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Chapter 28: Problem 28.41 (page 1147)

For D-arabinose:

  1. Draw its enantiomer.
  2. Draw an epimer at C3.
  3. Draw a diastereomer that is not an epimer.
  4. Draw a constitutional isomer that still contains a carbonyl group.

Short Answer

Expert verified

Answer

a.

b.

c.

d.

Step by step solution

01

Step-by-Step SolutionStep 1: Epimers

Epimers in stereochemistry are formed by swapping the groups at one or more than one stereocenters. From enantiomers, we get mirror images, and in the case of diastereomers, no mirror images are formed.

02

Enantiomer of D-arabinose

D-arabinose's enantiomer is drawn by swapping the groups (H and OH) attached to the three stereocenters of D-arabinose.

Enantiomer of D-arabinose

03

Epimer at C3

b. The epimer at C3 of D-arabinose is drawn by swapping the groups (H and OH) attached at C3 carbon.

04

Diastereomer that is not an epimer.

c. Diastereomer of D-arabinose is drawn so that it is not a mirror image of D-arabinose.


Diastereomer of D-arabinose

05

Constitutional isomer that still contains a carbonyl group.

d. A constitutional isomer of D-arabinose is drawn by changing the bonding patterns but keeping the same chemical formula.


Constitutional isomer of D-arabinose

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Most popular questions from this chapter

Question: Draw two possible epimers of D-erythrose. Name each of these compounds using Figure 28.4.

Question: Convert the ball-and-stick model to a Fischer projection.

Question:

a. Draw the enantiomer of D-fructose.

b. Draw an epimer of D-fructose at C4. What is the name of this compound?

c. Draw an epimer of D-fructose at C5. What is the name of this compound?

Question: Convert each aldohexose to the indicated anomer using a Haworth projection

The following isomerization reaction, drawn using D-glucose as starting material, occurs with all aldohexoses in the presence of base. Draw a stepwise mechanism that illustrates how each compound is formed.
See all solutions

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