When undergoes intramolecular aldol condensation, the major product formed is:
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Aldol condensations in the intramolecular form lead to the formation of cyclic products, often resulting in rings such as lactones or lactams depending on the structure of the reactants. Pay attention to the position of $\alpha$-hydrogens and the size of the cyclic product.
To determine the major product formed by the intramolecular aldol condensation of the given compound, we first need to understand the reaction mechanism involved.
Intramolecular Aldol Condensation: In this type of condensation, a molecule containing two carbonyl groups (usually a ketone and an aldehyde, or two aldehydes) undergoes a reaction to form a cyclic compound. The reaction involves the formation of an enolate ion, which then attacks the carbonyl carbon of the other group, followed by dehydration to form a double bond.
Identify the Enolizable Hydrogen: The compound shown in the image is a 1,5-diketone. It has enolizable hydrogens adjacent to each carbonyl group.
Formation of Enolate: The enolate ion can be formed at either end, but the formation of a stable 5 or 6-membered ring is preferred in intramolecular reactions.
Attack on Carbonyl Group: The enolate ion formed will attack the carbonyl group, leading to the formation of a 5 or 6-membered ring structure.
Dehydration: Finally, water is eliminated (dehydration) to form an α,β-unsaturated carbonyl compound.
Formation of the Product: In this specific reaction, a 5-membered ring is formed as it is more stable. The reaction proceeds through the following steps:
Formation of enolate ion at the alpha position next to one of the carbonyl groups.
Intramolecular nucleophilic attack of the enolate on the carbonyl carbon of the other end, forming a 5-membered ring.
Dehydration to form the double bond and yield the α,β-unsaturated ketone or aldehyde product.
The major product formed is a cyclic α,β-unsaturated carbonyl compound with a 5-membered ring. This corresponds to the option shown in the image above.
Thus, the correct answer is the third option.
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Intramolecular aldol condensation involves the reaction of a compound with both aldehyde or ketone groups present in the same molecule, leading to the formation of a cyclic product. The reaction proceeds via the formation of an enolate ion, which then attacks the carbonyl group, followed by dehydration to form a double bond.
The given compound is:
This compound contains two carbonyl groups, one ketone and one aldehyde. Intramolecular aldol condensation typically forms five or six-membered rings, as these are more stable. Let's see the step-by-step process:
Formation of Enolate: The α-hydrogen (next to the ketone carbonyl) is deprotonated to form an enolate ion.
Nucleophilic Attack: The enolate ion attacks the carbonyl carbon of the aldehyde group, forming a new carbon-carbon bond and resulting in a cyclic β-hydroxy ketone intermediate.
Dehydration: The β-hydroxy ketone undergoes dehydration (loss of water) to form an α,β-unsaturated ketone, which is more stable due to conjugation.
The major product formed is a cyclic compound, depicted in the image below:
This product is the result of an intramolecular aldol condensation leading to cyclization with the formation of a five-membered ring, which is commonly favored in such reactions.
undergoes intramolecular aldol condensation, the major product formed is:
