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

4. Alkanes and Cycloalkanes

Equatorial Preference

Organic Chemistry

4. Alkanes and Cycloalkanes

Equatorial Preference

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4:31 minutes

Problem 82

Textbook Question

The most stable form of glucose (blood sugar) is a six-membered ring in a chair conformation with its five substituents all in equatorial positions. Draw the most stable conformer of glucose by putting the OH groups and hydrogens on the appropriate bonds in the structure on the right.

Verified step by step guidance

Analyze the glucose derivative structure on the left. It is a six-membered ring (pyranose form) with hydroxyl (-OH) groups and a methoxy (-CH2OCH3) substituent attached to the ring.

Understand that the most stable chair conformation of glucose places bulky substituents, such as hydroxyl groups, in equatorial positions to minimize steric hindrance.

Examine the chair conformation template on the right. Identify the axial and equatorial positions for each carbon atom in the ring.

Place the hydroxyl (-OH) groups in equatorial positions on the chair conformation. For example, the -OH groups on carbons 2, 3, 4, and 6 should be positioned equatorially.

Position the methoxy (-CH2OCH3) group on carbon 1 in the equatorial position as well, ensuring all substituents are arranged to minimize steric strain and maximize stability.

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Key Concepts

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

Chair Conformation

The chair conformation is a three-dimensional representation of cyclohexane and its derivatives, where the carbon atoms are arranged in a staggered manner to minimize steric strain. In this conformation, substituents can occupy equatorial or axial positions, with equatorial positions generally being more stable due to reduced steric hindrance. Understanding this conformation is crucial for predicting the stability of cyclic sugars like glucose.

Equatorial vs. Axial Substituents

In a chair conformation, substituents can be positioned either equatorially or axially. Equatorial substituents extend outward from the ring, minimizing steric interactions with other groups, while axial substituents point up or down, potentially causing steric clashes. For glucose, placing hydroxyl (OH) groups in equatorial positions enhances stability, making it the preferred arrangement in its most stable form.

Glucose Structure and Isomerism

Glucose is a six-carbon aldose sugar that can exist in various forms, including linear and cyclic structures. The cyclic form predominantly exists as a pyranose (six-membered ring) and can have different stereoisomers based on the orientation of its hydroxyl groups. Recognizing the specific stereochemistry of glucose is essential for accurately drawing its most stable conformer and understanding its biological functions.

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Related Practice

Textbook Question

Draw the two chair conformers for each of the following and indicate which conformer is more stable:

c. trans-1-ethyl-2-isopropylcyclohexane

Textbook Question

Bromine is a larger atom than chlorine, but the equilibrium constants in Table 3.9 indicate that a chloro substituent has a greater preference for the equatorial position than does a bromo substituent. Suggest an explanation for this fact.

Textbook Question

Draw the most stable conformer of the following molecule. (A solid wedge points out of the plane of the paper toward the viewer. A hatched wedge points back from the plane of the paper away from the viewer.)

Textbook Question

Which has a higher percentage of the diequatorial-substituted conformer compared with the diaxialsubstituted conformer: trans-1,4-dimethylcyclohexane or cis-1-tert-butyl-3-methylcyclohexane?

Textbook Question

Draw two chair conformations of the molecules below. Indicate which is most stable.

(a)

Textbook Question

For each pair of conformations you drew in Assessment 3.41, indicate which is most stable.

Textbook Question

For each pair of conformations shown, choose which is most stable. If both are equally stable, then write 'no difference.' [If both conformations have the same number of axial substituents, choose the one with the smallest axial substituents.]

(e)

Textbook Question

(g)

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