1. Carbohydrates
Classification
Structure
Properties
Chemical Tests

2. Carbohydrates (glycans) have the following basic composition:
Monosaccharides - simple sugars with multiple OH groups. Based on number of carbons (3, 4, 5, 6), a monosaccharide is a triose, tetrose, pentose or hexose.
Disaccharides - 2 monosaccharides covalently linked.
Oligosaccharides - a few monosaccharides covalently linked.
Polysaccharides - polymers consisting of chains of monosaccharide or disaccharide units.

3. Classification Monosaccharides differ from each other in the number of carbons and in the arrangement of H and O
Aldose —carbonyl is a aldehyde
Ketose —carbonyl is a ketone

4. Smallest carbohydrates
D-glyceraldehyde
Dihydroxyacetone
In nature we rarely see monosaccharides
containing more than seven carbons

5. Monosaccharides Three Carbons = Triose Four Carbons = Tetrose
Five Carbons = Pentose
Six Carbons = Hexose

6. Learning Check
Identify each as tetrose, pentose or hexose, and as aldose or ketose
A B

7. Solution
A B
aldose, hexose ketose, pentose

10. In aqueous solutions monosaccharides form cyclic structures with an additional chiral center
This is related to the ability of alcohols to react with carbonyl groups to form hemiacetals or hemiketals

11. Cyclic Structures
Monosaccharides with 5-6 carbon atoms form cyclic structures
The hydroxyl group on C-5 reacts with the aldehyde group or ketone group
furanose
pyranose

13. Derivatives of monosaccharides
Sugar acids (aldonic acids)
Sugar alcohols or alditols
Deoxy sugars
Amino sugars
Glycosides

14. Uronic acids are found in glycosaminoglycans
D-Glucose
D-Gluconic acid
Mild enzymatic or chemical oxidation converts an aldose into an aldonic acid
Uronic acids are found in glycosaminoglycans

15. Amino sugars
The hydroxyl group at the 2 carbon is replaced by an amino group
-D-Glucosamine
-D-Galactosamine

16. Mild reduction leads to the formation alcohol sugars (polyhydroxy alcohols, alditols)
D-glucitol
(sorbitol)
D-Glucose

17. Glycosidic Bonds
The anomeric hydroxyl and a hydroxyl of another sugar or some other compound can join together, splitting out water to form a glycosidic bond:
R-OH + HO-R'  R-O-R' + H2O
E.g., methanol reacts with the anomeric OH on glucose to form methyl glucoside (methyl-glucopyranose).

18. Glycosides
Reaction of the anomeric hydroxyl with an hydroxyl of another compound (O-glycosidic bond)
N-glycosidic bonds are formed when the anomeric OH
reacts with an amino group (N-glycosidic bonds)

19. Glycosidic Bonds
The anomeric hydroxyl and a hydroxyl of another sugar or some other compound can join together, splitting out water to form a glycosidic bond:
R-OH + HO-R'  R-O-R' + H2O
E.g., methanol reacts with the anomeric OH on glucose to form methyl glucoside (methyl-glucopyranose).

20. Important Disaccharides
Maltose = Glucose + Glucose
Lactose = Glucose + Galactose
Sucrose = Glucose + Fructose

21. Maltose 
 -1,4-glycosidic bond
- D-Maltose

22. Lactose α-D-Lactose   -1,4-glycosidic bond
Formed by linking the Galactose C1 in the  configuration to the C4 of glucose
Lactose is a reducing sugar
 -1,4-glycosidic bond
α-D-Lactose
Galactose
Glucose 

23. Sucrose α, -1,2- glycosidic bond
D-Glucose
Formed by linking the Glucose 1 carbon in the  configuration to the Fructose 2 carbon in the  configuration
Sucrose is a non-reducing sugar
D-Fructose

24. Polysaccharides
Starch
Amylose
Amylopectin
Glycogen
Cellulose

26. Polysaccharides Polymers of D-Glucose
Plants store glucose as amylose or amylopectin, glucose polymers collectively called starch. Glucose storage in polymeric form minimizes osmotic effects.
Amylose is a glucose polymer with a(14) linkages. It adopts a helical conformation.

27. Amylose
Polymer with α-1,4 bonds
α-1,4 bonds

28. Amylopectin α-1,6 bond Polymer with α-1,4 and α-1,6 bonds branches

29. Glycogen, the glucose storage polymer in animals, is similar in structure to amylopectin. But glycogen has more a(16) branches.
The highly branched structure permits rapid release of glucose from glycogen stores, e.g., in muscle during exercise. The ability to rapidly mobilize glucose is more essential to animals than to plants.

30. Cellulose Linear polymer of glucose
Major structural component in plant cell walls
ß-1,4 bonds

31. Linear polymers are cross-linked by H-bonds

32. Learning Check Identify the polysaccharide in each as:
starch 2) glycogen 3) cellulose
A. B. C.

33. Solution
3) cellulose ) starch ) glycogen

34. General Tests for Carbohydrates
Molisch Test
Shows positive test for:
All carbohydrates. Monosaccharides give a rapid positive test. Disaccharides and polysaccharides react slower.
Reagents: concetrated H2SO4; -naphthol in 95% ethanol
Principle: hydrolysis, dehydration forming either a furfural or a 5-hydroxyfurfural; condensation with -naphthol.
Specificity: all types of CHO

35. Reactions:
The test reagent dehydrates pentoses to form furfural (a) and dehydrates hexoses to form 5-hydroxymethyl furfural (b). The furfurals further react with α-naphthol present in the test reagent to produce a purple product (reaction not shown). a) b)

36. positive test is indicated by:
The formation of a purple product at the interface of the two layers.
a negative test (left) and a positive test (right)

37. Anthrone test
Sugars react with the anthrone reagent under acidic conditions to yield a blue-green color.
Reagent: concentrated H2SO4; Anthrone (C14H10O) reagent
Principle involved: Hydrolysis, dehydration forming either a furfural or a 5- hydroxyfurfural; condensation of anthrone via anthranol intermediate
Specificity: all types of CHO

38. Positive result for anthrone test

39. Iodine test
Polysaccharides can trap iodine molecules and produce a deep blue-black product. The iodine test performed on monosaccharides and disaccharides does not yield deeply colored products. Starch gives a blue-black color.  A positive test for glycogen is a brown-blue color.  A negative test is the brown-yellow color of the test reagent.
Specificity: starch forms a violet color which disappears upon heating and reappears after cooling.
Glycogen- forms a red complex due to branching; disappears upon heating and reappears after cooling.

40. Principle:
-complexation
The use of Lugol's iodine reagent (IKI) is useful to distinguish starch and glycogen from other polysaccharides.  Lugol's iodine yields a blue-black color in the presence of starch.  Glycogen reacts with Lugol's reagent to give a brown-blue color.   Other polysaccharides and monosaccharides yield no color change; the test solution remains the characteristic brown-yellow of the reagent.  It is thought that starch and glycogen form helical coils.  Iodine atoms can then fit into the helices to form a starch-iodine or glycogen-iodine complex.  Starch in the form of amylose and amylopectin has less branches than glycogen.  This means that the helices of starch are longer than glycogen, therefore binding more iodine atoms.  The result is that the color produced by a starch-iodine complex is more intense than that obtained with a glycogen-iodine complex

42. MOLISCH CONDENSATION REACTION

43. ANTHRONE CONDENSATION REACTION

44. IODINE TEST I2
HEAT
COOL

45. Specific Reactions of Carbohydrates
Mucic Acid Test (C6H10O8)
– specific for galactose and lactose
Oxidation of most monosaccharides by nitric acid yields soluble dicarboxylic acids. However, oxidation of galactose yields an insoluble mucic acid.  Lactose will also yield a mucic acid, due to hydrolysis of the glycosidic linkage between its glucose and galactose subunits.  
-produces a rhombic crystal (broken glass)
Reagent: concentrated HNO3
Principle involved: 1,6 -oxidation of sugars whereby galactose – containing carbohydrates form a meso compound which upon standing yields a crystal.

46. CHO
CH2OH OH H
CO2H
glucose
Galactaric acid (aldaric acid)
Oxidation of CHO = aldonic acid
Oxidation of CH2OH = uronic acid
Oxidation of both = saccharic acid/aldaric acid

48. Benedict’s Test
Test for Reducing Sugars        
Alkaline solutions of copper are reduced by sugars having a free aldehyde or ketone group, with the formation of colored cuprous oxide.  Benedict's solution is composed of copper sulfate, sodium carbonate, and sodium citrate (pH 10.5).  The citrate will form soluble complex ions with Cu++, preventing the precipitation of CuCO3 in alkaline solutions. Benedict's reagent is a solution of copper sulfate, sodium hydroxide, and tartaric acid.                   
Reducing Sugars Carbohydrates that can undergo oxidation are called reducing sugars. All monosaccharides are reducing sugars and many disaccharides are also reducing sugars. Lactose and maltose are both reducing sugars, but sucrose is not.

49. Reagents: Na2CO3, CuSO4, Na citrate (less alkaline- lesser possibility of destroying sugar)
Principle involved: oxidation in less basic media

50. Aqueous sugar is mixed with Benedict's reagent, a solution of copper sulfate, sodium hydroxide, and tartaric acid. The mixture is heated. Carbohydrates which react with Benedict's reagent to reduce the blue copper (II) ion to form a brick red precipitate of copper (I) oxide are classified as reducing sugars.

51. Barfoed’s Test Shows positive test for:
Reducing monosaccharides. This reagent uses copper ions to detect reducing sugars in an acidic solution.  Barfoed's reagent is copper acetate in dilute acetic acid (pH 4.6).  distinguishes reducing monosaccharides and reducing disaccharides (mono forms Cu2O faster than di in acidic media!)
-   visible result: formation of brick red ppt. (Cu2O)
-   Reagents : Cu(Ac)2, acetic acid
Principle involved: oxidation of a reducing monosaccharide in acidic condition is faster than a disaccharide
Reactions:
Reducing monosaccharides are oxidized by the copper ion in solution to form a carboxylic acid and a reddish precipitate of copper (I) oxide within three minutes. Reducing disaccharides undergo the same reaction, but do so at a slower rate.

53. A positive test is indicated by:
The formation of a reddish precipitate within three minutes.
a negative test (left) and a positive test (right)

54. Bials Orcinol Test
Shows positive test for: Pentoses (xylose)
Reqagent: Bial's reagent uses orcinol, HCl, and FeCl3.  Orcinol forms colored condensation products with furfural generated by the dehydration of pentoses and pentosans.  It is necessary to use DILUTE sugar solutions with this test (0.02 M).
principle involved: dehydration forming furfural and condensation with orcinol
Reactions:
The test reagent dehydrates pentoses to form furfural. Furfural further reacts with orcinol and the iron ion present in the test reagent to produce a bluish product.

57. A positive test is indicated by:
The formation of a bluish product. All other colors indicate a negative result for pentoses. Note that hexoses generally react to form green, red, or brown products.
two negative tests (left, middle) and a positive test (right)

58. Seliwanoff’s Test
Seliwanoff's test is used to distinguish between aldehyde and ketone hexoses (carbohydrates containing 6 carbon atoms). A ketone hexose, also called a ketohexose, will form a deep red color when reacted with Seliwanoff's reagent. An aldehyde hexose, also called a aldohexose, will show a light pink color that takes a longer time to develop when reacted with Seliwanoff's reagent.
-         reagent: resorcinol, and HCl
-  principle involved: rapid dehydration forming 6-hydroxyfurfural and condensation reaction with resorcinol to yield a cherry red colored adduct
Shows positive test for:
Ketoses react quickly with Seliwanoff’s reagent form a deep red color. Aldoses react slowly with Seliwanoff’s to eventually produce a light pink solution. Thus, Seliwanoff’s test can be used to discriminate aldoses from ketoses. A positive test is the rapid evolution of a deep red color –ketoses.

59. Reactions:
The test reagent dehydrates ketohexoses to form 5-hydroxymethylfurfural. 5-hydroxymethylfurfural further reacts with resorcinol present in the test reagent to produce a red product within two minutes (reaction not shown). Aldohexoses react to form the same product, but do so more slowly.

61. A positive test is indicated by:
The formation of a red product.
a negative test (left) and a positive test (right)