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Réaction d'aldol croisée utilisant des bases faibles
Cette leçon traite de la réaction aldolique croisée utilisant des bases faibles. L'autocondensation d'un aldéhyde ayant de l'hydrogène α est évitée en…
Rappelons qu’une réaction efficace d’aldol croisé nécessite que l’un des deux composés carbonyles uniques réactifs soit dépourvu d’un α hydrogène, comme illustré dans la réaction du formaldéhyde avec un autre aldéhyde contenant un α-hydrogène.
L’auto-condensation de l’aldéhyde comprenant un α hydrogène est empêchée en l’ajoutant lentement au mélange réactionnel de formaldéhyde et d’une base faible.
Rappelez-vous que les hydroxydes et les alcoxydes avec des valeurs pKa relativement plus petites fonctionnent comme des bases faibles.
Dans cette réaction, l’ion hydroxyde déprotone l’hydrogène α de l’aldéhyde pour former un énolate nucléophile, qui attaque le formaldéhyde pour produire un seul produit d’aldol croisé.
De même, dans une solution d’un β-cétoester et d’une cétone, le β-cétoester avec un pKa plus faible est plus acide. Par conséquent, l’éthoxyde de sodium, une base faible, déprotone préférentiellement le β-cétoester, qui réagit ensuite avec la cétone pour former un seul produit à bon rendement.
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Q1: Why is an aldehyde with an α hydrogen added slowly in a crossed aldol reaction with formaldehyde?
Slow addition prevents self-condensation of the aldehyde containing an α hydrogen. By adding it gradually to a mixture of formaldehyde and weak base, the base preferentially deprotonates the formaldehyde's α carbon first, forming an enolate that attacks the slowly added aldehyde, yielding a single crossed aldol product rather than multiple byproducts.
Q2: What role do weak bases play in crossed aldol reactions?
Weak bases like hydroxide and alkoxide ions deprotonate the α hydrogen of carbonyl compounds to form nucleophilic enolates. These bases have relatively smaller pKa values, allowing them to selectively deprotonate the more acidic compound in a mixture, ensuring formation of a single enolate that attacks the electrophilic carbonyl partner.
Q3: How does pKa determine which compound reacts first in a crossed aldol with a β-ketoester and ketone?
The β-ketoester has a lower pKa than the ketone, making it more acidic. Sodium ethoxide, a weak base, preferentially deprotonates the more acidic β-ketoester to form its enolate. This enolate then attacks the ketone electrophile, producing a single crossed aldol product in good yield.
Q4: What is the difference between using formaldehyde and other aldehydes in crossed aldol reactions?
Formaldehyde lacks an α hydrogen, making it unable to undergo self-condensation or form an enolate. This property makes it an ideal electrophile in crossed aldol reactions. When paired with an aldehyde containing an α hydrogen, formaldehyde ensures selective formation of one crossed product without competing self-condensation pathways.
Q5: What is an enolate and how does it form in weak base-catalyzed aldol reactions?
An enolate is a nucleophilic carbanion formed when a weak base deprotonates the α hydrogen of a carbonyl compound. The resulting negative charge is stabilized by resonance with the carbonyl π system. In crossed aldol reactions, the enolate attacks the electrophilic carbonyl carbon of the second reactant, forming a new carbon-carbon bond.
Q6: Why does sodium ethoxide selectively deprotonate the β-ketoester over the ketone?
Sodium ethoxide is a weak base that selectively deprotonates the more acidic compound. The β-ketoester possesses a lower pKa than a simple ketone due to the electron-withdrawing ester group, which stabilizes the resulting enolate through resonance. This acidity difference ensures selective enolate formation from the β-ketoester.
Q7: What outcome results from using weak bases instead of strong bases in crossed aldol reactions?
Weak bases enable selective deprotonation based on relative acidity, allowing control over which carbonyl compound forms the enolate. This selectivity prevents unwanted self-condensation and side reactions, yielding a single crossed aldol product in good yield. Strong bases would deprotonate both compounds non-selectively, producing multiple products.