Jones M., Fleming S.A. Organic Chemistry

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22.5 The Fischer Determination of the Structure of D-Glucose 1159

Now we know all four stereoisomers of the D-aldopentoses, and we can construct

the

L series by drawing the mirror images of the D series. Doing so gives us all eight

aldopentoses.

Let’s complete the determination of the structures of the D-aldohexoses. D-Gulose

and

D-idose give the same osazone and must differ only in their configurations at

C(2) (Fig. 22.56).

There are only two remaining

D-aldopentoses. One of them, D-xylose, gives a

meso diacid on oxidation,whereas the other,

D-lyxose,gives an optically active diacid.

The structures must be as shown in Figure 22.55.

HNO

OHH

CHO

COOH

OHH

COOH

meso Diacid

D-Xylose

Optically active

diacid

D-Lyxose

OHH

HHO

CHO

COOH

HHO

COOH

OHH

FIGURE 22.55 D-Xylose gives a meso diacid and

D-lyxose gives an optically active diacid, which fixes

the structures of the remaining

D-aldopentoses.

OHH

CHO

D-Gulose

OHH

HHO

CHO

D-Idose

OHH

Same osazone

3 equiv.

PhNHNH

3 equiv.

PhNHNH

N NHPh

C(2) C(2)

FIGURE 22.56 D-Gulose and D-idose give the same osazone, which determines the structure of

D-idose and its mirror image L-idose.

1160 CHAPTER 22 Carbohydrates

OHH

CHO

D-Allose

HNO

OHH

HHO

CHO

D-Altrose

OHH

COOH

A meso diacid

OHH

HHO

COOH

An optically

active diacid

OHH

CHO

D-Ribose

HNO

Kiliani–Fischer

synthesis

FIGURE 22.57 The Kiliani–Fischer synthesis used to determine the structures of

D-allose, D-altrose, and their mirror-image L isomers.

OHH

CHO

D-Galactose

HNO

OHH

HHO

CHO

D-Talose

COOH

OHH

COOH

A meso diacid

OHH

HHO

COOH

An optically

active diacid

HNO

FIGURE 22.58 D-Galactose gives a meso diacid, and D-talose gives an optically active diacid.

The Kiliani–Fischer synthesis applied to D-ribose gives D-allose and

D-altrose, which must have the two structures shown in Figure 22.57. The two

structures can be assigned by noting that

D-allose gives an optically inactive,

meso diacid on treatment with nitric acid, whereas

D-altrose gives an optically

active diacid.

Of the two remaining

D-aldohexoses, D-galactose gives a meso diacid on

oxidation with nitric acid, but

D-talose gives an optically active diacid (Fig. 22.58).

Now we know the structures of all the aldohexoses.The structures of the eight

L isomers are simply the mirror images of the D series, and we have at last com-

pleted our task.

22.6 Special Topic: An Introduction to Di- and Polysaccharides 1161

22.6 Special Topic: An Introduction to Di- and

Polysaccharides

A great many sugars containing 12,18, and other multiples of 6 carbons are known.

Sugars that have the formula C

are called monosaccharides because they

are composed of a single sugar molecule. Sugars that come from the combination

of two sugar units, the most common being C

carbohydrates, are called disaccha-

rides and they have the formula of C

2n2

n1

because the two sugar units are

always joined by a dehydration process. Some of the disaccharides have common

names. For example, sucrose, lactose, and maltose are all disaccharides. Starch and

cellulose are polymers of simple repeating monosaccharide units, less many mol-

ecules of water. We call such biopolymers polysaccharides.

We will use lactose as an example of how the structure of a disaccharide can be

unraveled.The first important observation is that ()-lactose, a disaccharide having

the formula C

, can be hydrolyzed in acid to a pair of simple aldohexoses:

D-glucose and D-galactose (Fig. 22.59). This observation identifies the two mono-

saccharides making up ()-lactose.

Only this CH

O group...

Dilute acid does not affect

the other simple ethers

such as this one

The two sugars

must be attached

here, at the

glycosidic link

to C(1)

cat.

HCl

...is hydrolyzed to HO

ROH

cat.

HCl

OSugar

HOSugar

FIGURE 22.60 The fact that

()-lactose is hydrolyzed in dilute

acid to give two monosaccharides

means the monosaccharides must be

joined via a glycosidic linkage.

(+)-Lactose

OHH

CHO

D-Galactose

OHH

CHO

D-Glucose

FIGURE 22.59 In acid, ()-lactose

is hydrolyzed to

D-glucose and

D-galactose.These sugars are C(4)

epimers.

But this simple experiment gives us another piece of less obvious information.

Remember that only glycosidic linkages, the acetals,are cleaved in acid—simple ether

connections are not touched. We saw this distinction first in the hydrolysis of only

the glycosidic methoxy group of fully methylated

D-glucose (p. 1150). D-Glucose

and

D-galactose must therefore be attached via the C(1) hydroxyl group of one of

the sugars (Fig. 22.60). If the attachment were at any other position, dilute acid

would not break up the disaccharide.

1162 CHAPTER 22 Carbohydrates

We don’t (yet) know which

O is used to make this bond

Hemiacetal

link

cat.

HCl

(+)-Lactose in open form

COOH

D-Gluconic acid

Lactobionic acid

D-Galactopyranose

OHH

COOH

FIGURE 22.62 The carboxylic acid

group in lactobionic acid marks the

position of the aldehyde/hemiacetal

carbon in ()-lactose. The fact that

hydrolysis of lactobionic acid gives a

gluconic acid and not a galactonic

acid tells us that it is glucose that has

the aldehyde/hemiacetal group in

()-lactose.

FIGURE 22.61 The issues that must

be addressed to determine the

structure of ()-lactose.

We now need to know three details about the dimeric structure of lactose

(Fig. 22.61). The issues are:

1. Which sugar is the “attacher” (the ROH) and which is the “attachee” (receives

ROH at its anomeric carbon)?

2. Which hydroxyl group of the attacher is used to make the linkage to the anomeric

carbon of the attachee?

3. Are the two sugars linked with stereochemistry that is α or β?

1. Is this glucose

(equatorial OH) or

galactose (axial OH)?

2. To which

carbon is this bond

attached?

3. Is this linkage

α or β (axial or

equatorial)?

OSugar

WORKED PROBLEM 22.22 If the two monosaccharides in ()-lactose were

attached with C(1) of glucose as the acetal linkage and the galactose having the

free aldehyde/hemiacetal, what would the products of the experiment shown in

Figure 22.62 be?

()-Lactose can be oxidized by bromine in water to the improbably named

lactobionic acid (Fig. 22.62). Bromine in water oxidizes only the aldehydic or hemiac-

etal carbon of a sugar to give the carboxylic acid (p. 1144). In one of the partners in the

disaccharide there is no hemiacetal, and therefore no free aldehyde to be oxidized by

bromine/water. Acid hydrolysis of the lactobionic acid gives

D-galactopyranose and

D-gluconic acid.This result tells us that the aldehyde/hemiacetal group in ()-lactose

must belong to glucose,because it was the monosaccharide that was oxidized.Therefore

the galactose must be attached to glucose via an acetal linkage at the galactose C(1).

(continued)

22.6 Special Topic: An Introduction to Di- and Polysaccharides 1163

We do not yet

know which O

group is used to

make this bond

cat.

HCl

Hypothetical “reverse” lactobionic acid

Hypothetical “reverse” lactose

D-Glucose part D-Galactose part

D-Galactonic acid

OHH

HHO

OHH

COOH

D-Glucopyranose

Our next task is to determine which carbon of glucose is attached to C(1) of galac-

tose.To address this issue, we can use a Williamson ether synthesis to methylate all free

hydroxyl groups in ()-lactose (Fig.22.63). After methylation, the lactose is hydrolyzed

in acid to 2,3,4,6-tetramethyl

D-galactopyranose and 2,3,6-trimethyl D-glucopyranose.

ANSWER If the monosaccharides were attached this way, the galactose aldehyde

would be oxidized to , and hydrolysis of this hypothetical lactobionic acid

would give

D-galactonic acid and D-glucopyranose.

COOH

cat.

HCl

Three ?’s are methyl groups, but

one oxygen is used to make

the attachment to glucose

OCH

excess

OCH

The C(4) OH group was used

to attach the two sugars

OCH

2,3,4,6-Tetramethyl

D-galactopyranose

HO CH

C(1) of galactose

One of the oxygens in

glucose is used to make the

bond to C(1) of galactose

Galactose Glucose

O ?O

OCH

2,3,6-Trimethyl

D-glucopyranose

OCH

HO+

FIGURE 22.63 The position of the OH used to attach

glucose to C(1) of galactose can be determined

through a methylation followed by hydrolysis.

1164 CHAPTER 22 Carbohydrates

SUGAR SUBSTITUTES

The four major sugar substitutes vary widely in structure,

but all four are united by a single thread: Saccharin, cycla-

mate, aspartame, and sucralose were found not by a directed

search but by accident. Indeed, each of these four was dis-

covered when a chemist tasted a compound made for other

purposes. In fact, cyclamates were discovered because a ciga-

rette smoked in the lab tasted sweet. This observation may

seem like courting disaster to you—deliberately ingesting an

unknown laboratory chemical, not to mention smoking in a

chemistry lab—but it was quite common in the old days.

Even as recently as the 1960s it was normal to smell—

closely and carefully—just about everything one made. How

many chemists destroyed their sense of smell or succumbed

to the odor because of this practice, we will never know.

Now the laboratory is better ventilated and we gingerly waft

the fumes from a flask if we must test for an odor.

After 100 years of research and debate concerning sac-

charin (discovered in 1879), the consensus is that no ill

health effects are associated with moderate consumption.

Although cyclamate (discovered in 1937) is banned in the

United States, there is no evidence implicating it as a health

hazard according to the Food and Drug Administration

(FDA). Aspartame (discovered in 1965) may be the single

most studied food additive. It has been cleared for use,

except for phenylketonurics. People with this disease must

avoid phenylalanine, which is formed from the hydrolysis of

aspartame. Aspartame also releases methanol when heated,

but only extremely small amounts are produced. Sucralose

(discovered in 1976) has cleared all toxicology tests.

Although halogenated alkanes are known carcinogens, the

chlorinated sucralose is not lipophilic enough to be retained

in the body.

Lactose is found in milk and other dairy products, so it is often called milk sugar. Disaccharides such as lactose

need to be cleaved to monosaccharides in order to be absorbed into the bloodstream.The acetal bond in lactose is

cleaved by the enzyme lactase. Many adults are lactose intolerant because their bodies don’t make lactase.For such

individuals, consumed lactose doesn’t get absorbed into the bloodstream and the result is abdominal discomfort.

This hydrolysis liberates only the OH groups that in ()-lactose were tied up in

acetals; it leaves simple ether groups unchanged.It is the OH at C(4) of glucose that

is unmethylated, telling us that this is the glucose position con-

nected to galactose in the lactose molecule.

To determine whether the ()-lactose glycoside linkage is made

as the α or β anomer we turn to enzymes such as lactase

that have

evolved to cleave only β-glycosidic bonds,and to which α-linked dis-

accharides are inert. ()-Lactose is cleaved by lactase and therefore

must be β-linked.

H NMR spectroscopy can also be used to iden-

tify whether a disaccharide is linked as the α or β anomer.The hydro-

gen on the anomeric carbon is more deshielded than any of the other

hydrogens,and the C(1) equatorial hydrogen (δ 5.2 ppm) is further

downfield than the C(1) axial hydrogen (δ 4.6 ppm). Figure 22.64

shows the final three-dimensional representation of ()-lactose.

(+)-Lactose β linkage

WEB 3D

D-Galactose D-Glucose

FIGURE 22.64 The structure of the disaccharide

()-lactose.The monosaccharides

D-galactose and

D-glucose are joined by a β-glycosidic linkage between

C(1) of galactose and C(4) of glucose.

Cyclamate Aspartame

Saccharin Sucralose

HOOC

–

OCH

22.6 Special Topic: An Introduction to Di- and Polysaccharides 1165

PROBLEM 22.23 Draw a Fischer projection for ()-lactose. Hint: You will need

several long, “around the corner” bonds as illustrated in Figure 22.16.

PROBLEM 22.24 Cellobiose is a disaccharide cleaved by hydrolysis into two mol-

ecules of D-glucose. Methylation of cellobiose leads to an octamethyl derivative.

Hydrolysis with acid converts the octamethyl derivative into 2,3,6-trimethyl-

D-glucopyranose and 2,3,4,6-tetramethyl-D-glucopyranose. Cellobiose can be

cleaved enzymatically with lactase. There are actually two possible structures for

cellobiose. One has

H NMR signals at δ 4.5 and δ 4.7 ppm. The other isomer

has signals at δ 4.5 and δ 5.2 ppm. Draw the two structures of cellobiose.

PROBLEM 22.25 Red blood cells have glycoproteins (carbohydrates covalently

attached to a protein, Chapter 23) on the cell surface that play an important role

in a person’s immune system. There are three versions of the glycoproteins found

in humans. These variations lead to what are called blood types. You might have

blood type A, B, AB, or O.The differences are simply a matter of the number and

kinds of carbohydrates in the glycoprotein of your red blood cells. Other animals

have different glycoproteins associated with their red blood cells. Horses, for

example, have eight blood types.The carbohydrates involved in the human blood

types are shown. Type O is a trisaccharide composed of N-acetyl-

D-glucosamine,

D-galactose, and L-fucose. (Fucose is the common name for 6-deoxygalactose.)

Type B is a tetrasaccharide made up of the three monosaccharides of type O plus

D-galactose.Type A is also a tetrasaccharide that has the basic structure of type

O plus an N-acetyl-D-galactosamine.

Type O

Protein

Type B

Protein

(continued)

1166 CHAPTER 22 Carbohydrates

Type A

Protein

(a) Draw the Fischer projection of L-fucose. (b) Label the anomeric carbons as α

or β. (c) Are these saccharides reducing sugars? Why or why not?

D-Fructose

OHH

D-Glucopyranose (+)-Sucrose␤-D-Fructofuranose

FIGURE 22.65 In nonreducing sugars, there is no free aldehyde group. In the case of sucrose, attachment between the sugars is

between the anomeric carbons.This kind of attachment makes sucrose, which is table sugar, a nonreducing sugar.

Synthesis of disaccharides is possible using the protecting group methods dis-

cussed earlier (Fig. 22.43). We can use the free alcohol of one monosaccharide to

link with the hemiacetal of a protected second monosaccharide. The acetate-

In addition to lactase,the enzyme that cleaves the glycosidic bond in lactose, there

are many other glycosidase enzymes in Nature.They play a critical role in a variety

of processes related to our diet and our environment. For example, glycosidases are

necessary for the degradation of the polysaccharide biomass found in plants. The

chemistry a glycosidase facilitates is hydrolysis of the acetal linkage connecting two

sugars. The result is a disconnection at the anomeric carbon.

Sometimes two sugars are linked via an oxygen connecting the two anomeric car-

bons. Because these disaccharides are acetals on both sugar units, oxidation to an

aldonic acid (Fig.22.28) is not possible because there is no free aldehyde. Remember:

In aqueous conditions the cyclic acetals are not in equilibrium with the open form.

In contrast to reducing sugars,in which at least a small amount of an oxidizable free

aldehyde is present in solution, sugars containing no free aldehyde groups are nonre-

ducing sugars. An example is sucrose (common table sugar) shown in Figure 22.65,

in which the anomeric carbon C(1) of

D-glucose is linked via O to the anomeric

carbon C(2) of

D-fructose.

22.6 Special Topic: An Introduction to Di- and Polysaccharides 1167

protected sugar discussed in Problem 22.20 can be used to make disaccharides.

Benzyl protecting groups can also be used. An advantage to employing the benzyl

protecting group ( ) is that it can be removed from an oxygen (or nitrogen)

under the fairly mild conditions of hydrogenation (H

/Pd). Figure 22.66 uses this

protecting group method to make melibiose, a natural isomer of lactose. Melibiose

is a galactose–glucose disaccharide that uses the C(6) OH of D-glucose to attach to

C(1) of

D-galactose.

PhCH

cat.

HCl

PhCH

cat.

HCl

D-Glucopyranose

OCH

LiAlH

OCH

Ph O

excess

NaH

PhCH

excess

NaH

PhCH

Galactose

Glucose

cat.

HCl

D-Galactopyranose

cat.

HCl

Benzylated

galactopyranose

OCH

PhCH

OCH

PhCH

OCH

PhCH

OCH

PhCH

Benzylated

glucopyranoside

Benzylated

glucopyranoside

Benzylated

glucopyranose

Melibiose

PhCH

OCH

PhCH

FIGURE 22.66 Use of the benzyl

protecting group to make the

disaccharide melibiose.

1168 CHAPTER 22 Carbohydrates

22.7 Summary

New Concepts

(b) Starch amylose, which is poly-

D-glucose linked α

(a) Cellulose is poly-

D-glucose linked β

C(1)

C(4)

FIGURE 22.67 The two most

common natural polysaccharides:

(a) cellulose and (b) amylose.

right in the Fischer projection, the molecule is a

D sugar; if it is

on the left, the molecule is an

L sugar (Fig. 22.3).

When Emil Fischer worked out the structure of glucose in

the late 1800s, there was no way to tell absolute configuration.

Fischer could not tell the

D series from the enantiomeric L

series, so he assumed that the OH at C(5), the configurational

carbon, was on the right and worked out the relative configura-

tions of the other groups. As it turned out, he guessed correctly.

A number of times in this chapter we encounter reactions

in which a minor partner in an equilibrium reacts to give the

product. As long as equilibrium is reestablished, removal of the

minor partner can drive the reaction to completion (Fig. 22.14).

Carbohydrates, also called sugars or saccharides, are molecules

with the general formula of C

. The simplest carbohy-

drates, the monosaccharides, have an aldehyde at one end in the

open form of the molecule and a primary alcohol at the other

end. In solution, carbohydrates exist mainly as cyclic hemiacetals.

The stereochemical complications lead to the simplifying

tool of a Fischer projection in which a stylized drawing is used

to reduce complexity. The molecule is drawn vertically with the

aldehyde group at the top. Horizontal bonds are taken as com-

ing toward you and vertical bonds as retreating from you. The

stereogenic carbon closest to the primary alcohol group is called

the configurational carbon. If the OH on this carbon is on the

There is no reason that more than two sugars cannot be connected by glycosidic

bonds, and larger polysaccharides are common. The carbohydrate cellulose,for

instance, is a polysaccharide in which thousands of glucose units are connected lin-

early by β-glycosidic bonds between C(4) of one glucose and the anomeric C(1) of

a neighboring glucose (Fig. 22.67a). Cellulose, the stuff plants are made of, contains

most of the carbon on Earth! Cotton is 90% cellulose, and wood is about 50%.Most

mammals, including humans, are unable to break down cellulose. Starch is another

polysaccharide made up of glucose units. Starch is the main saccharide in our diet.

Corn, rice, potatoes, and wheat are sources of starch. Not all of the linkages in starch

are between C(4) and C(1), so it is described as a branched polysaccharide or as a

complex sugar. However, the linear polysaccharide amylose can be isolated from

starch (Fig.22.67b). Amylose has the same structure as cellulose except that the sug-

ars are connected through α linkages.