Transcript Camp 1 - TypePad

Frederick A. Bettelheim
William H. Brown
Mary K. Campbell
Shawn O. Farrell
www.cengage.com/chemistry/bettelheim
Chapter 20
Carbohydrates
William H. Brown • Beloit College
Carbohydrates
• Carbohydrates (or saccharides) consist of only
carbon, hydrogen and oxygen.
• Carbohydrates come primarily from plants, however
animals can also biosynthesize them
• The “Carbon Cycle” describes the processes by which
carbon is recycled on our planet
- Energy from the sun is stored in plants, which use
photosynthesis to convert carbon dioxide and water
to glucose and oxygen
- In the reverse process, energy is produced when
animals oxidize glucose during respiration
Photosynthesis
6CO2 + 6H2O + Energy
Respiration
C6H12O6 + 6O2
20-2
Simplified Carbon Cycle
20-3
Carbohydrates
Carbohydrate: A polyhydroxyaldehyde or
polyhydroxyketone, or a substance that gives these
compounds on hydrolysis.
Monosaccharide: A carbohydrate that cannot be
hydrolyzed to a simpler carbohydrate. Simple sugars.
Can’t be split into smaller carbohydrate units
- Examples: glucose, fructose, galactose, ribose
Monosaccharides have the general formula CnH2nOn,
where n varies from 3 to 8.
• Aldose: A monosaccharide containing an aldehyde
group.
• Ketose: A monosaccharide containing a ketone group.
20-4
Carbs
• Disaccharides are two monosaccharides bonded together
- Can be split into two monosaccharides using an acid or
enzyme catalyst
- Examples: sucrose (table sugar), lactose (milk sugar)
• Polysaccharides are polymers of monosaccharides
- Used for storage of carbohydrates
- Can be split into many monosaccharides with acid or
enzymes
- Examples: starch, cellulose, glycogen
20-5
Types of Carbs
20-6
Monosaccharides
•
•
•
•
•
Are chiral! The suffix -ose indicates that a molecule is a carbohydrate.
The prefixes tri-, tetra, penta, and so forth indicate the number of
carbon atoms in the chain.
Those containing an aldehyde group are classified as aldoses.
Those containing a ketone group are classified as ketoses.
There are only two trioses:
CHO
CH2 OH
CHOH
C= O
CH2 OH
CH2 OH
Gl yce ralde h yde
(an aldotri ose )
Di hydroxyace ton e
(a k e totri ose )
• Often aldo- and keto- are omitted and these compounds are
referred to simply as trioses.
• Although “triose” does not tell the nature of the carbonyl group, it
at least tells the number of carbons.
20-7
Monosacharides
Glyceraldehyde, the simplest aldose, contains one
stereocenter and exists as a pair of enantiomers.
CHO
CHO
H
C
OH
CH2 OH
HO
C
H
CH2 OH
20-8
Monosaccharides
Fischer projection: A two-dimensional representation for
showing the configuration of tetrahedral stereocenters.
• Horizontal lines represent bonds projecting forward
from the stereocenter.
• Vertical lines represent bonds projecting to the rear.
• Only the stereocenter is in the plane.
CHO
H
C
OH
CH2 OH
con vert to
a Fischer
projection
CHO
H
OH
CH2 OH
*Terminal carbonyl groups (aldehydes –CHO and carboxylic acids –COOH)
are written vertically on Fisher diagram
20-9
Monosacharides
In 1891, Emil Fischer made the arbitrary assignments of
D- and L- to the enantiomers of glyceraldehyde.
CHO
H
OH
CH2 OH
D-Glyce ral de h yde
25
[]D
= +13.5°
CHO
HO
H
CH2 OH
L-G lyce ral de h yde
25
[]D
= -13.5°
• D-monosaccharide: the -OH on its penultimate carbon
is on the right in a Fischer projection.
• L-monosaccharide: the -OH on its penultimate carbon
is on the left in a Fischer projection.
*Fischer got lucky. For sugars, D=R and rotates light
clockwise (+); L=S and counter (-). BUT, does the Fisher
20-10
Get a Model Kit
• Before building the molecule, draw the D- Glyceraldehyde
•
•
•
•
•
enantiomer (last slide) on your paper as both a wedge
diagram and a Fisher diagram. Answer the following:
1- Based solely on the Fisher diag, assign priority to the
groups (Ch 15 p.427). Reading the order of the groups, what
direction would you assign,clockwise or counterclockwise?__
2- Build the molecule, based on the Fisher. Is the carbonyl C
group pointing toward you or away from you?____ What about
the H and –OH? _______
3- Now, rotate the molecule, putting the lowest ranked group
away from you. Approximately, how many degrees did you
turn the molecule?____Where is the –OH after the rotation?
4- Is the direction now clockwise, or counter?_____________
5- In terms of ONLY seeing a (similar) Fisher diagram, what
20-11
will you have to do (in your head) before determining D or L?
D,L-Monosaccharides
• The most common D-tetroses and D-pentoses are:
CHO
H
OH
H
OH
CHO
HO
H
H
OH
CH2 OH
CH2 OH
D-Eryth rose D-Th re ose
CHO
H
OH
H
OH
H
OH
CHO
H
H
H
OH
H
OH
CH2 OH
CH2 OH
D-Ribose 2-De oxy-D-ri bos e
• The three most common D-hexoses are:
H
HO
To memorize these!
H
H
You do NOT have
*See table 20.1 and 20.2
CHO
OH
H
OH
OH
CH2 OH
D-G l ucos e
H
HO
HO
H
CHO
OH
H
H
OH
CH2 OH
D-G alactose
CH2 OH
C= O
HO
H
H OH
H OH
CH2 OH
D-Fructose
20-12
Sugars (Monosaccharides)
• All other sugars are classified based on
the position of the hydroxyl group farthest
away from the carbonyl, but not the one
on the “end”.
• Penultimate carbon- Point of reference
that refers to the next to last C atom in
the chain.
• See Table 20.1 and 20.2 (penultimate C
is in red)
20-13
Three Important Monosaccharides
• D-Glucose (aldose) is the most common monosaccharide
- Primary fuel for our cells, required for many tissues
- Main sources are fruits, vegetables, corn syrup and honey
- Blood glucose is maintained within a fairly small range
- Some glucose is stored as glycogen, excess is stored as fat
• D-Galactose (aldose) comes from hydrolysis of the
disaccharide lactose
- Used in cell membranes of central nervous system
- Converted by an enzyme into glucose for respiration (lack of
this enzyme causes galactosemia, which can cause retardation
in infants if not treated by complete removal from diet)
• D-Fructose (ketose) is the sweetest carbohydrate
- Converted by an enzyme into glucose for respiration
- Main sources are fruits and honey
20-14
- Also obtained from hydrolysis of the disaccharide sucrose
Structures of Glucose, D-Galactose and DFructose
•
Note that in nature, only the D enantiomers of sugars are used
• What is the relationship between D-glucose and L-glucose?
• What is the relationship between D-glucose and D-galactose?
• What is the relationship between D-glucose and D-fructose?
(*use these choices for each question: enantiomers,
diastereomers or constitutional isomers?)
H
O
O
H
H
O
CH2 OH
H
HO
OH
H
HO
H
H
H
OH
O
OH
HO
H
HO
H
H
OH
OH
H
OH
H
OH
HO
H
HO
H
OH
HO
H
H
CH2 OH
CH2 OH
D-Glucose
L-Glucose
CH2 OH
D-Galactose
H
CH2 OH
D-Fructose
20-15
Amino Sugars
Amino sugars contain an -NH2 group in place of an -OH
group.
• Only three amino sugars are common in nature: Dglucosamine, D-mannosamine, and D-galactosamine.
CHO
H N H2
HO H
H OH
H OH
CH2 OH
CHO
H 2N 2 H
HO H
H OH
H OH
CH2 OH
CHO
H N H2
HO H
HO 4 H
H OH
CH2 OH
H
HO
H
H
CHO O
N HCCH 3
H
OH
OH
CH 2 OH
D-Glu cosami n e D-Man n osam in e D-Gal actos ami n e N-Ace tyl-D(C -2 s tere ois ome r (C -4 s tere ois ome r glu cosami n e
of D-glu cos ami n e of D-gl u cos am in e )
20-16
Cyclic Structure
• Aldehydes and ketones react with alcohols to form
hemiacetals (Chapter 17).
• Cyclic hemiacetals form readily when the hydroxyl and
carbonyl groups are part of the same molecule and
their interaction can form a five- or six-membered ring.
O
4
1
H
red raw to show
-OH an d -CHO
clos e to each oth er
O-H
4-Hyd roxypentanal
1
4
O
H
C
H
O
H
O-H
O
A cyclic hemiacetal
Alcohol+Aldehyde (or alcohol+ketone)Hemiacetal
20-17
Cyclic Structures of Monosaccharides
• Recall that an alcohol can react with an aldehyde or ketone
to form a hemiacetal
• If the alcohol and aldehyde or ketone are in the same
molecule, a cyclic hemiacetal is formed
• Monosaccharides in solution are in equilibrium between the
open-chain and ring forms, and exist primarily in the ring
form
20-18
Haworth Projection
• A Haworth projection is a common way
of representing the cyclic structure of
monosaccharides with a simple threedimensional perspective
20-19
Drawing Haworth Structures for Cyclic Forms
• Step 1: Number the carbons in the chain and turn the Fischer
projection of the open-chain form clockwise 90 degrees
- Hydroxyl groups that were on the right are now on the bottom,
and hydroxyl groups that were on the left are now on the top
(they will stay on bottom or top in the Haworth structure)
• Step 2: Rotate around so that C-6 sticks up from C-5, and the
hydroxyl group on C-5 points towards C-1
• Step 3: Form the cyclic hemiacetal by bonding the hydroxyl O
to the carbonyl C and moving the hydroxyl H to the carbonyl O
• Note: For C-6 aldose sugars, the cyclic hemiacetal has a new
chiral carbon at C-1
- The two possible stereoisomers are called anomers
- The alpha anomer has the hydroxyl group down
20-20
- The beta anomer has the hydroxyl group up
Haworth Projections
• D-Glucose forms these two cyclic hemiacetals.
1
CHO
H
OH
HO
H
H
H
OH
5
OH
CH2 OH
D -Glucose
red raw to sh ow th e -OH
on carbon-5 close to the
aldeh yd e on carbon-1
CH2 OH
OH
H5
O
H
OH H C1
HO
H
H
OH
anomeric
anomeric
CH
OH
CH2 OH
2
carb on
carb on
O
O
H
H
H
OH (  )
H
H
+
OH
OH H
H
HO
HO
OH(  )
H
H OH
H OH
-D -Glucopyranose
-D -Glucopyranose
(-D -Glucose)
( -D -Glucos e )
20-21
Haworth Projections
• Groups bonded to the carbons of the ring then lie
either above or below the plane of the ring.
(*Remember your stereoisomer lab!)
• Stereoisomers that differ in configuration only at the
anomeric carbon are called anomers.
• In a ring, the –OH on the alpha anomer is down and
axial; Beta is up and equatorial. *Most Glucose in our
bodies is Beta because it’s farther away and more
stable
• The anomeric carbon of an aldose is C-1; that of the
most common ketoses is C-2 (because there is a
branch off of the “first” C and the C in the branch is
#1). See next slide.
20-22
Haworth Projections
D-Fructose (a 2-ketohexose) also forms a five-membered
cyclic hemiacetal.
HOCH2
5
1
O
H HO
CH2 OH
2
OH( )
H
HO
H
 -D -Fructofuranose
( - D -Fructos e)
1
CH2 OH
2
C=O
HO
H
H
OH
H 5 OH
CH2 OH
D -Fru ctose
HOCH2
5
O
H HO
H
OH ( )
2
CH2 OH
HO
H
1
 - D -Fru ctofu ran os e
(- D -Fructose)
*For test, know that 6 C ketose sugar (fructose) has anomeric C at #2
20-23
Haworth Projections
A six-membered hemiacetal ring is called a
pyranose, and a five-membered hemiacetal
ring is called a furanose because these ring
sizes correspond to the heterocyclic
compounds furan and pyran. (*DO NOT
have to know furanose for TEST)
O
O
Furan
Pyran
20-24
Chair Conformations
• For pyranoses, the six-membered ring is more accurately
represented as a chair conformation.
HO
HO
CH2 OH
O
anomeric
carbon
OH()
OH
 -D -Glu copyran os e
( - D -Glucos e)
HO
HO
*Which is equatorial and which is axial?
CH2 OH
OH
O
C
OH H
D -Glucos e
HO
HO
CH2 OH
O
HO
OH( )
- D -Glu copyran os e
(  - D -Glucose)
20-25
Chair Conformations
• In both Haworth projections and chair conformations, the
orientations of groups on carbons 1- 5 of -Dglucopyranose are up, down, up, down, and up.
6
CH2 OH
5
O OH()
H
H
4 OH
1
H
HO
H
3
2
H OH
-D -Glucop yranose
(Haw orth p rojection)
6
CH2 OH
4
HO
HO
O
5
3
2
OH 1
OH( )
- D -Glucopyranose
(ch air con formation)
20-26
Mutarotation
• Mutarotation: The change in specific rotation that
accompanies the equilibration of - and -anomers in
aqueous solution.
• Example: When either -D-glucose or -D-glucose is
dissolved in water, the specific rotation of the solution
gradually changes to an equilibrium value of +52.7°,
which corresponds to 64% beta and 36% alpha forms.
HO
HO
CH2 OH
O
OH
OH
-D -Glucopyranose
25
[] D = + 18.7°
HO
HO
CH2 OH
OH
O
C
HO
H
Open-chain form
HO
HO
CH2 OH
O
HO
OH
-D -Glucopyranose
25
[] D = +112°
*ONLY have to know that most glucose in body is Beta 20-27
Disaccharides
• Formation occurs by dehydration between 2 –OH’s of 2
monosaccharide monomers. One monomer removes –OH and
the other removes –H. The O that was not removed forms a
bond between the anomeric C’s. The new bond is called a
Glycosidic bond. The process is commonly called
dehydration synthesis or condensation.
• Maltose is glucose+glucose
• Sucrose is glucose+fructose
• Lactose is glucose+galactose
20-28
Disaccharides
Disaccharide: a carbohydrate containing two
monosaccharide units joined by a glycosidic bond.
Sucrose (table sugar)
• Sucrose is the most abundant disaccharide in the
biological world; it is obtained principally from the juice
of sugar cane and sugar beets. A water molecule is
removed (not shown)
CH2 OH
O
OH
1
HO
HO
OH
HO
OH
O
O
HO 2
CH2 OH
1
OH
HOCH2
a u n i t of-Dglu copyran ose
CH2 OH
O
HOCH2
O
HO
1
O
2
-1,2-glycosi dic bon d
a u n i t of-Dfru ctofu ran os e
CH2 OH
OH
1
20-29
Disaccharides
Maltose
• Present in malt, the juice from sprouted barley and
other cereal grains.
• Maltose consists of two units of D-glucopyranose
joined by an -1,4-glycosidic bond.
1
HOCH2 O
HO
CH2 OH
4
O
OH
OH
HO
OH
O
OH
HO
HO
-1,4-glycosi dic
bon d
CH2 OH
O
1
OH 4 CH2 OH
O
O
OH
HO
OH
20-30
Disaccharides
Lactose
• Lactose is the principal sugar present in milk; it makes
up about 5 to 8 percent of human milk and 4 to 6
percent of cow's milk.
• It consists of D-galactopyranose bonded by a -1,4glycosidic bond to carbon 4 of D-glucopyranose.
CH2 OH
O
CH2 OH
OH
O
OH
4
OH
OH
CH2 OH
-1,4-glycosid ic bond
O
4
O
1
OH
OH
HO
1
OH
O
HO
CH2 OH
O
OH
OH
OH
20-31
Physical Properties
Monosaccharides are colorless crystalline solids, very
soluble in water, but only slightly soluble in ethanol.
Sweetness relative to sucrose:
Sw eetnes s
Relative to
Carb oh yd rate
Sucrose
fru ctose
1.74
su cros e (table su gar) 1.00
honey
0.97
glucose
0.74
maltos e
0.33
galactose
0.32
lactose (milk s ugar)
0.16
Sw eetnes s
Relative to
Artificial
Sw eeten er
Sucrose
saccharin
450
aces ulfame-K
200
asp artame
180
su cralos e
600
20-32
Polysaccharides
Polysaccharide: A carbohydrate consisting of large
numbers of monosaccharide units joined by glycosidic
bonds.
Starch: A polymer of D-glucose.
• Starch can be separated into amylose and amylopectin.
• Amylose is composed of unbranched chains of up to
4000 D-glucose units joined by -1,4-glycosidic bonds.
• Amylopectin contains chains up to 10,000 D-glucose
units also joined by -1,4-glycosidic bonds; at branch
points, new chains of 24 to 30 units are started by 1,6-glycosidic bonds.
20-33
Polysaccharides
• Figure 20.3 Amylopectin, a branched polymer of
approximately 10,000 units of D-glucose joined by -1,4glycosidic bonds.
20-34
Polysaccharides
• Glycogen is the energy-reserve carbohydrate for animals.
• Glycogen is a branched polysaccharide of
approximately 106 glucose units joined by -1,4- and
-1,6-glycosidic bonds.
• The total amount of glycogen in the body of a wellnourished adult human is about 350 g, divided almost
equally between liver and muscle.
20-35
Polysaccharides
Cellulose is a linear polysaccharide of D-glucose units
joined by -1,4-glycosidic bonds.
• It has an average molecular weight of 400,000 g/mol,
corresponding to approximately 2200 glucose units per
molecule.
• Cellulose molecules act like stiff rods and align
themselves side by side into well-organized waterinsoluble fibers in which the OH groups form numerous
intermolecular hydrogen bonds.
• This arrangement of parallel chains in bundles gives
cellulose fibers their high mechanical strength.
• It is also the reason why cellulose is insoluble in water.
20-36
Polysaccharides
Figure 20.4 Cellulose is a linear polymer containing as
many as 3000 units of D-glucose joined by -1,4glycosidic bonds.
20-37
Polysaccharides
Cellulose (cont’d)
• Humans and other animals can not digest cellulose
because their digestive systems do not contain glycosidases, enzymes that catalyze the hydrolysis of
-glycosidic bonds.
• Termites have such bacteria in their intestines and can
use wood as their principal food.
• Ruminants (cud-chewing animals) and horses can also
digest grasses and hay.
• Instead, we have only -glucosidases; hence, the
polysaccharides we use as sources of glucose are
starch and glycogen.
• Many bacteria and microorganisms have glucosidases.
20-38
Chapter 20 Carbohydrates
End
Chapter 20
20-39