Textbook: McMurry 8th Ed. (2011)

Chapter 25: Biomolecules: Carbohydrates

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Individual Problems

Individual Problems

Problem # 317

α-D-Glucose is shown below. Draw its two chair forms. Which conformation is more stable? Explain.

I recommend using the common convention wedge = "up" and dash = "down."

Solution

The chair form on the left is more stable because it has no axial substituents, which lead to 1,3-diaxial interactions and higher energies. Glucose is the most stable of the hexose carbohydrates because it is the only one that has al substituents at C2-C5 in the equatorial position.

It's probably no coincidence that the most abundant sugar in the world (glucose) is also the sugar that is most stable!

Problem Details

MendelSet practice problem # 317 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Cycloalkanes and Cyclohexane Chair Forms, Carbohydrates

Question Type: Concept (explain why this is so..)

Keyword(s): cyclohexane chair forms

Problem # 715

α-D-glucose is shown below.

a) Is α-D-glucose an acetal, hemiacetal, ketal, or hemiketal?

b) Draw the carbonyl form of α-D-glucose.

Solution

a) You can spot the carbonyl carbon because it's the one with two oxygens bonded to it (oxidation state II).

The carbonyl carbon has an -OH (alcohol) bonded to it, and not an -OR (ether), so it's a hemiacetal or a kemiketal.

The carbonyl carbon also has a hydrogen bonded to it, so it must have come form an aldehyde (instead of ketone). So it's a hemiacetal. (remember: aldehyde -> acetal, ketone -> ketal).

b) The hemiacetal carbon's -OH (or -OR in the case of a ketal or acetal) is the oxygen that used to be the carbonyl C=O.