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.

Sample Problem 15.1 Monosaccharides Continued Study Check 15.1 Classify the following monosaccharide, erythrose, as an aldotetrose, ketotetrose, aldopentose, or ketopentose: Answer aldotetrose

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

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.

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

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°.

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.

Sample Problem 15.3 Drawing Haworth Structures for Sugars Continued Study Check 15.3 Draw the Haworth structure for α-D-mannose. Answer

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 Cu2+ 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?

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.

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) b. Type of glycosidic bond c. Name of disaccharide 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 α-D-galactose. When the —OH group on carbon 4 is below the plane, it is α-D-glucose. 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.

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

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.