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23
Carbohydrates
and Nucleic Acids
GOALS FOR
CHAPTER 23
Draw and identify the structures of glucose, its
anomers, and its epimers, as Fischer projections and as
chair conformations.
Correctly name monosaccharides and disaccharides,
and draw their structures from their names.
Predict the reactions of carbohydrates in acidic and
basic solutions, and with oxidizing and reducing agents.
Predict the reactions that convert their hydroxyl groups to
ethers or esters, and their carbonyl groups to acetals.
Draw the common types of glycosidic linkages, and identify these linkages in disaccharides and polysaccharides.
Recognize the structures of DNA and RNA, and draw
the structures of the common ribonucleotides and
deoxyribonucleotides.
Carbohydrates are the most abundant organic compounds in nature. Nearly all plants and
animals synthesize and metabolize carbohydrates, using them to store energy and deliver
it to their cells. Plants synthesize carbohydrates through photosynthesis, a complex
series of reactions that use sunlight as the energy source to convert carbon dioxide and
water into glucose and oxygen. Many molecules of glucose can be linked together to
form either starch for energy storage or cellulose to support the plant.
6 CO2 + 6 H2 O
light "
23-1
Introduction
6 O2 + C6 H12 O6 ¡ starch, cellulose + H2 O
glucose
Most living organisms oxidize glucose to carbon dioxide and water to provide the
energy needed by their cells. Plants can retrieve the glucose units from starch when
needed. In effect, starch is a plant’s storage unit for solar energy for later use. Animals
can also store glucose energy by linking many molecules together to form glycogen,
another form of starch. Cellulose makes up the cell walls of plants and forms their structural framework. Cellulose is the major component of wood, a strong yet supple material that supports the great weight of the oak, yet allows the willow to bend with the wind.
Almost every aspect of human life involves carbohydrates in one form or another.
Like other animals, we use the energy content of carbohydrates in our food to produce
and store energy in our cells. Clothing is made from cotton and linen, two forms of cellulose. Other fabrics are made by manipulating cellulose to convert it to the semisynthetic
fibers rayon and cellulose acetate. In the form of wood, we use cellulose to construct our
houses and as a fuel to heat them. Even this page is made from cellulose fibers.
Carbohydrate chemistry is one of the most interesting areas of organic chemistry.
Many chemists are employed by companies that use carbohydrates to make foods, building materials, and other consumer products. All biologists must understand carbohydrates, which play pivotal roles throughout the plant and animal kingdoms. At first
glance, the structures and reactions of carbohydrates may seem complicated. We will
1101
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CHAPTER 23
Carbohydrates and Nucleic Acids
learn how these structures and reactions are consistent and predictable, however, and we
can study carbohydrates as easily as we study the simplest organic compounds.
23-2
Classification of
Carbohydrates
Problem-solving Hint
The Fischer projection represents
each asymmetric carbon atom by a
cross, with the horizontal bonds
projecting toward the viewer and the
vertical bonds projecting away.
The carbon chain is arranged along
the vertical bonds, with the most
oxidized end (or carbon #1 in the
IUPAC name) at the top.
CHO
CHO
H
OH
CH2OH
= H
C* OH
CH2OH
For more than one asymmetric
carbon atom, the Fischer projection
represents a totally eclipsed
conformation. This is not the most
stable conformation, but it’s usually
the most symmetric conformation,
which is most helpful for comparing
stereochemistry.
The term carbohydrate arose because most sugars have molecular formulas
Cn1H2 O2m, suggesting that carbon atoms are combined in some way with water. In
fact, the empirical formula of most simple sugars is C1H2 O2. Chemists named these
compounds “hydrates of carbon” or “carbohydrates” because of these molecular formulas. Our modern definition of carbohydrates includes polyhydroxyaldehydes, polyhydroxyketones, and compounds that are easily hydrolyzed to them.
Monosaccharides, or simple sugars, are carbohydrates that cannot be hydrolyzed
to simpler compounds. Figure 23-1 shows Fischer projections of the monosaccharides
glucose and fructose. Glucose is a polyhydroxyaldehyde, and fructose is a polyhydroxy
ketone. Polyhydroxyaldehydes are called aldoses (ald- is for aldehyde and -ose is the
suffix for a sugar), and polyhydroxyketones are called ketoses (ket- for ketone, and -ose
for sugar).
We have used Fischer projections to draw the structures of glucose and fructose
because Fischer projections conveniently show the stereochemistry at all the asymmetric carbon atoms. The Fischer projection was originally developed by Emil Fischer, a
carbohydrate chemist who received the Nobel Prize for his proof of the structure of
glucose. Fischer developed this shorthand notation for drawing and comparing sugar
structures quickly and easily. We will use Fischer projections extensively in our work
with carbohydrates, so you may want to review them (Section 5-10) and make models
of the structures in Figure 23-1 to review the stereochemistry implied by these structures.
In aldoses, the aldehyde carbon is the most highly oxidized (and numbered 1 in the
IUPAC name), so it is always at the top of the Fischer projection. In ketoses, the
carbonyl group is usually the second carbon from the top.
PROBLEM 23-1
Draw the mirror images of glucose and fructose. Are glucose and fructose chiral? Do you
expect them to be optically active?
A disaccharide is a sugar that can be hydrolyzed to two monosaccharides. For
example, sucrose (“table sugar”) is a disaccharide that can be hydrolyzed to one molecule of glucose and one molecule of fructose.
+
1 sucrose
Fischer projections of sugars.
Glucose and fructose are
monosaccharides. Glucose is an
aldose (a sugar with an aldehyde
group), and fructose is a ketose
(a sugar with a ketone group).
Carbohydrate structures are
commonly drawn using
Fischer projections.
" 1 glucose + 1 fructose
Both monosaccharides and disaccharides are highly soluble in water, and most have
the characteristic sweet taste we associate with sugars.
Polysaccharides are carbohydrates that can be hydrolyzed to many monosaccharide
units. Polysaccharides are naturally occurring polymers (biopolymers) of carbohydrates.
They include starch and cellulose, both biopolymers of glucose. Starch is a polysaccharide whose carbohydrate units are easily added to store energy or removed to provide
CHO
FIGURE 23-1
H3 O
heat
CHO
H
C
OH
HO
C
H
H
C
OH
H
H
C
OH
H
H
or
CH2OH
HO
O
C
O
HO
C
H
OH
H
C
OH
H
OH
OH
H
C
OH
H
OH
OH
H
CH2OH
glucose
CH2OH
CH2OH
C
or
HO
CH2OH
H
CH2OH
fructose
23-3 Monosaccharides
1103
energy to cells. The polysaccharide cellulose is a major structural component of plants.
Hydrolysis of either starch or cellulose gives many molecules of glucose.
+
starch
" over 1000 glucose molecules
H3 O
heat
+
cellulose
H3 O
heat
" over 1000 glucose molecules
To understand the chemistry of these more complex carbohydrates, we must first
learn the principles of carbohydrate structure and reactions, using the simplest monosaccharides as examples. Then we will apply these principles to more complex disaccharides and polysaccharides. The chemistry of carbohydrates applies the chemistry of
alcohols, aldehydes, and ketones to these polyfunctional compounds. In general, the
chemistry of biomolecules can be predicted by applying the chemistry of simple organic
molecules with similar functional groups.
23-3A
23-3
Classification of Monosaccharides
Most sugars have their own specific common names, such as glucose, fructose, galactose, and mannose. These names are not systematic, although there are simple ways to
remember the common structures. We simplify the study of monosaccharides by grouping similar structures together. Three criteria guide the classification of monosaccharides:
1. Whether the sugar contains a ketone or an aldehyde group
2. The number of carbon atoms in the carbon chain
3. The stereochemical configuration of the asymmetric carbon atom farthest from the
carbonyl group
As we have seen, sugars with aldehyde groups are called aldoses, and those with
ketone groups are called ketoses. The number of carbon atoms in the sugar generally
ranges from three to seven, designated by the terms triose (three carbons), tetrose (four
carbons), pentose (five carbons), hexose (six carbons), and heptose (seven carbons).
Terms describing sugars often reflect these first two criteria. For example, glucose has
an aldehyde and contains six carbon atoms, so it is an aldohexose. Fructose also contains six carbon atoms, but it is a ketone, so it is called a ketohexose. Most ketoses have
the ketone on C2, the second carbon atom of the chain. The most common naturally
occurring sugars are aldohexoses and aldopentoses.
1
1
2
2
3
3
1
1
4
4
2
2
5
5
3
3
CHO
CHOH
CHOH
CHOH
CHOH
6
CH2OH
an aldohexose
CH2OH
C
O
CHOH
CHOH
CHOH
6
CH2OH
a ketohexose
CHO
CHOH
CHOH
4
CH2OH
an aldotetrose
CH2OH
C
O
CHOH
4
CH2OH
a ketotetrose
PROBLEM 23-2
(a) How many asymmetric carbon atoms are there in an aldotetrose? Draw all the aldotetrose stereoisomers.
(b) How many asymmetric carbons are there in a ketotetrose? Draw all the ketotetrose
stereoisomers.
(c) How many asymmetric carbons and stereoisomers are there for an aldohexose?
For a ketohexose?
Monosaccharides
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CHAPTER 23
Carbohydrates and Nucleic Acids
PROBLEM 23-3
(a) There is only one ketotriose, called dihydroxyacetone. Draw its structure.
(b) There is only one aldotriose, called glyceraldehyde. Draw the two enantiomers
of glyceraldehyde.
23-3B
The D and L Configurations of Sugars
Around 1880–1900, carbohydrate chemists made great strides in determining the
structures of natural and synthetic sugars. They found ways to build larger sugars out
of smaller ones, adding a carbon atom to convert a tetrose to a pentose and a pentose to a hexose. The opposite conversion, removing one carbon atom at a time
(called a degradation), was also developed. A degradation could convert a hexose to
a pentose, a pentose to a tetrose, and a tetrose to a triose. There is only one aldotriose,
glyceraldehyde.
These chemists noticed they could start with any of the naturally occurring sugars,
and degradation to glyceraldehyde always gave the dextrorotatory 1+2 enantiomer of
glyceraldehyde. Some synthetic sugars, on the other hand, degraded to the levorotatory 1-2 enantiomer of glyceraldehyde. Carbohydrate chemists started using the
Fischer–Rosanoff convention, which uses a D to designate the sugars that degrade to
1+2-glyceraldehyde and an L for those that degrade to 1-2-glyceraldehyde. Although
these chemists did not know the absolute configurations of any of these sugars, the D
and L relative configurations were useful to distinguish the naturally occurring D sugars
from their unnatural L enantiomers.
We now know the absolute configurations of 1+2- and 1-2-glyceraldehyde. These
structures serve as the configurational standards for all monosaccharides.
CHO
H
C
CHO
HO
OH
CH2OH
( + )-glyceraldehyde
D series of sugars
C
H
CH2OH
( – )-glyceraldehyde
L series of sugars
Figure 23-2 shows that degradation (covered in Section 23-14) removes the aldehyde carbon atom, and it is the bottom asymmetric carbon in the Fischer projection
(the asymmetric carbon farthest removed from the carbonyl group) that determines
which enantiomer of glyceraldehyde is formed by successive degradations.
CHO
CO2
H
C
OH
HO
C
H
H
C
H
C
CHO
degrade
CO2
HO
C
H
OH
H
C
OH
H
C
OH
OH
H
C
OH
H
C
OH
CH2OH
D-(+) -glucose
CHO
degrade
CH2OH
CH2OH
D-(–)-arabinose
D-(–)-erythrose
CO2
CHO
degrade
H
C
OH
CH2OH
D-(+)-glyceraldehyde
FIGURE 23-2
Degradation to glyceraldehyde. Degradation of an aldose removes the aldehyde carbon atom
to give a smaller sugar. Sugars of the D series give 1+2-glyceraldehyde on degradation to the
triose. Therefore, the OH group of the bottom asymmetric carbon atom of the D sugars must
be on the right in the Fischer projection.
23-3 Monosaccharides
We now know that the 1+2 enantiomer of glyceraldehyde has its OH group on
the right in the Fischer projection, as shown in Figure 23-2. Therefore, sugars of the
D series have the OH group of the bottom asymmetric carbon on the right in the
Fischer projection. Sugars of the L series have the OH group of the bottom asymmetric
carbon on the left. In the following examples, notice that the D or L configuration is
determined by the bottom asymmetric carbon, and the enantiomer of a D sugar is
always an L sugar.
CH2OH
CHO
HO
CHO
H
H
OH
H
HO
CH OH
C
O
O
CHO
H
C
OH
H
OH
HO
H
HO
H
H
OH
HO
H
H
CH OH
2
D-threose
CH2OH
CH OH
2
L-threose
CH OH
2
D-ribulose
CHO
OH
HO
H
H
OH
OH
HO
CH OH
2
L-ribulose
H
H
CH OH
2
D-xylose
2
L-xylose
As mentioned earlier, most naturally occurring sugars have the D configuration,
and most members of the D family of aldoses (up through six carbon atoms) are found
in nature. Figure 23-3 shows the D family of aldoses. Notice that the D or L configuration does not tell us which way a sugar rotates the plane of polarized light. This must
be determined by experiment. Some D sugars have 1+2 rotations, and others have
1-2 rotations.
CHO
H
OH
CH2OH
D-(+)-glyceraldehyde
CHO
CHO
H
OH
HO
H
OH
H
CH2OH
D-(–)-erythrose
CHO
OH
HO
H
OH
H
OH
HO
H
OH
H
OH
H
H
H
CH2OH
D-(–)-arabinose
CHO
CHO
HO
H
OH
H
OH
HO
H
OH
H
OH
H
OH
H
OH
HO
H
OH
H
OH
H
OH
H
OH
H
CH2OH
D-(+)-allose
CH2OH
D-(+)-altrose
HO
H
H
HO
H
HO
H
H
OH
HO
H
HO
H
H
OH
H
CH2OH
D-(+)-mannose
H
OH
CH2OH
D-(–)-gulose
HO
H
OH
CH2OH
D-(–)-lyxose
CHO
OH
CH2OH
D-(+)-glucose
H
OH
CHO
CHO
OH
H
OH
CH2OH
D-(+)-xylose
H
H
CHO
CHO
H
CHO
OH
CH2OH
D-(–)-threose
CHO
CH2OH
D-(–)-ribose
H
H
CHO
CHO
H
OH
HO
H
OH
HO
H
HO
H
H
HO
H
HO
H
OH
CH2OH
D-(–)-idose
FIGURE 23-3
The D family of aldoses. All these sugars occur naturally except for threose, lyxose, allose,
and gulose.
H
OH
CH2OH
D-(+)-galactose
H
OH
CH2OH
D-(+)-talose
1105
1106
CHAPTER 23
Carbohydrates and Nucleic Acids
On paper, the family tree of D aldoses (Figure 23-3) can be generated by starting with
D-1+2-glyceraldehyde and adding another carbon at the top to generate two aldotetroses:
erythrose with the OH group of the new asymmetric carbon on the right, and threose with
the new OH group on the left. Adding another carbon to these aldotetroses gives four
aldopentoses, and adding a sixth carbon gives eight aldohexoses.* In Section 23-15, we
describe the Kiliani–Fischer synthesis, which actually adds a carbon atom and generates
the pairs of elongated sugars just as we have drawn them in this family tree.
At the time the D and L system of relative configurations was introduced, chemists
could not determine the absolute configurations of chiral compounds. They decided to
draw the D series with the glyceraldehyde OH group on the right, and the L series with
it on the left. This guess later proved to be correct, so it was not necessary to revise all
the old structures.
Problem-solving Hint
Most naturally occurring sugars
are of the D series, with the OH
group of the bottom asymmetric
carbon on the right in the
Fischer projection.
PROBLEM 23-4
Draw and name the enantiomers of the sugars shown in Figure 23-2. Give the relative configuration (D or L) and the sign of the rotation in each case.
PROBLEM 23-5
Which configuration (R or S) does the bottom asymmetric carbon have for the
sugars? Which configuration for the L series?
23-4
Erythro and Threo
Diastereomers
D
series of
Erythrose is the aldotetrose with the OH groups of its two asymmetric carbons situated on
the same side of the Fischer projection, and threose is the diastereomer with the OH groups
on opposite sides of the Fischer projection. These names have evolved into a shorthand way
of naming diastereomers with two adjacent asymmetric carbon atoms. A diastereomer is
called erythro if its Fischer projection shows similar groups on the same side of the
molecule. It is called threo if similar groups are on opposite sides of the Fischer projection.
For example, syn dihydroxylation of trans-crotonic acid gives the two enantiomers of the threo diastereomer of 2,3-dihydroxybutanoic acid. The same reaction
with cis-crotonic acid gives the erythro diastereomer of the product.
COOH
COOH
CH3
C
H
C
H
H
OsO4/H2O2
OH
HO
COOH
trans-crotonic acid
H
HO
H
H
OH
CH3
CH3
(2R,3S)
(2S,3R)
threo-2,3-dihydroxybutanoic acid
COOH
CH3
C
H
COOH
C
H
cis-crotonic acid
OsO4/H2O2
COOH
H
OH
HO
H
H
OH
HO
H
CH3
CH3
(2R,3R)
(2S,3S)
erythro-2,3-dihydroxybutanoic acid
*Drawn in this order, the names of the four aldopentoses (ribose, arabinose, xylose, and lyxose) are remembered
by the mnemonic “Ribs are extra lean.” The mnemonic for the eight aldohexoses (allose, altrose, glucose, mannose,
gulose, idose, galactose, and talose) is “All altruists gladly make gum in gallon tanks.”
23-5 Epimers
1107
The terms erythro and threo are generally used only with molecules that do not
have symmetric ends. In symmetric molecules such as 2,3-dibromobutane and tartaric
acid, the terms meso and (d,l) are preferred because these terms indicate the diastereomer and tell whether or not it has an enantiomer. Figure 23-4 shows the proper use of
the terms erythro and threo for dissymmetric molecules, as well as the terms meso and
(d,l) for symmetric molecules.
CH2CH3
CH2CH3
CH3
H
Br
Br
H
H
Cl
H
Br
H
Br
H
OH
CH3
CH3
erythro
H
Cl
HO
H
CH3
threo
CH3
erythro
2,3 - dibromopentane
threo
3-chlorobutan-2-ol
CH3
CH3
CH3
COOH
COOH
H
Br
Br
H
H
OH
H
H
Br
H
Br
H
OH
HO
CH3
CH3
meso
(±) or (d,l )
2,3 -dibromobutane
COOH
meso
OH
FIGURE 23-4
Erythro and threo nomenclature.
The terms erythro and threo are used
with dissymmetric molecules whose
ends are different. The erythro
diastereomer is the one with similar
groups on the same side of the
Fischer projection, and the threo
diastereomer has similar groups on
opposite sides of the Fischer
projection. The terms meso and 1;2
[or (d,l)] are preferred with
symmetric molecules.
H
COOH
(±) or (d,l)
tartaric acid
PROBLEM 23-6
Draw Fischer projections for the enantiomers of threo-hexane-1,2,3-triol.
HOCH2 ¬ CH1OH2 ¬ CH1OH2 ¬ CH2 CH2 CH3
PROBLEM 23-7
The bronchodilator ephedrine is erythro-2-(methylamino)-1-phenylpropan-1-ol. The decongestant pseudoephedrine is threo-2-(methylamino)-1-phenylpropan-1-ol.
(a) Draw the four stereoisomers of 2-(methylamino)-1-phenylpropan-1-ol, either as Fischer
projections or as three-dimensional representations (dotted lines and wedges).
(b) Label ephedrine and pseudoephedrine. What is the relationship between them?
(c) Label the D and L isomers of ephedrine and pseudoephedrine using the
Fischer–Rosanoff convention.
(d) Both ephedrine and pseudoephedrine are commonly used as racemic mixtures.
Ephedrine is also available as the pure levorotatory 1-2 isomer (Biophedrine®), and
pseudoephedrine is also available as the more active 1+2 isomer (Sudafed®). Can you
label the 1-2 isomer of ephedrine and the 1+2 isomer of pseudoephedrine?
Many common sugars are closely related, differing only by the stereochemistry at a
single carbon atom. For example, glucose and mannose differ only at C2, the first
asymmetric carbon atom. Sugars that differ only by the stereochemistry at a single
carbon are called epimers, and the carbon atom where they differ is generally stated.
If the number of a carbon atom is not specified, it is assumed to be C2. Therefore,
glucose and mannose are “C2 epimers” or simply “epimers.” The C4 epimer of glucose is galactose, and the C2 epimer of erythrose is threose. These relationships are
shown in Figure 23-5.
23-5
Epimers
1108
CHAPTER 23
Carbohydrates and Nucleic Acids
1
CHO
1
C2 epimers
HO
2
H
H
2
OH
HO
3
H
HO
3
H
H
4
OH
H
4
H
5
OH
H
5
6
6
CH2OH
D-mannose
1
CHO
CHO
H
2
OH
HO
3
H
OH
HO
4
H
H
2
OH
HO
2
H
OH
H
5
OH
H
3
OH
H
3
OH
C4 epimers
6
CH2OH
D-glucose
1
4
CH2OH
D-galactose
CHO
1
C2 epimers
4
CH2OH
CHO
CH OH
2
D-threose
D-erythrose
FIGURE 23-5
Epimers are sugars that differ only by the stereochemistry at a single carbon atom. If the
number of the carbon atom is not specified, it is assumed to be C2.
PROBLEM 23-8
(a)
(b)
(c)
(d)
23-6
Cyclic Structures of
Monosaccharides
Draw D-allose, the C3 epimer of glucose.
Draw D-talose, the C2 epimer of D-galactose.
Draw D-idose, the C3 epimer of D-talose. Now compare your answers with Figure 23-3.
Draw the C4 “epimer” of D-xylose. Notice that this “epimer” is actually an L-series
sugar, and we have seen its enantiomer. Give the correct name for this L-series sugar.
Cyclic Hemiacetals In Chapter 18, we saw that an aldehyde reacts with one molecule of an alcohol to give a hemiacetal, and with a second molecule of the alcohol to
give an acetal. The hemiacetal is not as stable as the acetal, and most hemiacetals
decompose spontaneously to the aldehyde and the alcohol. Therefore, hemiacetals are
rarely isolated.
If the aldehyde group and the hydroxyl group are part of the same molecule, a
cyclic hemiacetal results. Cyclic hemiacetals are particularly stable if they result in fiveor six-membered rings. In fact, five- and six-membered cyclic hemiacetals are often
more stable than their open-chain forms.
MECHANISM 23-1 Formation of a Cyclic Hemiacetal
Step 1: Protonation of the carbonyl.
H
O
C
H+
H
Step 2: The OH group adds as a nucleophile.
H
O
C
H
O+
H
+
O
O
H
δ-hydroxyaldehyde
Step 3: Deprotonation gives a cyclic hemiacetal.
derived from
the OH group
H
O+
H2O
C
H
O
H
O
H
O H
cyclic hemiacetal
+
H3O+
derived from
the CHO group
C
H
O
H
1109
23-6 Cyclic Structures of Monosaccharides
The Cyclic Hemiacetal Form of Glucose Aldoses contain an aldehyde group and
several hydroxyl groups. The solid, crystalline form of an aldose is normally a cyclic
hemiacetal. In solution, the aldose exists as an equilibrium mixture of the cyclic
hemiacetal and the open-chain form. For most sugars, the equilibrium favors the
cyclic hemiacetal.
Aldohexoses such as glucose can form cyclic hemiacetals containing either fivemembered or six-membered rings. For most common aldohexoses, the equilibrium favors
six-membered rings with a hemiacetal linkage between the aldehyde carbon and the
hydroxyl group on C5. Figure 23-6 shows formation of the cyclic hemiacetal of glucose.
Notice that the hemiacetal has a new asymmetric carbon atom at C1. Figure 23-6 shows
both possibilities at C1: The hydroxyl group can be directed upward in the equatorial position, or it can be directed downward in the axial position. We discuss the stereochemistry
at C1 in more detail in Section 23-7.
The cyclic structure is often drawn initially in the Haworth projection, which
depicts the ring as being flat (of course, it is not). The Haworth projection is widely
used in biology texts, but most chemists prefer to use the more realistic chair conformation. Figure 23-6 shows the cyclic form of glucose both as a Haworth projection and
as a chair conformation.
Drawing Cyclic Monosaccharides Cyclic hemiacetal structures may seem complicated at first glance, but they can be drawn and recognized by following the process
illustrated in Figure 23-6.
1. Mentally lay the Fischer projection over on its right side. The groups that were
on the right in the Fischer projection are down in the cyclic structure, and the
groups that were on the left are up.
2. C5 and C6 curl back away from you. The C4 ¬ C5 bond must be rotated so that
the C5 hydroxyl group can form a part of the ring. For a sugar of the D series, this
rotation puts the terminal ¬ CH2 OH (C6 in glucose) upward.
3. Close the ring, and draw the result. Always draw the Haworth projection or chair
conformation with the oxygen at the back, right-hand corner, with C1 at the far
right. C1 is easily identified because it is the hemiacetal carbon—the only carbon
bonded to two oxygens. The hydroxyl group on C1 can be either up or down, as
discussed in Section 23-7.
1
CHO
HO
H
4
5
OH
6
H
H
=
CH2OH O
H C
1
H
OH
OH
4
OH
6
5
H
OH
HO
3
CH2OH
6
Fischer projection
H
HO 4
HO
H+
2
OH
H
on right side
6
CH2OH
5
H
3
H
C1 is the only carbon atom
bonded to 2 oxygens
O
H
2
OH
1
6
CH2OH
OH
H
chair conformation (all substituents equatorial)
FIGURE 23-6
Glucose exists almost entirely as its cyclic hemiacetal form.
H
4
HO
O ..
5
H
OH
3
H
H
H
H
O+
..
H
3
..
H
2
H C
1
4
H
5
CH2OH
HO
2
OH
O OH
H
OH
3
H
C6 rotated up
1
H
2
H
OH
Haworth projection
(OH on C1 up)
H
HO 4
HO
6
CH2OH
5
H
3
H
O
H
2
OH 1
OH
H
chair conformation (OH on C1 axial)
+ H+
1110
CHAPTER 23
Carbohydrates and Nucleic Acids
Chair conformations can be drawn by recognizing the differences between the sugar
in question and glucose. The following procedure is useful for drawing D-aldohexoses.
1. Draw the chair conformation puckered, as shown in Figure 23-6. The hemiacetal
carbon (C1) is drawn at far right (as the footrest), and the ring oxygen is at the
back, right corner.
2. Glucose has its substituents on alternating sides of the ring. In drawing the chair
conformation, just put all the ring substituents in equatorial positions.
3. To draw or recognize other common sugars, notice how they differ from glucose
and make the appropriate changes.
Problem-solving Hint
Learn to draw glucose, both in the
Fischer projection and in the chair
conformation (all substituents
equatorial). Draw other pyranoses
by noticing the differences from
glucose and changing the glucose
structure as needed.
Remember the epimers of
glucose (C2: mannose; C3: allose;
and C4: galactose). To recognize
other sugars, look for axial
substituents where they differ
from glucose.
SOLVED PROBLEM 23-1
Draw the cyclic hemiacetal forms of D-mannose and D-galactose both as chair conformations
and as Haworth projections. Mannose is the C2 epimer of glucose, and galactose is the C4
epimer of glucose.
SOLUTION
The chair conformations are easier to draw, so we will do them first. Draw the rings exactly
as we did for glucose in Figure 23-6. Number the carbon atoms, starting with the hemiacetal
carbon. Mannose is the C2 epimer of glucose, so the substituent on C2 is axial, while all the
others are equatorial as in glucose. Galactose is the C4 epimer of glucose, so its substituent on
C4 is axial.
H
OH 6
CH2OH
6
CH2OH
4
5
HO
C4
O
H
HO
OH
H
C2
3
H
Groups on the right in the Fischer
projection are down in the usual
cyclic structure, and groups that
were on the left in the Fischer
projection are up.
2
3
H
OH
OH 1
H
H
D-mannose
Problem-solving Hint
H
H
HO
1
O
5
H
OH
D-galactose
The simplest way to draw Haworth structures for these two sugars is to draw the chair
conformations and then draw the flat rings with the same substituents in the up and down positions. For practice, however, let’s lay down the Fischer projection for galactose. You should follow along with your molecular models.
1. Lay down the Fischer projection: right : down and left : up.
1 CHO
H
HO
HO
H
2
3
H
4
H
5
6
H
OH
HO
=
4
H
OH
6
5
CH2OH
OH
OH
3
1C
H
O
H
2
H
OH
CH2OH
D-galactose
2. Rotate the C4 ¬ C5 bond to put the C5 ¬ OH in place. (For a
¬ CH2 OH goes up.)
6
H
5
HO
4
H
OH
OH
3
H
6
CH2OH
H
2
OH
1C
O
CH2OH
5
OH
HO
H
4
H
H
OH
3
H
H
2
OH
O
C
1
H
D
sugar, the
1111
23-6 Cyclic Structures of Monosaccharides
3. Close the ring, and draw the final hemiacetal. The hydroxyl group on C1 can be
either up or down, as discussed in Section 23-7. Sometimes this ambiguous stereochemistry is symbolized by a wavy line.
equivocal
HO
H
CH2OH
O
H
OH
C
OH
H
H
OH
CH2OH
O
H
HO
or
H
H
H
OH
H
H
OH
HO
or
C
OH
H
CH2OH
O
H
OH
C
OH
H
H
OH
H
PROBLEM 23-9
Draw the Haworth projection for the cyclic structure of D-mannose by laying down the Fischer
projection.
PROBLEM 23-10
Allose is the C3 epimer of glucose. Draw the cyclic hemiacetal form of D-allose, first in the
chair conformation and then in the Haworth projection.
The Five-Membered Cyclic Hemiacetal Form of Fructose Not all sugars exist as sixmembered rings in their hemiacetal forms. Many aldopentoses and ketohexoses form
five-membered rings. The five-membered hemiacetal ring of fructose is shown in
Figure 23-7.* Five-membered rings are not puckered as much as six-membered rings,
so they are usually depicted as flat Haworth projections. The five-membered ring is
customarily drawn with the ring oxygen in back and the hemiacetal carbon (the one
bonded to two oxygens) on the right. The ¬ CH2 OH at the back left (C6) is in the up
position for D-series ketohexoses.
Pyranose and Furanose Names Cyclic structures of monosaccharides are named
according to their five- or six-membered rings. A six-membered cyclic hemiacetal is
called a pyranose, derived from the name of the six-membered cyclic ether pyran. A
five-membered cyclic hemiacetal is called a furanose, derived from the name of the
five-membered cyclic ether furan. For example, the six-membered ring of glucose is
called glucopyranose, and the five-membered ring of fructose is called fructofuranose.
The ring is still numbered as it is in the sugar.
2C
H
4
H
5
H
O
H
OH
OH
+
H
HO
6
H
+
CH 2 O..
O
H
..
HO
3
2OH
5
H
H4
OH
HO
2C
3
H
6
HOCH 2 O+
5
1CH
2OH
H
H4
OH
2
OH
HO C
3
6
HOCH 2 O
H 2O ..
..
1 CH
5
1CH
H
2OH
protonated cyclic form
6CH OH
2
D-fructose
FIGURE 23-7
Fructose forms a five-membered cyclic hemiacetal.* Five-membered rings are usually
represented as flat Haworth structures.
*Although the IUPAC has dropped the term “ketal” for the acetal of a ketone, most carbohydrate chemists still
use the term. Therefore, the cyclic hemiacetal of fructose is often called a hemiketal.
H
H4
OH
OH
2
HO C
3
1CH
H
cyclic form
2OH
1112
5
6
4
1
CHAPTER 23
H
3
2
O
pyran
HO
H
HO
H OH
2
1
3
Carbohydrates and Nucleic Acids
H
H
HO
4
OH
5
H
O 6 CH OH
2
6
CH2OH
4
5
H
HO
H
3
OH
H
5
2
O
H
D-glucopyranose
a pyranose
1
HOCH2 O
2 OH
5
H HO
CH2OH
3
H 4
1
OH H
3 4
4 3
OH
1
2
6
HO H
OH
H
O
HO 1 2
HOCH2O
furan
5
6
H
CH2OH
D-fructofuranose
a furanose
PROBLEM 23-11
Talose is the C4 epimer of mannose. Draw the chair conformation of D-talopyranose.
PROBLEM 23-12
(a) Figure 23-2 shows that the degradation of D-glucose gives D-arabinose, an aldopentose.
Arabinose is most stable in its furanose form. Draw D-arabinofuranose.
(b) Ribose, the C2 epimer of arabinose, is most stable in its furanose form. Draw
D-ribofuranose.
PROBLEM 23-13
The carbonyl group in D-galactose may be isomerized from C1 to C2 by brief treatment with
dilute base (by the enediol rearrangement, Section 23-8). The product is the C4 epimer of fructose. Draw the furanose structure of the product.
23-7
Anomers of
Monosaccharides;
Mutarotation
When a pyranose or furanose ring closes, the flat carbonyl group is converted to an
asymmetric carbon in the hemiacetal. Depending on which face of the (protonated)
carbonyl group is attacked, the hemiacetal ¬ OH group can be directed either up or
down. These two orientations of the hemiacetal ¬ OH group give diastereomeric
products called anomers. Figure 23-8 shows the anomers of glucose.
The hemiacetal carbon atom is called the anomeric carbon, easily identified as the
only carbon atom bonded to two oxygens. Its ¬ OH group is called the anomeric hydroxyl
group. The structure with the anomeric ¬ OH group down (axial) is called the a (alpha)
anomer, and the one with the anomeric ¬ OH group up (equatorial) is called the b (beta)
anomer. We can draw the a and b anomers of most aldohexoses by remembering that the
b form of glucose ( b -D-glucopyranose) has all its substituents in equatorial positions. To
draw an a anomer, simply move the anomeric ¬ OH group to the axial position.
anomeric carbon
H
HO 4
CH2OH
5
HO
H
6
H
3
H
O
H
2
CH2OH
HO 4
OH 1
H
5
HO
H
3
H
OH
α-D-glucopyranose
H
6
O
H
2
OH C
1
open-chain form
H
HO 4
O
H
6
CH2OH
5
HO
H
3
H
O
H
2
OH 1
H
OH
β-D-glucopyranose
FIGURE 23-8
The anomers of glucose. The hydroxyl group on the anomeric (hemiacetal) carbon is down
(axial) in the a anomer and up (equatorial) in the b anomer. The b anomer of glucose has all
its substituents in equatorial positions.
Another way to remember the anomers is to notice that the anomeric hydroxyl group
is trans to the terminal ¬ CH2 OH group in the a anomer, but it is cis in the b anomer.
This rule works for all sugars, from both the D and L series, as well as for furanoses.
Figure 23-9 shows the two anomers of fructose, whose anomeric carbon is C2. The
a anomer has the anomeric ¬ OH group down, trans to the terminal ¬ CH2 OH group,
while the b anomer has it up, cis to the terminal ¬ CH2 OH.
23-7 Anomers of Monosaccharides; Mutarotation
anomeric carbon
6
1
HOCH2 O
5
H
HO
cis = β
6
CH2OH
HOCH2 OH
5
2
H 4 3 OH
OH H
H
O
6
HOCH2 O
1
HO C 2 CH2OH
5
OH
HO 2
H 4 3 CH2OH
1
OH H
H 4 3
OH H
trans = α
α-D-fructofuranose
H
β-D-fructofuranose
FIGURE 23-9
The a anomer of fructose has the anomeric ¬ OH group down, trans to the terminal ¬ CH2 OH
group. The b anomer has the anomeric hydroxyl group up, cis to the terminal ¬ CH2 OH.
PROBLEM 23-14
Draw the following monosaccharides, using chair conformations for the pyranoses and Haworth
projections for the furanoses.
(a) a-D-mannopyranose (C2 epimer of glucose)
(b) b -D-galactopyranose (C4 epimer of glucose)
(c) b -D-allopyranose (C3 epimer of glucose)
(d) a-D-arabinofuranose
(e) b -D-ribofuranose (C2 epimer of arabinose)
Properties of Anomers: Mutarotation Because anomers are diastereomers, they
generally have different properties. For example, a-D-glucopyranose has a melting
point of 146 °C and a specific rotation of +112.2°, while b -D-glucopyranose has a
melting point of 150 °C and a specific rotation of +18.7°. When glucose is crystallized
from water at room temperature, pure crystalline a-D-glucopyranose results. If glucose is crystallized from water by letting the water evaporate at a temperature above
98 °C, crystals of pure b -D-glucopyranose are formed (Figure 23-10).
H
HO
HO
CH2OH
H
H
O
H
OH
H
HO
HO
H
HO
HO
H
H
H
H
O
H
OH C
HO
O
HO
H
equilibrium in solution
OH
H2O
H
equilibrium mixture of α and β
[α] = +52.6°
H
O
H
OH
OH
H
β anomer
crystallize above 98 °C
H
O
H
CH2OH
H
open-chain form
crystallize below 98 °C
CH2OH
H
H
OH
α anomer
H
CH2OH
H2O
HO
OH
pure α anomer
mp 146 °C, [α] = + 112.2°
FIGURE 23-10
An aqueous solution of D-glucose contains an equilibrium mixture of a-D-glucopyranose,
b -D-glucopyranose, and the intermediate open-chain form. Crystallization below 98 °C gives
the a anomer, and crystallization above 98 °C gives the b anomer.
HO
CH2OH
H
H
O
H
OH
OH
H
pure β anomer
mp 150 °C, [α] = + 18.7°
1113
1114
CHAPTER 23
Carbohydrates and Nucleic Acids
In each of these cases, all the glucose in the solution crystallizes as the favored
anomer. In the solution, the two anomers are in equilibrium through a small amount of
the open-chain form, and this equilibrium continues to supply more of the anomer that
is crystallizing out of solution.
When one of the pure glucose anomers dissolves in water, an interesting change in
the specific rotation is observed. When the a anomer dissolves, its specific rotation gradually decreases from an initial value of +112.2° to +52.6°. When the pure b anomer dissolves, its specific rotation gradually increases from +18.7° to the same value of +52.6°.
This change (“mutation”) in the specific rotation is called mutarotation. Mutarotation
occurs because the two anomers interconvert in solution. When either of the pure anomers
dissolves in water, its rotation gradually changes to an intermediate rotation that results
from equilibrium concentrations of the anomers. The specific rotation of glucose is usually listed as +52.6°, the value for the equilibrium mixture of anomers. The positive sign
of this rotation is the source of the name dextrose, an old common name for glucose.
SOLVED PROBLEM 23-2
Calculate how much of the a anomer and how much of the b anomer are present in an equilibrium mixture with a specific rotation of +52.6°.
SOLUTION
If the fraction of glucose present as the a anomer 1[a] = +112.2°2 is a, the fraction present
as the b anomer 1[a] = +18.7°2 is b, and the rotation of the mixture is +52.6°, we have
a1+112.2°2 + b1+18.7°2 = +52.6°
There is very little of the open-chain form present, so the fraction present as the a anomer
(a) plus the fraction present as the b anomer (b) should account for all the glucose:
a + b = 1 or b = 1 - a
Substituting 11 - a2 for b in the first equation, we have
a1112.2°2 + 11 - a2 118.7°2 = 52.6°
Solving this equation for a, we have a = 0.36, or 36%. Thus, b must be 11 - 0.362 = 0.64,
or 64%. The amounts of the two anomers present at equilibrium are
a anomer, 36%
b anomer, 64%
When we remember that the anomeric hydroxyl group is axial in the a anomer and equatorial
in the b anomer, it is reasonable that the more stable b anomer should predominate.
PROBLEM 23-15
Like glucose, galactose mutarotates when it dissolves in water. The specific rotation of
a-D-galactopyranose is +150.7°, and that of the b anomer is +52.8°. When either of the pure
anomers dissolves in water, the specific rotation gradually changes to +80.2°. Determine the
percentages of the two anomers present at equilibrium.
23-8
Reactions of
Monosaccharides:
Side Reactions
in Base
Sugars are multifunctional compounds that can undergo reactions typical of any of their
functional groups. Most sugars exist as cyclic hemiacetals, yet in solution they are in
equilibrium with their open-chain aldehyde or ketone forms. As a result, sugars undergo
most of the usual reactions of ketones, aldehydes, and alcohols. Reagents commonly
used with monofunctional compounds often give unwanted side reactions with sugars,
however. Carbohydrate chemists have developed reactions that work well with sugars
while avoiding the undesired side reactions. As we learn about the unique reactions of
simple sugars, we will often draw them as their open-chain forms because it is often the
small equilibrium amount of the open-chain form that reacts.
23-8 Reactions of Monosaccharides: Side Reactions in Base
1115
Epimerization and the Enediol Rearrangement One of the most important aspects of
sugar chemistry is the inability, in most cases, to use basic reagents because they cause
unwanted side reactions. Two common base-catalyzed side reactions are epimerization and enediol rearrangement.
Under basic conditions, the proton alpha to the aldehyde (or ketone) carbonyl group
is reversibly removed (shown in Mechanism 23-2). In the resulting enolate ion, C2 is
no longer asymmetric, and its stereochemistry is lost. Reprotonation can occur on either
face of the enolate, giving either configuration. The result is an equilibrium mixture of
the original sugar and its C2 epimer. Because a mixture of epimers results, this stereochemical change is called epimerization. The mechanism involves rapid base-catalyzed
equilibration of glucose to a mixture of glucose and its C2 epimer, mannose.
MECHANISM 23-2
Base-Catalyzed Epimerization of Glucose
Step 1: Abstraction of the a proton.
H
O
Step 2: Reprotonation on the other face.
O
H
C
HO –
H
HO
C
H
O
–
–OH
H
C
HO
O
H
C
C
OH
–
C
C
OH
OH
HO
H
H
H
C
HO
O
H
HO
H
H
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
CH2OH
CH2OH
CH2OH
D-glucose
+
–OH
CH2OH
enolate
D-mannose
PROBLEM 23-16
Propose a mechanism for the base-catalyzed epimerization of erythrose to a mixture of erythrose and threose.
Another base-catalyzed side reaction is the enediol rearrangement, which moves
the carbonyl group up and down the chain, as shown in Mechanism 23-3. If the enolate
ion (formed by removal of a proton on C2) reprotonates on the C1 oxygen, an enediol
intermediate results. Removal of a proton from the C2 oxygen and reprotonation on C1
gives fructose, a ketose.
MECHANISM 23-3
Base-Catalyzed Enediol Rearrangement
Step 1: Remove the a proton.
–
Step 2: Reprotonate on oxygen to give the enediol.
H
HO
O
C
C
H
HO
C
O–
H
C
OH
H
HO
H
C
+ H2O
C
OH
H
OH
HO
OH
H
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
CH2OH
D-glucose
CH2OH
enolate
CH2OH
enediol
(Continued)
1116
Carbohydrates and Nucleic Acids
CHAPTER 23
Step 3: Deprotonate the oxygen on C2.
H
Step 4: Reprotonate on C1 to give the ketose.
OH
H
+ –OH
C
C
C
C
OH
HO
H
H
OH
HO
O–
H
+ H2O
C
OH + –OH
C
O
HO
H
H
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
CH2OH
CH2OH
CH2OH
enediol
enolate
D-fructose
Under strongly basic conditions, the combination of enediol rearrangements and
epimerization leads to a complex mixture of sugars. Except when using protected sugars, most chemists doing sugar chemistry employ neutral or acidic reagents to avoid
these annoying side reactions.
PROBLEM 23-17
Show how C3 of fructose can epimerize under basic conditions.
PROBLEM 23-18
Show how another enediol rearrangement can move the carbonyl group from C2 in fructose to C3.
23-9
Like other aldehydes and ketones, aldoses and ketoses can be reduced to the corresponding polyalcohols, called sugar alcohols or alditols. The most common reagents are
sodium borohydride or catalytic hydrogenation using a nickel catalyst. Alditols are
named by adding the suffix -itol to the root name of the sugar. The following equation
shows the reduction of glucose to glucitol, sometimes called sorbitol.
Reduction of
Monosaccharides
H
C
H
H
CH2OH
HO
H
HO
O
HO
H
OH
OH
H
β- D-glucopyranose
H
CH2OH
O
OH
H
H
H2, Ni
HO
OH
H
H
OH
H
OH
H
OH
H
OH
CH2OH
open-chain aldehyde
CH OH
2
D-glucitol (D-sorbitol)
an alditol
Reduction of a ketose creates a new asymmetric carbon atom, formed in either of two
configurations, resulting in two epimers. Figure 23-11 shows how the reduction of
fructose gives a mixture of glucitol and mannitol.
Sugar alcohols are widely used in industry, primarily as food additives and sugar
substitutes. Glucitol has the common name sorbitol because it was first isolated from
23-10 Oxidation of Monosaccharides; Reducing Sugars
CH2OH
CH2OH
HOCH2 O
H HO
H
OH
OH H
α- D-fructofuranose
C
HO
CH2OH
O
H
H
NaBH4
HO
CH2OH
OH
HO
H
H
HO
H
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
CH2OH
open-chain ketone
CH OH
2
D-glucitol
1117
CH2OH
+
D-mannitol
a mixture of alditols
FIGURE 23-11
Reduction of fructose creates a new asymmetric carbon atom, which can have either
configuration. The products are a mixture of glucitol and mannitol.
the berries of the mountain ash, Sorbus aucuparia. Industrially, sorbitol is made by
catalytic hydrogenation of glucose. Sorbitol is used as a sugar substitute, a moistening agent, and a starting material for making vitamin C. Mannitol was first isolated
from plant exudates known as mannas (of Biblical fame), the origin of the names
mannose and mannitol. Mannitol is derived commercially from seaweed, or it can be
made by catalytic hydrogenation of mannose. Galactitol (dulcitol) also can be
obtained from plants, or it can be made by catalytic hydrogenation of galactose.
PROBLEM 23-19
When D-glucose is reduced with sodium borohydride, optically active glucitol results. When
optically active D-galactose is reduced, however, the product is optically inactive. Explain this
loss of optical activity.
PROBLEM 23-20
Emil Fischer synthesized L-gulose, an unusual aldohexose that reduces to give D-glucitol. Suggest
a structure for this L sugar, and show how L-gulose gives the same alditol as D-glucose. (Hint:
D-Glucitol has ¬ CH2 OH groups at both ends. Either of these primary alcohol groups might
have come from reduction of an aldehyde.)
Monosaccharides are oxidized by a variety of reagents. The aldehyde group of an aldose
oxidizes easily. Some reagents also selectively oxidize the terminal ¬ CH2 OH group
at the far end of the molecule. Oxidation is used to identify the functional groups of
a sugar, to help to determine its stereochemistry, and as part of a synthesis to convert
one sugar into another.
Bromine Water Bromine water oxidizes the aldehyde group of an aldose to a carboxylic acid. Bromine water is used for this oxidation because it does not oxidize
the alcohol groups and it does not oxidize ketoses. Also, bromine water is acidic and
does not cause epimerization or rearrangement of the carbonyl group. Because
bromine water oxidizes aldoses but not ketoses, it serves as a useful test to distinguish aldoses from ketoses. The product of bromine water oxidation is an aldonic
acid (older term: glyconic acid). For example, bromine water oxidizes glucose to
gluconic acid.
23-10
Oxidation of
Monosaccharides;
Reducing Sugars
1118
Carbohydrates and Nucleic Acids
CHAPTER 23
Example
aldehyde
O
CHO
acid
H
O
C
H
OH
COOH
OH
H
OH
C
H
Br2
HO
H
OH
H2O
H
OH
H
OH
H
OH
HO
Br2
(CHOH)n
(CHOH)n
H2O
CH2OH
CH2OH
aldose
aldonic acid
(glyconic acid)
CH2OH
H
CH2OH
glucose
gluconic acid
PROBLEM 23-21
Draw and name the products of bromine water oxidation of
(b) D-galactose
(a) D-mannose
(c)
D-fructose
Nitric Acid Nitric acid is a stronger oxidizing agent than bromine water, oxidizing both
the aldehyde group and the terminal ¬ CH2 OH group of an aldose to carboxylic acid
groups. The resulting dicarboxylic acid is called an aldaric acid (older terms: glycaric
acid or saccharic acid). For example, nitric acid oxidizes glucose to glucaric acid.
Example
COOH
CHO
HO
HNO3
(CHOH)n
alcohol CH2OH
Application: Blood Glucose Meters
Diabetics must monitor their blood
glucose levels several times a day. An
electronic glucose meter uses test
strips that contain electrodes bordering
a reaction chamber. A drop of blood
wicks up into the reaction chamber,
which is impregnated with glucose oxidase, an enzyme that specifically catalyzes oxidation of the aldehyde group
by oxygen in the air. The electrons liberated in this oxidation travel through the
electrodes and produce a current that is
directly proportional to the concentration of glucose in the blood sample.
H
HNO3
HO
H
H
OH
H
OH
H
OH
H
OH
acid COOH
aldose
aldaric acid
(glycaric acid)
CH2OH
COOH
glucose
glucaric acid
PROBLEM 23-22
Draw and name the products of nitric acid oxidation of
(b) D-galactose
(a) D-mannose
PROBLEM 23-23
Two sugars, A and B, are known to be glucose and galactose, but it is not certain which one
is which. On treatment with nitric acid, A gives an optically inactive aldaric acid, while B
gives an optically active aldaric acid. Which sugar is glucose, and which is galactose?
Tollens Test Tollens test detects aldehydes, which react with Tollens reagent to give
carboxylate ions and metallic silver, often in the form of a silver mirror on the inside
of the container.
O
C
OH
acid COOH
(CHOH)n
R
H
OH
H
aldehyde CHO
O
+
H + 2 Ag(NH3)2
aldehyde
–
OH
Tollens reagent
+
–
OH
R
C
O–
oxidized
acid anion
+
2 Ag
+
4 NH3 +
2 H2O
reduced
silver mirror
In its open-chain form, an aldose has an aldehyde group, which reacts with Tollens
reagent to give an aldonic acid and a silver mirror. This oxidation is not a good
23-11 Nonreducing Sugars: Formation of Glycosides
1119
synthesis of the aldonic acid, however, because Tollens reagent is strongly basic and
promotes epimerization and enediol rearrangements. Sugars that reduce Tollens reagent
to give a silver mirror are called reducing sugars.
acid COO–
aldehyde CHO
H
H
CH2OH
HO
O
H
HO
OH
OH
H
H
β -D-glucose
OH
H
Ag(NH3)+2 –OH
HO
H
OH
(Tollens reagent)
H
OH
H
OH
H
OH
HO
H
H
OH
CH2OH
NH4+
H
+
Ag
silver
mirror
CH2OH
open-chain form
gluconic acid
(+ side products)
Tollens test cannot distinguish between aldoses and ketoses because the basic Tollens
reagent promotes enediol rearrangements. Under basic conditions, the open-chain form
of a ketose can isomerize to an aldose, which reacts to give a positive Tollens test.
H
CH2OH
C
OH
O
H
C
C
–OH
O
C
R
OH
–OH
H
C
OH
Ag(NH3)2+ – OH
R
R
a ketose
O
enediol intermediate
HO
CH2OH acetal
O
H
H
H
OH H
–
NH4+
C
C
OH +
Ag
R
an aldose
positive Tollens test
What good is the Tollens test if it doesn’t distinguish between aldoses and ketoses? The
answer lies in the fact that Tollens reagent must react with the open-chain form of the
sugar, which has a free aldehyde or ketone. If the cyclic form cannot open to the free
carbonyl compound, the sugar does not react with Tollens reagent. Hemiacetals are easily
opened, but an acetal is stable under neutral or basic conditions (Section 18-17). If the
carbonyl group is in the form of a cyclic acetal, the cyclic form cannot open to the free
carbonyl compound, and the sugar gives a negative Tollens test (Figure 23-12).
H
H
O
23-11
Nonreducing
Sugars: Formation
of Glycosides
+ –
Ag(NH3)2 OH
no reaction
OR
OH
a glycoside
Examples of nonreducing sugars
H
HO
HO
CH2OH
H
H
acetal
O
H
OH
HOCH2 O
CH2OH
H HO
OCH3
H
methyl β-D-glucopyranoside
(or methyl β-D-glucoside)
H
OCH2CH3
OH H
ethyl α-D-fructofuranoside
(or ethyl α-D-fructoside)
acetal
FIGURE 23-12
Glycosides. Sugars that are full
acetals are stable to Tollens reagent
and are nonreducing sugars. Such
sugars are called glycosides.
1120
CHAPTER 23
Carbohydrates and Nucleic Acids
Sugars in the form of acetals are called glycosides, and their names end in the -oside
suffix. For example, a glycoside of glucose would be a glucoside, and if it were a
six-membered ring, it would be a glucopyranoside. Similarly, a glycoside of ribose
would be a riboside, and if it were a five-membered ring, it would be a ribofuranoside.
In general, a sugar whose name ends with the suffix -ose is a reducing sugar, and one
whose name ends with -oside is nonreducing. Because they exist as stable acetals
rather than hemiacetals, glycosides cannot spontaneously open to their open-chain
forms, and they do not mutarotate. They are locked in a particular anomeric form.
We can summarize by saying that Tollens test distinguishes between reducing sugars and nonreducing sugars: Reducing sugars (aldoses and ketoses) are hemiacetals, and
they mutarotate. Nonreducing sugars (glycosides) are acetals, and they do not mutarotate.
PROBLEM 23-24
Which of the following are reducing sugars? Comment on the common name sucrose for
table sugar.
(b) b -L-idopyranose (an aldohexose)
(a) methyl a-D-galactopyranoside
(d) ethyl b -D-ribofuranoside
(c) a-D-allopyranose
(e)
(f)
H
H
CH2OH O
CH2OH O
HO
HO
H
H
H
H
H
H
HO
HO
OH
OH
O
H
H
O
H CH2
HOCH2 O
O
HO
H
H HO
H
H
OH
HO
OH
CH2OH
OH H
H
H
sucrose
PROBLEM 23-25
Draw the structures of the compounds named in Problem 23-24 parts (a), (c), and (d). Allose
is the C3 epimer of glucose, and ribose is the C2 epimer of arabinose.
Formation of Glycosides Recall that aldehydes and ketones are converted to acetals
by treatment with an alcohol and a trace of acid catalyst (Section 18-17). These conditions also convert aldoses and ketoses to the acetals we call glycosides. Regardless of
the anomer used as the starting material, both anomers of the glycoside are formed
(as an equilibrium mixture) under these acidic conditions. The more stable anomer
predominates. For example, the acid-catalyzed reaction of glucose with methanol
gives a mixture of methyl glucosides.
H
H
CH2OH
HO
H
HO
O
CH3OH, H+
H
H
OH
H
α- D-glucopyranose
(either α or β )
OH
H2O, H+
CH2OH
HO
H
HO
O
H
H
OH
H
β glycosidic bond
H
OCH3
α glycosidic bond
aglycone
methyl α- D-glucopyranoside
+
CH2OH
HO
H
HO
O
H
OH
H
OCH3
H
aglycone
methyl β- D-glucopyranoside
Like other acetals, glycosides are stable to basic conditions, but they hydrolyze in aqueous acid to a free sugar and an alcohol. Glycosides are stable with basic reagents and
in basic solutions.
23-12 Ether and Ester Formation
1121
NH2
H
HO
aglycone
CH2OH
HO H
H
O
H
OH
H
H
OCH2CH3
aglycone
ethyl α-D-glucopyranoside
H
HO
N
HOCH2 O
H
aglycone
N
H
O
HO
HO H
H
H
OH OH
HO H
H
O
H
OH
O
H
H
cytidine, a nucleoside (Section 23-20)
CH2
HO
HO H
H
OH
C
O
CH
N
Ph
O
CH2OH
H
aglycone
CH2OH
HO H
H O
O
H
N C
NH
H
H3C
H
amygdalin
a component of laetrile, a controversial cancer drug
OH
salicin, from willow bark
RO
aglycone
O
H
O
H
H
H
CH2OH
CH2OH
C
H
CH2
protein
O
a glycoprotein N-glycoside
(showing the linkage from carbohydrate to protein)
FIGURE 23-13
Aglycones. The group bonded to the anomeric carbon of a glycoside is called an aglycone.
Some aglycones are bonded through an oxygen atom (a true acetal), and others are bonded
through other atoms such as nitrogen (an aminoglycoside).
An aglycone is the group bonded to the anomeric carbon atom of a glycoside. For
example, methanol is the aglycone in a methyl glycoside. Many aglycones are bonded
through an oxygen atom, but others are bonded through a nitrogen atom or some other heteroatom. Figure 23-13 shows the structures of some glycosides with interesting aglycones.
Disaccharides and polysaccharides are glycosides in which the alcohol forming
the glycoside bond is an ¬ OH group of another monosaccharide. We will consider
disaccharides and polysaccharides in Sections 23-17 and 22-18.
PROBLEM 23-26
The mechanism of glycoside formation is the same as the second part of the mechanism for
acetal formation. Propose a mechanism for the formation of methyl b -D-glucopyranoside.
PROBLEM 23-27
Show the products that result from hydrolysis of amygdalin in dilute acid. Can you suggest why
amygdalin might be toxic to tumor (and possibly other) cells?
PROBLEM 23-28
Treatment of either anomer of fructose with excess ethanol in the presence of a trace of HCl
gives a mixture of the a and b anomers of ethyl-D-fructofuranoside. Draw the starting materials, reagents, and products for this reaction. Circle the aglycone in each product.
Because they contain several hydroxyl groups, sugars are very soluble in water and
rather insoluble in organic solvents. Sugars are difficult to recrystallize from water
because they often form supersaturated syrups like honey and molasses. If the hydroxyl
groups are alkylated to form ethers, sugars behave like simpler organic compounds.
The ethers are soluble in organic solvents, and they are more easily purified by recrystallization and simple chromatographic methods.
Application: Diabetes
Many diabetics have long-standing elevated blood glucose levels. In the openchain form, glucose condenses with the
amino groups of proteins. This glycosylation of proteins may cause some of the
chronic effects of diabetes.
23-12
Ether and Ester
Formation
1122
Carbohydrates and Nucleic Acids
CHAPTER 23
R
H δ+ δ+
C I
H
H
O
H
sugar hydroxyl group
Example
δ–
δ+
Ag
O
R
δ+
Ag
+O
H
CH2OH
HO
R
O
polarized CH3I
H
HO
excess
CH3 I, Ag2O
O
H
OH
H
CH3
ether
H
H
–H +
CH3
CH3O
CH2OCH3
H
CH3O
H
CH3O
H
OH
α-D-glucopyranose
O
H
H
OCH3
methyl 2,3,4,6-tetra-O-methyl-α-D-glucopyranoside
FIGURE 23-14
Formation of methyl ethers. Treatment of an aldose or a ketose with methyl iodide and silver
oxide gives the totally methylated ether. If the conditions are carefully controlled, the
stereochemistry at the anomeric carbon is usually preserved.
Treating a sugar with methyl iodide and silver oxide converts the hydroxyl groups
to methyl ethers. Silver oxide polarizes the H3 C ¬ I bond, making the methyl carbon
strongly electrophilic. Attack by the carbohydrate ¬ OH group, followed by deprotonation, gives the ether. Figure 23-14 shows that the anomeric hydroxyl group is also converted to an ether. If the conditions are carefully controlled, the hemiacetal C ¬ O bond
is not broken, and the configuration at the anomeric carbon is preserved.
The Williamson ether synthesis is the most common method for forming simple
ethers, but it involves a strongly basic alkoxide ion. Under these basic conditions, a simple sugar would isomerize and decompose. A modified Williamson method may be used
if the sugar is first converted to a glycoside (by treatment with an alcohol and an acid catalyst). The glycoside is an acetal, stable to base. Treatment of a glycoside with sodium
hydroxide and methyl iodide or dimethyl sulfate gives the methylated carbohydrate.
H
H
CH2OH
HO
H
HO
O
H
OH
H
OCH3
methyl α- D-glucopyranoside
(stable to base)
CH3O
NaOH
O
H
CH3
O
S
CH2OCH3
H
CH3O
O
CH3
H
O
O
H
H
CH3O
OCH3
methyl 2,3,4,6-tetra-O-methyl-α- D-glucopyranoside
(dimethyl sulfate)
We can also easily convert hydroxyl groups to silyl ethers. Section 14-10B covered the use of the triisopropylsilyl (TIPS) protecting group for alcohols. Similarly,
sugars can be converted to their silyl ethers by treatment with a silyl chloride, such as
chlorotrimethylsilane (TMSCl), and a tertiary amine, such as triethylamine.
CH3
R
OH + CH3
Si
CH3
Cl
CH3
chlorotrimethylsilane
(CH3)3SiCl
(TMSCl)
Et3N
R
O
Si
CH3
Bu4N+ F–
H2O
R OH
+ CH3SiF
CH3
TMS ether
(R O TMS)
(deprotected)
Sugars are most commonly converted to their silyl ethers to make them easier
to handle and sufficiently volatile for gas chromatography and mass spectrometry.
1123
23-12 Ether and Ester Formation
For example, glucose would be more likely to char and decompose inside the injector of a gas chromatograph, rather than to vaporize and flow through the column
with the gas phase. The trimethylsilyl derivative of glucose is more volatile, however,
and it vaporizes at a low enough temperature to survive gas chromatography and
mass spectrometry.
H CH OH
2
HO
HO
H
O
H
Excess TMSCl
Et3N
OH
OH
H
H CH OTMS
2
O
H
TMSO
H
OTMS
TMSO
OTMS H
H
H
glucose
water soluble, not volatile
TMS derivative
organic-soluble, volatile
PROBLEM 23-29
Propose a mechanism for methylation of any one of the hydroxyl groups of methyl a-D-glucopyranoside, using NaOH and dimethyl sulfate.
PROBLEM 23-30
Draw the expected product of the reaction of the following sugars with excess methyl iodide
and silver oxide.
(a) a-D-fructofuranose
(b) b -D-galactopyranose
Application: Drug Excretion
Oxidation of glucose at C6 produces
glucuronic acid. In the body, a common
method for metabolizing drugs is to
attach glucuronic acid. The resulting
glucuronide derivative is water-soluble
and easily excreted in the urine.
H
COOH
HO
HO
Ester Formation Another way to convert sugars to easily handled derivatives is to
acylate the hydroxyl groups to form esters. Sugar esters are readily crystallized and
purified, and they dissolve in common organic solvents. Treatment with acetic anhydride and pyridine (as a mild basic catalyst) converts sugar hydroxyl groups to acetate
esters, as shown in Figure 23-15. This reaction acetylates all the hydroxyl groups,
including that of the hemiacetal on the anomeric carbon. The anomeric C ¬ O bond is
not broken in the acylation, and the stereochemistry of the anomeric carbon atom
is usually preserved. If we start with a pure a anomer or a pure b anomer, the product
is the corresponding anomer of the acetate.
C
O
C
CH3
H
R
sugar
O
C
O
R
CH3
R
H
HOCH2
H
H
OH
OH
HO
CH2OH
H
β-D-fructofuranose
excess
(CH3CO)2O
pyridine
O ..
acetate ester
O
O
C
CH3
H
Example
CH3
C
O
O
O
O
CH2
H
CH3
H
O
C
C
O
O
C
H
O
O
–..
.. O+
..O+
..
..
..
..
CH3
C
H
OH
glucuronic acid
..
O
CH3
CH3
.. ..
C
.. ..
R
O
O ..
O
OH
..
C
.. O–..
..
CH3
.. O ..
..
.. O ..
H
O
H
CH2O
CH3
O
C
H
CH3
O
penta-O-acetyl-β-D-fructofuranoside
FIGURE 23-15
Formation of acetate esters. Acetic anhydride and pyridine convert all the hydroxyl groups of
a sugar to acetate esters. The stereochemistry at the anomeric carbon is usually preserved.
CH3
+ HO
C
CH3
1124
CHAPTER 23
Carbohydrates and Nucleic Acids
PROBLEM 23-31
Predict the products formed when the following sugars react with excess acetic anhydride and
pyridine.
(a) a-D-glucopyranose
(b) b -D-ribofuranose
23-13
Before spectroscopy, one of the best ways to identify ketones and aldehydes was
conversion to crystalline hydrazones, especially phenylhydrazones and 2,4-dinitrophenylhydrazones (Section 18-16). In his exploratory work on sugar structures, Emil
Fischer often made and used phenylhydrazone derivatives. In fact, his constant use
of phenylhydrazine ultimately led to Fischer’s death in 1919 from chronic phenylhydrazine poisoning.
Reactions with
Phenylhydrazine:
Osazone Formation
R´
C
O +
H2N
R´
H+
NH
N
C
R
+
NH
H2O
R
ketone or
aldehyde
phenylhydrazone
phenylhydrazine
Sugars do not form the simple phenylhydrazone derivatives we might expect, however. Two molecules of phenylhydrazine condense with each molecule of the sugar to
give an osazone, in which both C1 and C2 have been converted to phenylhydrazones.
The term osazone is derived from the -ose suffix of a sugar and the last half of the word
hydrazone. Most osazones are easily crystallized, with sharp melting points. Melting
points of osazone derivatives provide valuable clues for the identification and comparison of sugars.
H
H
C
O
CHOH
excess H2N
NH
Ph
H+
(CHOH)n
C
N
NHPh
C
N
NHPh
(CHOH)n
CH2OH
CH2OH
aldose
osazone
H
H
CHOH
C
O
excess H2N
(CHOH)n
NH
Ph
H+
CH2OH
ketose
Problem-solving Hint
If two aldoses form the same
osazone, they are C2 epimers. If an
aldose and a ketose form the same
osazone, they have the same
structure at all carbons except
C1 and C2.
unaffected portion
C
N
NHPh
C
N
NHPh
(CHOH)n
CH2OH
osazone
In the formation of an osazone, both C1 and C2 are converted to phenylhydrazones.
Therefore, a ketose gives the same osazone as its related aldose. Also notice that the stereochemistry at C2 is lost in the phenylhydrazone. Thus, C2 epimers give the same osazone.
PROBLEM 23-32
(a) Show that D-glucose, D-mannose, and D-fructose all give the same osazone. Show the
structure and stereochemistry of this osazone.
(b) D-Talose is an aldohexose that gives the same osazone as D-galactose. Give the structure
of D-talose, and give the structure of its osazone.
23-15 Chain Lengthening: The Kiliani–Fischer Synthesis
In our discussion of D and L sugars, we briefly mentioned a method for shortening the
chain of an aldose by removing the aldehyde carbon at the top of the Fischer projection.
Such a reaction, removing one of the carbon atoms, is called a degradation.
The most common method used to shorten sugar chains is the Ruff degradation,
developed by Otto Ruff, a prominent German chemist around the turn of the twentieth
century. The Ruff degradation is a two-step process that begins with a bromine–water
oxidation of the aldose to its aldonic acid. Treatment of the aldonic acid with hydrogen
peroxide and ferric sulfate oxidizes the carboxyl group to CO2 and gives an aldose
with one less carbon atom. The Ruff degradation is used mainly for structure determination and synthesis of new sugars.
1125
23-14
Chain Shortening:
The Ruff
Degradation
Ruff degradation
CHO
H
OH
HO
H
H
H
OH
H
OH
HO
Br2
H2O
OH
H
OH
D-gluconic
CHO
H
HO
OH
CH2OH
D-arabinose
H2O2
Fe2(SO4)3
H
Br2
H2O
H
OH
H
OH
CH2OH
acid
D-arabinose
CO2
H
CHO
OH
OH
CH2OH
D-arabinonic
H
H
COOH
H
OH
HO
H
CH2OH
D-glucose
H
CHO
OH
H
CH2OH
HO
CO2
COOH
acid
H2O2
Fe2(SO4)3
H
OH
H
OH
CH2OH
D-erythrose
PROBLEM 23-33
Show that Ruff degradation of D-mannose gives the same aldopentose (D-arabinose) as does
D-glucose.
PROBLEM 23-34
D-Lyxose is formed by Ruff degradation of galactose. Give the structure of D-lyxose. Ruff
degradation of D-lyxose gives D-threose. Give the structure of D-threose.
PROBLEM 23-35
D-Altrose
is an aldohexose. Ruff degradation of D-altrose gives the same aldopentose as does
degradation of D-allose, the C3 epimer of glucose. Give the structure of D-altrose.
The Kiliani–Fischer synthesis lengthens an aldose carbon chain by adding one carbon
atom to the aldehyde end of the aldose. The result of this process is a chain-lengthened
sugar with a new carbon atom at C1 and the former aldehyde group (the former C1) now
at C2. This synthesis is useful both for determining the structure of existing sugars and
for synthesizing new sugars.
23-15
Chain Lengthening:
The Kiliani–Fischer
Synthesis
1126
CHAPTER 23
Carbohydrates and Nucleic Acids
The Kiliani–Fischer synthesis
C
CN
2
CHO
(CHOH)n
KCN
HCN
H
H
1
CHOH
H3O+
CHOH
H2
Pd/BaSO4
(CHOH)n
NH
1
C
O
2
CHOH
(CHOH)n
(CHOH)n
CH2OH
CH2OH
CH2OH
CH2OH
an aldose
a cyanohydrin
chain-lengthened imine
chain-lengthened aldose
The aldehyde carbon atom is made asymmetric in the first step with the formation
of the cyanohydrin. Two epimeric cyanohydrins result. For example, D-arabinose reacts
with HCN to give the following cyanohydrins.
CN
CN
CHO
HO
H
HO
H
H
OH
H
OH
C
KCN
HCN
OH
H
OH
+
CH2OH
D-arabinose
C
HO
H
H
CH2OH
HO
OH
H
H
H
OH
H
OH
CH2OH
two epimeric cyanohydrins
Hydrogenation of these cyanohydrins gives two imines, which hydrolyze to aldehydes. A poisoned catalyst of palladium on barium sulfate is used for the hydrogenation, to avoid overreduction.
H
C
H
C
N
OH
HO
H
H
OH
H
OH
H
H2
Pd/BaSO4
H
OH
HO
H
H
OH
H
OH
CH2OH
CHO
NH
H+
H2O
HO
OH
H
H
OH
H
OH
CH2OH
CH2OH
glucose
H
C
N
HO
H
HO
H
H
OH
H
OH
CH2OH
epimeric cyanohydrins
C
H2
Pd/BaSO4
NH
HO
H
HO
H
H
OH
H
OH
CH2OH
epimeric imines
CHO
H+
H2O
HO
H
HO
H
H
OH
H
OH
CH2OH
mannose
The Kiliani–Fischer synthesis accomplishes the opposite of the Ruff degradation.
Ruff degradation of either of two C2 epimers gives the same shortened aldose, and the
23-15 Chain Lengthening: The Kiliani–Fischer Synthesis
1127
Kiliani–Fischer synthesis converts this shortened aldose back into a mixture of the same
two C2 epimers. For example, glucose and mannose both undergo Ruff degradation
to give arabinose. Conversely, the Kiliani–Fischer synthesis converts arabinose into a
mixture of glucose and mannose.
PROBLEM 23-36
Ruff degradation of D-arabinose gives D-erythrose. The Kiliani–Fischer synthesis converts
D-erythrose to a mixture of D-arabinose and D-ribose. Draw out these reactions, and give the
structure of D-ribose.
PROBLEM 23-37
The Wohl degradation, an alternative to the Ruff degradation, is nearly the reverse of the
Kiliani–Fischer synthesis. The aldose carbonyl group is converted to the oxime, which is dehydrated by acetic anhydride to the nitrile (a cyanohydrin). Cyanohydrin formation is reversible,
and a basic hydrolysis allows the cyanohydrin to lose HCN. Using the following sequence of
reagents, give equations for the individual reactions in the Wohl degradation of D-arabinose to
D-erythrose. Mechanisms are not required.
(1) hydroxylamine hydrochloride
(2) acetic anhydride
(3) -OH, H2O
PROBLEM 23-38
On treatment with phenylhydrazine, aldohexoses A and B give the same osazone. On treatment
with warm nitric acid, A gives an optically inactive aldaric acid, but sugar B gives an optically
active aldaric acid. Sugars A and B are both degraded to aldopentose C, which gives an optically active aldaric acid on treatment with nitric acid. Aldopentose C is degraded to aldotetrose D, which gives optically active tartaric acid when it is treated with nitric acid. Aldotetrose
D is degraded to 1+2-glyceraldehyde. Deduce the structures of sugars A, B, C, and D, and use
Figure 23-3 to determine the correct names of these sugars.
PROBLEM 23-39
In 1891, Emil Fischer determined the structures of glucose and the seven other D-aldohexoses
using only simple chemical reactions and clever reasoning about stereochemistry and symmetry.
He received the Nobel Prize for this work in 1902. Fischer had determined that D-glucose is an
aldohexose, and he used Ruff degradations to degrade it to (+)-glyceraldehyde. Therefore, the eight
D-aldohexose structures shown in Figure 23-3 are the possible structures for glucose.
Pretend that no names are shown in Figure 23-3 except for glyceraldehyde, and use the
following results to prove which of these structures represent glucose, mannose, arabinose,
and erythrose.
(a) Upon Ruff degradation, glucose and mannose give the same aldopentose: arabinose.
Nitric acid oxidation of arabinose gives an optically active aldaric acid. What are the
two possible structures of arabinose?
(b) Upon Ruff degradation, arabinose gives the aldotetrose erythrose. Nitric acid oxidation
of erythrose gives an optically inactive aldaric acid, meso-tartaric acid. What is the
structure of erythrose?
(c) Which of the two possible structures of arabinose is correct? What are the possible
structures of glucose and mannose?
(d) Fischer’s genius was needed to distinguish between glucose and mannose. He developed a series of reactions to convert the aldehyde group of an aldose to an alcohol while
converting the terminal alcohol to an aldehyde. In effect, he swapped the functional
groups on the ends. When he interchanged the functional groups on D-mannose, he was
astonished to find that the product was still D-mannose. Show how this information
completes the proof of the mannose structure, and show how it implies the correct
glucose structure.
(e) When Fischer interchanged the functional groups on D-glucose, the product was an
unnatural L sugar. Show which unnatural sugar he must have formed, and show how it
completes the proof of the glucose structure.
Problem-solving Hint
In working this type of problem, it
is often easier to start with the
smallest structure mentioned (often
glyceraldehyde) and work backward
to larger structures. Write out all
possible structures and use the
clues to eliminate the wrong ones.
1128
CHAPTER 23
Carbohydrates and Nucleic Acids
23-16
Using methods similar to Fischer’s, the straight-chain form of any monosaccharide can
be worked out. As we have seen, however, monosaccharides exist mostly as cyclic pyranose or furanose hemiacetals. These hemiacetals are in equilibrium with the open-chain
forms, so sugars can react like hemiacetals or like ketones and aldehydes. How can
we freeze this equilibrium and determine the optimum ring size for any given sugar?
Sir Walter Haworth (inventor of the Haworth projection) used some simple chemistry
to determine the pyranose structure of glucose in 1926.
Glucose is converted to a pentamethyl derivative by treatment with excess methyl
iodide and silver oxide (Section 23-12). The five methyl groups are not the same, however. Four are methyl ethers, but one is the glycosidic methyl group of an acetal.
Determination of
Ring Size; Periodic
Acid Cleavage
of Sugars
ethers
part of an acetal
H
CH2OH O
H
H
OH
HO
HO
H
H CH OCH
3
2
CH3O
excess CH3I
Ag2O
OH
H
H
CH3O
H
O
OCH3
CH3O
H
H
methyl 2,3,4,6-tetra-O-methyl- β- D-glucoside
β-D-glucose
Acetals are easily hydrolyzed by dilute acid, but ethers are stable under these conditions.
Treatment of the pentamethyl glucose derivative with dilute acid hydrolyzes only the
acetal methyl group. Haworth determined that the free hydroxyl group is on C5 of the
hydrolyzed ether, showing that the cyclic form of glucose is a pyranose.
H CH OCH
3
2
CH3O
H
H 6CH OCH
3
2
4
O
H3O+
H
CH3O
OCH3
CH3O
H
5
CH3O
CH3O
H
3
H
H
pentamethyl derivative
H
1
O
2
1
CH3O
OH
H
2
OCH3
CH3O
3
H
H
4
H
5
H
free hemiacetal
6
Problem-solving Hint
Periodic acid cleaves each
carbon–carbon bond that joins two
carbon atoms that both bear OH
groups. As you break those bonds,
mentally replace each broken bond
with an OH group on either end. Any
carbon with two OH groups will lose
water and become a carbonyl group.
R
H
H
R
OH
H
OH HO
OH HO
CH2OCH3
open-chain form
2,3,4,6,-O-tetramethyl-D-glucose
PROBLEM 23-40
(a) Show the product that results when fructose is treated with an excess of methyl iodide
and silver oxide.
(b) Show what happens when the product of part (a) is hydrolyzed using dilute acid.
(c) Show what the results of parts (a) and (b) imply about the hemiacetal structure
of fructose.
R
OH
OH
OH
R
HO
R
H
C
C
H
H
R
O
OH
Periodic Acid Cleavage of Carbohydrates Another method used to determine the
size of carbohydrate rings is cleavage by periodic acid. Recall that periodic acid
cleaves vicinal diols to give two carbonyl compounds, either ketones or aldehydes,
depending on the substitution of the reactant (Section 11-11B).
OH
H
R
R
OCH3
R
OH
R
CHO
O
O
R
OH OH
vicinal diol
R
R´
+
HIO4
periodic acid
H
C
O +
O
C
R
R´
ketones and aldehydes
+
HIO3
+
H2O
1129
23-16 Determination of Ring Size; Periodic Acid Cleavage of Sugars
Because ether and acetal groups are unaffected, periodic acid cleavage of a glycoside
can help to determine the size of the ring. For example, periodic acid oxidation of methyl
b -D-glucopyranoside gives the following products. The structure of the fragment containing C4, C5, and C6 implies that the original glycoside was a six-membered ring
bonded through the C5 oxygen atom.
H
H 6 CH OH
2
4
HO
H
5
O
2 HIO4
2
H
HO
4
1
OH
3
H
C
O
6
CH2OH
5
OCH3
H
C
O 2
O
H
3
H
methyl β-D-glucopyranoside
H
C
4
O
O+
H3
1
OCH3
5
H
6
H
OH
CHO
OH
1
CHO
2
CHO
+ CH3OH
CH2OH
D-glyceraldehyde
O
H
3
C
OH
On the other hand, if glucose were a furanose (five-membered ring), the periodic
acid cleavage would give an entirely different set of products. Because glucose actually
exists as a pyranose (six-membered ring), these products are not observed.
HO
6
O
CH2
OCH3
HO
CH
5
4
H
O
OH
3
H
H
1
O
HIO4
6
CH2
CH
H
2
H
OH
OCH3
O
5
4
1
CH
3
O
HC
2
H3O+
H
5
CHO
1
CHO
4
CHOH
2
CHO
3
CHO
O
methyl β-D-glucofuranoside
6
H2C
+ CH3OH
O
(not observed)
PROBLEM 23-41
(a) Draw the reaction of methyl b -D-fructofuranoside with periodic acid, and predict the
products.
(b) Draw the structure of a hypothetical methyl b -D-fructopyranoside, and predict the products from periodic acid oxidation.
(c) The reaction of methyl b -D-glucopyranoside with periodic acid (shown above) gives
only the D-1+2 enantiomer of glyceraldehyde (among other products). If you oxidized
an aldohexose glycoside with periodic acid and one of the products was the L-1-2 enantiomer of glyceraldehyde, what would that tell you about the sugar?
SUMMARY
Problem-solving Hint
Cleavage occurs only between
two carbon atoms that bear
hydroxyl groups.
Reactions of Sugars
1. Undesirable rearrangements catalyzed by base (Section 23-8)
Because of these side reactions, basic reagents are rarely used with sugars.
a. Epimerization of the alpha carbon
CHO
CHO
H
OH
–OH
HO
H
(CHOH)n
(CHOH)n
CH2OH
CH2OH
(Continued )
1130
CHAPTER 23
Carbohydrates and Nucleic Acids
b. Enediol rearrangements
OH
H
C
CHO
H
HO
C
OH
H
CH2OH
–OH
OH
HO
H
C
C
O
H
HO
–OH
CH2OH
CH2OH
HO
C
–OH
CHOH
OH
C
–OH
O
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
CH2OH
CH2OH
glucose
CH2OH
enediol
CH2OH
fructose
enediol
2. Reduction (Section 23-9)
CHO
CH2OH
NaBH4
(CHOH)n
or H2 /Ni
CH2OH
CH2OH
aldose
alditol
3. Oxidation (Section 23-10)
a. To aldonic acids (glyconic acids) by bromine water
CHO
(CHOH)n
COOH
Br2
H2 O
CH2OH
aldose
aldonic acid
(CHOH)n
COOH
HNO3
CH2OH
aldaric acid
c. Tollens test for reducing sugars
CH2OH
CHO
C
O
(CHOH)n
COOH
aldose
or
(CHOH)n
CH2OH
b. To aldaric acids (glycaric acids) by nitric acid
CHO
CHOH
(CHOH)n
COO–
Ag(NH3)2OH
CHOH
(CHOH)n
(CHOH)n
(CHOH)n
CH2OH
CH2OH
CH2OH
aldose
ketose
+ rearrangement
+ Ag (silver mirror)
4. Glycoside formation (conversion to an acetal) (Section 23-11)
H
H
CH2OH
CH2OH
O
O
CH
OH
HO
HO
3
H
H
H+
H
H
H
OCH3
HO
HO
(α + β)
OH
OH
OH
H
H
H
(either anomer)
a methyl glycoside
(more stable anomer predominates)
CH2OH
etc.
1131
23-16 Determination of Ring Size; Periodic Acid Cleavage of Sugars
5. Alkylation to give ethers (Section 23-12)
H
CH2OH
O
HO
H
H
H
HO
OH
OH
H
H
CH2OH3
excess CH3I
CH3O
Ag2O
O
H
H
CH3O
H
CH3O
OCH3
H
(gives the same anomer as the starting material)
6. Acylation to give esters (Section 23-12)
H
CH2OH
O
HO
H
H
H
HO
OH
OH
H
H
excess Ac2O
CH2OAc
AcO
pyridine
Treatment with a silyl
chloride (such as TMSCl)
and a tertiary amine
converts sugars to their
silyl ethers. Fluoride
salts such as aqueous
Bu4NF hydrolyze the
silyl ethers.
O
H
H
AcO
H
AcO
OAc
H
(gives the same anomer as the starting materials)
7. Osazone formation (Section 23-13)
H
CHO
H
CHO
OH
HO
CH2OH
H
C
or
O
excess Ph
or
(CHOH)n
(CHOH)n
(CHOH)n
CH2OH
CH2OH
CH2OH
NHNH2
H+
C
N
NHPh
C
N
NHPh
(CHOH)n
CH2OH
either aldose epimer or ketose
osazone
8. Ruff degradation (Section 23-14)
CO2
CHO
CHOH
CHO
(1) Br2 / H2O
(2) H2O2 , Fe2(SO4)3
(CHOH)n
(CHOH)n
CH2OH
CH2OH
aldose
shortened aldose
9. Kiliani – Fischer synthesis (Section 23-15)
CHO
CHO
H
CHO
(CHOH)n
CH2OH
OH
HO
(CHOH)n
(1) HCN/KCN
(2) H2/Pd(BaSO4)
(3) H3O+
(CHOH)n
and
CH2OH
aldose
H
CH2OH
epimers of lengthened aldose
10. Periodic acid cleavage (Section 23-16)
H
H 6 CH OH
2
4
HO
HO
3
H
4
O
5
H
H
2
OH
2 HIO4
1
O
OCH3
O
H
methyl β-D-glucopyranoside
C
H
3
C
6
CH2OH
5
H
H
O
C
4
O
O+
H3
1
2
H
OH
OCH3
H
C HO
5
6
OH
1
2
C HO
CHO
+ CH3OH
CH2OH
D-glyceraldehyde
O
H
3
C
OH
1132
Carbohydrates and Nucleic Acids
CHAPTER 23
23-17
Disaccharides
As we have seen, the anomeric carbon of a sugar can react with the hydroxyl group of
an alcohol to give an acetal called a glycoside. If the hydroxyl group is part of another
sugar molecule, then the glycoside product is a disaccharide, a sugar composed of two
monosaccharide units (Figure 23-16).
In principle, the anomeric carbon can react with any of the hydroxyl groups of
another sugar to form a disaccharide. In naturally occurring disaccharides, however,
there are three common glycosidic bonding arrangements.
1. A 1,4¿ link. The anomeric carbon is bonded to the oxygen atom on C4 of the second
sugar. The prime symbol 1¿2 in 1,4¿ indicates that C4 is on the second sugar.
2. A 1,6¿ link. The anomeric carbon is bonded to the oxygen atom on C6 of the second sugar.
3. A 1,1¿ link. The anomeric carbon of the first sugar is bonded through an oxygen
atom to the anomeric carbon of the second sugar.
We will consider some naturally occurring disaccharides with these common
glycosidic linkages.
H
HO
5
4
HO
H
6
CH2OH
H
O
H
R
2
3
OH
H
OH
H+
OH
1
CH2OH
HO
H
HO
sugar 1
OR
H
glycosidic bond
a glycoside
H
HO
OH
H
H
O
H
6
CH2OH
5
4
HO
H
O
H
3
2
OH
H
1
OH
H
H
HO
sugar 2
–H2O
5
4
HO
1,4' linkage
6
CH2OH
H
3
H
O
H
H
6´
CH2OH
2
OH
1
H
O
5´
4´
HO
sugar 1
H
3´
glycosidic
bond
H
O
H
2´
OH
1´
OH
H
sugar 2
a disaccharide
FIGURE 23-16
Disaccharides. A sugar reacts with an alcohol to give an acetal called a glycoside. When the
alcohol is part of another sugar, the product is a disaccharide.
23-17A
The 1,4 œ Linkage: Cellobiose, Maltose, and Lactose
The most common glycosidic linkage is the 1,4¿ link. The anomeric carbon of one sugar
is bonded to the oxygen atom on C4 of the second ring.
Cellobiose: A B-1,4 œ Glucosidic Linkage Cellobiose, the disaccharide obtained by
partial hydrolysis of cellulose, contains a 1,4¿ linkage. In cellobiose, the anomeric
carbon of one glucose unit is linked through an equatorial 1b2 carbon–oxygen bond to
C4 of another glucose unit. This b-1,4¿ linkage from a glucose acetal is called a b-1,4¿
glucosidic linkage.
23-17 Disaccharides
Cellobiose, 4-O-(β- D -glucopyranosyl)-β-D -glucopyranose or 4-O-(β- D -glucopyranosyl)- D -glucopyranose
β-glucosidic linkage
6
H CH
2OH
4
HO
HO
5
H
H
2
O
OH
3
H 6´CH OH
2
O 4´
H
1
H HO
H
5´
H
β-glucosidic linkage
6
H CH
2OH
4
O
HO
OH
H
5
H
H
2
3
OH
2´
3´
HO
O
O 4´
OH
H
1´
H 6´CH OH
2
1
H HO
H
The complete name for cellobiose, 4-O-( b -D-glucopyranosyl)- b -D-glucopyranose,
gives its structure. This name says that a b -D-glucopyranose ring (the right-hand ring)
is substituted in its 4-position by an oxygen attached to a ( b -D-glucopyranosyl) ring,
drawn on the left. The name in parentheses says the substituent is a b-glucose, and the
-syl ending indicates that this ring is a glycoside. The left ring with the -syl ending is
an acetal and cannot mutarotate, while the right ring with the -ose ending is a hemiacetal and can mutarotate. Because cellobiose has a glucose unit in the hemiacetal form
(and therefore is in equilibrium with its open-chain aldehyde form), it is a reducing
sugar. Once again, the -ose ending indicates a mutarotating, reducing sugar.
Mutarotating sugars are often shown with a wavy line to the free anomeric hydroxyl
group, signifying that they can exist as an equilibrium mixture of the two anomers.
Their names are often given without specifying the stereochemistry of this mutarotating hydroxyl group, as in 4-O-( b -D-glucopyranosyl)-D-glucopyranose.
Maltose: An A-1,4 œ Glucosidic Linkage Maltose is a disaccharide formed when
starch is treated with sprouted barley, called malt. This malting process is the first step
in brewing beer, converting polysaccharides to disaccharides and monosaccharides
that ferment more easily. Like cellobiose, maltose contains a 1,4¿ glycosidic linkage
between two glucose units. The difference in maltose is that the stereochemistry of the
glucosidic linkage is a rather than b.
Maltose, 4-O-(α-D -glucopyranosyl)-D -glucopyranose
HO
H
O
H
HO
1
H
H 6´CH OH
2
OH
H
4´
O
α -1,4´ glucosidic linkage
5´
H
HO
3´
H
H
3´
H
Two alternative ways of drawing and naming cellobiose
H CH OH
2
5´
H
O
1´
2´
OH
OH
H
Like cellobiose, maltose has a free hemiacetal ring (on the right). This hemiacetal
is in equilibrium with its open-chain form, and it mutarotates and can exist in either the
a or b anomeric form. Because maltose exists in equilibrium with an open-chain
aldehyde, it reduces Tollens reagent, and maltose is a reducing sugar.
PROBLEM 23-42
Draw the structures of the individual mutarotating a and b anomers of maltose.
H
O
1´
2´
OH
H
OH
1133
1134
CHAPTER 23
Carbohydrates and Nucleic Acids
PROBLEM 23-43
Give an equation to show the reduction of Tollens reagent by maltose.
Lactose: A B-1,4 ¿ Galactosidic Linkage Lactose is similar to cellobiose, except that the
glycoside (the left ring) in lactose is galactose rather than glucose. Lactose is composed of
one galactose unit and one glucose unit. The two rings are linked by a b-glycosidic bond of
the galactose acetal to the 4-position on the glucose ring: a b-1,4¿ galactosidic linkage.
Lactose, 4-O-(β- D -galactopyranosyl)-D -glucopyranose
axial 4-hydroxyl group of galactose
OH
CH2OH
H
H
O
O
H
HO
H 6´CH OH
2
4´
1
OH
H
H
5´
H
HO
O
OH
H
H
OH
1´
2´
3´
H
β -galactosidic linkage
Lactose occurs naturally in the milk of mammals, including cows and humans.
Hydrolysis of lactose requires a b-galactosidase enzyme (sometimes called lactase).
Some humans synthesize a b-galactosidase, but others do not. This enzyme is present
in the digestive fluids of normal infants to hydrolyze their mother’s milk. Once the child
stops drinking milk, production of the enzyme gradually stops. In most parts of the
world, people do not use milk products after early childhood, and the adult population
can no longer digest lactose. Consumption of milk or milk products can cause digestive discomfort in lactose-intolerant people who lack the b-galactosidase enzyme.
Lactose-intolerant infants must drink soybean milk or another lactose-free formula.
PROBLEM 23-44
Does lactose mutarotate? Is it a reducing sugar? Explain. Draw the two anomeric forms
of lactose.
23-17B
The 1,6 œ Linkage: Gentiobiose
In addition to the common 1,4¿ glycosidic linkage, the 1,6¿ linkage is also found in naturally
occurring carbohydrates. In a 1,6¿ linkage, the anomeric carbon of one sugar is linked to the
oxygen of the terminal carbon (C6) of another. This linkage gives a different sort of stereochemical arrangement, because the hydroxyl group on C6 is one carbon atom removed from
the ring. Gentiobiose is a sugar with two glucose units joined by a b-1,6¿ glucosidic linkage.
Gentiobiose, 6-O-(β- D -glucopyranosyl)-D -glucopyranose
β -glucosidic linkage
H CH OH
2
HO
H
O
H
HO
1
OH
H
O
6´
CH2
H
H
4´
5´
HO
H
HO
3´
H
H
O
1´
2´
OH
H
OH
23-17 Disaccharides
1135
Although the 1,6¿ linkage is rare in disaccharides, it is commonly found as a branch
point in polysaccharides. For example, branching in amylopectin (insoluble starch)
occurs at 1,6¿ linkages, as discussed in Section 23-18B.
PROBLEM 23-45
Is gentiobiose a reducing sugar? Does it mutarotate? Explain your reasoning.
23-17C
Linkage of Two Anomeric Carbons: Sucrose
Some sugars are joined by a direct glycosidic linkage between their anomeric carbon
atoms: a 1,1¿ linkage. Sucrose (common table sugar), for example, is composed of one
glucose unit and one fructose unit bonded by an oxygen atom linking their anomeric
carbon atoms. (Because fructose is a ketose and its anomeric carbon is C2, this is actually a 1,2¿ linkage.) Notice that the linkage is in the a position with respect to the glucose ring and in the b position with respect to the fructose ring.
Sucrose, α-D -glucopyranosyl-β-D -fructofuranoside
(or β-D -fructofuranosyl-α-D -glucopyranoside)
H CH OH
2
HO
H
O
H
HO
H
OH
α -glycosidic linkage on glucose
H
HOCH2
H
O
O
H
OH
β -glycosidic linkage on fructose
HO
CH2OH
H
Both monosaccharide units in sucrose are present as acetals, or glycosides. Neither
ring is in equilibrium with its open-chain aldehyde or ketone form, so sucrose does not
reduce Tollens reagent and it cannot mutarotate. Because both units are glycosides, the
systematic name for sucrose can list either of the two glycosides as being a substituent
on the other. Both systematic names end in the -oside suffix, indicating a nonmutarotating, nonreducing sugar. Like many other common names, sucrose ends in the -ose
ending even though it is a nonreducing sugar. Common names are not reliable indicators of the properties of sugars.
Sucrose is hydrolyzed by enzymes called invertases, found in honeybees and yeasts,
that specifically hydrolyze the b -D-fructofuranoside linkage. The resulting mixture of
glucose and fructose is called invert sugar because hydrolysis converts the positive rotation [+66.5°] of sucrose to a negative rotation that is the average of glucose [+52.7°]
and fructose [-92.4°]. The most common form of invert sugar is honey, a supersaturated mixture of glucose and fructose hydrolyzed from sucrose by the invertase enzyme
of honeybees. Glucose and fructose were once called dextrose and levulose, respectively, according to their opposite signs of rotation.
SOLVED PROBLEM 23-3
An unknown carbohydrate of formula C12H22O11 reacts with Tollens reagent to form a silver
mirror. An a-glycosidase has no effect on the carbohydrate, but a b-galactosidase hydrolyzes it
to D-galactose and D-mannose. When the carbohydrate is methylated (using methyl iodide and
silver oxide) and then hydrolyzed with dilute HCl, the products are 2,3,4,6-tetra-O-methylgalactose
and 2,3,4-tri-O-methylmannose. Propose a structure for this unknown carbohydrate.
(Continued)
Application: Sucrose Stability
Sucrose (a nonreducing sugar) is not as
easily oxidized as a reducing sugar, so it
is much more useful for preserving
foods such as jams and jellies. A reducing sugar like glucose would oxidize
and spoil.
1136
CHAPTER 23
Carbohydrates and Nucleic Acids
Application: Blood Types
The primary differences among the
blood types O, A, B, and AB involve
antigenic carbohydrates on the surface
of the red blood cells. Type O cells have
no antigenic carbohydrates on the surface, while Type A has an antigenic
N-acetylgalactosamine, and Type B has
an antigenic galactose. Type AB
cells have both galactose and
N-acetylgalactosamine antigens.
Type O is called the “universal
donor” because the cells have no
antigen to provoke a deadly antigenantibody reaction. If the other blood
factors (Rh factor, for example) are
compatible, type O can be donated to
people with the other blood types.
H
H
HO 6CH OH
2
4
5
H
H
HO
H
O
2
O
1
OH
O
6´
CH2
H
H
OH
Ac
N-acetylgalactosamine
5´
HO
H
NH
H
4´
H
HO
The formula shows this is a disaccharide composed of two hexoses. Hydrolysis gives D-galactose
and D-mannose, identifying the two hexoses. Hydrolysis requires a b-galactosidase, showing that
galactose and mannose are linked by a b-galactosyl linkage. Since the original carbohydrate is a
reducing sugar, one of the hexoses must be in a free hemiacetal form. Galactose is present as a glycoside; thus mannose must be present in its hemiacetal form. The unknown carbohydrate must be
a 1b-galactosyl2-mannose.
The methylation/hydrolysis procedure shows the point of attachment of the glycosidic
bond to mannose and also confirms the size of the six-membered rings. In galactose, all the
hydroxyl groups are methylated except C1 and C5. C1 is the anomeric carbon, and the C5
oxygen is used to form the hemiacetal of the pyranose ring. In mannose, all the hydroxyl
groups are methylated except C1, C5, and C6. The C5 oxygen is used to form the pyranose ring
(the C6 oxygen would form a less stable seven-membered ring); therefore, the oxygen on C6
must be involved in the glycosidic linkage. The structure and systematic name are shown here.
3
HO OH
H
SOLUTION
H
HO
3´
O
OH
2´
1´
OH
H
H
H
6-O-(β-D-galactopyranosyl)-D-mannopyranose
PROBLEM 23-46
Trehalose is a nonreducing disaccharide 1C12 H 22 O112 isolated from the poisonous mushroom
Amanita muscaria. Treatment with an a-glucosidase converts trehalose to two molecules of glucose, but no reaction occurs when trehalose is treated with a b-glucosidase. When trehalose is
methylated by dimethyl sulfate in mild base and then hydrolyzed, the only product is 2,3,4,6-tetraO-methylglucose. Propose a complete structure and systematic name for trehalose.
PROBLEM 23-47
Raffinose is a trisaccharide 1C18 H32 O162 isolated from cottonseed meal. Raffinose does
not reduce Tollens reagent, and it does not mutarotate. Complete hydrolysis of raffinose
gives D-glucose, D-fructose, and D-galactose. When raffinose is treated with invertase, the
products are D-fructose and a reducing disaccharide called melibiose. Raffinose is unaffected by treatment with a b-galactosidase, but an a-galactosidase hydrolyzes it to D-galactose
and sucrose. When raffinose is treated with dimethyl sulfate and base followed by hydrolysis, the products are 2,3,4-tri-O-methylglucose, 1,3,4,6-tetra-O-methylfructose, and
2,3,4,6-tetra-O-methylgalactose. Determine the complete structures of raffinose and melibiose, and give a systematic name for melibiose.
24-18
Polysaccharides
Polysaccharides are carbohydrates that contain many monosaccharide units joined by
glycosidic bonds. They are one class of biopolymers, or naturally occurring polymers.
Smaller polysaccharides, containing about three to ten monosaccharide units, are sometimes called oligosaccharides. Most polysaccharides have hundreds or thousands of
simple sugar units linked together into long polymer chains. Except for units at the ends
of chains, all the anomeric carbon atoms of polysaccharides are involved in acetal glycosidic links. Therefore, polysaccharides give no noticeable reaction with Tollens
reagent, and they do not mutarotate.
23-18 Polysaccharides
1137
Cellulose
23-18A
Cellulose, a polymer of D-glucose, is the most abundant organic material. Cellulose is
synthesized by plants as a structural material to support the weight of the plant. Long
cellulose molecules, called microfibrils, are held in bundles by hydrogen bonding
between the many ¬ OH groups of the glucose rings. About 50% of dry wood and
about 90% of cotton fiber is cellulose.
H CH OH
2
O
H
H CH OH
2
O
4´
H
HO
O
1
OH
H
H
H
H CH OH
2
O
O
H
HO
H
OH
H
O
H
HO
H
O
OH
H
FIGURE 23-17
Partial structure of cellulose.
Cellulose is a b-1,4¿ polymer of
D-glucose, systematically named
poly11,4¿-O-b -D-glucopyranoside).
H
β-glucosidic linkage
Cellulose is composed of D-glucose units linked by b-1,4¿ glycosidic bonds. This
bonding arrangement (like that in cellobiose) is rather rigid and very stable, giving
cellulose desirable properties for a structural material. Figure 23-17 shows a partial
structure of cellulose.
Humans and other mammals lack the b-glucosidase enzyme needed to hydrolyze
cellulose, so they cannot use it directly for food. Several groups of bacteria and protozoa can hydrolyze cellulose, however. Termites and ruminants maintain colonies of
these bacteria in their digestive tracts. When a cow eats hay, these bacteria convert about
20% to 30% of the cellulose to digestible carbohydrates.
Rayon is a fiber made from cellulose that has been converted to a soluble derivative,
and then regenerated. In the common viscose process, wood pulp is treated with carbon
disulfide and sodium hydroxide to convert the free hydroxyl groups to xanthates, which are
soluble in water. The viscous solution (called viscose) is forced through a spinneret into an
aqueous sodium bisulfate solution, where a fiber of insoluble cellulose is regenerated.
Alternatively, the viscose solution can be extruded in sheets to give cellophane film. Rayon
and cotton are both cellulose, yet rayon thread can be much stronger because it consists of
long, continuously extruded fibers, rather than short cotton fibers spun together.
S
+
ROH
+
CS2
NaOH
RO
C
+
S– Na+
H2O
xanthate derivatives
(viscose)
cellulose
S
RO
C
S– Na+
extruded into solution
+
NaHSO4
H2O
ROH +
CS2
+
Na2SO4
rayon
(regenerated cellulose)
PROBLEM 23-48
Cellulose is converted to cellulose acetate by treatment with acetic anhydride and pyridine.
Cellulose acetate is soluble in common organic solvents, and it is easily dissolved and spun into
fibers. Show the structure of cellulose acetate.
23-18B
Starches: Amylose, Amylopectin, and Glycogen
Plants use starch granules for storing energy. When the granules are dried and ground
up, different types of starches can be separated by mixing them with hot water. About
20% of the starch is water-soluble amylose, and the remaining 80% is water-insoluble
The acoustic properties of cellulose
have never been surpassed by other
substances. Here, a luthier holds a
spruce front plate up to a light
to show how the thickness is
carefully graduated to enhance a
pleasing sound.
1138
CHAPTER 23
Carbohydrates and Nucleic Acids
FIGURE 23-18
Partial structure of amylose. Amylose
is an a-1,4¿ polymer of glucose,
systematically named poly11,4¿-O-aD-glucopyranoside). Amylose differs
from cellulose only in the
stereochemistry of the
glycosidic linkage.
H
O
HO
CH2OH
H
H
O
H
H H
OH
O
HO
CH2OH
H
H
O
H
1
OH
O
α-glucosidic
linkage
H H
CH2OH
4´
HO
H
H
Application: Dental Plaque
Oral bacteria convert glucose, fructose,
sucrose, and other common sugars into
a polysaccharide called dextran.
Dextran is an essential component of
the plaque that forms around teeth and
protects bacteria from the antibacterial
components in saliva. The dextran chain
consists of glucose molecules linked by
a-1,6 ¿ glucosidic linkages, with
branches at a-1,3 ¿ linkages. Candy makers use glucitol (“sorbitol”) and mannitol
to sweeten “sugarless” candies and
gum because bacteria cannot easily
convert these sugar alcohols to the
glucose they need to make dextran.
Application: Low-Carb Diets
Low-carbohydrate diets restrict the
intake of carbohydrates, sometimes
resulting in rapid weight loss. The
weight is lost because glycogen and
fatty acids are burned to maintain blood
glucose levels.
FIGURE 23-19
The starch-iodine complex of
amylose. The amylose helix forms a
blue charge-transfer complex with
molecular iodine.
O
H
OH
H
O
amylopectin. When starch is treated with dilute acid or appropriate enzymes, it is
progressively hydrolyzed to maltose and then to glucose.
Amylose Like cellulose, amylose is a linear polymer of glucose with 1,4¿ glycosidic
linkages. The difference is in the stereochemistry of the linkage. Amylose has a-1,4¿ links,
while cellulose has b-1,4¿ links. A partial structure of amylose is shown in Figure 23-18.
The subtle stereochemical difference between cellulose and amylose results in some
striking physical and chemical differences. The a linkage in amylose kinks the polymer
chain into a helical structure. This kinking increases hydrogen bonding with water and
lends additional solubility. As a result, amylose is soluble in water, and cellulose is not.
Cellulose is stiff and sturdy, but amylose is not. Unlike cellulose, amylose is an excellent food source. The a-1,4¿ glucosidic linkage is easily hydrolyzed by an a-glucosidase
enzyme, found in all animals.
The helical structure of amylose also serves as the basis for an interesting and useful reaction. The inside of the helix is just the right size and polarity to accept an iodine
1I22 molecule. When iodine is lodged within this helix, a deep blue starch-iodine complex results (Figure 23-19). This is the basis of the starch-iodide test for oxidizers. The
material to be tested is added to an aqueous solution of amylose and potassium iodide.
If the material is an oxidizer, some of the iodide 1I-2 is oxidized to iodine 1I22, which
forms the blue complex with amylose.
Amylopectin Amylopectin, the insoluble fraction of starch, is also primarily an
a-1,4¿ polymer of glucose. The difference between amylose and amylopectin lies in the
branched nature of amylopectin, with a branch point about every 20 to 30 glucose units.
Another chain starts at each branch point, connected to the main chain by an a-1,6¿ glycosidic linkage. A partial structure of amylopectin, including one branch point, is
shown in Figure 23-20.
Glycogen Glycogen is the carbohydrate that animals use to store glucose for readily
available energy. A large amount of glycogen is stored in the muscles themselves,
ready for immediate hydrolysis and metabolism. Additional glycogen is stored in the
I
I
23-18 Polysaccharides
H
CH2OH
O
H
HO
H
CH2OH
HO
H
H H
CH2OH
O
O
H
H
HO
O
H
CH2OH
O
H
HO
O
α-1,6´ glucosidic linkage
branch point
O
H
H H
OH
H
H
OH
H
H H
OH
H
O
H
OH
H
1139
O
CH2
H
HO
H
O
H
OH
H H
O
HO
CH2OH
H
H
O
H
OH
H
O
FIGURE 23-20
Partial structure of amylopectin. Amylopectin is a branched a-1,4¿ polymer of glucose.
At the branch points, there is a single a-1,6¿ linkage that provides the attachment point
for another chain. Glycogen has a similar structure, except that its branching is
more extensive.
liver, where it can be hydrolyzed to glucose for secretion into the bloodstream, providing an athlete with a “second wind.”
The structure of glycogen is similar to that of amylopectin, but with more extensive branching. The highly branched structure of glycogen leaves many end groups
available for quick hydrolysis to provide glucose needed for metabolism.
23-18C
Chitin: A Polymer of N-Acetylglucosamine
Chitin (pronounced ki¿-t’n, rhymes with Titan) forms the exoskeletons of insects.
In crustaceans, chitin forms a matrix that binds calcium carbonate crystals into the
exoskeleton. Chitin is different from the other carbohydrates we have studied. It is
a polymer of N-acetylglucosamine, an amino sugar (actually an amide) that is common in living organisms. In N-acetylglucosamine, the hydroxyl group on C2 of glucose is replaced by an amino group (forming glucosamine), and that amino group
is acetylated.
N-Acetylglucosamine, or 2-acetamido-2-deoxy-D -glucose
H 6 CH OH
2
4
5
HO
H
HO
3
H
O
1
2
NH
H
C
CH3
H
O
OH
This cicada is shedding its nymphal
exoskeleton. Chitin lends strength
and rigidity to the exoskeletons of
insects, but it cannot grow and
change shape with the insect.
1140
CHAPTER 23
Carbohydrates and Nucleic Acids
Application: Insecticide
Chitin synthase inhibitors are used commercially as insecticides because they
prevent the formation of a new exoskeleton and the shedding of the old one. The
insect becomes trapped in an old
exoskeleton that cannot grow. These
inhibitors are highly toxic to insects and
crustaceans, but relatively nontoxic to
mammals. The most common chitin synthase inhibitors are substituted benzoylureas such as diflubenzuron, which was
first registered as an insecticide in 1976.
Cl
F
O
O
N
H
N
H
Chitin is bonded like cellulose, except using N-acetylglucosamine instead of glucose.
Like other amides, N-acetylglucosamine forms exceptionally strong hydrogen bonds
between the amide carbonyl groups and N ¬ H protons. The glycosidic bonds are b-1,4¿
links, giving chitin structural rigidity, strength, and stability that exceed even that of cellulose. Unfortunately, this strong, rigid polymer cannot easily expand, so it must be shed
periodically by molting as the animal grows.
Chitin, or poly (1,4´-O-β-2-acetamido-2-deoxy- D-glucopyranoside),
a β-1,4-linked polymer of N-acetylglucosamine
H CH OH
2
O
H
O
H
HO
H
1
NH
C
F
diflubenzuron or DimilinTM
23-19
Nucleic Acids:
Introduction
HIV (the AIDS virus) is shown here
attacking a T-4 lymphocyte. HIV is
an RNA virus whose genetic material
must be translated to DNA before
inserting itself into the host cell’s
DNA. Several of the anti-AIDS drugs
are directed toward stopping this
reverse transcription of RNA to
DNA. (Magnification 1000X)
H CH OH
2
4´
CH3
H
O
H
O
HO
H CH OH
2
O
O
HO
H
H
β-glycosidic
linkage
NH
C
CH3
H
H
O
O
O
H
H
NH
C
H
O
CH3
Nucleic acids are substituted polymers of the aldopentose ribose that carry an organism’s genetic information. A tiny amount of DNA in a fertilized egg cell determines
the physical characteristics of the fully developed animal. The difference between a
frog and a human is encoded in a relatively small part of this DNA. Each cell carries
a complete set of genetic instructions that determine the type of cell, what its function will be, when it will grow and divide, and how it will synthesize all the structural
proteins, enzymes, fats, carbohydrates, and other substances the cell and the organism need to survive.
The two major classes of nucleic acids are ribonucleic acids (RNA) and
deoxyribonucleic acids (DNA). In a typical cell, DNA is found primarily in the nucleus,
where it carries the permanent genetic code. The molecules of DNA are huge, with
molecular weights up to 50 billion. When the cell divides, DNA replicates to form two
copies for the daughter cells. DNA is relatively stable, providing a medium for transmission of genetic information from one generation to the next.
RNA molecules are typically much smaller than DNA, and they are more easily
hydrolyzed and broken down. RNA commonly serves as a working copy of the nuclear
DNA being decoded. Nuclear DNA directs the synthesis of messenger RNA, which
leaves the nucleus to serve as a template for the construction of protein molecules in the
ribosomes. After it has served its purpose, the messenger RNA is then enzymatically
cleaved to its component parts, which become available for assembly into new RNA molecules to direct other syntheses.
The backbone of a nucleic acid is a polymer of ribofuranoside rings (fivemembered rings of the sugar ribose) linked by phosphate ester groups. Each ribose unit
carries a heterocyclic base that provides part of the information needed to specify a particular amino acid in protein synthesis. Figure 23-21 shows the ribose-phosphate backbone of RNA.
DNA and RNA each contain four monomers, called nucleotides, that differ in the
structure of the bases bonded to the ribose units. Yet this deceptively simple structure
encodes complex information just as the 0 and 1 bits used by a computer encode complex programs. First we consider the structure of individual nucleotides, then the bonding of these monomers into single-stranded nucleic acids, and finally the base pairing
that binds two strands into the double helix of nuclear DNA.
1141
23-20 Ribonucleosides and Ribonucleotides
FIGURE 23-21
Symbolically,
O
5´ end
O
A short segment of the RNA
polymer. Nucleic acids are assembled
on a backbone made up of
ribofuranoside units linked by
phosphate esters.
O
O–
P
O
O
P
O–
O
base1
CH2 O
H
H
O
HO
P
O–
H
base1
Ribose
H
O
O
O
O
O–
O
base2
CH2 O
H
H
O
HO
P
O–
H
O
P
base2
Ribose
H
O
O
O
P
O–
O
base3
CH2 O
H
H
base3
Ribose
H
H
HO
3´ end
Ribonucleosides are components of RNA based on glycosides of the furanose form of
D-ribose. We have seen (Section 23-11) that a glycoside may have an aglycone (the
substituent on the anomeric carbon) bonded by a nitrogen atom. A ribonucleoside is a
b -D-ribofuranoside (a b-glucosidase of D-ribofuranose) whose aglycone is a heterocyclic nitrogen base. The following structures show the open-chain and furanose forms
of ribose, and a ribonucleoside with a generic base bonded through a nitrogen atom.
H
OH
H
OH
H
OH
CH2 O
OH
H
H
H
H
OH OH
HO
HO
β- D-ribofuranose
CH2OH
Ribonucleosides
and Ribonucleotides
Application: Gout
Uric acid is one of the principal end
products of purine metabolism. Gout is
caused by elevated levels of uric acid in
the body, causing crystals of urate salts
to precipitate in the joints.
base
CHO
23-20
N
CH2 O
H
H
H
H
OH OH
O
H
N
a ribonucleoside
N
H
The four bases commonly found in RNA are divided into two classes: The monocyclic compounds cytosine and uracil are called pyrimidine bases because they resemble substituted pyrimidines, and the bicyclic compounds adenine and guanine are called
purine bases because they resemble the bicyclic heterocycle purine (Section 16-9C).
NH2
N
N
H
pyrimidine
O
N
N
NH
O
D-ribose
N
N
O
N
NH2
H
O
H
cytosine (C)
uracil (U)
pyrimidine bases
N
N
N
H
N
H
O
uric acid
O
H
N
N
N
N
H
adenine (A)
N
H
guanine (G)
purine bases
NH2
N
purine
N
N
1142
Carbohydrates and Nucleic Acids
CHAPTER 23
NH2
O
4
N3
5
HO
6
5´
CH2 O
4´
H
H
H
1´
2´
O
HO
6
5´
CH2 O
4´
H
H
H
7
N3
5
2
N1
H
OH OH
cytidine (C)
3´
H
4
2
N1
O
HO
1´
9
CH2 O
H
OH OH
uridine (U)
3´
H
N
5
N
4
H
H
O
7
6
N1
8
5´
4´
NH2
N
2
HO
3
1´
3´
9
CH2 O
H
OH OH
adenosine (A)
2´
H
5
N
4
H
H
H
6
N1
8
5´
4´
N
2
N
3
NH2
1´
H
OH OH
guanosine (G)
2´
3´
2´
FIGURE 23-22
The four common ribonucleosides are cytidine, uridine, adenosine, and guanosine.
When bonded to ribose through the circled nitrogen atoms, the four heterocyclic
bases make up the four ribonucleosides cytidine, uridine, adenosine, and guanosine
(Figure 23-22). Notice that the two ring systems (the base and the sugar) are numbered separately, and the carbons of the sugar are given primed numbers. For example, the 3¿ carbon of cytidine is C3 of the ribose ring.
PROBLEM 23-49
Cytosine, uracil, and guanine have tautomeric forms with aromatic hydroxyl groups. Draw
these tautomeric forms.
PROBLEM 23-50
(a) An aliphatic aminoglycoside is relatively stable to base, but it is quickly hydrolyzed by
dilute acid. Propose a mechanism for the acid-catalyzed hydrolysis.
R
R
HO
N
CH2 O
H
H
H
H
OH OH
H3O+
+
R2NH2 + sugar
an aliphatic riboside
(b) Ribonucleosides are not so easily hydrolyzed, requiring relatively strong acid. Using
your mechanism for part (a), show why cytidine and adenosine (for example) are not so
readily hydrolyzed. Explain why this stability is important for living organisms.
Ribonucleotides Ribonucleic acid consists of ribonucleosides bonded together
into a polymer. This polymer cannot be bonded by glycosidic linkages like those of
other polysaccharides because the glycosidic bonds are already used to attach the
heterocyclic bases. Instead, the ribonucleoside units are linked by phosphate esters.
The 5¿-hydroxyl group of each ribofuranoside is esterified by phosphoric acid. A
ribonucleoside that is phosphorylated at its 5¿ carbon is called a ribonucleotide (“tied”
to phosphate). The four common ribonucleotides, shown in Figure 23-23, are simply
phosphorylated versions of the four common ribonucleosides.
The phosphate groups of these ribonucleotides can exist in any of three ionization
states, depending on the pH of the solution. At the nearly neutral pH of most organisms
1pH = 7.42, there is one proton on the phosphate group. By convention, however, these
groups are usually written completely ionized.
O
O
HO
P
O
OH
in acid
ribose
–
O
P
O
O
ribose
OH
nearly neutral
–
O
P
O
ribose
O–
in base
(usually written)
23-21 The Structures of RNA and DNA
NH2
N
O
–O
O
P
O
5´
N
CH2 O
O–
–O
P
5´
H
OH
H
OH
H
H
OH
H
OH
H
cytidine monophosphate,
CMP (cytidylic acid)
N
CH2 O
O
O–
H
H
O
O
NH2
H
N
N
O
O
–O
P
5´
O
O–
H
H
OH
H
OH
H
uridine monophosphate,
UMP (uridylic acid)
N
CH2 O
O
N
N
1143
N
O
–O
adenosine monophosphate,
AMP (adenylic acid)
P
O
5´
N
CH2 O
O–
H
H
OH
H
OH
H
H
N
N
NH2
guanosine monophosphate,
GMP (guanidylic acid)
FIGURE 23-23
Four common ribonucleotides. These are ribonucleosides esterified by phosphoric acid at their
5¿ position, the ¬ CH2 OH at the end of the ribose chain.
Ribonucleic acid (RNA) and deoxyribonucleic acid (DNA) are both biopolymers of
nucleic acids, but they have minor structural differences that lead to major functional
differences. All living cells use DNA as the primary genetic material that is passed from
one generation to another. DNA directs and controls the synthesis of RNA, which serves
as a short-lived copy of part of the much larger DNA molecule. Then, the cellular
machinery translates the nucleotide sequence of the RNA molecule into a sequence of
amino acids needed to make a protein.
23-21A
23-21
The Structures of
RNA and DNA
The Structure of Ribonucleic Acid
Figure 23-24 shows how the individual ribonucleotide units are bonded into the RNA polymer. Each nucleotide has a phosphate group on its 5¿ carbon (the end carbon of ribose) and
a hydroxyl group on the 3¿ carbon. Two nucleotides are joined by a phosphate ester linkage
between the 5¿-phosphate group of one nucleotide and the 3¿-phosphate group of another.
The RNA polymer consists of many nucleotide units bonded this way, with a phosphate ester linking the 5¿ end of one nucleoside to the 3¿ end of another. A molecule of
RNA always has two ends (unless it is in the form of a large ring). One end has a free
3¿ group, and the other end has a free 5¿ group. We refer to the ends as the 3¿ end and
the 5¿ end, and we refer to directions of replication as the 3¿ : 5¿ direction and the
5¿ : 3¿ direction. Figures 23-21 and 23-24 show short segments of RNA with the 3¿ end
and the 5¿ end labeled.
5´ end
OH
O
OH
O–
P
O
O
O
5´
5´
CH2 O base1
4´
H
H
3´-hydroxyl
CH2 O base1
1´
H
H
H
OH
3´
H
2´
OH
O
5´-phosphate
O–
P
O–
P
O
O
CH2 O base2
CH2 O base2
H
H
OH + H2O
5´
5´
4´
H
3´
O
(–H2O)
OH
O
O–
P
H
3´
OH
H
1´
H
OH
2´
H
H
3´
OH
3´ end
H
H
OH
FIGURE 23-24
Phosphate linkage of nucleotides in
RNA. Two nucleotides are joined by
a phosphate linkage between the
5¿-phosphate group of one and the
3¿-hydroxyl group of the other.
1144
CHAPTER 23
Carbohydrates and Nucleic Acids
Deoxyribose and the Structure
of Deoxyribonucleic Acid
23-21B
All our descriptions of ribonucleosides, ribonucleotides, and ribonucleic acid also apply
to the components of DNA. The principal difference between RNA and DNA is the
presence of D-2-deoxyribose as the sugar in DNA instead of the D-ribose found in RNA.
The prefix deoxy- means that an oxygen atom is missing, and the number 2 means it is
missing from C2.
1
base
CHO
no OH
H
2
H
H
3
OH
H
4
OH
5
5
HO
CH2OH
HO
CH2 O
OH
4
H
H 1
2
H
H 3
OH H no OH
β- D-2-deoxyribofuranose
D-2-deoxyribose
N
CH2 O
H
H
H
H
OH H
a deoxyribonucleoside
Another key difference between RNA and DNA is the presence of thymine in DNA
instead of the uracil in RNA. Thymine is simply uracil with an additional methyl group.
The four common bases of DNA are cytosine, thymine, adenine, and guanine.
additional CH3
NH2
O
N
N
H
H3C
O
N
N
H
O
N
N
N
N
H
H
cytosine (C)
thymine (T)
pyrimidine bases
O
NH2
H
N
N
N
NH2
N
H
adenine (A)
guanine (G)
purine bases
These four bases are incorporated into deoxyribonucleosides and deoxyribonucleotides
similar to the bases in ribonucleosides and ribonucleotides. The following structures
show the common nucleosides that make up DNA. The corresponding nucleotides are
simply the same structures with phosphate groups at the 5¿ positions.
The structure of the DNA polymer is similar to that of RNA, except there are no
hydroxyl groups on the 2¿ carbon atoms of the ribose rings. The alternating deoxyribose
rings and phosphates act as the backbone, while the bases attached to the deoxyribose
units carry the genetic information. The sequence of nucleotides is called the primary
structure of the DNA strand.
Four common deoxyribonucleosides that make up DNA
H3C
N
N
HOCH2 O
H
H
H
H
OH H
deoxycytidine
NH2
O
NH2
O
H
N
N
HOCH2 O
H
H
H
H
OH H
deoxythymidine
O
N
N
HOCH2 O
H
H
H
H
OH H
deoxyadenosine
N
N
O
H
N
N
HOCH2 O
H
H
H
H
OH H
N
N
deoxyguanosine
NH2
23-21 The Structures of RNA and DNA
23-21C
Base Pairing
Having discussed the primary structure of DNA and RNA, we now consider how the
nucleotide sequence is reproduced or transcribed into another molecule. This information transfer takes place by an interesting hydrogen-bonding interaction between specific pairs of bases.
Each pyrimidine base forms a stable hydrogen-bonded pair with only one of the two
purine bases (Figure 23-25). Cytosine forms a base pair, joined by three hydrogen bonds,
with guanine. Thymine (or uracil in RNA) forms a base pair with adenine, joined by two
hydrogen bonds. Guanine is said to be complementary to cytosine, and adenine is complementary to thymine. This base pairing was first suspected in 1950, when Erwin
Chargaff of Columbia University noticed that various DNAs, taken from a wide variety of species, had about equal amounts of adenine and thymine and about equal amounts
of guanine and cytosine.
N
ribose
H
N
O
H
H
N
O
G
C
N
N
N
N
H
guanine
O
H
N
N
H
CH3
N
N
N
N
H
ribose
cytosine
ribose
N
H
ribose
O
N
adenine
A
thymine
T
FIGURE 23-25
Base pairing in DNA and RNA. Each purine base forms a stable hydrogen-bonded pair with a
specific pyrimidine base. Guanine forms a base pair with three hydrogen bonds to cytosine,
and adenine forms a base pair with two hydrogen bonds to thymine (or uracil in RNA). The
electrostatic potential maps show that hydrogen bonding takes place between electron-poor
hydrogen atoms (blue and purple regions) and electron-rich nitrogen or oxygen atoms
(red regions). (In these drawings, “ribose” means b -D-2-deoxyribofuranoside in DNA and
b -D-ribofuranoside in RNA.)
PROBLEM 23-51
All of the rings of the four heterocyclic bases are aromatic. This is more apparent when the polar
resonance forms of the amide groups are drawn, as is done for thymine at the right. Redraw the
hydrogen-bonded guanine-cytosine and adenine-thymine pairs shown in Figure 23-25, using the
polar resonance forms of the amides. Show how these forms help to explain why the hydrogen
bonds involved in these pairings are particularly strong. Remember that a hydrogen bond arises
between an electron-deficient hydrogen atom and an electron-rich pair of nonbonding electrons.
CH3
–O
N+
H
+
O–
thymine
23-21D
The Double Helix of DNA
In 1953, James D. Watson and Francis C. Crick used X-ray diffraction patterns of DNA
fibers to determine the molecular structure and conformation of DNA. They found that
DNA contains two complementary polynucleotide chains held together by hydrogen
N
ribose
1145
1146
CHAPTER 23
Carbohydrates and Nucleic Acids
O–
3´end
HO
ribose
O
P
O–
O
ribose
O
P
O
T
ribose
O–
P
ribose
O
P
O–
O
P
OH
O
C
O
ribose
5´end
T
A
O
O
P
O
G
O
O
O–
G
O
O
ribose
C
O
P
O
O
A
HO
O–
O
O
ribose
O–
O
P
O–
O
ribose
OH
3´end
5´end
FIGURE 23-26
Antiparallel strands of DNA. DNA usually consists of two complementary strands, with all the
base pairs hydrogen bonded together. The two strands are antiparallel, running in opposite
directions. (In these drawings of DNA, “ribose” means b -D-2-deoxyribofuranoside.)
bonds between the paired bases. Figure 23-26 shows a portion of the double strand of
DNA, with each base paired with its complement. The two strands are antiparallel:
One strand is arranged 3¿ : 5¿ from left to right, while the other runs in the opposite
direction, 5¿ : 3¿ from left to right.
Watson and Crick also found that the two complementary strands of DNA are coiled
into a helical conformation about 20 Å in diameter, with both chains coiled around the
same axis. The helix makes a complete turn for every ten residues, or about one turn in
every 34 Å of length. Figure 23-27 shows the double helix of DNA. In this drawing, the
two sugar-phosphate backbones form the vertical double helix with the heterocyclic
bases stacked horizontally in the center. Attractive stacking forces between the pi clouds
of the aromatic pyrimidine and purine bases are substantial, further helping to stabilize
the helical arrangement.
When DNA undergoes replication (in preparation for cell division), an enzyme
uncoils part of the double strand. Individual nucleotides naturally hydrogen bond to their
complements on the uncoiled part of the original strand, and a DNA polymerase enzyme
couples the nucleotides to form a new strand. This process is depicted schematically in
strand I:
5' end
strand II:
3' end
A ... T
T ... A
G ...
... C
C ...
...G
A ... T
G ...
... C
T ... A
A ... T
C ...
...G
G...
...C
34 A
...
C...G
A... T
T ... A
G ...
... C
T ...A
FIGURE 23-27
Double helix of DNA. Two
complementary strands are joined by
hydrogen bonds between the base
pairs. This double strand coils into a
helical arrangement.
strand I:
3' end
strand II:
5' end
23-22 Additional Functions of Nucleotides
FIGURE 23-28
5´ end
3´ end
...
A ...
T
...
...G
C...
...
...C
G...
... A
T ...
... T
A ...
...
...G
C...
... T
A ...
daughter
strand
uncoiling
G
3´ end
... T
A ...
5´ end
... T
A ...
... T
A ...
...
...
G...
C
... A
T ...
...
...
...
G
C
...
...G
C...
......T
A ...
... ...
G ...
C
T ... A
...
......C
G ...
... G
C ...
daughter
strand
...T
A...
...
...
G...C
... A
T ...
...
...
G...C
......T
A...
... C...
G ...
T ......
A
... C
G...
5´ end
Replication of the double strand of
DNA. A new strand is assembled on
each of the original strands, with the
DNA polymerase enzyme forming
the phosphate ester bonds of
the backbone.
parent strands
C
1147
..A
T....
3´ end
... A
T ...
5´ end
Electron micrograph of doublestranded DNA that has partially
uncoiled to show the individual
strands. (Magnification 13,000X)
3´ end
Figure 23-28. A similar process transcribes DNA into a complementary molecule of
messenger RNA for use by ribosomes as a template for protein synthesis.
A great deal is known about replication of DNA and translation of the DNA/RNA
sequence of bases into proteins. These exciting aspects of nucleic acid chemistry are part
of the field of molecular biology, and they are covered in detail in biochemistry courses.
We generally think of nucleotides as the monomers that form DNA and RNA, yet these
versatile biomolecules serve a variety of additional functions. Here we briefly consider
a few additional uses of nucleotides.
AMP: A Regulatory Hormone Adenosine monophosphate (AMP) also occurs in a
cyclic form, where the 3¿- and 5¿-hydroxyl groups are both esterified by the same
phosphate group. This cyclic AMP is involved in transmitting and amplifying the
chemical signals of other hormones.
NH2
7
N
O
–O
P
5
N
O
CH2 O
O–
9N
N
1
2
N
H
OH
H
OH
adenosine monophosphate (AMP)
N
CH2 O
3
H
H
4
Additional
Functions of
Nucleotides
NH2
6
8
23-22
O
O
P
H
H
O
H
OH
H
O–
cyclic AMP
N
N
Application: Genetic Disease
Adenosine deaminase replaces the
C6 amino group with a hydroxyl group,
an important step in purine metabolism.
A genetic deficiency of the enzyme
causes a severe immunodeficiency,
called “baby in a bubble syndrome,”
because the child must live in a
sterile environment.
1148
CHAPTER 23
Carbohydrates and Nucleic Acids
NAD: A Coenzyme Nicotinamide adenine dinucleotide (NAD) is one of the principal
oxidation–reduction reagents in biological systems. This nucleotide has the structure
of two D-ribose rings (a dinucleotide) linked by their 5¿ phosphates. The aglycone of
one ribose is nicotinamide, and the aglycone of the other is adenine. A dietary deficiency of nicotinic acid (niacin) leads to the disease called pellagra, caused by the
inability to synthesize enough nicotinamide adenine dinucleotide.
O
C
O
C
O
OH
C
N+
NH2
O
N
N
nicotinic acid
(niacin)
O
nicotinamide
N
O
OH
H
OH
NH2
P
O–
O
CH2 O
adenine
N
N
H
H
OH
H
OH
H
N
NAD+
Application: Insomnia
Adenosine may also act as a neurotransmitter that induces sleep. Caffeine
blocks the adenosine receptor, resulting
in wakefulness or insomnia.
nicotinamide adenine dinucleotide
The following equation shows how NAD+ serves as the oxidizing agent in the biological oxidation of an alcohol. Just the nicotinamide portion of NAD shown takes part in the
reaction. The enzyme that catalyzes this reaction is called alcohol dehydrogenase (ADH).
O
H
H
NH2
C
O
ADH enzyme
H3C
+
H
+
H+
N
sugar
sugar
NAD+
ethanol
NH2
+
C
N
H
O
H H
C
H +
C
H
N
N
H
H3C
H
H
O
N
N
CH2 O
P
–O
NH2
O
NH2
acetaldehyde
NADH (reduced)
ATP: An Energy Source When glucose is oxidized in the living cell, the energy
released is used to synthesize adenosine triphosphate (ATP), an anhydride of phosphoric acid. As with most anhydrides, hydrolysis of ATP is highly exothermic. The
hydrolysis products are adenosine diphosphate (ADP) and inorganic phosphate.
NH2
NH2
N
O
HO
P
O
O
–
O
P
O
–
O
P
O
N
CH2 O
–
O
H
H
OH
H
OH
H
adenosine triphosphate (ATP)
N
N
O
N
O
+
H2O
HO
P
O
–
O
N
O
P
O
N
CH2 O
–
O
H
H
OH
H
OH
H
adenosine diphosphate (ADP)
Δ H° = –31 kJ/mol (–7.3 kcal/mol)
N
O
+
–
O
OH
P
–
O
phosphate
Essential Terms
1149
The highly exothermic nature of ATP hydrolysis is largely explained by the heats of
hydration of the products. ADP is hydrated about as well as ATP, but inorganic phosphate
has a large heat of hydration. Hydrolysis also reduces the electrostatic repulsion of the three
negatively charged phosphate groups in ATP. Hydrolysis of adenosine triphosphate
(ATP) liberates 31 kJ (7.3 kcal) of energy per mole of ATP. This is the energy that muscle cells use to contract and all cells use to drive their endothermic chemical processes.
ESSENTIAL PROBLEM-SOLVING SKILLS IN CHAPTER 23
Each skill is followed by problem numbers exemplifying that particular skill.
Draw the Fischer projections and the chair conformations of the anomers and
epimers of glucose from memory. Identify and name these sugars based on how they
differ from the structure of glucose.
Correctly name monosaccharides and disaccharides, and draw their structures
from their names.
Predict which carbohydrates mutarotate, which reduce Tollens reagent, and which
undergo epimerization and isomerization under basic conditions. (Those with free
hemiacetals will, but glycosides with full acetals will not.)
Problems 23-52 and 53
Problems 23-53, 54, 55, 59, 60,
and 61
Problems 23-58, 62, 63, and 66
Predict the reactions of carbohydrates in acidic and basic solutions, and with
oxidizing and reducing agents. Predict the reactions that convert their hydroxyl groups
to ethers or esters, and their carbonyl groups to acetals.
Problems 23-56, 57, 58, 63, 64, 65,
and 66
Determine the structure of an unknown carbohydrate based on its reactions.
Determine its ring size from the methylation and periodic acid cleavage reactions.
Problems 23-57, 63, 64, 65, and 66
Draw the common types of glycosidic linkages, and identify these linkages in
disaccharides and polysaccharides.
Problems 23-59, 60, 61, and 64
Recognize the structures of DNA and RNA, and draw the structures of the common
ribonucleotides and deoxyribonucleotides.
Problems 23-69, 70, and 73
ESSENTIAL TERMS
aglycone
A nonsugar residue bonded to the anomeric carbon of a glycoside (the acetal form of a sugar).
Aglycones are commonly bonded to the sugar through oxygen or nitrogen. (p. 1121)
(glycaric acid, saccharic acid) A dicarboxylic acid formed by oxidation of both end carbon
atoms of a monosaccharide. (p. 1118)
(sugar alcohol) A polyalcohol formed by reduction of the carbonyl group of a
monosaccharide. (p. 1116)
(glyconic acid) A monocarboxylic acid formed by oxidation of the aldehyde group of an
aldose. (p. 1117)
A monosaccharide containing an aldehyde carbonyl group. (p. 1103)
A sugar (such as glucosamine) in which a hydroxyl group is replaced by an amino group. (p. 1139)
The hemiacetal carbon in the cyclic form of a sugar (the carbonyl carbon in the open-chain
form). The anomeric carbon is easily identified because it is the only carbon with two bonds
to oxygen atoms. (p. 1112)
Sugar stereoisomers that differ in configuration only at the anomeric carbon. Anomers are
classified as a or b depending on whether the anomeric hydroxyl group (or the aglycone in
a glycoside) is trans 1a2 or cis 1b2 to the terminal ¬ CH2 OH. (p. 1112)
aldaric acid
alditol
aldonic acid
aldose
amino sugar
anomeric carbon
anomers
anomeric carbon
H
H
6
CH2OH
HO 4
5
HO
H
3
H
O
H
2
OH 1
OH
α-D-glucopyranose
HO 4
H
H
6
CH2OH
5
HO
H
3
H
O
H
2
OH C
1
open-chain form
H
HO 4
O
H
6
CH2OH
5
HO
H
3
H
O
H
2
OH 1
H
β-D-glucopyranose
OH
1150
CHAPTER 23
carbohydrates
cellulose
chitin
degradation
deoxyribonucleic acid
deoxy sugar
dextrose
D series of sugars
disaccharide
enediol rearrangement
epimers
erythro and threo
Carbohydrates and Nucleic Acids
(sugars) Polyhydroxy aldehydes and ketones, including their derivatives and polymers.
Many have formula Cn1H2 O2m from which they received the name “hydrates of carbon”
or “carbohydrates.” (p. 1102)
A linear b-1,4¿ polymer of D-glucopyranose. Cellulose forms the cell walls of plants and is the
major constituent of wood and cotton. (p. 1137)
A b-1,4¿ polymer of N-acetylglucosamine that lends strength and rigidity to the exoskeletons
of insects and crustaceans. (p. 1139)
A reaction that causes loss of a carbon atom. (p. 1125)
(DNA) A biopolymer of deoxyribonucleotides that serves as a template for the synthesis of
ribonucleic acid. DNA is also the template for its own replication, through uncoiling and the
pairing and enzymatic linking of complementary bases. (p. 1140)
A sugar in which a hydroxyl group is replaced by a hydrogen. Deoxy sugars are recognized by
the presence of a methylene group or a methyl group. (p. 1144)
The common dextrorotatory isomer of glucose, D-1+2-glucose. (p. 1114)
All sugars whose asymmetric carbon atom farthest from the carbonyl group has the same configuration as the asymmetric carbon atom in D-1+2-glyceraldehyde. Most naturally occurring
sugars are members of the D series. (p. 1105)
A carbohydrate whose hydrolysis gives two monosaccharide molecules. (p. 1132)
(Lobry de Bruyn–Alberta van Ekenstein reaction) A base-catalyzed tautomerization that interconverts aldoses and ketoses with an enediol as an intermediate. This enolization also epimerizes C2 and other carbon atoms. (p. 1115)
Two diastereomeric sugars differing only in the configuration at a single asymmetric carbon
atom. The epimeric carbon atom is usually specified, as in “C4 epimers.” If no epimeric
carbon is specified, it is assumed to be C2. The interconversion of epimers is called
epimerization. (pp. 1107, 1115)
Diastereomers having similar groups on the same side (erythro) or on opposite sides (threo) of
the Fischer projection. This terminology was adapted from the names of the aldotetroses
erythrose and threose. (p. 1106)
CHO
H
OH
Br
H
HO
H
OH
Br
H
H
CH2OH
D-erythrose
furanose
furanoside
glucoside
glycoside
glycosidic linkage
glucosidic linkage:
galactosidic linkage:
Haworth projection
ketose
Kiliani–Fischer synthesis
L
series of sugars
monosaccharide
mutarotation
CHO
COOH
CH2CH3
erythro-2,3-dibromopentanoic acid
CH3
H
H
OH
OH
Cl
H
CH2OH
D-threose
CH3
threo-3-chlorobutan-2-ol
A five-membered cyclic hemiacetal form of a sugar. (p. 1111)
A five-membered cyclic glycoside. (p. 1119)
A glycoside derived from glucose. (p. 1120)
A cyclic acetal form of a sugar. Glycosides are stable to base, and they are nonreducing sugars.
Glycosides are generally furanosides (five-membered) or pyranosides (six-membered), and
they exist in anomeric a and b forms. (p. 1119)
A general term for an acetal bond from an anomeric carbon joining two monosaccharide
units. (pp. 1120, 1132)
A glycosidic linkage using an acetal bond from the anomeric carbon of glucose.
A glycosidic linkage using an acetal bond from the anomeric carbon of galactose.
A flat-ring representation of a cyclic sugar. The Haworth projection does not show the axial and
equatorial positions of a pyranose, but it does show the cis and trans relationships. (p. 1109)
A monosaccharide containing a ketone carbonyl group. (p. 1103)
A method for elongating an aldose at the aldehyde end. The aldose is converted into two
epimeric aldoses with an additional carbon atom. For example, Kiliani–Fischer synthesis converts D-arabinose to a mixture of D-glucose and D-mannose. (p. 1125)
All sugars whose asymmetric carbon atom farthest from the carbonyl group has the same configuration as the asymmetric carbon atom in L-1-2-glyceraldehyde. Sugars of the L series are
not common in nature. (p. 1105)
A carbohydrate that does not undergo hydrolysis of glycosidic bonds to give smaller sugar
molecules. (p. 1102)
A spontaneous change in optical rotation that occurs when a pure anomer of a sugar in its
hemiacetal form equilibrates with the other anomer to give an equilibrium mixture with an
averaged value of the optical rotation. (p. 1112)
1151
Study Problems
An N-glycoside of b -D-ribofuranose or b -D-deoxyribofuranose, where the aglycone is one of
several derivatives of pyrimidine or purine. (p. 1141)
A 5¿-phosphate ester of a nucleoside. (p. 1141)
nucleoside
nucleotide
NH2
N
O
–O
P
O
O
5´
N
CH2 O
O–
H
H
OH
H
OH
H
N
O
O
cytidine monophosphate,
CMP (cytidylic acid)
–O
P
NH2
H
O
5´
N
CH2 O
O–
H
H
OH
H
OH
H
N
O
O
uridine monophosphate,
UMP (uridylic acid)
–O
P
O
5´
N
CH2 O
O–
H
H
OH
H
OH
H
O
N
N
adenosine monophosphate,
AMP (adenylic acid)
N
O
–O
P
O
5´
N
CH2 O
O–
H
H
OH
H
OH
H
H
N
N
NH2
guanosine monophosphate,
GMP (guanidylic acid)
The four common ribonucleotides of RNA
oligosaccharide
osazone
polysaccharide
primary structure
pyranose
pyranoside
rayon
reducing sugar
ribonucleic acid
ribonucleotide
Ruff degradation
starches
amylose:
amylopectin:
glycogen:
sugar
Tollens test
A carbohydrate whose hydrolysis gives about two to ten monosaccharide units, but not as
many as a polysaccharide. (p. 1136)
The product, containing two phenylhydrazone groups, that results from reaction of a reducing
sugar with phenylhydrazine. (p. 1124)
A carbohydrate whose hydrolysis gives many monosaccharide molecules. (p. 1136)
The primary structure of a nucleic acid is the sequence of nucleotides forming the polymer.
This sequence determines the genetic characteristics of the nucleic acid. (p. 1144)
A six-membered cyclic hemiacetal form of a sugar. (p. 1111)
A six-membered cyclic glycoside. (p. 1119)
A commercial fiber made from regenerated cellulose. (p. 1137)
Any sugar that gives a positive Tollens test. Both ketoses and aldoses (in their hemiacetal
forms) give positive Tollens tests. (p. 1119)
(RNA) A biopolymer of ribonucleotides that controls the synthesis of proteins. The synthesis
of RNA is generally controlled by and patterned after DNA in the cell. (p. 1143)
The 5¿-phosphate ester of a ribonucleoside, a component of RNA based on b -D-ribofuranose
and containing one of four heterocyclic bases as the aglycone. (p. 1142)
A method for shortening the chain of an aldose by one carbon atom by treatment with bromine
water, followed by hydrogen peroxide and Fe21SO423. (p. 1125)
A class of a-1,4¿ polymers of glucose used for carbohydrate storage in plants and
animals. (p. 1137)
A linear a-1,4¿ polymer of D-glucopyranose used for carbohydrate storage in plants.
A branched a-1,4¿ polymer of D-glucopyranose used for carbohydrate storage in plants.
Branching occurs at a-1,6¿ glycosidic linkages.
An extensively branched a-1,4¿ polymer of D-glucopyranose used for carbohydrate storage in
animals. Branching occurs at a-1,6¿ glycosidic linkages.
(saccharide) Any carbohydrate, regardless of structure, complexity, or taste. A simple sugar is
a monosaccharide. (p. 1102)
A test for reducing sugars, employing the same silver–ammonia complex used as a test for
aldehydes. A positive test gives a silver precipitate, often in the form of a silver mirror. Tollens
reagent is basic, and it promotes enediol rearrangements that interconvert ketoses and aldoses.
Therefore, both aldoses and ketoses give positive Tollens tests if they are in their hemiacetal
forms, in equilibrium with open-chain carbonyl structures. (p. 1118)
STUDY PROBLEMS
23-52
23-53
Glucose is the most abundant monosaccharide. From memory, draw glucose in
(a) the Fischer projection of the open chain
(b) the most stable chair conformation of the most stable pyranose anomer
(c) the Haworth projection of the most stable pyranose anomer
Without referring to the chapter, draw the chair conformations of
(a) b -D-mannopyranose (the C2 epimer of glucose)
(b) a-D-allopyranose (the C3 epimer of glucose)
(c) b -D-galactopyranose (the C4 epimer of glucose)
(d) N-acetylglucosamine, glucose with the C2 oxygen atom replaced by an acetylated amino group
1152
23-54
23-55
Carbohydrates and Nucleic Acids
CHAPTER 23
Use Figure 23-3 (the D family of aldoses) to name the following aldoses.
(b) the C3 epimer of D-mannose
(c) the C3 epimer of D-threose
(a) the C2 epimer of D-arabinose
(e) the C5 epimer of D-glucose
(d) the enantiomer of D-galactose
Classify the following monosaccharides. (Examples: D-aldohexose, L-ketotetrose.)
(a) 1+2-glucose
(b) 1-2-arabinose
(c) L-fructose
(d)
(e)
CHO
C
H
HO
H
H
OH
OH
H
OH
HO
CHO
H
O
HO
H
(f)
CH2OH
HO
H
CH2OH
(+)-threose
CH2OH
H
OH
(–)-ribulose
CH2OH
23-57
23-58
23-59
23-60
23-61
CHO
H
HO
NHCOCH3
H
H
OH
H
OH
CH2OH
(+)-gulose
23-56
(g)
N-acetylglucosamine
(a) Give the products expected when 1+2-glyceraldehyde reacts with HCN.
(b) What is the relationship between the products? How might they be separated?
(c) Are the products optically active? Explain.
The relative configurations of the stereoisomers of tartaric acid were established by the following syntheses:
HCN "
(1) D-1+2-glyceraldehyde
diastereomers A and B (separated)
(2) Hydrolysis of A and B using aqueous Ba1OH22 gave C and D, respectively.
(3) HNO3 oxidation of C and D gave 1-2-tartaric acid and meso-tartaric acid, respectively.
(a) You know the absolute configuration of D-1+2-glyceraldehyde. Use Fischer projections to show the absolute configurations of products A, B, C, and D.
(b) Show the absolute configurations of the three stereoisomers of tartaric acid: (+)-tartaric acid, (-)-tartaric acid, and
meso-tartaric acid.
Predict the products obtained when D-galactose reacts with each reagent.
(a) Br2 and H2O
(b) NaOH, H2O
(c) CH3OH, H +
(d) Ag(NH3)2+ - OH
(e) H2, Ni
(f) excess Ac2 O and pyridine (g) excess CH3 I, Ag2 O
(h) NaBH4
(i) Br2, H2 O, then H2 O2 and Fe21SO423
(j) HCN, then H3 O+, then Na(Hg) (k) excess HIO4
Draw the following sugar derivatives.
(b) 2,3,4,6-tetra-O-methyl-D-mannopyranose
(a) methyl b -D-glucopyranoside
(d) methyl 2,3,4,6-tetra-O-methyl-b -D-galactopyranoside
(c) 1,3,6-tri-O-methyl-D-fructofuranose
Draw the structures (using chair conformations of pyranoses) of the following disaccharides.
(a) 4-O-(a-D-glucopyranosyl)-D-galactopyranose
(b) a-D-fructofuranosyl- b -D-mannopyranoside
(c) 6-O-( b -D-galactopyranosyl)-D-glucopyranose
Give the complete systematic name for each structure.
OCH3
HOCH2 O
H
(a)
OCH3
HO
CH2OH
H
OH
CH2OH
HO
H O
H
HO
23-62
CH2
H
H
(d)
HO
H
HO
H
OH
(b) HO
CH3O
H
(c) HOCH2 O
H
H
CH2OH O
H
OH
H
OH
H
H
O
OH
OH
H
H
CH2OH O
H
OH
H
NH
H
H
C O
CH3
Which of the sugars mentioned in Problems 23-59, 23-60, and 23-61 are reducing sugars? Which ones would undergo
mutarotation?
Study Problems
23-63
23-64
23-65
23-66
23-67
After a series of Kiliani–Fischer syntheses on 1+2-glyceraldehyde, an unknown sugar is isolated from the reaction mixture. The following experimental information is obtained:
(1) Molecular formula C6 H12 O6.
(2) Undergoes mutarotation.
(3) Reacts with bromine water to give an aldonic acid.
(4) Reacts with phenylhydrazine to give an osazone, mp 178 °C.
(5) Reacts with HNO3 to give an optically active aldaric acid.
(6) Ruff degradation followed by HNO3 oxidation gives an optically inactive aldaric acid.
(7) Two Ruff degradations followed by HNO3 oxidation give meso-tartaric acid.
(8) Formation of the methyl glycoside (using CH3 OH and HCl), followed by periodic acid oxidation, gives a mixture
of products that includes 1+ 2-glyceraldehyde.
(a) Draw a Fischer projection for the open-chain form of this unknown sugar. Use Figure 23-3 to name the sugar.
(b) Draw the most stable conformation of the most stable cyclic hemiacetal form of this sugar, and give the structure
a complete systematic name.
An unknown reducing disaccharide is found to be unaffected by invertase enzymes. Treatment with an a-galactosidase
cleaves the disaccharide to give one molecule of D-fructose and one molecule of D-galactose. When the disaccharide is
treated with excess iodomethane and silver oxide and then hydrolyzed in dilute acid, the products are 2,3,4,6-tetra-O-methylgalactose and 1,3,4-tri-O-methylfructose. Propose a structure for this disaccharide, and give its complete systematic name.
(a) Which of the D-aldopentoses will give optically active aldaric acids on oxidation with HNO3?
(b) Which of the D-aldotetroses will give optically active aldaric acids on oxidation with HNO3?
(c) Sugar X is known to be a D-aldohexose. On oxidation with HNO3, X gives an optically inactive aldaric acid. When X
is degraded to an aldopentose, oxidation of the aldopentose gives an optically active aldaric acid. Determine the structure of X.
(d) Even though sugar X gives an optically inactive aldaric acid, the pentose formed by degradation gives an optically
active aldaric acid. Does this finding contradict the principle that optically inactive reagents cannot form optically
active products?
(e) Show what product results if the aldopentose formed from degradation of X is further degraded to an aldotetrose.
Does HNO3 oxidize this aldotetrose to an optically active aldaric acid?
When the gum of the shrub Sterculia setigera is subjected to acidic hydrolysis, one of the water-soluble components of the
hydrolysate is found to be tagatose. The following information is known about tagatose:
(1) Molecular formula C6 H12 O6.
(2) Undergoes mutarotation.
(3) Does not react with bromine water.
(4) Reduces Tollens reagent to give D-galactonic acid and D-talonic acid.
(5) Methylation of tagatose (using excess CH3 I and Ag2 O) followed by acidic hydrolysis gives 1,3,4,5-tetra-O-methyltagatose.
(a) Draw a Fischer projection structure for the open-chain form of tagatose.
(b) Draw the most stable conformation of the most stable cyclic hemiacetal form of tagatose.
An important protecting group developed specifically for polyhydroxy compounds like nucleosides is the tetraisopropyldisiloxanyl group, abbreviated TIPDS, that can protect two alcohol groups in a molecule.
i-Pr
i-Pr
Si
Cl
HO
i-Pr
O
Si
Cl
TIPDS chloride
*23-68
1153
i-Pr
base
O
+
Et3N
H
H
OH
H
OH
H
a ribonucleoside
(a) The TIPDS group is somewhat hindered around the Si atoms by the isopropyl groups. Which OH is more likely to
react first with TIPDS chloride? Show the product with the TIPDS group on one oxygen.
(b) Once the TIPDS group is attached at the first oxygen, it reaches around to the next closest oxygen. Show the final
product with two oxygens protected.
(c) The unprotected hydroxyl group can now undergo reactions without affecting the protected oxygens. Show the product after the protected nucleoside from (b) is treated with tosyl chloride and pyridine, followed by NaBr, ending with
deprotection with Bu4NF.
6
Some protecting groups can block two OH groups of a carbohydrate at the
Ph
O
same time. One such group is shown here, protecting the 4-OH and 6-OH
groups of b -D-glucose.
O
4
O
(a) What type of functional group is involved in this blocking group?
HO
(b) What did glucose react with to form this protected compound?
OH
OH
(Continued )
1154
23-69
23-70
Carbohydrates and Nucleic Acids
CHAPTER 23
(c) When this blocking group is added to glucose, a new chiral center is formed. Where is it?
Draw the stereoisomer that has the other configuration at this chiral center. What is the relationship between these two
stereoisomers of the protected compound?
(d) Which of the two stereoisomers in part (c) do you expect to be the major product? Why?
(e) A similar protecting group, called an acetonide, can block reac- HO
O base
tion at the 2¿ and 3¿ oxygens of a ribonucleoside. This protected
O
H+
derivative is formed by the reaction of the nucleoside with ace+
H2O + ?
3´
2´
tone under acid catalysis. From this information, draw the protected product formed by the reaction.
OH OH
Draw the structures of the following nucleotides.
(a) guanosine triphosphate (GTP)
(b) deoxycytidine monophosphate (dCMP)
(c) cyclic guanosine monophosphate (cGMP)
Draw the structure of a four-residue segment of DNA with the following sequence.
13¿ end2 G-T-A-C 15¿ end2
23-71
Erwin Chargaff’s discovery that DNA contains equimolar amounts of guanine and cytosine and also equimolar amounts of
adenine and thymine has come to be known as Chargaff’s rule:
G = C and A = T
23-72
*23-73
*23-74
(a) Does Chargaff’s rule imply that equal amounts of guanine and adenine are present in DNA? That is, does G = A?
(b) Does Chargaff’s rule imply that the sum of the purine residues equals the sum of the pyrimidine residues? That is,
does A + G = C + T?
(c) Does Chargaff’s rule apply only to double-stranded DNA, or would it also apply to each individual strand if the double helical strand were separated into its two complementary strands?
Retroviruses like HIV, the pathogen responsible for AIDS, incorporate an RNA template that is copied into DNA during
infection. The reverse transcriptase enzyme that copies RNA into DNA is relatively nonselective and error-prone, leading
to a high mutation rate. Its lack of selectivity is exploited by the anti-HIV drug AZT 13¿-azido-2¿,3¿-dideoxythymidine2,
which becomes phosphorylated and is incorporated by reverse transcriptase into DNA, where it acts as a chain terminator.
Mammalian DNA polymerases are more selective, having a low affinity for AZT, so its toxicity is relatively low.
(a) Draw the structures of AZT and natural deoxythymidine.
(b) Draw the structure of AZT 5¿-triphosphate, the derivative that inhibits reverse transcriptase.
Exposure to nitrous acid (see Section 19-16), sometimes found in cells, can convert cytosine to uracil.
(a) Propose a mechanism for this conversion.
(b) Explain how this conversion would be mutagenic upon replication.
(c) DNA generally includes thymine, rather than uracil (found in RNA). Based on this fact, explain why the nitrous
acid-induced mutation of cytosine to uracil is more easily repaired in DNA than it is in RNA.
H. G. Khorana won the Nobel Prize in Medicine in 1968 for developing the synthesis of DNA and RNA and for helping to
unravel the genetic code. Part of the chemistry he developed was the use of selective protecting groups for the 5¿ OH group
of nucleosides.
HO
base
O
5´
3´
OH
OCH3
2´
OH
Ph
Ph
Ph
C
OR
Ph
trityl, triphenylmethyl ether
H3CO
C
OR
Ph
MMT, monomethoxytrityl ether
H3CO
C
OR
Ph
DMT, dimethoxytrityl ether
The trityl ether derivative of just the 5¿ OH group is obtained by reaction of the nucleoside with trityl chloride, MMT
chloride, or DMT chloride and a base like Et3 N. The trityl ether derivative can be removed in dilute aqueous acid. DMT
derivatives hydrolyze fastest, followed by MMT derivatives, and trityl derivatives slowest.
(a) Draw the product with the trityl derivative on the 5¿ oxygen.
(b) Explain why the trityl derivative is selective for the 5¿ OH group. Why doesn’t it react at 2¿ or 3¿?
(c) Why is the DMT group easiest to remove under dilute acid conditions? Why does the solution instantly turn orange
when acid is added to a DMT derivative?
COO–
x
eli
α-h
24
Amino Acids,
Peptides,
and Proteins
+NH
3
GOALS FOR
CHAPTER 24
Name amino acids and peptides, and draw the
structures from their names. Explain why the naturally
occurring amino acids are called L-amino acids.
Use information from terminal residue analysis and
partial hydrolysis to determine the structure of an
unknown peptide.
Identify which amino acids are acidic, which are basic,
and which are neutral. Use the isoelectric point to predict
the charge on an amino acid at a given pH.
Identify the levels of protein structure, and explain how
a protein’s structure affects its properties.
Show how to synthesize amino acids from simpler
compounds, and show how to combine amino acids in the
proper sequence to synthesize a peptide.
Proteins are the most abundant organic molecules in animals, playing important roles
in all aspects of cell structure and function. Proteins are biopolymers of A-amino acids,
so named because the amino group is bonded to the a carbon atom, next to the carbonyl group. The physical and chemical properties of a protein are determined by its
constituent amino acids. The individual amino acid subunits are joined by amide linkages called peptide bonds. Figure 24-1 (next page) shows the general structure of an
a-amino acid and a protein.
Proteins have an amazing range of structural and catalytic properties as a result of
their varying amino acid composition. Because of this versatility, proteins serve an
astonishing variety of functions in living organisms. Some of the functions of the major
classes of proteins are outlined in Table 24-1.
TABLE 24-1
24-1
Introduction
Examples of Protein Functions
Class of Protein
Example
Function of Example
structural proteins
collagen, keratin
strengthen tendons, skin, hair, nails
enzymes
DNA polymerase
replicates and repairs DNA
transport proteins
hemoglobin
transports O2 to the cells
contractile proteins
actin, myosin
cause contraction of muscles
protective proteins
antibodies
complex with foreign proteins
hormones
insulin
regulates glucose metabolism
toxins
snake venoms
incapacitate prey
1155
1156
Amino Acids, Peptides, and Proteins
CHAPTER 24
α carbon atom
O
H2N
CH
α-amino group
C
OH
R
side chain
an α-amino acid
O
O
H2N
CH
C
OH H2N
CH
OH H2N
C
CH
CH2OH
CH3
alanine
O
O
serine
C
OH H2N
CH
C
H
CH2SH
glycine
cysteine
O
OH H2N
CH
C
OH
CH(CH3)2
valine
several individual amino acids
peptide bonds
O
O
NH
CH
C
NH
CH
NH
C
CH2OH
CH3
O
O
CH
C
NH
H
CH
C
CH2SH
O
NH
CH
C
CH(CH3)2
a short section of a protein
FIGURE 24-1
Structure of a general protein and its constituent amino acids. The amino acids are joined by
amide linkages called peptide bonds.
The study of proteins is one of the major branches of biochemistry, and there is no
clear division between the organic chemistry of proteins and their biochemistry. In this
chapter, we begin the study of proteins by learning about their constituents, the amino
acids. We also discuss how amino acid monomers are linked into the protein polymer,
and how the properties of a protein depend on those of its constituent amino acids.
These concepts are needed for the further study of protein structure and function in a
biochemistry course.
24-2
Structure and
Stereochemistry of
the a-Amino Acids
The term amino acid might mean any molecule containing both an amino group
and any type of acid group; however, the term is almost always used to refer to an
a-amino carboxylic acid. The simplest a-amino acid is aminoacetic acid, called
glycine. Other common amino acids have side chains (symbolized by R)
substituted on the a carbon atom. For example, alanine is the amino acid with a
methyl side chain.
O
H2N
CH2
C
O
OH
H 2N
CH
C
O
OH
R
glycine
a substituted amino acid
H2N
CH
C
OH
CH3
alanine (R = CH3)
Except for glycine, the a-amino acids are all chiral. In all of the chiral amino
acids, the chirality center is the asymmetric a carbon atom. Nearly all the naturally
occurring amino acids are found to have the (S) configuration at the a carbon atom.
Figure 24-2 shows a Fischer projection of the (S) enantiomer of alanine, with the
carbon chain along the vertical and the carbonyl carbon at the top. Notice that the
configuration of (S)-alanine is similar to that of L-1-2-glyceraldehyde, with the amino
group on the left in the Fischer projection. Because their stereochemistry is similar
24-2 Structure and Stereochemistry of the a-Amino Acids
H2N
COOH
CHO
COOH
C
C
C
H
CH3
HO
COOH
H2N
H
CH2OH
H2N
CHO
HO
H
CH3
Almost all the naturally occurring
amino acids have the (S)
configuration. They are called
L-amino acids because their
stereochemistry resembles that of
L-1-2-glyceraldehyde.
R
H
H2N
CH2OH
L-alanine
(S )-alanine
FIGURE 24-2
COOH
H
H
R
L-(–)-glyceraldehyde
an L-amino acid
(S) configuration
(S )-glyceraldehyde
to that of L-1-2-glyceraldehyde, the naturally occurring (S)-amino acids are classified as L-amino acids.
Although D-amino acids are occasionally found in nature, we usually assume the
amino acids under discussion are the common L-amino acids. Remember once again that
the D and L nomenclature, like the R and S designation, gives the configuration of the
asymmetric carbon atom. It does not imply the sign of the optical rotation, 1+2 or 1-2,
which must be determined experimentally.
Amino acids combine many of the properties and reactions of both amines and
carboxylic acids. The combination of a basic amino group and an acidic carboxyl group
in the same molecule also results in some unique properties and reactions. The side
chains of some amino acids have additional functional groups that lend interesting properties and undergo reactions of their own.
24-2A
Application: Antibiotics
Bacteria require specific enzymes,
called racemases, to interconvert D- and
L-amino acids. Mammals do not use
D-amino acids, so compounds that block
racemases do not affect mammals and
show promise as antibiotics.
The Standard Amino Acids of Proteins
The standard amino acids are 20 common a-amino acids that are found in nearly all
proteins. The standard amino acids differ from each other in the structure of the side
chains bonded to their a carbon atoms. All the standard amino acids are L-amino acids.
Table 24-2 shows the 20 standard amino acids, grouped according to the chemical
properties of their side chains. Each amino acid is given a three-letter abbreviation and
a one-letter symbol (green) for use in writing protein structures.
TABLE 24-2
Name
The Standard Amino Acids
Symbol
Abbreviation
Structure
Functional Group
in Side Chain
Isoelectric
Point
CH
COOH
none
6.0
COOH
alkyl group
6.0
COOH
alkyl group
6.0
alkyl group
6.0
alkyl group
6.0
side chain is nonpolar, H or alkyl
glycine
G
Gly
H2N
alanine
A
Ala
H2N
H
CH
CH3
*valine
V
Val
H2N
CH
CH
CH3
CH3
*leucine
L
Leu
H2N
CH
CH2
COOH
CH
CH3
CH3
*isoleucine
I
Ile
1157
H2N
CH
COOH
CH3
CH
CH2CH3
(Continued )
1158
CHAPTER 24
TABLE 24-2
Amino Acids, Peptides, and Proteins
(continued)
Name
Symbol
*phenylalanine
Structure
Functional Group
in Side Chain
CH
aromatic group
5.5
rigid cyclic structure
6.3
hydroxyl group
5.7
hydroxyl group
5.6
phenolic — OH group
5.7
thiol
5.0
sulfide
5.7
amide
5.4
amide
5.7
indole
5.9
carboxylic acid
2.8
carboxylic acid
3.2
Abbreviation
Phe
F
H2N
COOH
Isoelectric
Point
CH2
proline
HN
Pro
P
CH
COOH
CH2
H2C
CH2
side chain contains an
serine
S
OH
Ser
H2N
CH
COOH
CH2
*threonine
tyrosine
T
Y
Thr
Tyr
OH
H2N
CH
COOH
HO
CH
CH3
H2N
CH
COOH
CH2
OH
side chain contains sulfur
cysteine
C
H2N
Cys
CH
COOH
CH2
*methionine
M
Met
H2N
SH
CH
COOH
CH2
CH2
S
CH3
side chain contains nonbasic nitrogen
asparagine
N
Asn
H2N
CH
COOH
CH2
C
NH2
O
glutamine
Q
Gln
H2N
CH
COOH
CH2
CH2
C
NH2
O
*tryptophan
W
H2N
Trp
CH
COOH
CH2
N
H
side chain is acidic
aspartic acid
D
Asp
H2N
CH
COOH
CH2
glutamic acid
E
Glu
H2N
COOH
CH
COOH
CH2
CH2
COOH
24-2 Structure and Stereochemistry of the a-Amino Acids
TABLE 24-2
Name
(continued)
Symbol
Abbreviation
Functional Group
in Side Chain
Structure
Isoelectric
Point
side chain is basic
*lysine
K
*arginine
R
H2N
Lys
H2N
Arg
CH
COOH
CH2
CH2
CH
COOH
CH2
CH2
amino group
CH2
CH2
NH2
guanidino group
CH2
NH
9.7
C
10.8
NH2
NH
*histidine
H
H2N
His
CH
COOH
imidazole ring
CH2
NH
N
*essential amino acid
Notice in Table 24-2 how proline is different from the other standard amino acids.
Its amino group is fixed in a ring with its a carbon atom. This cyclic structure lends
additional strength and rigidity to proline-containing peptides.
COOH
proline
N
H
H
α carbon
α-amino group
PROBLEM 24-1
Draw three-dimensional representations of the following amino acids.
(b) L-histidine
(c) D-serine
(a) L-phenylalanine
(d)
L-tryptophan
PROBLEM 24-2
Most naturally occurring amino acids have chirality centers (the asymmetric a carbon atoms)
that are named (S) by the Cahn–Ingold–Prelog convention (Section 5-3). The common naturally occurring form of cysteine has a chirality center that is named (R), however.
(a) What is the relationship between (R)-cysteine and (S)-alanine? Do they have the opposite
three-dimensional configuration (as the names might suggest) or the same configuration?
(b) (S)-alanine is an L-amino acid (Figure 24-2). Is (R)-cysteine a D-amino acid or an
L-amino acid?
24-2B
Essential Amino Acids
Humans can synthesize about half of the amino acids needed to make proteins. Other
amino acids, called the essential amino acids, must be provided in the diet. The
ten essential amino acids, starred 1*2 in Table 24-2, are the following:
arginine (Arg)
threonine (Thr)
lysine (Lys)
valine (Val)
phenylalanine (Phe)
tryptophan (Trp)
methionine (Met)
histidine (His)
leucine (Leu)
isoleucine (Ile)
7.6
1159
1160
CHAPTER 24
Amino Acids, Peptides, and Proteins
Application: Gelatin
Gelatin is made from collagen, which is
a structural protein composed primarily
of glycine, proline, and hydroxyproline.
As a result, gelatin has low nutritional
value because it lacks many of the
essential amino acids.
Proteins that provide all the essential amino acids in about the right proportions
for human nutrition are called complete proteins. Examples of complete proteins are
those in meat, fish, milk, and eggs. About 50 g of complete protein per day is adequate
for adult humans.
Proteins that are severely deficient in one or more of the essential amino acids are
called incomplete proteins. If the protein in a person’s diet comes mostly from one
incomplete source, the amount of human protein that can be synthesized is limited by
the amounts of the deficient amino acids. Plant proteins are generally incomplete. Rice,
corn, and wheat are all deficient in lysine. Rice also lacks threonine, and corn also lacks
tryptophan. Beans, peas, and other legumes have the most complete proteins among
the common plants, but they are deficient in methionine.
Vegetarians can achieve an adequate intake of the essential amino acids if they eat
many different plant foods. Plant proteins can be chosen to be complementary, with
some foods supplying amino acids that others lack. An alternative is to supplement the
vegetarian diet with a rich source of complete protein such as milk or eggs.
PROBLEM 24-3
The herbicide glyphosate (Roundup®) kills plants by inhibiting an enzyme needed for synthesis of phenylalanine. Deprived of phenylalanine, the plant cannot make the proteins it needs,
and it gradually weakens and dies. Although a small amount of glyphosate is deadly to a plant,
its human toxicity is quite low. Suggest why this powerful herbicide has little effect on humans.
Rare and Unusual Amino Acids
24-2C
In addition to the standard amino acids, other amino acids are found in protein in smaller
quantities. For example, 4-hydroxyproline and 5-hydroxylysine are hydroxylated versions of standard amino acids. These are called rare amino acids, even though they are
commonly found in collagen.
OH
H
4
5
3
1
2
N
COOH
6
H2N
CH2
H
3
4
5
CH
CH2
2
CH
CH2
OH
NH2
H
4-hydroxyproline
1
COOH
5-hydroxylysine
Some of the less common D enantiomers of amino acids are also found in nature.
For example, D-glutamic acid is found in the cell walls of many bacteria, and D-serine
is found in earthworms. Some naturally occurring amino acids are not a-amino acids:
g-Aminobutyric acid (GABA) is one of the neurotransmitters in the brain, and b-alanine
is a constituent of the vitamin pantothenic acid.
COOH
H
COOH
H
NH2
CH CH COOH
2
2
D-glutamic
24-3
Acid–Base Properties
of Amino Acids
acid
γ
NH2
CH2
CH OH
NH2
2
D-serine
β
CH2
α
CH2
COOH
γ-aminobutyric acid
β
CH2
NH2
α
CH2
COOH
β-alanine
Although we commonly write amino acids with an intact carboxyl 1 ¬ COOH2 group
and amino 1 ¬ NH22 group, their actual structure is ionic and depends on the pH. The
carboxyl group loses a proton, giving a carboxylate ion, and the amino group is protonated to an ammonium ion. This structure is called a dipolar ion or a zwitterion
(German for “dipolar ion”).
24-3 Acid–Base Properties of Amino Acids
O
H2N
CH
O
C
+
OH
H3N
R
CH
C
O–
R
uncharged structure
(minor component)
dipolar ion, or zwitterion
(major component)
The dipolar nature of amino acids gives them some unusual properties:
1. Amino acids have high melting points, generally over 200 °C.
+
H3 N ¬ CH2 ¬ COOglycine, mp 262 °C
2. Amino acids are more soluble in water than they are in ether, dichloromethane,
and other common organic solvents.
3. Amino acids have much larger dipole moments 1m2 than simple amines or
simple acids.
+
H3 N ¬ CH2 ¬ COO-
CH3 ¬ CH2 ¬ CH2 ¬ NH2
glycine, m = 14 D
propylamine, m = 1.4 D
CH3 ¬ CH2 ¬ COOH
propionic acid, m = 1.7 D
4. Amino acids are less acidic than most carboxylic acids and less basic than most
amines. In fact, the acidic part of the amino acid molecule is the ¬ NH3+ group,
not a ¬ COOH group. The basic part is the ¬ COO - group, and not a free
¬ NH2 group.
R
R
COOH
pKa = 5
R
NH2
pKb = 4
+
H3N
CH
COO–
pKa = 10
pKb = 12
Because amino acids contain both acidic 1 ¬ NH3+ 2 and basic 1 ¬ COO-2 groups,
they are amphoteric (having both acidic and basic properties). The predominant form
of the amino acid depends on the pH of the solution. In an acidic solution, the
¬ COO- group is protonated to a free ¬ COOH group, and the molecule has an
overall positive charge. As the pH is raised, the ¬ COOH loses its proton at about
pH 2. This point is called pKa1, the first acid-dissociation constant. As the pH is raised
further, the ¬ NH3+ group loses its proton at about pH 9 or 10. This point is called
pKa2, the second acid-dissociation constant. Above this pH, the molecule has
an overall negative charge.
+
H3N
CH
COOH
–OH
H+
CH
COO–
–OH
H+
R
R
cationic in acid
+
H3N
pKa1 ≈ 2
neutral
H2N
CH
COO–
R
pKa2 ≈ 9–10
anionic in base
Figure 24-3 shows a titration curve for glycine. The curve starts at the bottom left,
where glycine is entirely in its cationic form. Base is slowly added, and the pH is
recorded. At pH 2.3, half of the cationic form has been converted to the zwitterionic
form. At pH 6.0, essentially all the glycine is in the zwitterionic form. At pH 9.6, half
of the zwitterionic form has been converted to the basic form. From this graph, we can
see that glycine is mostly in the cationic form at pH values below 2.3, mostly in the zwitterionic form at pH values between 2.3 and 9.6, and mostly in the anionic form at pH
values above 9.6. By varying the pH of the solution, we can control the charge on the
molecule. This ability to control the charge of an amino acid is useful for separating and
identifying amino acids by electrophoresis, as described in Section 24-4.
1161
1162
CHAPTER 24
Amino Acids, Peptides, and Proteins
FIGURE 24-3
0.5
12
1
1.5
2
O
..
A titration curve for glycine. The pH
controls the charge on glycine: cationic
below pH 2.3; zwitterionic between
pH 2.3 and 9.6; and anionic above
pH 9.6. The isoelectric pH is 6.0.
H2N
CH2
C
O–
anionic above pH 9.6
10
pKa2 = 9.6
8
O
pH
+
H3N
Isoelectric
point = 6.0
6
CH2
C
O–
zwitterionic near the
isoelectric point
4
O
2
+
H3N
pKa1 = 2.3
CH2
C
OH
cationic below pH 2.3
0
0.5
1
2
1.5
Equivalents of –OH added
24-4
Isoelectric Points
and Electrophoresis
+
H3N
CH
An amino acid bears a positive charge in acidic solution (low pH) and a negative
charge in basic solution (high pH). There must be an intermediate pH where the
amino acid is evenly balanced between the two forms, as the dipolar zwitterion
with a net charge of zero. This pH is called the isoelectric pH or the isoelectric
point, abbreviated pI.
COOH
–OH
H+
+
H3N
CH
COO–
R
R
low pH
(cationic in acid)
–OH
H+
H2N
CH
COO–
R
isoelectric pH
(neutral)
high pH
(anionic in base)
The isoelectric points of the standard amino acids are given in Table 24-2. Notice
that the isoelectric pH depends on the amino acid structure in a predictable way.
acidic amino acids:
neutral amino acids:
basic amino acids:
aspartic acid (2.8), glutamic acid (3.2)
(5.0 to 6.3)
lysine (9.7), arginine (10.8), histidine (7.6)
The side chains of aspartic acid and glutamic acid contain acidic carboxyl groups. These
amino acids have acidic isoelectric points around pH 3. An acidic solution is needed to
prevent deprotonation of the second carboxylic acid group and to keep the amino acid
in its neutral isoelectric state.
Basic amino acids (histidine, lysine, and arginine) have isoelectric points at pH
values of 7.6, 9.7, and 10.8, respectively. These values reflect the weak basicity of the
imidazole ring, the intermediate basicity of an amino group, and the strong basicity of
the guanidino group. A basic solution is needed in each case to prevent protonation of
the basic side chain to keep the amino acid electrically neutral.
The other amino acids are considered neutral, with no strongly acidic or basic side
chains. Their isoelectric points are slightly acidic (from about 5 to 6) because the
¬ NH3+ group is slightly more acidic than the ¬ COO- group is basic.
24-4 Isoelectric Points and Electrophoresis
1163
PROBLEM 24-4
Draw the structure of the predominant form of
(a) isoleucine at pH 11
(b) proline at pH 2
(c) arginine at pH 7
(d) glutamic acid at pH 7
(e) a mixture of alanine, lysine, and aspartic acid at (i) pH 6; (ii) pH 11; (iii) pH 2
PROBLEM 24-5
Problem-solving Hint
Draw the resonance forms of a protonated guanidino group, and explain why arginine has such
a strongly basic isoelectric point.
PROBLEM 24-6
Although tryptophan contains a heterocyclic amine, it is considered a neutral amino acid.
Explain why the indole nitrogen of tryptophan is more weakly basic than one of the imidazole
nitrogens of histidine.
Electrophoresis uses differences in isoelectric points to separate mixtures of amino
acids (Figure 24-4). A streak of the amino acid mixture is placed in the center of a layer
of acrylamide gel or a piece of filter paper wet with a buffer solution. Two electrodes
are placed in contact with the edges of the gel or paper, and a potential of several thousand volts is applied across the electrodes. Positively charged (cationic) amino acids are
attracted to the negative electrode (the cathode), and negatively charged (anionic) amino
acids are attracted to the positive electrode (the anode). An amino acid at its isoelectric
point has no net charge, so it does not move.
As an example, consider a mixture of alanine, lysine, and aspartic acid in a buffer solution at pH 6. Alanine is at its isoelectric point, in its dipolar zwitterionic form with a net
charge of zero. A pH of 6 is more acidic than the isoelectric pH for lysine (9.7), so lysine
is in the cationic form. Aspartic acid has an isoelectric pH of 2.8, so it is in the anionic form.
–
Beginning
power
supply
+
cathode
anode
–
+
wet with pH 6 buffer solution
streak containing Ala, Lys, Asp
–
End
cathode
–
power
supply
+
anode
+
Asp– moves toward the positive charge
Ala does not move
Lys+ moves toward the negative charge
FIGURE 24-4
A simplified picture of the electrophoretic separation of alanine, lysine, and aspartic acid at pH 6.
Cationic lysine is attracted to the cathode; anionic aspartic acid is attracted to the anode. Alanine
is at its isoelectric point, so it does not move.
At its isoelectric point (pI), an amino
acid has a net charge of zero, with
NH3+ and COO- balancing each
other. In more acidic solution
(lower pH), the carboxyl group
becomes protonated and the net
charge is positive. In more basic
solution (higher pH), the amino
group loses its proton and the net
charge is negative.
1164
CHAPTER 24
Amino Acids, Peptides, and Proteins
Structure at pH 6
+
H3N
CH
+
COO–
H3N
CH3
CH
(CH2)4
+
COO–
H3N
NH+3
lysine (charge +1)
alanine (charge 0)
CH
COO–
CH2
COO–
aspartic acid (charge –1)
When a voltage is applied to a mixture of alanine, lysine, and aspartic acid at pH 6, alanine does not move. Lysine moves toward the negatively charged cathode, and aspartic
acid moves toward the positively charged anode (Figure 24-4). After a period of time,
the separated amino acids are recovered by cutting the paper or scraping the bands out
of the gel. If electrophoresis is being used as an analytical technique (to determine the
amino acids present in the mixture), the paper or gel is treated with a reagent such as ninhydrin (Section 24-7C) to make the bands visible. Then the amino acids are identified
by comparing their positions with those of standards.
PROBLEM 24-7
Draw the electrophoretic separation of Ala, Lys, and Asp at pH 9.7.
PROBLEM 24-8
Draw the electrophoretic separation of Trp, Cys, and His at pH 6.0.
24-5
Naturally occurring amino acids can be obtained by hydrolyzing proteins and separating the amino acid mixture. Even so, it is often less expensive to synthesize the pure
amino acid. In some cases, an unusual amino acid or an unnatural enantiomer is needed,
and it must be synthesized. In this chapter, we consider four methods for making amino
acids. All these methods are extensions of reactions we have already studied.
Synthesis of
Amino Acids
24-5A
Reductive Amination
Reductive amination of ketones and aldehydes is one of the best methods for synthesizing amines (Section 19-18). It also forms amino acids. When an a-ketoacid is treated with
ammonia, the ketone reacts to form an imine. The imine is reduced to an amine by hydrogen and a palladium catalyst. Under these conditions, the carboxylic acid is not reduced.
O
R
C
N
excess NH3
COOH
α-ketoacid
R
H
NH2
COO– +NH4
C
imine
H2
R
Pd
CH
COO–
α-amino acid
This entire synthesis is accomplished in one step by treating the a-ketoacid with
ammonia and hydrogen in the presence of a palladium catalyst. The product is a
racemic a-amino acid. The following reaction shows the synthesis of racemic phenylalanine from 3-phenyl-2-oxopropanoic acid.
O
Ph
CH2
C
NH2
COOH
3-phenyl-2-oxopropanoic acid
NH3, H2
Pd
Ph
CH2
CH
COO– +NH4
(D,L)-phenylalanine (ammonium salt)
(30%)
We call reductive amination a biomimetic (“mimicking the biological process”)
synthesis because it resembles the biological synthesis of amino acids. The biosynthesis begins with reductive amination of a-ketoglutaric acid (an intermediate in the
1165
24-5 Synthesis of Amino Acids
metabolism of carbohydrates), using ammonium ion as the aminating agent and NADH
as the reducing agent. The product of this enzyme-catalyzed reaction is the pure L enantiomer of glutamic acid.
H H
O
HOOC
NH2
C
–
C
CH2CH2
O
COO
+
+
NH4 +
α-ketoglutaric acid
+
enzyme
HOOC
+ H+
N
H
NH3
CH2CH2
CH
L-glutamic
acid
C
COO– +
N
sugar
NADH
NAD+
Biosynthesis of other amino acids uses L-glutamic acid as the source of the
amino group. Such a reaction, moving an amino group from one molecule to another,
is called a transamination, and the enzymes that catalyze these reactions are called
transaminases. For example, the following reaction shows the biosynthesis of aspartic acid using glutamic acid as the nitrogen source. Once again, the enzyme-catalyzed
biosynthesis gives the pure L enantiomer of the product.
O
NH3
HOOC
CH2CH2
L-glutamic
CH
COO–
HOOC
CH2CH2
+
transaminase
+
O
HOOC
CH2
C
COO–
C
α-ketoglutaric acid
acid
+
NH3
COO–
HOOC
oxaloacetic acid
CH2
CH
L-aspartic
acid
COO–
PROBLEM 24-9
Show how the following amino acids might be formed in the laboratory by reductive amination of the appropriate a-ketoacid.
(a) alanine
(b) leucine
(c) serine
(d) glutamine
24-5B
Amination of an a-Halo Acid
The Hell–Volhard–Zelinsky reaction (Section 22-6) is an effective method for introducing bromine at the a position of a carboxylic acid. The racemic a-bromo acid is converted
to a racemic a-amino acid by direct amination, using a large excess of ammonia.
O
R
CH2
C
carboxylic acid
Br
OH
(1) Br2/PBr3
(2) H2O
R
NH2
+
sugar
+
CH
O
C
α-bromo acid
NH2 O
OH
NH3
(large excess)
R
In Section 19-11, we saw that direct alkylation is often a poor synthesis of amines, giving large amounts of overalkylated products. In this case, however, the reaction gives
acceptable yields because a large excess of ammonia is used, making ammonia the
nucleophile that is most likely to displace bromine. Also, the adjacent carboxylate ion
in the product reduces the nucleophilicity of the amino group. The following sequence
shows bromination of 3-phenylpropanoic acid, followed by displacement of bromide ion,
to form the ammonium salt of racemic phenylalanine.
CH
C
O
O– +NH4
(D,L)-α-amino acid
(ammonium salt)
+ H2O
1166
CHAPTER 24
Amino Acids, Peptides, and Proteins
NH2
Br
Ph
CH2
CH2
COOH
3-phenylpropanoic acid
(1) Br2/PBr3
Ph
(2) H2O
CH2
CH
excess NH3
COOH
Ph
CH2 CH COO– +NH4
(D,L)-phenylalanine (salt)
(30–50%)
PROBLEM 24-10
Show how you would use bromination followed by amination to synthesize the following
amino acids.
(a) glycine
(b) leucine
(c) glutamic acid
24-5C
The Gabriel–Malonic Ester Synthesis
One of the best methods of amino acid synthesis is a combination of the Gabriel synthesis of amines (Section 19-20) with the malonic ester synthesis of carboxylic acids
(Section 22-16). The conventional malonic ester synthesis involves alkylation of diethyl
malonate, followed by hydrolysis and decarboxylation to give an alkylated acetic acid.
temporary ester group
COOEt
O
H
C
C
CO2
COOEt
O
(1)–OEt
OEt
(2) RX
H
C
C
H3O+, heat
OEt
H
R
H
H
O
C
C
OH
R
malonic ester
alkylated acetic acid
To adapt this synthesis to making amino acids, we begin with a malonic ester that contains an a-amino group. The amino group is protected as a non-nucleophilic amide to
prevent it from attacking the alkylating agent (RX).
The Gabriel–malonic ester synthesis begins with N-phthalimidomalonic ester. Think
of N-phthalimidomalonic ester as a molecule of glycine (aminoacetic acid) with the
amino group protected as an amide (a phthalimide in this case) to keep it from acting
as a nucleophile. The acid is protected as an ethyl ester, and the a position is further activated by the additional (temporary) ester group of diethyl malonate.
temporary ester group
O
N
O
CH
COOEt
N-phthalimidomalonic ester
COOEt
O
O
COOEt
=
N
O
C
C
H
O
Et
protected acid
glycine
protected amine
Just as the malonic ester synthesis gives substituted acetic acids, the N-phthalimidomalonic
ester synthesis gives substituted aminoacetic acids: a-amino acids.
N-Phthalimidomalonic ester is alkylated in the same way as malonic ester. When the
alkylated N-phthalimidomalonic ester is hydrolyzed, the phthalimido group is hydrolyzed
along with the ester groups. The product is an alkylated aminomalonic acid. Decarboxylation
gives a racemic a-amino acid.
1167
24-5 Synthesis of Amino Acids
The Gabriel – malonic ester synthesis
temporary ester group
O
COOEt
N
(1) base
(2) R X
CH
COOEt
N
COOEt
O
CO2
O
C
R
+
H3N
COOEt
O
N-phthalimidomalonic ester
COOH
H3O+
C
H
+
heat
R
H3N
C
COOH
alkylated
R
COOH
α -amino acid
hydrolyzed
The Gabriel–malonic ester synthesis is used to make many amino acids that cannot be formed by direct amination of haloacids. The following example shows the synthesis of methionine, which is formed in very poor yield by direct amination.
O
O
COOEt
N
(1) NaOEt
(2) Cl CH2CH2SCH3
CH
N
COOEt
O
COOEt
C
H
H3O+
CH2CH2SCH3
+
H3N
heat
COOEt
O
CH2CH2SCH3
C
COOH
(D,L)-methionine
(50%)
PROBLEM 24-11
Show how the Gabriel–malonic ester synthesis could be used to make
(a) valine
(b) phenylalanine
(c) glutamic acid
(d) leucine
PROBLEM 24-12
COOEt
O
O
The Gabriel–malonic ester synthesis uses an aminomalonic ester with the amino
group protected as a phthalimide. A variation has the amino group protected as an acetamido group. Propose how you might use an acetamidomalonic ester synthesis to make
phenylalanine.
CH3
C
N
C
H
H
C
O
acetamidomalonic ester
24-5D
The Strecker Synthesis
The first known synthesis of an amino acid occurred in 1850 in the laboratory of Adolph
Strecker in Tübingen, Germany. Strecker added acetaldehyde to an aqueous solution of
ammonia and HCN. The product was a-amino propionitrile, which Strecker hydrolyzed
to racemic alanine.
The Strecker synthesis of alanine
O
CH3
C
+
NH2
H
acetaldehyde
+
NH3 +
HCN
H2O
CH3
C
H
C
N
α-amino propionitrile
H3
O+
NH3
CH3
C
H
COOH
(D,L)-alanine
(60%)
The Strecker synthesis can form a large number of amino acids from appropriate
aldehydes. The mechanism is shown next. First, the aldehyde reacts with ammonia to give
an imine. The imine is a nitrogen analogue of a carbonyl group, and it is electrophilic
when protonated. Attack of cyanide ion on the protonated imine gives the a-amino nitrile.
This mechanism is similar to that for formation of a cyanohydrin (Section 18-14), except
that in the Strecker synthesis cyanide ion attacks an imine rather than the aldehyde itself.
Et
1168
CHAPTER 24
Amino Acids, Peptides, and Proteins
Step 1: The aldehyde reacts with ammonia to form the imine (mechanism in Section 18-15)
O
R
H +
C
H+
NH3
R
aldehyde
N
H
C
H +
H2O
imine
Step 2: Cyanide ion attacks the imine.
H
H
R
C
+
H
C
H
N
N
H
H
CN
R
imine
NH2
R
H
C
–
CN
CN
α -amino nitrile
In a separate step, hydrolysis of the a-amino nitrile (Section 21-7D) gives an
a-amino acid.
R
H2 N
CH
C
N
R
H3O+
+
H3N
α-amino nitrile
CH
COOH
α-amino acid (acidic form)
SOLVED PROBLEM 24-1
Show how you would use a Strecker synthesis to make isoleucine.
SOLUTION
Isoleucine has a sec-butyl group for its side chain. Remember that CH3 ¬ CHO undergoes Strecker synthesis to give alanine, with CH3
as the side chain. Therefore, sec-butyl ¬ CHO should give isoleucine.
CH3
CH3CH2CH
O
C
H
sec-butyl CHO
(2-methylbutanal)
CH3 NH2
NH3, HCN
CH3CH2CH
H2O
C
H
C
N
CH3 +NH3
H3O+
CH3CH2CH
C
COOH
(D,L)-isoleucine
PROBLEM 24-13
Problem-solving Hint
In the malonic ester synthesis, use
the side chain of the desired amino
acid (must be a good SN2 substrate)
to alkylate the ester. In the Strecker
synthesis, the aldehyde carbon
becomes the a carbon of the amino
acid: begin with [side chain] ¬ CHO.
SUMMARY
(a) Show how you would use a Strecker synthesis to make phenylalanine.
(b) Propose a mechanism for each step in the synthesis in part (a).
PROBLEM 24-14
Show how you would use a Strecker synthesis to make
(a) leucine
(b) valine
(c) aspartic acid
Syntheses of Amino Acids
1. Reductive amination (Section 24-5A)
O
R
C
N
COOH
α-ketoacid
excess NH3
R
C
H
COO– +NH4
imine
NH2
H2
Pd
H
R
CH
COO–
α-amino acid
1169
24-6 Resolution of Amino Acids
2. Amination of an α-haloacid (Section 24-5B)
O
R
C
CH2
OH
(2) H2O
R
NH2 O
O
Br
(1) Br2/PBr3
CH
C
NH3
OH
R
(large excess)
CH
α-bromo acid
carboxylic acid
O– +NH
C
(D,L)-α-amino salt
(ammonium salt)
3. The Gabriel – malonic ester synthesis (Section 24-5C)
temporary ester group
O
CO2
O
COOEt
N
(1) base
(2) R X
CH
N
COOEt
O
COOEt
C
R
+
H3N
heat
C
H
+
heat
R
H3N
COOH
COOEt
O
N-phthalimidomalonic ester
COOH
H3O+
α-amino acid
4. The Strecker synthesis (Section 24-5D)
NH2
O
R
C
H
aldehyde
+
NH3
+
HCN
H2O
R
C
H
C
N
H3
α-amino nitrile
All the laboratory syntheses of amino acids described in Section 24-5 produce racemic
products. In most cases, only the L enantiomers are biologically active. The D enantiomers may even be toxic. Pure L enantiomers are needed for peptide synthesis if the
product is to have the activity of the natural material. Therefore, we must be able to
resolve a racemic amino acid into its enantiomers.
In many cases, amino acids can be resolved by the methods we have already discussed (Section 5-16). If a racemic amino acid is converted to a salt with an optically
pure chiral acid or base, two diastereomeric salts are formed. These salts can be separated by physical means such as selective crystallization or chromatography. Pure
enantiomers are then regenerated from the separated diastereomeric salts. Strychnine and
brucine are naturally occurring optically active bases, and tartaric acid is used as an
optically active acid for resolving racemic mixtures.
Enzymatic resolution is also used to separate the enantiomers of amino acids.
Enzymes are chiral molecules with specific catalytic activities. For example, when
an acylated amino acid is treated with an enzyme like hog kidney acylase or carboxypeptidase, the enzyme cleaves the acyl group from just the molecules having the
natural (L) configuration. The enzyme does not recognize D-amino acids, so they are
unaffected. The resulting mixture of acylated D-amino acid and deacylated L-amino
acid is easily separated. Figure 24-5 shows how this selective enzymatic deacylation
is accomplished.
PROBLEM 24-15
Suggest how you would separate the free L-amino acid from its acylated
Figure 24-5.
D
O+
enantiomer in
R
COOH
hydrolyzed
alkylated
C
+NH
3
R
C
H
COOH
α-amino acid
24-6
Resolution of
Amino Acids
1170
Amino Acids, Peptides, and Proteins
CHAPTER 24
COOH
H2N
C
O
H
CH3
C
CH3C
C
COOH
H
H2N
C
R
L
acylase
NH2
H
COOH
O
C
C
NH
CH3
H
R
is deacylated
COOH
O
C
C
NH
CH3
R
acid
D
racemic amino acid
H
R
2O
R
D-amino
NH
) )
acid
COOH
H
C
O
R
L-amino
COOH
acylated
is unaffected
(easily separated mixture)
FIGURE 24-5
Selective enzymatic deacylation. An acylase enzyme (such as hog kidney acylase or
carboxypeptidase) deacylates only the natural L-amino acid.
24-7
Reactions
of Amino Acids
Amino acids undergo many of the standard reactions of both amines and carboxylic
acids. Conditions for some of these reactions must be carefully selected, however, so
that the amino group does not interfere with a carboxyl group reaction, and vice versa.
We will consider two of the most useful reactions, esterification of the carboxyl group
and acylation of the amino group. These reactions are often used to protect either the
carboxyl group or the amino group while the other group is being modified or coupled
to another amino acid. Amino acids also undergo reactions that are specific to the
a-amino acid structure. One of these unique amino acid reactions is the formation of a
colored product on treatment with ninhydrin, discussed in Section 24-7C.
Esterification of the Carboxyl Group
24-7A
Like monofunctional carboxylic acids, amino acids are esterified by treatment with a
large excess of an alcohol and an acidic catalyst (often gaseous HCl). Under these
acidic conditions, the amino group is present in its protonated 1 ¬ NH3+2 form, so it
does not interfere with esterification. The following example illustrates esterification
of an amino acid.
Cl–
O
+
H2N
H2C
CH
CH2
C
O–
CH2
+
Ph
CH2
OH
HCl
H2N
H2C
CH
CH2
proline
O
C
O
CH2Ph
CH2
proline benzyl ester
(90%)
Esters of amino acids are often used as protected derivatives to prevent the carboxyl
group from reacting in some undesired manner. Methyl, ethyl, and benzyl esters are
the most common protecting groups. Aqueous acid hydrolyzes the ester and regenerates the free amino acid.
O
+
H3N
CH
C
CH2
Ph
OCH2CH3
phenylalanine ethyl ester
H3O+
O
+
H3N
CH
C
CH2
Ph
phenylalanine
OH +
CH3CH2
OH
24-7 Reactions of Amino Acids
1171
Benzyl esters are particularly useful as protecting groups because they can be removed
either by acidic hydrolysis or by neutral hydrogenolysis (“breaking apart by addition
of hydrogen”). Catalytic hydrogenation cleaves the benzyl ester, converting the benzyl
group to toluene and leaving the deprotected amino acid. Although the mechanism of
this hydrogenolysis is not well known, it apparently hinges on the ease of formation of
benzylic intermediates.
O
+
H3N
CH
O
H2, Pd
OCH2
C
+
H3N
Ph
CH2
CH
C
CH2
phenylalanine benzyl ester
+
OH
CH3
Ph
phenylalanine
toluene
PROBLEM 24-16
Application: Allergy
Decarboxylation is an important reaction
of amino acids in many biological
processes. Histamine, which causes
runny noses and itchy eyes, is synthesized in the body by decarboxylation of
histidine. The enzyme that catalyzes this
reaction is called histidine decarboxylase.
Propose a mechanism for the acid-catalyzed hydrolysis of phenylalanine ethyl ester.
PROBLEM 24-17
Give equations for the formation and hydrogenolysis of glutamine benzyl ester.
CH2CH2NH2
Acylation of the Amino Group: Formation of Amides
24-7B
Just as an alcohol esterifies the carboxyl group of an amino acid, an acylating agent
converts the amino group to an amide. Acylation of the amino group is often done to
protect it from unwanted nucleophilic reactions. A wide variety of acid chlorides and
anhydrides are used for acylation. Benzyl chloroformate acylates the amino group to give
a benzyloxycarbonyl derivative, often used as a protecting group in peptide synthesis
(Section 24-10).
O
H2N
CH
CH3
COOH
CH2
NH
(
O
CH3
C
)
2
C
O
N
N-acetylhistidine
O
COOH
CH2CH(CH3)2
leucine
COOH
NH
histidine
CH
CH
CH2
(acetic anhydride)
N
H2N
NH
PhCH2OC
O
Cl
(benzyl chloroformate)
PhCH2O
C
NH
CH
COOH
CH2CH(CH3)2
N-benzyloxycarbonyl leucine
(90%)
The amino group of the N-benzyloxycarbonyl derivative is protected as the amide half
of a carbamate ester (a urethane, Section 21-16), which is more easily hydrolyzed than
most other amides. In addition, the ester half of this urethane is a benzyl ester that
undergoes hydrogenolysis. Catalytic hydrogenolysis of the N-benzyloxycarbonyl amino
acid gives an unstable carbamic acid that quickly decarboxylates to give the deprotected
amino acid.
NH
N
histamine
1172
Amino Acids, Peptides, and Proteins
CHAPTER 24
O
CH2
O
C
H
N
CH
COOH
H2, Pd
CH3
HO
O
H
C
N
CH
CH2
CH2
CH(CH3)2
CH(CH3)2
toluene
N-benzyloxycarbonyl leucine
CO2
COOH
H2N
CH
COOH
CH2
CH(CH3)2
a carbamic acid
leucine
PROBLEM 24-18
Give equations for the formation and hydrogenolysis of N-benzyloxycarbonyl methionine.
24-7C
Reaction with Ninhydrin
Ninhydrin is a common reagent for visualizing spots or bands of amino acids that have
been separated by chromatography or electrophoresis. When ninhydrin reacts with an
amino acid, one of the products is a deep violet, resonance-stabilized anion called
Ruhemann’s purple. Ninhydrin produces this same purple dye regardless of the structure of the original amino acid. The side chain of the amino acid is lost as an aldehyde.
Reaction of an amino acid with ninhydrin
O
H2N
OH
COOH + 2
CH
O–
O
pyridine
+ CO2
N
OH
R
+ R
O
amino acid
O
ninhydrin
CHO
O
Ruhemann’s purple
The reaction of amino acids with ninhydrin can detect amino acids on a wide variety of substrates. For example, if a kidnapper touches a ransom note with his fingers,
the dermal ridges on his fingers leave traces of amino acids from skin secretions.
Treatment of the paper with ninhydrin and pyridine causes these secretions to turn
purple, forming a visible fingerprint.
PROBLEM 24-19
Use resonance forms to show delocalization of the negative charge in the Ruhemann’s purple anion.
SUMMARY
Reactions of Amino Acids
1. Esterification of the carboxyl group (Section 24-7A)
R
+
H3N
CH
O
C
O–
+
amino acid
R´
H+
OH
+
H3N
alcohol
R
O
CH
C
O
R´
R
O
CH
C
+
H2O
amino ester
2. Acylation of the amino group: formation of amides (Section 24-7B)
H2N
R
O
CH
C
amino acid
O
OH
+
R´
C
O
X
acylating agent
R´
C
NH
acylated amino acid
OH
+
H
X
24-8 Structure and Nomenclature of Peptides and Proteins
1173
3. Reaction with ninhydrin (Section 24-7C)
H2N
CH
OH
+
COOH
O–
O
O
pyridine
2
N
OH
R
O
O
ninhydrin
amino acid
+
R
+
CO2
O
CHO
Ruhemann’s purple
4. Formation of peptide bonds (Sections 24-10 and 24-11)
peptide bond
+
H3N
O
CH
C
O–
+
+
H3N
R1
O
CH
C
O–
O
+
loss of H2O
H3N
CH
R2
C
O
NH
R1
CH
C
O–
R2
Amino acids also undergo many other common reactions of amines and acids.
24-8A
24-8
Peptide Structure
The most important reaction of amino acids is the formation of peptide bonds. Amines
and acids can condense, with the loss of water, to form amides. Industrial processes
often make amides simply by mixing the acid and the amine, then heating the mixture
to drive off water.
O
O
R
OH +
C
acid
H2N
R´
R
amine
C
Structure and
Nomenclature of
Peptides and
Proteins
O
+
O– H3N
R´
heat
R
C
salt
NH
R´ +
H2O
amide
Recall from Section 21-13 that amides are the most stable acid derivatives. This stability is partly due to the strong resonance interaction between the nonbonding electrons
on nitrogen and the carbonyl group. The amide nitrogen is no longer a strong base, and
the C ¬ N bond has restricted rotation because of its partial double-bond character.
Figure 24-6 shows the resonance forms we use to explain the partial double-bond character and restricted rotation of an amide bond. In a peptide, this partial double-bond
character results in six atoms being held rather rigidly in a plane.
Having both an amino group and a carboxyl group, an amino acid is ideally suited
to form an amide linkage. Under the proper conditions, the amino group of one molecule condenses with the carboxyl group of another. The product is an amide called a
peptide bond
O
O
C
R
R
N
H
C
R
FIGURE 24-6
–
+
R
N
H
amide plane
Resonance stabilization of an amide
accounts for its enhanced stability,
the weak basicity of the nitrogen
atom, and the restricted rotation of
the C ¬ N bond. In a peptide, the
amide bond is called a peptide
bond. It holds six atoms in a plane:
the C and O of the carbonyl, the
N and its H, and the two associated
a carbon atoms.
1174
Amino Acids, Peptides, and Proteins
CHAPTER 24
dipeptide because it consists of two amino acids. The amide linkage between the amino
acids is called a peptide bond. Although it has a special name, a peptide bond is just
like other amide bonds we have studied.
peptide bond
R2
O
+
+
C
H3N
–
C
O
+
+
C
C
H3N
R1
H
O
H
O–
N
C
R1 H
O
C
H
O
H
C
C
H3N
loss of H2O
R2
O–
In this manner, any number of amino acids can be bonded in a continuous chain.
A peptide is a compound containing two or more amino acids linked by amide bonds
between the amino group of each amino acid and the carboxyl group of the neighboring amino acid. Each amino acid unit in the peptide is called a residue. A polypeptide
is a peptide containing many amino acid residues but usually having a molecular weight
of less than about 5000. Proteins contain more amino acid units, with molecular
weights ranging from about 5000 to about 40,000,000. The term oligopeptide is occasionally used for peptides containing about four to ten amino acid residues. Figure 24-7
shows the structure of the nonapeptide bradykinin, a human hormone that helps to
control blood pressure.
The end of the peptide with the free amino group 1 ¬ NH3+2 is called the N-terminal
end or the N terminus, and the end with the free carboxyl group 1 ¬ COO-2 is called the
C-terminal end or the C terminus. Peptide structures are generally drawn with the N terminus at the left and the C terminus at the right, as bradykinin is drawn in Figure 24-7.
C terminus
N terminus
O
O
+
H 3N
CH
C
N
CH
C
O
N
CH
C
O
NH
CH
H
C
O
NH
CH
C
O
NH
CH2
C
N
CH
C
NH
CH2
NH
+
H2N
CH
O
O
CH
NH
+
H2N
NH2
Pro
Gly
Phe
C
O–
NH
OH
Pro
CH
CH2
C
Arg
C
O
Ser
Pro
Phe
C
NH2
Arg
FIGURE 24-7
The human hormone bradykinin is a nonapeptide with a free ¬ NH3+ at its N terminus and a
free ¬ COO- at its C terminus.
24-8B
Peptide Nomenclature
The names of peptides reflect the names of the amino acid residues involved in the
amide linkages, beginning at the N terminus. All except the last are given the -yl suffix
of acyl groups. For example, the following dipeptide is named alanylserine. The alanine
residue has the -yl suffix because it has acylated the nitrogen of serine.
O
+
H3N
CH
C
O
NH
CH
C
CH2OH
CH3
alanyl
serine
Ala-Ser
O–
24-8 Structure and Nomenclature of Peptides and Proteins
1175
Bradykinin (Figure 24-7) is named as follows (without any spaces):
arginyl prolyl prolyl glycyl phenylalanyl seryl prolyl phenylalanyl arginine
This is a cumbersome and awkward name. A shorthand system is more convenient,
representing each amino acid by its three-letter abbreviation. These abbreviations, given
in Table 24-2, are generally the first three letters of the name. Once again, the amino acids
are arranged from the N terminus at the left to the C terminus at the right. Bradykinin
has the following abbreviated name:
Arg-Pro-Pro-Gly-Phe-Ser-Pro-Phe-Arg
Single-letter symbols (also given in Table 24-2) are becoming widely used as well.
Using single letters, we symbolize bradykinin by
RPPGFSPFR
PROBLEM 24-20
Draw the complete structures of the following peptides:
(a) Thr-Phe-Met
(b) serylarginylglycylphenylalanine
(c) IMQDK
(d) ELVIS
Disulfide Linkages
24-8C
Amide linkages (peptide bonds) form the backbone of the amino acid chains we call peptides and proteins. A second kind of covalent bond is possible between any cysteine
residues present. Cysteine residues can form disulfide bridges (also called disulfide
linkages) that can join two chains or link a single chain into a ring.
Mild oxidation joins two molecules of a thiol into a disulfide, forming a disulfide
linkage between the two thiol molecules. This reaction is reversible, and a mild reduction cleaves the disulfide.
R ¬ SH + HS ¬ R
two molecules of thiol
[oxidation]
IRRRJ
[reduction]
R ¬ S ¬ S ¬ R + H2O
disulfide
Similarly, two cysteine sulfhydryl 1 ¬ SH2 groups are oxidized to give a disulfidelinked pair of amino acids. This disulfide-linked dimer of cysteine is called cystine.
Figure 24-8 shows formation of a cystine disulfide bridge linking two peptide chains.
Two cysteine residues may form a disulfide bridge within a single peptide chain,
making a ring. Figure 24-9 shows the structure of human oxytocin, a peptide hormone
that causes contraction of uterine smooth muscle and induces labor. Oxytocin is a
nonapeptide with two cysteine residues (at positions 1 and 6) linking part of the molecule
peptide
chain
O
NH
CH
C
CH2
SH
CH
C
O
two cysteine residues
CH
C
CH2
S
+ H2O
S
[H]
(reduce)
CH2
NH
NH
[O]
(oxidize)
SH
peptide
chain
O
CH2
NH
CH
C
FIGURE 24-8
O
Cystine, a dimer of cysteine, results
when two cysteine residues are
oxidized to form a disulfide bridge.
cystine disulfide bridge
1176
CHAPTER 24
Amino Acids, Peptides, and Proteins
O
O
CH3
CH
CH2CH2C
C
CH3CH2
CH
NH
O
CH
C
NH
O
C
O
H2N
O
NH
CH2 CH
HO
NH2
CH2
CH
NH
C
C
NH
CH
CH2
S
CH2
S
CH
NH2
C
O
O
C
N
CH
O
N terminus
NH
C
O
CH
C
CH2
O
H3C
H
NH2
C
C terminus
(amide form)
CH3
Gln
Tyr
Asn
S S
Cys
Pro
Gly NH2
Leu
.
Cys
CH
CH
cystine disulfide bridge
Ile
NH
N terminus
C terminus (amide form)
FIGURE 24-9
Application: Hormone
Orexin A (from the Greek orexis,
“appetite”) is a 33 amino acid neuropeptide connected by two disulfide bridges.
Orexin A is a powerful stimulant for food
intake and gastric juice secretion.
Scientists are studying orexin A to learn
more about the regulation of appetite
and eating, hoping to learn more about
causes and potential treatments for
anorexia nervosa.
Structure of human oxytocin. A disulfide linkage holds part of the molecule in a large ring.
in a large ring. In drawing the structure of a complicated peptide, arrows are often used
to connect the amino acids, showing the direction from N terminus to C terminus.
Notice that the C terminus of oxytocin is a primary amide 1Gly # NH22 rather than a
free carboxyl group.
Figure 24-10 shows the structure of insulin, a more complex peptide hormone that
regulates glucose metabolism. Insulin is composed of two separate peptide chains, the
A chain, containing 21 amino acid residues, and the B chain, containing 30. The A and
B chains are joined at two positions by disulfide bridges, and the A chain has an additional disulfide bond that holds six amino acid residues in a ring. The C-terminal amino
acids of both chains occur as primary amides.
A chain
N terminus
Ile
Val
Glu
Gln
Cys
S
S
Cys
Cys
Leu
Val
Gln
His
Leu
Leu
Glu
Asn
Tyr
Gly
Ser
Asn NH2
S
His
Glu
Ala
Leu
Tyr
Leu
Val
Cys
Gly
Leu
Glu
Arg
NH2 Ala
.
B chain
Cys
S
Val
Cys
Gln
Ser
S
Asn
Tyr
Val
Ala
S
Ser
.
Gly
C terminus
Lys
Pro
Thr
Tyr
Phe
Phe
Phe
C terminus
N terminus
FIGURE 24-10
Structure of insulin. Two chains are joined at two positions by disulfide bridges, and a third
disulfide bond holds the A chain in a ring.
Gly
24-9 Peptide Structure Determination
1177
Disulfide bridges are commonly manipulated in the process of giving hair a
permanent wave. Hair is composed of protein, which is made rigid and tough partly by
disulfide bonds. When hair is treated with a solution of a thiol such as 2-mercaptoethanol
1HS ¬ CH2 ¬ CH2 ¬ OH2, the disulfide bridges are reduced and cleaved. The hair is
wrapped around curlers, and the disulfide bonds are allowed to re-form, either by air oxidation or by application of a neutralizer. The disulfide bonds re-form in new positions,
holding the hair in the bent conformation enforced by the curlers.
Insulin is a relatively simple protein, yet it is a complicated organic structure. How is
it possible to determine the complete structure of a protein with hundreds of amino acid
residues and a molecular weight of many thousands? Chemists have developed clever
ways to determine the exact sequence of amino acids in a protein. We will consider
some of the most common methods.
24-9A
24-9
Peptide Structure
Determination
Cleavage of Disulfide Linkages
The first step in structure determination is to break all the disulfide bonds, opening any
disulfide-linked rings and separating the individual peptide chains. The individual
peptide chains are then purified and analyzed separately.
Cystine bridges are easily cleaved by reducing them to the thiol (cysteine) form.
These reduced cysteine residues have a tendency to reoxidize and re-form disulfide
bridges, however. A more permanent cleavage involves oxidizing the disulfide linkages with peroxyformic acid (Figure 24-11). This oxidation converts the disulfide
bridges to sulfonic acid 1 ¬ SO3H2 groups. The oxidized cysteine units are called
cysteic acid residues.
24-9B
Determination of the Amino Acid Composition
Once the disulfide bridges have been broken and the individual peptide chains have
been separated and purified, the structure of each chain must be determined. The first
step is to determine which amino acids are present and in what proportions. To analyze
O
NH
CH
O
C
NH
CH2
NH
H
C
CH2 cysteic acid
O
S
CH
C
SO3H
OOH
S
SO3H
CH2
CH2
cysteic acid
CH
C
CH
C
NH
O
O
SO3H
S S
SO3H
O
H
C
OOH
FIGURE 24-11
S
S
S
S
SO3H
SO3H
SO3H
HO3S
Oxidation of a protein by
peroxyformic acid cleaves all the
disulfide linkages by oxidizing
cystine to cysteic acid.
1178
CHAPTER 24
Amino Acids, Peptides, and Proteins
FIGURE 24-12
buffer
solution
hydrolysate
ion-exchange resin
ninhydrin
solution
different amino acids
move at different
speeds
light
photocell
intensity of absorption
In an amino acid analyzer, the
hydrolysate passes through an
ion-exchange column. The solution
emerging from the column is treated
with ninhydrin, and its absorbance is
recorded as a function of time. Each
amino acid is identified by the
retention time required to pass
through the column.
time
recorder
waste
the amino acid composition, the peptide chain is completely hydrolyzed by boiling it
for 24 hours in 6 M HCl. The resulting mixture of amino acids (the hydrolysate) is
placed on the column of an amino acid analyzer, diagrammed in Figure 24-12.
In the amino acid analyzer, the components of the hydrolysate are dissolved in an
aqueous buffer solution and separated by passing them down an ion-exchange column.
The solution emerging from the column is mixed with ninhydrin, which reacts with
amino acids to give the purple ninhydrin color. The absorption of light is recorded and
printed out as a function of time.
The time required for each amino acid to pass through the column (its retention
time) depends on how strongly that amino acid interacts with the ion-exchange resin.
The retention time of each amino acid is known from standardization with pure amino
acids. The amino acids present in the sample are identified by comparing their retention times with the known values. The area under each peak is nearly proportional to
the amount of the amino acid producing that peak, so we can determine the relative
amounts of amino acids present.
Figure 24-13 shows a standard trace of an equimolar mixture of amino acids, followed by the trace produced by the hydrolysate from human bradykinin (Arg-Pro-ProGly-Phe-Ser-Pro-Phe-Arg).
rg
A
is
H
Ty
r
Ph
e
Ly
s
o
G
ly
A
la
Cy
s
V
al
M
e
Ilet
Le
u
Pr
A
sp
Th
Se r
r
G
lu
standard
FIGURE 24-13
rg
A
time
Ph
e
ly
G
o
Pr
Se
r
bradykinin
absorption
Use of an amino acid analyzer to
determine the composition of human
bradykinin. The bradykinin peaks for
Pro, Arg, and Phe are larger than
those in the standard equimolar
mixture because bradykinin has three
Pro residues, two Arg residues, and
two Phe residues.
24-9 Peptide Structure Determination
1179
Sequencing the Peptide: Terminal Residue Analysis The amino acid analyzer determines the amino acids present in a peptide, but it does not reveal their sequence: the
order in which they are linked together. The peptide sequence is destroyed in the
hydrolysis step. To determine the amino acid sequence, we must cleave just one amino
acid from the chain and leave the rest of the chain intact. The cleaved amino acid can
be separated and identified, and the process can be repeated on the rest of the chain.
The amino acid may be cleaved from either end of the peptide (either the N terminus
or the C terminus), and we will consider one method used for each end. This general
method for peptide sequencing is called terminal residue analysis.
Sequencing from the N Terminus: The Edman
Degradation
24-9C
The most efficient method for sequencing peptides is the Edman degradation. A peptide
is treated with phenyl isothiocyanate, followed by acid hydrolysis. The products are
the shortened peptide chain and a heterocyclic derivative of the N-terminal amino acid
called a phenylthiohydantoin.
This reaction takes place in three stages. First, the free amino group of the N-terminal
amino acid reacts with phenylisothiocyanate to form a phenylthiourea. Second, the
phenylthiourea cyclizes to a thiazolinone and expels the shortened peptide chain. Third,
the thiazolinone isomerizes to the more stable phenylthiohydantoin.
Step 1: Nucleophilic attack by the free amino group on phenyl isothiocyanate,
followed by a proton transfer, gives a phenylthiourea.
Ph
N
C
S
H2N
Ph
CH
R
1
C
NH
N
peptide
S
C
H2N
O
+
–
Ph
CH
R
C
1
NH
peptide
NH
C
HN
O
S
CH
C
1
O
R
NH
peptide
a phenylthiourea
Step 2: Treatment with HCl induces cyclization to a thiazolinone and expulsion of the
shortened peptide chain.
+
S
C
R
1
C
HN
HN
CH
NHPh
NHPh
NHPh
C
NH
peptide
H
C
+
O
R
1
C
C
S
C
NHPh
S
N
NH
peptide
H
C
R
OH
C
1
+
NH2
peptide
O
H
S
N
H
C
R
C
1
O
H
H2O
protonated phenylthiourea
Step 3: In acid, the thiazolinone isomerizes to the more stable phenylthiohydantoin.
S
NHPh
N
S
HCl
C
HN
H
R1
O
thiazolinone
C
R1
N
Ph
C
O
a phenylthiohydantoin
The phenylthiohydantoin derivative is identified by chromatography, by comparing it with phenylthiohydantoin derivatives of the standard amino acids. This gives the
identity of the original N-terminal amino acid. The rest of the peptide is cleaved intact,
and further Edman degradations are used to identify additional amino acids in the
a thiazolinone
+ H2N
peptide
+ H3O+
1180
CHAPTER 24
Amino Acids, Peptides, and Proteins
chain. This process is well suited to automation, and several types of automatic
sequencers have been developed.
Figure 24-14 shows the first two steps in the sequencing of oxytocin. Before
sequencing, the oxytocin sample is treated with peroxyformic acid to convert the disulfide bridge to cysteic acid residues.
Step 1: Cleavage and determination of the N-terminal amino acid
S
O
CH
C
NH
Tyr
Ile
Gln
peptide
(1) Ph
N
C
C
S
..
..
H2N
HN
(2) H3O+
N
CH
CH2
Ph + H2N
Tyr
Ile
Gln
peptide
C
O
CH2SO3H
cysteic acid
phenylthiohydantoin
SO3H
cysteic acid
Step 2: Cleavage and determination of the second amino acid (the new N-terminal amino acid)
S
O
HO
CH
C
NH
Ile
Gln
peptide
(1) Ph
N
C
(2) H3O+
S
C
..
..
H2N
HN
CH
CH2
HO
CH2
Ph + H2N
N
Ile
Gln
peptide
C
O
tyrosine phenylthiohydantoin
FIGURE 24-14
The first two steps in sequencing oxytocin. Each Edman degradation cleaves the N-terminal
amino acid and forms its phenylthiohydantoin derivative. The shortened peptide is available
for the next step.
In theory, Edman degradations could sequence a peptide of any length. In practice, however, the repeated cycles of degradation cause some internal hydrolysis of
the peptide, with loss of sample and accumulation of by-products. After about 30 cycles
of degradation, further accurate analysis becomes impossible. A small peptide such
as bradykinin can be completely determined by Edman degradation, but larger proteins must be broken into smaller fragments (Section 24-9E) before they can be completely sequenced.
PROBLEM 24-21
Draw the structure of the phenylthiohydantoin derivatives of
(a) alanine
(b) tryptophan
(c) lysine
(d) proline
PROBLEM 24-22
Show the third and fourth steps in the sequencing of oxytocin. Use Figure 24-14 as a guide.
PROBLEM 24-23
The Sanger method for N-terminus determination is a less common alternative to the Edman degradation. In the Sanger method,
the peptide is treated with the Sanger reagent, 2,4-dinitrofluorobenzene, and then hydrolyzed by reaction with 6 M aqueous HCl. The
N-terminal amino acid is recovered as its 2,4-dinitrophenyl derivative and identified.
24-9 Peptide Structure Determination
1181
The Sanger method
O
O2N
F
+
H2N
CH
C
NH
peptide
R1
NO2
2,4-dinitrofluorobenzene
(Sanger reagent)
peptide
O
O2N
NH
CH
C
NH
peptide
6 M HCl, heat
O2N
R1
NO2
CH
NH
R1
NO2
+
2,4-dinitrophenyl derivative
derivative
COOH
amino acids
(a) Propose a mechanism for the reaction of the N terminus of the peptide with 2,4-dinitrofluorobenzene.
(b) Explain why the Edman degradation is usually preferred over the Sanger method.
Application: Blood Clotting
The selective enzymatic cleavage of
proteins is critical to many biological
processes. For example, the clotting of
blood depends on the enzyme thrombin
cleaving fibrinogen at specific points
to produce fibrin, the protein that
forms a clot.
O
peptide
NH
CH
Rn – 1
C
24-9D
C-Terminal Residue Analysis
There is no efficient method for sequencing several amino acids of a peptide starting
from the C terminus. In many cases, however, the C-terminal amino acid can be identified using the enzyme carboxypeptidase, which cleaves the C-terminal peptide bond.
The products are the free C-terminal amino acid and a shortened peptide. Further
reaction cleaves the second amino acid that has now become the new C terminus of
the shortened peptide. Eventually, the entire peptide is hydrolyzed to its individual
amino acids.
O
O
NH
CH
C
OH
carboxypeptidase
H 2O
peptide
NH
Rn
CH
C
O
OH + H2N
Rn – 1
(further cleavage)
A peptide is incubated with the carboxypeptidase enzyme, and the appearance of
free amino acids is monitored. In theory, the amino acid whose concentration increases
first should be the C terminus, and the next amino acid to appear should be the second
residue from the end. In practice, different amino acids are cleaved at different rates,
making it difficult to determine amino acids past the C terminus and occasionally the
second residue in the chain.
24-9E Breaking the Peptide into Shorter Chains: Partial
Hydrolysis
Before a large protein can be sequenced, it must be broken into smaller chains, not
longer than about 30 amino acids. Each of these shortened chains is sequenced, and
then the entire structure of the protein is deduced by fitting the short chains together like
pieces of a jigsaw puzzle.
Partial cleavage can be accomplished either by using dilute acid with a shortened
reaction time or by using enzymes, such as trypsin and chymotrypsin, that break bonds
between specific amino acids. The acid-catalyzed cleavage is not very selective, leading to a mixture of short fragments resulting from cleavage at various positions. Enzymes
are more selective, giving cleavage at predictable points in the chain.
TRYPSIN: Cleaves the chain at the carboxyl groups of the basic amino acids
lysine and arginine.
CHYMOTRYPSIN: Cleaves the chain at the carboxyl groups of the aromatic
amino acids phenylalanine, tyrosine, and tryptophan.
CH
C
Rn
free amino acid
OH
1182
CHAPTER 24
Amino Acids, Peptides, and Proteins
Let’s use oxytocin (Figure 24-9) as an example to illustrate the use of partial hydrolysis. Oxytocin could be sequenced directly by C-terminal analysis and a series of
Edman degradations, but it provides a simple example of how a structure can be pieced
together from fragments. Acid-catalyzed partial hydrolysis of oxytocin (after cleavage
of the disulfide bridge) gives a mixture that includes the following peptides:
Ile-Gln-Asn-Cys
Gln-Asn-Cys-Pro
Pro-Leu-Gly # NH2
Cys-Tyr-Ile-Gln-Asn
Cys-Pro-Leu-Gly
When we match the overlapping regions of these fragments, the complete sequence of
oxytocin appears:
Application: Meat Tenderizer
Proteolytic (protein-cleaving) enzymes
also have applications in consumer
products. For example, papain (from
papaya extract) serves as a meat tenderizer. It cleaves the fibrous proteins,
making the meat less tough.
Cys-Tyr-Ile-Gln-Asn
Ile-Gln-Asn-Cys
Gln-Asn-Cys-Pro
Cys-Pro-Leu-Gly
Pro-Leu-Gly # NH2
Complete structure
Cys-Tyr-Ile-Gln-Asn-Cys-Pro-Leu-Gly # NH2
The two Cys residues in oxytocin may be involved in disulfide bridges, either linking
two of these peptide units or forming a ring. By measuring the molecular weight of
oxytocin, we can show that it contains just one of these peptide units; therefore, the
Cys residues must link the molecule in a ring.
PROBLEM 24-24
Show where trypsin and chymotrypsin would cleave the following peptide.
Tyr-Ile-Gln-Arg-Leu-Gly-Phe-Lys-Asn-Trp-Phe-Gly-Ala-Lys-Gly-Gln-Gln # NH2
PROBLEM 24-25
After treatment with peroxyformic acid, the peptide hormone vasopressin is partially
hydrolyzed. The following fragments are recovered. Propose a structure for vasopressin.
Phe-Gln-Asn
Asn-Cys-Pro-Arg
24-10
Solution-Phase
Peptide Synthesis
24-10A
Pro-Arg-Gly # NH2
Tyr-Phe-Gln-Asn
Cys-Tyr-Phe
Introduction
Total synthesis of peptides is rarely an economical method for their commercial production. Important peptides are usually derived from biological sources. For example,
insulin for diabetics was originally taken from pork pancreas. Now, recombinant DNA
techniques have improved the quality and availability of peptide pharmaceuticals. It is
possible to extract the piece of DNA that contains the code for a particular protein,
insert it into a bacterium, and induce the bacterium to produce the protein. Strains of
Escherichia coli have been developed to produce human insulin that avoids dangerous
reactions in people who are allergic to pork products.
Laboratory peptide synthesis is still an important area of chemistry, however,
for two reasons: If the synthetic peptide is the same as the natural peptide, it proves
the structure is correct; and the synthesis provides a larger amount of the material for
further biological testing. Also, synthetic peptides can be made with altered amino
acid sequences to compare their biological activity with the natural peptides. These
comparisons can point out the critical areas of the peptide, which may suggest causes
and treatments for genetic diseases involving similar abnormal peptides.
24-10 Solution-Phase Peptide Synthesis
1183
Peptide synthesis requires the formation of amide bonds between the proper amino
acids in the proper sequence. With simple acids and amines, we would form an amide
bond simply by converting the acid to an activated derivative (such as an acyl halide or
anhydride) and adding the amine.
O
R
O
X +
C
H2N
R´
R
C
NH
R´
+
H
X
(X is a good leaving group, preferably electron-withdrawing)
Amide formation is not so easy with amino acids, however. Each amino acid has
both an amino group and a carboxyl group. If we activate the carboxyl group, it reacts
with its own amino group. If we mix some amino acids and add a reagent to make them
couple, they form every conceivable sequence. Also, some amino acids have side chains
that might interfere with peptide formation. For example, glutamic acid has an extra carboxyl group, and lysine has an extra amino group. As a result, peptide synthesis always
involves both activating reagents to form the correct peptide bonds and protecting groups
to block formation of incorrect bonds.
Chemists have developed many ways of synthesizing peptides, falling into two
major groups. The solution-phase method involves adding reagents to solutions of growing peptide chains and purifying the products as needed. The solid-phase method
involves adding reagents to growing peptide chains bonded to solid polymer particles.
Many different reagents are available for each of these methods, but we will consider
only one set of reagents for the solution-phase method and one set for the solid-phase
method. The general principles are the same regardless of the specific reagents.
24-10B
Solution-Phase Method
Consider the structure of alanylvalylphenylalanine, a simple tripeptide:
O
H2N
CH
C
O
NH
CH3
CH
C
O
NH
C
OH
CH2Ph
CH(CH3)2
alanyl
CH
valyl
Ala-Val-Phe
phenylalanine
Solution-phase peptide synthesis begins at the N terminus and ends at the C terminus,
or left to right as we draw the peptide. The first major step is to couple the carboxyl
group of alanine to the amino group of valine. This cannot be done simply by activating
the carboxyl group of alanine and adding valine. If we activated the carboxyl group of
alanine, it would react with another molecule of alanine.
To prevent side reactions, the amino group of alanine must be protected to make it
nonnucleophilic. In Section 24-7B, we saw that an amino acid reacts with benzyl chloroformate (also called benzyloxycarbonyl chloride) to form a urethane, or carbamate
ester, that is easily removed at the end of the synthesis. This protecting group has been
used for many years, and it has acquired several names. It is called the benzyloxycarbonyl
group, the carbobenzoxy group (Cbz), or simply the Z group (abbreviated Z).
Preliminary step: Protect the amino group with Z.
Z group
O
CH2
O
C
O
Cl +
H2N
CH
C
CH3
benzyl chloroformate
Z-Cl
alanine
Ala
O
OH
Et3N
CH2
O
C
O
NH
CH
C
CH3
benzyloxycarbonyl alanine
Z-Ala
OH
+ Et3NHCl
1184
CHAPTER 24
Amino Acids, Peptides, and Proteins
The amino group in Z-Ala is protected as the nonnucleophilic amide half of a
carbamate ester. The carboxyl group can be activated without reacting with the protected amino group. Treatment with ethyl chloroformate converts the carboxyl group
to a mixed anhydride of the amino acid and carbonic acid. It is strongly activated
toward nucleophilic attack.
Step 1: Activate the carboxyl group with ethyl chloroformate.
anhydride of carbonic acid
O
Z
NHCH
O
C
+
OH
Cl
C
O
OCH2CH3
Z
NHCH
CH3
O
C
O
OCH2CH3
C
+
HCl
CH3
protected alanine
ethyl chloroformate
mixed anhydride
When the second amino acid (valine) is added to the protected, activated alanine,
the nucleophilic amino group of valine attacks the activated carbonyl of alanine, displacing the anhydride and forming a peptide bond. (Some procedures use an ester of the new
amino acid to avoid competing reactions from its carboxylate group.)
Step 2: Form an amide bond to couple the next amino acid.
O
Z
NHCH
C
O
C
O
O
O
OCH2CH3 + H2N
CH
C
Z
OH
NHCH
C
NH
CH
C
OH
+ CO2
+ CH3CH2OH
CH(CH3 )2
Z-Ala-Val
CH3
CH(CH3 )2
valine
CH3
protected, activated alanine
O
PROBLEM 24-26
Give complete mechanisms for the formation of Z-Ala, its activation by ethyl chloroformate,
and the coupling with valine.
At this point, we have the N-protected dipeptide Z-Ala-Val. Phenylalanine must be
added to the C terminus to complete the Ala-Val-Phe tripeptide. Activation of the valine
carboxyl group, followed by addition of phenylalanine, gives the protected tripeptide.
Step 1: Activate the carboxyl group with ethyl chloroformate.
Z
O
O
NHCHCNHCH
C
CH3
O
OH + Cl
C
OEt
Z
O
O
NHCHCNHCH
C
CH(CH3)2
Ala
CH3
Val
Ala
O
O
C
OEt + HCl
CH(CH3)2
Val
Step 2: Form an amide bond to couple the next amino acid.
O
Z
Ala
NHCH
C
CH(CH3)2
Val
O
O
C
O
OEt + H2N
CH
C
CH2 Ph
phenylalanine
O
OH
Z
Ala
NHCH
H3C
C
O
NH
CH
CH CH3
CH2
Z-Ala-Val-Phe
C
OH + CO2 + EtOH
Ph
To make a larger peptide, repeat these two steps for the addition of each amino
acid residue:
1. Activate the C terminus of the growing peptide by reaction with ethyl chloroformate.
2. Couple the next amino acid.
24-11 Solid-Phase Peptide Synthesis
1185
The final step in the solution-phase synthesis is to deprotect the N terminus of the
completed peptide. The N-terminal amide bond must be cleaved without breaking any
of the peptide bonds in the product. Fortunately, the benzyloxycarbonyl group is partly
an amide and partly a benzyl ester, and hydrogenolysis of the benzyl ester takes place
under mild conditions that do not cleave the peptide bonds. This mild cleavage is the
reason for using the benzyloxycarbonyl group (as opposed to some other acyl group)
to protect the N terminus.
Final step: Remove the protecting group.
O
CH2
O
C
O
NHCHC
O
Val
CH3
Z-Ala-Val-Phe
Phe
H2, Pd
H2NCHC
Val
Phe + CO2
+
Ph
CH3
CH3
Ala-Val-Phe
PROBLEM 24-27
Show how you would synthesize Ala-Val-Phe-Gly-Leu starting with Z-Ala-Val-Phe.
Problem-solving Hint
Remember that classical (solutionphase) peptide synthesis:
PROBLEM 24-28
Show how the solution-phase synthesis would be used to synthesize Ile-Gly-Asn.
The solution-phase method works well for small peptides, and many peptides
have been synthesized by this process. A large number of chemical reactions and
purifications are required even for a small peptide, however. Although the individual yields are excellent, with a large peptide, the overall yield becomes so small as
to be unusable, and several months (or years) are required to complete so many steps.
The large amounts of time required and the low overall yields are due largely to the
purification steps. For larger peptides and proteins, solid-phase peptide synthesis is
usually preferred.
In 1962, Robert Bruce Merrifield of Rockefeller University developed a method for synthesizing peptides without having to purify the intermediates. He did this by attaching
the growing peptide chains to solid polystyrene beads. After each amino acid is added, the
excess reagents are washed away by rinsing the beads with solvent. This ingenious method
lends itself to automation, and Merrifield built a machine that can add several amino acid
units while running unattended. Using this machine, Merrifield synthesized ribonuclease
(124 amino acids) in just six weeks, obtaining an overall yield of 17%. Merrifield’s work
in solid-phase peptide synthesis won the Nobel Prize in Chemistry in 1984.
24-11A
The Individual Reactions
Three reactions are crucial for solid-phase peptide synthesis. These reactions attach the
first amino acid to the solid support, protect each amino group until its time to react, and
form the peptide bonds between the amino acids.
Attaching the Peptide to the Solid Support The greatest difference between solutionphase and solid-phase peptide synthesis is that solid-phase synthesis is done in the
opposite direction: starting with the C terminus and going toward the N terminus,
right to left as we write the peptide. The first step is to attach the last amino acid (the
C terminus) to the solid support.
1. Goes N : C. Protect the
N terminus (Z group) first,
deprotect last.
2. Couple each amino acid by
activating the C terminus (ethyl
chloroformate), then adding the
new amino acid.
24-11
Solid-Phase Peptide
Synthesis
1186
Amino Acids, Peptides, and Proteins
CHAPTER 24
The solid support is a special polystyrene bead in which some of the aromatic rings
have chloromethyl groups. This polymer, often called the Merrifield resin, is made by
copolymerizing styrene with a few percent of p-(chloromethyl)styrene.
Formation of the Merrifield resin
CH2Cl
CH2Cl
CH2Cl
+
=
H
H
C
C
C
H
C
H
H
CH
CH2
CH2
CH
CH2
P
H
p-(chloromethyl)styrene
styrene
CH
polymer
abbreviation
Like other benzyl halides, the chloromethyl groups on the polymer are reactive
toward SN2 attack. The carboxyl group of an N-protected amino acid displaces chloride,
giving an amino acid ester of the polymer. In effect, the polymer serves as the alcohol
part of an ester protecting group for the carboxyl end of the C-terminal amino acid.
The amino group must be protected, or it would attack the chloromethyl groups.
Attachment of the C-terminal amino acid
O
protecting
group
NH
CH
C
H
O
–
O
H
C
protecting
group
Cl
NH
CH
R
C
Cl–
CH2
O
R
P
P
Once the C-terminal amino acid is fixed to the polymer, the chain is built on the
amino group of this amino acid.
Using the tert-Butyloxycarbonyl (Boc) Protecting Group The benzyloxycarbonyl
group (the Z group) cannot be used with the solid-phase process because the Z group is
removed by hydrogenolysis in contact with a solid catalyst. A polymer-bound peptide
cannot achieve the intimate contact with a solid catalyst required for hydrogenolysis.
The N-protecting group used in the Merrifield procedure is the tert-butyloxycarbonyl
group, abbreviated Boc or t-Boc. The Boc group is similar to the Z group, except that it
has a tert-butyl group in place of the benzyl group. Like other tert-butyl esters, the Boc
protecting group is easily removed under acidic conditions.
The acid chloride of the Boc group is unstable, so we use the anhydride, di-tertbutyldicarbonate, to attach the group to the amino acid.
Protection of the amino group as its Boc derivative
O
CH3
CH3
C
O
C
O
O
C
O
CH3
C
CH3
di-tert-butyldicarbonate
O
CH3
CH3
CH3 + H2N
CH
COOH
R
amino acid
CH3
C
O
C
NH
CH3
CH
COOH
R
Boc-amino acid
CH3
+ CO2 + CH3
C
CH3
OH
24-11 Solid-Phase Peptide Synthesis
The Boc group is easily cleaved by brief treatment with trifluoroacetic acid (TFA),
CF3COOH. Loss of a relatively stable tert-butyl cation from the protonated ester gives an
unstable carbamic acid. Decarboxylation of the carbamic acid gives the deprotected amino
group of the amino acid. Loss of a proton from the tert-butyl cation gives isobutylene.
CH3
CH3
C
O
C
O+ H
CH3
O
NH
CH
CH3
COOH
CF3COOH
O
C
CH3
C+
CH3
+
O
CH
COOH
R
Boc-amino acid
O
NH
CH3
R
CH3
C
protonated
H
C
NH
CH
CH3
CH3
+
H3N
COOH
COOH + CH2
CH
CH3
R
R
free amino acid
a carbamic acid
+ CO2
C
isobutylene
People who synthesize peptides generally do not make their own Boc-protected
amino acids. Because they use all their amino acids in protected form, they buy and
use commercially available Boc amino acids.
Use of DCC as a Peptide Coupling Agent The final reaction needed for the Merrifield
procedure is the peptide bond-forming condensation. When a mixture of an amine and
an acid is treated with N,N¿-dicyclohexylcarbodiimide (abbreviated DCC), the amine
and the acid couple to form an amide. The molecule of water lost in this condensation
converts DCC to N,N¿-dicyclohexylurea (DCU).
O
R
C
O
O–
+
+ H 3N
acid
R´ +
amine
N
C
N
R
C
N,N´-dicyclohexylcarbodiimide (DCC)
NH
R´ +
H
O
H
N
C
N
N,N´-dicyclohexylurea (DCU)
amide
The mechanism for DCC coupling is not as complicated as it may seem. The carboxylate ion adds to the strongly electrophilic carbon of the diimide, giving an activated
acyl derivative of the acid. This activated derivative reacts readily with the amine to give
the amide. In the final step, DCU serves as an excellent leaving group. The cyclohexane
rings are miniaturized for clarity.
Formation of an activated acyl derivative
N
C
N
O
R
O
R
C
O
C
N
O
H
C
N
–
N
O
+
NH2
R´
R
–
C
O
activated
H2N
R´
C
NH
Coupling with the amine and loss of DCU
O
R
C
N
O
O
C
R
NH
R´
NH2
R´
–
C
+
NH2
N
O
C
R
NH
C
O
N
O
+
H
O
C
NH
N
R´
–
H
R C NHR´
amide (peptide)
+ DCU
1187
1188
Amino Acids, Peptides, and Proteins
CHAPTER 24
At the completion of the synthesis, the ester bond to the polymer is cleaved by
anhydrous HF. Because this is an ester bond, it is more easily cleaved than the amide
bonds of the peptide.
Cleavage of the finished peptide
CH2F
O
peptide
C
O
O
HF
CH2
peptide
C
OH +
P
P
PROBLEM 24-29
Propose a mechanism for the coupling of acetic acid and aniline using DCC as a coupling agent.
Problem-solving Hint
Remember that solid-phase peptide
synthesis:
1. Goes C : N. Attach the Bocprotected C terminus to the
bead first.
2. Couple each amino acid by
removing (TFA) the Boc group
from the N terminus, then add
the next Boc-protected amino
acid with DCC.
3. Cleave (HF) the finished peptide
from the bead.
Now we consider an example to illustrate how these procedures are combined in
the Merrifield solid-phase peptide synthesis.
An Example of Solid-Phase Peptide Synthesis
24-11B
For easy comparison of the solution-phase and solid-phase methods, we will consider
the synthesis of the same tripeptide we made using the solution-phase method.
Ala-Val-Phe
The solid-phase synthesis is carried out in the direction opposite that of the
solution-phase synthesis. The first step is attachment of the N-protected C-terminal
amino acid (Boc-phenylalanine) to the polymer.
Me3C
O
O
O
O
C
NH
CH
Ph
CH2
Boc
C
O– + CH2
Me3C
Cl
O
C
Boc
Boc-Phe
O
NH
CH
C
Ph
CH2
O
Boc-Phe— P
P
CH2
P
Trifluoroacetic acid (TFA) cleaves the Boc protecting group of phenylalanine so that
its amino group can be coupled with the next amino acid.
O
Me3C
O
Boc
C
O
NH
CH
Ph
CH2
Boc-Phe— P
C
O
CH2
CF3COOH
(TFA)
O
+
H3N
Ph
P
CH
C
CH3
O
CH2 + CH2
CH3
CH2
Phe— P
+ CO2
C
P
The second amino acid (valine) is added in its N-protected Boc form so that it cannot couple with itself. Addition of DCC couples the valine carboxyl group with the free
¬ NH2 group of phenylalanine.
1189
24-11 Solid-Phase Peptide Synthesis
O
O
Boc
NH
CH
C
+
O– + H3N
(CH3)2CH
CH
Ph
C
O
CH2
Boc
NH
C
CH
(CH3)2CH
CH2
Phe— P
Boc-Val
O
O
DCC
NH
CH
Ph
CH2
C
CH2 + DCU
O
Boc-Val-Phe— P
P
P
To couple the final amino acid (alanine), the chain is first deprotected by treatment
with trifluoroacetic acid. Then the N-protected Boc-alanine and DCC are added.
Step 1: Deprotection
O
O
Boc
NH
CH
C
(CH3)2CH
NH
CH
Ph
CH2
C
O
CH2
CF3COOH
(TFA)
O
+
H3N
CH
C
(CH3)2CH
P
Boc-Val-Phe— P
O
NH
CH
Ph
CH2
C
CH3
O
CH2 + CH3
C
CH2
+ CO2
P
Val-Phe— P
Step 2: Coupling
O
+
H3N
CH
C
(CH3)2CH
Boc
O
O
NH
CH
Ph
CH2
C
O
CH2
NH
CH
CH3
DCC
C
O–
Boc
NH
C
CH
NH
CH3
CH
C
(CH3)2CH
P
Val-Phe— P
O
O
O
NH
CH
C
Ph
CH2
O
CH2 + DCU
P
Boc-Ala-Val-Phe— P
If we were making a longer peptide, the addition of each subsequent amino acid
would require the repetition of two steps:
1. Use trifluoroacetic acid to deprotect the amino group at the end of the growing chain.
2. Add the next Boc-amino acid, using DCC as a coupling agent.
Once the peptide is completed, the final Boc protecting group must be removed, and
the peptide must be cleaved from the polymer. Anhydrous HF cleaves the ester linkage
that bonds the peptide to the polymer, and it also removes the Boc protecting group. In
our example, the following reaction occurs:
O
Boc
NH
CH
CH3
C
O
O
NH
CH
C
(CH3)2CH
NH
CH
Ph
CH2
C
O
O
CH2
HF
O
O
+
H3N
CH
C
CH3
NH
CH
C
NH
CH
Ph
CH2
(CH3)2CH
C
OH
Ala-Val-Phe
Boc-Ala-Val-Phe— P
P
CH3
+ CO2 + CH3
C
CH2 +
P
CH2F
1190
CHAPTER 24
Amino Acids, Peptides, and Proteins
PROBLEM 24-30
Show how you would synthesize Leu-Gly-Ala-Val-Phe starting with Boc-Ala-Val-Phe— P .
PROBLEM 24-31
Show how solid-phase peptide synthesis would be used to make Ile-Gly-Asn.
24-12
Classification
of Proteins
Proteins may be classified according to their chemical composition, their shape, or their
function. Protein composition and function are treated in detail in a biochemistry course.
For now, we briefly survey the types of proteins and their general classifications.
Proteins are grouped into simple and conjugated proteins according to their chemical composition. Simple proteins are those that hydrolyze to give only amino acids.
All the protein structures we have considered so far are simple proteins. Examples are
insulin, ribonuclease, oxytocin, and bradykinin. Conjugated proteins are bonded to a
nonprotein prosthetic group such as a sugar, a nucleic acid, a lipid, or some other
group. Table 24-3 lists some examples of conjugated proteins.
TABLE 24-3
Classes of Conjugated Proteins
Class
Prosthetic Group
glycoproteins
carbohydrates
Examples
g-globulin, interferon
nucleoproteins
nucleic acids
ribosomes, viruses
lipoproteins
fats, cholesterol
high-density lipoprotein
metalloproteins
a complexed metal
hemoglobin, cytochromes
Proteins are classified as fibrous or globular depending on whether they form long
filaments or coil up on themselves. Fibrous proteins are stringy, tough, and usually
insoluble in water. They function primarily as structural parts of the organism. Examples
of fibrous proteins are a-keratin in hooves and fingernails, and collagen in tendons.
Globular proteins are folded into roughly spherical shapes. They usually function as
enzymes, hormones, or transport proteins. Enzymes are protein-containing biological
catalysts; an example is ribonuclease, which cleaves RNA. Hormones help to regulate
processes in the body. An example is insulin, which regulates glucose levels in the blood
and its uptake by cells. Transport proteins bind to specific molecules and transport them
in the blood or through the cell membrane. An example is hemoglobin, which transports
oxygen in the blood from the lungs to the tissues.
24-13
Levels of
Protein Structure
24-13A
Primary Structure
Up to now, we have discussed the primary structure of proteins. The primary structure
is the covalently bonded structure of the molecule. This definition includes the sequence
of amino acids, together with any disulfide bridges. All the properties of the protein
are determined, directly or indirectly, by the primary structure. Any folding, hydrogen
bonding, or catalytic activity depends on the proper primary structure.
24-13B
Secondary Structure
Although we often think of peptide chains as linear structures, they tend to form
orderly hydrogen-bonded arrangements. In particular, the carbonyl oxygen atoms
24-13 Levels of Protein Structure
1191
R
C
O
C
H
H
N
R
H
N
C
CH
C
H
N
R
O
N
CH
O
C
N
O C O
H
HC
R
C
H
O
C
CH
O
H N
N
CH
C O
R
CH HC R
R
N
H
O
C
H
O
C
O
N
C = gray
N = blue
O = red
R = green
FIGURE 24-15
The a helical arrangement. The peptide chain curls into a helix so that each peptide carbonyl
group is hydrogen-bonded to an N ¬ H hydrogen on the next turn of the helix. Side chains are
symbolized by green atoms in the space-filling structure.
form hydrogen bonds with the amide 1N ¬ H2 hydrogens. This tendency leads to
orderly patterns of hydrogen bonding: the A helix and the pleated sheet. These
hydrogen-bonded arrangements, if present, are called the secondary structure of
the protein.
When a peptide chain winds into a helical coil, each carbonyl oxygen can hydrogenbond with an N ¬ H hydrogen on the next turn of the coil. Many proteins wind into an
a helix (a helix that looks like the thread on a right-handed screw) with the side chains
positioned on the outside of the helix. For example, the fibrous protein a keratin is
arranged in the a-helical structure, and most globular proteins contain segments of a
helix. Figure 24-15 shows the a-helical arrangement.
Segments of peptides can also form orderly arrangements of hydrogen bonds
by lining up side-by-side. In this arrangement, each carbonyl group on one chain
forms a hydrogen bond with an N ¬ H hydrogen on an adjacent chain. This arrangement may involve many peptide molecules lined up side-by-side, resulting in a twodimensional sheet. The bond angles between amino acid units are such that the sheet
is pleated (creased), with the amino acid side chains arranged on alternating sides
of the sheet. Silk fibroin, the principal fibrous protein in the silks of insects and
arachnids, has a pleated-sheet secondary structure. Figure 24-16 shows the pleatedsheet structure.
...
R
R
H
N
CH
R
CH
C
H
O
O
H
C
C
N
O
H
O
R
R
C
CH
R
R
CH
N
CH
R
H
N
CH
N
...
O
H
C
H
O
O
H
N
O
H
R
N
O
H
H
O
CH
R
R
C
CH
N
CH
R
R
CH
N
CH
C
C
N
CH
N
C
C
CH
O
H
...
C
CH
C
...
N
CH
R
O
N
CH
...
O
C
N
...
H
R
CH
C
R
H
...
H
N
CH
O
...
O
...
R
N
...
R
C
C
R
H
...
CH
N
CH
O
...
R
H
Spider web is composed mostly of
fibroin, a protein with pleated-sheet
secondary structure. The pleatedsheet arrangement allows for multiple
hydrogen bonds between molecules,
conferring great strength.
O
C
C
N
O
H
CH
R
FIGURE 24-16
The pleated-sheet arrangement. Each
peptide carbonyl group is hydrogenbonded to an N ¬ H hydrogen on an
adjacent peptide chain.
1192
CHAPTER 24
Amino Acids, Peptides, and Proteins
A protein may or may not have the same secondary structure throughout its length.
Some parts may be curled into an a helix, while other parts are lined up in a pleated
sheet. Parts of the chain may have no orderly secondary structure at all. Such a structureless region is called a random coil. Most globular proteins, for example, contain segments of a helix or pleated sheet separated by kinks of random coil, allowing the
molecule to fold into its globular shape.
24-13C
Tertiary structures of proteins are
determined by X-ray crystallography.
A single crystal of the protein is
bombarded with X rays, whose
wavelengths are appropriate to be
diffracted by the regular atomic
spacings in the crystal. A computer
then determines the locations of the
atoms in the crystal.
Tertiary Structure
The tertiary structure of a protein is its complete three-dimensional conformation.
Think of the secondary structure as a spatial pattern in a local region of the molecule. Parts of the protein may have the a-helical structure, while other parts may
have the pleated-sheet structure, and still other parts may be random coils. The tertiary structure includes all the secondary structure and all the kinks and folds in
between. The tertiary structure of a typical globular protein is represented in
Figure 24-17.
Coiling of an enzyme can give three-dimensional shapes that produce important catalytic effects. Polar, hydrophilic (water-loving) side chains are oriented toward the outside of the globule. Nonpolar, hydrophobic (water-hating) groups are arranged toward
the interior. Coiling in the proper conformation creates an enzyme’s active site, the
region that binds the substrate and catalyzes the reaction. A reaction taking place at the
active site in the interior of an enzyme may occur under essentially anhydrous, nonpolar conditions—while the whole system is dissolved in water!
random coil
C terminus
COO–
x
eli
αh
FIGURE 24-17
+NH
The tertiary structure of a typical
globular protein includes segments
of a helix with segments of random
coil at the points where the helix
is folded.
3
24-13D
N terminus
Quaternary Structure
Quaternary structure refers to the association of two or more peptide chains in the
complete protein. Not all proteins have quaternary structure. The ones that do are those
that associate together in their active form. For example, hemoglobin, the oxygen carrier
in mammalian blood, consists of four peptide chains fitted together to form a globular
protein. Figure 24-18 summarizes the four levels of protein structure.
24-14 Protein Denaturation
Ile
N
Gln
Tyr
Asn
Cys
Leu
primary structure
tertiary structure
Gly NH2
N
R
H
N
CH
CH
Pro
C
N
H O
H O
C
C
CH
R
.
S S
O
C
O
R
Cys
H
secondary structure
FIGURE 24-18
A schematic comparison of the levels
of protein structure. Primary structure
is the covalently bonded structure,
including the amino acid sequence
and any disulfide bridges. Secondary
structure refers to the areas of
a helix, pleated sheet, or random
coil. Tertiary structure refers to the
overall conformation of the molecule.
Quaternary structure refers to the
association of two or more peptide
chains in the active protein.
quaternary structure
For a protein to be biologically active, it must have the correct structure at all levels. The
sequence of amino acids must be right, with the correct disulfide bridges linking the
cysteines on the chains. The secondary and tertiary structures are important, as well.
The protein must be folded into its natural conformation, with the appropriate areas of
a helix and pleated sheet. For an enzyme, the active site must have the right conformation, with the necessary side-chain functional groups in the correct positions. Conjugated
proteins must have the right prosthetic groups, and multichain proteins must have the
right combination of individual peptides.
With the exception of the covalent primary structure, all these levels of structure are
maintained by weak solvation and hydrogen-bonding forces. Small changes in the environment can cause a chemical or conformational change resulting in denaturation:
disruption of the normal structure and loss of biological activity. Many factors can cause
denaturation, but the most common ones are heat and pH.
24-14A
1193
24-14
Protein
Denaturation
Reversible and Irreversible Denaturation
The cooking of egg white is an example of protein denaturation by high temperature.
Egg white contains soluble globular proteins called albumins. When egg white is heated,
the albumins unfold and coagulate to produce a solid rubbery mass. Different proteins
have different abilities to resist the denaturing effect of heat. Egg albumin is quite sensitive to heat, but bacteria that live in geothermal hot springs have developed proteins
that retain their activity in boiling water.
When a protein is subjected to an acidic pH, some of the side-chain carboxyl
groups become protonated and lose their ionic charge. Conformational changes result,
leading to denaturation. In a basic solution, amino groups become deprotonated,
similarly losing their ionic charge, causing conformational changes and denaturation.
Milk turns sour because of the bacterial conversion of carbohydrates to lactic
acid. When the pH becomes strongly acidic, soluble proteins in milk are denatured
Irreversible denaturation of egg
albumin. The egg white does not
become clear and runny again
when it cools.
1194
CHAPTER 24
Amino Acids, Peptides, and Proteins
and precipitate. This process is called curdling. Some proteins are more resistant to acidic
and basic conditions than others. For example, most digestive enzymes such as amylase and
trypsin remain active under acidic conditions in the stomach, even at a pH of about 1.
In many cases, denaturation is irreversible. When cooked egg white is cooled, it does
not become uncooked. Curdled milk does not uncurdle when it is neutralized.
Denaturation may be reversible, however, if the protein has undergone only mild denaturing conditions. For example, a protein can be salted out of solution by a high salt concentration, which denatures and precipitates the protein. When the precipitated protein
is redissolved in a solution with a lower salt concentration, it usually regains its activity together with its natural conformation.
24-14B
Micrograph of normal human brain
tissue. The nuclei of neurons appear
as dark spots.
Brain tissue of a patient infected with
vCJD. Note the formation of (white)
vacuole spaces and (dark, irregular)
plaques of prion protein.
(Magnification 200X)
Prion Diseases
Up through 1980, people thought that all infectious diseases were caused by microbes
of some sort. They knew about diseases caused by viruses, bacteria, protozoa, and fungi.
There were some strange diseases, however, for which no one had isolated and cultured
the pathogen. Creutzfeldt–Jakob Disease (CJD) in humans, scrapie in sheep, and
transmissible encephalopathy in mink (TME) all involved a slow, gradual loss of mental function and eventual death. The brains of the victims all showed unusual plaques
of amyloid protein surrounded by spongelike tissue.
Workers studying these diseases thought there was an infectious agent involved (as
opposed to genetic or environmental causes) because they knew that scrapie and TME
could be spread by feeding healthy animals the ground-up remains of sick animals.
They had also studied kuru, a disease much like CJD among tribes where family members showed their respect for the dead by eating their brains. These diseases were generally attributed to “slow viruses” that were yet to be isolated.
In the 1980s, neurologist Stanley B. Prusiner (of the University of California at
San Francisco) made a homogenate of scrapie-infected sheep brains and systematically
separated out all the cell fragments, bacteria, and viruses, and found that the remaining
material was still infectious. He separated out the proteins and found a protein fraction
that was still infectious. He suggested that scrapie (and presumably similar diseases) is
caused by a protein infectious agent that he called prion protein. This conclusion contradicted the established principle that contagious diseases require a living pathogen.
Many skeptical workers repeated Prusiner’s work in hopes of finding viral contaminants in the infectious fractions, and most of them finally came to the same conclusion.
Prusiner received the 1998 Nobel Prize in Medicine or Physiology for this work.
Since Prusiner’s work, prion diseases have become more important because of their
threat to humans. Beginning in 1996, some cows in the United Kingdom developed
“mad cow disease” and would threaten other animals, wave their heads, fall down, and
eventually die. The disease, called bovine spongiform encephalopathy, or BSE, was
probably transmitted to cattle by feeding them the remains of scrapie-infected sheep.
The most frightening aspect of the BSE outbreak was that people could contract a fatal
disease, called new-variant Creutzfeldt–Jakob Disease (vCJD) from eating the infected
meat. Since that time, a similar disease, called chronic wasting disease, or CWD, has
been found in wild deer and elk in the Rocky Mountains. All of these (presumed) prion
diseases are now classified as transmissible spongiform encephalopathies, or TSEs.
The most widely accepted theory of prion diseases suggests that the infectious prion
protein has the same primary structure as a normal protein found in nerve cells, but it
differs in its tertiary structure. In effect, it is a misfolded, denatured version of a normal protein that polymerizes to form the amyloid protein plaques seen in the brains of
infected animals. When an animal ingests infected food, the polymerized protein resists
digestion. Because it is simply a misfolded version of a normal protein, the infectious
prion does not provoke the host’s immune system to attack the pathogen.
When the abnormal prion interacts with the normal version of the protein on the
membranes of nerve cells, the abnormal protein somehow induces the normal molecules to change their shape. This is the part of the process we know the least about.
Essential Terms
1195
(We might think of it like crystallization, in which a seed crystal induces other molecules to crystallize in the same conformation and crystal form.) These newly misfolded
protein molecules then induce more molecules to change shape. The polymerized abnormal protein cannot be broken down by the usual protease enzymes, so it builds up in the
brain and causes the plaques and spongy tissue associated with TSEs.
We once thought that a protein with the correct primary structure, placed in the
right physiological solution, would naturally fold into the correct tertiary structure and
stay that way. We were wrong. We now know that protein folding is a carefully controlled
process in which enzymes and chaperone proteins promote correct folding as the protein is synthesized. Prion diseases have shown that there are many factors that cause
proteins to fold into natural or unnatural conformations, and that the folding of the protein can have major effects on its biological properties within an organism.
ESSENTIAL PROBLEM-SOLVING SKILLS IN CHAPTER 24
Each skill is followed by problem numbers exemplifying that particular skill.
Correctly name amino acids and peptides, and draw the structures from
their names.
Use perspective drawings and Fischer projections to show the stereochemistry
of D- and L-amino acids. Explain why the naturally occurring amino acids are called
L-amino acids.
Explain which amino acids are acidic, which are basic, and which are neutral. Use
the isoelectric point to predict whether a given amino acid will be positively charged,
negatively charged, or neutral at a given pH.
Problems 24-33, 40, and 41
Problems 24-37 and 53
Problems 24-32 and 40
Show how to make a given amino acid using one of the following syntheses:
Reductive amination, HVZ followed by ammonia, the Gabriel–malonic ester synthesis,
and the Strecker synthesis.
Problems 24-34, 35, 37, 38,
and 39
Predict products of the acylation and esterification of amino acids, and their
reaction with ninhydrin.
Problems 24-34 and 36
Use information from terminal residue analysis and partial hydrolysis to determine
the structure of an unknown peptide.
Problems 24-42, 43, 46, 50, and 51
Show how you would use solution-phase synthesis or solid-phase synthesis to
make a given peptide. Use appropriate protecting groups to prevent unwanted
couplings.
Problems 24-44, 45, and 52
Discuss and identify the four levels of protein structure (primary, secondary,
tertiary, and quaternary). Explain how the structure of a protein affects its properties
and how denaturation changes the structure.
ESSENTIAL TERMS
active site
amino acid
biomimetic synthesis
complete proteins
The region of an enzyme that binds the substrate and catalyzes the reaction. (p. 1192)
Literally, any molecule containing both an amino group ( ¬ NH22 and a carboxyl group
1 ¬ COOH2. The term usually means an A-amino acid, with the amino group on the carbon
atom next to the carboxyl group. (p. 1155)
A laboratory synthesis that is patterned after a biological synthesis. For example, the synthesis
of amino acids by reductive amination resembles the biosynthesis of glutamic acid. (p. 1164)
Proteins that provide all the essential amino acids in about the right proportions for human
nutrition. Examples include those in meat, fish, milk, and eggs. Incomplete proteins are
severely deficient in one or more of the essential amino acids. Most plant proteins are
incomplete. (p. 1160)
1196
CHAPTER 24
conjugated protein
C terminus
denaturation
disulfide linkage
Edman degradation
electrophoresis
enzymatic resolution
enzyme
essential amino acids
fibrous proteins
globular proteins
A helix
hydrogenolysis
Amino Acids, Peptides, and Proteins
A protein that contains a nonprotein prosthetic group such as a sugar, nucleic acid, lipid, or
metal ion. (p. 1190)
(C-terminal end) The end of the peptide chain with a free or derivatized carboxyl group.
As the peptide is written, the C terminus is usually on the right. The amino group of the
C-terminal amino acid links it to the rest of the peptide. (p. 1174)
An unnatural alteration of the conformation or the ionic state of a protein. Denaturation generally results in precipitation of the protein and loss of its biological activity. Denaturation may
be reversible, as in salting out a protein, or irreversible, as in cooking an egg. (p. 1193)
(disulfide bridge) A bond between two cysteine residues formed by mild oxidation of their
thiol groups to a disulfide. (p. 1175)
A method for removing and identifying the N-terminal amino acid from a peptide without
destroying the rest of the peptide chain. The peptide is treated with phenyliso thiocyanate,
followed by a mild acid hydrolysis to convert the N-terminal amino acid to its phenylthiohydantoin derivative. The Edman degradation can be used repeatedly to determine the sequence
of many residues beginning at the N terminus. (p. 1179)
A procedure for separating charged molecules by their migration in a strong electric field.
The direction and rate of migration are governed largely by the average charge on the
molecules. (p. 1163)
The use of enzymes to separate enantiomers. For example, the enantiomers of an amino acid
can be acylated and then treated with hog kidney acylase. The enzyme hydrolyzes the acyl
group from the natural L-amino acid, but it does not react with the D-amino acid. The resulting
mixture of the free L-amino acid and the acylated D-amino acid is easily separated. (p. 1169)
A protein-containing biological catalyst. Many enzymes also include prosthetic groups, nonprotein constituents that are essential to the enzyme’s catalytic activity. (p. 1190)
Ten standard amino acids that are not biosynthesized by humans and must be provided in
the diet. (p. 1159)
A class of proteins that are stringy, tough, threadlike, and usually insoluble in water. (p. 1190)
A class of proteins that are relatively spherical in shape. Globular proteins generally have
lower molecular weights and are more soluble in water than fibrous proteins. (p. 1190)
A helical peptide conformation in which the carbonyl groups on one turn of the helix are
hydrogen-bonded to N ¬ H hydrogens on the next turn. Extensive hydrogen bonding stabilizes this helical arrangement. (p. 1191)
Cleavage of a bond by the addition of hydrogen. For example, catalytic hydrogenolysis
cleaves benzyl esters. (p. 1171)
O
R
C
O
O
H2, Pd
CH2
R
C
benzyl ester
isoelectric point, pl
L-amino
acid
H
CH3
L-alanine
(S)-alanine
peptide bonds
H
CH2
toluene
(isoelectric pH) The pH at which an amino acid (or protein) does not move under electrophoresis. This is the pH where the average charge on its molecules is zero, with most of the
molecules in their zwitterionic form. (p. 1162)
An amino acid having a stereochemical configuration similar to that of L-1-2-glyceraldehyde.
Most naturally occurring amino acids have the L configuration. (p. 1157)
H2N
oligopeptide
peptide
H +
acid
COOH
N terminus
O
COOH
CHO
HO
H
CH2OH
L-(–)-glyceraldehyde
(S)-glyceraldehyde
H
H2N
R
an L-amino acid
(S) configuration
(N-terminal end) The end of the peptide chain with a free or derivatized amino group. As the
peptide is written, the N terminus is usually on the left. The carboxyl group of the N-terminal
amino acid links it to the rest of the peptide. (p. 1174)
A small polypeptide, containing about four to ten amino acid residues. (p. 1174)
Any polymer of amino acids linked by amide bonds between the amino group of each amino
acid and the carboxyl group of the neighboring amino acid. The terms dipeptide, tripeptide,
etc. may specify the number of amino acids in the peptide. (p. 1174)
Amide linkages between amino acids. (pp. 1155, 1174)
1197
Essential Terms
pleated sheet
polypeptide
primary structure
prion protein
prosthetic group
protein
quaternary structure
random coil
residue
Sanger method
secondary structure
sequence
simple proteins
solid-phase peptide synthesis
solution-phase peptide synthesis
standard amino acids
Strecker synthesis
A two-dimensional peptide conformation with the peptide chains lined up side-by-side. The
carbonyl groups on each peptide chain are hydrogen-bonded to N ¬ H hydrogens on the adjacent chain, and the side chains are arranged on alternating sides of the sheet. (p. 1191)
A peptide containing many amino acid residues. Although proteins are polypeptides, the
term polypeptide is commonly used for molecules with lower molecular weights than
proteins. (p. 1174)
The covalently bonded structure of a protein; the sequence of amino acids, together with any
disulfide bridges. (p. 1190)
A protein infectious agent that is thought to promote misfolding and polymerization of normal
protein molecules, leading to amyloid plaques and destruction of nerve tissue. (p. 1194)
The nonprotein part of a conjugated protein. Examples of prosthetic groups are sugars, lipids,
nucleic acids, and metal complexes. (p. 1190)
A biopolymer of amino acids. Proteins are polypeptides with molecular weights higher than
about 5000 amu. (p. 1190)
The association of two or more peptide chains into a composite protein. (p. 1192)
A type of protein secondary structure where the chain is neither curled into an a helix nor
lined up in a pleated sheet. In a globular protein, the kinks that fold the molecule into its globular shape are usually segments of random coil. (p. 1192)
An amino acid unit of a peptide. (p. 1174)
A method for determining the N-terminal amino acid of a peptide. The peptide is treated with
2,4-dinitrofluorobenzene (Sanger’s reagent), then completely hydrolyzed. The derivatized amino
acid is easily identified, but the rest of the peptide is destroyed in the hydrolysis. (p. 1180)
The local hydrogen-bonded arrangement of a protein. The secondary structure is generally the
a helix, pleated sheet, or random coil. (p. 1190)
As a noun, the order in which amino acids are linked together in a peptide. As a verb, to determine the sequence of a peptide. (p. 1179)
Proteins composed of only amino acids (having no prosthetic groups). (p. 1190)
A method in which the C-terminal amino acid is attached to a solid support (polystyrene
beads) and the peptide is synthesized in the C : N direction by successive coupling of
protected amino acids. When the peptide is complete, it is cleaved from the solid
support. (p. 1185)
(classical peptide synthesis) Any of several methods in which protected amino acids are
coupled in solution in the correct sequence to give a desired peptide. Most of these methods
proceed in the N : C direction. (p. 1183)
The 20 a-amino acids found in nearly all naturally occurring proteins. (p. 1157)
Synthesis of a-amino acids by reaction of an aldehyde with ammonia and cyanide ion,
followed by hydrolysis of the intermediate a-amino nitrile. (p. 1167)
O
R
C
NH2
H
+
NH3
+
HCN
H2O
R
C
H
C
N
H3
+NH
3
O+
R
H
COOH
α-amino nitrile
aldehyde
C
α-amino acid
Strecker synthesis of an amino acid
terminal residue analysis
tertiary structure
transamination
zwitterion
Sequencing a peptide by removing and identifying the residue at the N terminus or at the
C terminus. (p. 1179)
The complete three-dimensional conformation of a protein. (p. 1192)
Transfer of an amino group from one molecule to another. Transamination is a common
method for the biosynthesis of amino acids, often involving glutamic acid as the source of the
amino group. (p. 1165)
(dipolar ion) A structure with an overall charge of zero but having a positively charged
substituent and a negatively charged substituent. Most amino acids exist in zwitterionic forms.
(p. 1160)
O
H2N
CH
C
O
OH
R
uncharged structure
(minor component)
+
H3N
CH
C
O–
R
dipolar ion, or zwitterion
(major component)
1198
CHAPTER 24
Amino Acids, Peptides, and Proteins
STUDY PROBLEMS
24-32
24-33
(a) The isoelectric point (pI) of phenylalanine is pH 5.5. Draw the structure of the major form of phenylalanine at pH values of 1, 5.5, and 11.
(b) The isoelectric point of histidine is pH 7.6. Draw the structures of the major forms of histidine at pH values of 1, 4,
7.6, and 11. Explain why the nitrogen in the histidine ring is a weaker base than the a-amino group.
(c) The isoelectric point of glutamic acid is pH 3.2. Draw the structures of the major forms of glutamic acid at pH values of 1, 3.2, 7, and 11. Explain why the side-chain carboxylic acid is a weaker acid than the acid group next to the
a-carbon atom.
Draw the complete structure of the following peptide.
24-34
Predict the products of the following reactions.
Ser-Gln-Met # NH2
O
(a) Ile
O
OH
OH
+
pyridine
heat
(b) Ph
CH2
O
C
CH3
NH
CH
COOH
H2, Pd
O
(c) Lys + excess 1CH3CO22O ¡
(d) (D,L)-proline
(1) excess Ac2O
"
(2) hog kidney acylase, H2O
CHO
(e) CH CH
3
2
CH
CH3
NH3, HCN
(f) product from part (e)
(g) 4-methylpentanoic acid + Br2 >PBr3 ¡
24-35
H3O +
H2O
"
(h) product from part 1g2 + excess NH3 ¡
Show how you would synthesize any of the standard amino acids from each starting material. You may use any
necessary reagents.
O
(a) (CH3)2CH
C
COOH
(b) CH3
CH
CH2
COOH
CH2CH3
(c) 1CH322CH ¬ CH2 ¬ CHO
24-36
24-37
(d)
CH2Br
Show how you would convert alanine to the following derivatives. Show the structure of the product in each case.
(a) alanine isopropyl ester
(b) N-benzoylalanine
(c) N-benzyloxycarbonyl alanine
(d) tert-butyloxycarbonyl alanine
Suggest a method for the synthesis of the unnatural D enantiomer of alanine from the readily available L enantiomer of
lactic acid.
CH3 ¬ CHOH ¬ COOH
lactic acid
24-38
24-39
24-40
24-41
Show how you would use the Gabriel–malonic ester synthesis to make histidine. What stereochemistry would you expect
in your synthetic product?
Show how you would use the Strecker synthesis to make tryptophan. What stereochemistry would you expect in your synthetic product?
Write the complete structures for the following peptides. Tell whether each peptide is acidic, basic, or neutral.
(a) methionylthreonine
(b) threonylmethionine
(c) arginylaspartyllysine
(d) Glu-Cys-Gln
The following structure is drawn in an unconventional manner.
O
CH3
CH3CH2
CH
CH
NH
CONH2
(a) Label the N terminus and the C terminus.
(c) Identify and label each amino acid present.
C
O
CH
CH2CH2
NH
CO
C
NH
CH2NH2
(b) Label the peptide bonds.
(d) Give the full name and the abbreviated name.
Study Problems
24-42
24-43
Aspartame (Nutrasweet®) is a remarkably sweet-tasting dipeptide ester. Complete hydrolysis of aspartame gives phenyl
alanine, aspartic acid, and methanol. Mild incubation with carboxypeptidase has no effect on aspartame. Treatment of
aspartame with phenyl isothiocyanate, followed by mild hydrolysis, gives the phenylthiohydantoin of aspartic acid.
Propose a structure for aspartame.
A molecular weight determination has shown that an unknown peptide is a pentapeptide, and an amino acid analysis shows
that it contains the following residues: one Gly, two Ala, one Met, one Phe. Treatment of the original pentapeptide with
carboxypeptidase gives alanine as the first free amino acid released. Sequential treatment of the pentapeptide with phenyl
isothiocyanate followed by mild hydrolysis gives the following derivatives:
first time
second time
S
NH
O
24-45
24-46
H
Ph
CH2Ph
S
N
O
NH
Ph
CH3
H
N
NH
O
Propose a structure for the unknown pentapeptide.
Show the steps and intermediates in the synthesis of Leu-Ala-Phe
(a) by the solution-phase process.
(b) by the solid-phase process.
Using classical solution-phase techniques, show how you would synthesize Ala-Val and then combine it with Ile-Leu-Phe
to give Ile-Leu-Phe-Ala-Val.
Peptides often have functional groups other than free amino groups at the N terminus and other than carboxyl groups at
the C terminus.
(a) A tetrapeptide is hydrolyzed by heating with 6 M HCl, and the hydrolysate is found to contain Ala, Phe, Val, and Glu.
When the hydrolysate is neutralized, the odor of ammonia is detected. Explain where this ammonia might have been
incorporated in the original peptide.
(b) The tripeptide thyrotropic hormone releasing factor (TRF) has the full name pyroglutamylhistidylprolinamide. The
structure appears here. Explain the functional groups at the N terminus and at the C terminus.
H2
C
H2C
H2
C
O
O
CH
C
O
C
H
C
N
H
N
C
H2N
N
N
N
CH2
CH2
HC
CH2
H
24-47
third time
S
N
Ph
24-44
1199
C
O
H
(c) On acidic hydrolysis, an unknown pentapeptide gives glycine, alanine, valine, leucine, and isoleucine. No odor of ammonia is detected when the hydrolysate is neutralized. Reaction with phenyl isothiocyanate followed by mild hydrolysis
gives no phenylthiohydantoin derivative. Incubation with carboxypeptidase has no effect. Explain these findings.
Lipoic acid is often found near the active sites of enzymes, usually bound to the peptide by a long, flexible amide linkage
with a lysine residue.
NH
O
COOH
CH
C
N
S
S
S
lipoic acid
S
C
O
H
bound to lysine residue
(a) Is lipoic acid a mild oxidizing agent or a mild reducing agent? Draw it in both its oxidized and reduced forms.
(b) Show how lipoic acid might react with two Cys residues to form a disulfide bridge.
(c) Give a balanced equation for the hypothetical oxidation or reduction, as you predicted in part (a), of an aldehyde by
lipoic acid.
O
R
C
COOH
H +
S
S
H2O
1200
24-48
24-49
24-50
24-51
CHAPTER 24
Amino Acids, Peptides, and Proteins
Histidine is an important catalytic residue found at the active sites of many enzymes. In many cases, histidine appears to
remove protons or to transfer protons from one location to another.
(a) Show which nitrogen atom of the histidine heterocycle is basic and which is not.
(b) Use resonance forms to show why the protonated form of histidine is a particularly stable cation.
(c) Show the structure that results when histidine accepts a proton on the basic nitrogen of the heterocycle and then is
deprotonated on the other heterocyclic nitrogen. Explain how histidine might function as a pipeline to transfer protons
between sites within an enzyme and its substrate.
Metabolism of arginine produces urea and the rare amino acid ornithine. Ornithine has an isoelectric point close to 10.
Propose a structure for ornithine.
Glutathione (GSH) is a tripeptide that serves as a mild reducing agent to detoxify peroxides and maintain the cysteine
residues of hemoglobin and other red blood cell proteins in the reduced state. Complete hydrolysis of glutathione gives
Gly, Glu, and Cys. Treatment of glutathione with carboxypeptidase gives glycine as the first free amino acid released.
Treatment of glutathione with 2,4-dinitrofluorobenzene (Sanger reagent, Problem 24-23, page 1180), followed by complete hydrolysis, gives the 2,4-dinitrophenyl derivative of glutamic acid. Treatment of glutathione with phenyl isothiocyanate does not give a recognizable phenylthiohydantoin, however.
(a) Propose a structure for glutathione consistent with this information. Why would glutathione fail to give a normal product from Edman degradation, even though it gives a normal product from the Sanger reagent followed by hydrolysis?
(b) Oxidation of glutathione forms glutathione disulfide (GSSG). Propose a structure for glutathione disulfide, and write a
balanced equation for the reaction of glutathione with hydrogen peroxide.
Complete hydrolysis of an unknown basic decapeptide gives Gly, Ala, Leu, Ile, Phe, Tyr, Glu, Arg, Lys, and Ser. Terminal
residue analysis shows that the N terminus is Ala and the C terminus is Ile. Incubation of the decapeptide with chymotrypsin gives two tripeptides, A and B, and a tetrapeptide, C. Amino acid analysis shows that peptide A contains Gly,
Glu, Tyr, and NH3; peptide B contains Ala, Phe, and Lys; and peptide C contains Leu, Ile, Ser, and Arg. Terminal residue
analysis gives the following results.
N terminus
24-52
C terminus
A
Gln
Tyr
B
Ala
Phe
C
Arg
Ile
Incubation of the decapeptide with trypsin gives a dipeptide D, a pentapeptide E, and a tripeptide F. Terminal residue
analysis of F shows that the N terminus is Ser, and the C terminus is Ile. Propose a structure for the decapeptide and for
fragments A through F.
There are many methods for activating a carboxylic acid in preparation for coupling with an amine. The following method
converts the acid to an N-hydroxysuccinimide (NHS) ester.
+
R
F3C
OH
O
O
O
O
O
N
Et3N
O
O
R
+
O N
F3C
OH
O
O
NHS ester
24-53
(a) Explain why an NHS ester is much more reactive than a simple alkyl ester.
(b) Propose a mechanism for the reaction shown.
(c) Propose a mechanism for the reaction of the NHS ester with an amine, R ¬ NH2.
Sometimes chemists need the unnatural D enantiomer of an amino acid, often as part of a drug or an insecticide. Most
L-amino acids are isolated from proteins, but the D-amino acids are rarely found in natural proteins. D-amino acids can be
synthesized from the corresponding L-amino acids. The following synthetic scheme is one of the possible methods.
COOH
H2N
R
H
COOH
NaNO2
HCl
intermediate 1
NaN3
intermediate 2
L configuration
(a) Draw the structures of intermediates 1 and 2 in this scheme.
(b) How do we know that the product is entirely the unnatural D configuration?
H2
Pd
R
H
NH2
D configuration