Química Orgânica 2 · Exam review
Substituição acílica, aminas e sais de diazônio
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- Azo coupling of benzenediazonium with 2-naphthoxide
- Balz–Schiemann: 4-fluorotoluene from 4-methylbenzenediazonium tetrafluoroborate
- Boc protection of benzylamine with di-tert-butyl dicarbonate
- Phenol from benzenediazonium and warm water
- DMAP-catalysed acetylation of cyclohexanol
- Releasing benzylamine from N-benzylphthalimide with hydrazine
- Hinsberg test: propan-1-amine and benzenesulfonyl chloride
- Hofmann elimination opens the piperidine ring
- Acid-catalysed lactonisation of 2-(hydroxymethyl)benzoic acid
Ver todos os 77 problemas de substituição acílica, aminas e sais de diazônio →
O que você vai aprender
- Rank carboxylic acid derivatives by reactivity and decide which interconversions go directly downhill.
- Draw nucleophilic acyl substitution and hydrolysis mechanisms in acid and in base with the acids and bases actually present.
- Predict and explain amine basicity from hybridization, resonance, induction and solvation, and use pKaH values in calculations.
- Choose a clean route to a primary, secondary or tertiary amine: Gabriel synthesis, reductive amination or amide reduction.
- Form arenediazonium ions and convert them into phenols, aryl halides, nitriles, arenes and azo compounds.
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The reactivity ladder
Carboxylic acid derivatives react with nucleophiles in the order acid chloride > acid anhydride > ester ≈ carboxylic acid > amide > carboxylate. Two effects set the order. First, the group on the carbonyl carbon donates its lone pair into the C=O π system: chlorine barely donates, an ester oxygen donates moderately and an amide nitrogen donates strongly, so the amide carbonyl carbon is the least electrophilic. Second, the group must leave: the weaker its conjugate base, the better. Chloride (HCl, pKa about −7) and carboxylate (pKa about 5) leave easily, alkoxide (pKa about 16) less so, and an amide anion (NH₃, pKa about 38) hardly at all. Drawn left to right: acetyl chloride, acetic anhydride, methyl acetate and acetamide.
Addition, then elimination
Nucleophilic acyl substitution goes by addition, then elimination, not by a one-step SN2 at the carbonyl carbon. The nucleophile adds to the carbonyl carbon and the C=O π pair moves onto oxygen, giving a tetrahedral intermediate; the oxygen lone pair then re-forms C=O and the best leaving group departs. A derivative can therefore be turned directly into any derivative below it on the ladder, but not into one above it: an acid chloride gives an anhydride, ester or amide, while an amide does not give an ester or acid chloride by simple exchange. Reactions that release HCl need a base (a second equivalent of amine, triethylamine or aqueous NaOH, as in the Schotten–Baumann method) to stop the nucleophile being protonated. Amines out-compete alcohols for an acylating agent, and anions (phenoxide, alkoxide) out-compete the neutral O–H compounds. A nucleophilic catalyst such as DMAP is first acylated itself, giving an N-acylpyridinium ion that transfers the acyl group to an alcohol far faster than the anhydride could. Di-tert-butyl dicarbonate (Boc₂O) is an anhydride whose leaving group, tert-butyl carbonate, takes a proton and then falls apart into CO₂ and tert-butanol, which is why Boc protection of amines runs to completion.
Hydrolysis in acid and in base
In aqueous base, hydroxide adds to the ester, the alkoxide leaves, and the carboxylic acid formed is deprotonated at once. The carboxylate cannot be attacked again, so saponification goes to completion, consumes one hydroxide per ester plus one for every other acidic group present (a phenol released from an aryl ester, for example), and the free acid appears only when the solution is acidified. The oxygen from hydroxide ends in the carboxylate; the alkoxy oxygen leaves in the alcohol. In aqueous acid, hydronium protonates the carbonyl oxygen, water adds, proton transfers turn the OR group into a neutral alcohol leaving group, and loss of the alcohol and a proton gives the acid. Every step is reversible, so a large excess of water drives the hydrolysis; the same steps run backwards when a hydroxy acid closes to a five-membered lactone, which is favoured because the ring closure is intramolecular. Nitriles are hydrated in strong acid through a nitrilium ion and an imidic acid to the amide. Amides need hot concentrated acid or base; in acid the amine released is protonated to an ammonium ion, which pulls the reaction forward. Shown: methyl benzoate and the benzoate ion it gives in NaOH.
How basic is an amine?
Amine basicity is quoted as the pKa of the conjugate acid, pKaH: the higher the pKaH, the stronger the base, and pKb = 14.00 − pKaH in water. Alkylamines have pKaH about 10–11. Aniline (4.6) is far weaker because its lone pair is delocalised into the ring; pyridine (5.2) holds its lone pair in an sp² orbital with more s character; an amide (about −0.5) is protonated on oxygen, not nitrogen, because the nitrogen lone pair is part of the C=O π system. Pyrrole is not basic at nitrogen at all, since that lone pair is part of its aromatic sextet. Drawn: ethylamine, aniline, pyridine and acetamide. At a given pH, Henderson–Hasselbalch gives the ratio of protonated to free amine: [BH⁺]/[B] = 10^(pKaH − pH).
Making amines cleanly
Alkylating ammonia with an alkyl halide gives mixtures, because each amine formed is at least as nucleophilic as the one before. Three routes avoid this. In the Gabriel synthesis the phthalimide anion (phthalimide pKa about 8.3) is alkylated once by SN2 on a primary or unhindered secondary halide, and hydrazine then releases the primary amine as it closes the six-membered ring of phthalhydrazide. In reductive amination an aldehyde or ketone and an amine form an iminium ion at pH about 5–6, and NaBH₃CN reduces the iminium ion much faster than the carbonyl compound. Reducing an amide with LiAlH₄ turns its C=O into CH₂ and keeps every N-substituent. Quaternary ammonium hydroxides, made by exhaustive methylation, undergo Hofmann elimination by E2 to give mainly the less substituted alkene; with a cyclic amine the first elimination only opens the ring. In the Hinsberg test, benzenesulfonyl chloride converts a primary amine into a sulfonamide whose remaining N–H (pKa about 10–11) dissolves in aqueous NaOH, while a secondary amine gives an insoluble sulfonamide with no N–H. Drawn: phthalimide.
Arenediazonium ions
A primary arylamine with NaNO₂ and HCl at 0–5 °C gives an arenediazonium ion, ArN₂⁺. The nitrosonium ion (NO⁺) nitrosates the amine nitrogen, proton transfers give a diazohydroxide, and loss of water gives the N≡N group. N₂ is an outstanding leaving group, so the diazonium group can be replaced: CuCl, CuBr or CuCN (Sandmeyer) give ArCl, ArBr or ArCN; KI gives ArI; HBF₄ then heat gives ArF; warm aqueous acid gives a phenol; H₃PO₂ replaces it by H. As a weak electrophile the diazonium ion also couples with strongly activated rings to give azo dyes: phenoxides react para to O⁻ (2-naphthoxide at C1), and tertiary arylamines para to nitrogen. A primary arylamine such as aniline attacks through its nitrogen instead, giving a triazene, Ar–NH–N=N–Ar. Alkyl diazonium ions lose N₂ at once, so only aryl diazonium salts are useful intermediates. Drawn: the benzenediazonium ion.
What to carry forward
Read every acyl reaction as addition then elimination, and check it runs down the reactivity ladder. Use the real acid or base of the medium for each proton transfer. Judge amine basicity by where the lone pair is and how well the ammonium ion is stabilised. Make amines by routes that cannot over-alkylate, and treat ArN₂⁺ as an aryl group carrying a superb leaving group.
Problemas de prática (77)
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Desenhe o mecanismo 37
Diazonium salts
- Azo coupling of benzenediazonium with 2-naphthoxideDraw the attack of C1 of 2-naphthoxide on the terminal nitrogen of benzenediazonium and the deprotonation that restores aromaticity, and explain why coupling occurs at C1.
- Balz–Schiemann: 4-fluorotoluene from 4-methylbenzenediazonium tetrafluoroborateDraw the loss of N₂ from 4-methylbenzenediazonium and the transfer of fluoride from tetrafluoroborate to the aryl cation that gives 4-fluorotoluene.
- Phenol from benzenediazonium and warm waterDraw the loss of N2 from the benzenediazonium ion, the capture of the aryl cation by water and the final deprotonation that gives phenol.
- Diazoaminobenzene from benzenediazonium and anilineDraw the attack of aniline's nitrogen on the terminal nitrogen of benzenediazonium and the deprotonation by acetate that give 1,3-diphenyltriazene.
Acyl substitution reactivity
- Boc protection of benzylamine with di-tert-butyl dicarbonateDraw the addition of benzylamine to di-tert-butyl dicarbonate, the loss of tert-butyl carbonate, the proton transfer that gives the Boc carbamate, and the breakdown of tert-butyl hydrogen carbonate to CO₂ and tert-butanol.
- DMAP-catalysed acetylation of cyclohexanolDraw how DMAP acts as a nucleophilic catalyst: it is acylated by acetic anhydride to an N-acetylpyridinium ion, which transfers the acetyl group to cyclohexanol and releases DMAP again.
- Phenyl benzoate by the Schotten–Baumann methodDraw how aqueous hydroxide turns phenol into phenoxide and how phenoxide adds to benzoyl chloride and expels chloride to give phenyl benzoate.
- Ring opening of succinic anhydride by ammoniaDraw how ammonia opens the five-membered ring of succinic anhydride by addition–elimination, with the second ammonia as the base, to give ammonium succinamate.
- Phenyl acetate by pyridine-catalysed acetylation of phenolDraw pyridine attacking acetyl chloride to form the N-acetylpyridinium ion, phenol acylation by that ion, and deprotonation that gives phenyl acetate.
- N,N-Diethylbenzamide from benzoyl chloride and diethylamineDraw the addition–elimination of diethylamine on benzoyl chloride and the proton transfer to a second equivalent of amine that gives N,N-diethylbenzamide.
- Benzamide by ammonolysis of methyl benzoateDraw ammonia adding to methyl benzoate, the proton transfer to a second ammonia, loss of methoxide, and proton return that gives benzamide.
- N-Acetylglycine from glycine and acetic anhydrideDraw the acetylation of glycine by acetic anhydride: amine addition to an anhydride carbonyl, loss of acetate, and deprotonation by acetate to give N-acetylglycine.
Making and using amines
- Releasing benzylamine from N-benzylphthalimide with hydrazineDraw the two acyl substitutions by hydrazine that free benzylamine from N-benzylphthalimide and trap both carbonyls in the six-membered ring of phthalhydrazide.
- Hinsberg test: propan-1-amine and benzenesulfonyl chlorideDraw the sulfonylation of a primary amine and the two deprotonations by hydroxide that leave the N-propylbenzenesulfonamide anion dissolved in aqueous base.
- Hofmann elimination opens the piperidine ringDraw the N-methylation of N-methylpiperidine and the Hofmann E2 elimination of the 1,1-dimethylpiperidinium ion that opens the six-membered ring to N,N-dimethylpent-4-en-1-amine.
- Reductive amination of benzaldehyde with morpholineDraw how benzaldehyde and the secondary amine morpholine form an iminium ion through a carbinolamine, with acetic acid and acetate moving the protons, and how cyanoborohydride reduces it to 4-benzylmorpholine.
- 3-(Methylamino)propanenitrile by aza-Michael additionDraw the conjugate addition of methylamine to the β-carbon of acrylonitrile and the proton transfers through methanol that give 3-(methylamino)propanenitrile.
- Aniline from benzoyl chloride by the Curtius rearrangementDraw benzoyl azide formation, its Curtius rearrangement to phenyl isocyanate with loss of N2, and the hydrolysis and decarboxylation that release aniline.
- Pent-1-ene by Hofmann elimination of pentan-2-amineExcess CH3I with NaOH first converts pentan-2-amine into N,N-dimethylpentan-2-amine. Draw the third SN2 methylation to the quaternary ammonium salt and the Hofmann E2 of the quaternary ammonium hydroxide that gives pent-1-ene, the less substituted alkene.
- N-Cyclohexylaniline by reductive amination of cyclohexanone with anilineDraw iminium formation from cyclohexanone and aniline under acetic acid catalysis and its reduction by cyanoborohydride to N-cyclohexylaniline.
Hydrolysis in acid and base
- Acid-catalysed lactonisation of 2-(hydroxymethyl)benzoic acidDraw the acid-catalysed ring closure of 2-(hydroxymethyl)benzoic acid to the five-membered lactone phthalide, with hydronium and water as the only acid and base.
- Hydration of benzonitrile to benzamide in aqueous acidDraw the acid-catalysed addition of water to benzonitrile through the nitrilium ion and the imidic acid, and the tautomerisation that gives benzamide.
- Saponification of phenyl acetateDraw the hydroxide addition and phenoxide loss that cleave phenyl acetate, and the two proton transfers that leave acetate and phenoxide, consuming two equivalents of hydroxide.
Reatividade na substituição acílica
- N-Acetilação do álcool 4-aminobenzílico com anidrido acéticoDesenhe a N-acetilação quimiosseletiva do álcool 4-aminobenzílico com anidrido acético, terminando com o acetato removendo o próton N–H.
- Acoplamento com DCC do ácido 3-fenilpropanoico com morfolinaDesenhe como a DCC converte o ácido 3-fenilpropanoico em uma O-acilisoureia e como a morfolina então forma a 4-(3-fenilpropanoil)morfolina com perda de diciclo-hexilureia.
- 1-Propanoilpirrolidina a partir de cloreto de propanoílaDesenhe a adição–eliminação da pirrolidina ao cloreto de propanoíla e a desprotonação pela trietilamina que fornece a 1-propanoilpirrolidina.
- Cloreto de butanoíla a partir de ácido butanoico e cloreto de tionilaDesenhe como o cloreto de tionila converte o ácido butanoico em cloreto de butanoíla por meio de um grupo abandonador clorossulfito que se decompõe em SO2 e cloreto.
Hidrólise em meio ácido e básico
- Hidrólise ácida da N-metilbenzamida a ácido benzoicoDesenhe a hidrólise da N-metilbenzamida a ácido benzoico catalisada por ácido, incluindo a protonação que transforma o grupo metilamino em grupo abandonador.
- Hidrólise básica da fenilacetonitrila a fenilacetatoDesenhe a hidrólise da fenilacetonitrila promovida por hidróxido, passando pela fenilacetamida, até fenilacetato e amônia.
- Abertura do anel da N-metilsuccinimida por hidróxidoDesenhe como um equivalente de hidróxido abre a N-metilsuccinimida a 4-(metilamino)-4-oxobutanoato e explique por que a amida formada sobrevive.
- Saponificação da γ-butirolactona com hidróxido de sódioDesenhe a abertura do anel da γ-butirolactona pelo hidróxido e as transferências de próton que terminam no 4-hidroxibutanoato.
Sais de diazônio
- Acoplamento azo do 4-nitrobenzenodiazônio com fenolDesenhe o acoplamento azo do fenóxido com o 4-nitrobenzenodiazônio na posição para e a rearomatização que fornece o (E)-4-[(4-nitrofenil)diazenil]fenol.
- Diazotação da 4-metoxianilina com ácido nitrosoDesenhe como o ácido nitroso e o HCl convertem a 4-metoxianilina em cloreto de 4-metoxibenzenodiazônio, passando pelo íon nitrosônio e por uma N-nitrosamina.
Preparo e uso de aminas
- Alquilação de Gabriel da ftalimida com bromoacetato de etilaDesenhe a desprotonação da ftalimida pelo carbonato e o ataque SN2 do ânion imideto ao bromoacetato de etila.
- Rearranjo de Hofmann da 3-metilbutanamida a isobutilaminaDesenhe o rearranjo de Hofmann da 3-metilbutanamida a 2-metilpropan-1-amina, da N-bromação, passando pelo isocianato, até a perda de CO2.
- Redução da piperidin-2-ona a piperidina com LiAlH4Desenhe a redução da lactama piperidin-2-ona a piperidina com LiAlH4, mostrando por que é o oxigênio, e não o nitrogênio, que sai do intermediário tetraédrico.
- Aminação redutiva da acetofenona com metilaminaDesenhe a formação do íon imínio a partir da acetofenona e da metilamina em ácido acético e sua redução pelo cianoboro-hidreto a N-metil-1-feniletan-1-amina.
Preveja o produto 8
Acyl substitution reactivity
- Acetylating 2-aminoethanolPredict which nucleophile of an amino alcohol is acylated first by an anhydride.
- Which conversion runs directly?Use the reactivity ladder to decide which acyl substitution occurs on simple mixing.
- Anhydride + aminePredict both products when an acid anhydride acylates a primary amine.
Hydrolysis in acid and base
- Which group does mild base cleave?Predict which acyl group of an ester–amide is hydrolysed by mild aqueous base and in what protonation state the product exists.
- Open a lactone with hydroxidePredict the ring-opened product of saponifying a lactone.
Diazonium salts
- Diazotisation then potassium iodidePredict the product when an arenediazonium salt is treated with potassium iodide and keep the other ring substituent unchanged.
- Product of a Sandmeyer reaction with CuCNPredict the product of a Sandmeyer cyanation and place the new group at the former diazonium carbon.
Making and using amines
- Hofmann elimination of pentan-2-aminePredict the major alkene from a Hofmann elimination.
Classificar 2
Making and using amines
- What the Gabriel synthesis cannot makeRecognise that the Gabriel synthesis needs an SN2-capable alkyl halide.
Amine basicity
- Which nitrogen of nicotine is protonated first?Identify the more basic nitrogen in a molecule with an sp³ amine and an sp² pyridine nitrogen.
Comparar 6
Acyl substitution reactivity
- Rank acetyl derivatives by reactivityOrder an amide, an ester, an anhydride and an acid chloride by their reactivity toward nucleophilic acyl substitution.
- Ester or Weinreb amide for a ketone?Compare how an ester and a Weinreb amide react with one equivalent of a Grignard reagent.
- Thioester or ester: which acylates faster?Explain why a thioester is a better acyl donor than the corresponding oxoester.
Amine basicity
- Rank four nitrogen compounds by basicityOrder an amide, an arylamine, a benzylic amine and an alkylamine by basicity from where the nitrogen lone pair is.
- Rank para-substituted anilines by basicityOrder para-substituted anilines by basicity from the resonance and inductive effects of the substituent.
- Rank four nitrogen heterocycles by basicityRank pyrrole, pyridine, imidazole and piperidine by basicity.
Calcular 5
Hydrolysis in acid and base
- Hydroxide consumed by 4-acetoxybenzoic acidCount every acidic proton and ester cleavage to find how much hydroxide a multifunctional compound consumes on saponification.
- Hydroxide for a triesterCount the hydroxide equivalents consumed by complete saponification of a triglyceride model.
Amine basicity
- pKb of methylamineConvert the pKa of an ammonium ion into the pKb of the amine.
- How much of an amine drug is protonated in blood?Use Henderson–Hasselbalch to find what percentage of an amine is protonated at a given pH.
- Neutral fraction of an amine at pH 7.8Calculate the percentage of an amine present as the neutral base at a given pH.
Encontre o erro 1
- Find the error in an acid hydrolysis mechanismIdentify the step in a proposed acid-catalysed ester hydrolysis that is impossible in the stated medium.
Interprete as evidências 3
Hydrolysis in acid and base
- Follow the labelled oxygen from waterUse acyl–oxygen cleavage to predict where an oxygen label from the solvent ends up after saponification.
Misto: nomenclatura, espectros
- Use o IV para distinguir uma amida primária de um ésterAtribua um espectro de IV a uma amida primária usando a posição da C=O e o número de estiramentos N–H.
- Identifique o grupo responsável por uma banda fina em 2250 cm⁻¹ no IVAtribua uma banda fina de IV em 2250 cm⁻¹ a um grupo nitrila C≡N.
Trabalhe de trás para frente 1
- Choose partners for a reductive aminationWork backward from a secondary amine to the carbonyl compound and amine that give it by reductive amination.
Planeje uma síntese 5
Making and using amines
- Choose the amide for a LiAlH₄ routeChoose the amide whose LiAlH₄ reduction gives a specified secondary amine, keeping track of which carbon becomes CH₂.
- Route to 2-phenylethan-1-amineChoose a route to a primary amine that also adds one carbon.
Diazonium salts
- Plan a route to 1,3,5-tribromobenzeneUse an amino group as a removable directing group in a multistep arene synthesis.
Hydrolysis in acid and base
- Conditions to hydrolyze an amideChoose conditions that actually hydrolyze an unactivated amide.
Sais de diazônio
- Escolha o reagente que forma um fluoreto de arilaEscolha a sequência de reagentes que substitui um grupo diazônio por flúor.
Explique por quê 9
Acyl substitution reactivity
- Why is an acid chloride so much more reactive than an amide?Explain the reactivity gap between an acid chloride and an amide from π donation and leaving-group ability.
Hydrolysis in acid and base
- Why amides resist hydroxideExplain why a tertiary amide survives conditions that saponify an ester.
- Why saponification goes to completionExplain why ester hydrolysis in base is irreversible while acid-catalysed hydrolysis is an equilibrium.
Diazonium salts
- Why aryl diazonium salts can be usedExplain why arenediazonium ions are useful intermediates while alkyldiazonium ions are not.
Amine basicity
- Why DMAP is a much stronger base than pyridineExplain how resonance donation from a para amino group raises the basicity of a pyridine nitrogen.
- Why trimethylamine breaks the trend in waterExplain the basicity order of the methylamines in water from solvation of their ammonium ions.
Making and using amines
- Why the Gabriel synthesis stops at one alkyl groupExplain how the phthalimide nitrogen prevents the over-alkylation seen with ammonia.
Sais de diazônio
- Por que a diazotação é feita a 0–5 °CExplique por que os sais de arenodiazônio são preparados e mantidos a 0–5 °C.
Preparo e uso de aminas
- Por que a aminação redutiva supera a alquilação no preparo de aminasExplique por que a aminação redutiva fornece uma amina secundária limpa, enquanto a alquilação direta com um haleto de alquila não fornece.
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