHBr —ÉN., cHO cH,O CH,CH,NH, cHO cH,O cH,O 3,4,5-Trimethoxybenzyl 243,4,5-Trimethoxyphenyl)- 2-(3,4,5-Trimethoxyphenylethylamine bromide ethanenitrile (mescaline) 22.45 Reductive amination of a ketone with methy lamine yields a secondary amine. Methamphetamine is N-methyl-1-pheny1-2-propanamine. 9 NHcH, Ho, Ni ( Scnten, + CHNH, —D» ( Senenen, Benzyl methyl Methylamine N-Methyl-1-phenyl- ketone 2-propanamine (methamphetamine) 22.46 There is no obvious reason why the dimethylamino group in 4-(N, N-dimethy lamino)pyridine should be appreciably more basic than it is in N,N-dimethylaniline; it is the ring nitrogen of Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website Back| AMINES 649 4-(N,N-dimethy lamino)pyridine that is more basic. Note that protonation of the ring nitrogen per- mits delocalization of the dimethylamino lone pair and dispersal of the positive charge. CN(CH) “N(CHs) Most stable protonated form of 4-(N,N-dimethylamino)pyridine 22.47 The 'H NMR spectrum of each isomer shows peaks corresponding to five aromatic protons, so com- pounds A and B each contain a monosubstituted benzene ring. Only four compounds of molecular formula C,H,,N meet this requirement. C;HCHNHCH, — C,H)NHCH,CH, CoHsÇHCH, C;H,CH,CH,NH, NH, N-Methylbenzylamine N-Ethylaniline 1-Phenylethylamine 2-Phenylethylamine Neither 'H NMR spectrum is consistent with N-methylbenzylamine, which would have two singlets due to the methyl and methylene groups. Likewise, the spectra are not consistent with N-ethylaniline, which would exhibit the characteristic triplet-quartet pattern of an ethyl group. Al- though a quartet occurs in the spectrum of compound A, it corresponds to only one proton, not the two thatan ethyl group requires. The one-proton quartet in compound A arises from anH—C—CH, unit. Compound A is 1-phenylethylamine. Doublet (ô 1.2 ppm) Hs CH—C—NH; À 'Singlet (ô 1.3 ppm) Quartet (ô 3.9 ppm) Compound B has an !H NMR spectrum that fits 2-phenylethylamine. Singlet (ô 1.1 ppm) Pairof triplets A, atô 2.75 ppm and 2.95 ppm 22.48 Only the unshared electron pair on nitrogen that is not part of the 77 electron cloud of the aromatic system will be available for protonation. Treatment of 5-methyl-y-carboline with acid will give the salt shown. cH cH PONTO w ONO / ps N 5-Methyl-y-carboline H 22.49 Write the structural formulas for the two possible compounds given in the problem and consider how their 2C NMR spectra will differ from each other. Both will exhibit their CH, carbons at high field signal, but they differ in the positions of their CH, and quaternary carbons. A carbon bonded to Forward] MainMenu| TOC] StudyGuideTOC| StudentOLC| | MHHE Website 650 SELF-TEST Back| Forward AMINES nitrogen is more shielded than one bonded to oxygen, because nitrogen is less electronegative than oxygen. Lowerfield Higher field Higher field Lower field signal signal signal signal (CH) CHANH, (CH CHOH OH NH, 1-Amino-2-methyl-2-propanol | 2-Amino-2-methyl-I-propanol In one isomer the lowest field signal is a quaternary carbon; in the other it is a CH, group. The spec- trum shown in Figure 22.10 shows the lowest field signal as a CH, group. The compound is there- fore 2-amino-2-methyl-1-propanol, (CHfCHOH. NH, This compound cannot be prepared by reaction of ammonia with an epoxide, because in basic so- lution nucleophiles attack epoxides at the less hindered carbon, and therefore epoxide ring opening will give 1-amino-2-methyl-2-propanol rather than 2-amino-2-methyl-1-propanol. —. (CENGER, + NH, (CH CCHNH, > 0H 2,2-Dimethyloxirane Ammonia 1-Amino-2-methyl-2-propanol PART A A-1. Give an acceptable name for each of the following. Identify each compound as a primary, secondary, or tertiary amine. cH, I (a) CHCH.SCH, () NHCH,CH,CH, NH, Br (b) [nnca, A-2. Provide the correct structure of the reagent omitted from each of the following reactions: ! (a) CHCHBr Sam CHCHNH, 3H0 1 (6) CHCHBr Sam CH,CH;CH,NH, 3H0 * o 1? NH (e) CHCHBr Sm COHCHNH, + NH o MainMenu) TOC] StudyGuideTOC| Student OLC| | MHHE Website AMINES B-4. B-5. B-6. B-7. B-8. Back| Forward Main Menul 653 Which of the following arylamines will not form a diazonium salt on reaction with sodium nitrite in hydrochloric acid? (a) m-Ethylaniline (b) 4-Chloro-2-nitroaniline (c) | p-Aminoacetophenone (d) N-Ethyl-2-methylaniline The amines shown are isomers. Choose the one with the lowest boiling point. ' om N. N. É) Dom () CH, (a) (b) (c) (d) Which of the following is the strongest acid? [ [ (a) He (d) H=NH H H H H H | NE NH N » oo H H NE N. O The reaction NHCH,CH, + CHI (excess) —Ss 9 gives as final product (a) A primary amine (b) A secondary amine (c) A tertiary amine (d) A quaternary ammonium salt A substance is soluble in dilute aqueous HCI and has a single peak in the region 3200-3500 cm'! in its infrared spectrum. Which of the following best fits the data? (a) (net, (e) (men, (b) (mm, (d) ( com TOC| Study Guide TOC] Student OLC| | MHHE Website 654 AMINES B-9. Identify product D in the following reaction sequence: cH, | K,Cr,0,. H,SO, soct, (CH)NH 1. LiAIH,, diethyl ether CHGCH.CHOH HO, heat A (mol) 2.H,0 cH, q q (a) CHÇCHSC=N (d) CHGCHLCHN(CHy, cH, cH, cH, NCHy, q, (b) CHICHACHNCH;) (e) CH;CCH,CHN(CH,), cH, cH, OH cH, O | (o) CHÇCHCN(CH)» cH, B-10. Which one of the following is the best catalyst for the reaction shown? KCN CH(CH),CH;Br — e CH(CH),CH,CN f (rena ( )rnmcen, (rena, aim (a) (c) (e) = Om Omo (b) (d) B-11. What will be the major product of each of the two reactions shown? 1. CH,CH,CHCH, heat “N(CHy), “OH E cH,CH=CHCH, + CH,CH;CH=CH, 2. CH,CH,CHCH, + CH,CH,ONa —teat | ” ? be (a) 1x, 2x (b) 1x,2y (9) Iy,2x (d) 1y,2y Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website AMINES 655 B-12. Which sequence represents the best synthesis of 4-isopropylIbenzonitrile? Cen )c=N 4Isopropylbenzonitrile (a) 1. Benzene + (CH;),CHCI, AICL,;; 2. Br,, FeBr;; 3. KCN (b) 1. Benzene + (CH,),CHCI, AICL; 2. HNO,, H,SO,; 3. Fe, HCI; 4. NaOH; 5. NaNO,, HCI, H,0; 6. CuCN (c) 1. Benzene + (CH),CHCI, AICI;; 2. HNO,, H,SO,; 3. Fe, HCI; 4. NaOH; 5. KCN (d) 1. Benzene + HNO,, H,SO,; 2. (CH,),CHCI, AICI;; 3. Fe, HCI; 4. NaOH; 5. NaNO,, HCI, H,0; 6. CuCN (e) 1. Benzene + HNO,, H,SO,; 2. Fe, HCI; 3. NaOH; 4. NaNO,, HCI, H,O; 5. CuCN; 6. (CH, CHCI, AICL, B-13. The major products from the following sequence of reactions are CH Ago heat 9 (CH;)CHCH,N(CH,CH;), TH VT (a) (CH), CHCH,NH, + H,C=CH, (b) (CH),NCH,CH, + H;C=C(CH;), cH, (c) (CH; CHCH,NCH,CH, + H,C=CH, (d) (CH)NCH.CH; 1 + H,C—CH, (e) None of these combinations of products is correct. B-14. Which compound yields an N-nitrosoamine after treatment with nitrous acid (NaNO,, HCI)? (a) (Senna, (d) nc Sm, í (b) (5x ) (e) CNH, (e) (Snes Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website