Síntese da tropinona de Robinson: faça você mesmo o movimento de elétrons
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Na síntese de Robinson de 1917, o sucinodialdeído, a metilamina e o ácido acetona-dicarboxílico se organizam em tropinona em água, perto de pH 5. Duas reações de Mannich unem enóis a carbonos de imínio para construir a bicicleta em ponte; o aquecimento conforme necessário remove os dois grupos carboxila como CO₂, deixando a cetona central intacta.
Desenhe todas as 54 setas nas 18 cenas de reação acima, com três perguntas pelo caminho. Cada seta é verificada à medida que você a desenha, e a Dica revela o próximo movimento aos poucos.
Transforme suas próprias anotações em exercícios como este
Com o plano estudante do Arrow Lab, você pode fotografar uma reação das suas anotações de aula ou do seu livro-texto. O Arrow Lab lê os materiais de partida e os reagentes, você confere essa leitura, e ele monta o mecanismo como um exercício igual ao de cima: você desenha cada seta, recebe dicas quando travar, e vê na hora se cada movimento está correto.
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A síntese da tropinona de Robinson, etapa por etapa
A estratégia da dupla Mannich constrói duas ligações C–C ao adicionar um enol neutro a um íon imínio, primeiro entre moléculas e depois dentro da cadeia já conectada. O ramo desenhado usa o diácido ceto livre: cada grupo β-cetoácido perde CO₂ por meio de um deslocamento cíclico de elétrons, gerando um enol que deve retornar à cetona. Algumas transferências de próton e desidratações são agrupadas em cenas, e não em eventos elementares isolados. O desenho final identifica a tropinona como base livre neutra, embora seu nitrogênio esteja predominantemente protonado na mistura de pH 5.
As 22 etapas da demonstração, com as palavras do próprio exercício. Tente desenhar cada uma acima antes de lê-las.
Choose the first carbonyl
The dialdehyde supplies the two future bridgeheads; the central ketone is retained in tropinone.
Add methylamine to an aldehyde
A small equilibrium population of free methylamine attacks even though much of the amine is protonated near pH 5. The tetrahedral adduct has ammonium and alkoxide charges, not a neutral nitrogen with four bonds.
Relay a proton to make the hemiaminal
A solvent relay exchanges an ammonium proton for a water proton, giving a neutral hemiaminal. No direct, unsupported nitrogen-to-oxygen proton jump is required.
Dehydrate to the first iminium
Acid activation and water loss are grouped into one scene; the short-lived protonated hemiaminal is not isolated. The iminium nitrogen has four bond orders and a positive charge, making its carbon a strong electrophile.
Choose the carbon nucleophile
Carboxyl groups can be ionized in the real mixture, but that does not make the reacting carbon an enolate; the drawn reactive branch uses the free diacid and its neutral enol.
Enolize the keto diacid
This water-mediated proton relay summarizes keto–enol equilibration, not an isolated enolate or a claim about its rate-limiting elementary step. The two sides of the starting keto diacid are equivalent.
Make the first Mannich carbon–carbon bond
The first Mannich reaction joins two starting molecules. Positive charge moves from nitrogen to the protonated ketone oxygen; the connected chain is redrawn on the future tropane template without changing any other bonds.
Recover the ketone after the first Mannich reaction
The secondary amine and ketone are neutral again; the explicit water has regenerated hydronium.
Capture the remaining aldehyde intramolecularly
The tethered amine closes a five-membered nitrogen-containing ring. It still carries one hydrogen, so this tetrahedral adduct is an ammonium alkoxide, not yet a tertiary amine.
Relay the remaining nitrogen proton
The second neutral hemiaminal now has a tertiary nitrogen with three carbon bonds and one lone pair. The solvent relay conserves every explicit hydrogen.
Dehydrate to the cyclic iminium
Again, hydroxyl protonation and water departure are grouped. Nitrogen is positively charged; the remaining side of the ketone must now become the carbon nucleophile.
Enolize the other side of the ketone
The same net solvent relay supplies the enol on the side that can reach the cyclic iminium. Enolization elsewhere is reversible; this branch is trapped by formation of the bicyclic framework.
Close the tropane bicycle by a second Mannich reaction
The intramolecular Mannich reaction completes the 8-azabicyclo[3.2.1]octane skeleton. Both nitrogen-bridge bonds point toward the viewer and the bridgehead hydrogens point away; the temporary carboxyl-bearing stereocentres are not assigned.
Neutralize the bicyclic ketone oxygen
The bicycle is complete, but the two extra carboxyl groups remain. Each is part of a β-keto acid and can leave as carbon dioxide after the reacting carboxyl group is in its free-acid form.
Why a β-keto acid?
The free diacid is drawn for this branch; real carboxylate populations must equilibrate with the reactive acid before this specific proton-coupled pathway.
Release the first carbon dioxide
Three electron pairs move around a six-atom cyclic arrangement. Carbon–carbon cleavage releases neutral CO₂ and produces an enol, not the ketone directly; the carboxyl carbon and both of its oxygens remain visible.
Protonate the first decarboxylation enol at carbon
Carbon protonation replaces the lost carboxyl substituent with hydrogen and gives a protonated ketone. The other carboxyl group must wait until this carbonyl is restored.
Restore the ketone before the second CO₂ loss
This monoacid is still a β-keto acid. Restoring the ketone supplies the proton acceptor and electron sink for the second decarboxylation.
Release the second carbon dioxide
The second β-keto acid undergoes the same cyclic fragmentation. Both auxiliary carboxyl carbons have now left as CO₂, while all seven carbons of the bicyclic skeleton remain.
Protonate the final enol at carbon
The last missing carbon–hydrogen bond forms as the enol becomes a protonated ketone. Neither tautomerization changes either bridgehead configuration.
Regenerate acid and reveal tropinone
Hydronium is regenerated and the ketone is neutral. The free-base drawing identifies the covalent product; in the pH 5 reaction mixture its tertiary nitrogen is predominantly protonated.
Tropinone: two rings from three starting molecules
The product is (1R,5S)-tropinone, 8-methyl-8-azabicyclo[3.2.1]octan-3-one. Despite its two bridgehead stereocentres, tropinone is achiral (meso overall), not a resolved enantiomer: the symmetric bridges relate the two centres. Wedges place both nitrogen-bridge bonds toward the viewer and hashes place the bridgehead hydrogens away. Two Mannich carbon–carbon bonds and two nitrogen–carbon bonds assembled the bicycle; two aldehyde oxygens left in two waters and the two carboxyl groups left as two CO₂ molecules. The solvent water and hydronium are retained for explicit proton accounting. The neutral free base is shown for identification, not as the predominant protonation state at pH 5.
Como o mecanismo é desenhado
Robinson’s tropinone synthesis (1917): succinaldehyde, methylamine and acetonedicarboxylic acid (the free diacid, not its calcium salt or an ester), in water near pH 5, with warming for decarboxylation as needed. The drawn branch uses reactive neutral amine and free-acid species, not the dominant solution speciation. Water and one hydronium ion enter only when they transfer protons; its spectator counterion is omitted. Water relays summarize keto–enol equilibration, and hydroxyl protonation plus dehydration share a scene; these are bookkeeping groupings, not assertions of elementary kinetics. Reversible hydration, competing condensations and temporary carboxyl-bearing stereochemistry are not followed. The final free base is drawn for identification; nitrogen protonation in the acidic solution is not a separate step. The linked heavy-atom 3D panel uses seven milestone shapes and shows the nearest milestone, not every intermediate.