1559T_Ch25_439-45310/20/055:59APa9e439 EQA 25 Heterocydes:Heteroatoms in Cycic Organic Compounds The material in this chapter falls into two broad cates matic and aromatic heterocyeles You have already seen many examples of nonaromatic hetero ycles (see the section references at the start of the text chapter).There will be new ones,especially compounds containing nitrogen atoms in three fourorfive Heteroatoms may also be present in aromatic rings,and compounds with this feature are the subject of the last six sections of the chapter.In general,the properties of heterocyclic compounds will be predictable from prnciples you've en:They will be s r to acyclic compoun tic. atoms unless (1 Outline of the Chapter 25-1 Naming the Heterocycles 25-2 Nonaromatic Heterocycles 25-3,25-4 Aromatic Heterocyclopentadienes:Pyrrole,Furan,and Thiophene 25-5,25-6 Pyridine,an Azabenzene The most common monocyclic heteroaromatic compounds. 25-7 Quinoline and Isoquinoline:The Benzpyridines 25-8 Alkaloids Even more rings! Keys to the Chapter 25-1.Naming the Het erocycles universally accepted common names for the aromatic systems. enclature for nonaromatic heterocycles but uses the 25-2.Nonaromatic Hete ocycles aae ontaining nitrogen and sulfur illustra
25 Heterocycles: Heteroatoms in Cyclic Organic Compounds The material in this chapter falls into two broad categories: nonaromatic and aromatic heterocycles. You have already seen many examples of nonaromatic heterocycles (see the section references at the start of the text chapter). There will be new ones, especially compounds containing nitrogen atoms in three-, four-, or fivemembered rings. Heteroatoms may also be present in aromatic rings, and compounds with this feature are the subject of the last six sections of the chapter. In general, the properties of heterocyclic compounds will be predictable from principles you’ve already seen: They will be similar to acyclic compounds with similar heteroatoms unless (1) the ring is strained or (2) the ring is aromatic. When the ring is aromatic, the effect of the heteroatom on its chemistry will be important, and this will be a new topic to which you will need to apply your knowledge of inductive and resonance effects. Outline of the Chapter 25-1 Naming the Heterocycles 25-2 Nonaromatic Heterocycles 25-3, 25-4 Aromatic Heterocyclopentadienes: Pyrrole, Furan, and Thiophene 25-5, 25-6 Pyridine, an Azabenzene The most common monocyclic heteroaromatic compounds. 25-7 Quinoline and Isoquinoline: The Benzpyridines 25-8 Alkaloids Even more rings! Keys to the Chapter 25-1. Naming the Heterocycles Note simply that the text sticks to strict systematic nomenclature for nonaromatic heterocycles but uses the universally accepted common names for the aromatic systems. 25-2. Nonaromatic Heterocycles This section generalizes what you have previously learned concerning the preparation and reactions of cyclic ethers. Rings containing nitrogen and sulfur illustrate the principal similarity with their oxygen relatives: Ring opening to relieve strain governs the reactivity of the smaller rings (three or four atoms). 439 1559T_ch25_439-453 10/20/05 5:59 AM Page 439
1559Tch25_439-45310/20/055:59 M Page4a0 EQA 440.chapter 25 HETEROCYCLES:HETEROATOMS IN CYCLC ORGANIC COMPOUNDS 25-3 through 25-7.Aromatic Hete rocycles tainin determine whethera lone pair of a heteroatom is part of the cyclicsystem:If only single bonds link the het e l lepair from he n of the two lone pairson the heter atom is art of thesystem:The other is in an sporbital and has nothin to do with the lecule's aromaticity at a I the hete om is y bonded to another ring ato om,as i 2 structures illustrated in these scctions as you can They are aromatic,so you know what the ans wers should be inations of carbonyl c and eu no fundamentally new chemistry here.Reactions of these systems show a blending of benzene-aromatic chem istry and the logs with the same heteroatom.Furan is a good exa &cioddioadeeChemsLHmguheiagembeodeaconshecrehoihcrpopaieconpateas should help you organize them for study and problem-solving. Solutions to Problems 0 26.(a) (b CoHs CoHs HN (ds、s (e)2-Formylfuran or furan-2-carbaldehyde:(f)N-methylpyrrole or 1-methylpyrrole: (g)quinoline-4-carboxylic acid:(h)2.3-dimethylthiophene 27.(a b)〈NHCH H (e)A bit tricky.so follow this mechanism OCH-CHs +6:0n :0::0H cH % →CH.CCH,CCH3 CH.CCH.CCH+CH.CH2OH CH
25-3 through 25-7. Aromatic Heterocycles Five- and six-membered heterocyclic compounds make up the vast majority of aromatic heteroatom-containing systems. Counting electrons in the systems of these rings is occasionally confusing. Here is a simple way to determine whether a lone pair of a heteroatom is part of the cyclic system: If only single bonds link the heteroatom to its neighbors in the ring, then one lone pair from the heteroatom may be in a p orbital and become part of the cyclic system. Examples are pyrrole, furan, and thiophene. Notice that in the latter two, only one of the two lone pairs on the heteroatom is part of the system: The other is in an sp2 orbital and has nothing to do with the molecule’s aromaticity at all. If the heteroatom is doubly bonded to another ring atom, as is the case with pyridine, a lone pair on it will be in an sp2 orbital and will never count toward the cyclic system in the molecule. Try to apply these rules to as many structures illustrated in these sections as you can: They are all aromatic, so you know what the answers should be. Typically, syntheses of the aromatic heterocycles utilize combinations of carbonyl condensation reactions and 1,2- or 1,4-additions of enolate and heteroatom nucleophiles to , -unsaturated carbonyl compounds— no fundamentally new chemistry here. Reactions of these systems show a blending of benzene–aromatic chemistry and the chemistry of nonaromatic analogs with the same heteroatom. Furan is a good example: It undergoes electrophilic substitution (aromatic chemistry), ring cleavage in acid (ether chemistry), and Diels-Alder cycloaddition (diene chemistry). Putting the large number of reactions here into their appropriate compartments should help you organize them for study and problem-solving. Solutions to Problems 26. (a) (b) (c) (d) (e) 2-Formylfuran or furan-2-carbaldehyde; (f) N-methylpyrrole or 1-methylpyrrole; (g) quinoline-4-carboxylic acid; (h) 2,3-dimethylthiophene 27. (a) (b) (c) A bit tricky, so follow this mechanism. H H2O OCH2CH3 CH3 CH3 CH3 O OCH2CH3 CH3 CH3 CH3 O H CH3CCH2CCH3 CH3 CH2CH3 O OH CH3CCH2CCH3 CH3CH2OH CH3 O OH H CH2OCH2CH3 NHCH3 H H H OH CCH2CH2CH3 O S S O S HN O C6H5 C6H5 O 440 • Chapter 25 HETEROCYCLES: HETEROATOMS IN CYCLIC ORGANIC COMPOUNDS 1559T_ch25_439-453 10/20/05 5:59 AM Page 440
1559T_ch25_439-45310/20/055:59 AM Page441 ⊕ EQA Soutionso Problems441 0 28.(a)Ch.CH-CNH COOH C HsCH-CNH HO CH NH COOH C.H.CH.( CH-S CH Protein-NHH 0 (b)CoHsCH2CNH Formed viaa parallel mechanism,with Has the HOOC ing for antibiotic properties. 29.Use the Lewis acids to activate the ring oxygen in (a)and (b). o+-sncla (e)Use the Lewis acid to activate the anhydride and form an acylium cation.similar to the first step in Friedel-Crafts alkanoylation. 0 0 CH.C-0-CCH,=CHCo*CH.c-0-MgBra-and MgBr2- CH:C-0-MgBr2-=CHC-0-MgBr Br
28. (a) (b) Formed via a parallel mechanism, with H2O as the nucleophile. This product, penicilloic acid, no longer possesses the necessary strained azacyclobutanone ring for reaction with bacterial protein. It therefore lacks any antibiotic properties. 29. Use the Lewis acids to activate the ring oxygen in (a) and (b). (a) (b) (c) Use the Lewis acid to activate the anhydride and form an acylium cation, similar to the first step in Friedel-Crafts alkanoylation. and MgBr2 O CH3C MgBr Br CH3C O O O MgBr2 MgBr2 CH3CO O CH3C O CCH3 CH3C O O O C6H5 BF3 Hydrolysis C6H5 OBF4 product O CH3CH2CH2CH2 Li Hydrolysis SnCl4 CH2OSnCl4 CH2OH O (A type of FriedelCrafts reaction) CH3 CH3 S N H HOOC CH C6H5CH2CNH O COOH CH3 CH3 S C N H Protein NH O Penicilloyl protein CH C6H5CH2CNH O Ring-opening relieves strain COOH CH3 CH3 S H O N NH Protein C6H5CH2CNH O Protein–NH2 COOH CH3 CH3 S N O Most reactive toward nucleophiles C6H5CH2CNH O Solutions to Problems • 441 1559T_ch25_439-453 10/20/05 5:59 AM Page 441
1559r.eh25_439-45310/20/055:59 M Page442 442 chapter 25 HETEROCYCLES:HETEROATOMS IN CYCLC ORGANIC COMPOUNDS Br 一 30.The order of basicity is the reverse of the order of the acidity of the conjugate acids (pk values shown below the structures). Bases NH*>H2O 00 rone “”心 Alvdouble bondsone lone pairbclecnsall areomtie All have stronger bases than pyrrole
Then, the acylium ion converts the ether oxygen into a good leaving group and allows SN2 displacement by bromide to occur. 30. The order of basicity is the reverse of the order of the acidity of the conjugate acids (pKa values shown below the structures). Bases Conjugate acids 31. All have two double bonds plus one lone pair in a p orbital 6 electrons, so all are aromatic. All have sp2 -hybridized lone pairs on nitrogen, not tied up in the aromatic system, and therefore available to act in a Lewis-base manner. Pyrrole lacks an sp2 -hybridized lone pair; therefore all the compounds above are stronger bases than pyrrole. 32. (a) (b) O CH3 N N N H p sp2 N N H p sp2 S p N sp2 N sp2 O p H3O NH4 H2O Weakest acid H H Strongest acid pKa 4.4 0.0 5.3 9.2 15.7 N H N H H2O Weakest base Strongest base HO NH3 N H N CH3 CH3 Br O O C H3C H H CH3 O O C Configuration is inverted here (SN2) Br H3C H3C H H 442 • Chapter 25 HETEROCYCLES: HETEROATOMS IN CYCLIC ORGANIC COMPOUNDS 1559T_ch25_439-453 10/20/05 5:59 AM Page 442
1559T_ch25_439-45310/20/055:59 AM Page443 中 EQA Solutions to Problems.443 33.Abbreviated mechanism 0H0) CC.Hs + 一CH,OOCCH CHCOOCHS CH.OOCCHNCH.COOCH CH.CO CH.CO Synthesis CH.OOCCHI.SCH-COOCHCHOOCSCOOCHO Pdluc 34.There are three factors to keep in mind:(1)the inherent preference of these compounds for h work the same way as in benzene). (a)Two conflicting preferences m二 In this case.a mixture might be expected: The first is actually the major product:the directing effect of the activating ring oxygen to C5 wins out over that of the moderately deactivating COOCH group to C4. (b)Easier
33. Abbreviated mechanism Synthesis 34. There are three factors to keep in mind: (1) the inherent preference of these compounds for substitution at C2 over C3, (2) the much greater reactivity of all of them compared with benzene, and (3) directing effects of substituent groups (which work the same way as in benzene). These are toughies! (a) Two conflicting preferences In this case, a mixture might be expected: The first is actually the major product; the directing effect of the activating ring oxygen to C5 wins out over that of the moderately deactivating COOCH3 group to C4. (b) Easier NO2 S CH3 C5 is strongly activated CH3 o, p-group o, p-group Ring preference S Cl O COOCH3 5 Cl O COOCH3 4 O COOCH3 m-directing group Preference of ring NaOCH3 NaOH, H2O HC CH3OOCCH2SCH2COOCH3 CH3OOC S COOCH3 O CH O product 2 H2O CH3CO H H product N OH OH CH3OOC COOCH3 C6H5 C6H5 C6H5C CH3CO O CC6H5 O CH3OOCCHNCH2COOCH3 C6H5C CH3CO OH N CC6H5 O CH3OOCCH CHCOOCH3 Solutions to Problems • 443 1559T_ch25_439-453 10/20/05 5:59 AM Page 443
1559T_ch25_439-45310/20/055:59 AM Page444 EQA 444.chapter 25 HETEROCYCLES:HETEROATOMS IN CYCLC ORGANIC COMPOUNDS (c)Tricky.If this (CHa)2CH (d)Easy o.p-grou Br Ring preferenc (e)Now you have to work from scratch!Compare attacks at various carbons Poor!(N at top lacks octe 9-g ood resonance fom
(c) Tricky. If this were benzene, the Friedel-Crafts reaction would not work at all because of the presence of the COCH3 group. It does proceed here, because the heterocycle is much more reactive. The ketone substituent is complexed by AlCl3 during the reaction, making it even more strongly deactivating and meta-directing, however. The overall result is slow formation of (d) Easy (e) Now you have to work from scratch! Compare attacks at various carbons. C2 C4 C5 N N H E H Three good resonance forms N N H E H N N H E H N N H N N H E H E H Only two resonance forms N N H E H N N H E H N N H E H Poor! (N at top lacks octet) Br o, p-group o, p-group Ring preference Ring preference C1 is doubly preferred. S S Br Br C O (CH3)2CH CH3 O 444 • Chapter 25 HETEROCYCLES: HETEROATOMS IN CYCLIC ORGANIC COMPOUNDS 1559T_ch25_439-453 10/20/05 5:59 AM Page 444
1559T_ch25439-45310/20/055:59 AM Page445 EQA Solutions to Problems.445 attack by typical e ophiles.In this pa ular example, reaction at C2 gives a symmetrical intermediate with two equivalent resonance forms. E=CoHsN2" 35.(a b)Diels-Alder: 0 d)CH.CH-CH.CH-CC.Hs e CH、CH 36.(a)CH 女CH 00 (b)Hantzsch CoH CCHCOCH CH,HC-O.NH CH,CH00C、入 COOCHCH0 .C0.△ →product 入CH5
Rule out C4 (only two resonance forms for cation). Rule out C2 next, because attack results in an electron sextet and a positive charge on one of the electronegative N atoms. Thus, C5 is the site of attack by typical electrophiles. In this particular example, however, the major product is due to diazo coupling at C2, because under the basic conditions the imidazole anion is attacked, and reaction at C2 gives a symmetrical intermediate with two equivalent resonance forms. E C6H5N2 35. (a) (b) Diels-Alder: (c) (d) (e) 36. (a) (b) Hantzsch product 1. HNO3, H2SO4 2. KOH, H2O 3. CaO, C6H5 C6H5 CH3CH2OOC COOCH2CH3 N H O O C6H5CCH2COCH2CH3 H2C O, NH3 O O CH3 CH3 CH3 CH3 CH3 CH3 CH3 CH3 O P2O5, N C(CH3)3 CH3CH2CH2CH2CC6H5 O O C C6H5 S via N SH O O O O S O O O O S N SO3H (CH3)2N N N N N E E H N N E H N N E H N H N Tautomerizes N N Product Solutions to Problems • 445 1559T_ch25_439-453 10/20/05 5:59 AM Page 445
1559T.ch25.439-45310/20/055:59 M Page4a6 EQA 446.chapter 25 HETEROCYCLES:HETEROATOMS IN CYCLC ORGANIC COMPOUNDS (e)Paal-Knorr 0 0 HCCH-CHCH NH product CHs CHs 0 CH.CCHs 37 0 CH,CHO O OCH CH 0 ⊕ CH OCH.CH,B Compare synthesis of furans,etc.,from 1.4-diketones 38.Protonation of keto C00褥09
(c) Paal-Knorr (d) 37. Compare synthesis of furans, etc., from 1,4-diketones. 38. Protonation of ketones and aldehydes can give rise to electrophiles capable of aromatic substitution: Then, H OH N H N H H H HO N H H HO H H O H OH H OH CH2 CH3CH2OC O O COCH2CH3 O O O C C C CH2 HCl, H2O (With ester hydrolysis) HOOC O O COOH CH3CCH3 O CH3CH2O CH3CH2O OCH2CH3 OCH2CH3 O O O C O C C C NaOCH2CH3, CH3CH2OH Double Claisen condensation O O H CH3 CH3 CH3 CH3 S P2S5, O O product CH3 CH3 HCCH CHCH NH3 446 • Chapter 25 HETEROCYCLES: HETEROATOMS IN CYCLIC ORGANIC COMPOUNDS 1559T_ch25_439-453 10/20/05 5:59 AM Page 446
Next.pr tonation of the hydroxy pyrole ing enables it to leave.The carbocation that results can underg 39.Asix-electron cycloaddition takes you directly to the product: R'C-N-0: a-d 40.(a)H2.Pt (b)Step (lactam)formation of azacyclopropane first.followed by intramolecuar amid SO-CI
Next, protonation of the hydroxy group enables it to leave. The carbocation that results can undergo substitution with a second pyrrole ring: 39. A six-electron cycloaddition takes you directly to the product: 40. (a) H2, Pt (b) Stepwise, ring opening of azacyclopropane first, followed by intramolecular amide (lactam) formation. (c) 41. NHS HO, H2O, O O NHCCH3 O N Product NH2 NHCCH3 CH3CCl, py ClSO3H 1. NaNH2, NH3 2. H, H2O O O NHCCH3 SO2Cl O (Chapter 15, Problem 47) N N NH2 N NH COCH2CH3 CH3CH2OH CO2CH2CH3 O O NH2 N H H N N N NH (b) N O O R C R R R N RC CR N H N H H H
1559T_ch25439-45310/20/055:59 AM Page44g ⊕ EQA Solutions to Problems.449 the product: -0 aromatic substitution.take place on the benzene ring.which is more reactive toward electrophiles than is the pyridine ring.Without the extra ethanoyl group substitution would be ex- te ca of th (see resonance forms),leaving substitution at position 8 as the most likely outcome. Substitution at C5
46. The hydrogen atoms on the nitrogen are relatively acidic because of resonance stabilization of the anion that results from deprotonation. Conjugate addition followed by intramolecular aldol condensation gives the product: 47. To answer the second question first, the process will be slower than nitration on quinoline itself. Nitration is an electrophilic substitution, and the alkanoyl group is a strong deactivating group toward electrophilic aromatic substitution. Substitution will take place on the benzene ring, which is more reactive toward electrophiles than is the pyridine ring. Without the extra ethanoyl group substitution would be expected to occur at positions 5 and 8, because the cationic intermediates from substitution at these positions do not disrupt the aromaticity of the pyridine ring. However, the extra substituent deactivates position 5 (see resonance forms), leaving substitution at position 8 as the most likely outcome. Substitution at C5: O CH3 N E H Very poor O CH3 N O E H CH3 N E H O CH3 N O E H CH3 N E H Better (leave pyridine ring intact) O CH3 8 5 N E Aldol condensation O CH3 N H Base O NH2 H O NH H O H O CH3 O CH3 N H Solutions to Problems • 449 1559T_ch25_439-453 10/20/05 5:59 AM Page 449