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

Previous problemNext problem

1:07 minutes

Problem 62b

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.

Verified step by step guidance

Understand the Tollens test: The Tollens test is used to identify aldehydes. Tollens reagent contains [Ag(NH₃)₂]⁺, which oxidizes aldehydes to carboxylic acids, reducing Ag⁺ to metallic silver in the process.

Analyze the structure of glucose: Glucose exists in equilibrium between its open-chain form (which contains an aldehyde group) and its cyclic hemiacetal forms (α and β anomers). The open-chain form is the reactive species for the Tollens test.

Consider the equilibrium: Although the cyclic hemiacetal forms are more stable and predominate at equilibrium, a small amount of the open-chain form is always present in solution due to the dynamic equilibrium between the forms.

Determine the reactivity: The presence of the open-chain form, even in small amounts, means that glucose has an aldehyde group available to react with Tollens reagent. This aldehyde group can be oxidized, leading to a positive Tollens test.

Conclude the explanation: Since the open-chain form of glucose is in equilibrium with the cyclic forms, and the aldehyde group in the open-chain form can reduce Tollens reagent, an aqueous solution of glucose will give a positive Tollens test.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above

Video duration:

Play a video:

Was this helpful?

Key Concepts

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

Cyclic Hemiacetal Formation

Cyclic hemiacetals are formed when an aldehyde reacts with an alcohol, resulting in a ring structure. In the case of glucose, the open-chain form can cyclize to form a six-membered ring (pyranose) or a five-membered ring (furanose). This transformation is crucial because the cyclic form is more stable and predominates in solution, influencing the reactivity of glucose.

Tollens' Reagent

Tollens' reagent is a solution of silver nitrate in ammonia, used to test for aldehydes. It oxidizes aldehydes to carboxylic acids while reducing silver ions to metallic silver, resulting in a positive test indicated by a silver mirror. Understanding the reactivity of glucose in its cyclic form is essential to determine if it can reduce Tollens' reagent.

Equilibrium and Reactivity

In the context of glucose, the equilibrium between the open-chain and cyclic forms affects its reactivity. Although the cyclic form is more stable and predominates, a small amount of the open-chain form is always present, which is the reactive species that can reduce Tollens' reagent. Thus, the presence of this minor component is key to understanding glucose's ability to give a positive Tollens test.

Watch next

Master Monosaccharides - Forming Cyclic Hemiacetals with a bite sized video explanation from Johnny

Start learning

Related Videos

Related Practice

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.

(a) Propose a mechanism for the cyclization, assuming a trace of acid is present.

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).

Textbook Question

Without referring to the chapter, draw the chair conformations of