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Síntese Avançada · Modern methods

Reatividade heteroaromática

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O que você vai aprender

  • Rank heteroarenes by basicity from where the nitrogen lone pair sits, using approximate pKaH values (pyrrole ≈ −3.8, pyridine 5.2, imidazole 7.0, DMAP 9.6).
  • Predict the site of electrophilic substitution on pyridine, pyridine N-oxide, pyrrole, furan, thiophene, indole and quinoline from the stability of the σ-complex.
  • Draw nucleophilic aromatic substitution at C2 and C4 of halopyridines and haloquinolines, and explain Chichibabin amination and Minisci radical alkylation.
  • Plan metalation of heteroarenes: direct C2 lithiation of five-membered rings and N-protected indoles, LDA-directed lithiation of halopyridines and halogen–metal exchange at −78 °C.
  • Build pyrroles, furans, thiophenes, pyrazoles, indoles and pyridines by the Paal–Knorr, Knorr, Fischer and Hantzsch syntheses, and aromatise Hantzsch 1,4-dihydropyridines with an oxidant.

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Where the nitrogen lone pair sits decides basicity

In pyridine the nitrogen lone pair lies in an sp² orbital in the ring plane, outside the aromatic π sextet, so protonation costs no aromaticity: pyridine has pKaH 5.2. It is a far weaker base than piperidine (pKaH 11.1) because an sp² lone pair is held closer to the nucleus. In pyrrole the nitrogen lone pair is one of the six π electrons. Protonating it would destroy the aromaticity, so pyrrole is barely basic (pKaH ≈ −3.8), and strong acid protonates it on carbon instead, mainly at C2, which leaves a conjugated iminium ion. Imidazole has one nitrogen of each kind; the pyridine-like nitrogen is protonated (pKaH 7.0) and the cation is stabilised by two equivalent amidinium contributors. A donor at C4 raises basicity further: 4-(dimethylamino)pyridine, DMAP, has pKaH 9.6. Electron-withdrawing groups lower it: 2-chloropyridine has pKaH 0.7 and pyridine N-oxide about 0.8. Remember that a pKaH gap is a logarithm: [B]/[BH⁺] = 10^(pH − pKaH).

Pyridine resists electrophiles; its N-oxide does not

Pyridine is π-deficient: the electronegative nitrogen withdraws electron density from every ring carbon. Pyridine itself is about as unreactive as nitrobenzene, and the strong acids used for nitration or sulfonation protonate the nitrogen first, leaving a pyridinium ring that is less reactive still. Substitution needs forcing conditions (Br₂ in oleum at about 130 °C) and goes to C3, because attack at C2 or C4 would put the positive charge of the σ-complex on an electron-poor nitrogen bearing only six electrons. Converting pyridine to its N-oxide (with a peracid) reverses the picture: the N–O⁻ oxygen donates into the ring, so nitration (HNO₃/H₂SO₄) occurs at C4. The N-oxide also activates C2 toward nucleophiles: POCl₃ converts it to 2-chloropyridine, and PCl₃ removes the oxygen at the end of a sequence.

Nucleophiles attack C2 and C4

The same electron deficiency makes C2 and C4 of pyridine electrophilic. A halide at C2 or C4 is replaced by alkoxides, thiolates and amines by addition–elimination: the anionic intermediate puts its negative charge on nitrogen, like the Meisenheimer complex of a nitroarene. A halide at C3 has no such contributor and is inert under the same conditions. Fluoride is the fastest leaving group because the addition step, not loss of the halide, limits the rate. In 4,7-dichloroquinoline only the C4 chlorine, para to the ring nitrogen, is displaced by amines; the C7 chlorine sits on the carbocyclic ring. Even hydride can be the leaving group: in the Chichibabin reaction sodium amide adds to C2 of pyridine on heating and expelled hydride removes a proton from the new amino group to release H₂, giving 2-aminopyridine.

Minisci alkylation: radicals meet pyridinium ions

Alkyl radicals are nucleophilic: their singly occupied orbital is high in energy, so they add fastest to electron-poor π systems. Protonating a pyridine in acid lowers its LUMO, and a tertiary or secondary alkyl radical then adds to C2 or C4 of the pyridinium ion. In the classic conditions a carboxylic acid, AgNO₃ and ammonium persulfate in aqueous acid generate the radical: silver(II) oxidises the carboxylate, and the acyloxyl radical loses CO₂, so RCO₂H delivers R•. After addition the adduct loses a hydrogen atom from the attacked carbon (by hydrogen-atom transfer to sulfate radical anion, or by oxidation and loss of a proton) to restore the aromatic ring. The electron transfers cannot be drawn with arrows, but decarboxylation, addition and hydrogen-atom loss can be drawn with fishhooks. With lepidine (4-methylquinoline) C4 is blocked and the tert-butyl group goes to C2. Electron-rich arenes such as anisole are poor partners: the polarity is mismatched.

Pyrrole, furan and thiophene: fast substitution at C2

The five-membered heteroarenes are π-excessive: six π electrons over five atoms. They react with electrophiles far faster than benzene, in the order pyrrole > furan > thiophene ≫ benzene, because nitrogen donates its lone pair best, oxygen holds it more tightly and sulfur's 3p orbital overlaps poorly. They therefore need mild reagents: acetyl nitrate rather than mixed acid, NBS rather than Br₂/FeBr₃, Ac₂O with a mild Lewis acid, or even an acid chloride alone for pyrrole. Strong protic acid makes pyrrole and furan polymerise. Attack at C2 gives a σ-complex with three contributors, one with the charge on the heteroatom; attack at C3 gives only two, so C2 wins. A 2-substituted ring reacts at C5, the remaining α-position. Aromaticity runs the other way: approximate resonance energies are benzene 150, thiophene 122, pyrrole 90 and furan 67 kJ/mol, which is why furan also behaves as a diene and adds maleic anhydride in a Diels–Alder reaction, while thiophene does not. A bulky N-triisopropylsilyl group blocks C2 of pyrrole and steers electrophiles to C3.

Indole: C3 for electrophiles, N1 and C2 for bases

Indole reacts with electrophiles at C3, not C2. Attack at C3 puts the positive charge on C2 next to nitrogen, which shares it as an iminium ion while the benzene ring stays intact; attack at C2 can only delocalise the charge onto nitrogen by disrupting the benzene sextet. The Mannich reaction with formaldehyde and dimethylamine in acetic acid gives gramine, 3-(dimethylaminomethyl)indole, and the Vilsmeier reagent (DMF/POCl₃) gives an iminium salt that aqueous base hydrolyses to indole-3-carbaldehyde. Gramine is a 3-indolylmethyl donor: after N-methylation of the amine, the indole nitrogen pushes out trimethylamine to form a 3-methyleneindolenine, which nucleophiles such as cyanide attack at the CH₂ group. The N–H is weakly acidic (pKa about 21 in DMSO), so NaH in DMF gives the indolyl anion, which an alkyl halide traps on nitrogen. Once nitrogen is protected or alkylated, n-butyllithium or LDA removes the C2 proton, the most acidic C–H because it sits next to the electronegative nitrogen, and the 2-lithioindole reacts with electrophiles.

Metalating heteroarenes

Furan, thiophene and N-substituted pyrroles are deprotonated at C2 by n-BuLi, again because that C–H is next to the heteroatom. To put a lithium anywhere else, start from a bromide: halogen–metal exchange with n-BuLi at −78 °C is faster than deprotonation or addition, so 3-bromothiophene gives 3-thienyllithium and 3-bromopyridine gives 3-pyridyllithium. Pyridines need care, because alkyllithiums add to the C=N bond. A hindered amide base such as LDA at −78 °C removes a ring proton instead; a halogen or alkoxy group acidifies the neighbouring C–H, so 2-chloropyridine is lithiated at C3 and the aryllithium is trapped with an electrophile such as I₂.

Quinoline and isoquinoline

Benzo-fusion splits the work between two rings. Electrophiles, which in strong acid meet the N-protonated cation, attack the carbocyclic ring: nitration of quinoline gives about equal amounts of 5- and 8-nitroquinoline, and isoquinoline gives mainly 5-nitroisoquinoline. Nucleophiles attack the pyridine ring at the positions conjugated with nitrogen: C2 and C4 of quinoline, and C1 of isoquinoline, which is the most reactive because the anionic intermediate keeps the benzene ring intact. Sodium amide converts isoquinoline into 1-aminoisoquinoline. Quinoline (pKaH 4.9) and isoquinoline (pKaH 5.4) are bases of about the same strength as pyridine.

Building the rings

Many heteroarenes are made by condensing carbonyl compounds with a heteroatom source. In the Paal–Knorr synthesis the four carbons of a 1,4-dicarbonyl compound become C2–C5 of the ring: acid alone closes a furan, a primary amine or ammonia gives a pyrrole (in weak acid only, because strong acid ties the amine up as its ammonium salt and the furan forms instead), and Lawesson's reagent or P₄S₁₀ gives a thiophene. In the Knorr pyrazole synthesis a hydrazine supplies N–N and a 1,3-dicarbonyl compound the three carbons. In the Fischer indole synthesis an arylhydrazone undergoes a [3,3] shift: the hydrazone carbon becomes C2, its α-carbon becomes C3, NH₃ is lost, and ring substituents keep their positions relative to the nitrogen-bearing carbon, so a para substituent ends up at indole C5. The Hantzsch synthesis gives a 1,4-dihydropyridine (drawn here), which is not yet aromatic: an oxidant such as DDQ or HNO₃ removes the C4 hydrogen as hydride and the N–H as a proton to give the pyridine.

What to carry forward

Ask where the nitrogen lone pair is and how the σ-complex or anionic intermediate carries its charge. Pyridine-type rings are basic, resist electrophiles (C3 when forced) and accept nucleophiles and nucleophilic radicals at C2 and C4; N-oxides switch electrophiles to C4. Pyrrole, furan and thiophene take electrophiles at C2 under mild conditions; indole takes them at C3. Bases remove the C–H next to the heteroatom, LDA avoids addition to pyridines, and halogen–metal exchange places lithium where deprotonation cannot. Ring syntheses map carbonyl carbons onto ring atoms: count the atoms, then assign each to its component.

Problemas de prática (40)

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Desenhe o mecanismo 12

Pyridines: substitution and N-oxides

  • Chichibabin amination of pyridineDraw the Chichibabin amination of pyridine: addition of amide ion at C2, loss of hydride to restore the aromatic ring, H₂ formation and protonation of the aminopyridine anion on work-up.
  • Nitration of pyridine N-oxide at C4Draw the nitration of pyridine N-oxide at C4, showing how the N-oxide oxygen donates electrons to stabilise the σ-complex.
  • Pyridine N-oxide to 2-chloropyridine with POCl₃Draw the conversion of pyridine N-oxide into 2-chloropyridine: activation of the oxygen by phosphorus, chloride addition at C2 and elimination that restores aromaticity.
  • SNAr of 2-fluoropyridine with sodium ethoxideDraw the addition–elimination substitution of fluoride at C2 of pyridine by ethoxide, placing the negative charge of the intermediate on the ring nitrogen.

Pyrrole, furan, thiophene and indole

  • Acetylation of furan at C2Draw the electrophilic acetylation of furan at C2, showing the ring oxygen stabilising the σ-complex and the rearomatising deprotonation.
  • Mannich reaction of indole: gramineDraw the Mannich aminomethylation of indole at C3 by the dimethylmethyleneiminium ion and the rearomatising deprotonation that gives gramine.
  • Bromination of 3-methylthiophene with NBSDraw the polar bromination of 3-methylthiophene at C2 by NBS and explain why C2 is preferred over C5.
  • Vilsmeier–Haack formylation of indole at C3Draw the Vilsmeier–Haack formylation of indole: C3 attack on the chloroiminium ion, rearomatisation, loss of chloride to the iminium salt, and its hydrolysis on basic work-up to indole-3-carbaldehyde.

Building the rings

  • Aromatising a Hantzsch 1,4-dihydropyridine with DDQDraw the hydride transfer from C4 of a Hantzsch 1,4-dihydropyridine to DDQ and the proton transfer that complete the oxidation to the pyridine.

Metalation and radical substitution

  • LDA lithiation of 2-chloropyridine at C3 and iodinationDraw the LDA deprotonation of 2-chloropyridine next to chlorine and the trapping of the 3-pyridyl anion by iodine.
  • Minisci tert-butylation of lepidineDraw the radical steps of the Minisci reaction: decarboxylation of the pivaloyloxyl radical, addition of the nucleophilic tert-butyl radical to C2 of the protonated quinoline, and rearomatisation, then explain the role of the acid.
  • 2-Lithiation of thiophene and carboxylationDraw the deprotonation of thiophene at C2 by n-butyllithium, the addition of 2-thienyllithium to carbon dioxide and the acidic work-up that gives thiophene-2-carboxylic acid.

Preveja o produto 11

Metalation and radical substitution

  • Halogen–metal exchange on 3-bromothiophenePredict the product of halogen–lithium exchange of 3-bromothiophene at −78 °C followed by formylation with DMF.
  • Lithiating an N-protected indolePredict where a base metalates an N-sulfonyl indole and which product the lithio species gives with iodomethane.

Pyridines: substitution and N-oxides

  • Which chlorine does morpholine replace?Predict which chlorine of 4,7-dichloroquinoline is replaced by an amine nucleophile.
  • Chichibabin reaction of isoquinolinePredict the site of Chichibabin amination of isoquinoline from the stability of the anionic adduct.
  • Forcing bromination of pyridinePredict the regiochemistry of electrophilic bromination of pyridine under forcing acidic conditions.

Building the rings

  • Where does the methoxy group end up?Map the substituent of a para-substituted arylhydrazine onto the indole ring formed in a Fischer synthesis.
  • Fischer indole synthesis from propanalPredict the indole formed from an aldehyde phenylhydrazone by mapping the carbonyl and α-carbons onto C2 and C3.
  • Paal–Knorr synthesis of a thiophenePredict the heteroarene formed from a 1,4-diketone and a sulfurising reagent in the Paal–Knorr synthesis.

Pyrrole, furan, thiophene and indole

  • Gramine as a 3-indolylmethyl donorPredict the product of cyanide displacement on gramine methiodide and explain the elimination–addition pathway.
  • Alkylating indole with NaH and iodomethanePredict the site of methylation of indole after deprotonation by sodium hydride in DMF.

Basicity and aromaticity

  • Where does pyrrole take a proton?Identify the most stable conjugate acid of pyrrole by comparing protonation at nitrogen, C2 and C3.

Comparar 4

Metalation and radical substitution

  • Which arene suits a Minisci reaction?Identify which arene a nucleophilic alkyl radical attacks fastest under acidic Minisci conditions.

Pyrrole, furan, thiophene and indole

  • Rank arenes toward electrophilesRank pyrrole, furan, thiophene and benzene by their rate of electrophilic aromatic substitution.

Basicity and aromaticity

  • Rank nitrogen heterocycles by basicityRank pyrrole, pyridine, imidazole and piperidine by basicity from where each nitrogen lone pair sits.

Pyridines: substitution and N-oxides

  • Rank aryl chlorides toward methoxideRank chlorobenzene and three chloropyridines by their rate of nucleophilic aromatic substitution with methoxide.

Calcular 3

Basicity and aromaticity

  • How much more basic is imidazole than pyridine?Convert the difference in pKaH between imidazole and pyridine into a ratio of base strengths.
  • How much pyridine is protonated at pH 4.2?Calculate the percentage of pyridine present as pyridinium at a given pH from its pKaH.

Pyrrole, furan, thiophene and indole

  • Count the contributors of pyrrole's σ-complexCount the resonance contributors of the σ-complex formed when an electrophile attacks C2 of pyrrole.

Encontre o erro 2

Building the rings

  • Furan instead of pyrroleDiagnose why a Paal–Knorr pyrrole synthesis run in strong acid gives a furan, and choose the correction.

Pyridines: substitution and N-oxides

  • Find the flaw: SNAr at C3Diagnose why a planned methoxide substitution of 3-chloropyridine fails, using the charge distribution of the addition intermediate.

Interprete as evidências 1

  • Which acetylmethylfuran formed?Use furan ring coupling constants to identify the regiochemistry of acetylation of 2-methylfuran.

Trabalhe de trás para frente 2

Building the rings

  • Partners for an N-phenylpyrazoleIdentify the 1,3-dicarbonyl compound and hydrazine that combine to give a 1-aryl-3,5-dialkylpyrazole.

Metalation and radical substitution

  • Choose the radical sourceWork back from a 2-alkylquinoline to the carboxylic acid that supplies the alkyl radical in a Minisci reaction.

Planeje uma síntese 2

Metalation and radical substitution

  • Plan: 3-bromopyridine to pyridine-3-carbaldehydeChoose conditions that convert 3-bromopyridine into pyridine-3-carbaldehyde by halogen–metal exchange.

Pyridines: substitution and N-oxides

  • Plan a synthesis of 4-nitropyridineChoose a reagent sequence that uses the N-oxide to place a nitro group at C4 of pyridine.

Explique por quê 3

Basicity and aromaticity

  • Furan reacts as a diene; thiophene does notExplain why furan undergoes Diels–Alder reactions with maleic anhydride while thiophene does not, using aromatic stabilisation energies.
  • Why is imidazole more basic than pyridine?Explain why imidazole is a stronger base than pyridine using the delocalisation of charge in its conjugate acid.

Metalation and radical substitution

  • Why LDA for pyridines?Explain why a hindered amide base, not n-butyllithium, is used to deprotonate a pyridine ring.

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