Química Orgânica 2 · Condensação de Claisen e adição conjugada
One pot, two Mannich reactions: tropinone
Condições: 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.
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O mecanismo, passo a passo
Próximo mostra as setas curvas de cada passo e as explica. Os passos escritos estão abaixo.
Ler todos os 22 passos escritos
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.
As setas curvas
- The amine lone pair bonds to an aldehyde carbon.
- The carbonyl π pair moves onto oxygen.
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.
As setas curvas
- Water accepts one nitrogen proton.
- The N–H pair returns to nitrogen.
- The alkoxide accepts a proton from the same water relay.
- The water O–H pair stays on its oxygen.
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.
As setas curvas
- The hemiaminal oxygen accepts a hydronium proton.
- Hydronium gives its O–H pair back to oxygen.
- The nitrogen lone pair forms the C=N π bond.
- The C–O pair leaves with oxygen as water.
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.
As setas curvas
- Water accepts the drawn α proton.
- The α C–H pair forms the enol C=C bond.
- The carbonyl π pair forms an O–H bond to the relay proton.
- The water O–H pair stays on water oxygen.
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.
As setas curvas
- The enol π pair bonds its α carbon to the iminium carbon.
- The iminium π pair returns to nitrogen.
- The enol oxygen lone pair rebuilds the carbonyl.
Recover the ketone after the first Mannich reaction
The secondary amine and ketone are neutral again; the explicit water has regenerated hydronium.
As setas curvas
- Water accepts the protonated ketone oxygen proton.
- The O–H pair stays on the ketone oxygen.
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.
As setas curvas
- The secondary amine lone pair attacks the remaining aldehyde carbon.
- The aldehyde π pair moves onto oxygen.
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.
As setas curvas
- Water accepts the remaining nitrogen proton.
- The N–H pair restores the nitrogen lone pair.
- The alkoxide accepts the proton carried by the water relay.
- The relay O–H pair returns to water oxygen.
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.
As setas curvas
- The hemiaminal hydroxyl accepts the hydronium proton.
- The hydronium O–H pair returns to oxygen.
- The nitrogen lone pair forms the second iminium π bond.
- The C–O pair leaves with the second water molecule.
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.
As setas curvas
- Water accepts the α proton on the unlinked side of the ketone.
- The C–H pair forms the second enol π bond.
- The carbonyl π pair forms the enol O–H bond.
- The relay O–H pair stays on water oxygen.
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.
As setas curvas
- The enol π pair forms the second Mannich carbon–carbon bond.
- The iminium π pair returns to nitrogen.
- The enol oxygen lone pair rebuilds the ketone π bond.
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.
As setas curvas
- Water accepts the protonated carbonyl oxygen proton.
- The O–H pair returns to the ketone oxygen.
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.
As setas curvas
- The ketone π pair forms an O–H bond to the first acid proton.
- The acid O–H pair forms the second carbon dioxide π bond.
- The breaking carbon–carbon σ pair forms the enol C=C bond.
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.
As setas curvas
- The enol oxygen lone pair rebuilds the carbonyl π bond.
- The enol π pair bonds the α carbon to a hydronium proton.
- The hydronium O–H pair returns to oxygen.
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.
As setas curvas
- Water takes the proton from the positively charged ketone oxygen.
- The O–H pair stays on the ketone oxygen.
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.
As setas curvas
- The ketone π pair accepts the remaining carboxylic acid proton.
- The acid O–H pair completes the second carbon dioxide molecule.
- The second breaking carbon–carbon σ pair forms an enol π bond.
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.
As setas curvas
- The enol oxygen lone pair forms the final carbonyl π bond.
- The enol π pair takes a hydronium proton at carbon.
- The hydronium O–H pair stays on oxygen.
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.
As setas curvas
- Water accepts the final carbonyl oxygen proton.
- The O–H pair stays on the ketone oxygen.
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.
O que você vai praticar
Build the tropane bicycle from three starting molecules through two hemiaminal–iminium sequences, two enol Mannich reactions and two β-keto-acid decarboxylations with explicit proton accounting.
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