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Ch. 23 - Carbohydrates and Nucleic Acids
Wade - Organic Chemistry 9th Edition
Wade9th EditionOrganic ChemistryISBN: 9780135213728당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Wade 9th EditionCh. 23 - Carbohydrates and Nucleic Acids문제 19a
23장, 문제 19a

Except for the Tollens test, basic aqueous conditions are generally avoided with sugars because they lead to fast isomerizations.
(a) Under basic conditions, the proton alpha to the aldehyde (or ketone) carbonyl group is reversibly removed, and the resulting enolate ion is no longer asymmetric. Reprotonation can occur on either face of the enolate, giving either the original structure or its epimer. Because a mixture of epimers results, this process is called epimerization. Propose a mechanism for the base-catalyzed equilibration of glucose to a mixture of glucose and its C2 epimer, mannose.

검증된 단계별 안내
1
Step 1: Recognize the key concept of epimerization. Under basic conditions, the alpha proton (a hydrogen atom attached to the carbon adjacent to the carbonyl group) is acidic and can be removed by a base, forming an enolate ion. This enolate ion is planar and can be reprotonated on either face, leading to the formation of two different stereoisomers (epimers).
Step 2: Identify the starting molecule, glucose, and its C2 epimer, mannose. Glucose and mannose differ only in the stereochemistry at the C2 carbon. In glucose, the hydroxyl group at C2 is on the right (D-configuration), while in mannose, it is on the left.
Step 3: Write the mechanism for the base-catalyzed reaction. First, the base (e.g., OH⁻) abstracts the alpha proton from the carbon adjacent to the aldehyde group in glucose, forming an enolate ion. Represent this step using a curved arrow mechanism to show the movement of electrons.
Step 4: Show the planar enolate intermediate. The enolate ion is no longer asymmetric, as the sp² hybridized carbon is planar. This allows for reprotonation to occur on either face of the enolate, leading to two possible products: the original glucose or its epimer, mannose.
Step 5: Illustrate the reprotonation step. Use curved arrows to show the addition of a proton (H⁺) to the enolate ion. Reprotonation on one face regenerates glucose, while reprotonation on the opposite face forms mannose. This results in an equilibrium mixture of glucose and mannose under basic conditions.

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주요 개념

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Enolate Ion Formation

Enolate ions are formed when a proton is removed from the alpha carbon of a carbonyl compound, creating a resonance-stabilized anion. This process is crucial in organic reactions, particularly in aldol reactions and isomerizations, as it allows for the formation of new bonds. The enolate can react with electrophiles or undergo reprotonation, leading to different structural outcomes.
추천 영상:
02:26
Formation of Enolates

Epimerization

Epimerization is the process by which one epimer is converted into another through the reversible interconversion of stereocenters. In the context of sugars, this occurs when the configuration at a specific carbon atom (such as C2 in glucose) is altered, resulting in a mixture of epimers. This reaction is significant in carbohydrate chemistry as it affects the properties and reactivity of sugars.
추천 영상:
02:57
General Reaction

Base-Catalyzed Equilibration

Base-catalyzed equilibration refers to the process where a base facilitates the conversion of one isomer to another, often through the formation of an enolate intermediate. In the case of sugars, this mechanism allows for the interconversion between different anomeric forms or epimers under basic conditions. Understanding this concept is essential for predicting the outcomes of reactions involving carbohydrates.
추천 영상:
01:57
Base Catalyzed
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교과서 질문

Except for the Tollens test, basic aqueous conditions are generally avoided with sugars because they lead to fast isomerizations.

(b) Propose a mechanism for the isomerization of a ketose to an aldose, via the enediol intermediate, shown immediately above. Note that the enediol has two –OH protons, and removing one or the other gives two different enolate ions.

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교과서 질문

Which of the following are reducing sugars? Comment on the common name sucrose for table sugar.

(e)

(f)

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교과서 질문

Draw and name the products of bromine water oxidation of

(a) D-mannose

(b) D-galactose

(c) D-fructose

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교과서 질문

Two sugars, A and B, are known to be glucose and galactose, but it is not certain which one is which. On treatment with nitric acid, A gives an optically inactive aldaric acid, while B gives an optically active aldaric acid. Which sugar is glucose, and which is galactose?

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교과서 질문

Draw and name the products of nitric acid oxidation of

(a) D-mannose

(b) D-galactose

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교과서 질문

Which of the following are reducing sugars? Comment on the common name sucrose for table sugar.

(c) α-D-allopyranose

(d) ethyl β-D-ribofuranoside

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