Help with Aldol Condensation - Organic Chemistry
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Determine the major product of the given intramolecular aldol reaction.
Determine the major product of the given intramolecular aldol reaction.
Keep in mind the following principles: Cyclization is favored when a five/six-member ring may be formed. Addition at an aldehyde is favored relative to the same reaction at a ketone.
As a result, abstraction of a hydrogen bound to carbon 6 (an alpha-carbon) is favored since the resulting carbanion may attack the aldehyde (carbon 1) to form a six-member ring, resulting in compound II. Compound I results from abstracting a hydrogen from carbon 2, generating a carbanion which may then attack the ketone. Based on the latter of the above principles, this is a minor product.
Keep in mind the following principles: Cyclization is favored when a five/six-member ring may be formed. Addition at an aldehyde is favored relative to the same reaction at a ketone.
As a result, abstraction of a hydrogen bound to carbon 6 (an alpha-carbon) is favored since the resulting carbanion may attack the aldehyde (carbon 1) to form a six-member ring, resulting in compound II. Compound I results from abstracting a hydrogen from carbon 2, generating a carbanion which may then attack the ketone. Based on the latter of the above principles, this is a minor product.
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What is the final organic product of the reaction shown?
What is the final organic product of the reaction shown?
First step: Friedel-Crafts acylation of benzene
Second step: Formation of enolate
Third step: aldol addition (enolate attacks carbonyl carbon in benzaldehyde)
Fourth step: neutralization of anion and dehydration forming alkene
First step: Friedel-Crafts acylation of benzene
Second step: Formation of enolate
Third step: aldol addition (enolate attacks carbonyl carbon in benzaldehyde)
Fourth step: neutralization of anion and dehydration forming alkene
Compare your answer with the correct one above
Determine the major product of the given intramolecular aldol reaction.
Determine the major product of the given intramolecular aldol reaction.
Keep in mind the following principles: Cyclization is favored when a five/six-member ring may be formed. Addition at an aldehyde is favored relative to the same reaction at a ketone.
As a result, abstraction of a hydrogen bound to carbon 6 (an alpha-carbon) is favored since the resulting carbanion may attack the aldehyde (carbon 1) to form a six-member ring, resulting in compound II. Compound I results from abstracting a hydrogen from carbon 2, generating a carbanion which may then attack the ketone. Based on the latter of the above principles, this is a minor product.
Keep in mind the following principles: Cyclization is favored when a five/six-member ring may be formed. Addition at an aldehyde is favored relative to the same reaction at a ketone.
As a result, abstraction of a hydrogen bound to carbon 6 (an alpha-carbon) is favored since the resulting carbanion may attack the aldehyde (carbon 1) to form a six-member ring, resulting in compound II. Compound I results from abstracting a hydrogen from carbon 2, generating a carbanion which may then attack the ketone. Based on the latter of the above principles, this is a minor product.
Compare your answer with the correct one above
What is the final organic product of the reaction shown?
What is the final organic product of the reaction shown?
First step: Friedel-Crafts acylation of benzene
Second step: Formation of enolate
Third step: aldol addition (enolate attacks carbonyl carbon in benzaldehyde)
Fourth step: neutralization of anion and dehydration forming alkene
First step: Friedel-Crafts acylation of benzene
Second step: Formation of enolate
Third step: aldol addition (enolate attacks carbonyl carbon in benzaldehyde)
Fourth step: neutralization of anion and dehydration forming alkene
Compare your answer with the correct one above
Determine the major product of the given intramolecular aldol reaction.
Determine the major product of the given intramolecular aldol reaction.
Keep in mind the following principles: Cyclization is favored when a five/six-member ring may be formed. Addition at an aldehyde is favored relative to the same reaction at a ketone.
As a result, abstraction of a hydrogen bound to carbon 6 (an alpha-carbon) is favored since the resulting carbanion may attack the aldehyde (carbon 1) to form a six-member ring, resulting in compound II. Compound I results from abstracting a hydrogen from carbon 2, generating a carbanion which may then attack the ketone. Based on the latter of the above principles, this is a minor product.
Keep in mind the following principles: Cyclization is favored when a five/six-member ring may be formed. Addition at an aldehyde is favored relative to the same reaction at a ketone.
As a result, abstraction of a hydrogen bound to carbon 6 (an alpha-carbon) is favored since the resulting carbanion may attack the aldehyde (carbon 1) to form a six-member ring, resulting in compound II. Compound I results from abstracting a hydrogen from carbon 2, generating a carbanion which may then attack the ketone. Based on the latter of the above principles, this is a minor product.
Compare your answer with the correct one above
What is the final organic product of the reaction shown?
What is the final organic product of the reaction shown?
First step: Friedel-Crafts acylation of benzene
Second step: Formation of enolate
Third step: aldol addition (enolate attacks carbonyl carbon in benzaldehyde)
Fourth step: neutralization of anion and dehydration forming alkene
First step: Friedel-Crafts acylation of benzene
Second step: Formation of enolate
Third step: aldol addition (enolate attacks carbonyl carbon in benzaldehyde)
Fourth step: neutralization of anion and dehydration forming alkene
Compare your answer with the correct one above
Determine the major product of the given intramolecular aldol reaction.
Determine the major product of the given intramolecular aldol reaction.
Keep in mind the following principles: Cyclization is favored when a five/six-member ring may be formed. Addition at an aldehyde is favored relative to the same reaction at a ketone.
As a result, abstraction of a hydrogen bound to carbon 6 (an alpha-carbon) is favored since the resulting carbanion may attack the aldehyde (carbon 1) to form a six-member ring, resulting in compound II. Compound I results from abstracting a hydrogen from carbon 2, generating a carbanion which may then attack the ketone. Based on the latter of the above principles, this is a minor product.
Keep in mind the following principles: Cyclization is favored when a five/six-member ring may be formed. Addition at an aldehyde is favored relative to the same reaction at a ketone.
As a result, abstraction of a hydrogen bound to carbon 6 (an alpha-carbon) is favored since the resulting carbanion may attack the aldehyde (carbon 1) to form a six-member ring, resulting in compound II. Compound I results from abstracting a hydrogen from carbon 2, generating a carbanion which may then attack the ketone. Based on the latter of the above principles, this is a minor product.
Compare your answer with the correct one above
What is the final organic product of the reaction shown?
What is the final organic product of the reaction shown?
First step: Friedel-Crafts acylation of benzene
Second step: Formation of enolate
Third step: aldol addition (enolate attacks carbonyl carbon in benzaldehyde)
Fourth step: neutralization of anion and dehydration forming alkene
First step: Friedel-Crafts acylation of benzene
Second step: Formation of enolate
Third step: aldol addition (enolate attacks carbonyl carbon in benzaldehyde)
Fourth step: neutralization of anion and dehydration forming alkene
Compare your answer with the correct one above