General orrganic and biological chemistry structures off liffe 5th by karen timberlake15 worked examples

Sample Problem 15.1 Monosaccharides
Classify each of the following monosaccharides as an aldopentose, ketopentose, aldohexose, or ketohexose:

Solution
a. Ribulose has five carbon atoms (pentose) and is a ketone, which makes it a ketopentose.
b. Glucose has six carbon atoms (hexose) and is an aldehyde, which makes it an aldohexose.

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 15.1 Monosaccharides
Continued

Study Check 15.1
Classify the following monosaccharide, erythrose, as an aldotetrose, ketotetrose, aldopentose, or ketopentose:

Answer
aldotetrose

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 15.2 Fischer Projections for Monosaccharides
Ribulose, which is used in various brands of artificial sweeteners, has the following Fischer projection:

Identify the compound as D- or L-ribulose.

Solution

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 15.2 Fischer Projections for Monosaccharides
Continued
Step 1

Number the carbon chain starting at the top of the Fischer projection.

Step 2

Locate the chiral carbon farthest from the top of the Fischer projection. The chiral carbon farthest from the top is carbon 4.

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 15.2 Fischer Projections for Monosaccharides
Continued
Step 3

Identify the position of the —OH group as D- or L-. In this Fischer projection, the —OH group is drawn

on the right of carbon 4, which makes it D-ribulose.

Study Check 15.2
Draw and name the Fischer projection for the mirror image of the ribulose in Sample Problem 15.2.

Answer

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 15.3 Drawing Haworth Structures for Sugars
D-Mannose, a carbohydrate found in immunoglobulins, has the following Fischer projection. Draw the Haworth structure for β-d-mannose.

Solution

Step 1

Turn the Fischer projection clockwise by 90°.

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.

Sample Problem 15.3 Drawing Haworth Structures for Sugars
Continued
Step 2

Fold the horizontal carbon chain into a hexagon, rotate the groups on carbon 5, and bond the
O on carbon 5 to carbon 1.

Step 3

Draw the new —OH group on carbon 1 above the ring to give the β anomer.

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 15.3 Drawing Haworth Structures for Sugars
Continued

Study Check 15.3
Draw the Haworth structure for α-D-mannose.

Answer

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.

Sample Problem 15.4 Reducing Sugars
Why is D-glucose called a reducing sugar?

Solution
The aldehyde group with an adjacent hydroxyl of D-glucose is easily oxidized by Benedict’s reagent. A carbohydrate that reduces Cu

2+

+
to Cu is called a reducing sugar.

Study Check 15.4
A solution containing a tablet of Benedict’s reagent turns brick red with a urine sample.
According to Table 15.1, what might this result indicate?

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 15.4 Reducing Sugars
Continued

Answer
The brick-red color of the Benedict’s reagent shows a high level of reducing sugar (probably glucose) in the urine, which may indicate type 2 diabetes.

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 15.5 Glycosidic Bonds in Disaccharides
Melibiose is a disaccharide that is 30 times sweeter than sucrose.
a.

What are the monosaccharide units in melibiose?

b.

What type of glycosidic bond links the monosaccharides?

c.

Identify the structure as α- or β-melibiose.

Solution
a.

First monosaccharide (left)
Second monosaccharide (right)

When the —OH group on carbon 4 is above the plane, it is D-galactose. When the —OH group on carbon 1 is below the plane, it is αD-galactose.

b.

Type of glycosidic bond

When the —OH group on carbon 4 is below the plane, it is α-D-glucose.

c.

Name of disaccharide

The —OH group at carbon 1 of α-D-galactose bonds with the —OH group on carbon 6 of glucose, which makes it an α(1 → 6)glycosidic bond.
The —OH group on carbon 1 of glucose is below the plane, which
α-melibiose.

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 15.5 Glycosidic Bonds in Disaccharides
Continued

Study Check 15.5
Cellobiose is a disaccharide composed of two D-glucose molecules connected by a β(1 → 4)-glycosidic linkage. Draw the Haworth structure for β-cellobiose.

Answer

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.

Sample Problem 15.6 tructures of Polysaccharides
Identify the polysaccharide described by each of the following:
a. a polysaccharide that is stored in the liver and muscle tissues
b. an unbranched polysaccharide containing β(1 → 4)-glycosidic bonds
c. a starch containing α(1 → 4)- and α(1 → 6)-glycosidic bonds

Solution
a. glycogen

b. cellulose

c. amylopectin, glycogen

Study Check 15.6
Cellulose and amylose are both unbranched glucose polymers. How do they differ?

Answer
Cellulose contains glucose units connected by β(1 → 4)-glycosidic bonds, whereas the glucose units in amylose are connected by α(1 → 4)-glycosidic bonds.

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.