CONCEPT MAP: THE GENERATION OF BIOCHEMICAL ENERGY
21.4 Structure of Glucose and Other Monosaccharides
d-Glucose, also called dextrose or blood sugar, is the most abundant of all monosaccha- rides and has the most important function. In nearly all living organisms, d-glucose serves as a source of energy to fuel biochemical reactions. It is stored as starch in plants and glycogen in animals (Section 21.9). Our discussion here of the structure of d- glucose illustrates a major point about the structure of monosaccharides: Although they can
S E C T I O N 2 1 . 4 Structure of Glucose and Other Monosaccharides 665 be written with the carbon atoms in a straight chain, monosaccharides with five or six
carbon atoms exist primarily in their cyclic forms when in solution, as they are found in living organisms.
Recall from Section 16.7 that the key to recognizing the hemiacetal is a carbon atom bonded to both an iOH group and an iOR group.
6 5
5
4 4
3
3 2
1
5
D-Glucose open-chain form (Fischer projection)
Turn on side Coil CH2OH
to the back
Rotate
Close ring
Anomeric, hemiacetal C atom CH2OH
6
O H
C
H OH
3
2
H OH
H OH
HO H
6
6 5
CH2OH
2 β-D-Glucose 4
α-D-Glucose 1
1 O CH2OH
OH OH
OH OH H
OH
4
3 2
1 O OH CH2OH
OH OH
OH
H
O
OH H OH
OH
3 6 5
CH2OH
2
4 1
OH O
OH OH H OH
▲ Figure 21.3
The structure of d-glucose.
d-Glucose can exist as an open-chain polyhydroxy aldehyde or as a pair of cyclic hemiacetals. The cyclic forms differ only at C1, where the iOH group is either on the opposite side of the six- membered ring from the CH2OH 1a2 or on the same side 1b2. To convert the Fischer projection into the six-membered ring formula, the Fischer projection is laid down with C1 to the right and the other end curled around at the back. Then the single bond between C4 and C5 is rotated so that the iCH2OH group is vertical. Finally, the hemiacetal OiR bond is formed by connecting oxy- gen from the iOH group on C5 to C1, and the hemiacetal OiH group is placed on C1. (H’s on carbons 2–5 are omitted here for clarity.)
Look at the Fischer projection of d-glucose at the top left-hand corner of Figure 21.3, and notice the locations of the aldehyde group and the hydroxyl groups. You have seen that aldehydes and ketones react reversibly with alcohols to yield hemiacetals as shown below.
H R
H R
O O
An aldehyde An alcohol A hemiacetal
+ O R′ O R′
C C
H H
Since glucose has alcohol hydroxyl groups and an aldehyde carbonyl group in the same molecule, internal hemiacetal formation is possible. The aldehyde carbonyl group at carbon 1 (C1) and the hydroxyl group at carbon 5 (C5) in glucose react to form a six- membered ring that is a hemiacetal. Ketones undergo internal hemiacetal formation as well; in ketones, the reacting carbonyl group is on C2. Monosaccharides with five or six carbon atoms form rings in this manner.
The three structures at the top in Figure 21.3 show how to picture the C5-hydroxyl and the C1-aldehyde group approaching each other for hemiacetal formation. When visualized in this manner, Fischer projections are converted to cyclic structures that (like the Fischer projections) can be interpreted consistently because the same relative arrangements of the groups on the chiral carbon atoms are maintained.
In the cyclic structures at the bottom of Figure 21.3, note how the iOH group on carbon 3, which is on the left in the Fischer projection, points up in the cyclic struc- ture, and iOH groups that are on the right on carbons 2 and 4 point down. When cyclic structures (called Haworth projections) are drawn as shown in Figure 21.3, such relationships are always maintained. Note also that the iCH2OH group in d sugars is always above the plane of the ring.
The hemiacetal carbon atom (C1) in the cyclic structures, like that in other hemiac- etals, is bonded to two oxygen atoms (one in iOH and one in the ring). This carbon is chiral. As a result, there are two cyclic forms of glucose, known as the a and b forms. To see the difference, compare the locations of the hemiacetal iOH groups on C1 in the two bottom structures in Figure 21.3. In the b form, the hydroxyl at C1 points up and is on the same side of the ring as the iCH2OH group at C5. In the a form, the hydroxyl at C1 points down and is on the opposite side of the ring from the iCH2OH group.
Cyclic monosaccharides that differ only in the positions of substituents at carbon 1 are known as anomers, and carbon 1 is said to be an anomeric carbon atom. It is the carbonyl carbon atom (C1 in an aldose and C2 in a ketose) that is now bonded to two O atoms. Note that the a and b anomers of a given sugar are not optical isomers because they are not mirror images.
Although the structural difference between anomers appears small, it has enormous biological consequences. For example, this one small change in structure accounts for the vast difference between the digestibility of starch, which we can digest, and that of cellulose, which we cannot digest (Section 21.9).
Ordinary crystalline glucose is entirely in the cyclic a form. Once dissolved in water, however, equilibrium is established among the open-chain form and the two anomers. The optical rotation of a freshly made solution of a-d-glucose gradually changes from its original value until it reaches a constant value that represents the optical activity of the equilibrium mixture. A solution of b-d-glucose or a mixture of the a and b forms also undergoes this gradual change in rotation, known as mutarota- tion, until the ring opening and closing reactions come to the following equilibrium:
Anomers Cyclic sugars that differ only in positions of substituents at the hemiacetal carbon (the anomeric car- bon); a the form has the iOH on the opposite side from the iCH2OH; the b form has the iOH on the same side as the iCH2OH.
Anomeric carbon atom The hemi- acetal C atom in a cyclic sugar; the C atom bonded to an iOH group and an O in the ring.
Mutarotation Change in rotation of plane-polarized light resulting from the equilibrium between cyclic ano- mers and the open-chain form of a sugar.
Open-chain D-Glucose
(64%) (0.02%)
(36%)
β-D-Glucose α-D-Glucose
HO C H
H
C C H
C H
C H
H OH
OH C H O
6 5 4 3 2 1
OH OH
6 5 4
3 2
1
CH2OH O H OH OH
OH OH
6 5 4
3 2
1
CH2OH
OH H
O OH
OH OH
All monosaccharides with five or six carbon atoms establish similar equilibria, but with different percentages of the different forms present.
Monosaccharide Structures—Summary
• Monosaccharides are polyhydroxy aldehydes or ketones.
• Monosaccharides have three to seven carbon atoms, and a maximum of 2n possible stereoisomers, where n is the number of chiral carbon atoms.
• d and l enantiomers differ in the orientation of the iOH group on the chiral carbon atom farthest from the carbonyl. In Fischer projections, d sugars have this
iOH on the right and l sugars have this iOH on the left.
• d-Glucose (and other 6-carbon aldoses) forms cyclic hemiacetals conventionally repre- sented (as in Figure 21.3) so that iOH groups on chiral carbons on the left in Fischer projections point up and those on the right in Fischer projections point down.
H OH
CH2OH
D-Glucose
Enantiomers
C O H
C
C H C OH
H OH
HO H
C
CH2OH
L-Glucose C
O H
C
C H
H C
H HO
HO HO
OH H
C
S E C T I O N 2 1 . 4 Structure of Glucose and Other Monosaccharides 667
• In glucose, the hemiacetal carbon (the anomeric carbon) is chiral, and a and b anomers differ in the orientation of the iOH groups on this carbon. The a anomer has the
iOH on the opposite side from the iCH2OH, and the b anomer has the iOH on the same side as the iCH2OH.
Worked Example 21.3 Converting Fisher Projections to Cyclic Hemiacetals
The open-chain form of d-altrose, an aldohexose isomer of glucose, has the follow- ing structure. Draw d-altrose in its cyclic hemiacetal form:
C H H
C O OH
D-Altrose OH H C H
H C HO
OH H C OH H C
SOLUTION
First, coil d-altrose into a circular shape by mentally grasping the end farthest from the carbonyl group and bending it backward into the plane of the paper:
C H H
C OH
OH H C H
H C HO
OH H C H
C Coil up
O 6
6 5 4 3 2 1
5 4
3 2
C H
O
1
OH
CH2OH OH
OH HO
HO
Next, rotate the bottom of the structure around the single bond between C4 and C5 so that the iCH2OH group at the end of the chain points up and the iOH group on C5 points toward the aldehyde carbonyl group on the right:
Rotate
6 5
3 5 6 4
2
C H
O
1
CH2OH OH
OH HO
OH
3 4
2
C H
O
1
OH OH
OH HO
CH2OH
Finally, add the iOH group at C5 to the carbonyl C“O to form a hemiacetal ring. The new iOH group formed on C1 can be either up 1b2 or down 1a2:
β
+
α 6
5
3 4
2
C H
O
1
OH OH OH HO
CH2OH
CH2OH O
OH HO
OH
CH2OH O
OH
OH OH
HO
OH
Anomers
3 6 5
2 α-D-Glucose 4
β-D-Glucose 1 1
CH2OH O
OH OH H OH
OH
3 6 5
2 4
CH2OH O
OH H OH OH
OH
PROBLEM 21.8
d-Talose, a constituent of certain antibiotics, has the open-chain structure shown below. Draw d-talose in its cyclic hemiacetal form.
C H H
C O OH
D-Talose H C H
H C HO
H C
OH OH
OH H C
PROBLEM 21.9
The cyclic structure of d-idose, an aldohexose, is shown below. Convert this to the straight-chain Fischer projection structure.
D-Idose 6 5 4
3 2
1
CH2OH
O OH HO
OH OH
Worked Example 21.4 Identifying Sugars and Sugar Derivatives in Polysaccharides Framycetin, a topical antibiotic, is a four-ring molecule consisting of several amino- glycosides—sugars that have some of the iOH groups on the sugars replaced by
iNH2 groups—and another ring, with oxygen links between the rings. What sugar or other molecule is each ring derived from?
NH2
NH2
NH2
NH2 NH2
H2N
HO HO
HO HO
OH
OH OH
O O
O O
O O
H H
H
H H
H Ring 1
Ring 2
Ring 3 Ring 4
ANALYSIS Look at each ring carefully. Ring 2 does not include an O. It cannot be a sugar. Rings 1, 3, and 4 all contain O as a ring member. Imagine the rings as underivatized sugars, that is with iOH groups instead of iNH2 groups; count the number of carbon atoms in each sugar and draw the sugar form to help iden- tify the sugar.
SOLUTION
Ring 2 has six carbon atoms and no oxygen atoms as part of the ring; it is not a sugar, but is a cyclohexane derivative. Rings 1 and 4 are derived from the aldohexose, glucose, while ring 3 is derived from the aldopentose, ribose.
S E C T I O N 2 1 . 5 Some Important Monosaccharides 669
KEY CONCEPT PROBLEM 21.10
Neomycin is an antibiotic used in topical applications to inhibit the growth of bac- teria. It is an aminoglycoside, that is, some of the iOH groups on the sugars have been replaced by iNH2 or R groups. The four rings that constitute neomycin are joined by glycosidic bonds and two of the rings are amino sugars. In the structure shown, identify (a) the amino sugar rings by number, (b) the unmodified sugar ring structure, and (c) the non-sugar ring structure. List how many carbon atoms are in each ring.
NH2 NH2
H2N H2N