Table of contents

1. A Review of General Chemistry5h 5m
2. Molecular Representations1h 14m
3. Acids and Bases2h 46m
4. Alkanes and Cycloalkanes4h 17m
5. Chirality3h 39m
6. Thermodynamics and Kinetics1h 22m
7. Substitution Reactions1h 48m
8. Elimination Reactions2h 30m
9. Alkenes and Alkynes2h 9m
10. Addition Reactions3h 19m
11. Radical Reactions1h 58m
12. Alcohols, Ethers, Epoxides and Thiols2h 42m
13. Alcohols and Carbonyl Compounds2h 17m
14. Synthetic Techniques1h 26m
15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
16. Conjugated Systems6h 13m
17. Ultraviolet Spectroscopy51m
18. Aromaticity2h 34m
19. Reactions of Aromatics: EAS and Beyond5h 1m
20. Phenols55m
21. Aldehydes and Ketones: Nucleophilic Addition4h 56m
22. Carboxylic Acid Derivatives: NAS2h 51m
23. The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides1h 10m
24. Enolate Chemistry: Reactions at the Alpha-Carbon1h 53m
25. Condensation Chemistry2h 9m
26. Amines1h 43m
27. Heterocycles2h 0m
28. Carbohydrates5h 53m
29. Amino Acids4h 20m
30. Peptides and Proteins2h 42m
31. Catalysis in Organic Reactions1h 30m
32. Lipids 2h 50m
33. The Organic Chemistry of Metabolic Pathways2h 52m
34. Nucleic Acids1h 32m
35. Transition Metals6h 14m
36. Synthetic Polymers1h 49m

28. Carbohydrates

Monosaccharides - Forming Cyclic Hemiacetals

Organic Chemistry

28. Carbohydrates

Monosaccharides - Forming Cyclic Hemiacetals

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3:59 minutes

Problem 62a

Textbook Question

Two structures for the sugar glucose are shown on page 914. Interconversion of the open-chain and cyclic hemiacetal forms is catalyzed by either acid or base.
(a) Propose a mechanism for the cyclization, assuming a trace of acid is present.

Verified step by step guidance

Step 1: Understand the problem. The question involves two parts: (a) proposing a mechanism for the cyclization of glucose into its cyclic hemiacetal form in the presence of acid, and (b) determining whether the open-chain form of glucose can reduce Tollens' reagent and explaining why.

Step 2: For part (a), recall that the cyclization of glucose involves the reaction of the aldehyde group (in the open-chain form) with a hydroxyl group on the same molecule to form a hemiacetal. In the presence of a trace of acid, the mechanism involves protonation of the carbonyl oxygen, making the carbonyl carbon more electrophilic. This facilitates nucleophilic attack by the hydroxyl group. Write the mechanism step-by-step: (1) Protonation of the carbonyl oxygen, (2) Nucleophilic attack by the hydroxyl group, (3) Formation of the cyclic structure, and (4) Deprotonation to yield the cyclic hemiacetal.

Step 3: For part (b), recall that Tollens' reagent is a mild oxidizing agent that reacts with aldehydes to form carboxylic acids, producing a silver mirror as a positive test. Even though the cyclic hemiacetal form of glucose is more stable and predominant at equilibrium, the open-chain form is in equilibrium with the cyclic form. This means that a small amount of the open-chain form is always present in solution. Since the open-chain form contains an aldehyde group, it can reduce Tollens' reagent.

Step 4: Explain why the equilibrium between the cyclic and open-chain forms allows the Tollens test to be positive. The interconversion between the two forms ensures that as the open-chain form reacts with Tollens' reagent, more of the cyclic form converts to the open-chain form to maintain equilibrium. This continuous supply of the open-chain form allows the reaction with Tollens' reagent to proceed.

Step 5: Summarize the key points: (a) The mechanism for cyclization involves acid-catalyzed protonation, nucleophilic attack, and deprotonation to form the cyclic hemiacetal. (b) An aqueous solution of glucose will give a positive Tollens test because the open-chain form, which contains the aldehyde group, is in equilibrium with the cyclic form and can reduce Tollens' reagent.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Cyclization of Sugars

Cyclization refers to the process where an open-chain form of a sugar molecule, like glucose, forms a cyclic structure. This occurs when a hydroxyl group reacts with the carbonyl group, leading to the formation of a hemiacetal. In the presence of acid, this reaction is facilitated as the acid protonates the carbonyl oxygen, making it more electrophilic and promoting the nucleophilic attack by the hydroxyl group.

Stability of Cyclic Hemiacetals

Cyclic hemiacetals, such as the cyclic form of glucose, are generally more stable than their open-chain counterparts due to the formation of a five- or six-membered ring, which minimizes steric strain and allows for favorable intramolecular interactions. This stability means that at equilibrium, the concentration of the cyclic form is significantly higher than that of the open-chain form, influencing the reactivity of the sugar in various chemical tests.

Tollens' Test

Tollens' test is a qualitative test used to identify aldehydes, which can reduce silver ions in Tollens' reagent to metallic silver. In the case of glucose, although the cyclic form is predominant, it can still interconvert to the open-chain form, which contains an aldehyde group. However, if the equilibrium favors the cyclic form significantly, the amount of open-chain glucose available to reduce Tollens' reagent may be insufficient, potentially leading to a negative test result.

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Textbook Question

4-Hydroxy- and 5-hydroxyaldehydes exist primarily as cyclic hemiacetals. Draw the structure of the cyclic hemiacetal formed by each of the following:

c. 5-hydroxypentanal

Textbook Question

4-Hydroxy- and 5-hydroxyaldehydes exist primarily as cyclic hemiacetals. Draw the structure of the cyclic hemiacetal formed by each of the following:

d. 4-hydroxyheptanal

Textbook Question

Draw the structures of the compound methyl α-D-galactopyranoside.

Allose is the C3 epimer of glucose, and ribose is the C2 epimer of arabinose.

Textbook Question

Draw the structures of the compound ethyl β-D-ribofuranoside.

Allose is the C3 epimer of glucose, and ribose is the C2 epimer of arabinose.

Textbook Question

Two structures for the sugar glucose are shown on page 914. Interconversion of the open-chain and cyclic hemiacetal forms is catalyzed by either acid or base.

(b) The cyclic hemiacetal is more stable than the open-chain form, so very little of the open-chain form is present at equilibrium. Will an aqueous solution of glucose reduce Tollens reagent and give a positive Tollens test? Explain.

Textbook Question

Draw the structures of the compound α-D-allopyranose.

Allose is the C3 epimer of glucose, and ribose is the C2 epimer of arabinose.

Textbook Question

Without referring to the chapter, draw the chair conformations of

(a) β-D-mannopyranose (the C2 epimer of glucose).

Textbook Question

Without referring to the chapter, draw the chair conformations of

(b) α-D-allopyranose (the C3 epimer of glucose).

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