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

Monosaccharide

Organic Chemistry

28. Carbohydrates

Monosaccharide

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3:19 minutes

Problem 42

Textbook Question

A student isolated a monosaccharide and determined that it had a molecular weight of 150. Much to his surprise, he found that it was not optically active. What is the structure of the monosaccharide?

Verified step by step guidance

Determine the molecular formula of the monosaccharide: Monosaccharides generally follow the formula \( C_nH_{2n}O_n \). Given the molecular weight of 150, calculate \( n \) by solving \( 12n + 2n + 16n = 150 \), where 12, 2, and 16 are the atomic weights of carbon, hydrogen, and oxygen, respectively.

Analyze the optical activity: The lack of optical activity suggests that the monosaccharide is achiral. This means the molecule either lacks a chiral center or is a meso compound (a molecule with chiral centers but an internal plane of symmetry).

Consider the possible structures: For a monosaccharide with \( n \) carbons, identify possible achiral structures. For example, a meso compound like D-glucose or D-mannose in their open-chain forms could be candidates if they exhibit symmetry.

Verify the structure: Check if the proposed structure satisfies the molecular formula and the condition of being optically inactive. For example, a meso compound such as D-glucose in its cyclic form (pyranose or furanose) could exhibit internal symmetry, making it optically inactive.

Conclude the structure: Based on the molecular weight, achirality, and structural analysis, identify the specific monosaccharide that fits all criteria. Ensure the structure is consistent with the chemical and physical properties described in the problem.

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Key Concepts

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

Monosaccharide Structure

Monosaccharides are the simplest form of carbohydrates, consisting of single sugar units. They typically have a backbone of carbon atoms, with hydroxyl groups (-OH) attached, and a carbonyl group (C=O) that can be either an aldehyde or a ketone. The structure can be linear or cyclic, with cyclic forms being more common in solution.

Optical Activity

Optical activity refers to the ability of a compound to rotate plane-polarized light, which is a characteristic of chiral molecules. A molecule is chiral if it has non-superimposable mirror images, typically due to the presence of one or more asymmetric carbon atoms. If a monosaccharide is not optically active, it suggests that it may have a symmetrical structure, leading to no net rotation of light.

Molecular Weight and Composition

The molecular weight of a compound is the sum of the atomic weights of all atoms in its molecular formula. For monosaccharides, the general formula is (CH2O)n, where n is the number of carbon atoms. A molecular weight of 150 suggests a monosaccharide with a specific composition, likely indicating a structure with five carbons (C5H10O5) or six carbons (C6H12O6) that may be symmetrical, thus explaining the lack of optical activity.

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